% Use a classic PDF cross-reference table for downstream stamping. \ifdefined\pdfobjcompresslevel\pdfobjcompresslevel=0\relax\fi \documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern,microtype} \usepackage[margin=25mm]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools,mathrsfs} \usepackage{enumitem,booktabs,array,xurl,needspace,longtable,capt-of} \usepackage{xcolor,multicol} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \hypersetup{bookmarksdepth=2,pdftitle={Prime factors of fusion-category dimensions: weak group-theoreticity, rank and decomposition},pdfauthor={Sebastian Burciu, Jingcheng Dong, Sebastien Palcoux}} \allowdisplaybreaks[2] \setlength{\emergencystretch}{2em} \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{question}[theorem]{Question} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{jointideallemma}{Lemma} \renewcommand{\thejointideallemma}{\thetheorem\alph{jointideallemma}} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{example}[theorem]{Example} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\C}{\mathcal C} \newcommand{\D}{\mathcal D} \newcommand{\E}{\mathcal E} \newcommand{\M}{\mathcal M} \newcommand{\Pcat}{\mathcal P} \newcommand{\Qcat}{\mathcal Q} \newcommand{\Z}{\mathcal Z} \newcommand{\one}{\mathbf1} \renewcommand{\Vec}{\mathrm{Vec}} \newcommand{\Rep}{\operatorname{Rep}} \newcommand{\FPdim}{\operatorname{FPdim}} \newcommand{\Irr}{\operatorname{Irr}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\Inv}{\operatorname{Inv}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\ad}{\mathrm{ad}} \newcommand{\pt}{\mathrm{pt}} \newcommand{\rev}{\mathrm{rev}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\hs}{\widehat s} \newcommand{\K}{K_0} \newcommand{\kk}{\mathbb C} \newcommand{\R}{\mathbb R} \newcommand{\ZZ}{\mathbb Z} \newcommand{\QQ}{\mathbb Q} \newcommand{\CC}{\mathbb C} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\lcm}{\operatorname{lcm}} \newcommand{\sVec}{\mathrm{sVec}} \DeclareMathOperator{\ord}{ord} \newcommand{\OO}{\mathcal O} \title{Prime factors of fusion-category dimensions:\\ weak group-theoreticity, rank and decomposition} \author{Sebastian Burciu \and Jingcheng Dong \and S\'ebastien Palcoux} \date{} \newcommand{\cost}{\operatorname{cost}} \newcommand{\WGT}{\mathrm{WGT}} \newcommand{\RankLink}[2]{\href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/rank/#1}{#2}} \newcommand{\RankFile}[1]{\RankLink{#1}{\nolinkurl{#1}}} \newcommand{\contentsgroup}[1]{\phantomsection\addcontentsline{toc}{part}{#1}} \setcounter{tocdepth}{1} \makeatletter \newcommand{\papercontents}{\clearpage\begingroup\normalsize \parskip=1.5pt\relax \pdfbookmark[1]{Contents}{wholecontents}\section*{Contents} \def\@pnumwidth{2.0em}\def\@tocrmarg{2.8em} \def\l@section{\@dottedtocline{1}{0pt}{2.2em}} \renewcommand*\l@part[2]{\par\addvspace{.55\baselineskip}\noindent{\small\bfseries ##1}\par\nobreak} \@starttoc{toc}\endgroup\clearpage} \makeatother \begin{document} \maketitle \begin{abstract} Let $D$ be the Frobenius--Perron dimension of a weakly integral fusion category and $Q(D)$ its square-full part. For a divisor-minimal braided obstruction to weak group-theoreticity, we bound the conductors of character fields by formal codegrees and relate residue blocks to primary twists through local Weyl operators. For primes $p{\raggedright\arraybackslash}p{.21\linewidth}@{\hspace{.02\linewidth}}>{\raggedright\arraybackslash}p{.49\linewidth}@{\hspace{.02\linewidth}}>{\raggedright\arraybackslash}p{.25\linewidth}@{}}\toprule Prime support & Generating forms or operation & Where the conditions enter\\\midrule Two or three repeated primes & $p^Aq^B$; $p^Aq^Br^C$ with one small exponent, bounded triples, quartic and quintic strips, or a sextic at a smaller prime & Sections~\ref{sec:three} and~\ref{sec:three-ordered}--\ref{fam:Q:sec:1}; refinements in Section~\ref{found:group:uniform}\\ Four repeated primes & $P^A(uvw)^3$, $(uv)^3P^AQ^B$, $P^aQ^Bu^3v^2$ and $P^AQ^B(uv)^2$, including height, residue and dyadic tests & Sections~\ref{sec:two-squared}--\ref{sub:residue-lift}\\ Five repeated primes & $p^4(uvw)^3t^2$, $(puvwt)^3$ and the exact minimal-obstruction restriction at three specified valuations & Section~\ref{fam:W:sec:1}\\ Lifts and more primes & Raising designated squares to cubes; adjoining $r^B$ to a covered $N$; $P^AR^B\prod t_i^{c_i}$; $p^A m$ at fixed cofactor & Sections~\ref{sec:incidence}, \ref{fam:W:sec:1} and~\ref{ext:section}\\ Recognition and boundaries & $p^a q^b r^4$ with a principal-block degree condition; minimal obstructions with three exponents bounded by $6,6,4$ & Theorem~\ref{fam:Q:n:quart:recognition}; Section~\ref{sec:boundary}\\ Without braiding & Covered $D^2$, ordered $uvp^Aq^B$, $p^Aq^Br$ and square-free $D$ on at most five primes & Theorems~\ref{thm:intro-two-free} and~\ref{thm:three}; Corollary~\ref{thm:sq30030}\\ Rank with prescribed dimension & $D=qM^2$ by parity, square class, prime support or largest degree; $qU^2p^{2e}$; two isolated primes; a prime denominator; three primes with $q=10$ & Sections~\ref{mrk:packets}--\ref{mrk:certificates}; Corollary~\ref{fam:rank:two-isolated}\\ Decomposition and solvability & Prime components, signed graphs at $L,L^2$ ($L$ square-free), modular and slightly degenerate solvability tests & Section~\ref{sec:coprime-core}; Proposition~\ref{prop:solv-square-divisor}\\\bottomrule \end{tabular}\end{minipage}\end{center} \papercontents \contentsgroup{Character tables, local blocks and Weyl columns} \section{Reduction to a simple character table}\label{fam:M:sec:2}\label{sec:wgt-preliminaries} Use the positive spherical structure of a weakly integral category over $\CC$, so $D=\FPdim(\C)$ and $d_X=\FPdim(X)$ are its categorical dimensions \cite[Propositions~8.23 and~8.24]{ENOannals}. Write $\C_{\pt},\C_{\ad},\C',U(\C)$ for its pointed subcategory, adjoint subcategory, M\"uger center and universal grading group. We call a category \emph{perfect} if its only invertible simple is the unit, and \emph{fusion-simple} if its only fusion subcategories are trivial or the whole category. The corresponding ring conditions concern invertible basis elements and based subrings. \begin{lemma}[Minimal counterexample]\label{fam:M:2.1}\label{lem:minimal}\label{lem:proper-divisor-reduction}\label{lem:minimal-counterexample-simple} Suppose every braided category of dimension a proper divisor of $D$ is WGT. A counterexample $\C$ at $D$ is integral, equals $\C_{\ad}$, and has one of two forms: \begin{enumerate}[label=\textup{(\roman*)},leftmargin=*] \item $\C$ is modular, perfect and fusion-simple; \item $\C'=\C_{\pt}=\sVec$, and the only fusion subcategories are $\Vec,\sVec,\C$. The ring modulo the transparent fermion is perfect and simple. \end{enumerate} \end{lemma} \begin{proof} A nontrivial universal grading reduces the question to its neutral component. A proper nontrivial nondegenerate subcategory splits off with its centralizer. A nontrivial Tannakian $\Rep(G)$ reduces to the neutral de-equivariantization, of dimension $D/(|G||H|)$ for grading support $H\le G$. In each case the smaller dimensions divide $D$ and WGT passes back by extension, equivariantization or tensor product \cite[Propositions~8.15 and~8.20]{ENOannals}, \cite[Proposition~4.1]{ENOsolvable}, \cite[Theorems~3.13(i) and~4.44, Proposition~4.26]{DGNO}, \cite[Proposition~4.1]{NataleWGT}. Thus $\C=\C_{\ad}$ is integral by the dimension grading \cite[Theorem~3.10]{GNnil}, and its symmetric center is $\Vec$ or $\sVec$. Every proper subcategory is integral and WGT, hence pointed in the absence of Tannakian subcategories \cite[Theorem~3.1]{NataleCore}. It thus lies in $\C_{\pt}$, which is trivial in the modular case by $(\C_{\ad})'=\C_{\pt}$, and is $\sVec$ in the slightly degenerate case by \cite[Lemma~7.1]{NataleWGT}. In the latter case pull back based subrings from the fermion quotient. \end{proof} We use \emph{ordinary modular table} for the character table of an integral modular category, and \emph{positive reduced table} for the table of an integral slightly degenerate category after identifying its transparent fermion with the unit. A \emph{minimal table} is one of the two categorical tables in Lemma~\ref{lem:minimal}, with the simplicity and perfectness conditions stated there. In particular, its twists, traces and indicators come from the category. Statements formulated for based rings or symmetric integral character data alone use the algebraic hypotheses specified in those statements. Put $\Delta=D$ in case (i) and $\Delta=D/2$ in case (ii). In (ii) use the fusion ring $R=K_0(\C)/(f-\one)$ for the transparent fermion $f$. The twist gives $fX\not\simeq X,X^*$. Duality-compatible orbit representatives containing $\one$ give structure constants $N_{XY}^Z+N_{XY}^{fZ}$ and unit coefficient $\delta_{Y,X^*}$. These two rings are instances of the following definition. \begin{definition}[Symmetric integral character data]\label{def:symmetric-integral-data} Let $R$ be a commutative fusion ring with basis $[X]$, duality involution and unit $0$. Its trace $\tau$ is the coefficient of the unit; $\tau([X][Y]^*)=\delta_{XY}$ makes $a\mapsto\tau(aa^*)$ positive definite when $*$ is extended conjugate-linearly to the complex algebra. Fix a complete bijection from basis labels $Y$ to the irreducible characters $\chi_Y$ and require $\chi_0(X)=d_X\in\mathbb Z_{>0}$. Put $\Delta=\sum_Xd_X^2$. We require that all $\chi_Y(X)$ lie in the integer ring of a common cyclotomic field, that $S_{XY}=d_Y\chi_Y(X)$ is symmetric, and that \begin{equation}\label{eq:wgt-modular-data} \begin{gathered}S_{\one X}=d_X,\quad S^\dagger S=\Delta I,\quad S_{X,Y^*}=\overline{S_{XY}},\\ \chi_Y(X)=S_{XY}/d_Y,\quad f_Y=\|\chi_Y\|^2=\Delta/d_Y^2\in\ZZ.\end{gathered} \end{equation} The involution and orthogonality above give the coefficient trace \begin{equation}\label{eq:fusion-coefficient-trace} \tau(B)=\sum_Y\chi_Y(B)/f_Y. \end{equation} Galois conjugation acts on the complete set of characters. Since it commutes with duality and fixes the rational $f_Y$, the induced permutation $\pi_\sigma$ preserves $d_Y$. Symmetry therefore gives the unsigned covariance \begin{equation}\label{eq:galois-permutations} \sigma S_{XW}=S_{\pi_\sigma X,W}=S_{X,\pi_\sigma W}. \end{equation} \end{definition} We use $I_s=\{X:d_X=s\}$ and $n_s=|I_s|$. A twist, a congruence representation, and a categorical indicator are additional inputs in the rows of Table~\ref{tab:proof-inputs} where they occur. For the modular table these identities are standard \cite[Sections~8.13--8.14]{EGNO}. For the reduced table, the normalized columns give all characters by \cite[Proposition~3.3]{YuSlightlyDegenerate} (see \cite[Proposition~2.7]{BGNPWSuper} for the unitary case). The ratio $\Delta/d_Y^2$ is an algebraic integer by \cite[Corollary~3.4]{YuSlightlyDegenerate}, hence an integer here. \begin{table}[ht] \centering\small \caption{Hypotheses used in passing from character data to dimension and rank bounds.} \label{tab:proof-inputs} \begin{tabular}{@{}p{.27\linewidth}p{.66\linewidth}@{}}\toprule Input & Consequence and use\\\midrule Fusion ring with cyclotomic character values & A rational codegree $f$ bounds the different and gives conductor dividing $2f$ (Theorem~\ref{lem:codifferent}, Corollary~\ref{cor:field-conductor}).\\ Such a ring with symmetric integral character data & Unsigned Galois covariance in both indices and character orthogonality sharpen the conductor bound to $f$ (Theorem~\ref{por:th:exact-conductor}).\\ Integral braided category & The LCM energy estimate of Proposition~\ref{prop:braided-lcm} includes the contribution of the symmetric center; the modular case uses the estimate of Dong and Palcoux \cite{DPLCM}.\\ Modular category or category with transparent fermion & Rotations and congruence give principal-block orthogonality (Theorem~\ref{thm:block}); at two the positive reduced table uses squared twists.\\ Integral modular category & The congruence action connects character fields to joint twist components (Theorem~\ref{found:C:7.1}). At the first open dimension, its orbit and phase restrictions enter the rank bound (Theorem~\ref{por:ranktheorem}).\\ \bottomrule \end{tabular} \end{table} \begin{proposition}[LCM energy and prime companions]\label{fam:M:2.2}\label{prop:braided-lcm} Let $\C$ be integral braided, with its positive ribbon structure, and let $S$ be its unnormalized matrix. For $X$ simple set $h_X=\sum_{T\in\Irr(\C')}N_{XX^*}^{T}d_T$. Then \[ \sum_{Y:S_{XY}\ne0}\lcm(d_X,d_Y)^2\le D h_X. \] If $X$ centralizes $\C_{\pt}$, is not transparent, and $p\mid d_X$ while $p\nmid\FPdim(\C_{\pt})$, there is a noninvertible $Y$ with $p\nmid d_Y$, $S_{XY}\ne0$, and $\FPdim(\C_{\pt})d_X^2+\lcm(d_X,d_Y)^2\le Dh_X$. \end{proposition} \begin{proof} The Hopf-link characters $a_Y(X)=S_{XY}/d_Y$ are ring characters \cite[Proposition~8.13.11]{EGNO}. B\'ezout applied to $a_Y(X)$ and $a_X(Y)$ shows $S_{XY}/\lcm(d_X,d_Y)$ is integral, as in the modular criterion of Dong--Palcoux \cite[Lemma~2.7 and the proof of Theorem~1.1]{DPLCM}. The regular element $R=\sum_Y d_Y[Y]$ satisfies $[T]R=d_T R$. Accordingly $a_T(R)$ is $D$ when $a_T=d$ and zero otherwise. In the positive structure $a_T=d$ means precisely $T\in\C'$ by the double-braiding criterion \cite[Proposition~8.20.5(i)]{EGNO}. Symmetry and $a_Y(X^*)=\overline{a_Y(X)}$ give \[ \sum_Y |S_{XY}|^2 =\sum_Y d_Y^2a_Y(XX^*)=\sum_T N_{XX^*}^{T}d_Ta_T(R)=Dh_X. \] Average this rational equality over a cyclotomic splitting field. Every nonzero integral normalized entry has mean squared modulus at least one by its norm. This proves the first assertion. For the second, $a_X(R)=0$ and the invertibles contribute $\FPdim(\C_{\pt})$. Reduction at a prime above $p$ forces a noninvertible $p$-free hit; retain it and all invertibles in the first inequality. \end{proof} In a modular category $h_X=1$. In a slightly degenerate category the same holds for the full table: $gX\not\simeq X$ excludes the transparent fermion from $XX^*$. Passage to the reduced table halves the full energy sum and the number of invertible columns. \begin{lemma}[Averaging over a character orbit]\label{fam:M:2.3}\label{deep:field-energy}\label{r8:trace-norm}\label{fam:Q:2.1}\label{fam:Q:n:q12:orbit} In ordinary or positive reduced symmetric integral data, let $H$ be a subgroup of the cyclotomic Galois group, $e=|H\cdot Z|$ and $A_H(x)=|H|^{-1}\sum_{\sigma\in H}\sigma(x)$. Then \[ f_Z/e=\sum_X |A_H(\chi_Z(X))|^2,\qquad 0=\sum_X d_X A_H(\chi_Z(X))\quad (Z\ne\one). \] For the full Galois group write $K_X=\QQ(\chi_X)$, $F_X=\QQ(\chi_Z(X))$ and $T_X=\Tr_{F_X/\QQ}(\chi_Z(X))$. Grouping columns into orbits gives \begin{align} f_Z&=\sum_{[X]}[K_Z:F_X][K_X:F_X]T_X^2,\label{eq:r8-tracenorm}\\ 0&=\sum_{[X]}d_X[K_X:F_X]T_X\qquad(Z\ne\one).\label{eq:r8-tracedim} \end{align} For any $H$, let $A$ be a set of nonunits whose entries $z_X=S_{XZ}/\lcm(d_X,d_Z)$ are fixed by $H$. With $\mathsf M(z)$ the mean of $|\sigma z|^2$ over all conjugates, we have \[ e\left(d_Z^2+\sum_{X\in A}\lcm(d_X,d_Z)^2\mathsf M(z_X)\right)\le\Delta . \] \end{lemma} \begin{proof} The $e$ conjugates of the row $\chi_Z$ are orthogonal with squared norm $f_Z$, and their average at $X$ is $A_H(\chi_Z(X))$ by \eqref{eq:galois-permutations}. Averaging preserves orthogonality to the dimension character. For the full group, the value at each of the $[K_X:\QQ]$ conjugates of $X$ is $T_X/[F_X:\QQ]$. For the inequality multiply the norm identity by $d_Z^2$, retain the vacuum and $A$, then take the full Galois mean. The Galois group is abelian, so the conjugates of these entries are also fixed by $H$. \end{proof} Write $GZ$, $I_p Z$, $I_p^{\rm tame}Z$ for the full, inertia and chosen tame inertia orbits in the splitting field; at odd $p$ the tame subgroup is the Teichm\"uller subgroup of the primary factor. Conjugates of an entry $z$ are indexed by $\operatorname{Gal}(\QQ(z)/\QQ)$, independently of the orbit of its row label. Thus $A_H$ is a normalized trace, $T_X$ an unnormalized field trace, and $\mathsf M$ a mean squared modulus. In particular, for pairwise disjoint nonvacuum sets $A_i$ of $H$-fixed entries and lower bounds $c_i\le\lcm(d_X,d_Z)^2\mathsf M(z_X)$ for $X\in A_i$, the last inequality reads \begin{equation}\label{eq:disjoint-energy} |HZ|\left(d_Z^2+\sum_i|A_i|c_i\right)\le\Delta . \end{equation} This applies to subsets of columns, even when several columns have the same entry. Two choices of $H$ will recur. For one nonunit $X$, take $H$ fixing $L=\QQ(z_X)$ and $A$ equal to the full Galois orbit of $X$. Then $e=[K_Z:L]$ and the sum is $[K_X:\QQ]\lcm(d_X,d_Z)^2\mathsf M(z_X)$. For inertia at $s$, all columns with $K_X$ unramified at $s$ have fixed entries. If $Z$ is principal at $t\ne s$ and $t\nmid d_X$, then $\chi_Z(X)\equiv d_X$ above $t$ gives $z_X\ne0$, hence $\mathsf M(z_X)\ge1$ by the algebraic norm bound. This applies simultaneously to any set of such columns. \begin{lemma}[Zeros and square-fullness]\label{fam:M:2.4}\label{lem:tables}\label{lem:galois-row-bound}\label{lem:perfect-data-squarefull}\label{lem:strict-eigenvalues}\label{lem:odd-codegrees-duality} In perfect symmetric integral data, every nonunit row contains a zero and $\Delta$ is square-full. For $s=d_X,t=d_Y$, $S_{XY}/\lcm(s,t)$ is an algebraic integer. In a simple perfect table $S_{XY}=0$ whenever $X,Y$ are nonunit and $(s,t)=1$; in particular no nonunit dimension is a prime power. If $\Delta$ is odd then duality has no nonunit fixed point. \end{lemma} \begin{proof} Apply the B\'ezout and Galois-mean bounds of Proposition~\ref{prop:braided-lcm} to either character table. If a nonunit row of dimension $s$ had no zero, averaging its norm would give \[ \Delta/s^2\ \ge\ \sum_Y (d_Y/\gcd(s,d_Y))^2 \ >\ \sum_Y d_Y^2/s^2=\Delta/s^2; \] the strict inequality holds already at $\one$. For ordinary modular categories the row-vanishing theorem is due to Gelaki, Naidu and Nikshych \cite[Theorem~6.1]{GNNcenters}. Square-fullness in the perfect integral modular case is \cite[Corollary~10.10]{BP}, and the coprime-degree and prime-power obstructions in the integral modular case follow from Etingof, Nikshych and Ostrik \cite[Lemma~7.1 and Corollary~7.2]{ENOsolvable}. For the fusion-ring vanishing theorem see \cite[Theorem~1.2]{BP}. Now $P=\prod_X XX^*$ vanishes on every non-Perron character and hence, by \eqref{eq:fusion-coefficient-trace}, $\tau(P)=\prod_X d_X^2/\Delta$ is an integer. Since $d_X^2\mid\Delta$, no prime can occur only once in $\Delta$. For a non-Perron character, the labels at which $|\chi(X)|=d_X$ form a based subring: equality in the triangle inequality for a product propagates to each constituent, and the set is closed under duality. Simplicity therefore makes the inequality strict off the unit, since equality everywhere would force codegree $\Delta$. For coprime $s,t>1$, all conjugates of $S_{XY}/(st)$ have modulus strictly below one; its norm forces it to be zero. On the other hand a nonunit row satisfies \begin{equation}\label{eq:reduced-unit-orthogonality} 1+\sum_{Y\ne\one}d_Y\chi_X(Y)=0. \end{equation} Modulo any prime dividing $\Delta$, the row must meet a nonunit label of degree prime to that prime. A prime-power row cannot do so. In odd dimension all $1/f_X$ reduce to one at two; hence \eqref{eq:fusion-coefficient-trace} gives $\tau(B)\equiv\Tr(N_B)\pmod2$ for $B\in R$. Apply this congruence to $B=X^2$ and then to $B=X$, using $\Tr(N_X^2)\equiv\Tr(N_X)\pmod2$. It gives $\delta_{X,X^*}\equiv\delta_{X,\one}\pmod2$. \end{proof} \begin{lemma}[Small ordinary codegrees]\label{lem:small-ordinary-codegrees} In a nontrivial simple perfect integral modular category every formal codegree is at least five. \end{lemma} \begin{proof} Use the unnormalized modular matrix $S$ with $S_{0Y}=d_Y$. For a nonunit $X$ put $n=d_X$, $f=D/n^2$ and $u_Y=S_{XY}/n$. Write $H=\{Y\ne\one:u_Y\ne0\}$. Orthogonality gives $\sum_{Y\in H} d_Yu_Y=-1$ and $\sum_{Y\in H}|u_Y|^2=f-1$. Thus the degrees in $H$ have gcd one. The $(X,0)$ entry of the modular relation $STS=p^+T^{-1}ST^{-1}$, where $|p^+|=\sqrt D$, gives \[ \sqrt f\,n\le 1+\sum_{Y\in H}d_Y|u_Y|. \] Symmetry and algebraic integrality of the character values give $u_Y\in(d_Y/\gcd(d_Y,n))\mathcal O$, where $\mathcal O$ denotes the algebraic integers. Averaging the squared-norm identity over the cyclotomic Galois group, each nonzero term contributes at least $(d_Y/\gcd(d_Y,n))^2$, by the arithmetic--geometric mean inequality applied to its conjugates. If $f\le4$, we conclude that $|H|\le f-1$ and every $d_Y$ in $H$ divides $n$. Perfection and the gcd condition require at least two terms. If there are two, both degrees are proper divisors of $n$. Cauchy's inequality then gives $\sqrt f\,n\le1+(n/2)\sqrt{2(f-1)}$, impossible for $f=3,4$ and $n\ge2$. For three terms we have $f=4$, and all three entries are roots of unity: each has mean square one, so all its conjugates have modulus one. At most one of the degrees can be $n$, by the gcd condition. If one is $n$, the others cannot both equal $n/2$ since $n/2>1$. Ordering the proper divisors therefore gives $2n\le1+11n/6$. This forces $n\le6$. A nonunit degree in our category has at least two distinct prime factors by Lemma~\ref{lem:tables}; hence $n=6$, and its only divisor admissible as a nonunit degree is $6$, contradicting the gcd condition. The vacuum codegree is $D$ and is larger than four, since such a category contains a nonunit of degree at least six. \end{proof} \subsection{Fields and residue classes of characters}\label{fam:M:sec:2.1} For an irreducible representation $E$ of a fusion ring $R$, put $K=\QQ(\chi_E(X):X)$ and let $e_E$ be its central primitive idempotent. The formal codegree $f_E$ is the Schur element of $\tau$ \cite[Proposition~1.15]{ChlSchur}. Thus $\dim(E) f_E=\sum_X|\chi_E(X)|^2$ and \cite[Lemma~2.3]{OstrikFormalCodegrees} \begin{equation}\label{eq:character-formal-codegree-general} \sum_X\chi_E(X^*)[X]=f_Ee_E . \end{equation} For a categorified ring $K$ is cyclotomic in the sense of lying in a cyclotomic field \cite[Theorem~8.51]{ENOannals}; the reduced characters also lift. \begin{theorem}[The different divides a rational formal codegree]\label{fam:M:2.5}\label{lem:codifferent} For an irreducible representation of a categorified fusion ring, or of a fusion ring with cyclotomic character values, a rational formal codegree $f$ is a positive integer. The different of its character field satisfies \begin{equation}\label{eq:character-different-divisibility} \mathfrak D_K\mid f\mathcal O_K . \end{equation} Only primes dividing $f$ ramify. \end{theorem} \begin{proof} For $a\in\mathfrak D_K^{-1}$ the element $\sum_X\Tr_{K/\QQ}(a\chi_E(X^*))[X]$ belongs to $R$. By \eqref{eq:character-formal-codegree-general} it equals $f\sum_{\sigma\in\Gal(K/\QQ)}\sigma(a)e_{\sigma E}$, with distinct conjugate characters. Its eigenvalue $af$ is integral. Thus $f\mathfrak D_K^{-1}\subseteq\mathcal O_K$, and $a=1$ gives the integrality of $f$. \end{proof} \begin{corollary}[Conductor of a character field]\label{fam:M:2.6}\label{cor:field-conductor} If a cyclotomic character field $K$ has rational formal codegree $f$, its conductor divides $2f$. \end{corollary} \begin{proof} Work after a finite unramified extension of $\QQ_p$, so the ramified part $E$ of $K$ lies in the totally ramified cyclotomic extension $L$ of minimal conductor $p^m$. Let $e=[E:F]$, where $F$ is the unramified base, and let $d_E$ be the different exponent at $E$. Theorem~\ref{lem:codifferent} gives $d_E/e\le v_p(f)$. The derivative of $\Phi_{p^m}$ gives $d_L/[L:F]=m-1/(p-1)$. For odd $p$ and $e=e_0p^s>1$, $e_0\mid p-1$, cyclic wild inertia gives $m=s+1$ and $L/E$ is tame. Thus $(d_E+1)/e=(d_L+1)/[L:F]$, or \begin{equation}\label{eq:odd-normalized-different} \delta_p(e):=d_E/e=s+1-\frac{1-p^{-s}}{p-1}-\frac1e . \end{equation} This also gives zero at $e=1$. For $s=0$, $e>1$ it is positive, and for $s\ge1$ it lies strictly between $s$ and $s+1$. Hence $p^m\mid f$. At two, conductor four gives normalized different one. For $m\ge3$, minimality excludes $1+2^{m-1}$ from $\Gal(L/E)$. Every nontrivial subgroup of $\langle5\rangle$ contains that element, so $[L:E]\le2$. The possible quadratic generators are $-1$ and $2^{m-1}-1$. For either generator $\sigma$, a primitive $2^m$th root $\zeta$ satisfies $v_L(\zeta-\sigma\zeta)=2$; this is the relative different exponent. Transitivity therefore gives $d_E/e\ge m-1-2/2^{m-1}>m-2$, whence $2^{m-1}\mid f$. \end{proof} By symmetry the field of $S_{XY}/\lcm(d_X,d_Y)$ lies in both character fields; its different divides both codegrees by transitivity, and the next theorem sharpens its conductor bound. In particular, the field of a character of $p$-power codegree lies in a $p$-power cyclotomic field by Kronecker--Weber and the conductor bound (for odd $p$, adjoining a factor of two to an odd level does not change the field). Total ramification there gives a unique prime over $p$ in the character field \cite[Lemma~3.1.13 and Theorem~7.5.5]{Sharifi}. \begin{theorem}[Conductor at an individual label]\label{por:th:exact-conductor} In symmetric integral character data (Definition~\ref{def:symmetric-integral-data}), write $K_Z$ for the field generated by the character $\chi_Z$. Then the conductor of $K_Z$ divides $f_Z$. In particular the conductor of the lcm-normalized $XY$ crossing divides $\gcd(f_X,f_Y)$. In ordinary data, if the dyadic part of a twist has order $2^v$ with $v\ge4$, then $2^{v-1}\mid f_X$ for its label $X$. \end{theorem} \begin{proof} The different bound (Corollary~\ref{cor:field-conductor}) already puts $K_Z$ in a cyclotomic field of conductor dividing $2f_Z$. It remains to remove its last possible dyadic level. Put $m=2^{v_2(f_Z)}$. There is nothing to check for $m=1$. Take $L=F(\zeta)$ with $F$ an odd cyclotomic field large enough and $\zeta$ primitive of order $2m$. Suppose the dyadic conductor is $2m$. In $G=(\mathbb Z/2m)^\times$ the stabilizer $H$ of the row does not contain $e=1+m$, since the fixed field of $e$ is $F(\zeta^2)$. Write $b_Y=\chi_Z(Y)=\sum_{k=0}^{m-1}z_{kY}\zeta^k$. The coefficients belong to $\mathcal O_F$ by coprimality of conductors. The orthogonality of the $m$ odd powers of $\zeta$ gives \[ z_{1Y}=m^{-1}\sum_{g\in G}\zeta^{-g}\chi_{gZ}(Y). \] Different rows in this sum are orthogonal; the squared norm of each is $f_Z$. Thus \[ \sum_Y|z_{1Y}|^2=\frac{f_Z}{m^2}\sum_{h\in H}\sum_{g\in G}\zeta^{g(h-1)} =\frac{f_Z}{m}. \] Indeed the inner geometric sum vanishes unless $h=1$ or $h=e$, and only $1$ belongs to $H$. Lift $e$ to the common splitting field, trivially on $F$. Its permutation of the argument labels sends $z_{1Y}$ to $-z_{1Y}$. Every cycle on the nonzero coefficients therefore has even length and contributes to $2\mathcal O$ in the squared norm. This contradicts oddness of $f_Z/m$. Symmetry gives the crossing assertion. For the twist assertion, Theorem~\ref{thm:twistdiv} puts $\mathbb Q(\zeta_{2^{v-1}})^+$ in $K_X$ by the square action whenever $v\ge4$. Its conductor is $2^{v-1}$; the normalization scalar has order at most eight, hence does not affect this conclusion. \end{proof} \paragraph{Local projectors.} Fix a prime $\mathfrak p$ of a splitting field over $p$. The full residue classes (or $p$-blocks) group the characters of the commutative order $R$ with equal reductions on every basis element. Over a complete unramified coefficient ring $\mathcal O$ containing the residues in its residue field, the special-fiber idempotents lift. Hence for a union $L$ of full blocks, $e_L=\sum_{X\in L}e_X$ lies in $R\otimes\mathcal O$, and for locally integral $B$, \begin{equation}\label{eq:local} \tau(e_LB)=\sum_{X\in L}\chi_X(B)/f_X\in\mathcal O . \end{equation} The character transform $\Phi(B)=(\chi_z(B))_z$ sends $e_z$ to the coordinate vector $v_z$ and intertwines multiplication by $e_L$ with $P_L=\sum_{z\in L}v_zv_z^*$, the coordinate projection. We keep the algebra element $e_L$ in trace identities and the operator $P_L$ in matrix identities. \begin{lemma}[Principal residues at every prime]\label{lem:principal-all-places} In the data of Definition~\ref{def:symmetric-integral-data}, if $Z$ is principal at one place over $p$, then $\chi_Z(X)\equiv d_X$ at every place over $p$ for every $X$. The congruence may be divided by a rational $p$-unit. In particular, a locally integral algebraic number $u$ which vanishes at all places above $p$ belongs to $p\mathcal O_{K,(p)}$ if its field $K$ is unramified at $p$. \end{lemma} \begin{proof} For any $\sigma$, $\sigma\chi_Z(X)=\chi_Z(\pi_\sigma X)$ and $d_{\pi_\sigma X}=d_X$, so the congruence survives transport to each conjugate place. The last assertion follows from $p\mathcal O_{K,(p)}=\prod_{\mathfrak p\mid p}\mathfrak p\mathcal O_{K,(p)}$. \end{proof} \begin{theorem}[Support in the principal residue class]\label{fam:M:2.7}\label{thm:principal-support}\label{lem:prime-power-support} In symmetric integral character data, for any label set $\mathcal A$ put $L=\{X:\chi_X(W)\equiv d_W\pmod{\mathfrak p}\ \text{for every } W\in\mathcal A\}$. If $f_Z=p^a$, the support of $\chi_Z$ lies in $L$; it equals $L$ when $\mathcal A=I_{d_Z}$ and $p\nmid d_Z$. For any $Z$ and any Galois-stable $\mathcal A$, $\sum_{X\in L}|\chi_Z(X)|^2$ is an integer divisible by $p^{v_p(f_Z)}$. \end{theorem} \begin{proof} Write $\alpha$ for the sum. Symmetry gives $\alpha/f_Z=\tau(e_LZZ^*)\in\mathcal O$. If $\mathcal A$ is Galois-stable, so is $L$ at the fixed prime $\mathfrak p$: by \eqref{eq:galois-permutations}, permuting $X$ permutes the tests on $W$ and preserves $d_W$. Thus $\alpha\in\ZZ$ with the asserted divisibility. Apply this to the set of \emph{all} labels. Its principal block contains the unit, so $1\le\alpha\le f_Z=p^a$ forces $\alpha=f_Z$. The support lies in this block and hence in every $L$. Conversely for $\mathcal A=I_{d_Z}$, $X\in L$ gives $\chi_X(Z)\equiv d_Z$, nonzero if $p\nmid d_Z$. \end{proof} \paragraph{Blocks of defect one.} After a sufficiently large finite unramified base change with fraction field $F$, integer ring $\mathcal O$ and residue field $k$, the field factors of a block correspond to its full inertia orbits and are totally ramified. An orbit is \emph{tame} when wild inertia fixes it pointwise. The following observation explains why a single character of small defect can determine the entire block. \begin{lemma}[A block with a conductor of length one]\label{fam:M:2.9}\label{lem:conductor-node} Let $A$ be a commutative local symmetric $\mathcal O$-order with normalization $M=\prod_i\mathcal O_i$, where all residue fields are $k$. Suppose its perfect trace on factor $i$ is $\Tr_{E_i/F}/f_i$, and put $c_i=e_i v_p(f_i)-d_i$, where $e_i,d_i$ are its ramification index and different exponent. A factor with $c_i=0$ is the whole block and $A=\mathcal O_i$. If $c_i=1$, there are precisely two factors, both with $c=1$, and $A=\{(x,y)\in M:\bar x=\bar y\}$. \end{lemma} \begin{proof} Since $A$ is self-dual, the conductor $(A:M)$ is the trace dual $M^\#=\prod_i f_i\mathfrak D_i^{-1}$. A unit conductor on one factor puts its idempotent in the local ring $A$. If $c_i=1$, the conductor contains its maximal ideal $\mathfrak m_i$ on that factor alone. Writing $B$ for the projection to the other factors, we have $A=\mathcal O_i\times_k B$: common residues are necessary, and the copy of $\mathfrak m_i$ lifts every compatible pair. There must be another factor, since otherwise $A=\mathcal O_i$ has conductor exponent zero. The perfect trace identifies $A$ with $\Hom_{\mathcal O}(A,\mathcal O)$, so $A$ is Gorenstein. The fiber-product criterion \cite[Fact~2.6 and Corollary~2.7]{NSTV} makes $B$ a discrete valuation ring. The conductor of this two-factor order is the product of the two maximal ideals. \end{proof} \begin{proposition}[The block of a defect-one character]\label{fam:M:2.10} \label{seq:odd-defect-one}\label{lem:class}\label{prop:principal-defect-one-general}\label{lem:one-defect-block}\label{lem:two-adic-defect-one-rank}\label{lem:even-rank-defect-one} In a local symmetric order as above, with cyclotomic characters and rational integral codegrees, suppose a character has defect $v_p(f)=1$. Except for a single ramified quadratic branch at two, the block has two inertia orbits and is the fiber product of their valuation rings over $k$. In particular no unramified higher-defect character belongs to this block. For odd $p$ both orbits are tame and of defect one. At $p=3$ their lengths in symmetric integral character data are one and two. At two, if the given branch is unramified and the other has degree $e$, the block has rank $1+e$ and the other defect is $1+v_2(e)$. Consequently uniform defect one at two forces rank two; in an even-rank dyadic block of rank greater than two all defects are at least two. \end{proposition} \begin{proof} At odd $p$ the given branch is tame by Corollary~\ref{cor:field-conductor}, so $e_i-d_i=1$; an unramified defect-one branch at two has the same conductor length. By Lemma~\ref{lem:conductor-node} the other branch satisfies $v_p(f)=(d_E+1)/e$, which gives defect one if it is unramified. At odd $p$ the right side for wild degree $p^s>1$ lies strictly between $s$ and $s+1$ by \eqref{eq:odd-normalized-different}; hence both branches are tame of defect one. At three their lengths are one or two. The integral identity trace $\sum e_i/f_i=\sum e_i d_{X_i}^2/\Delta$ forces the sum of the lengths to vanish modulo three, since after canceling the common powers of three each unit square is one there. At two a ramified defect-one branch satisfies $2(e_i-1)\le d_i\le e_i$ by the first two ramification groups and the different bound. Hence $e_i=d_i=2$ and the conductor has length zero. In the two-branch case the identity trace $1/f_1+e/f_2$ is integral; its first summand has valuation $-1$, so the other does too. Since $e$ is a power of two, even rank $1+e$ forces $e=1$. A defect-zero branch is unramified with conductor length zero, hence has rank one by itself. \end{proof} \section{Prime blocks and primary twists}\label{fam:M:sec:3}\label{sec:blocks} Write $s=S/\sqrt D$, $T=\operatorname{diag}(\theta_i)$ and $N=\ord(T)$ for ordinary modular data. Let $v_i$ denote a coordinate vector, $e_i$ a character idempotent, and $0$ the unit label. A twist decomposes uniquely into primary parts $\theta_{i,p}$; we call $i$ \emph{$p$-active} if $\theta_{i,p}\ne1$. For a weakly integral modular category use the lift $W=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\mapsto s$, $U=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\mapsto t=T/\gamma$, where $\gamma^3=p^+/\sqrt D$. We may take $\gamma$ of two-power order, equal to one for a center. The representation factors through $\mathrm{SL}_2(\mathbb Z/M)$, $M=\ord(t)$ \cite[(1-10), Theorem~II(i), Proposition~6.7(ii) and Theorem~6.10]{DLNCongruence}. A subscript on $W,U$ denotes its local factor under the Chinese remainder isomorphism. For integral slightly degenerate $\mathcal B$ put $\Delta=\FPdim(\mathcal B)/2$ and let $g$ be its transparent fermion. With the positive spherical structure, $\mathcal B$ is pseudo-unitary, so the extension theorem of Johnson-Freyd and Reutter \cite[Corollary~1.3]{JFR} gives a minimal modular extension of dimension $4\Delta$, with matrices $s,T$. The Weyl matrix interchanges $v_i\mapsto v_{gi}$ and $v_i\mapsto\epsilon_i v_i$, with $\epsilon_i$ the fermion-monodromy sign. On their common $+1$ space, the vectors $(v_i+v_{gi})/\sqrt2$ give the normalized reduced matrix $\widehat s$. Table~\ref{tab:fermion-dictionary} gives the full normalization. Even twist powers and odd local factors preserve this space. Their Bruhat and Galois identities therefore restrict, with the same local Galois sign since the two dimension normalizations differ by a rational factor. \begin{lemma}[Rotation with a central insertion]\label{lem:rotation-congruences} Let $\C$ be a weakly integral modular category with its positive spherical structure. For a simple object $C$ of its center and $k\ge1$ put $L_k^C=\nu^C_{k,1}$. Choose a cyclotomic integer ring $\mathcal O$ containing the coefficients being used. The additive functional $L_k^C$ extends $\mathcal O$-linearly to $\mathcal O\otimes K_0(\C)$, with values in algebraic integers. We extend scalars to the splitting field when evaluating $L_k^C$ on an idempotent. \begin{enumerate}[label=\textup{(\roman*)},leftmargin=*] \item If $p$ is prime, $p\mid k$, and $y$ belongs to this algebra, then at each prime above $p$, \[ L_k^C(y)^p\equiv L_{k/p}^C(y^p). \] \item Suppose $\theta_C=1$ and $A$ is a genuine object. Put $m=N_{p'}$, choose $b\ge\max(1,v_p(N))$ with $p^b\equiv1\pmod m$, and let $n\ge2b$. Then \begin{equation}\label{eq:large-power} L_m^C(A^{p^n})\equiv L_{mp^b}^C(A^{p^{n-b}}) \pmod {p^{n-2b+1}\overline{\ZZ}_p}. \end{equation} \end{enumerate} For $C=(i,j^*)$ under $\Z(\C)\simeq\C\boxtimes\C^{\rev}$ write $L_k^{ij}=L_k^C$. With $A_k=sT^ks^{-1}$ one has \begin{align} L_k^{ij}(f_z e_z)&=f_z(A_k)_{iz}\overline{(A_k)_{jz}}, \label{eq:mixed-rotation}\\ L_k^{ii}(e_z)&=|(A_k)_{iz}|^2.\label{eq:indicator-entry} \end{align} \end{lemma} \begin{proof} The $k$th power of the rotation is the scalar $\theta_C^{-1}$, so its eigenvalues are roots of unity. Its trace is additive in the object by \cite[Corollary~5.5]{NSind}, whose coprime-index hypothesis holds for $(k,1)$. On a genuine object, reduction of the trace to the $p$th power gives the trace of the $p$th power of the rotation. Grouping $p$ strands identifies the latter with $L_{k/p}^C(A^{\otimes p})$ by \cite[Lemmas~2.5, 2.7 and Proposition~2.8]{NSind}. The Frobenius map in the commutative residue fusion algebra now gives (i) for integral virtual coefficients. For (ii) the two sides are traces of the powers $p^n,p^{n-b}$ of the same rotation on $mp^n$ strands. Its eigenvalues have order dividing $mp^n$ because $\theta_C=1$. All its power traces lie in $\QQ(\zeta_N)$ by \cite[Proposition~5.7]{NSind} and \cite[Theorem~7.7]{NSfs}, so its eigenvalues have constant multiplicity on orbits over that field. The indicated powers agree on the prime-to-$p$ part and on roots whose $p$-part has order at most $p^{n-b}$. For a root with $p$-part of order $p^t>p^{n-b}$, use the subgroup acting trivially on $\mu_m$ and by units $1\bmod p^b$ on the $p$-part. It fixes $\QQ(\zeta_N)$, its orbit has size $p^{t-b}$, and each of the two powers is constant on that orbit, since $t\le n$. Thus its contribution to the difference is divisible by $p^{n-2b+1}$. Finally \cite[Proposition~6.1]{NSind} gives $L_k^{ij}([c])=\sum_{a,b}s_{ia}s_{jb}(\theta_a/\theta_b)^kN_{ab}^c$. Contract with $f_ze_z=\sum_c\overline{\chi_z(c)}[c]$. Multiplicativity replaces $N_{ab}^c$ by $\overline{\chi_z(a)\chi_z(b)}$ after summing on $c$; replacing $b$ by $b^*$ in the second sum gives the two conjugate matrix factors, with $s_{0z}^{-2}=f_z$. \end{proof} \begin{theorem}[Block orthogonality]\label{fam:M:3.1}\label{thm:block}\label{thm:zero}\label{lem:reduced-block}\label{lem:reduced-two-block} For a full $p$-block $B$ of ordinary modular data, \[ \sum_{z\in B}\overline{s_{zi}}s_{zj}=0 \qquad(\theta_{i,p}\ne\theta_{j,p}). \] In reduced integral data the same holds for $\widehat s$ at odd $p$, and at two whenever $\theta_{i,2}^2\ne\theta_{j,2}^2$. In particular, in the ordinary table \begin{equation}\label{eq:zero} p\nmid f_z\in\ZZ,\quad p\mid\ord(\theta_j) \quad\Longrightarrow\quad S_{zj}=0 . \end{equation} \end{theorem} \begin{proof} Write $L_k^i=L_k^{ii}$. The insertion $(i,i^*)$ has twist one because dual labels have the same twist. Apply Lemma~\ref{lem:rotation-congruences}(ii) with $m=N_{p'}$. \emph{Separating blocks.} In the mod-$p$ fusion ring, residues with a given kernel and image $\mathbb F_{p^h}$ are Frobenius conjugate. Denote their blocks in Frobenius order by $B_0,\ldots,B_{h-1}$ (all conjugates occur by local Galois symmetry). For $a\in\mathbb F_{p^h}^{\times}$, lift the Chinese-remainder element with residue $a$ at this kernel and zero at all others to a genuine object, taking nonnegative basis coefficients. Choose $h\mid b,n$. As $n$ tends to infinity, both powers of the object in \eqref{eq:large-power} converge in character coordinates to $\sum_j[a]^{p^j}e_{B_j}$, where brackets denote Teichm\"uller lifts. Hence $L_m^i$ and $L_{mp^b}^i$ agree on that sum. The characters $a\mapsto[a]^{p^j}$ of $\mathbb F_{p^h}^\times$ are distinct, so the functionals agree on each $e_{B_j}$. As $N\mid mp^b$, \eqref{eq:indicator-entry} gives \begin{equation}\label{eq:block-projector-mass} \sum_{z\in B}|(sT^ms^{-1})_{iz}|^2=\mathbf1_{i\in B}. \end{equation} Thus $sT^ms^{-1}$ commutes with the coordinate projector of $B$. Conjugating back gives the assertion because $T^m$ distinguishes the $p$-primary twists. If $p\nmid f_z$, the idempotent $e_z$ is locally integral, so $\{z\}$ is a block. Taking $i=0$ gives \eqref{eq:zero}. \emph{Passage to a fermion quotient.} In its minimal modular extension select all labels whose characters on the even subring have the prescribed residue of $B$. Their projector $P$ is a sum of ordinary full-block projectors: equality of residues on the full ring implies equality on the even part. Since $z$ and $gz$ agree on the even subring, $P$ preserves the positive fermion subspace, on which it restricts to the reduced block projector. Put $T_p=\operatorname{diag}(\theta_{i,p})$. The ordinary result gives $[P,sT_ps^{-1}]=0$. Restrict this identity to the positive subspace at odd $p$, or its squared-operator version at two: $s$ preserves that subspace and the members of a fermion pair have equal odd primary twists and equal squared dyadic twists. This gives precisely the two reduced assertions. \end{proof} \begin{lemma}[Local modular action on residue blocks]\label{fam:M:3.2}\label{lem:local-block-invariance} Every full $p$-block subspace is invariant under the $p$-local modular group, in ordinary weakly integral data and also in positive reduced data at odd $p$. In particular $W_p$ preserves each block. \end{lemma} \begin{proof} Its coordinate projector commutes with $U_p$, which is diagonal, and with $sU_ps^{-1}$ by Theorem~\ref{thm:block}, since $U_p$ differs from $T_p$ by a scalar. These elementary unipotents generate the local factor. \end{proof} \subsection{When two primes never occur in the same twist}\label{fam:M:sec:3.1} \begin{lemma}[Twist primes seen by a row]\label{fam:M:3.3}\label{lem:complement} In ordinary weakly integral modular data, if the local Weyl element $W_p$ preserves the line of $z$ then $p\nmid f_z$. In particular: \begin{enumerate} \item $p\mid f_z$ exactly when the $z$-row hits a $p$-active column; \item if no twist contains both $p$ and $q$, every $p$-active label has $q$-unit codegree. \end{enumerate} Every prime in the twist of $z$ divides $f_z$. \end{lemma} \begin{proof} We may assume $p\mid D$ and hence $p$ divides the level by categorical Cauchy. Choose $k$ divisible by its $p$-part and congruent to one on the complementary level. The identity $U(WUW^{-1})U=W^{-1}$ on the complementary factors identifies $sT^ks^{-1}$, up to diagonal phases on the left and right, with their inverse Weyl. If $W_p$ preserves the line of $z$, the absolute value of the resulting $(0,z)$ entry is $|s_{0z}|=1/\sqrt{f_z}$. Thus for $y=f_ze_z$, \eqref{eq:indicator-entry} gives $L_k^0(y)=1$. By Lemma~\ref{lem:rotation-congruences}(i), \[ L_k^0(y)^p\equiv L_{k/p}^0(y^p) =f_z^{p-1}L_{k/p}^0(y) \] at every prime above $p$. Thus $p\nmid f_z$. Now if the row misses all active columns, then $sT_p s^{-1}v_z=v_z$. The local upper and lower unipotents both preserve the line; they generate the local group, so $W_p$ also preserves it. On this line the two unipotents have equal scalars because they are conjugate. Comparing with $sT_p s^{-1}v_z=v_z$ gives $\theta_{z,p}=1$. Together with \eqref{eq:zero} these facts give (1) and the twist assertion. For (2) the $q$-local group preserves every nontrivial $p$-twist fiber. There its upper unipotent and all its local conjugates act by scalars since the $q$-twist is trivial. Thus $W_q$ preserves each coordinate line. \end{proof} \begin{proposition}[Cauchy and vanishing at odd primes]\label{fam:M:3.4}\label{lem:reduced-cauchy} For an integral slightly degenerate category and any odd $p\mid\Delta$, some simple has $p$-active twist. All such labels have zero $S$-entry on rows of $p$-unit codegree. \end{proposition} \begin{proof} If all even labels of a minimal extension were $p$-inactive, the $p$-local upper unipotent would fix the positive fermion subspace pointwise (the anomaly is dyadic). The lower unipotent would too, since $s$ preserves this subspace. The fixed space of the local factor is invariant under the entire modular representation and contains $v_0+v_g$, hence also $T(v_0+v_g)=v_0-v_g$ and $sv_0$. Every coordinate of $sv_0$ is nonzero. Thus the $p$-primary twist would be trivial everywhere, contrary to Cauchy's theorem in the extension \cite[Theorem~3.9]{BNRWFinite}. Vanishing follows from block orthogonality on the singleton block of a $p$-unit codegree. \end{proof} \subsection{Twist orders and individual codegrees}\label{fam:M:sec:3.2} \begin{theorem}[Twist orders and formal codegrees]\label{fam:M:3.5}\label{thm:twistdiv} In a weakly integral modular category put $m_z=\ord(\theta_z)$ and $f_z=D/d_z^2$. Then \[ (m_z)_{\rm odd}\mid f_z,\qquad m_z\mid \begin{cases}4f_z&\text{in the integral case},\\8f_z&\text{in general},\end{cases} \qquad \ord(\theta_x/\theta_y)\mid4\lcm(f_x,f_y),\qquad N\mid4D. \] The odd-part assertion holds also for an integral slightly degenerate category with reduced codegree $(D/2)/d_z^2$. \end{theorem} \begin{proof} The lift $t=T/\gamma$ has $\ord(\gamma)\mid8$ in the integral case and $\ord(\gamma)\mid16$ in general \cite[Proposition~6.7 and Theorem~6.10]{DLNCongruence}. If $p^k$ is the exact $p$-part of $\ord(t_{zz})$, each automorphism fixing $K_z$ fixes $z$, so its square fixes $t_{zz}$ \cite[Theorem~II]{DLNCongruence}. At odd $p$ its exponent is $\pm1\bmod p^k$; at two and $k\ge4$ it is $\pm1\bmod 2^{k-1}$. Thus $K_z$ contains the real subfield $F$ at conductor $p^a$, with $a=k$ for odd $p$ and $a=k-1$ at two. The conductor of $F$ is $p^a$ unless $F=\QQ$. By Corollary~\ref{cor:field-conductor}, at odd $p$ we obtain $p^k\mid f_z$; the exception $p^k=3$ follows from Lemma~\ref{lem:complement}. At two and $k\ge4$ we get $2^{k-2}\mid f_z$. For an actual twist whose two-part exceeds $\ord(\gamma)$ this gives the same estimate. Otherwise use $2\mid f_z$ for every nontrivial actual two-part, again by Lemma~\ref{lem:complement}. This gives the two objectwise bounds. For a ratio with two-part $2^k$, $k\ge4$, one of $t_{xx},t_{yy}$ has two-part at least $2^k$, so $2^{k-2}\mid\lcm(f_x,f_y)$. For $1\le k\le3$ at least one actual twist has even order. Combined with the odd-part result this proves the ratio bound; taking $y=0$ gives $N\mid4D$ since the codegrees divide $D$. In a minimal modular extension each original label has codegree four times its reduced codegree, giving the last assertion. \end{proof} For modular categories of prime-power exponent, Czenky, Plavnik and Schopieray bound the exponent by the norm of the global dimension, up to a factor four at two \cite[main theorem, p.~2]{CPSexponent}. Here integrality of $D$ allows arbitrary exponent and gives an individual bound. Ising ($D=4$, $d_z^2=2$, $m_z=16$) attains the weakly integral factor eight and the global factor four. By passage to the center and \cite[Theorem~5.5]{NSfs}, the Frobenius--Schur exponent of any weakly integral category of dimension $n$ divides $4n^2$, and its odd part divides $n^2$. \subsection{Mass of a primary twist sector}\label{fam:M:sec:3.3}\label{sec:twist-mass} In an integral modular category write $\mu(A)=D^{-1}\sum_{X\in A}d_X^2$ for a set of labels $A$. \begin{lemma}[Mass and overlap of two label sets]\label{fam:M:3.6}\label{mass} For sets of labels $A,B$ put $a=\mu(A)$, $b=\mu(B)$. Then \[ D^{-1}\sum_{X\in A,Y\in B} d_Xd_Y|s_{XY}| \ \ge\ ab-\sqrt{a(1-a)b(1-b)}. \] In particular $S_{A,B}=0$ implies $a+b\le1$, and hence the union of the $A$-row supports has mass at least $a$. For $m$ labels $H$ of common degree $z$ and codegree $f=D/z^2$, taking $A=B=H$ gives $\sum_{X,Y\in H}|S_{XY}|/z\ge\sqrt f\,m(2m/f-1)$. \end{lemma} \begin{proof} The ranges of the projections $P=s^*P_As$, $Q=TsP_Bs^*T^*$ have orthonormal bases $s^*v_X,Tsv_Y$. Their cross inner products have moduli $|s_{XY}|$ by $sTs=\xi T^{-1}sT^{-1}$, $|\xi|=1$. Expanding $\langle Pv_0,Qv_0\rangle$ in these bases bounds its modulus by the displayed sum. The vacuum coordinates of $Pv_0,Qv_0$ are $a,b$, equal to their squared norms. On their components orthogonal to $v_0$, Cauchy--Schwarz gives the claimed lower bound. \end{proof} \begin{proposition}[Mass from a generating simple]\label{fam:M:3.7}\label{th:fs} Put $L_p=\sum_{p\mid\ord(\theta_i)}d_i^2$. If $X$ generates $\C$ and $L_p>0$, then $L_p\ge(p-1)D/(2p d_X)$. In a fusion-simple integral modular category one can take the smallest nonunit degree whenever $p\mid D$. \end{proposition} \begin{proof} Set $n=N_{p'}$, $t_i=\theta_i^n$. The indicator formula \cite[Theorem~7.5]{NSfs} gives \[ \nu_n(X)=D^{-1}\sum_{i,j}N_{ij}^{X}d_i d_j\,t_i/t_j\in\ZZ : \] it is an algebraic integer in both the $p$-power cyclotomic field of these phases and the prime-to-$p$ field of the $n$-fold rotation. The weights sum to $Dd_X$. Equality $\nu_n(X)=d_X$ would force equal phases on every edge $i^*\to j$ of multiplication by $X$ (duality preserves twists). Since $X$ generates and $X^*$ reverses the edges, every phase would be one. Thus $d_X-\nu_n(X)\ge1$. The mean of a nontrivial $p$-power root is $-1/(p-1)$ at order $p$ and zero otherwise. After Galois averaging, each unequal-phase pair contributes at most $p/(p-1)$ times its normalized weight to this deficit. Its first or second label is active; either choice has total weight $L_pd_X$ by reciprocity. This proves the bound. Categorical Cauchy gives $L_p>0$ for $p\mid D$ \cite[Theorem~3.9]{BNRWFinite}. \end{proof} \begin{proposition}[Odd-valuation slices]\label{fam:M:3.8}\label{prop:blockwise-odd-slice}\label{prop:odd-slice} Suppose $p$ and $v_p(D)$ are odd. Let $B$ be a full $p$-block or the whole label set. Let $x,y$ have trivial $p$-twists, with $y$ fixed by $p$-inertia. For a complementary twist $\eta$ put \[ M_j=\sum_{\substack{i\in B,\ \theta_{i,p'}=\eta\\ \ord(\theta_{i,p})=p^j}}s_{xi}\overline{s_{yi}}. \] Then $M_1=(p-1)M_0$, also after summing over $\eta$. The same formula holds in integral positive reduced data with $D$ replaced by $\Delta$, provided the complementary fiber is taken up to the fermion sign. Thus on each such fiber the squared-dimension mass of order-$p$ twists is $p-1$ times the inactive mass, and the active proportion is at least $1-1/p$. For the entire label set, activity at all primes in any set $J$ of odd primes of odd valuation has proportion at least $\prod_{p\in J}(1-1/p)$, even conditional on fixed twist components outside $J$. \end{proposition} \begin{proof} Let $E$ project onto $B\cap\{i:\theta_{i,p'}=\eta\}$. It commutes with the local $p$-factor by Lemma~\ref{lem:local-block-invariance} and commutation of distinct factors; so does $Q=sEs^{-1}$ since $s$ normalizes that factor. Write $u=T_p,w,h_t$ for its upper unipotent, Weyl and torus. For a unit $n$ modulo the local level $p^a$, the $n$th weighted moment of the $p$-twist on the fiber is \[ (Qwu^nw^{-1})_{xy}=(Qu^{-1/n}wh_{-n}u^{-1/n})_{xy} =\Big(\frac{-n}{p}\Big)(Qw)_{xy}. \] The middle identity is Bruhat. The outer unipotents act trivially at $x,y$. Since inertia fixes $y$, the torus acts there by the Galois sign of $d_y/\sqrt D$ \cite[Theorem~II and Remark~4.5]{DLNCongruence}; at odd $v_p(D)$ this is the displayed quadratic character, from $\sqrt{(-1)^{(p-1)/2}p}$. Average over $n$: the right side vanishes, while the left gives $M_0-M_1/(p-1)$ by the means of $p$-power roots. For the reduced formula, use the restricted projector of Theorem~\ref{thm:block} and the complementary twists modulo sign. The odd local group acts on the positive subspace, and its torus sign is the same since the two dimension normalizations differ by two. At $x=y=0$ the weights are $d_i^2/D$ or $d_i^2/\Delta$. The active bound holds on each complementary fiber and thus on every union of them; iteration over $J$ proves the product bound. \end{proof} \begin{corollary}[The two masses in a defect-one block]\label{fam:M:3.9} \label{cor:defect-one-orbit-masses} Let $B$ be a full $p$-residue block of an integral modular category, where $p$ is odd, and assume that $v_p(f_i)=1$ for every $i\in B$. Then $B$ consists of one inertia orbit with trivial $p$-primary twist and one inertia orbit with $p$-primary twist of order $p$. If their respective lengths and dimensions are $a,r$ and $b,s$, then \[ b s^2=(p-1)a r^2. \] \end{corollary} \begin{proof} The codegree identity implies that $v_p(D)$ is odd. Proposition~\ref{lem:class} gives exactly two inertia orbits; their lengths divide $p-1$ and the dimension is constant on each. Theorem~\ref{thm:twistdiv} bounds all their $p$-primary twist orders by $p$. Inertia preserves these orders. Proposition~\ref{prop:blockwise-odd-slice}, with the complementary condition omitted, says that both the trivial and the order-$p$ masses are nonzero and have ratio $p-1$. Thus one orbit is of each kind and the displayed equation follows. \end{proof} \begin{lemma}[The visible class at defect one]\label{lem:quintic-common-class} In an ordinary integral modular table, let $t,s$ be distinct odd primes with $s\nmid t-1$. Suppose $\mathcal U$ is a nonempty set of $s$-free labels of $t$-defect one and $\mathcal V$ a nonempty set of $t$-defect-one labels in $P_s$. All these rows see only $1$ and a single common square class of primitive $t$th roots as $t$-phases. \end{lemma} \begin{proof} A defect-one block has an inactive and an active orbit; the latter has one square class of primitive phases. Their lengths $a_0,a_1\mid t-1$ and degrees $d_0,d_1$ satisfy $a_1d_1^2=(t-1)a_0d_0^2$ (Proposition~\ref{fam:M:2.10}, Corollary~\ref{fam:M:3.9}). Thus the two degrees have equal $s$-valuation. In each block of a label in $\mathcal U$, all active representatives have $s$-unit degree, so every $V\in\mathcal V$ hits them by principality. Its row can see only phases in its own block: the local Weyl preserves the block and the complementary Weyl preserves $t$-phases. The active classes of every such pair of blocks therefore agree. \end{proof} \section{Principal defects and fermion quotients}\label{fam:M:sec:4}\label{sec:three} The conductor of a residue block bounds its principal degrees. Paired with orthogonality, this excludes a third repeated prime of exponent at most three. Two further tools will also be needed at higher exponents: comparison of the two fermion quotients at two, and positive residues of a common character value at distinct primes. \subsection{The first possible principal defect}\label{fam:M:sec:4.1} \begin{corollary}[Principal valuation bound]\label{fam:M:4.1} \label{cor:principal-valuation-bound}\label{cor:principal-cubic} If $v_p(\Delta)\ge2$, every ordinary or reduced principal-block row satisfies \[ v_p(f_X)\ge2,\qquad v_p(d_X)\le\lfloor(v_p(\Delta)-2)/2\rfloor . \] In particular the defect is at least three at odd valuation, and at valuation two or three every principal degree is $p$-free. \end{corollary} \begin{proof} The dimension character is unramified of defect $v_p(\Delta)$. Proposition~\ref{lem:one-defect-block} excludes defect one; defect zero has an integral idempotent separating it from the dimension character. Now use $d_X^2f_X=\Delta$. \end{proof} \begin{proposition}[Prime avoidance on a principal block]\label{fam:M:4.2} \label{prop:principal-support}\label{prop:principal-common-divisor} Let $P$ be a full principal $p$-block of ordinary integral modular data, or of positive reduced data at an odd prime. Let $E=P$; in ordinary data one may also intersect $P$ with the fiber of trivial prime-to-$p$ twist. Assume that the unit is the only invertible in $E$. Fix a label $X$ and a prime $s\ne p$. If $X$ is $p$-active, or if $p$ and $v_p(\Delta)$ are odd and $s\nmid p-1$, then at each prime over $s$ there is a nonunit $Y\in E$ for which $d_Y\chi_X(Y)$ has nonzero residue. In particular $s\nmid d_Y$. \end{proposition} \begin{proof} The block projector commutes with the local group (Lemma~\ref{lem:local-block-invariance}); so does the indicated complementary projector. Conjugation by the global Weyl normalizes the local group. For an active $X$ the $X,0$ entry of the conjugate projector is zero, whence $\sum_{Y\in E}d_Y\chi_X(Y)=0$. For an inactive $X$ in the second case the slice identity (Proposition~\ref{prop:odd-slice}) is \[ \sum_{\substack{Y\in E\\\ord(\theta_{Y,p})=p}}d_Y\chi_X(Y) =(p-1)\sum_{\substack{Y\in E\\\theta_{Y,p}=1}}d_Y\chi_X(Y). \] In either identity the vacuum contribution is a unit at $s$, so a nonunit term must survive modulo each prime over $s$. \end{proof} In either setting of the proposition, at valuation two or three the nonunit degrees in $P_p$ of a perfect table are $p$-free and have gcd one: apply prime avoidance to an active column at every other prime. For larger defects an index inequality bounds the total ramification of a block. \begin{proposition}[A mixed-defect block bound]\label{fam:M:4.3} \label{prop:mixed-discriminant-budget} Let $A$ be a commutative local symmetric order over an unramified $p$-adic integer ring, with separable cyclotomic generic algebra, rational positive integral codegrees, and a common character residue. After an unramified extension write its generic factors as $E_j/K$, $1\le j\le t$, with totally ramified degrees $e_j=u_jp^{s_j}$, $u_j\mid p-1$. Put $a_j=v_p(f_j)$. Then \[ \sum_{j=1}^t\bigl(p^{s_j}-(a_j-1)e_j\bigr)\le2. \] At two this becomes $\sum_j(2-a_j)e_j\le2$. \end{proposition} \begin{proof} Put $M=\prod_j\OO_{E_j}$. The trace on $E_j$ is $\Tr_{E_j/K}/f_j$, so the discriminant-index formula and self-duality of $A$ give $2\operatorname{length}(M/A)=\sum_j(a_je_j-d_j)$. Common residues impose $t-1$ independent conditions on $M$; the first two ramification groups give $d_j\ge e_j+p^{s_j}-2$. Combining yields the bound, with $e_j=2^{s_j}$ at two. \end{proof} \subsection{The cubic exponent at two}\label{fam:M:sec:4.2} \label{subsec:cubic-two-signed} \begin{lemma}[An integral minimal modular extension]\label{fam:M:4.4} \label{lem:integral-minimal-extension} Let $\mathcal B$ be integral and slightly degenerate, with $\dim\mathcal B=D>2$, pointed subcategory $\sVec=\langle g\rangle$, and no other proper nontrivial fusion subcategory. Every minimal modular extension has pointed subcategory $\sVec$, adjoint subcategory $\mathcal B$, and precisely the proper fusion subcategories $\Vec,\sVec,\mathcal B$. Moreover one can choose an integral minimal modular extension. In such an extension, noninvertible simples of coprime dimensions have zero $S$-entry. \end{lemma} \begin{proof} By the extension theorem of Johnson-Freyd and Reutter \cite[Corollary~1.3]{JFR} there is a minimal modular extension $\M_0$, of dimension $2D$, with neutral component $\mathcal B$ for fermion monodromy. An odd invertible $h$ would give a pointed subcategory of order four whose bicharacter has trivial radical: $b(g,g)=1$, $b(g,h)=-1$, and in the cyclic case $b(h,h)^2=-1$. Splitting it off gives $\mathcal B=\sVec\boxtimes\mathcal N$ with $\dim\mathcal N=D/2>1$, impossible. Hence $(\M_0)_{\pt}=\sVec$ and $(\M_0)_{\ad}=\mathcal B$. A proper subcategory crossing the grading has twice the dimension of its intersection with $\mathcal B$, hence dimension two or four. It is pointed or Ising. The pointed case has an odd invertible. In the Ising case the subcategory is nondegenerate by \cite[Corollary~B.12]{DGNO} and splits off with fermion $g$, giving the same contradiction. Normalize $t_X=\theta_X/\zeta$ by choosing a two-power root $\zeta$ with $\zeta^3=p^+/\sqrt{2D}$ \cite[Theorem~6.10]{DLNCongruence}. At an odd prime any inertia element $\sigma$ fixes $t_0$. The Galois image of the unit is invertible and its normalized twist is $\sigma^2(t_0)=t_0$; thus it is the unit, since $\theta_g=-1$ \cite[Theorem~II]{DLNCongruence}. The vacuum-column ratios $d_X$ are consequently unramified at every odd prime. Since $d_X^2$ is an integer, $d_X$ is an integer or $\sqrt2$ times one. In the nonintegral case the integral subcategory is exactly $\mathcal B$. Let $\mathcal I$ be Ising with fermion $\psi$ and noninvertible $\sigma$. Condense the diagonal boson $g\boxtimes\psi$ in $\M_0\boxtimes\mathcal I$: the local modules have dimension $(2D)4/2^2=2D$ and are nondegenerate \cite[Propositions~4.26, 4.55(i) and~4.56(i)--(ii)]{DGNO}. Free modules on $X\boxtimes\one$, $X\in\mathcal B$, embed $\mathcal B$ braided fully faithfully since the nontrivial algebra summand contributes no morphisms by adjunction. These give all even pairs after identifying boson translates. An odd pair $(Y,\sigma)$ has dimension $2n$; a free orbit gives one simple of dimension $2n$, and a fixed orbit gives two of dimension $n$ since a cyclic stabilizer has only one-dimensional projective irreducibles. This extension is integral. For noninvertible $X$ in an integral extension, $\langle XX^*\rangle=\mathcal B$: containment in $\sVec$ gives $d_X^2\le2$ since $1,g$ each occur at most once. A nonzero entry on coprime-dimensional noninvertibles would imply projective centrality \cite[Proposition~8.22.5 and Lemma~9.15.1]{EGNO}, hence membership of one in $\mathcal B'=\sVec$, impossible. \end{proof} \begin{lemma}[The two fermion quotients]\label{fam:M:4.5} \label{lem:fermion-quotients} Let $\M$ be integral modular with fermion $g$ and $\mathcal B=\langle g\rangle'$. Put \[ R_\pm=\K(\mathcal B)/(g\mp1). \] These are symmetric integral orders with identical special fibers at two. Their blocks selected by the dimension residue have the same rank. Restrictions of modular characters have precisely the fermion orbits as fibers and belong to $R_+$ or $R_-$ according as their labels are even or odd. For either quotient their codegrees are \begin{equation}\label{eq:fermion-restriction-codegree} f_X^\pm= \begin{cases}f_X/2&gX=X,\\ f_X/4&gX\ne X,\end{cases} \qquad f_X=\dim(\M)/d_X^2. \end{equation} A restricted character has the dimension residue if and only if the full modular character does. Let $P$ be the full principal two-block of $\M$. Then \begin{equation}\label{eq:principal-fermion-masses} \sum_{X\in P\cap\mathcal B}d_X^2 =\sum_{X\in P\setminus\mathcal B}d_X^2. \end{equation} If $\dim\mathcal B=2^{s+1}m$ with $s\in\{2,3\}$ and $m$ odd, the positive principal degrees are odd and its rank $k$ satisfies $2^s\mid k$. Every full principal degree is odd or twice odd, and an even-dimensional principal label lies outside $\mathcal B$. \end{lemma} \begin{proof} Choose duality-compatible orbit representatives in $\mathcal B$ ($gX\not\simeq X,X^*$ by twists). In either quotient the unit-coefficient trace satisfies $\tau(XY^*)=\delta_{XY}$, and the orders coincide modulo two, giving equal ranks of corresponding local blocks. If $\epsilon_Z$ is the fermion monodromy, then $\chi_{gX}(Z)=\epsilon_Z\chi_X(Z)$. Orthogonality with the even projector gives \begin{equation}\label{eq:graded-orthogonality} \sum_{Z\in\Irr(\mathcal B)}\chi_X(Z)\overline{\chi_Y(Z)} =\tfrac12 f_X(\delta_{XY}+\delta_{X,gY}). \end{equation} This identifies the restriction fibers. Every character of the even subalgebra extends by semisimplicity, its value on $g$ selects the quotient, and summing over representatives halves the displayed norm, giving \eqref{eq:fermion-restriction-codegree}. These rational codegrees are integral Casimir eigenvalues. A restricted dimension residue extends to the full ring: each odd $Y$ has $Y^2$ even and squaring is injective in the residue field. The principal labels $1,g$ have opposite dyadic twists, giving \eqref{eq:principal-fermion-masses} by block orthogonality. Corollary~\ref{cor:principal-valuation-bound} at dimensions $2^s m$, $2^{s+2}m$ gives the claimed degree parities. Finally the positive block idempotent has integral trace $\sum_{Z\in P_+}d_Z^2/(2^sm)$ and each odd square equals one modulo $2^s$ for $s=2,3$. Thus $2^s\mid k$. \end{proof} \paragraph{Normalization in the fermion extension.} Let $g$ be the transparent fermion of $\mathcal B$, and $\mathcal M$ a minimal modular extension. On its coordinate space put $Fv_X=v_{gX}$ and $Gv_X=\epsilon_X v_X$, where $\epsilon_X$ is the monodromy sign. For an odd orbit $O$ set $v_O=|O|^{-1/2}\sum_{Y\in O}v_Y$. A fixed odd label therefore has $v_O=v_Y$. \begin{center}\small\begin{minipage}{\linewidth} \captionof{table}{Dimension, codegree and matrix conventions. In the lower panel $s=S^{\mathcal M}/(2\sqrt\Delta)$.}\label{tab:fermion-dictionary} \centering \begin{tabular}{@{}lll@{}}\toprule Character data & Dimension convention & Codegree\\\midrule Ordinary modular & $D=\FPdim(\C)$, $s=S/\sqrt D$ & $D/d_X^2$\\ Full modular extension & $\FPdim(\mathcal M)=4\Delta$ & $4\Delta/d_Y^2$\\ Positive quotient $R_+$ & $\Delta=\FPdim(\mathcal B)/2$ & $\Delta/d_X^2$\\ Signed order $R_-$ & Same even source $\mathcal B$ of dimension $2\Delta$ & $2\Delta/(|O|d_Y^2)$, $Y\in O$\\\midrule Space $(F,G)$ & Orthonormal basis & Action of $s$\\\midrule $A=(+,+)$ & $(v_X+v_{gX})/\sqrt2$, one per even pair & $\widehat s_{[X],[Z]}=2s_{XZ}$\\ $B=(-,+)$ & $(v_X-v_{gX})/\sqrt2$, same choices & $s:B\longrightarrow C$\\ $C=(+,-)$ & $v_O$, one per odd orbit & $s:C\longrightarrow B$\\\bottomrule \end{tabular}\end{minipage}\end{center} The signed codegree uses the unit-coefficient trace on $R_-$ in \eqref{eq:fermion-restriction-codegree}. Expanding the basis vectors with $S_{gX,Y}=\epsilon_Y S_{XY}$ gives the $B,C$ entry $\sqrt{2|O|}\,s_{XY}$ for $Y\in O$. Opposite twists on an even pair make $T$ exchange $A$ and $B$. Equal twists on odd mates make $T$ preserve $C$. Both $T^2$ and the odd primary twists preserve all three spaces; the dyadic observable on $A$ is the squared dyadic twist. The integral applications below specify when the extension is integral. \begin{lemma}[Principal lifts at odd primes]\label{lem:odd-principal-lift} Fix a place above an odd prime $p$. A positive principal pair in $R_+$ has a representative whose full character agrees with dimension modulo that place. Both representatives are principal if every odd degree vanishes in the residue field; otherwise exactly one is principal. \end{lemma} \begin{proof} For a lift $X$, write $a_Y=\overline{\chi_X(Y)}$ and $b_Y=\overline{d_Y}$ in the residue field (bars here denote reduction). The two maps agree on the even sector. For odd $Y,Z$, the products $Y^2,YZ$ are even, so $a_Y^2=b_Y^2$ and $a_Ya_Z=b_Yb_Z$. If all $b_Y$ vanish, so do all $a_Y$. Otherwise fix $Z$ with $b_Z\ne0$. Then $a_Z/b_Z=\pm1$ and every $a_Y$ equals the same sign times $b_Y$. Passing to $gX$ multiplies every odd evaluation by $-1$ and fixes the even ones, proving the assertion. \end{proof} \begin{lemma}[Unramified signed characters from sign-only twists]\label{fam:M:4.6} \label{lem:signed-unramified-sign-twists} Let $\M$ be an integral modular category with fermion $g$, and let $\mathcal B=\langle g\rangle'$. If the two-primary part of every actual twist of $\mathcal B$ is $1$ or $-1$, then every character field of $\K(\mathcal B)$ is unramified at two. In particular this holds for the signed quotient $\K(\mathcal B)/(g+1)$. \end{lemma} \begin{proof} Let $\sigma$ belong to inertia at two. The Galois identities \cite[Theorem~II]{DLNCongruence} give $\sigma\chi_Y=\chi_{\widehat\sigma(Y)}$ and $t_{\widehat\sigma(Y)}=\sigma^2(t_Y)$, $t_Y=\theta_Y/\zeta$. Since the dimension character is rational, $\widehat\sigma(0)=0$ and $\sigma^2(\zeta)=\zeta$. For even $Y,Z$ the balancing formula expresses $S_{YZ}$ in integral dimensions and actual even-sector twists, all in an odd cyclotomic field. Thus $Y$ and $\widehat\sigma(Y)$ have the same restricted character and lie in one fermion orbit by Lemma~\ref{lem:fermion-quotients}. Their normalized twists agree, while $Y,gY$ have opposite twists. Hence $\widehat\sigma(Y)=Y$. Symmetry of $S$ and rationality of dimensions now give $\sigma(\chi_X(Y))=\chi_X(Y)$ on all even $Y$ for every $X$. Every character of the even subalgebra extends. \end{proof} \begin{proposition}[Odd supports in the principal two-block]\label{fam:M:4.7} \label{three:odd-reduced}\label{lem:fermion-no-pair-degree}\label{lem:principal-fermion-two}\label{prop:cubic-two-exclusion} Choose an integral minimal extension $\M$ of a slightly degenerate minimal counterexample $\mathcal B$. Its positive reduced dimension is even. Two odd primes $p,q$ of valuation two in $\dim\mathcal B$ cannot occur as a simple degree $pq$. If all principal two-block degrees are odd, their noninvertible supports intersect pairwise with empty total intersection. With exactly three odd primes each of their three pairs must occur as a support. When $v_2(\dim\mathcal B)=3$ all principal degrees are indeed odd. Thus this two-valuation is excluded if the odd part has at most two prime divisors, or three with two valuations equal to two. \end{proposition} \begin{proof} If $\Delta=\dim\mathcal B/2$ were odd, the positive dimension idempotent would be integral at two. Thus the full principal block would have only $1,g$ in its even sector by Lemma~\ref{lem:fermion-quotients}. Its odd mass would be two by \eqref{eq:principal-fermion-masses}, forcing an odd invertible, contrary to Lemma~\ref{lem:integral-minimal-extension}. A label of degree $pq$ has odd residue, hence every noninvertible principal two-block label meets $\{p,q\}$ in its support by coprime vanishing. Choose even $p$- and $q$-active columns $Y,Z$ by reduced Cauchy. A $p$-bearing label has $p$-unit codegree and vanishes at $Y$; likewise at $q,Z$. Thus $1,g$ are their only common principal hits. Replacing $Z$ by $gZ$ makes the dyadic phases distinct while both contributions stay positive, contrary to Theorem~\ref{thm:block}. If all principal degrees are odd, any two noninvertible labels meet by principality and coprime vanishing. They have no common prime: modulo that prime the mass identity \eqref{eq:principal-fermion-masses} would read $2=0$, from the even invertibles $1,g$ alone. This also excludes an empty family. On two primes intersecting supports have a common prime; on three, empty total intersection forces all three pairs. The squared pair would then occur as a degree, contrary to the first assertion. If $v_2(\dim\mathcal B)=3$, Lemma~\ref{lem:fermion-quotients} gives signed principal rank divisible by four. An even principal label would be odd-sector of degree twice odd and signed defect zero or one by \eqref{eq:fermion-restriction-codegree}, contradicting Proposition~\ref{lem:one-defect-block}. \end{proof} \subsection{Minimum positive mass and exposed faces}\label{mass:section} Fix a prime $s$, an integer $a\ge1$, $L=s^a$, and $\zeta$ primitive of order $L$. Identify a coset of $\mu_s$ with its indices modulo $L/s$. Choose $m$ coprime to $s$. At each $j\bmod L$ permit the phases $\zeta^j\eta$ where $\eta$ runs over the primitive roots of orders in a specified subset $H_j$ of the divisors of $m$. These sets may vary with $j$ and may be empty. Let $\mu$ be the M\"obius function and $\varphi$ Euler's totient. For $\epsilon=\pm1$ put \[ b_j^\epsilon=\min\{\varphi(n):n\in H_j,\ \mu(n)=\epsilon\}, \qquad \min\varnothing=+\infty . \] Representations have nonnegative rational coefficients. Their mass is the sum of these coefficients. \begin{theorem}\label{mass:thm:mass} For rational $c_j$ and $x=\sum_j c_j\zeta^j$, the least mass of a representation by permitted phases is \[ \sum_{C\in\mu_L/\mu_s}\ \min_{t\in\{c_j:j\in C\}} \sum_{j\in C}\bigl[b_j^+(c_j-t)_++b_j^-(t-c_j)_+\bigr]. \] Here $0\cdot\infty=0$, and an infinite minimum means that no representation exists. Every finite minimum is attained and gives an explicit word. \end{theorem} \begin{proof} Average a representation over the complementary Galois group; it fixes $x$, preserves mass, and acts transitively on the primitive roots of each allowed order. Their normalized trace is $\mu(n)/\varphi(n)$. If the resulting rational coefficient at $j$ is $u>0$, its minimum cost is $b_j^+u$. Indeed no orbit contributes more per unit mass than $1/b_j^+$; equality comes from a minimizing orbit, since the sum over it is $1$. For $u<0$ use an orbit of sum $-1$ and obtain $b_j^-(-u)$. An orbit of sum zero cannot reduce the minimum. The only rational relations among the $L$ indicated powers of $\zeta$ are the sums of the $\mu_s$-cosets: these $L/s$ relations have disjoint supports and $L-\varphi(L)=L/s$. Hence the possible coefficients on $C$ are exactly $c_j-t$. The nonnegative piecewise linear cost attains any finite minimum at a breakpoint $c_j$, since feasibility endpoints also occur there. Different cosets are independent. \end{proof} Say that $E\subset\mu_L$ is exposed if some rational linear functional on the cyclotomic field takes value one at the indicated pure roots and strictly less than one on all other permitted phases. Require $1\in H_j$ at every marked position. \begin{theorem}\label{mass:thm:expose} $E$ is exposed if and only if in every coset $C$, \[ |E\cap C| < \sum_{j\in C\setminus E}b_j^- . \] \end{theorem} \begin{proof} Average an exposing functional over the complementary Galois group. It then takes the value $t_j\mu(n)/\varphi(n)$ on the primitive order-$n$ orbit above $j$. The rational values $t_j$ satisfy precisely $\sum_{j\in C}t_j=0$, by the coset relations and linear disjointness. At a marked root $t_j=1$ works, since every nonidentity orbit has mean less than one. At an unmarked position with finite $b_j^-$, strictness requires $t_j>-b_j^-$. Summing gives necessity, including the case of a completely marked coset. Conversely choose $t_j\le0$ on the unmarked positions with total $-|E\cap C|$. With a finite, positive total capacity, distribute this negative sum proportionally to the capacities. If an infinite capacity occurs, place it all there. For an unmarked coset take zeros throughout. The strict inequality makes every unmarked value less than one, including orbits with zero or positive mean. Composing the resulting functional on $\QQ(\zeta)$ with normalized relative trace defines the required functional on the whole field. \end{proof} For $H_j$ the full set of divisors of $m>1$, $b_j^-=\ell-1$ for $\ell$ the smallest prime dividing $m$. The criterion specializes to $|E\cap C|1$, every complementary component equals $1$. \item In either case, after absorbing permitted signs, the coefficient vector of the expression is a convex combination of integral optima on the signed configuration. With $N=1$ this assertion holds directly. If all integral optima on its allowed columns agree, its multiplicities are therefore determined. \item More generally, let a real-linear functional on the real scalar extension of the cyclotomic lattice be at most the cost on every signed column. If its value on the target equals the mass $g$, every positively weighted letter of an optimal expression attains equality. This conclusion applies to any actual balancing expression on the alphabet in question. \end{enumerate} \end{lemma} \begin{proof} Use $\ZZ[\mu_{MN}]=R\otimes_\ZZ\ZZ[\mu_N]$ by coprimality and equality of degrees. An integral functional $\lambda$ on the second factor fixing $1$ induces an $R$-linear contraction fixing the prescribed residues: for $p\mid M$ the radical of $(p)$ at either level is generated by $1-\zeta_{p^{v_p(M)}}$, since the prime-to-$p$ extension is unramified. The contraction thus gives an integral target in $R$ with the same congruences. In an integral basis of $R$, the columns for $\mu_M$ are unimodular by Beck--Ho\c{s}ten \cite[Theorems~7 and~13]{BH}; Theorem~7 reduces to square-free levels. Adjoining negative columns preserves unimodularity. Consequently every vertex of $Ax=y$, $x\ge0$, is integral for integral $y$ (solve in independent columns). Positive costs make every nonempty optimal face bounded. In particular a rational signed representation of a contracted target has cost at least $g$. For $c=1$, to eliminate a particular nonsign complementary root $a$, use a basis in each prime-power factor obtained by omitting one root in every prime-order coset. Keep $1$ and, in a factor where the component of $a$ is not a sign, also keep that component. This is possible at two because the two positions then lie in different cosets. Taking the coefficient of $1$ in the product basis kills $a$ and takes values $0,\pm1$ on all roots. Positive mass at $a$ would give cost strictly below $g$. For the asymmetric assertion choose independently in each $u$-power factor of $N$ the functional taking value $1$ at $1$, $-1$ at one uniformly chosen nonidentity $u$th root and zero elsewhere. It respects every $\mu_u$-coset relation. On a term with a nontrivial $u$-component the tensor product gives expected signed cost at most $c/(u-1)<1$; the other terms cost at most one in expectation. A positive coefficient at such a term would make some contraction cost less than $g$. Only retained signed columns remain in either case, so the original expression belongs to their bounded optimal face. Its zero coordinates stay zero at every vertex in a convex decomposition. Finally, equality after summing a termwise functional inequality with nonnegative weights forces equality on each term with positive weight. \end{proof} \subsection{Short cyclotomic sums and complementary principal labels}\label{fam:M:sec:4.3} A cyclotomic integer has \emph{weight at most $g$} if it is a nonnegative rational combination of roots of unity of total coefficient at most $g$. \begin{proposition}[Inertia forced by several positive residues]\label{fam:M:4.8} \label{deep:positive-inertia} Let $z$ be a cyclotomic integer, all of whose conjugates have modulus strictly less than a positive integer $g$. Suppose $z\equiv g$ at every prime above each prime in a finite set $P$. For every nonempty proper subset $R\subset P$ let $H$ be the subgroup of $\Gal(\QQ(z)/\QQ)$ generated by the inertia groups at $s\in R$. Then \[ \prod_{s\in R}s < |H|g. \] In particular any $s\ge g$ in a pair of prescribed positive residues ramifies. If $z\ne0$ vanishes at every prime above $s$, then $s\ge g$ forces ramification at $s$. \end{proposition} \begin{proof} Put $e=|H|$, $M=\prod_{s\in R}s$ and $w=\sum_{\sigma\in H}\sigma z$. Its field is unramified at the primes in $R$, and $w\equiv eg$ at every prime above $P$. Therefore $(w-eg)/M$ is integral. If $M\ge eg$, every conjugate of $1+(w-eg)/M$ has modulus strictly less than one, forcing $w=eg-M$. A prime of $P\setminus R$ would divide $M$. In the last assertion every prime of $\QQ(z)$ above $s$ divides $(z)$. If $s$ were unramified, their product would be $(s)$, so $z/s$ would be a nonzero algebraic integer of house less than one. Lemma~\ref{lem:principal-all-places} supplies the all-places hypothesis when the residue comes from a principal row. \end{proof} \begin{theorem}[Short cyclotomic sums with one or two positive residues]\label{fam:M:4.9}\label{first:balance}\label{seq:two-positive}\label{three:word} Let $g$ be a positive integer and let the cyclotomic integer $z$ have weight at most $g$. Suppose $z\equiv g$ at every prime over a prime $r>g$. If $r>2g$, then $z$ is an unsigned integral sum of exactly $g$ roots of $r$-power order, unique at any common $r$-power level. Suppose also that $qg$ one then has \[ z=h+q\sum_{i=1}^m\xi_i,\qquad h+qm=g, \] with $h,m\ge0$ and the $\xi_i$ nonidentity $r$-power roots, except possibly when $r=g+1$ and $q\mid g-1$. In that exceptional case another possibility is \[ z=-\sum_{t\in\mathbb F_r^\times}\eta_t\zeta_r^t, \] where the $q$-power roots $\eta_t$ consist of one copy of $1$ and $(g-1)/q$ complete cosets of $\mu_q$, with repetitions allowed. In particular $z=g$ if $g2g$ alone the signed augmentation is $g$ modulo $r$ and lies in $[-g,g]$, forcing $g$ positive signs. For two primes erase the $r$-parts, obtaining $w$. Its field is unramified at $r$, so $(w-g)/r$ is integral. All conjugates of $1+(w-g)/r$ lie in the unit disk, meeting its boundary only at $1$; the norm therefore gives $w=g$ or $g-r$. Erase instead the $q$-parts, retaining signed multiplicities $c_i$ in the $r$-power roots. The erased sum is $g$ modulo $q$ in this unramified ring. Thus $c_i-g\delta_{i0}$ is constant modulo $q$ on each $\mu_r$-coset. If every coset has a vacant position after marking index zero, these constants vanish. The augmentation, also the residue of $w$ above $q$, is then $g$ modulo $q$; this excludes $w=g-r$. Equality $w=g$ forces every erased summand to equal $1$, hence the original word is unsigned and $r$-primary. Each nonidentity multiplicity is divisible by $q$ (for $q=2$ signed and unsigned multiplicities have the same parity). The only other possibility uses all $g=r-1$ positions, once each, on the nonidentity $r$th roots. Their signs are $-g$ modulo $q$. This excludes $q=2$ and a positive sign, and gives $q\mid g-1$. As $q$ is odd, the negative summands cannot give $w=g$. Hence $w=-1$ and $\sum_t\eta_t=1$. The $\mu_q$-coset relations show that, after removing one copy of $1$, the multiplicities are nonnegative and constant on each coset. For the reduction to signed words choose a common cyclotomic level containing the original phases and all retained primes. Apply Lemma~\ref{lem:cyclotomic-optimum} with both signs of cost one. Its integral contraction preserves the prescribed residues at all places over the retained primes. The alternatives just proved have minimum signed weight $g$: for one prime the augmentation forces it; for two primes a vacant slot forces $g$ positive signs, while the exceptional full coset already uses $r-1=g$ slots. For any complementary component other than a sign choose the contraction which annihilates it. Its positive occurrence in the original expression would leave weight strictly below $g$, a contradiction. Thus the complementary components are signs, and unimodularity on the retained alphabet applies to $z$ itself. In the unsigned expression, tame $r$-units stabilizing $z$ permute the $m$ nonidentity roots freely, counted after division by $q$. For $m>0$ their stabilizer order divides $m$. In the exception, independence of the nonidentity $r$th roots over the $q$-power field makes $\eta_t$ constant on the free orbits of a tame stabilizer. Its order divides $\sum_t\eta_t=1$ and is therefore one. At $q\le g<2q$, strict inequality $|z|M^2/(hpq)$ also suffices. In both cases $c$ is unrestricted. In \eqref{eq:three-separation}, if $p,q,r$ are all odd, one may instead use half the left side and the residues defined with $2rp,2rq$. \end{proposition} \begin{proof} In a minimal table all three primes remain. The principal block at the odd prime $h$ has an $r$-free nonunit $X$ by prime avoidance applied to an active column. Its degree $s$ divides $M/h$ by Corollary~\ref{cor:principal-valuation-bound} and square-divisibility. Every $r$-free nonunit degree is a multiple of $pq$ dividing $M$, and an $r$-divisible degree occurs since $\Delta\mid\prod_Xd_X^2$. Proposition~\ref{prop:prime-free-gcd} gives $rpq\le sM\le M^2/h$. The nonunit terms of $XX^*$ have degrees divisible by $pq,pr$ or $qr$. Thus $s^2-1=pqA+rpB+rqC$ with nonnegative coefficients. As $pq\mid s$, reduction modulo $q,p$ gives $B\ge u$, $C\ge v$. This contradicts \eqref{eq:three-separation}. In an odd table pair dual constituents before the same reduction to obtain the variant. \end{proof} \paragraph{Positive defect and active columns.} The perfect double $\Z(\Vec_{A_5\times A_5})$ illustrates the role of defect zero in \eqref{eq:zero}. Its invertibles are trivial since $A_5$ is centerless and perfect. Take the pure charge $Z$ of degree $4\cdot3$ (standard and rotation representations), and flux $Y$ of an involution in the second factor with centralizer character $-1$ on that involution. Then $d_Y=15$, $\theta_Y=-1$ and $S_{ZY}=4\cdot15\cdot(-1)=-60$ by the charge--flux formula \cite[Section~3.1, p.~4189]{NNW}, since a half-turn has rotation trace $-1$. \subsection{Common divisors and the primes of a minimal table}\label{fam:M:sec:4.5} Balancing expresses $\chi_Y(X)$ as a positive rational sum of roots of unity of total weight $d_X$. In the reduced table use the actual simple lifts in the even ribbon category. Simple degrees are integral, so Galois symmetry gives $\chi_Y(X^\sigma)=\sigma\chi_Y(X)$ at the same row $Y$. Thus Galois permutes the based ring multiplicatively and preserves degrees and duals. In an odd ordinary table the twists, and hence all balancing phases, have odd order by the categorical Cauchy theorem \cite[Theorem~3.9]{BNRWFinite}. \begin{lemma}[Common constituents and difference sets]\label{fam:Q:2.2}\label{fam:Q:n:lim:diff} Fix a nonunit row $Y$ of a minimal table. Say a set of admissible phases is \emph{exposed} if a rational linear functional on the common cyclotomic field takes value one on it and strictly less than one on the other phases. If the characters of $UU^*$ and $VV^*$ at $Y$ have full-weight positive expressions supported on exposed sets $E,F$, for nonunits $U,V$, their adjoints share a nonunit whose balancing phases at $Y$ belong to $E\cap F$. In particular, suppose $X$ is a nonunit and $\chi_Y(X)$ is a scalar root of unity times a positive rational sum of $s$-power roots of total weight $d_X$, for an odd prime $s$. Let $A$ be their exponent set at a common $s$-power level and $T=A-A$. If each $\mu_s$-coset contains fewer than $s/2$ positions of $T$, then $T\cap uT$ contains a nonzero exponent for every $s$-adic unit $u$. In an odd ordinary table, fewer than $2s/3$ positions per coset suffice. \end{lemma} \begin{proof} Commutativity and rigidity give $\langle UU^*,VV^*\rangle=\langle UV,UV\rangle$. If only the unit were common, it would occur once in each adjoint, so $UV$ would be simple. This also implies simplicity for a product of chosen lifts in the positive quotient. Its double braiding would be scalar, contrary to strictness at $U,V$. Take a common nonunit and use its same balancing expression in both decompositions. Each exposing functional attains the full weight, so every phase in that expression lies in both sets. Theorem~\ref{mass:thm:expose} exposes the primary difference set under the stated coset bounds, using normalized complementary trace (whose minimum on odd roots is $-1/2$). The full expressions for $XX^*$ and $X^u(X^u)^*$ have supports in $T,uT$, since the scalar cancels. If zero were their only intersection, the common nonunit would have character equal to its degree at $Y$, again contrary to strictness. \end{proof} \paragraph{The complementary phase at a failed exposure.} If the criterion in Theorem~\ref{mass:thm:expose} fails on a coset with $k$ marked positions, let $v$ count the unmarked positions with $b_j^-b(g)$, so $p>2g$. Theorem~\ref{first:balance} writes $z$ as a sum of $g$ positive $p$-power roots. Rescaling by $d_X/g$ gives a positive word for $\chi_Y(X)$ of full weight $d_X$. A single position contradicts strictness. If $g=2$, the two positions have difference $t\ne0$ of $p$-power order, and $\{0,\pm t\}$ meets its double only at zero. Since $3

1,\\ 1+(p-1)/e_0 & s>0 . \end{cases} \] In particular $E(x)\ge p^2$ if $e=1$ and $E(x)\ge p^{s+1}$ if $s>0$. For odd $p$, in a $p$-power cyclotomic field a prescribed integer residue $0\le k0$ the inertia conductor is $p^{s+1}$. With $\zeta=\zeta_{p^{s+1}}$ and $F$ as above, the relative integral basis over $F(\zeta_p)$ gives a unique expansion $x=\sum_{j=0}^{p^s-1}\zeta^j\gamma_j$. The terms are orthogonal for the relative trace, hence $\mathsf M(x)=\sum_j\mathsf M(\gamma_j)$. Inertia fixing $x$ has order $g=(p-1)/e_0$; it acts freely on nonzero indices and preserves coefficient means, since the coefficients are conjugate up to $p$th roots of unity. Some nonzero-index orbit must occur by wild ramification. Two such orbits cost at least $2g\ge g+1$, and one together with $\gamma_0\ne0$ costs $g+1$. For a single orbit with $\gamma_0=0$, all its coefficients have the same residue at each prime over $p$. Their residues sum to zero, so each vanishes there since $p\nmid g$. The preceding tame estimate (or the unramified estimate) gives mean at least $2p/(p-1)$ for each. Their sum is at least $g+1$. Finally $[\QQ(x):\QQ]\ge e$ gives the energy bounds. For the prescribed residue in $\QQ(\zeta_p)$ choose integer coefficients $c_j$ with $\sum c_j=k$. The same quadratic form gives $\mathsf M(x)=(p\sum c_j^2-k^2)/(p-1)\ge k(p-k)/(p-1)$. At higher level the orthogonal expansion above adds the means of its coefficients. Their least residues sum to $k$ modulo $p$, and $u(p-u)+v(p-v)\ge w(p-w)$ for $w\equiv u+v\bmod p$ in $[0,p)$. Iteration gives the claim. \end{proof} The bounds $2p$ and $p^{s+1}$ are sharp. For any subgroup $H\le\mathbb F_p^\times$ of index $e>1$ and distinct cosets $aH,bH$, $\sum_{j\in aH}\zeta_p^j-\sum_{j\in bH}\zeta_p^j$ vanishes above $p$. Its coefficient sum is zero, its stabilizer is $H$, and its energy is $2p$. For every $s\ge1$ take the Teichm\"uller subgroup $H\subset(\ZZ/p^{s+1}\ZZ)^\times$ of order $p-1$. The integer $1+\sum_{j\in H}\zeta_{p^{s+1}}^j$ vanishes above $p$ and has mean $p$, by the distinct orthogonal indices modulo $p^s$. Its nonconstant word has length $p-1$ and misses a slot on every $\mu_p$-coset, so its stabilizer is precisely $H$. Its energy is $p^{s+1}$. \begin{corollary}[Conductor and row budget]\label{fam:M:5.5}\label{cor:table-defect-filtration} \label{prop:defect-filtration-energy}\label{cor:mixed-block-general-energy} In a symmetric integral table fix an odd prime $p$ and put $c=v_p(\Delta)$ and $t_X=v_p(f_X)=c-2v_p(d_X)$. A row field has zero $p$-conductor exponent when $t_X=0$ and conductor exponent at most $t_X$ otherwise. The same holds for an entry field with $t_X$ replaced by $t=\min(t_z,t_X)$. When $t\ge1$ its wild inertia has order at most $p^{t-1}$. Suppose the entries $\chi_z(X)$ on $p$-bearing degrees vanish modulo $p$ (at every prime above $p$). This holds if the block of $z$ contains a $p$-unit degree. It also holds if $p\mid d_z$ and each nonzero entry on such a degree has $X\in P_p$ with $v_p(d_X)\ge v_p(d_z)$. For each $p$-bearing degree $d$ let $u_d=(d/\gcd(d,d_z))_{p'}$. Partition the nonzero entries $\chi_z(X)/u_d$ at degree $d$ into Galois orbits $O$ counted with multiplicity. Write $h_O=|O|$, $e_O=e_{0,O}p^{s_O}$ for the degree and ramification index of the entry field. Then \begin{equation}\label{eq:merged-row-budget} \sum_{d:\,p\mid d}\ \sum_{O\text{ at }d} u_d^2 h_O W_p(e_O)\ \le\ f_z-1 . \end{equation} \end{corollary} \begin{proof} Corollary~\ref{cor:field-conductor} also shows that a codegree-four row takes values in $\QQ(\zeta_8)$. The stated conductor bounds follow from the same result, applied to either row since their normalized entry lies in both fields. The bound from the different is attained at conductor $p^t$: the full cyclotomic field has normalized different $t-1/(p-1)$. On a $p$-bearing degree $d$, every character of $p$-unit degree vanishes above $p$ by lcm integrality. The common residue of its block gives the same vanishing for $\chi_z$. In the second case use symmetry: $\chi_z(X)=(d_X/d_z)\chi_X(z)$, with $\chi_X(z)\equiv d_z$ and $d_X/d_z$ integral at $p$. In both cases Galois covariance gives vanishing on all conjugates. Division by the integral lcm scale $u_d$, prime to $p$, preserves both integrality and vanishing. Sum the energy bounds over the orbits and reserve the unit's squared contribution one. \end{proof} If all the scales under consideration are at least $u$, a nonzero orbit costs at least $p^2u^2$ when unramified, $2pu^2$ when nontrivially tame, and $p^{s+1}u^2$ at wild depth $s$. \begin{corollary}[Small budgets and forced hits]\label{fam:M:5.6}\label{cor:defect-two-energy-stratum} \label{cor:three-prime-mixed-energy} For a row satisfying this vanishing condition, if $f_z=p^2m$, $p\nmid m$, at most $m-1$ nonzero orbits on $p$-bearing degrees are unramified or wild in total; the wild depth is at most one. For $m=1$ all such orbits are nontrivially tame and number at most $(p-1)/2$. For any prime $p$, suppose there are just three dimension primes $p,q,r$, every nonunit degree involves at least two, and the nonunit row $z$ has zero residue on its supported $p$-bearing degrees at every prime above $p$. Then \[ f_z\ge [K_z:\QQ]+2p^2 . \] \end{corollary} \begin{proof} For the first assertion divide the budget $p^2m-1$ by $p^2$ or by $2p$ and use the conductor bound at $t_z=2$. For the second apply \eqref{eq:r8-tracedim} modulo $q$ and $r$. Its vacuum term is one, so each reduction supplies an orbit of nonzero trace on a nonunit degree prime to the respective prime. The two orbits are distinct and both degrees are $p$-bearing. Each trace is a nonzero integer divisible by $p$. Their contributions to \eqref{eq:r8-tracenorm} together with the vacuum give the asserted bound. \end{proof} \subsection{Degrees at a prime-square codegree}\label{fam:M:sec:5.3} \begin{proposition}[Degrees supporting a nonzero trace]\label{fam:M:5.7} \label{prop:quartic-two-codegree4}\label{prop:quartic-prime-top-codegree} \label{prop:quartic-five-codegree25} Let $Z\ne\one$ have formal codegree $p^2$ in a perfect ordinary or positive reduced integral table. The nonunit degrees at which $\chi_Z$ has nonzero absolute trace all divide \[ M=d_Z/\gcd(p,d_Z) \] and have greatest common divisor one. If the table is fusion-simple, then $M$ has at least three distinct prime divisors; when it has exactly three, each pair is the prime support of some such degree. In particular at a quartic prime of a three-prime minimal table there is no row of codegree $p^2$, for any $p$. \end{proposition} \begin{proof} By \eqref{eq:r8-tracenorm} a nonzero trace $T_X$ at $X\ne\one$ is prime to $p$: a nonzero multiple of $p$ would exhaust the norm before counting the unit. Now $d_X^2\mid\Delta=p^2d_Z^2$, so $d_X/\gcd(d_X,d_Z)$ is one or $p$. In the latter case lcm integrality makes $\chi_Z(X)$, and hence its trace, a multiple of $p$. Thus $d_X\mid d_Z$. Moreover $X\in P_p$ by Theorem~\ref{thm:principal-support}. If $v_p(d_X)=v_p(d_Z)>0$, then $\chi_Z(X)=(d_X/d_Z)\chi_X(Z)$ vanishes above $p$ by principality. Lemma~\ref{lem:principal-all-places} gives this vanishing for every conjugate, hence $p\mid T_X$. Hence $d_X\mid M$. The dimension identity \eqref{eq:r8-tracedim} expresses $-1$ as an integral combination of the indicated degrees. In the fusion-simple case no nonunit has prime-power degree. For each prime dividing $M$ some indicated degree omits it; this proves both assertions about prime supports. Finally $v_p(\Delta)=4$ gives $v_p(d_Z)=1$, so with only two other dimension primes $M$ cannot have three. \end{proof} \section{Local Weyl columns at separated primes}\label{fam:M:sec:6}\label{subsec:sqvac} For the vacuum estimates let $\C$ be an integral modular category of square dimension $D=n^2$ with its positive spherical structure. Write $s=S/n$ and use the lift $t=T/\gamma$ of Section~\ref{sec:blocks}, with $\gamma^3=\xi=p^+/n$ and $\gamma$ of two-power order. For a set $J$ of primes of the level write $W_J=\prod_{p\in J}W_p$, with $\gamma_J$ and $\theta_{Z,J}$ the $J$-parts of the corresponding roots. Put $M_m(Z)=\sum_Yd_Y\chi_Z(Y)\theta_Y^m$. \begin{theorem}[The degree-normalized vacuum column]\label{por:th:weyl-div} For $m\ge1$ and all $Z$, $|M_m(Z)|^2/D$ is an algebraic integer. The vector $w=W_Jv_0=W_J^{-1}v_0$ is rational and \[ k_{J,Z}=n w_Z/d_Z\in\ZZ . \] If $m$ is one on the primary levels in $J$ and zero on the others, then $M_m(Z)=n\gamma_J^3\theta_{Z,J}^{-1} k_{J,Z}$. Every prime $p\notin J$ dividing $f_Z$ divides $k_{J,Z}$. \end{theorem} \begin{proof} In the congruence lift, $\sigma_a(\rho(g))=\rho(A_agA_a^{-1})$ for $A_a=\operatorname{diag}(a,1)$ by \cite[Theorem~II and Remark~4.5]{DLNCongruence}. The torus permutations have positive signs and fix the vacuum, since its entries $d_i/n$ are positive rational. Conjugating $W_J$ introduces a torus on its right; the torus $W_J^2$ also fixes the vacuum. This gives both assertions about $w$. In \eqref{eq:mixed-rotation} take $i=j=0$. Since $\theta_{Y^*}=\theta_Y$, we have $(sT^ms^{-1})_{0Z}=(sT^ms^{-1})_{Z0}=d_ZM_m(Z)/D$. Consequently $L_m^{00}(f_Ze_Z)=|M_m(Z)|^2/D$ is an algebraic integer. For the indicated CRT exponent, the local identity $U(WUW^{-1})U=W^{-1}$ in each selected factor gives \[ (sT^ms^{-1}v_0)_Z=\gamma_J^3\theta_{Z,J}^{-1}w_Z . \] This proves the formula for $M_m(Z)$. The rational number $k_{J,Z}$ has integral square and is therefore an integer. If $p\notin J$ choose the same CRT exponent divisible by $p$. Lemma~\ref{lem:rotation-congruences}(i) applied to $y=f_Ze_Z$ gives $L_m^{00}(y)^p\equiv f_Z^{p-1}L_{m/p}^{00}(y)$ above $p$. Here $L_m^{00}(y)=k_{J,Z}^2$, proving the divisibility. \end{proof} \begin{lemma}[Vacuum entry and local support]\label{fam:M:6.1}\label{lem:local-vacuum-column} For the CRT exponent of Theorem~\ref{por:th:weyl-div}, \[ n^{-2}\sum_i d_i^2\theta_i^m=\xi^m(W_J)_{00},\qquad nw\in\ZZ^{\Irr(\C)} . \] If $p\mid n$ lies outside $J$, then $p\mid nw_i$ for each $i$; thus $w_i$ is $p$-integral when $v_p(n)=1$. For $J=\{q\}$ the support is contained in the principal $q$-block. \end{lemma} \begin{proof} Set $Z=0$ in the theorem and use $\gamma_J^3=\xi^m$. At any label $nw_i=d_i k_{J,i}$; if $p\nmid d_i$, then $p\mid f_i$ and the theorem supplies the factor $p$. The principal $q$-block subspace is invariant under $W_q$ by Lemma~\ref{lem:local-block-invariance} and contains $v_0$. \end{proof} \begin{proposition}[Weyl columns on separated prime sets]\label{fam:M:6.2}\label{deep:cluster}\label{deep:separated-inertia} Partition the primes in the modular level of an integral modular category of dimension $n^2$ into sets $J$ such that each simple has twist supported in at most one set. Write $\gamma_J$ for the $J$-part of the anomaly normalization $\gamma$, $n_J$ for the $J$-part of $n$, and $V_J=\gamma_J^{-3}W_J$. Then \begin{equation}\label{eq:weylsplit} (V_I-1)(V_J-1)=0\quad(I\ne J),\qquad s=\xi\Bigl(1+\sum_J(V_J-1)\Bigr),\qquad n_J W_J v_0\in\mathbb Z^{\Irr(\C)} . \end{equation} If $p\in J$ and $v_p(d_X)0$ (and $j_t1$), $m_tj_t\equiv-\epsilon\pmod t$, and $\sum_{t\in T}j_t/t\le1+1/n$. Every $t$-active label has degree $m_t$, and their number is at least $(t-1)j_t$. One has \begin{equation}\label{eq:isolated-deficit} tj_t=\sum_{Y\ t\text{-active}}(1-\theta_{Y,t}). \end{equation} For odd $n$, each $j_t$ is even. Each $a_t$ is integral, supported exactly on the nonunit $t$-free degrees, which belong to $P_t$. The column identities are \begin{align} a_t(X)&\equiv d_X/m_t\pmod t,& \sum_{X\ne0} d_Xa_t(X)&=(n+\epsilon)j_t,\nonumber\\ \sum_{X\ne0}a_t(X)^2&=2tj_t-j_t^2,& \sum_{X\ne0}a_u(X)a_v(X)&=-j_uj_v\quad(u\ne v).\label{eq:budget} \end{align} In particular, after the active labels are removed the residual squared norm is at most $B_t=j_t(t+1-j_t)$. \end{proposition} \begin{proof} Use the singleton clusters $\{t\}$, $t\in T$, and their complement $H$. A $t$-active label $Y$ has $f_Y=t^2$, by complementarity and $v_t(f_Y)=2-2v_t(d_Y)>0$. The equality $s=\xi W_t$ on $E_t$ gives $(W_t)_{Y0}=1/(t\xi)$; rationality gives the common sign. The vacuum Gauss formula gives \eqref{eq:isolated-deficit}, so $j_t>0$. The mean of a nontrivial $t$-power root is at least $-1/(t-1)$, proving the count. For odd $n$ the active labels come in distinct equal-twist dual pairs, and $t$ is odd. Write $z_t=(W_t-I)v_0$. By \eqref{eq:weylsplit}, $s=W_H+\epsilon\sum_{t\in T}(W_t-I)$, with $W_H=I$ when $H$ is empty and $\epsilon=1$ in that case. Unitarity at the vacuum gives the deficit-sum bound. Multiply the split vacuum column by $n$ and reduce modulo $t$ to get both congruences. In particular $j_t\ne t$, so $j_t1$. The scaled column is integral and its vacuum entry is nonzero, so every nonunit coordinate has absolute value $v$, which the real part of this identity precludes. At level $v$ the cyclotomic relation forces constant multiplicity on the nontrivial phases, hence multiplicity one. Since $\gamma$ has two-power order, the $v$-part of $\ord(T/\gamma)$ is exactly $v$. The $Y_u,0$ entry of the local Bruhat identity $WU^kW^{-1}=U^{-1/k}Wh_{-k}U^{-1/k}$ for $k\ne0$ gives $\sum_{t\ne0}(\zeta_v^{kt}-1)\chi_{Y_u}(Y_t)=\xi v\zeta_v^{-u/k}$: the torus fixes the vacuum. Summing over $k\ne0$ gives $\sum_t\chi_{Y_u}(Y_t)=\xi$; Fourier inversion gives the displayed entries. On the active coordinates their Gram matrix has diagonal $v(v-1)-1$ and off-diagonal $-v-1$ by Parseval. The full rows have Gram matrix $v^2I$, so their inactive restrictions coincide, with squared norm $v+1$. Galois permutes the active rows, making their common inactive entries rational integers. Principality gives the congruences. Removing the unit from the norm and from orthogonality against the dimension character gives \eqref{eq:singleton}. Duality preserves each phase and so fixes its unique label; square twist covariance gives precisely the indicated real field in $\QQ(\zeta_v)$. On an inactive $X$, the vector $v_X-b_X v_0$ and its inverse Weyl image vanish on the active sector. Both local unipotents, and hence the local torus, fix it. The torus fixes $v_0$ as well, so $X$ is inertia-fixed. \end{proof} \begin{lemma}[Column norm on isolated row sets]\label{fam:M:6.5}\label{lem:singleton-energy}\label{lem:local-integer-column}\label{new:rank-budget} In the setting of Proposition~\ref{second:two-columns}, let $X$ be any nonunit label of degree $d$ and $[d]_t$ its centered residue at $t$. For $k=|[d]_t|$ put $E_t=j_tk(t-k)/t^2$. Then $\sum_{t\in T}E_t<1$. At an unramified $t$ one may replace $E_t$ by $(t-1)j_t[d]_t^2/t^2$. If also $t\ge5$, every $t$-active $Z$ has $\chi_Z(X)=[d]_t$. For such an inactive $X$, \[ (W_t-I)v_X=[d]_t(W_t-I)v_0,\qquad a_t(X)=-\epsilon j_t[d]_t . \] Any set $L$ of nonunit inactive $t$-unramified labels therefore satisfies $1+\sum_{X\in L}[d_X]_t^2\le\lfloor(t+1)/j_t\rfloor$. In particular, for every isolated odd singleton, $j_t\mid t-1$ or $j_t\mid t+1$. \end{lemma} \begin{proof} Each active $Z$ is $t$-principal, of codegree $t^2$. Its value on $X$ lies in a $t$-power cyclotomic field and reduces to $d$ at $t$. Theorem~\ref{lem:singleton-residue-energy} and the active count give at least $E_t$ in the Galois mean of the normalized $X$-column norm. Distinct active sectors are disjoint and the vacuum also contributes. Apply the estimate at the active degree $m_t=n/t$, with $k=|[m_t]_t|$. Then $01$ are prime to three. Their reciprocals sum to at most $1/2+1/4+1/5$, so the three summands contribute at least $41/20>2+2/m$, contradicting the identity. \end{proof} \begin{lemma}[An isolated quartic prime]\label{fam:M:6.8}\label{fourm:heightone} Let $q$ be an isolated prime in an ordinary minimal table of dimension $n^2$, where $n=q^2N$ and $q\nmid N$. If at least two simple degrees (counted with multiplicity) have $q$-valuation one, then at least $q(q-1)$ active labels have degree $N$, and some active $J$ has degree $qN$. The row of each such $J$ vanishes on all $q$-bearing degrees, and necessarily $q\ge5$. \end{lemma} \begin{proof} Write $A_0,A_1$ for the active labels of degrees $N,qN$. These are all the possibilities by Lemma~\ref{lem:complement}. Put $m=|A_0|$, $\delta_i=\sum_{Z\in A_i}(1-\theta_{Z,q})$ and $b=q^2W_q v_0$. Proposition~\ref{deep:cluster} makes $b$ integral, $v_q(b_X)=v_q(d_X)$ for $v_q(d_X)<2$, and all height-one labels $q$-principal. The vacuum Gauss formula gives, for a root of unity $\lambda$, \[ \lambda b_0=q^2-\delta_0/q^2-\delta_1 . \] For such a label $X$, every $Z\in A_0$ has $S_{XZ}\ne0$ by principality, and $S_{XZ}/(qN)$ is integral by lcm divisibility. Their Galois-mean contribution to the normalized row norm is at least $m/q^2$ besides the vacuum; thus $m0$. All conjugates of the algebraic integer $1-\delta_0/q^2$ are strict convex averages of $1$ and nontrivial roots of unity, hence have modulus less than one. Therefore $\delta_0=q^2$. The mean of $1-\theta$ for a nontrivial $q$-power root is at most $q/(q-1)$, so $m\ge q(q-1)$. If $A_1$ were empty, $|b_0|=q^2-1$; with the two height-one coordinates this costs at least $(q^2-1)^2+2q^2>\|b\|^2=q^4$. Finally let $J\in A_1$. Its row spends at least $m\ge q(q-1)$ on $A_0$ by principality. For height-one $X$, $\chi_J(X)=(d_X/d_J)\chi_X(J)$ vanishes above $q$, since $X\in P_q$ and $d_X/d_J$ is a $q$-unit. This holds on all conjugates. A nonzero orbit would cost at least $2q$ by Theorem~\ref{thm:cyclotomic-residue-energy}, exceeding the remaining norm $f_J-m\le q$. Height two is excluded by defect-zero vanishing at the active column $J$. By Lemma~\ref{new:trace} this row also hits at least three distinct proper $q$-free degrees in $P_q$. Together with $A_0$ and the vacuum they cost at least $q(q-1)+4$, forcing $q\ge5$. \end{proof} \subsection{A denominator bound for local Weyl columns}\label{fam:W:sec:2} Fix a set $J$ of prime divisors of the level, and write $T_J=\operatorname{diag}(\theta_{z,J})$, $U_J=\prod_{q\in J}U_q$, so $T_J=\gamma_JU_J$. A $J$-active label has $\theta_{z,J}\ne1$. \begin{theorem}[Denominators of a complementary Weyl product]\label{fam:W:2.1}\label{fam:W:n:thm:foreign-column} For $p\notin J$ put $a=v_p(f_z)$. At every place above $p$, normalized by $v(p)=1$, a nonzero entry $w_{iz}$ of $W_J^{\pm1}$ satisfies \[ v(w_{iz})\ge \begin{cases} 0 & a=0,\\ -\lfloor(a-1)/2\rfloor & a\ge1 . \end{cases} \] If $v_p(f_z)\le2$ for every $J$-active label, both matrices are integral at all places above $p$. \end{theorem} \begin{proof} Take $m>0$ equal to one on the primary levels in $J$ and zero on the others, also requiring $p\mid m$. Put $A=sT^ms^{-1}$ and $w=W_J^{-1}$. Since the selected factors commute, multiplying their Bruhat identities gives $U_J(W_JU_J W_J^{-1})U_J=W_J^{-1}$. Thus $A$ is $w$ up to left and right diagonal roots of unity. Moreover, Galois conjugation by $A_k=\operatorname{diag}(k,1)$ as in Theorem~\ref{por:th:weyl-div} fixes $w$ whenever $k$ is one on all the selected levels. Its entries therefore belong to the cyclotomic field of conductor dividing the product of these levels, unramified at $p$. For $k=-1$ the same conjugation gives $\bar w=w^{-1}=Hw$, with $H=W_J^2$ the product of the tori at $-1$. It is an involutive permutation matrix in this normalization. At $j=Hi$, $\overline{w_{jz}}=w_{iz}$. Equation~\eqref{eq:mixed-rotation} makes $f_z A_{iz}\overline{A_{jz}}$ integral. If $p\mid f_z$, Lemma~\ref{lem:rotation-congruences}(i) and $(f_ze_z)^p=f_z^{p-1}(f_ze_z)$ give strictly positive valuation. Hence $a+2v(w_{iz})\ge0$, strictly if $a>0$. Unramifiedness makes $v(w_{iz})$ an integer. This gives the asserted floor bound and also applies to $W_J=Hw$. For the last assertion take $R$ to be the intersection of the valuation rings above $p$ in a common cyclotomic field. The columns $u_z=W_Jv_z$ for $J$-active $z$ and their adjoints lie over $R$, since conjugation permutes these places. Unitarity gives \[ W_JT_JW_J^{-1}=I+\sum_{z\ J\text{-active}}(\theta_{z,J}-1)u_zu_z^* . \] This is integral over $R$ entrywise. Replacing $T_J$ by $U_J$ changes a unit scalar. The product Bruhat identity now gives $W_J^{-1}$ over $R$, and its adjoint gives $W_J$. \end{proof} \begin{lemma}[The tetrahedral Weyl matrix]\label{lem:tetrahedral-weyl} Let $u$ be a $3$-cycle and $w$ a nonidentity element of the normal Klein four-group $K$ in $A_4$. A unitary representation of $A_4$ of dimension at most four is a sum of characters of $A_4/K$, or consists of one tetrahedral summand $T$ and at most one such character. In the latter case $T$ has one line at each $u$-eigenvalue. In a suitable orthonormal basis of these lines the matrix of $w$ on $T$ is $2J/3-I$, where $J$ has all entries one. On any character summand $w$ acts by one. If $T$ occurs and the $u$-fixed space is two-dimensional with orthonormal basis $v,v'$, and $x=\|P_Tv\|^2$, then \[ |\langle v',wv\rangle|^2=\tfrac{16}{9}x(1-x). \] \end{lemma} \begin{proof} The three nontrivial characters of $K$ are cycled by $u$. Choose a unit vector $b$ of the unique nontrivial $K$-character taking the value $+1$ on $w$, if this character occurs. Then $b,ub,u^2b$ lie in mutually orthogonal character spaces and span an invariant copy of $T$. Its orthogonal complement has dimension at most one. If no such character occurs, $K$ acts trivially throughout. On the displayed basis $w$ has signs $(1,-1,-1)$. For $\omega=\exp(2\pi i/3)$, the vectors $3^{-1/2}\sum_{j=0}^2\omega^{-kj}u^jb$ diagonalize $u$. The projector onto $\CC b$ has matrix $J/3$ in that basis, giving the formula for $w$. On the two-dimensional fixed space the compression of $w$ is $I-4P_T/3$, where $P_T$ restricts there to a rank-one projection. Its off-diagonal entry gives the asserted square. \end{proof} \subsection{Local Weyl coefficients and twist masses}\label{por:tools} The following applications use the notation and degree-normalized integers of Theorem~\ref{por:th:weyl-div}. \begin{proposition}\label{por:prop:separated-high-row} Let $p,q$ be separated twist primes in a perfect fusion-simple integral modular category of square dimension. At a nonunit $Z$ suppose $a=p^{v_p(n/d_Z)}>1$, $b=q^{v_q(n/d_Z)}>1$, and all other primes of $n$ occur to full height in $d_Z$. Let $L_p,L_q$ bound the degrees of $p$-active and $q$-active labels, respectively. Put $f=f_Z$. Then \[ 2L_p(f-1)\ge nb^2,\qquad 2L_q(f-1)\ge na^2. \] \end{proposition} \begin{proof} Separation forces full $q$-height on $p$-active degrees and full $p$-height on $q$-active degrees. The deficiencies at both primes, together with full height elsewhere, therefore give $\theta_Z=1$. Each other active prime misses the row $Z$ by defect-zero vanishing. Write $b_Y=\chi_Z(Y)$ and put \[ H_t=\sum_Y d_Y b_Y(\theta_{Y,t}-1)\quad(t=p,q). \] By $\sum d_Yb_Y=0$, Theorem~\ref{por:th:weyl-div} gives $H_t=n\gamma_t^3k_{t,Z}$. Every nonzero hit with nontrivial twist has only one of the two components, hence $H_p+H_q=n\xi$ by the vacuum modular relation, where $|\xi|=1$ and $\xi$ is a root of unity. At $p$-active labels, $q^{v_q(n)}\mid d_Y$ and $b\mid b_Y$ by lcm integrality. Thus $b\mid k_{p,Z}$; similarly $a\mid k_{q,Z}$. Neither integer vanishes: the other would then have absolute value one, violating its divisibility. On the $p$-active labels, a union of Galois orbits, the integral entries $b_Y/b$ satisfy $\sum |b_Y/b|\le\sum |b_Y/b|^2$. Indeed every nonzero orbit has mean square at least one by the absolute norm, followed by Cauchy--Schwarz. Consequently \[ nb\le |H_p|\le2L_p\sum_{Y\ p\text{-active}}|b_Y| \le 2L_p(f-1)/b , \] with one unit of squared norm reserved for the vacuum. Interchanging $p,q$ proves the other inequality. \end{proof} \paragraph{A two-variable integer test.} The magnitude bounds can be combined with the phase equation. Under the hypotheses of Proposition~\ref{por:prop:separated-high-row}, there are nonzero integers $u\in b\ZZ$ and $v\in a\ZZ$ such that \[ u+v=1,\qquad \frac n2\left(\frac{b|u|}{L_p}+\frac{a|v|}{L_q}\right) \le f-1 . \] In particular $\xi=\pm1$ in this situation. Indeed if both primes are odd, $\gamma_p=\gamma_q=1$ so the phase equation is $k_{p,Z}+k_{q,Z}=\xi$, an integer of modulus one. If $p=2$, it is $\xi k_{p,Z}+k_{q,Z}=\xi$; as $k_{q,Z}\ne0$ both sides force $\xi$ real. The other case follows on interchanging the primes. Absorb these signs into $u,v$ to obtain $u+v=1$. The estimate for $H_p$ in the proof gives a squared row norm of at least $nb|u|/(2L_p)$ on the $p$-active sector, and likewise for $q$. The two sectors are disjoint and omit the vacuum. This proves the test. For fixed $a,b$ the minimum on the left is elementary: solve $u\equiv0\pmod b$, $u\equiv1\pmod a$, and take the solution on each side of $[0,1]$ nearest to it. Thus the test applies to prime decomposition forms without choosing individual simple labels. The same argument bounds the total number of deficient rows. Fix the primes and $n$ in Proposition~\ref{por:prop:separated-high-row}, but let $Z$ range over all nonunit labels with the two indicated deficiencies and full height elsewhere. Since $k_{p,Z}$ is a nonzero multiple of $q^{v_q(n/d_Z)}$, the unit norm of $W_pv_0$ gives \[ \sum_Z p^{2v_p(d_Z)}0$. Dividing $D$ by $ez^2c^2$ gives the strict bound. The hit set is a union of full Galois orbits since \eqref{eq:galois-permutations} and the rational zero test preserve it. \end{proof} \contentsgroup{Prime families and dimension bounds} \section{Three ordered prime powers}\label{fam:M:sec:7}\label{sec:three-ordered} Write $p0$. The two normalized entries $z_Y=S_{VY}/\lcm(d_V,d_Y)$, $Y=A,B$, belong to $K_V$ and have strict house less than $h$. At $A$ one has $t^i z_A=\chi_A(V)\equiv d_V\equiv0$ at every place above $h$: Galois covariance permutes $V$ through labels of the same degree (Lemma~\ref{lem:principal-all-places}), and $t$ is an $h$-unit. Yet $z_A\ne0$ since $V\in P_t$ and $t\nmid d_A$. At $B$ the principalities of $V,B$ give residue $h$ at both $t$ and $k$. The zero-residue and two-residue assertions of Proposition~\ref{deep:positive-inertia} force ramification of $K_V$ at $h$ and at $k>h$, respectively, contradicting separation. \end{proof} \subsection{Twist phases at maximal height}\label{fam:M:sec:7.2} We first record the common indicator computation for the odd and dyadic phase arguments. Fix a prime $p$, write $W_o$ for the complementary Weyl, $w=W_ov_0$, and suppose a Galois-stable label set $J$ satisfies $|w_j|^2=(h/D)d_j^2\mathbf1_J(j)$ for an integer $h$. Choose $m$ zero on the $p$-level and one on the complement. The complementary upper--lower--upper identity gives $|(sT^ms^{-1})_{0j}|=|w_j|$. Thus \[ b_X=\nu_m(X)=\frac hD\sum_{j\in J}d_j^2\chi_j(X), \qquad b=\sum_X b_X[X^*]. \] These indicators are algebraic integers by \eqref{eq:indicator-entry} and rational by Galois symmetry. The $b_X$ are therefore integers, constant on character orbits by symmetry of $S$. Orthogonality gives $\chi_j(b)=h\mathbf1_J(j)$, so $b^2=hb$. Also $b=b^*$, $b_0=\|w\|^2=1$ and $\|b\|^2=h$. In each application $0\in J$, giving $\dim b=h$. \begin{lemma}[An odd-order phase cannot be missing]\label{fam:M:7.4}\label{seq:missing-odd} Let an integral perfect modular category have odd dimension $D$, with $v_p(D)$ odd. If all $p$-primary twists have order at most $p$, every $p$th root of unity occurs. \end{lemma} \begin{proof} Write $U,W$ for the $p$-local upper unipotent and Weyl, $P$ for the projection onto $\ker(U-I)$. Then $Pw=w$, $Ww=sv_0$, and the local torus acts on $w$ by the Legendre character as in Proposition~\ref{prop:odd-slice}. For $1\le tp$, their visible $p$-phases are contained in $1$ and one common primitive square class. A deeper phase, or the other square class by Lemma~\ref{seq:missing-odd}, is missed. At $p=2$, Lemma~\ref{seq:dyadic-pair} gives $1$ and one primitive fourth root with a representative of the witness degree at both phases. Opposite witnesses hit these representatives by principality, so the nontrivial phase agrees. Lemma~\ref{seq:two-phase} and perfectness supply a phase outside the two allowed. Apply Lemma~\ref{lem:visible-phase-transfer} to this nonempty $p$-fiber, first at $q$ and then at $r$. Its labels have pure $p$-power codegree. Now suppose $p$ is odd. The orbit $\mathcal O$ of $Z$ has at least $p-1$ elements, since $K_Z$ contains $\mathbb Q(\zeta_p)^+$ by twist covariance and is nonreal by oddness. If $r\le2p^{k-1}$ the bound follows from $H_p(r)\ge2$. Otherwise reduce the orthogonality of the $Z$-column to the vacuum on $P_p$ at $r$. It gives a hit $X$ of degree $p^h q^i$, $1\le h\le k-1$ by principality. Pair it with a $p^kr^l$ witness in $P_r$. The normalized entry has residue $p^h$ at $r>2p^h$ and strict house below $p^h$. Theorem~\ref{three:word} and Lemma~\ref{lem:short-tame-orbit} give an $r$-inertia orbit of $X$ of size $e\ge(r-1)/H_{p^h}(r)$. Now $K_Z$ is unramified at $r$ by its codegree. At the nonzero entry $S_{ZX}$ all $e$ inertia conjugates of $X$ therefore agree on $Z$. The field-energy bound (Lemma~\ref{deep:field-energy}) and $|\mathcal O|\ge p-1$ give \[ e(p-1)\lcm(d_Z,d_X)^2<\Delta,\qquad (p-1)(r-1)< H_{p^h}(r)p^{a-2h}. \] Here $d_Z=p^j\sqrt{q^b r^c}$ and $d_X=p^h q^i$, so the first inequality gives $e(p-1)p^k$, every braided category of dimension dividing $M$ is WGT. If $q>p^{k-1}$, the same conclusion holds as soon as it holds on every divisor of \[ p^A q^{2\lfloor B/2\rfloor}r^{2\lfloor C/2\rfloor}d. \] Under this second gap one may take $B=5$ and arbitrary $C$. One may also take arbitrary $B,C$ for odd $p$ and $A\ge4$ when $(p-1)(r-1)\ge H_p(r)p^{A-2}$. Without a gap hypothesis the caps $A=B=5$ always suffice. More precisely, if $p^aq^br^c$ is the square-full part of a divisor-minimal counterexample, then $q\le p^{\lfloor a/2\rfloor}$, and $q>p^{\lfloor a/2\rfloor-1}$ forces both $b,c$ even. \end{theorem} \begin{proof} Take a least counterexample among the divisors of $M$. Its three exponents $a,b,c$ satisfy $a,b\ge5$, $c\ge4$ by Theorems~\ref{thm:three} and~\ref{seq:quartic-any}. With $l=\lfloor a/2\rfloor$ one has $q\le p^l$ by Lemma~\ref{three:complementary}, proving the first assertion. We claim that $q>p^{l-1}$ forces $b,c$ even. In the ordinary table this is Theorem~\ref{three:height}. In a reduced table $p=2$ by Proposition~\ref{three:odd-reduced}; an even $a$ contradicts Corollary~\ref{prop:reduced-quartic-two-exclusion}, and for odd $a$ apply Proposition~\ref{seq:reduceddyadic5}. Now $l\le k$: a gap at $k-1$ makes both actual exponents even, so the rounded caps suffice. For $B=5$ the rounded $q$-exponent is four, covered by Theorem~\ref{seq:quartic-any}. When $p$ is odd, the table is ordinary by Proposition~\ref{three:odd-reduced}, and Theorem~\ref{three:height} forces $(p-1)(r-1)p$ automatically, so every uncovered choice has $A+B\ge11$. \end{proof} \subsection{The largest quartic prime}\label{fam:M:sec:7.4} At $v_q(\Delta)=5$, reducing unit orthogonality on $P_q$ at the omitted prime forces degrees $qp^i$ and $qr^j$. More precisely, for any $X$ of degree $q^j r^b$ with $j,b>0$, some $Y\in P_q$ of degree $qp^a$ satisfies, with $z=S_{XY}/\lcm(d_X,d_Y)$, \begin{equation}\label{three:trace} r\nmid [K_Y:\QQ(z)]\Tr_{\QQ(z)/\QQ}(z). \end{equation} Indeed use block orthogonality if $X$ is $q$-active and the odd-valuation slice of Proposition~\ref{prop:principal-support} otherwise. The respective vacuum coefficients $1,-(q-1)$ are units at $r$. Some $r$-free orbit in $P_q$ thus contributes nontrivially modulo $r$; the principal height bound gives degree $qp^a$. Its contribution before the slice coefficient is $qp^{2a}[K_Y:\QQ(z)]\Tr(z)$ by \eqref{eq:galois-permutations}. For $j=1$, $r^b z=\chi_Y(X)\equiv0$ at every prime over $q$, so the trace is a nonzero multiple of $q$ and $\QQ(z)$ ramifies there (otherwise $z/q$ is integral of house less than one). If $r\nmid f_X$, then $\QQ(z)\subseteq K_X$ is unramified at $r$ and \eqref{three:trace} excludes wild $r$-ramification in $K_Y$. Finally, when $v_r(\Delta)=4$, a $p$-free nonunit in $P_r$ has degree $q^2r$: degree $qr$ paired with $qp^a\in P_q$ violates Theorem~\ref{three:word} at $g=q$. \begin{proposition}[Complementary fields at a quintic middle prime]\label{fam:M:7.12}\label{three:sextic-gate} Let $k\ge3$ and consider $D\mid p^{2k}q^5r^c d$, $pg$, the two fields contain $\QQ(\zeta_r)$, except that when $r=g+1=q+2$ the guaranteed common subfield is $\QQ(\zeta_r)^+$. \end{proposition} \begin{proof} Theorem~\ref{three:least-height} forces $b=5$ and excludes every $a\le2k-1$. In the reduced table $p=2$ and $p$-heights are at most $k-1$. In the ordinary table a $p$-active fiber contains a $q$-active column by Lemma~\ref{lem:odd-valuation-joint-twist}. Its $r$-free hit on $P_q$ has $p$-height at most $k-1$ by defect-zero vanishing. In either case Lemma~\ref{three:complementary} and $q>p^{k-2}$ force height $k-1$ there and at least $k-1$ on every opposite label. The principal $q$-height bound gives $d_X=gq$. Apply Theorem~\ref{three:word} to each opposite pairing with weight $g$; for general $r>g$ the proof gives tame orbit size at least $(r-1)/\lfloor g/q\rfloor$. \end{proof} At $2k=6$ the exceptional equality $p^2=q+1$, $r=q+2$ forces $(p,q,r)=(2,3,5)$. In the exceptional word, if $1$ occurs once, its position forces each stabilizer to fix $\zeta_5$. Otherwise the coefficients are $1,1,\omega,\bar\omega$ for a primitive cube root $\omega$. Unless the two positions of $1$ are opposite modulo five, every stabilizer fixes $\zeta_5$; if opposite, its only nonidentity element is complex conjugation. Thus the common entry field contains $\QQ(\zeta_5)$ or equals $\QQ(\zeta_{15})^+$, and always contains $\QQ(\sqrt5)$. The forced degree $p^{k-1}q$ also satisfies the tensor-square inequality of Proposition~\ref{prop:three-separation}. For $p=2$ and $k=3$, Theorem~\ref{aux:dyadic-six} covers all odd pairs except $(3,5)$, with both companion exponents unrestricted. A related descent observation holds at any odd prime $r$. Every WGT braided category of dimension divisible by $r^3$ contains a nontrivial Tannakian subcategory. Indeed, a category without one would equal its core. By \cite[Theorem~1.1]{NataleCore} it would then be pointed or a product of a pointed category and an Ising category. The pointed $r$-primary part would have order at least $r^3$. Its quadratic form has a nonzero isotropic element: a cyclic subgroup of order at least $r^2$ has an isotropic subgroup of order $r$, and in the elementary abelian case a quadratic form over $\mathbb F_r$ in at least three variables has a nonzero zero. This gives the required Tannakian subcategory. In particular this applies to the unordered $(6,5,4)$ family by Theorem~\ref{fam:Q:1.1} whenever an odd cube survives in the dimension. For arbitrary companion exponents in $p^Aq^Br^4d$, a Tannakian $\Rep(G)$ with $r\mid |G|$ suffices for WGT by descent: the neutral de-equivariantization has $r$-valuation at most three, so Theorem~\ref{thm:three} applies. \subsection{Dyadic blocks of a fermion centralizer}\label{sec:fermion-block-transfer} Let $\M$ be an integral modular category of dimension $D=4\Delta$ with its positive spherical structure, a fermion $g$ and even sector $\mathcal B=\langle g\rangle'$. Use the spaces $A,B,C$ and the signed order of Lemma~\ref{lem:fermion-quotients} and its normalization table. Write $K=W_2|_A$, $W_o=W_{2'}$, $U=\lambda T_2$, $\lambda^8=1$, and $\epsilon_Y=\theta_{Y,2}^2$ on an even pair. The dyadic Weyl interchanges $B,C$, while $U$ interchanges $A,B$ and preserves $C$. \begin{lemma}[Positive defect one and the dyadic level]\label{lem:positive-defect-one-phases} Suppose $v_2(D)$ is odd, and let $E$ be a positive two-block containing a character of dyadic defect one. Its squared twists are $+1,-1$, with equal sums of squared degrees at the two signs. Its rows vanish on pairs of every other squared twist. If the actual dyadic twists of \emph{all} even labels have order dividing four, the full dyadic level divides eight, and $E$ consists of two labels of equal degree, both unramified at two. Under this twist bound, each full character whose field is contained in a dyadic cyclotomic field takes values in $\ZZ[i]$ on even objects. \end{lemma} \begin{proof} The residue on the even ring determines the full residue by squaring. The projector onto the lifted block commutes with the dyadic group by Lemma~\ref{lem:local-block-invariance}. Its restriction to $A$ projects onto $E$ and commutes with $K$. Thus $\widehat s^{-1}\CC^E=W_o^{-1}\CC^E$ has nonzero vacuum projection, and $E$ contains squared phase $+1$. Apply the Bruhat mass identity \eqref{eq:dyadic-bruhat-mass} from Lemma~\ref{seq:dyadic-pair} on $A$. The operators $U(2),L(2j),H_{1+4j}$ ($j$ odd), the projector and its $\widehat s$-conjugate all preserve $A$. Here $L(t)=WU(-t)W^{-1}$. The torus negates the symmetrized vacuum ($v_2(D)$ is odd and both vacuum labels have rational characters). Its weights on $E$ are $d_Y^2/\Delta$. Averaging $\theta_{Y,2}^{-2j}$ over odd $j$ gives $+1,-1,0$ at squared twist $+1,-1$, or higher order, respectively. Thus the two signed masses are equal and nonzero. By Proposition~\ref{lem:one-defect-block}, $E$ has at most two inertia orbits. The two signs already occupy distinct orbits, so no higher order occurs. Since $W_o$ preserves squared dyadic twist, the equality of subspaces above gives the support assertion. Now assume the global bound on the even twists. Balancing puts even-even $S$-entries in $\QQ(i,\zeta_m)$ for some odd $m$. A dyadic Galois unit $l\equiv1\pmod4$, trivial on the odd factor, fixes the restricted character of every even label. Each even fermion pair is consequently fixed as a set by Lemma~\ref{lem:fermion-quotients}. The square action on actual twists fixes each representative's twist, and opposite twists distinguish the representatives. Symmetry of $S$ and rationality of degrees now fix the values of every character on even objects. In particular an odd fermion orbit is fixed as a set. Its members have equal twists, so the unit $l=5$ gives $\theta_{Y,2}^{24}=1$ also on odd labels. The normalization scalar cancels in the square action since the vacuum is fixed. All actual dyadic twists have order dividing eight, as does $U$ since $\lambda^8=1$. For a purely dyadic character the fixed field of the units $1\bmod4$ lies in $\QQ(i)$, giving the asserted integrality. All positive characters have dyadic ramification index at most two. The two-branch defect-one block has an unramified branch by Proposition~\ref{lem:one-defect-block}. The other has length one or two, and length two would equate an integer square to twice another integer square by the mass equality. Thus each sign has one label, of the same degree. \end{proof} \begin{lemma}[Signed blocks at dyadic level dividing eight]\label{lem:signed-dyadic-pair} Suppose $v_2(D)$ is odd and all even actual dyadic twists have order dividing four. Then $U^4$ is scalar on $C$. If a positive two-block contains a defect-one character of degree $d$, the full block over it contains two odd fermion orbits, both free. Their actual dyadic twists have ratio $\pm i$ and their degrees $u,v$ satisfy \[ u^2+v^2=2d^2,\qquad v_2(u)=v_2(v)=v_2(d). \] If the entire odd sector has just two dyadic twists with ratio $\pm i$, then $R_+$ has a nontrivial invertible basis element. \end{lemma} \begin{proof} The level bound just proved uses only the bound on even actual twists. Put $Z=T_2^2|_A$, $F=KZK^{-1}$. These are involutions. The torus $H_5$ fixes the even labels as a permutation and negates $\sqrt D$, hence $H_5=-I$ on $A$. Modulo eight the identity $U(2)L(2)=L(2/5)H_5U(2/5)$ gives $ZF=-FZ$. Since $H_3=H_5H_{-1}=-K^2$ on $A$, it commutes with both. The factorization \[ (WU)^{-1}U^2(WU)=U(-1)L(-2)U(1) =L(-2/3)H_3U(2/3)=L(2)H_3U(6) \] gives the exact restriction \begin{equation}\label{eq:fermion-dyadic-transfer} (WU)^{-1}U^2(WU)|_A=\lambda^4 F H_3 Z . \end{equation} Its square is scalar since $H_3^2=I$, and $WU$ maps $A$ onto $C$. Over a positive defect-one block, Lemma~\ref{lem:positive-defect-one-phases} gives two labels of degree $d$, both unramified at two. Their full block contains four even labels, and $H_3$ acts scalarly on the two-dimensional positive space. Since $WU$ preserves the full block, its odd symmetric space also has dimension two. The invertible operator $FZ$ anticommutes with $Z$, so its two eigenvalues are opposites. By \eqref{eq:fermion-dyadic-transfer} the squares of the two odd twists differ by a sign. In the signed order $R_-$ with its unit-coefficient trace, the two corresponding characters have equal dyadic defect if inertia interchanges them. If inertia fixes both, their local order over the completed maximal unramified base $\mathcal O$ is a suborder of $\mathcal O^2$ containing the diagonal. The ideal of coordinate differences is $2^h\mathcal O$ for some $h\ge1$. Thus the order consists of pairs congruent modulo $2^h$, with conductor $2^h$ on each branch. By self-duality the conductor on each unramified branch is its codegree ideal (Lemma~\ref{lem:conductor-node}). Equal codegree valuations in \eqref{eq:fermion-restriction-codegree} give the same fixed/free type by parity, then $v_2(u)=v_2(v)$. Column orthogonality of $1,g$ on the full block equates its odd mass to $4d^2$. If both labels were fixed, $u^2+v^2=4d^2$ would have odd two-valuation on the left and even on the right. Thus both orbits are free and $u^2+v^2=2d^2$, which also gives $v_2(u)=v_2(d)$. If the only odd spins are $\delta_1,\delta_2$ with ratio $\pm i$, on $C$ one has $U=aI+bU^2$, where, writing $x=\lambda\delta_1$, $y=\lambda\delta_2$, one has $a=xy/(x+y)$, $b=1/(x+y)$ and $|a|=|b|=1/\sqrt2$. Restrict $WUW=U^{-1}WU^{-1}$ to $U A$, and multiply by $U^{-1}$ on the left. Since $WU A=C$ and $W^2$ is central, on $A$ this gives \[ U^{-2}K =aK^2+b\,U^{-1}L(-2)U K^2. \] Here $U^{-2}=\lambda^{-2}Z$, and \eqref{eq:fermion-dyadic-transfer} gives $U^{-1}L(-2)U=cF H_3 Z$ with $|c|=1$. Since $H_3=H_5 H_{-1}=-K^2$ and $K^4=I$, it follows that $K=a'ZK^2+b'F$ with $|a'|=|b'|=1/\sqrt2$. For the projection $P=(1+Z)/2$ this yields $PKP=a'P K^2 P$. Put $w=W_o v_{[1]}$ in $A$. We have $Pw=w$, while $K^2w$ is a unit scalar multiple of $w$ by the torus action on the vacuum and commutation of local factors. Since $\widehat s v_{[1]}=Kw$, $|w_{[j]}|^2=2d_j^2/\Delta$ on $\epsilon_j=1$ and vanishes otherwise. Lift back to the full vacuum $w^{\rm full}=W_o v_0$. This is supported on the even labels with actual dyadic twist one: $W_o$ preserves monodromy sector and commutes with $T_2$. On every even pair the reduced coefficient is $w^{\rm full}_j+w^{\rm full}_{gj}$ because $W_o$ commutes with the fermion permutation. Exactly one representative of each positive-sign pair can contribute, so its squared weight is again $2d_j^2/\Delta$. Take $m$ zero on the dyadic level and one on the odd complementary level. The indicator formula at the beginning of Subsection~\ref{fam:M:sec:7.2} gives, for even $Y$, \[ \nu_m(Y)=\sum_{j\in\Irr(\mathcal M)} |w^{\rm full}_j|^2\chi_j(Y) =\frac1\Delta\sum_{[j]}(1+\epsilon_j)d_j^2 \chi_{[j]}([Y]). \] This is the coefficient at $[Y]^*$ of the reduced element $1+J$, where $J=\sum_{[j]}\epsilon_j e_{[j]}$. It is an algebraic integer by the cyclotomic integrality of indicators \cite[Theorem~5.1]{NSpivot} and rational by Galois covariance of the sign. Thus $J$ has integral coefficients. It is self-dual with $J^2=1$, so has squared based-ring norm one, and its dimension is $\epsilon_1=1$. This gives a positive invertible basis element in $R_+$. Both signs occur by anticommutation, so it differs from the unit. \end{proof} \section{Three repeated primes with a quartic or quintic exponent}\label{fam:Q:sec:1} \begin{theorem}[Three bounded exponents]\label{fam:Q:1.1}\label{fam:F:3.2}\label{fam:F:n:754}\label{fam:Q:n:q13:654} Let $p,q,r$ be any distinct primes, and let $d$ be square-free and coprime to $pqr$. Every weakly integral braided fusion category of dimension dividing $p^7q^5r^4 d$ is WGT. \end{theorem} The exponents in this theorem may be assigned to the primes in any order. Throughout this section, we write $pC(e). \label{fam:Q:n:test} \end{equation} \begin{theorem}\label{fam:Q:1.2}\label{fam:Q:n:quart:main} Let $p3q$ suffices for the two tests at $m=3$. More generally, for each $p$, the whole interval $qp^{\lfloor A/2\rfloor-2}$. This holds automatically for $A=6,7$. The even case combines the ordinary and fermion reductions in Sections~\ref{fam:Q:sec:6}--\ref{fam:F:sec:3}. The unrestricted quartic problem remains open. In a divisor-minimal obstruction all three repeated primes survive and the square-free factor disappears. A third exponent at most three is already covered (Lemmas~\ref{fam:M:2.1}, \ref{fam:M:2.4}, Proposition~\ref{fam:M:4.7} and Theorem~\ref{fam:M:4.12}). Thus $v_r(\Delta)=4$ in the quartic problem, including the positive table, whose dimension is half the categorical dimension. A $q$-free or $p$-free nonunit in $P_r$ then has degree $rp^i$ or $rq^j$ with $i,j>0$, by the principal height bound and prime avoidance (Lemma~\ref{fam:M:4.10}). We use \emph{house} for the largest absolute value among all conjugates. \subsection{Character values and tensor products}\label{fam:Q:sec:2} \begin{lemma}[An odd word of prime weight]\label{fam:Q:2.4}\label{fam:Q:n:lem:oddword} Let $tr$. Apply Lemma~\ref{lem:cyclotomic-optimum} at this asymmetric cost with retained primes $t,r$. The lower bound just proved uses only the two residues, so it applies to every contracted integral target. A nontrivial spectator component at an odd prime $u$ has expected cost at most $3/(2(u-1))<1$. Thus every positive summand is supported at $t,r$. It remains to minimize on the unsigned two-prime roots alone. Again the configuration is unimodular and an integral minimum exists. Each optimum has length $r$. Erasing $r$-parts now gives a sum $w$ of $r$ unsigned $t$-power roots, divisible by $r$. It is nonzero since a vanishing unsigned sum has length divisible by $t$. The integral norm makes $w/r$ a root of unity. Equality in the triangle inequality makes all the $t$-parts the same root $\eta$. The remaining unsigned $r$-word has augmentation $r$ and coefficients $c_i-r\delta_{i0}$ constant mod $t$ on each $\mu_r$-coset. A full nonidentity coset makes the word zero. If all $r-1$ nonidentity positions of $\mu_r$ are occupied, the extra term cannot be outside the coset or duplicate one position, by these congruences. The complete coset again gives the prohibited zero at $t$. All constants therefore vanish. Strict house excludes the word supported only at one. This proves the formula. Every optimal vertex has the same $\eta$, since the ratio of two such roots lies in $\mathbb Q(\zeta_{t^a})\cap\mathbb Q(\zeta_{r^b})=\mathbb Q$. The remaining coefficients are unique: every nonzero coset has a vacant position, and a reduced unsigned $r$-word has a unique coefficient vector. The original expression is a convex combination of those optimal vertices. \end{proof} \begin{lemma}[A single residue in odd order]\label{fam:Q:2.5}\label{fam:Q:n:q12:positive} Let $s$ be prime and let $z=\sum_i c_i\xi_i$ be a cyclotomic integer with $c_i\in\mathbb Q_{\ge0}$ and $\sum_i c_i=g\in\mathbb Z_{>0}$. Discard the zero coefficients; all assertions about terms concern $c_i>0$. Suppose every prime dividing the orders of the remaining roots is at least $\ell\ge3$, and $z\equiv g$ at all primes above $s>g$ in the common field. If $(s-g)(\ell-1)>g$, all terms of this expression have $s$-power order, and $z$ has an integral unsigned $s$-word of length $g$. \end{lemma} \begin{proof} Choose a cost $c$ with $\max(1,g/(s-g))g$, while augmentation $g$ forces exactly $g$ positive terms. Integral optima exist for the signed one-prime configuration by unimodularity. Thus any loss of cost contradicts its minimum, proving the assertion on the original terms and the integral word. \end{proof} \subsection{A prime-weight crossing with one zero residue}\label{found:C:5} Write $\mu_{s^\infty}$ for the roots of $s$-power order. Let a word of weight $r$ mean a positive rational sum of roots of unity whose coefficients sum to $r$. All residue conditions below are at every prime over the indicated rational prime. \begin{proposition}\label{found:C:5.1}\label{found:C:key:new:zero} Let $q1$. This is the second possibility. In particular no signed integral word of length below $r$ has the required residues. This lower bound required only the residues: strict house entered solely to exclude $m=0$ in the first alternative. Lemma~\ref{lem:cyclotomic-optimum} with both signed costs one therefore removes every spectator except a sign and gives an integral optimal word on $q,r$. In the first possibility there are at most $1+\lfloor(r-1)/q\rfloor 0$, $\eta$ of $q$-power order and each $\xi_j$ of $r$-power order. \item All $p$-free nonunits of $P_q$ have $r$-height two. The $q$- and $r$-active labels are disjoint, and a normalized entry belongs to a $p$-power cyclotomic field. Its canonical positive coefficient vector $(c_i)$ has length $L\le r$ with $L\equiv r\pmod p$. On every $\mu_p$-coset the entries $c_i-r\delta_{i0}$ are constant modulo $q$. \end{enumerate} \end{proposition} \begin{proof}[Proof of Proposition~\ref{fam:Q:n:quart:alternative} and Theorem~\ref{fam:Q:n:quart:main}] Suppose a minimal counterexample exists. Its table is ordinary of odd dimension. Choose $X\in P_r$ of degree $rp^i$. Every $p$-free nonunit $Y\in P_q$ has degree $q^j r^v$ with $v=1$ or $2$. Suppose $v=1$. The normalized entry $z=S_{XY}/\lcm(d_X,d_Y)$ has positive odd-order weight $r$ and strict house less than $r$. Its residues are $r$ at both $r$ and $q$: use $\chi_X(Y)=q^jz$ at $r$, and $\chi_Y(X)=p^iz$ at $q$. Lemma~\ref{fam:Q:n:lem:oddword}, with $t=q$, gives at most $m+1$ positions of $r$-power order up to translation. This proves the first alternative when such a label exists. Under the hypotheses of Theorem~\ref{fam:Q:n:quart:main}, the support bound holds at $r$. Lemma~\ref{fam:Q:n:lem:j11-descent}, applied after rescaling by $\lcm(d_X,d_Y)/d_Y$, gives a contradiction. We may therefore assume every such $Y$ has $v=2$. Its formal codegree $f_Y=\Delta/d_Y^2$ is an $r$-unit, so its row vanishes on $r$-active columns (Theorem~\ref{fam:M:3.1}). If a column were both $q$- and $r$-active, orthogonality to the vacuum on $P_q$, reduced modulo $p$, would give a $p$-free nonunit hit (Proposition~\ref{fam:M:4.2}). This proves that $q$- and $r$-activity are disjoint. By Proposition~\ref{fam:M:6.2} no character field then ramifies at both $q$ and $r$. If $r$ were also separated from $p$, it would be an isolated quartic prime, excluded in three-prime ordinary data by Lemma~\ref{fam:M:7.2}. Fix a joint $p,r$-active column $J$. Orthogonality on $P_p$ on the trivial prime-to-$p$ twist fiber, reduced modulo $q$, gives a $q$-free nonunit $V\in P_p$ hitting $J$. It has degree $p^e r$: both primes divide the degree by the exclusion of prime powers, and $r^2$ is impossible at a hit of $J$ by defect-zero vanishing. Take $A\in P_r$ of degree $rq^i$ and a $p$-free nonunit $B\in P_q$, hence $d_B=r^2q^j$. At $V,A$ the normalized entry has weight $r$ and is nonzero by $V\in P_p$ since $p\nmid d_A$. Its residue at each prime above $r$ is zero by $A\in P_r$, since $\chi_A(V)=p^e S_{VA}/\lcm(d_V,d_A)$. If this value were unramified at $r$, division of the normalized entry by $r$ would give a nonzero algebraic integer of house less than one. Thus $K_V=\QQ(\chi_V)$ ramifies at $r$ and is unramified at $q$. The field $K_B$ is unramified at $r$ by its codegree (Theorem~\ref{fam:M:2.5}). Now form $z'=S_{VB}/\lcm(d_V,d_B)$. It lies in $K_V\cap K_B$, so is unramified at both $q$ and $r$. It belongs to a field of $p$-power conductor: the entry fields are cyclotomic with no other conductor primes, and the inertia groups of the other two factors fix $z'$. It has weight $r$, strict house less than $r$ and residue $r$ at both $p$ and $q$ by the principalities of $V,B$. Average its balancing expression over the $q,r$-cyclotomic factor. The negative mean of a complementary root has magnitude at most $1/(q-1)$. Replace a negative coefficient on a $p$-power root by positive coefficients on the other $p-1$ positions of its $\mu_p$-coset. This gives a positive rational $p$-word of weight at most $r$. Subtract the minimum coefficient on each coset. The remaining coefficients $c_i$ are nonnegative integers, since $z'$ also has an integral expression and two such vectors differ by a constant on each coset. Write $L=\sum c_i\le r$. Reduction at $p$ gives $L\equiv r\pmod p$. Reduction at $q$ makes $c_i-r\delta_{i0}$ constant modulo $q$ on each $\mu_p$-coset. This proves the second alternative, without any size hypothesis. To finish the WGT theorem assume \eqref{fam:Q:n:test}. Suppose zero is the only vacant position in the identity coset and put $\rho=r-mq$. Each of its $p-1$ other coefficients is $q-\rho$ modulo $q$, and elsewhere all coefficients are multiples of $q$. Thus $L=(p-1)(q-\rho)+qn$ with $0\le n\le m$. Reduction at $p$ gives $n\equiv m+1\pmod p$, impossible since $m+1(k-1)q$ with $q\ge11$ give the corresponding prime cutoffs; for $k=6$ use $q\ge17$. For general $M$, choose $T_k$ large enough that Lemma~\ref{fam:Q:2.3} excludes $k$ positions at primes $s>T_k$; such a choice exists since $k$ is fixed. It suffices that $q\ge\max_{2\le k\le M}\lceil T_k/(k-1)\rceil$. \end{proof} \subsection{Small exponents and even dimensions}\label{fam:Q:sec:4}\label{fam:Q:n:sec:small} \begin{theorem}[A bounded exponent and a large companion prime]\label{fam:Q:4.1}\label{fam:Q:n:quart:five} Let $t,u,r$ be distinct primes with $r$ odd, and $c\ge4$. Put $w=t^{\lfloor(c-2)/2\rfloor}$ and $e=w(w-1)/2$. If \[ 1+2eC(e), \] then every braided fusion category of dimension dividing $t^c u^A r^B d$ is WGT for arbitrary $A,B$ and square-free $d$ coprime to $tur$. For three odd primes one may replace $r/2$ by $2r/3$. \end{theorem} \begin{proof} We work in a minimal table with all three primes repeated. Except when $t=2$ and the table is reduced, take an $r$-free nonunit of $P_t$ and a $u$-free nonunit of $P_r$. Their degrees have shapes $t^i u^j$ and $t^h r^k$ and their gcd $g$ is at most $w$ by the principal valuation bound at $t$. Their normalized entry has residue $g$ at $r$ and weight $g$. The stated inequalities imply $r>2w$ (for odd data $w\ge3$). Thus the single-residue theorem gives an unsigned $r$-word on at most $w$ positions, contrary to Lemma~\ref{fam:Q:n:lem:j11-descent}. In the reduced dyadic case exchange $u,r$ if necessary to make $r$ the larger, preserving the inequalities since $w$ is fixed. Use complementary witnesses in $P_u$ and $P_r$; their common $2$-part satisfies $g\le2^{\lfloor(c-1)/2\rfloor}\le2w$. It gives both positive residues $g$ at $u,r$. Here the general phase inequality implies $r>2w+1$ since $w\ge2$. Hence Theorem~\ref{fam:M:4.9} gives the unsigned $r$-word with at most $1+\lfloor g/u\rfloor\le 1+\lfloor 2w/3\rfloor\le w$ positions. The same lemma finishes. \end{proof} For $t=3$ and $c=5$ this gives all $r\ge17$ in either parity, and all $r\ge11$ for three odd primes: $w=3$, $e=3$ and $C(3)=4$. The first separation in the proof above also gives a parity reduction when the second separation is unavailable. We record a version that includes a dyadic least prime. \begin{proposition}\label{fam:Q:4.2}\label{fam:Q:n:quart:parity} Let $p2: &1+2e<2r/3,\quad (r-1)/2>C(e),\\ p=2: &q\nmid(r-1)(r-2),\quad 1+2eC(e). \end{array} \] In particular both tests hold when $q+2G$, all divisors of $p^a q^b r^4d$ are covered (Theorem~\ref{fam:M:7.11}). Otherwise, with $m_0=\lfloor G/q\rfloor$, $e_0=m_0(m_0+1)/2$, they are covered whenever \[ r>G+1,\qquad 1+2e_0C(e_0). \] Indeed take complementary witnesses of degrees $p^i q^j$ in $P_q$ and $p^t r$ in $P_r$. Their common weight $g=p^{\min(i,t)}$ is at most $G$. The two-residue theorem (Theorem~\ref{fam:M:4.9}) writes the normalized entry as a positive $r$-word on at most $1+\lfloor g/q\rfloor$ positions when $r>g+1$. Lemma~\ref{fam:Q:n:lem:j11-descent} gives the contradiction. In odd dimension the same test may interchange $p$ and $q$, using the ordinary principal block at $p$. \begin{corollary}[A middle-prime test independent of the largest prime]\label{fam:Q:4.3}\label{fam:Q:n:q13:midtest} Let $pq$, with $B$ arbitrary. \end{corollary} \begin{proof} If $M=0$, use Theorem~\ref{fam:M:7.11}. For $M>0$, the preceding large-prime test and Theorem~\ref{fam:Q:n:quart:main} both use $r$-words on $1+m$ positions for $m=\lfloor\min(r,G)/q\rfloor$: if $rG$, the former applies since the odd numbers $r,G$ differ by at least two. Thus $1\le m\le M$ and $r>mq$. For $m=1$ the bound at $r$ is immediate as $r\ge7$. For $m\ge2$ set $e=m(m+1)/2$. At $h\le e$ one has $(h-1)/g(h)\le m$: if $g(h)mq\ge m(m+1)^2$ verifies both inequalities. The equality $r=G$ is impossible for the distinct primes. \end{proof} \begin{proposition}[Transport of deep quartic twists]\label{fam:Q:4.4} Let an ordinary or positive reduced minimal three-prime table at $p0$, the number $N$ of degree-$n$ exceptional labels satisfies $2N\ge c(p-1)p^k$, where $c=2$ in odd dimension and $c=1$ otherwise. For $U\in P_p$ of $p$-height $j$, an inertia orbit on $U$ of length $e$ at a prime $s\ne p$ then satisfies \[ c(p-1)e p^{2j}<2p^{k+1}. \] \end{corollary} \begin{proof} Higher strata in the congruence vanish under either hypothesis: $p\mid L_i$ gives $p^3\mid p^{2i}(L_i+pM_i)$ for $i\ge1$. Thus a nonzero $L_0+pM_0$ is at least $cp^{k+1}$, by parity in odd dimension. Its contribution to $N$ is $L_0+(p-1)M_0/2\ge(p-1)(L_0+pM_0)/(2p)$. Each of these labels has codegree $p^{2k+1}$, so its field is unramified at $s$. Lemma~\ref{fam:Q:2.1} gives $eN(p^jn)^2\max(t,u)$ and $v_t(D)=5$. Suppose the $s$-free nonunits of $P_u$ and the $u$-free nonunits of $P_s$ all have $t$-height two. Some $s$-free nonunit of $P_t$ is fixed by $s$-inertia. \end{proposition} \begin{proof} Suppose all $s$-free nonunits of $P_t$ are moved. Call the two sets in the hypotheses the witnesses. They exist by prime avoidance. Each witness has $t$-defect one. Lemma~\ref{lem:quintic-common-class} gives a single visible quadratic class for both sets. Each $s$-active or $u$-active label hits a corresponding witness by principal-block orthogonality modulo the omitted prime (Proposition~\ref{fam:M:4.2}). All their $t$-phases are therefore $1$ or in the chosen class. There is a label outside these phases: either a deeper twist occurs or Lemma~\ref{fam:M:7.4} gives the opposite primitive class. Lemma~\ref{lem:visible-phase-transfer} applied at $s,u$ gives every label $Z$ of such a fiber pure $t$-power codegree. It follows that $D=t^5n^2$ and the possible codegrees $t,t^3,t^5$ have degrees $t^2n,tn,n$, respectively. Each exceptional Galois orbit contains at least $t-1$ labels: square covariance gives at least $(t-1)/2$ distinct phases, and both duals occur at each phase in odd dimension. The row of $Z$ is supported on $P_t$. The dimension--trace equation reduced at each outer prime gives separate nonunit orbits of degrees $tu^i$ and $ts^j$ with nonzero entry trace. Each contributes at least one to the trace norm, while the vacuum contributes $[K_Z:\QQ]\ge t-1$. Hence $f_Z\ne t$. If an exceptional label had degree $n$, take an $s$-free nonunit $U\in P_t$. It has degree $tu^i$ and inertia orbit length $e\ge2$ by assumption. Corollary~\ref{fam:Q:6.7}, with $k=2$, $j=1$ and $c=2$, would give $(t-1)ep$. Lemma~\ref{fam:Q:n:q12:positive} gives an unsigned $r$-word of length $pq$, the two exponents $A$ and $C$ may both be arbitrary in $p^A q^5 r^C d$. \end{proposition} \begin{proof} In a minimal table with three primes the middle valuation is exactly five by Theorem~\ref{fam:M:7.3}. First a $p$-free nonunit $A\in P_r$ has $q$-height two: height one would give weight $q$ with a complementary $qp^i\in P_q$, with residue $q$ at both $q,r$. By Theorem~\ref{fam:M:4.9} the house would be $q$, contradicting strictness. Suppose now that an $r$-free nonunit $U\in P_p$ has degree $p^i q$. Compare two character values on this one row. Choose $A\in P_r$ as above, of degree $q^2 r^l$, and a $p$-free $B\in P_q$ of degree $qr^k$. Each crossing has common weight $q$. At $U,A$ the residues at $p,r$ are both $q$, so Theorem~\ref{fam:M:4.9} gives an unsigned $r$-word on at most $1+j$ positions; its signed alternative would require $r=q+1$. At $U,B$ the residues at $p,q$ are $q,0$. Lemma~\ref{fam:Q:n:lem:oddword} gives, up to a scalar, an unsigned $q$-word on at most $1+j$ positions. Rescaling each normalized crossing gives the corresponding character value of full weight $d_A$ or $d_B$. The adjoints have full words on $r$-power and $q$-power roots, respectively, each on at most $1+j(j+1)<2q/3$ positions. Both supports are exposed and meet only at $1$, contradicting Lemma~\ref{fam:Q:2.2}. Thus every $r$-free nonunit of $P_p$ has $q$-height two. For any $r$-free nonunit $X\in P_q$, $d_X=qp^i$. Pair it with such a $p$-free witness of $P_r$. Their common weight is $q$ and the residue at $r$ is $q$. If $(r-q)(p-1)>q$, Lemma~\ref{fam:Q:n:q12:positive} gives an unsigned $r$-word of length $qp^{\lfloor A/2\rfloor-2}$. Every weakly integral braided fusion category of dimension dividing $p^Aq^5r^4d$ is WGT. In particular $A=6$ and $A=7$ work for all odd prime triples. \end{theorem} \begin{proof} A divisor-minimal failure has table of dimension $p^\alpha q^5r^4$, by the smaller-quartic and cubic-exponent theorems (Theorems~\ref{fam:M:7.3} and~\ref{fam:M:4.12}). Theorem~\ref{fam:Q:n:q12:unified} gives $h=\lfloor\alpha/2\rfloor\ge3$. Consider opposite nonunits of $P_q,P_r$, respectively $r$-free and $q$-free. Their common $p$-part is at least $q$ by Lemma~\ref{fam:M:4.10}. Since $q>p^{h-2}$, their individual $p$-heights are $h-1$ or $h$. They cannot all have height $h$. At $\alpha=2h$ all these witnesses would have $p$-unit codegree. But a $p$-active fiber contains a $q$-active column by the odd $q$-valuation, and orthogonality on $P_q$ modulo $r$ gives a witness hitting it. At $\alpha=2h+1$ every witness would have $p$-defect one. As in Theorem~\ref{fam:M:7.9}, its two orbit degrees agree: their squared ratio is $a_0$ or $2a_0$ with $a_0\mid p-1$, and the primes of the degrees are all at least $p$. The opposite witnesses thus see the same primitive quadratic class, as in Proposition~\ref{fam:Q:n:q13:common}. The missing-phase argument gives a row of pure $p$-codegree, which would force the $q$-valuation to be even. Consequently there exists an opposite pair of common weight $G=p^{h-1}$. Write $M=\lfloor G/q\rfloor$, so $1\le MG$ the two-residue theorem gives an $r$-word on at most $1+M$ positions: $r=G+1$ is impossible for two odd numbers. For $rmq\ge m(m+1)^2$. For $m\ge2$ the estimates of Corollary~\ref{fam:Q:n:q13:midtest} give both support inequalities at $r$, and for $m=1$ they hold since $r\ge7$. This contradicts Lemma~\ref{fam:Q:n:lem:j11-descent}. In particular $h\ge4$ is excluded, since then $q>p^2\ge(M+1)^2$. It remains that $h=3$ and $q<(M+1)^2$. Since $q\ge5$ we must have $M\ge2$, so $q0$. Suppose $N$ labels have odd degree $n$ divisible by $q^i$ and pure dyadic codegree $f$. Then the $r$-inertia orbit size $e$ of $U$ satisfies $e\ge(r-1)/2$ and $4eN0$. \end{proof} \begin{proposition}[An ordinary quintic at two]\label{fam:Q:5.8}\label{fam:Q:n:q12:ordinary-five} There is no ordinary minimal table of dimension $2^5q^b r^c$ for odd primes $q3$, let $\gcd(d_V,d_B)=3$. Set $z=S_{VB}/\lcm(d_V,d_B)$. If $z\equiv1$ at every prime above two and $z\equiv0$ above three, then $z=\lambda\sqrt{-3}$ for a dyadic root of unity $\lambda$. \end{lemma} \begin{proof} A signed integral word of length at most three represents $z$: at a common level $2^a3^br^c$ the root lattice on the odd factors has a totally unimodular root configuration by \cite[Theorems~3 and~13]{BH}. A basis of the $2$-power lattice consists of one member of each opposite pair, so tensoring gives disjoint signed copies of the same unimodular matrix. Integral minimum length follows. Erase the $3$-parts of the word. The remaining sum $w$ is divisible by three in its unramified ring and has house at most three. If $w=0$ the only possibility is a cancelling pair (a vanishing triangle would require the prime three in the complement). Thus $z=\lambda(\alpha-\beta)$ with $\lambda$ prime-to-three and $\alpha,\beta$ of $3$-power order. If $w\ne0$, Kronecker and equality in the triangle inequality give $z=\lambda(\alpha+\beta+\gamma)$ with the same conventions. In either case $\lambda$ has trivial $r$-part: invert its $r$-part while fixing the other factors and reduce both values modulo two. They are both $1$ and the reduction is faithful on the odd-order roots. Erase the dyadic factor modulo two. In the pair case the vector of $\alpha,\beta,1$ modulo two is a union of complete $\mu_3$-cosets by the cyclotomic relations. Its odd cardinality is at most three, so it is the identity coset and $\{\alpha,\beta\}=\{\omega,\omega^{-1}\}$ for a primitive cube root $\omega$. In the triple case the vector of $\alpha,\beta,\gamma,1$ has even cardinality at most four and is a union of complete triples. Thus it is empty and the word is $\lambda(1+2\xi)$ for a $3$-power root $\xi$. Strict house excludes $\xi=1$. At order at least nine, the adjoint support $1,\xi,\xi^{-1}$ has one mark in each occupied $\mu_3$-coset. Its dilation by the $3$-adic unit four meets it only at one. Lemma~\ref{fam:Q:n:lim:diff} after scaling to the character value at $B$ gives a contradiction. Thus $\xi=\omega^{\pm1}$ and in both branches $z=\pm\lambda(\omega-\omega^{-1})$. \end{proof} \begin{proposition}[Transfer at two and three]\label{fam:Q:6.2} In an ordinary or positive reduced minimal three-prime table at $2,3,r$, a row $U\in P_2$ of degree $2^i3$ forces every odd-degree label to have the same $3$-primary twist as $U$. No such row exists when $v_3(\Delta)$ is odd. The same implications hold in a full integral modular table if there is a subset $E\subseteq P_3$ with the following properties. It is invariant under ternary inversion $\iota$, contains the vacuum as its only invertible, and supports restricted orthogonality at unequal ternary phases and the odd-valuation mass identity. At each odd-degree nonunit of $E$, the crossing with $U$ is negated by $\iota$ and is nonzero. \end{proposition} \begin{proof} Let $\iota$ invert the $3$-power roots and fix the other primary factors, and put $H=\chi_U+\chi_{U^\iota}$. The two row labels have the same $3$-twist by Galois square covariance. Each odd nonunit $B\in E=P_3$ in the minimal tables has degree $3^h r^j$ with $h,j>0$. The normalized crossing has weight three and residues $1,0$ at $2,3$, respectively, by principality. Lemma~\ref{fam:Q:6.1} therefore gives the required negation and nonvanishing. In all cases $H(B)=0$ at each such $B$, and $B^\iota\ne B$. On every even $B$ both terms in $H$ vanish modulo any prime above two by principality. If an odd-degree $J$ has different $3$-twist, block orthogonality gives $\sum_{B\in E}\chi_J(B)\overline{H(B)}=0$. The vacuum contributes two. On every even $B$, lcm divisibility makes $\chi_J(B)$ divisible by two, so all other nonzero contributions have dyadic valuation greater than one. This proves the transfer; in particular every odd-degree label is $3$-inactive by the vacuum. Now the odd labels of $E$ number one modulo two: all except the vacuum are paired by $\iota$. At odd $v_3(\Delta)$ the principal-block mass identity of Prop.~\ref{fam:M:3.8} reads \[ \sum_{B\in E,\ \mathrm{ord}(\theta_{B,3})=3} d_B^2 =2\sum_{B\in E,\ \theta_{B,3}=1}d_B^2. \] The left is zero modulo four whereas the right is two. \end{proof} \begin{lemma}[Three-primary phases at odd valuation]\label{fam:Q:6.3} If an ordinary or positive reduced minimal table has odd $v_3(\Delta)$, then some $3$-twist lies outside $\{1,\eta\}$ for each primitive cube root $\eta$. \end{lemma} \begin{proof} If all phases are in $\{1,\eta\}$, the first paragraph of the proof of Lemma~\ref{fam:M:7.4} and its immediately preceding indicator construction (with $p=h=3$) give an integral self-adjoint virtual element $b$ with $b^2=3b$, $b_1=1$, $\dim b=3$. In the ordinary table this uses the nontrivial local Legendre sign from $v_3(\Delta)$ odd, and applies also in even dimension; odd dimension in that lemma is used subsequently for the nonreal character orbit. For clarity, the missing Fourier coefficient gives $w=-G Psv_0$ on the inactive phase projector, $|G|^2=3$. The indicator construction gives $\chi_j(b)=3\mathbf1_{\theta_{j,3}=1}$, rational integral coefficients and the stated normalization. For a positive reduced table we obtain the same element as follows. In the integral minimal modular extension let $g$ be the fermion and $A$ the positive space spanned by $(v_Y+v_{gY})/\sqrt2$ for even fermion pairs. Write $W_o$ for the product of the local Weyls away from three and $w=W_o v_0$. It interchanges the fermion permutation and its monodromy sign. Therefore the coordinates of $W_o(v_0+v_g)/\sqrt2$ in the pair basis are $2w_Y$, with $w_Y=w_{gY}$ for even $Y$. For $m$ zero on the ternary level and one on its complement, the complementary Weyl identity and the ordinary indicator formula give, for even $V$, \[ \nu_m(V)+\nu_m(gV)= \sum_{[Y]}|2w_Y|^2\,\widehat\chi_{[Y]}([V]). \] Indeed the odd-sector contributions cancel and each even pair contributes twice the same character. If all ternary phases in the positive table lie in $\{1,\eta\}$, the proof of Lemma~\ref{fam:M:7.4} applies its ternary Bruhat identity on $A$ and gives $|2w_Y|^2=3d_Y^2/\Delta$ at inactive pairs and zero elsewhere. The displayed sum is rational by Galois invariance and is an algebraic integer as a sum of indicators. It therefore defines the same integral virtual element in the positive ring, with norm squared three and dimension three. Let $M=N_b$; $M^2=3M$ and $M$ is symmetric integral. At a diagonal $a=1$ or $2$ the off-diagonal squared row norm is $a(3-a)=2$, so the row has two neighbors with signs. Rows at $0,3$ are isolated. The unit component is a cycle of length three or four: a longer cycle has nonneighbors with a unique length-two path, contradicting the matrix equation. A triangle has trace three (the trace is a multiple of three and the unit diagonal is one), hence rank one and all three diagonals one. Its positive dimension eigenvector forces equal dimensions, contradicting perfectness. On a four-cycle let $A$ oppose the vacuum. Adjacent diagonals sum to three, so the two neighbors $Y,Z$ have diagonal two. Their dimension equations give $d_Y,d_Z\in\{d_A-1,d_A+1\}$. Coprime vanishing gives $\chi_A(Y)=\chi_A(Z)=0$, impossible because $b=1\pm Y\pm Z$ and the only eigenvalues of $b$ are zero and three. \end{proof} \begin{theorem}[A pure ternary codegree]\label{fam:Q:6.4}\label{fam:Q:n:quart:pure} In an ordinary minimal table of dimension $2^a3^5r^4$ with $r>3$, the exponent $a$ is even. For some primitive cube root $\eta$, the set of labels with $3$-primary twist outside $\{1,\eta\}$ is nonempty, and every such label has formal codegree $27$ or $243$. \end{theorem} \begin{proof} By prime avoidance, $P_2$ contains $r$-free nonunits. Their degrees have positive $2$- and $3$-heights, and hence $3$-height exactly two by the transfer proposition. The $2$-free nonunits of $P_r$ likewise have degree $9r$ (Section~\ref{fam:M:sec:7.4}). All these witnesses have ternary defect one. By Lemma~\ref{lem:quintic-common-class} their rows see only $1$ and a common primitive cube root $\eta$. By principal orthogonality modulo the omitted prime, each $2$-active or $r$-active column hits one of the witnesses. Their $3$-phases therefore lie in $\{1,\eta\}$. The preceding lemma gives a nonempty $3$-phase fiber outside the pair. Lemma~\ref{lem:visible-phase-transfer} at $2,r$ gives every row $Z$ on it $2$- and $r$-unit codegree, so $a$ is even, and $f_Z=3^{5-2j}$ for $0\le j\le2$. The value three is impossible: orthogonality to the vacuum reduced at each of $2,3,r$ requires three distinct nonunit hits, since a hit omitting two primes would have prime-power degree. The Galois-averaged squared norm of each nonzero algebraic integer is at least one, so the vacuum and these three hits already exceed three. (Equivalently average the entire row norm over the splitting field.) \end{proof} \begin{lemma}[The trace restriction at codegree $27$]\label{fam:Q:6.5}\label{fam:Q:n:quart:twentyseven} Suppose there is an ordinary minimal table of dimension $2^a3^5r^4$ and a label $Z$ of pure codegree $27$, so $a$ is even and $d_Z=2^{a/2}3r^2$. There is a unique Galois orbit of $r$-free nonunits with a nonzero absolute trace on this row. For its representative $U$ write $d_U=2^i3$. Its field equals $K_Z$, is of $3$-power conductor, and \[ 9\cdot2^i\equiv\pm1\pmod r . \] There is likewise exactly one odd nonunit orbit contributing nonzero trace, of degree $3r^j$, $j=1,2$, with field $K_Z$. \end{lemma} \begin{proof} Every hit lies in $P_3$ by Theorem~\ref{fam:M:2.7}. The integral trace--dimension relation of Lemma~\ref{fam:M:2.3} at $Z$ reduced modulo $r$ and modulo two gives a separate trace-contributing $r$-free and odd nonunit orbit. Every such label $U$ has exactly $3$-height one by the principal valuation bound and absence of prime-power nonunit degrees. Then $\chi_U(Z)$ vanishes above three by principality, and $\chi_Z(U)=(d_U/d_Z)\chi_U(Z)$ does too since the ratio is a $3$-unit. This holds at every prime above three, so its absolute field trace is an integer multiple of three (reduce the sum of all conjugates). For $F=\QQ(\chi_Z(U))$ the exact contribution to the orbit trace norm is $[K_U:F][K_Z:F]\operatorname{Tr}_{F/\QQ}(\chi_Z(U))^2$ by Lemma~\ref{fam:M:2.3}. A nonzero contribution is at least nine. Three such orbits are impossible by the positive vacuum term, and either relative field index exceeding one would likewise cost at least $18+9+1>27$. Thus for each of these two orbits the entry field equals both row fields, and the corresponding absolute trace is $\pm3$. The dimension trace identity modulo $r$ now reads $1\pm3d_U=0\bmod r$ for the unique $r$-free orbit. \end{proof} \begin{proposition}[The odd witness at the ternary quintic]\label{fam:Q:6.8} Let an ordinary minimal table have dimension $2^a3^5r^4$, with $r>3$. If $r\ge11$, no such table exists, for any $a$. If $r=7$, all exceptional ternary phases of Theorem~\ref{fam:Q:n:quart:pure} have codegree $243$; every $r$-free nonunit of $P_3$ has tame $r$-inertia orbit of size exactly two, and every $r$-active label has even degree. \end{proposition} \begin{proof} Take any $r$-free nonunit $A\in P_3$ and any odd nonunit $Y\in P_r$. Their degrees have the forms $3\cdot2^i$ and $9r$. Their normalized crossing $z$ has weight three and strict house less than three, and $z\equiv3$ above $r$ by $r$-principality. If $K_A$ is $r$-unramified and $r\ge7$, the algebraic integer $(z-3)/r$ has house less than one. This would give $z=3$, which is impossible. In particular the $r$-free trace witness of Lemma~\ref{fam:Q:n:quart:twentyseven} cannot exist, so every exceptional label has codegree $243$. For $r\ge7$ the single-residue theorem gives an unsigned $r$-word of length three. Lemma~\ref{lem:short-tame-orbit} gives orbit length at least $(r-1)/3$. By Corollary~\ref{fam:Q:n:quart:orbitcount}, applied with $j=1$ and $k=2$, the length is strictly less than three. This excludes every $r\ge11$. At seven every such $A$ has full and tame inertia orbit size two. The word at any odd nonunit $Y\in P_7$ therefore consists of a single order-three orbit of roots; wild inertia fixes the word, and its orbits on the three terms are singletons, so all terms have order seven. Thus $z$ is one of the two quadratic periods, whose sum is $-1$. Let $A'$ be the conjugate under inversion at seven. Every label in $P_3$ is $7$-inactive: an exceptional row of codegree $243$ has degree prime to three, hence hits all of $P_3$ by principality and only $7$-inactive labels by its $7$-unit codegree. Thus $A,A'$ are $7$-inactive. The function $H=3(\chi_A+\chi_{A'})+\dim$ equals $7$ at the vacuum and vanishes on every odd nonunit $Y\in P_7$ since $\chi_A(Y)+\chi_{A'}(Y)=-d_Y/3$. Orthogonality on $P_7$ to any $7$-active row $J$ now precludes $d_J$ odd: at even $Y$ the factor $\chi_J(Y)$ would be even by lcm integrality, leaving the odd vacuum term. \end{proof} \begin{theorem}[A sextic dyadic and a quintic companion]\label{fam:Q:6.9}\label{fam:Q:n:quart:sextic} Let $2G+1$, such a counterexample cannot occur. \end{proposition} \begin{proof} In the positive table, pair an $r$-free nonunit of $P_q$ with a $q$-free nonunit of $P_r$. Such witnesses exist by prime avoidance, and the common weight $g$ is a power of two between $q$ and $2^k$ by Lemma~\ref{fam:M:4.10} and square divisibility. If either witness has dyadic height less than $k$, then $q\le g\le G<2q$ and $r>g+1$. The positive residues at $q,r$ and strict house force the normalized entry to be $g-q+q\xi$ with $\xi$ a nonidentity $r$-primary root by Theorem~\ref{fam:M:4.9}. This two-point word has three differences, and the difference-set argument (Lemma~\ref{fam:Q:n:lim:diff}) excludes it for $r\ge7$: its support is exposed since $3g+1$. Thus all complementary witnesses have maximal height $k$ and singleton positive two-blocks. Corollary~\ref{cor:complementary-singleton-spins} contradicts odd $b$. \end{proof} \subsection{A quintic valuation at three in the fermion extension}\label{fam:F:sec:3} \begin{proposition}\label{fam:F:3.1}\label{fam:F:n:lifted} A slightly degenerate minimal obstruction cannot have dimension $2^a3^5r^c$ for any prime $r>3$ and positive exponents $a,c$. \end{proposition} \begin{proof} Use the integral minimal extension $\mathcal M$ of $\mathcal B$ with fermion $g$; a missing repeated prime is covered by the two-prime reduction. Its only invertibles are $1,g$. For a noninvertible $X$, the constituents of $XX^*$ generate $\mathcal B$, whose centralizer is $\langle g\rangle$. Thus strict house and coprime vanishing hold for all full noninvertibles (Lemma~\ref{fam:M:4.4}). None can have prime-power degree $t^i$: for $s\in\{3,r\}\setminus\{t\}$ a $t$-free nonunit of positive $P_s$ exists by prime avoidance, and its full principal lift (Lemma~\ref{lem:odd-principal-lift}) would have a nonzero coprime crossing. We first exclude $U\in P_2(\mathcal M)$ of degree $2^i3$, $i>0$. Put \[ E=P_3(\mathcal M)\cap\{B:\theta_{B,3'}=1\}, \] where $3'$ denotes the entire prime-to-three factor. The only invertible in $E$ is the vacuum, since $\theta_g=-1$. Evaluation at $g$ makes every label of $P_3(\mathcal M)$ even. The projector onto $E$ commutes with the local ternary modular group, as do its two factors. Block orthogonality at unequal ternary phases and the odd-valuation mass identity hold on $E$ by Lemma~\ref{fam:M:3.2}, Proposition~\ref{fam:M:3.8} and the proof of Proposition~\ref{fam:M:4.2}. An odd-degree nonunit $B\in E$ has degree $3r^j$ with $j>0$, by the exclusion of prime powers and the principal height bound at valuation five. The number $z=S_{UB}/\lcm(d_U,d_B)$ has weight three, residue one above two by $U\in P_2$, and residue zero above three by $B\in P_3$. The weight-three argument of Lemma~\ref{fam:Q:6.1} gives $z=\lambda\sqrt{-3}$ with $\lambda$ a dyadic root of unity. To justify its use in the full extension, the cyclotomic calculation requires only the three level primes and strict house. Its remaining difference-set step uses the adjoints of $B$ and its Galois translate, evaluated at $U$. Both fusion labels are even, and tensoring by $g$ switches the sign of their twists, so neither $BB^*$ nor $B^\sigma(B^\sigma)^*$ contains $g$. The vacuum is their only possible invertible constituent. All noninvertible constituents satisfy the strictness just proved. These are the two uses of simplicity in Lemma~\ref{fam:Q:2.2}. Ternary inversion preserves $E$ by square covariance and negates these nonzero crossings. Proposition~\ref{fam:Q:6.2} on $E$ now excludes $U$, since $v_3(\dim\mathcal M)=5$. Every $r$-free noninvertible $U\in P_2(\mathcal M)$ consequently has $v_3(d_U)=2$. Indeed its degree has both primes $2,3$ and $d_U^2\mid\dim\mathcal M$. We need a full $r$-principal even label $V$ of odd degree with $v_3(d_V)=2$. Take a $2$-free positive nonunit from $P_r$ and lift it as above. If its degree were $3r^j$, compare it in the positive table with an $r$-free nonunit of $P_3$, necessarily of degree $3\cdot2^k$. The normalized crossing has weight $3$ and residue $3$ above both $3$ and $r$. Indeed if $R$ is the $P_3$ witness the entry is $\chi_R(V)/r^j$, whose denominator is a unit at three. The equality case of Theorem~\ref{fam:M:4.9} gives modulus three, contrary to strictness. Both $V$ and the $r$-free noninvertibles of $P_2(\mathcal M)$ have ternary defect one. Lemma~\ref{lem:quintic-common-class} supplies a common visible primitive phase $\eta$. In every even pair $J,gJ$ at least one representative has nontrivial dyadic twist. For it, orthogonality to the vacuum on the full $P_2$ yields an $r$-free noninvertible hit, since modulo $r$ the invertibles $1,g$ together contribute two. Its row sees only $1,\eta$ at three. The two representatives have the same ternary twist, so every positive label has ternary phase $1$ or $\eta$. This contradicts Lemma~\ref{fam:Q:6.3} in the positive table. \end{proof} \begin{proof}[Proof of Theorem~\ref{fam:Q:1.1}] If the quartic is below a larger repeated prime, Theorem~\ref{fam:M:7.3} applies. For a largest quartic and a least quintic use Theorem~\ref{fam:Q:1.3}. The remaining ordering has a middle quintic. In odd dimension Theorem~\ref{fam:Q:5.6} applies at exponent seven. For a dyadic least prime and $q\ge5$, use Theorem~\ref{fam:M:7.11} with $2^{\lfloor7/2\rfloor-1}=40$, satisfying \[ h(h-1)\ge(H-1)k,\qquad k=\min\{j\ge1:j(j+1)\ge h\}. \] Equality imposes two more restrictions. \begin{proposition}\label{por:pr:designrigid} At equality $h(h-1)=(H-1)k$, the following hold. \begin{enumerate} \item If $k\ge2$, then $h1$ dividing $H$ and every prime $\ell\nmid b$ for which $-1$ is a power of $\ell$ modulo $b$, $v_\ell(n)$ is even. \end{enumerate} \end{proposition} \begin{proof} Use the minimizing word and its $h$ opposite difference pairs $A$ from Lemma~\ref{fam:Q:2.3}. The tame multiplier group modulo sign is cyclic of order $H$, acts freely on the pairs, and satisfies $|A\cap uA|\ge k$ for $u\ne1$. Equality in the count forces every nonidentity overlap to equal $k$ and all pairs to lie in one free orbit: pairs from different orbits would contribute to no overlap. Identify the orbit with the group $G$. Then $A$ is a cyclic $(H,h,k)$ difference set. Suppose $h=k(k+1)$ and fix $u\ne1$. Let $B=A\cap uA$. The common word on $\{0\}\cup\{\pm b:b\in B\}$ has at least $h$ differences. The upper bound is exactly $k(k+1)=h$, with the possible differences given by $b,2b$ and $b\pm b'$ for different members of $B$. Thus all these pairs are distinct and are realized; write $A'$ for the resulting $h$-pair mask. It too is minimizing and lies in a single free orbit. Since it contains $B$, that orbit is the original one. Comparing the two words yields $c_g=|A'\cap gA|\ge k$ for every $g\in G$. The difference-set equations and double counting give \[ \sum_g(c_g-k)=h-k,\qquad \sum_g(c_g-k)^2=(h-k)^2 . \] Indeed $\sum_g c_g=h^2$ and $\sum_g c_g(c_g-1)=kh(h-1)$. All excesses are nonnegative, so one equals $h-k$. In particular $A'=gA$ for some $g$. Both $B$ and $2B$ lie in $A'$. Multiplication by two on the original free orbit is consequently one tame multiplier $v$: at one pair it agrees up to sign modulo its order with a tame unit, and every pair on this orbit has the same order. Here $v\ne1$ since $r>3$. Therefore $B\subseteq A'\cap v^{-1}A'$ and equality follows from the common cardinality $k$. Every intersection of two distinct translates of $A$ is consequently a translate of $A\cap v^{-1}A$. The $H$ translates of $A$ are the blocks of a symmetric design with $H$ points and $H$ blocks, and any two points lie in exactly $k$ blocks. Let $t$ be the stabilizer size of a block intersection and $s$ the number of blocks containing that intersection. All intersections form one orbit of length $H/t$. Two blocks containing the intersection meet exactly in it. Counting ordered pairs of distinct blocks gives $s(s-1)=t(H-1)$. But two points of the intersection give $s\le k$, while $H-1=h(h-1)/k>k(k-1)$. This proves the first assertion. For the second, choose a character $\chi$ of $G$ of order $b$ and put $\alpha=\sum_{a\in A}\chi(a)$. The difference-set equations give $\alpha\overline\alpha=n$. In $\mathbb Q(\zeta_b)$, $\ell$ is unramified and the decomposition group of each prime over $\ell$ contains the Frobenius $\ell$. Thus it contains complex conjugation. At that prime the valuations of $\alpha,\overline\alpha$ agree and the valuation of $n$ equals $v_\ell(n)$. \end{proof} \section{Two squared primes and their local columns}\label{fam:M:sec:8}\label{sec:four}\label{sec:two-squared} \begin{theorem}[Heights at the two higher primes]\label{fam:M:8.1}\label{thm:higher-two-squares}\label{main:four}\label{seq:four-five} Let $p0$. Moreover every $p$-free nonunit of $P_q$ is divisible by $uv$, and such a label exists. Every $q$-active label has degree divisible by $puv$, and conversely every $p$-active label has degree divisible by $quv$. \end{proposition} \begin{proof} A $t$-active column hits only $t$-free degrees. Reducing its orthogonality to the vacuum modulo $p$ or $q$, and excluding nonunit prime-power degrees, gives the two asserted degrees. A $p$-free nonunit of $P_q$ must hit both $p^i u$ and $p^{i'}v$, hence is divisible by $uv$. Orthogonality on $P_q$ of a $q$-active column modulo $p$ supplies such a label. If the column were also $t$-active, the required hit would have $t$-unit codegree, a contradiction. This separates $u,v$ from $p,q$. A column active at both $u,v$ has all nonunit hits divisible by $pq$, contrary to vacuum orthogonality modulo $p$. Complementary localization now gives codegree $t^2$ on the $t$-active labels; indeed its only prime is $t$ and its positive valuation is $2-2v_t(d)$. This gives even $A,B$ and degree $n/t$. Lemma~\ref{lem:isolated-dyadic-square} gives $t\ge5$. Compare a $q$-active column $Z$ with a $u$-active column on $P_q$. Their $q$-phases differ. Any nonunit common hit is $u$-free by defect zero and hence divisible by $p$, by the preceding description of $p$-free nonunits in $P_q$. If $p\nmid d_Z$, lcm divisibility makes every nonunit contribution vanish modulo $p$, contradicting the vacuum term one. Complementary localization gives $uv\mid d_Z$. Interchange $p,q$ for the other assertion. \end{proof} If three positions are squared, applying the proposition to each pair isolates all four primes and gives all three pair degrees. \begin{corollary}[Rank from two isolated primes] \label{fam:rank:two-isolated} In an ordinary minimal table of dimension $(p^h q^k uv)^2$ with four distinct primes and isolated singletons $u,v$, the rank satisfies \[ R\ge 7+(u-1)j_u+(v-1)j_v . \] If the dimension is odd, then $R\ge13+u j_u+v j_v$. Here $h,k\ge2$ and $j_u,j_v$ are the two vacuum deficits. For a bound depending also on the heights put $x=p^h$, $y=q^k$, $n=xyuv$, $c=q\lceil4/q\rceil$. Then \begin{equation}\label{mrk:eq:two-singletons} R\ge L+2-\max\{16/x^2+16/y^2,\ 16/(xy)^2+16/c^2\}, \end{equation} where \[ \begin{aligned} L={}&17-16/n^2+(u-1)(1-16/u^2)+(v-1)(1-16/v^2)\\ &+4-16(x^{-2}+y^{-2})(u^{-2}+v^{-2})+1-16/(uv)^2. \end{aligned} \] \end{corollary} \begin{proof} Put $n=p^h q^k uv$. By Proposition~\ref{second:two-columns} the $u$- and $v$-active sets have sizes at least $(u-1)j_u,(v-1)j_v$, and their degrees are $n/u,n/v$. The mixed inner product \eqref{eq:budget} gives a nonunit whose degree omits both $u$ and $v$. Proposition~\ref{prop:isolation} gives four further degrees with prime supports $\{p,u\},\{q,u\},\{p,v\},\{q,v\}$ and a $p$-free degree divisible by $uv$. These six distinct nonunit degrees and the unit lie outside the two active sets. In odd dimension, Theorem~\ref{main:full-multiplicity} supplies $uj_u,vj_v$ labels of the two large degrees. Each of the six other degrees occurs on a distinct dual pair by Lemma~\ref{lem:odd-codegrees-duality}. For the height-dependent estimate let $\delta_Z=1-16(d_Z/n)^2$. Each $\delta_Z$ is nonnegative by the small-codegree bound and Theorem~\ref{thm:no-nine}, and $R=16+\sum_Z \delta_Z$. The vacuum, the two active sectors, the four degrees supported on one higher prime and one singleton, and the common nonunit omitting $u,v$ contribute at least $L$ after adding the initial $16$. There are also a $p$-free and a $q$-free nonunit divisible by $uv$, by Proposition~\ref{prop:isolation}. If distinct, they contribute at least $2-16/x^2-16/y^2$. If equal, their degree is $uv$. A $q$-active label has degree divisible by $puv$, and its denominator is a multiple of $q$ by Theorem~\ref{thm:twistdiv}. Its denominator is at least four, hence at least $c$. Together these two labels contribute at least $2-16/(xy)^2-16/c^2$. \end{proof} \begin{lemma}[Two squared primes in the reduced table]\label{fam:M:8.4}\label{seq:two-odd-squares}\label{lem:fermion-pair} A slightly degenerate minimal counterexample cannot have dimension $2^a q^b u^2v^2$, with $q,u,v$ distinct odd primes, $a\ge1$, $b\ge2$. \end{lemma} \begin{proof} In the reduced table choose a $u$-active column by Proposition~\ref{lem:reduced-cauchy}. Its nonunit hits are $u$-free; orthogonality modulo two gives an odd one, of degree $q^iv$, $i>0$. Likewise $q^ju$ occurs. Lift both to the integral minimal extension with fermion $g$. A noninvertible $q$-free label of its principal two-block hits both lifts, hence has $uv\mid d_X$ by coprime vanishing. Take a $u$-active even label $Y$, choosing between $Y,gY$ to make its two-primary phase nontrivial. On the full principal two-block its vacuum orthogonality modulo $q$ has no noninvertible contribution: a $q$-free one has $u$-unit codegree. The two invertibles contribute one each, contradicting $q\nmid2$. \end{proof} For four surviving repeated primes, this excludes the reduced case with two specified squared source positions: a squared two would leave two-valuation one in the reduced dimension, and a square-free two leaves an odd reduced dimension (Lemma~\ref{lem:tables} and Proposition~\ref{three:odd-reduced}). \subsection{The two singleton columns}\label{fam:M:sec:8.1} For the rest of this section, a \emph{common} label is a nonunit whose degree omits both specified singletons. For four primes write $n=Cuv$, $C=p^h q^k$. We call the degrees $p^iw,q^jw$ of Proposition~\ref{prop:isolation} the \emph{axes at $w$}. \begin{lemma}[Inertia from principal residues]\label{fam:M:8.5}\label{fourm:one-residue} Let $z=S_{XY}/\operatorname{lcm}(d_X,d_Y)$ be a normalized entry on two nonunits, $g=\gcd(d_X,d_Y)$, $m$ the least dimension prime, and $b=m/(m-1)$. Suppose $z\equiv g$ at every prime above $t$, as occurs for $Y\in P_t$ and $t\nmid d_X$. If $e$ is the ramification index of $\mathbb Q(z)$ at $t$, then $t\le beg$. More generally $st\le beg$ when $z\equiv g$ also above a distinct prime $s$ unramified in $\mathbb Q(z)$. The $t$-inertia orbit of each label has length at least \[ L(t,g)=\begin{cases}(t-1)/\max\{e:e\mid t-1,\ e\le g\},&t>2g,\\ 1,&t\le2g.\end{cases} \] \end{lemma} \begin{proof} Take $s=1$ for the first assertion. The sum $w$ of the inertia translates of $z$ has $(w-eg)/(st)$ integral: $(z-g)/s$ is integral, and its inertia trace vanishes at every prime over $t$ in the unramified fixed field. If $st>eg$, strict house $|z|2g$ Theorem~\ref{three:word} gives an unsigned $t$-word of length $g$. Both character fields contain $z$; apply Lemma~\ref{lem:short-tame-orbit}. \end{proof} For a four-prime root with $h,k\ge1$, write $I$ for the nonunit labels of degrees prime to $uv$. Each such degree is $d=p^iq^j$, $i,j>0$, and its field is unramified at one of $u,v$ by Proposition~\ref{deep:cluster}. At that endpoint $t$, the opposite axes are $t$-principal, and their gcds with $d$ are at most $p^i,q^j$. Lemma~\ref{fourm:one-residue} yields the first bound below. The unramified estimate in Lemma~\ref{lem:singleton-energy} gives the second: \begin{equation}\label{eq:firstestimate} t\le\frac{m}{m-1}\min(p^i,q^j),\qquad (t-1)j_t[d]_t^21+C^{-1}, \] no such table exists. For a center and $C\equiv1\pmod{rs}$ the weaker hypothesis $H\ge1$ suffices. In particular the obstruction applies whenever $A=B=2$ and $C\equiv\pm1\pmod{rs}$. \end{proposition} \begin{proof} Put $m=sj_r+rj_s$ and let $\rho$ be the least positive residue of $\epsilon C^{-1}$ modulo $rs$. By the two-column identities, $m\equiv-\rho\pmod{rs}$ and $11$ forces $\rho=1$ and the right side is strictly below $2rs(1+C^{-1})$ (below $2rs$ for a center, where $\epsilon=1$). Use $r^2+s^2\ge2rs$. For $A=B=2$, $H\ge(1-1/4)^2+(1-1/9)^2>1+1/36\ge1+1/C$. \end{proof} \begin{proposition}[Bounds for the other three primes]\label{fam:M:8.7}\label{note:finite} Under the hypotheses and notation of the two-column proposition, let $ubu$ its field would ramify at $v$ and $q$, contrary to isolation. This gives the last bound by symmetry. If $u=m$, every other prime is at least $u+2>bu$, impossible. A prescribed repeated prime with bounded exponent bounds $L$, then $u,v$ and the other prime by \eqref{eq:finite}. A prescribed $v$ bounds $u$ and both other primes. For fixed $u$, either $v\le2u$ or both repeated primes are at most $2u$; bounded exponents then bound $L$ and $v$. \end{proof} \begin{lemma}[Multiplicity at an isolated odd singleton]\label{fam:M:8.8}\label{seq:no-odd-singleton-five} Suppose an ordinary minimal table has odd root $n$ and an isolated prime $t\ge5$ of root-valuation one. If $H$ distinct proper nonunit $t$-free degrees occur, then \[ H\min\{2j_t^2,\,2^{v_2(t-1)}\}\ \le\ B_t=j_t(t+1-j_t). \] In particular, no isolated singleton of an odd minimal square root can be five. \end{lemma} \begin{proof} An unramified label and its distinct dual cost at least $2j_t^2$ by Lemma~\ref{lem:local-integer-column}. A ramified label has a pure $t$-power field by isolation and Proposition~\ref{second:two-columns}. Its field is nonreal since the dual is distinct. Its stabilizer in the cyclic cyclotomic Galois group omits the unique involution, so its orbit size is a multiple of $2^{v_2(t-1)}$. Every translate costs at least one in squared norm. Sum these bounds on the $H$ proper degrees. At $t=5$ Lemma~\ref{new:trace} gives $H\ge3$. Both even deficits $j_t=2,4$ would require twelve units, exceeding $B_t=8$. \end{proof} \begin{lemma}[A unique complementary degree]\label{fam:M:8.9}\label{seq:five-larger}\label{new:only-common-pair} Let $u,v\ge5$ be distinct isolated odd singletons in an ordinary minimal table of square root $n$. Suppose $D=n/(uv)$ is its only nonunit degree prime to $uv$, and the signed coordinates of these labels are centered residues at $u,v$. If $u1$. At most one endpoint ramifies at each label. At an unramified endpoint its deficit divides its signed coordinate; both $j_t\mid A$ and $j_w\mid B$ would force $M\le1$. Thus one endpoint, say $w$, is unramified at every common label. A $t$-active row $Z$ and a $w$-active row now give $1+[D]_w\sum_{d_X=D}\chi_Z(X)=0$. Integrality gives $[D]_w=\pm1$, hence $|B|=j_w$ and $j_t=M|A|$. Reducing at $t$ gives $MD\equiv\pm1\pmod t$. Since $M\mid j_t$ and $j_t\mid t-1$ or $t+1$ by Lemma~\ref{lem:singleton-energy}, $M\mid t\pm1$ and the centered residue of $D$ has magnitude at least $(t-1)/M$. If $|A|\ge2$, the active rows at $t$ charge at least \[ |A|(t-1)\bigl(t-(t-1)/M\bigr)/t^2\ \ge\ 1-1/t^2 \] to the $D$-column norm. At $w$, $|[D]_w|=1$ charges at least $j_w(w-1)/w^2$. This is greater than $1/t^2$ since $tj_w=|1-wA|\ge2w-1$. The two sets of active rows are disjoint. Thus $|A|=1$, $M=j_t$, and $w=tj_w\pm1$. Parity gives $j_w\ge2$ even and $w>t$. For odd $n$, a common label and its distinct dual share their signed coordinate by rationality; its square is at most $B_t/2$, which forces centering at each endpoint. \end{proof} \begin{corollary}[Three omitted primes]\label{fam:M:8.10}\label{lem:singleton-triple} Let $u2v&\Longrightarrow& \mathcal T_{u,U}(q,v,\{p^iv:1\le i\le h\}),\\ v>2p^h&\Longrightarrow& \mathcal T_{u,U}(v,p^h,\{p^iq^j:1\le i\le h,\ 1\le j\le k,\ u\le2\min(p^i,q^j)\}). \end{array} \] In particular $u\le2p^h$, $v\le p^h U+1$, $q\le vU+1$, and $q\le2u$ when $v>2u$. \end{lemma} \begin{proof} Both tests are the signed dimension trace at $u$, reduced at the omitted prime $P$. Only its $P$-free nonunit terms survive. For $P=q>2v$ these are the axes $p^iv$: comparison with a $q$-principal $p$-free label of degree $q^juv$ has gcd $v$. For $P=v>2p^h$ these are the common degrees; comparison with a $v$-free $p$-axis has gcd at most $p^h$. In either case positive multiplicity forces $P$-ramification by Lemma~\ref{fourm:one-residue}, hence $u$-unramifiedness by isolation. Thus $a_u=-\epsilon j_u[d]_u$, and in the second case $u\le2\min(p^i,q^j)$ by \eqref{eq:firstestimate}. Cancel $j_u6$ or $h=k=3$. \end{theorem} \begin{proof} Order $u2u$. For $t=u,v$ use $1\le j_t(t-1)j$ if $jm>t+1$, and $F=t$ if $j=1$ and $m>t+1$. \end{lemma} \begin{proof} For a proper $t$-free nonunit degree $d$ put $I=m/d$ and $A=[I^{-1}]_t$. Since $I>1$ and $IA\equiv1\pmod t$, positive $IA$ is at least $t+1$, and negative $IA$ is at most $1-t$. Hence \[ -A^2/(t-1)\le A/I\le A^2/(t+1). \] Removing the $F$ full labels leaves trace $(t+\epsilon/m)j-F$ and norm $2tj-j^2-F$ by Proposition~\ref{second:two-columns}. Summing the two bounds over the remaining labels gives the interval. \end{proof} \begin{theorem}[The two ramified orbits at an isolated singleton]\label{fam:M:9.2}\label{nov:bruhat}\label{nov:three-deficit}\label{fourm:twoorbits}\label{main:full-multiplicity} Let $t$ be an isolated singleton in an odd ordinary minimal root $n$, and write $m=n/t$, $j=j_t$. Exactly $tj$ labels have degree $m$. The inactive $t$-ramified labels form two orbits of size $j$, one of degree $m$ and one of a proper degree $d_*5$ by $R\le2j$, and $t=5$ is already excluded by Lemma~\ref{seq:no-odd-singleton-five}. \emph{Parity in the vacuum column.} The vector $x_0-(L/e)v_0$ and its inverse Weyl image vanish on active coordinates, so it is locally invariant. Its vacuum coordinate gives $ea_t(X)=-\epsilon jL$. Also $\sum c_{ia}^2=e-1+jL^2/e\equiv -jL\equiv0\pmod2$. Hence $jL^2/e$ is an odd integer. Since $|jL^2/e|=|L a_t(X)|$, both factors are odd and $\nu_2(j)=\nu_2(e)=\nu_2(t-1)$. Each active signed coordinate is one by Proposition~\ref{second:two-columns}; the $(t-1)j$ active labels thus leave exactly $j(t+1-j)$ in the inactive squared norm. Dividing by $j$, unramified terms $j[d_X]_t^2$ are even and each ramified orbit contributes the odd integer $ea_t(X)^2/j=jL^2/e$. Since $t+1-j$ is even, the number of ramified orbits is zero or two. In the latter case $R\le2j$ and $e\ge j$ force $e=j$ on both. Equality in the row minima forces $r=1$ or $e-1$ and one coefficient in each row to differ from the others by $\pm1$. The periods sum to $-1$, so the evaluations are integers plus signed periods, with trace $L=-\epsilon a_t(X)$ congruent to $\pm1\bmod j$. \emph{The full and proper degrees.} Lemma~\ref{new:trace} supplies at least three distinct proper $t$-free nonunit degrees. None is active and at most two can ramify, by the orbit count above. Choose a label $X$ of an unramified one. Its normalized diagonal entry is unramified at $t$ and has positive residue $d_X$ there because $X\in P_t$. Lemma~\ref{fourm:one-residue} gives $t\le b d_X\le bm/3\le m/2$, using the odd index $m/d_X>1$. Thus $jm>t+1$. Lemma~\ref{main:singleton-trace-interval} supplies a full label beyond the $(t-1)j$ active ones. An inactive full label has codegree $t^2$ and nonreal character field unramified away from $t$, so it ramifies at $t$. The orbit count therefore gives two inactive ramified orbits of size $j$. If both were full, the lower trace bound would give $j(1-\epsilon/m)\le j(t-1-j)/(t-1)$, again impossible. Thus exactly one is full. Subtracting the $tj$ full labels from the two identities and using $a_t(d)=-\epsilon j[d]_t$ on unramified dual pairs gives the displayed residual identities. Their norm and the oddness of $(t-1)/j$ give the congruence and the bound on $k_t$. Since $k_t\ge3$, the odd index $(t-1)/j$ is at least five. For distinct proper degrees $x,y$, at least one of the labels is unramified at $t$. Their normalized entry has the positive principal residue $\gcd(x,y)$ at $t$; Lemma~\ref{fourm:one-residue} gives the bound. All labels of degree $d_*$ belong to the ramified orbit: their common coordinate is odd while an unramified coordinate is divisible by the even $j$. Finally $K_Y$ is pure at $t$; if $d_X\notin\{d_*,m\}$ its character field is unramified at $t$ (by defect zero when $t\mid d_X$). The two fields intersect in $\mathbb Q$, giving the rational entry. \end{proof} \begin{proposition}[Two residues at an isolated prime]\label{fam:M:9.3}\label{prop:singleton-two-residues} In an odd ordinary minimal table let $s$ be an isolated singleton. Put $b=m/(m-1)$ for the least dimension prime $m$. Suppose $X,Y$ are $s$-free, $s$-inactive nonunits whose normalized entry $z$ has positive residue $g=\gcd(d_X,d_Y)$ at every prime over a second prime $a\ne s$. Then \[ \min(a,s)bu$, since $j_v\ge2$, $u\ge5$ and $b\le3/2$. Some common degree therefore has a prime-power factor at most $J$. Under divisor descent, an odd reduced table is excluded by Proposition~\ref{three:odd-reduced}. Loss of a repeated prime leaves the three-prime small-exponent case, and a higher root height below two is covered by Proposition~\ref{seq:one-unrestricted}. Otherwise the singletons persist and $J$ only decreases. \end{proof} \subsection{Isolation of a quartic prime}\label{fam:M:sec:9.3} \begin{theorem}[Full isolation at four odd primes]\label{fam:M:9.14}\label{fourm:isoheight}\label{fourm:isoquart} Let $n=p^h q^k uv$ be an odd ordinary minimal root, $h,k\ge1$. If all four primes are isolated then $uv$ is absent and neither $h$ nor $k$ equals two. \end{theorem} \begin{proof} Proposition~\ref{seq:one-unrestricted} excludes height one. With $h,k\ge2$, full isolation and the occurrence of $uv$ would make $p^h q^k$ the sole common degree by Proposition~\ref{seq:sextic-fully-separated}, contrary to Proposition~\ref{main:gap-all-heights}. Either height equal to two forces $uv$ by the same isolation proposition. \end{proof} \subsection{The two higher prime powers in increasing order}\label{fam:M:sec:9.4} Close singletons rule out the full common degree, independently of the two higher exponents. This elementary estimate has two applications: it treats a quartic valuation at the least prime with arbitrary companion height, and later it applies to the odd sextic root. \begin{theorem}[The full common degree is absent]\label{fam:M:9.15}\label{new:no-full-common} Let $n=Cuv$ be an odd ordinary minimal root with $C$ the product of two prime powers of exponents at least two. Let $p$ be the least of the four primes and $b=p/(p-1)$. If $u\le bp^2$ and $up-2$. Oddness gives $|a_t|=p$. Hence $t>p^2$, $\delta=p+1$, and $t=p(p+1)\pm1$: here $t-1\le p\delta<2t-1$. If $t=v$, then $u=p^2$ or $p^2-2$, contrary to primality and $u\ge p^2-1$. The choice $t=u=p^2+p-1$ gives composite $v=p(p+2)$. The remaining inverse pair is \[ (u,v)=(p^2+p+1,(p+1)^2+1),\qquad (a_u,a_v)=(-p,p+1). \] The $C$-orbit is $v$-unramified, so $j_v\mid p+1$. Since $v-1=(p+1)^2$, the two-orbit theorem excludes all inactive $v$-ramification, contrary to the compulsory inactive full orbit at $v$ in Theorem~\ref{main:full-multiplicity}. \end{proof} \begin{theorem}[A quartic exponent at the smaller higher prime]\label{fam:M:9.16} \label{seq:four-high-common}\label{leastquart}\label{fourm:leastquart-smallboth} Let $qbq$, write $x=q^2p^i$. For $ibq$, hence ramifies there. Every proper common label is therefore unramified at both singletons. The mixed Gram identity has a ramified common label by Corollary~\ref{r6:twopart}, so $C$ occurs. Proposition~\ref{main:gap-all-heights} gives $u\le v\le bq^2$ and $v\le bu$, so Theorem~\ref{new:no-full-common} excludes $C$. \end{proof} \begin{proposition}[An unrestricted power at the least odd prime]\label{fam:M:9.17} \label{r6:least-unrestricted} Let $n=p^h q^2uv$ be an odd ordinary minimal root, for distinct primes with $p$ least, $ub^2P$. By \eqref{eq:firstestimate} every common degree is $L=p^h q$ or $C=p^h q^2$. Proposition~\ref{main:gap-all-heights} forces $L$ to occur. It is $q$-principal since the $p$-valuation is full and the vacuum denominator has a $q$-part. Hence $uv$ is absent. By \eqref{eq:firstestimate}, $q\ge u/b>bP$. A $q$-free nonunit in $P_p$ has degree $p^iuv$, $1\le i1$ at every $L$-label. They are therefore unramified at both singletons. At each $t$ their comparison with both opposite axes gives $t\le bq$ and full $p$-axis height since $t>bP$. The mixed column from Theorem~\ref{fourm:isoquart} has, by Proposition~\ref{fourm:mixed-degrees}, a $t$-free hit in $P_q$ with deficient heights. Both common degrees and all $p$-axes have full $p$-height, so the hit has degree $Y=p^l q w$, $1\le lb^2P/2$ (also for $p=3$). The bounds increase with $h$; smaller roots with $h'\le2$ or $k'\le1$ descend by Theorem~\ref{fourm:leastquart-smallboth} or Proposition~\ref{seq:one-unrestricted}. \end{proof} \begin{corollary}[Reduction to an odd sextic root]\label{fam:M:9.18}\label{cor:four-before-sextic} Let $pbp$. The bound in Proposition~\ref{main:gap-all-heights} therefore restricts common degrees to $A,B,L,C$. A common label with full $p$-valuation and deficient $q$-valuation is $q$-principal, and symmetrically with $p,q$ exchanged, by Proposition~\ref{deep:cluster} for the cluster $\{p,q\}$. Proposition~\ref{prop:isolation} supplies a $p$-free nonunit in $P_q$ whose degree contains $uv$; since its $q$-height is deficient, that degree is $quv$ whenever $uv$ is absent. Theorem~\ref{fourm:isoquart} supplies a mixed $p,q$-active row $Z$. We use one trace test throughout. Let $T$ be active at $t\in\{u,v\}$ and let $w=uv/t$. Suppose all proper common hits of $Z,T$, on the whole table or on $P_q$, have degree $d$. Then \[ [d]_t=\pm1\quad\text{if unramified at }t,\qquad a_t(d)=\pm1\quad\text{if ramified at }t. \] Indeed ramification status is constant on a proper degree by Corollary~\ref{deep:ram-parity}. In the first case $\chi_T(X)=[d]_t$ on every hit. In the second, $K_X$ is pure at $t$ and $K_Z$ is unramified there, as purity would contradict the square Galois action on its nontrivial $q$-twist ($q>3$). Thus $\chi_Z(X)$ is integral rational and constant on the orbit. The orbit sum of $\chi_T(X)$ is $-\epsilon a_t(d)$ by Theorem~\ref{fourm:twoorbits}, and the principal block is Galois stable. In both cases orthogonality makes the indicated integer divide the vacuum term one. By Proposition~\ref{main:gap-all-heights}, $v\le bu$ and at least one of $A,B,L$ occurs. \begin{lemma}[Axis heights forced by a ramified common degree]\label{fam:M:12.2} \label{fourm:ram-B-axes} If a label of degree $B$ or $L$ ramifies at a singleton $t$, every $p$-axis has full height. For $L$ the $q$-axes are full as well, no degree $pqs$, $s=u,v$, occurs, and $qw\le bp^3$, $w=uv/t$. \end{lemma} \begin{proof} The gap bound gives $u\ge(p-1)\delta+1\ge2p-1>bp$; also $q>bp$. Write $D$ for the ramified label. It is principal at $a=p$ for $D=B$ and at $a=q$ for $D=L$. For a label $X$ of degree $ds$, $s=t,w$, consider its normalized pairing $z$ with $D$. For the axis and witness degrees considered below, $ds$ differs from both $d_D$ and $Cw=n/t$. Thus $z$ is rational by Theorem~\ref{main:full-multiplicity}. Now let $s'=uv/s$. If $D=B$ and $d=p^i$, $i\le2$, then $p\mid z$ by $p$-principality and $z/p\equiv p^{i-1}\pmod{s'}$, impossible since $s'>bp$ by the strict house and balancing mean. If $D=L$ and $d=p^i$ ($i=1,2$), both $q$ and $s'$ divide $p^i-z$, giving $qs'\le bp^i$. For $d=p^iq$ ($i=0,1$), $q\mid z$ and $z/q\equiv p^i\pmod{s'}$ give $s'\le bp^i$. Again $q,s'>bp$ exclude these possibilities. In particular the rational pairing of $L$ with the compulsory $p^3t$ gives $qw\le bp^3$. \end{proof} \begin{corollary}[The least prime three]\label{fam:M:12.3}\label{r6:sextic-three} An odd ordinary minimal root $27q^2uv$ is impossible. \end{corollary} \begin{proof} Propositions~\ref{r6:least-unrestricted} and~\ref{main:gap-all-heights} give $u\le20$, $ubp^2$. The only possible common degrees are $L$ and $C$ by \eqref{eq:firstestimate}, and $L$ occurs by Proposition~\ref{main:gap-all-heights}. Comparison of $L$ at an unramified endpoint gives $q\ge u/b>p^2$. An $L$-label ramified at $t$ would give $qw\le bp^3$ by Lemma~\ref{fourm:ram-B-axes}, contrary to $q>p^2$ and $w>bp$. Thus $L$ is unramified at both singletons; then $v\le bq$ by the $q$-axis comparison. The mixed Gram identity and even multiplicity exclude $L$ alone, so $C$ occurs. The centered inverse coordinates $A_u=[v^{-1}]_u$ and $A_v=[u^{-1}]_v$ on $C$ satisfy $vA_u+uA_v=1$. Hence $C$ is ramified at the odd coefficient endpoint $t$, and unramified at the even endpoint $w$. If $q\le bp^2$, its orbit cost gives $A_t^2\le t+1-j_t\le t-1$, and $\delta A_t\equiv\pm1\pmod t$ with $\delta$ even implies $\delta\ge\sqrt{t-1}>p$. Yet $q\le p^2+p+1$ and $q\le bp^2bp^2$. The comparison of $L$ with the $p$-axes forces all of them to have full height since $t>b p^2$. On $P_q$ a mixed column, reduced at $t$, therefore needs a hit $Y_l=p^lqs$ with $1\le l\le2$, $s=uv/t$. Comparing it at $q$ with $p^3t$ forces both to be $q$-ramified, hence unramified at $t$. The even quantity in \eqref{eq:four-product} for $L,Y_l$ is congruent to $p^l$. As $t>p^2$, this requires $p^l>3t/4$, forcing $l=2$ and $v<4p^2/3$. Now $\delta\le u/(p-1)<2p$ so $p\nmid\delta$. At the even endpoint $w$, the identity $pY_2=sL$ shows that $p[Y_2]_w\pm\delta[L]_w$ is a multiple of $w$. It is nonzero because $0<|[L]_w|<\sqrt w/2p^2\ge25$. This excludes the second alternative. \end{proof} \begin{lemma}[The odd sextic root]\label{fam:M:12.9}\label{fourm:common-fields} No ordinary minimal table has root $p^3q^2uv$ with $p\ge5$ least. \end{lemma} \begin{proof} The possible common degrees are exhausted by the following cases. The principal memberships listed in the middle column are the ones forced by the complementary full height in the $p,q$ cluster. \[ \begin{array}{c|c|l} \text{degree}&\text{membership}&\text{use in the elimination}\\ \hline C=p^3q^2 & \text{full at both} & \text{excluded by the gap bound}\\ A=p^2q &\text{neither height full}&\text{cannot occur alone; forces }B\\ B=p^2q^2&P_p & \text{cannot occur alone; must supply the ramified orbit}\\ L=p^3q&P_q&\text{if }A\text{ occurs, unramified at both ends} \end{array} \] \emph{Removing the full degree.} These bounds give $\delta\le u/(p-1)\le p^3/(p-1)^2 \max\left\{\frac{\sqrt s}2,\frac{s+1}{\delta+3}\right\}. \end{equation} The first inequality follows from parity. For the square-root bound $(s-p)/\sqrt s$ increases with $s$. At $s_0=(p-1)\delta+1\le p^2$ one has $2(s_0-p)/\delta=2(p-1)(1-1/\delta)$, greater than $p$ for $\delta\ge4$ and equal to $p-1>\sqrt{2p-1}$ for $\delta=2$. The second bound says $3s>(p+1)\delta+3p$. Using $s\ge s_0$ gives this for $\delta\ge4$; at $\delta=2$ equality could occur only for $p=5,s=9$, which is composite. \emph{The sole-$A$ case.} Put $F=C/A=pq$. Suppose $A$ occurs alone. Use the two endpoints and signed coordinates from Lemma~\ref{fourm:sole}. The coordinate congruence $a_t\equiv A/(Cw)=1/(Fw)\pmod t$ now gives $F\equiv\pm\delta^{-1}\pmod t$. Since $A=pF$ we obtain $p j_t\equiv\pm\delta\pmod t$. At $w$ one has $C=A^2/p\equiv1/p$ by $[A]_w=\pm1$. The vacuum congruence $j_wCt\equiv\pm1\pmod w$ therefore gives $\delta j_w\equiv\pm p\pmod w$. Since $pj_t\ge2p>\delta$, the first congruence and parity give $pj_t\ge2t-\delta>p\delta$, so $j_t\ge\delta+2$. At $w$, \eqref{fourm:residue-distance} gives $j_w(\delta+3)>w+1$, contrary to $j_w(1+j_t)\le w+1$. Finally, if $uv$ occurred, full-height principality would exclude both $p^2q^2$ and $p^3q$ and force $A$. The exclusion of the full common degree $C$ would then make $A$ the sole common degree, as just ruled out. \emph{When $A$ occurs.} We first record a restriction common to $D=B,L$. If an $A$- or $D$-label ramifies at a singleton $t$, with partner $w$, then its normalized pairing $z=S_{A,D}/D$ with any label of the other degree is rational. Indeed the degrees are distinct, neither is full at $t$, and the rational-entry assertion of Theorem~\ref{main:full-multiplicity} applies. For $D=L$ a rational pairing is itself impossible. The integer $z$ is divisible by $q$ since $L\in P_q$ and $\chi_L(A)=z$. Both labels are principal and units at $u,v$, so $z/q\equiv p^2\pmod{uv}$. The house and balancing mean give $-p^2/(p-1)\le z/qb^2p$, so $q\ge uv/(bp)>bv$. Finally $B\in P_p$ excludes the coprime degree $uv$; the complementary $p$-free witness in $P_q$ has degree $quv$. Comparing it at $q$ with $p^is$ for $s=u,v$ gives $q$-inertia at least $q/(bs)>1$ on the corner. Isolation of either singleton then excludes ramification there. For the rest of the proof we will repeatedly use two consequences at an endpoint $s$, with $s'=uv/s$. Once the $p$-axes are full, all proper common hits of $Z$ and an $s$-active row on $P_q$ have degrees \[ A,\qquad Y_1=pqs',\qquad Y_2=p^2qs'. \] Indeed the two heights are deficient, $p$-free nonunits carry $uv$, and a hit omits $s$. A $Y_1$-hit is $s$-unramified: its comparison in $P_q$ with $p^3s$ has gcd $p$, and $q>bp$ forces $q$-ramification. Also if a common label of degree $x$ is $s$-unramified, an $s$-unramified $Y_l$ is excluded whenever $xY_l/(Cs')$ is an odd integer at most $3s/4$, by \eqref{eq:four-product}. For $x=A$ these quotients are $1,p$; for $x=L$ they are $p,p^2$. Throughout $p<3s/4$. \emph{The sole-$B$ case.} Under the sole-degree hypothesis, Lemma~\ref{fourm:sole} provides endpoints $t,w$ with $a_t(B)=\pm1$, $[B]_w=\pm1$, $j_t B\equiv\pm1\pmod t$ and $j_w(1+j_t)\le w+1$. As $C=pB$, the first congruence gives $pw\equiv\pm1\pmod t$. All $p$-axes have full height by Lemma~\ref{fourm:ram-B-axes}. Since $B$ is the sole common degree, the proper $P_q$-hits at $t$ are $Y_1$ (unramified) or $Y_2$. If a $Y_2$-hit were unramified at $t$, its dual-pair budget would give $0(p-1)\delta$, and $p\ge5$. Their sum and absolute difference are even and less than $2t$, forcing $j_t=\delta^2$. Finally $p\delta\equiv\pm1\pmod t$ and $0bv$ by the two cases for $A$. All $p$-axes are unramified at both singletons and have full height. Indeed if $B$ ramifies at either singleton this follows from Lemma~\ref{fourm:ram-B-axes}; otherwise apply \eqref{eq:four-product} to $B,p^i r$ at $s=uv/r$ for $r=u,v$. At $i=1,2$ the quotient $p^{i-1}\le p<3s/4$ is forbidden. We claim that $A$ is unramified at both endpoints. Suppose instead it ramifies at $t$. All $A$-labels are then pure at $t$, unramified at $w=uv/t$, and $L$ is absent by the $A,L$ comparison above. None can belong to $P_q$: its rational normalized pairing with $p^3t$ would give two positive residues $p^2$ at $q,w$, hence $qw\le bp^2$, contrary to $q>bv$. The $P_q$-witnesses are now $Y_1,Y_2$ at $w$. Since $A$ is unramified there, the quotient test excludes all unramified $Y_l$, including every $Y_1$. Only ramified $Y_2$ remain. Their trace criterion gives $a_w(Y_2)=\pm1$, or $pq\equiv\pm1\pmod w$. Hence $|[A]_w|\ge p-1$ as $w\ge2p-1$. But its unramified dual-pair budget gives $|[A]_w|<\sqrt w/2\le bp/2t+1$ by \eqref{fourm:residue-distance}. The $p^3w$-axis has at least $\delta$ labels, all $t$-unramified. By the $q$-principal comparison with $quv$ its multiplicity is at least $(q-1)/w$ for $q>2w$; for $q\le2w$ this bound follows from duality. Thus its even multiplicity is at least $t/(bp)-1/w>\delta-2$: use $t>(p-1)\delta$ if $\delta\le p-1$, and $t\ge p^2+2$ by primality if $\delta=p+1$. The vacuum, this axis and an $A$ dual pair give $j_t(\delta+3)\le t+1$ in the unramified budget, a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:higher-two-squares}] The root heights $i,j$ in Corollary~\ref{cor:four-before-sextic} satisfy $i\le h$, $j\le k$. Its even case uses the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/dyadic_total_height.py}{dyadic enumeration} of Theorem~\ref{fourm:even-total}. The corollary gives each asserted bound except possibly $h+k\ge6$ at an odd root $p^3q^2uv$. Such a root is excluded by Corollary~\ref{r6:sextic-three} at $p=3$ and by Lemma~\ref{fourm:common-fields} at $p\ge5$. \end{proof} \section{Four repeated primes with an unrestricted exponent}\label{fam:W:sec:5}\label{fam:W:n:sec:four-next} \begin{theorem}\label{fam:W:1.2}\label{fam:W:n:thm:unbounded} Let $P,Q,u,v$ be distinct primes with $Q$ greater than at least one of $P,u,v$. For every nonnegative integer $A$, the integer $P^5Q^A u^3v^2$ is covered. \end{theorem} Write a four-prime ordinary minimal root as $P^h Q^k uv$, with $uv+1$, the full-multiplicity bound of Lemma~\ref{main:singleton-trace-interval} reserves a full label beyond the $(t-1)j_t$ full labels already reserved. Its signed coordinate is one. The residual squared budget is $j_t(t+1-j_t)-1$ and the degree trace is $(j_t-1)m_t+\epsilon j_t$. Further full labels remain allowed. Minimize each axis cost at its trace residue and impose the common norm and mixed constraints of \eqref{fourm:budgettest} with these residual budgets. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/unbounded_four.py}{recurrence} gives 279 vacuum residue rows, 53 after both axis budgets, and two after the common Gram sum. Both rows have $q=3$, $u=5$, $v=7$, $j_u=1$, $\epsilon=-1$. Their $(j_v,3^k\bmod35)$ values are $(2,17)$ and $(1,27)$. One calculation rules out both rows. At five the remaining squared norm is four and the degree trace is $-1$. Both axis types and a common degree must still occur, so there are exactly four remaining labels, each of signed coordinate $[d_X]_5=\pm1$. A $3$-free one can only have degree $14$: the alternative $28$ has $|[28]_5|=2$. Its signed trace contribution is $-14$. At least one such axis is compulsory, and the trace modulo three requires a second one. There are exactly two since the other axis and a common degree are also compulsory. The other axis has degree $7\cdot3^b$ with $b$ odd because $|[7\cdot3^b]_5|=1$. Every inactive label is unramified at this unit-deficit singleton. A common degree $2^i3^a$ must therefore satisfy $5\le2\cdot 2^i$ by the opposite axis, hence $i=2$. Its residue has magnitude one precisely for even $a\ge2$. The trace would give \[ \pm4\cdot3^a\ \pm7\cdot3^b=27. \] As the exponents have opposite parity, the $3$-valuation gives $\min(a,b)=3$. Hence $b=3$, $a\ge4$, and $4\cdot3^{a-3}$ would have to be $6$ or $8$. This is impossible. For any specified dyadic root height $H\ge2$, replace four by $2^H$ in the cap and window and allow all powers $2^i$ with $1\le i\le H$. Again $2^Hq^ku>v+1$ forces the extra full label. The algorithm thus tests all opposite heights at once for each input $H$. If $QB,\ k\bmod d\in R\}$, with $k_0=\infty$ if $R$ is empty. Then $P^A Q^{2k_0-1}u^3v^2e$ and all its divisors are covered, with $\infty$ denoting an unrestricted cap. On the two-square box the higher valuation of a divisor-minimal obstruction is exactly $2k$. Theorem~\ref{fam:M:10.3} then lifts either singleton to a cube. In particular one can start with $B=0$ and arbitrary $A$. At one endpoint put $G_t=\mathbb F_t^\times/\langle-1,P\rangle$. If the odd part of $t-1$ is a prime $\ell\ge5$, then $H_t=\ell$. Thus $\bar Q^k=\bar\ell/\bar w$ in $G_t$, which has no solutions or one class $k=c\pmod d$, where $d=\ord_{G_t}(\bar Q)$. In the latter case every $k_0\ge2$ in that class gives the lift \[ B(A,2k_0,2,2)\ \Longrightarrow\ B(A,2k_0+2d-1,3,2) \] for the same prime assignments. For example, let $u=113$, $v\equiv14$, $P\equiv1$, and $Q\equiv3$ modulo $113$. The only allowed $H_u$ is $7$. The element $3$ has order $112$, with $3^{44}=56$ and $3^{100}=57$ modulo $113$. Thus $14\cdot3^k\equiv\pm7$ is equivalent to $k\equiv44\pmod {56}$. Consequently the box $(A,87,3,2)$ is covered for every $A$, and for every $b\ge0$ one obtains the transfer \[ (A,88+112b,2,2)\ \Longrightarrow\ (A,199+112b,3,2). \] The primes $(P,Q,u,v)=(227,3,113,127)$ realize this assignment, with $Q$ least and $127/113<3/2$. The two-endpoint test can be sharper. If $u=449$, $v=463$ and $P,Q$ have the residues of $23,13$ modulo $uv$, the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/residue_classes.py}{calculation in the finite quotient} gives $d=168$ and $R=\{42,46,126,130\}$, proving $(A,83,3,2)$ for arbitrary $A$. If $Q$ instead has the residues of $67$, then $R$ is empty. Already at $449$ the only allowed $H$ is $7$ and $23^{112},67^{112},7^{112},463^{112}$ are $1,1,1,-1$ respectively. Also $(-1)^{112}=1$. Thus the two higher exponents are both unrestricted in this family. The set $R$ can be computed from discrete logarithms at the selected endpoints and elementary linear congruences. This eliminates $h$ by the generalized Chinese remainder theorem and never enumerates source dimensions or any finite range of the two heights. \paragraph{Arbitrarily prescribed heights.} Fix an odd prime $Q$, an integer $L\ge2$ and a distinct odd prime $t$. Fix the residue of $P$ modulo $t$, of multiplicative order $m$. For all but at most $2m(L-1)\tau(t-1)$ nonzero residues of the other singleton $w$ modulo $t$, the box $(A,2L+1,3,2)$ is covered for every $A$, for distinct odd primes $P,Q,t,w$. Here $\tau$ counts positive divisors and the cube may be put at either singleton. Indeed, for $2\le k\le L$ the vacuum congruence restricts $w$ to the residues $\epsilon H P^{-h}Q^{-k}$, with $H\mid t-1$ odd, $H\ge5$, and $h$ taken modulo $m$. This proves the count. The divisor descent above gives the asserted box. The residue criterion excludes prime assignments even within the ratio bound of Proposition~\ref{main:gap-all-heights}. Fix $Q,L$ and let $t>Q$ tend to infinity through primes. Take $P\equiv1\pmod t$, so that $m=1$. The prime number theorem gives $\sim t/((Q-1)\log t)$ primes $w$ in $(t,Qt/(Q-1)]$, all in distinct classes modulo $t$. At most $2(L-1)\tau(t-1)\le 4(L-1)\sqrt t$ are exceptional. The proportion covered by this test therefore tends to one in this interval. One may choose any prime $P>w$ in the class $1$ modulo $t$ by Dirichlet's theorem. Then $Q$ is the least prime and $w\le Qt/(Q-1)$ as required. Other necessary conditions may also exclude some or all of these assignments. \subsection{Finite windows with one unbounded height}\label{sec:finite-four-windows} \paragraph{The odd least-prime window.} For the feasibility problem following \eqref{fourm:budgettest} at odd least prime $P$ and root $P^hQ^2uv$, all three primes $Q,u,v$ are free. For $h\ge3$ use the window of Propositions~\ref{fam:M:8.7} and \ref{fam:M:9.17}, Lemma~\ref{fam:M:8.14}. Its first limit is \[ u\le\begin{cases} (9P^{h-1}+2)/4,&P=3,\\ \bigl(1+P^2/(2(P-1)^2)\bigr)P^{h-1},&P\ge5. \end{cases} \] Put $U=\max(2,\lfloor(u+1)/j_u\rfloor-1)$; then $v\le P^hU+1$ and $Q\le vU+1$. Remove $5,7,13$ as isolated singleton positions by Lemma~\ref{fam:M:8.8} and Corollary~\ref{fam:M:9.6}. Proposition~\ref{main:gap-all-heights} gives $v\le Pu/(P-1)$. The deficits are even, with $3j_t\le t+1$ and $(t-1)/j_t$ odd (Theorem~\ref{fam:M:9.2}); each proper degree occurs in dual pairs. At height $h=4$ the two budgets and the sign of one mixed contribution suffice. For given $u,v,j_u,j_v,\epsilon$, solve \[ Q^2\equiv-\epsilon(P^4v j_u)^{-1}\pmod u,\qquad Q^2\equiv-\epsilon(P^4u j_v)^{-1}\pmod v . \] Combine root pairs by CRT and retain primes in the window. At each $t$ reserve twice the smallest admissible squared cost on each of the two axes (odd duality). Since the mixed sum is $-j_uj_v<0$, some admissible common degree $y$ must have $A_u(y)A_v(y)<0$, fit both residual budgets with multiplicity two, and pass the unramified-endpoint test in Lemma~\ref{fourm:dyadicbudget}. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/new_odd.py}{enumeration} implements these necessary tests and gives \[ \begin{array}{c|rrrrrrr} P&3&5&7&11&13&17&19\\\hline \text{vacuum classes}&14&59&181&1338&2706&8065&12977\\ \text{positive axis budgets}&0&3&0&1&0&1&1\\ \text{negative common term}&0&0&0&0&0&0&0 \end{array} \] Thus the other primes are unrestricted at each of these least octic primes. The test can be applied at any other odd $P$ and any root height $h\ge3$ in $P^hQ^2uv$, replacing fourth powers by $h$th powers and using the corresponding proven window. \paragraph{A general prime-size bound and the dyadic case.} Let $P,Q,u,v$ be distinct primes, $H\ge2$ and $A\ge0$. Set $b=P/(P-1)$ and $c=1$ for $P=2$, $c=2$ for odd $P$. The bounds on the two axes imply that \[ Q>1+\frac{b^3}{c}P^{2H} \quad\Longrightarrow\quad P^{2H+1}Q^A u^3v^2e\ \text{is covered}. \] For the two-square base, Corollary~\ref{fam:M:9.18} gives a putative ordinary minimal root $P^hQ^k uv$, with $P$ least and $2\le h\le H$, $k\ge2$. After ordering $ubu$, then $v\le bu$ and hence $Q\le bu^2/c+1$. If $Q\le bu$ this bound holds as well. Substitution gives the contradiction, and Theorem~\ref{fam:M:10.3}, raises either of the two squared positions to a cube. For $H=3,4$, consider all ordinary minimal roots $2^Hq^k uv$, with $q,u,v$ distinct odd primes and $k\ge2$. Run the periodic vacuum, two axis trace and signed Gram tests of Section~\ref{fam:W:sec:5}, with cap $2^H$ and all positive powers of $q$. Record just the $q$ values among the surviving classes, denoting their set by $S_H$. The finite recurrence gives \[ \begin{array}{r|rrrrr} H & \text{vacuum}&\text{axes}&\text{Gram}& |S_H|&\max S_H\\\hline 3&3967&582&98&9&37\\ 4&87793&5704&1638&39&283 \end{array} \] Here $S_3=\{3,5,7,11,13,17,19,31,37\}$ is contained in $S_4$. Two independent calculations give the same sets. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/prime_projection.py}{first} enumerates residues by multiplication until repetition. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/four_primes/independent_projection.py}{second} reconstructs the vacuum residue by CRT, checks membership in the cyclic group generated by $q$, and retains minimal costs for each partial Gram sum. Allowing exponent signatures for all positive exponents includes every signature with actual exponent at most $k$. Thus an omitted $q$ rules out all $k$ and all prime assignments to $u,v$. Consequently $2^{2H+1}q^Au^3v^2e$ is covered for $q\notin S_H$, arbitrary $A$ and coprime square-free $e$. Apply the two-square descent of Section~\ref{fam:W:sec:5.1}: heights one and two are already covered, and $S_3\subseteq S_4$ covers height three when $H=4$. The periodic test excludes the remaining height, and Theorem~\ref{thm:cubic-lift} raises either square. \paragraph{Unbounded opposite height at three.} There is an odd version of the same projection with no bound on the opposite height. Consider an odd ordinary minimal root $P^hQ^k uv$, $P$ least, $u3$, arbitrary $A\ge0$, any further distinct primes $u,v$, and coprime square-free $e$. An even singleton cannot yield an ordinary minimal obstruction by Corollary~\ref{fam:M:9.18}. \par Consequently an ordinary obstruction to the unordered $(8,4,2,2)$ family must be odd, with the octic prime least and at least $23$. The four dimension thresholds do not require the finite windows of this section. \section{Five repeated primes and local Weyl columns}\label{fam:W:sec:1} \begin{theorem}\label{fam:W:1.1}\label{fam:W:n:thm:main} For arbitrary distinct primes $p,a,b,c,d$, the integer $p^4(abcd)^2$ is covered. Any three of the four squared exponents can be raised to three. \end{theorem} The square of a product of five primes gives the starting case. Its proof also yields the unbraided solvability assertion of Theorem~\ref{thm:intro-two-free}. We then pass to the higher exponents by studying the supports of principal witnesses. \begin{lemma}[Intersecting degree supports]\label{lem:intersecting-degree-edges} A family of pairwise-intersecting two-element sets is contained in a star or in a triangle. In the graph of two-position degrees, disjoint edges are impossible whenever their two labels are principal at a common prime omitted by both. \end{lemma} \begin{proof} Given edges $ab,ac$, either all contain $a$, or there is an edge $bc$. In the latter case any edge meeting all three lies in the triangle. The cases with at most one edge are immediate. In the stated application disjoint edges give coprime degrees, hence zero crossing; common principality at an omitted prime gives a nonzero residue. \end{proof} \subsection{The square of a product of five primes}\label{fam:M:sec:10.2}\label{seq:five-section} \begin{theorem}\label{fam:M:10.4}\label{seq:five-squares} Let $n$ be the product of five distinct primes. Every weakly integral braided fusion category of dimension dividing $n^2d$, with $d$ square-free and coprime to $n$, is WGT. \end{theorem} \begin{proof} At a least counterexample among its divisors, losing a repeated prime is covered by Theorems~\ref{thm:three} and~\ref{thm:four-cubics}. The reduced case has either two-valuation one or odd reduced dimension, excluded by Lemma~\ref{lem:tables} and Proposition~\ref{three:odd-reduced}, respectively. Thus the table is ordinary with square root $n$. Every isolated singleton is at least five by Lemma~\ref{new:trace}. Suppose first some twist contains two primes $a,b$. Put $C=ab$ and let $uh$. \end{proof} \begin{corollary}[Solvability without a braiding assumption]\label{fam:M:10.5}\label{thm:sq30030} Every fusion category of square-free dimension with at most five prime divisors is solvable; if integral it is group-theoretical. For four distinct primes with $\min(u,v)<\min(p,q)$, every fusion category of dimension dividing $uvp^a q^b$ is solvable, with arbitrary $a,b\ge0$. In particular this includes $2pq\ell^h$ for any distinct odd primes $p,q,\ell$. \end{corollary} \begin{proof} Pass to the Drinfeld center, whose dimension is the square of the given dimension. The five-square theorem applies in the first case. In the second, Corollary~\ref{cor:four-before-sextic} excludes every minimal obstruction, since the least prime would have to be $p$ or $q$. WGT descends from the center by Morita invariance and closure under subcategories \cite[Proposition~4.1]{ENOsolvable}. In both cases $4\nmid\FPdim(\C)$. To prove solvability, and for later use at the odd cutoffs, recall the group-extension argument. A WGT category is Morita equivalent to a nilpotent category built by extensions; the group orders multiply to its dimension. If $4$ does not divide the dimension, each group has odd order or order twice odd. In the latter case the sign of the regular permutation representation gives an odd-order kernel of index two. The odd-order theorem \cite[p.~775]{FT} and \cite[Proposition~4.5(i)]{ENOsolvable} now give solvability. In integral square-free dimension use also \cite[Proposition~9.1]{ENOsolvable}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:intro-two-free}] For $p\beta A$ forces $K_X$ to ramify at $q$. If $y\ne q,x$ is an isolated singleton and $K_X$ is unramified there, then $q\le\beta A y$. \end{lemma} \begin{proof} Choose an $x$-free nonunit of $P_q$ by prime avoidance. It hits $X$ by $q$-principality. If $e$ is the inertia index at $q$ of the normalized entry field, Lemma~\ref{fam:M:8.5} gives $q\le\beta e\gcd(d_X,d_Y)\le\beta eA$. All inertia translates of $X$ are proper $y$-free labels. If unramified at $y$, there are at most $y$ of them by the displayed bound. Their number is at least $e$ because the entry lies in $K_X$. \end{proof} \subsubsection{Which primes can mix}\label{fam:F:sec:5.1} \begin{proposition}[The unique possible mixed pair]\label{fam:F:5.3}\label{fam:F:5.4}\label{fam:F:5.5} In an ordinary minimal table with square root $p^2abcd$, at most one pair of distinct primes can mix. If $p,a$ mix, the other three primes $b\beta p$ its field ramifies by Lemma~\ref{fam:F:5.2}. These primes belong to distinct twist clusters, so at most one can ramify. If $p$ were least and odd, each companion would exceed $\beta p=p^2/(p-1)$, a contradiction. Now suppose $u,v$ are squared and mix. The preceding case forbids either from mixing also with $p$. On the mixed column $J$ every hit omits $u,v$ by defect-zero vanishing. Vacuum orthogonality on $P_u$ reduced successively at $p,c,d$ forces supports $cd,pc,pd$ in $P_u$, since the lesser supports have prime-power degree. The same holds on $P_v$. For any prime $t\notin\{c,d\}$ every nonunit of $P_t$ hits the $t$-free degree $cd$, so contains $c$ or $d$. A $t,c$-mixed column would require a $d$-free $P_t$-hit, by vacuum orthogonality, whereas all its hits omit $c$. The same argument separates $d$ from $t$. Finally a $c,d$-mixed column would share only the vacuum hit with $J$. Thus $c,d$ are isolated and at least five (Lemma~\ref{fam:M:6.6}). The mixed norm at the isolated columns $c,d$ gives a common nonunit omitting both by Proposition~\ref{fam:M:6.3}. Any such $X$ must contain $u,v$: otherwise it would have to hit a coprime $cd$ from the corresponding principal block. If its $p$-height were less than two, the complementary cluster $\{p,u,v\}$ puts $X$ in $P_p$ by full height at $u,v$ (Proposition~\ref{fam:M:6.2}), again impossible at the $p$-free degree $cd$. Thus $D=p^2uv$ is the unique common degree. The prime $p$ is isolated by the first case and the isolation of $c,d$. A witness of degree $p^ic$ on $P_u$ or $P_v$ cannot have $i=1$: it would then belong to $P_p$ and have to hit a coprime $p$-free nonunit of $P_c$. The same argument at $d$ supplies the two squared axes in the assertion. Finally, a $p$-free nonunit of $P_t$ must meet both $p^2t$ witnesses in the opposite principal blocks whenever it omits $u$ or $v$. Thus it contains both $u,v$, and it must also meet $p^2w$ by $t$-principality. This forces $uvw$. Prime avoidance gives its existence. \end{proof} \subsubsection{A quartic at two}\label{fam:F:sec:5.2} \begin{proposition}[Separation among four odd squares]\label{fam:F:5.7}\label{fam:F:n:f14:oddsep} Let an ordinary minimal table have dimension $16M^2$, with $M$ the product of four distinct odd primes. No two odd primes occur in the same twist. No degree has the form $2a$ for $a\mid M$ prime. A degree $4a$ can occur only if at most one of the other prime factors $s$ of $M$ satisfies $s\ge11$ or mixes with two and satisfies $s\ge5$. \end{proposition} \begin{proof} Suppose a column is active at odd primes $ah$, Proposition~\ref{fam:M:4.8} forces ramification at $c$, a contradiction. For $p=2$ there cannot be two edges $4a,4b$. Their weight-four crossing would ramify at both other singleton primes, each at least five, by the same two-residue result. As these primes are isolated this is impossible. Thus $4a$ is the unique edge on four and every edge meets $a$. At most one other odd prime is smaller than $a$ by the three-singleton result. On the other hand Proposition~\ref{fam:F:n:f14:oddsep} allows at most one of the other three to exceed seven, so the other two are $5,7$, both smaller than $a$. This excludes the dyadic quartic. \subsection{Three isolated primes in odd dimension}\label{fam:C:sec:2} For the rest of the odd five-prime argument put $H=p^2$ and $\beta=m/(m-1)$, where $m$ is the least prime. The following calculation allows an arbitrary complementary factor $G$. \begin{lemma}[Centered Weyl column for a separated set of odd primes]\label{fam:C:2.1}\label{fam:C:n:centered} In an odd ordinary minimal table of square root $n=HN$, let $H>1$ be the full part of $n$ supported on a nonempty set of odd dimension primes which is separated from its complement: every twist is supported on one side or the other. Put $J=H(1-(W_H)_{00})$ and $a_H(X)=\epsilon H(W_H)_{X0}$ for nonunits, with $\epsilon=\xi$. Then $J$ is a positive even integer and \[ JN\equiv-\epsilon\pmod H,\qquad a_H(X)=[d_X/N]_H. \] In particular its nonunit support is exactly the degrees deficient at some prime of $H$. No assumption on the internal prime powers or internal separation in $H$ is needed. \end{lemma} \begin{proof} The split Weyl column is integral after scaling each local column by its part of $n$ (Proposition~\ref{fam:M:6.2}); reduction of the vacuum and nonvacuum entries modulo $H$ gives the displayed congruences before centering. Every $H$-active label $Z$ has degree $Nk_Z$ with $k_Z\mid H$ by complementarity; on its active subspace $s=\epsilon W_H$, so $a_H(Z)=k_Z$. Here $k_Z1$, and $\gcd(G,pcd)=1$. For each pair of the three positions $Q,c,d$, suppose that every nonunit degree prime to both equals the full complementary divisor: $QG,Gd,Gc$ for the pairs $(c,d),(Q,c),(Q,d)$, respectively. Then degree $Qc$ cannot occur. \end{proposition} \begin{proof} For $H=Q,c,d$ let $z_H=(W_H-I)v_0$. Its vacuum coordinate is nonzero and its other coordinates are supported precisely on the degrees deficient at $H$, by Lemma~\ref{fam:C:n:centered}. For $H=Q$ this means precisely the $p$-free degrees, by Lemma~\ref{fam:F:5.9}. The vectors for distinct positions are orthogonal by \eqref{eq:weylsplit}, so every pair has a common nonunit coordinate to cancel the vacuum product. The pairs $(c,d)$ and $(Q,d)$ therefore supply the degrees $QG$ and $Gc$, respectively. At $c,d$ the only common nonunit degree is $QG=n/(cd)$. Since $n$ is odd, Lemma~\ref{fam:M:8.9} gives $d=cj_d\pm1$, with $j_d$ even and positive by Theorem~\ref{main:full-multiplicity}. Hence $d\ge2c-1>3c/2$. If $Qc$ occurred, $Qc$ and $Gc$ would be distinct proper $d$-free nonunit degrees, with greatest common divisor $c$. The same theorem gives $d\le\ell c/(\ell-1)\le3c/2$ for the least dimension prime $\ell$, a contradiction. \end{proof} \begin{corollary}[Odd roots with five repeated primes]\label{fam:C:2.3}\label{fam:C:n:f15:odd-mixed} In an odd ordinary minimal root $n=p^2uvcd$ at five distinct primes, no two squared primes mix. Under full isolation there cannot be exactly two degrees of the form $p^2x$ with $x$ a singleton. \end{corollary} \begin{proof} If $u,v$ mix, put $G=uv$ and order the isolated singletons $cs$, Lemma~\ref{fam:M:8.9} gives signed $s$-coordinate $\pm1$ at $n/(sw)$. This odd coordinate forces ramification at $s$. These are distinct proper degrees, whereas Theorem~\ref{main:full-multiplicity} allows only one proper ramified orbit. This proves the first assertion. Finally choose any two $H$-edges. Their normalized crossing has weight $H$ and positive residues at the two omitted singleton primes. The two-residue estimate at an isolated prime, applied in both orders, puts each omitted prime below $\beta H/2$. \end{proof} \begin{proposition}[Crossings of two quartic axes]\label{fam:C:3.2}\label{fam:C:n:matching} In an odd, fully isolated minimal root $p^2abcd$, let $Hx,Hy$ be two occurring degrees, where $H=p^2$. For the complementary companions $z,w$ one has $zw\le\beta H$. In particular $p$ cannot be least. \end{proposition} \begin{proof} The crossing has weight $H$ and positive residue $H$ at $z,w$. If nonrational, its field ramifies at a prime $t$ common to the two character fields. By their codegrees $t\in\{z,w\}$. Both degrees would belong to the sole proper $t$-ramified orbit of Theorem~\ref{main:full-multiplicity}, impossible since they differ. The crossing is therefore an integer $H-kzw$, $k\ge1$, and its balancing mean is at least $-H/(m-1)$, where $m$ is the least prime (Lemma~\ref{fam:M:8.5}). This yields the inequality. If $p$ is least, $zw\ge(p+2)^2>p^3/(p-1)=\beta H$, whereas Proposition~\ref{fam:C:3.1} supplies two axes. \end{proof} \begin{proposition}[No remaining quartic axes in odd dimension]\label{fam:C:3.3} An odd ordinary minimal table of root $p^2abcd$ cannot have all five primes isolated and at least three occurring axes $p^2t$. \end{proposition} \begin{proof} Put $H=p^2$ and $N=abcd$. No ordinary pair can be an edge: it would miss one of the at least three $H$-edges. For each companion $t$, prime avoidance on $P_t$ forces a $p$-free nonunit degree $E_t=N/t$, since no proper subproduct of the other three companions can occur. This is the only nonunit degree omitting both $p,t$. Proposition~\ref{fam:C:n:matching} excludes $p$ least. Order the companions $a\beta a$. The three axes are $Hb,Hc,Hd$ and $d\le\beta b$; from $c\ge b+2$ it follows that $b\ge2a-1$. Fix a companion $t$. Its unique proper ramified orbit is given by Theorem~\ref{main:full-multiplicity}. An $E_t$-label has codegree $(Ht)^2$, so its nonreal character field ramifies at $p$ or $t$, and at only one by isolation. Its signed coordinate $B_t=[H^{-1}]_t$ has odd parity exactly in the latter case (Corollary~\ref{fam:M:9.5}); the choice is common to all its labels. If $Ht$ occurs, its nonreal field ramifies at another companion $u$, by its codegree. Were $E_t$ ramified at $t$, the unique proper ramified degrees at $t,u$ would be $E_t,Ht$ respectively. The former contains $u$ and the latter $t$, leaving no common ramified orbit at either endpoint, contrary to Corollary~\ref{fam:M:9.4}. Consequently all $E_b,E_c,E_d$ ramify at $p$. Choose a $p$-active row $Z$ and a $t$-active row $T$. Their only possible common nonunit hits have degree $E_t$, by defect-zero vanishing and full heights. The rows are orthogonal with vacuum contribution one. If $E_t$ ramifies at $p$, the constant evaluation on $T$ is $[E_t]_t$; it therefore divides one and $|B_t|=j_t$. If $E_t$ ramifies at $t$, its labels form one orbit. The evaluations on $Z$ are constant integers, since the two character fields are pure at distinct primes; the evaluations on $T$ sum to $-\epsilon B_t$ by Theorem~\ref{fam:M:9.2}. Thus $|B_t|=1$. Since $HB_t\equiv1\pmod t$, we have $Hj_t\equiv\pm1\pmod t$ in the first case and $Hj_t\equiv\pm j_t\pmod t$ in the second. The fully separated Weyl sum on the vacuum, with the trivial auxiliary two-primary factor, gives \[ H=J+\sum_{t\in\{a,b,c,d\}}Hj_t/t+\epsilon/N,\qquad J=H(1-(W_p)_{00})\in\mathbb Z \] (Lemma~\ref{fam:C:2.1}). If $E_a$ also ramifies at $p$, an integer equals $\pm1/a\pm1/b\pm1/c\pm1/d+\epsilon/N$. This has magnitude less than one, so it vanishes. Multiplication by the odd $N$ gives four odd terms and the extra odd $\epsilon$, an impossibility. If $E_a$ ramifies at $a$, the integer instead equals $\pm j_a/a\pm1/b\pm1/c\pm1/d+\epsilon/N$. It again has magnitude less than one, since $3j_a\le a+1$, $a\ge11$ and $b\ge2a-1$. But it is nonzero: $j_a/a\ge2/a>3/(2a-1)+1/N$. This exhausts the possibilities. \end{proof} \subsection{The remaining odd configuration}\label{fam:C:sec:4} \begin{proposition}[The three isolated primes beside a mixed quartic]\label{fam:C:4.1} Let the odd ordinary minimal table have root $p^2abcd$, with $a$ mixing with $p$ and $b\beta c$. Thus $pb$ occurs and $d\le\beta b$. Absence of either other low degree would force $d\ge2b-1>\beta b$. The least prime cannot be $b$, since $d\ge b+2$, or $p$, by Proposition~\ref{fam:F:5.3}; hence it is $a$. Finally $b+2\le d\le ab/(a-1)$ and parity give $b\ge2a-1$. \end{proof} After loss of a repeated prime or a drop of the quartic valuation use Theorems~\ref{fam:M:10.3}, \ref{fam:M:4.12} and~\ref{fam:F:2.1}. The reduced core was excluded in Section~\ref{fam:F:sec:4}. For the ordinary root $p^2abcd$ the preceding witness arguments leave the following cases. \[ \begin{array}{l|l} \text{support or parity}&\text{obstruction or remaining argument}\\ \hline p=2&\text{degree edges at the dyadic quartic (Section~\ref{fam:F:sec:5.4})}\\ \text{odd, two squared primes mix}&\text{three isolated corners (Corollary~\ref{fam:C:2.3})}\\ \text{odd, no two primes mix}&\text{quartic axes (Propositions~\ref{fam:C:3.1}--\ref{fam:C:3.3})}\\ \text{odd, the quartic mixes}&\text{three-prime cycle below}\\ 2\text{ squared}&\text{the dyadic partner cases of Section~\ref{fam:W:sec:4}} \end{array} \] \subsection{The odd five-prime root}\label{fam:W:sec:3} Suppose all five primes are odd. Permute the companion primes so that $at\ge b. \] The window $d-b\le b/(a-1)$ now gives $b>(a-1)^2$. If $t\to r$ is low-only, the incoming edge $s\to t$ carries a high degree. Its low degree $pr$ is unramified at $t$. The outgoing congruence $par\equiv\pm1\pmod t$ gives $a|[pr]_t|\ge t-1$, since the only integer multiple of $a$ congruent to $\pm1$ with magnitude less than $t-1$ would be $\pm1$. Its dual-pair budget gives $2j_t[pr]_t^2\le t+1-j_t\le t-1$. Thus $t\le1+a^2/(2j_t)\le1+a^2/4$. But $t\ge b>(a-1)^2>1+a^2/4$ for $a\ge3$. It follows that every edge carries a high degree. \paragraph{The cyclic divisibility contradiction.} At $s\to t\to r$ let \[ A=[s^{-1}]_t,\qquad B=[r^{-1}]_t, \qquad w=[rs]_t . \] The incoming high degree $Hr$ has even coordinate $A$ at $t$ and multiplicity $j_s$; the outgoing high degree $Hs$ has odd coordinate $B$ at $t$ and multiplicity $j_t$. The degree $rs$ contributes at least $2j_t^2w^2$ to the residual norm at $t$. Pairing this cost first with the incoming high cost $j_s A^2$ and then with the outgoing high cost $j_t B^2$ gives \[ |Aw|\le\frac{t+1-j_t}{2\sqrt{2j_s}}<\frac t4, \qquad |Bw|\le\frac{t+1-j_t}{2\sqrt{2j_t}}<\frac t4 . \] Here $j_s,j_t\ge2$. The vertex window gives $2/3\le r/t,s/t\le3/2$, strictly inside $(1/4,7/4)$. For $x=r,s$ the only representative of $x$ modulo $t$ that can have magnitude less than $t/4$ is $x-t$: all $x+kt$ with $k\ge0$ exceed $t/4$, and all those with $k\le-2$ are less than $-t/4$. As $Aw\equiv r$ and $Bw\equiv s$ modulo $t$, this gives $Aw=r-t$ and $Bw=s-t$. As $A$ is even, $B$ odd, and $w\ne0$, \[ \nu_2(r-t)>\nu_2(s-t). \] Applied at the three vertices of the directed cycle this gives a strict cyclic inequality. The odd minimal table is impossible. \subsection{A squared prime two}\label{fam:W:sec:4} The remaining root has a squared two in the dimension and an odd quartic prime $p$. Two primes \emph{mix} if the order of some simple twist is divisible by both. There is one mixing pair and it contains two. Its partner is $p$ or another squared prime $u$ (Proposition~\ref{fam:F:5.3} gives at most one pair, and Lemma~\ref{fam:M:6.6} excludes an isolated squared two). Let $P_\ell$ be the principal residue block at $\ell$. In a perfect fusion-simple table nonunit labels of coprime degrees have zero $S$-entry, while a label in $P_\ell$ hits every label whose degree is prime to $\ell$ (Sections~\ref{fam:M:sec:2}--\ref{fam:M:sec:4}). \paragraph{The odd denominator.} Suppose first that two mixes with $p$. Write $n=2p^2bcd$ and order the three isolated primes as $bB(W)$; one may choose the smallest convenient monomial among the allowed pairs $(p,t)$. Every factor of $W$ is a companion preceding $r$ in the order $t_1,\ldots,t_k,R$. Thus the hypotheses can be satisfied successively by choosing primes. The certificate records the terminal theorem or the pair $(p,t)$ for every branch. A hypothesis $r>B(V)$ covers a recorded $W$ whenever $V\ge W$ throughout the prescribed order and heights. For monomials $V,W$, write $V/W$ as a product of powers of $t_1$ and of the ratios $t_i/t_{i-1}$. The exponents in this expression are the successive suffix sums of the exponent differences. Nonnegative suffix sums suffice; for affine exponents with independent $h_i\ge1$ this is checked by checking their slopes and their values at $h_i=1$. Replaying these checks for every branch proves the divisor assertion by Lemma~\ref{ext:lem:two-free-crossing} and the minimal-counterexample reduction, without substituting individual primes. \end{proof} \begin{corollary}\label{ext:cor:five-two-free} Let $P,R,u,v,w$ be distinct primes, with $u(uv-1)(uv)^2$, every weakly integral braided fusion category whose dimension divides \[ P^A R^B(uvw)^3 e \] is weakly group-theoretical, for arbitrary $A,B\ge0$ and any square-free $e$ coprime to the five primes. \end{corollary} \begin{proof} Apply Proposition~\ref{ext:prop:crossing-certificates} with three companion caps equal to three. On any nonterminal branch both $P,R$ and at least two companions survive: otherwise the terminal theorem for at most four primes applies. In either table take $r=R$, $p=P$ and $t$ the largest surviving companion, necessarily odd. Its valuation is at most three, so the certificate removes its whole height; each other companion has height at most one. Hence $W\le uv$ and $B(W)\le B(uv)=(uv-1)(uv)^2$. \end{proof} \subsection{One new prime of arbitrary height} \begin{proposition}\label{ext:gap:adjoin} Let $p_1<\cdots\frac{T^2}{p_1p_k}, \] then $Nr^b$ is covered for every $b\ge0$. For $k=2$ the weaker inequality $r>T^2/(p_1p_2^2)$ suffices. \end{proposition} \begin{proof} Fix $b,e$ and take a counterexample of least dimension $D$ among their allowed divisors. Its ordinary or positive reduced table has dimension $\Delta=D$ or $D/2$ and a perfect, simple, integral based ring. Moreover $\Delta$ is square-full, $d_Y^2\mid\Delta$ for each label $Y$, and $\Delta\mid\prod_Yd_Y^2$ by Lemmas~\ref{fam:M:2.1} and~\ref{fam:M:2.4}. If $v_r(D)\le1$, include the possible factor $r$ in the square-free decoration of the covered prefix. Thus $r^2\mid\Delta$, since $r$ is odd. Dong's two-prime theorem applies unless at least two prefix primes are repeated in $D$. All repeated prefix primes remain in $\Delta$, where valuation one is impossible. Their largest, $h$, is odd. Put $T'=\prod_i p_i^{\lfloor v_{p_i}(\Delta)/2\rfloor}$. Choose an $h$-active column by categorical Cauchy, using (Proposition~\ref{fam:M:3.4}) in the reduced case. Prime avoidance produces an $r$-free nonunit $X$ in the principal $h$-block, so $d_X\le T'/h$ by Corollary~\ref{fam:M:4.1} and Proposition~\ref{fam:M:4.2}. Every $r$-free degree is at most $T'$. Some degree is divisible by $r$ since $\Delta\mid\prod_Yd_Y^2$. Each such degree is at least $rp_1$, since nonunit degrees cannot be prime powers (Lemma~\ref{fam:M:2.4}). The bound $T'^2/h\le T^2/p_k$ holds also when $h0}$ and every other eigenvalue belongs to $\mathbb Z\cap(-m,m)$. Then $v_p(\Delta)\le v_p((2m-1)!)$. \end{lemma} \begin{proof} The polynomial $\prod_{j=1-m}^{m-1}(B-j)$ equals $(2m-1)!e_0$, by evaluation at every character. Its unit coefficient is $(2m-1)!/\Delta$ and is integral at the place by the integral structure constants. \end{proof} \begin{theorem}\label{found:D:2.1}\label{found:D:key:th:exponent} In a minimal table let $p$ divide $\Delta$ and write $\Delta=p^a m$ with $a=v_p(\Delta)\ge1$. Put \[ u(m)=\prod_{\ell\mid m}\ell^{\lfloor v_\ell(m)/2\rfloor}, \qquad c=2^{\mathbf1_{3\mid m}},\qquad t=(p-1)c . \] There is a nonunit $X$ in the full principal $p$-block with $p\nmid d_X[K_X:\QQ]$. Every such $X$ satisfies \[ a\le v_p\big((2t d_X-1)!\big)<2c\,d_X\le2c\,u(m). \] The bound $a\le v_p((2t d_X-1)!)$ also holds for a specified nonunit $X$ whose field is tame at odd $p$, or unramified at $p=2$. If $K_X$ is unramified at odd $p$, one may use $t=c$ in the factorial. \end{theorem} \begin{proof} The unit coefficient of the $p$-local principal-block idempotent is $\Delta^{-1}\sum_{X\in P_p}d_X^2$. Thus $p^a\mid\sum_{X\in P_p}d_X^2$, so this sum vanishes modulo $p$. The block is Galois-stable and the unit orbit contributes one. Some other orbit has degree and size prime to $p$. Its field is tame at odd $p$, and unramified at two if $p=2$, since the wild inertia or the entire dyadic inertia, respectively, is a $p$-group. Also $d_X\mid u(m)$. Fix a primitive $\ell$th root $\eta_\ell$ for every $\ell\ge5$ dividing $m$. On the $\ell$-power roots prescribe a linear functional $L_\ell$ taking $1$ at $1$, $-1$ at $\eta_\ell$, and zero elsewhere. These values sum to zero on each coset of $\mu_\ell$, exactly the cyclotomic relations at a prime power. Form their tensor product with the normalized trace on the dyadic factor, $(p-1)$ times the normalized trace on the $p$-factor for odd $p$, and twice the normalized trace on a complementary three-factor. Include the dyadic factor just once if $p=2$. Call the result $F$. Then $F(1)=t$, and $F$ is integer-valued on cyclotomic integers: check on roots, using the normalized traces $1,-1/(\ell-1),0$ at orders $1,\ell,\ell^k$ with $k>1$. For unramified odd $p$, use the prime-to-$p$ conductor of $K_X$ and extend by normalized trace at $p$ to evaluate balancing words. Integrality on $K_X$ is preserved and $t=c$. On $K_X$ the functional is a sum of embeddings with coefficients integral at $p$. Indeed, take a cyclotomic field containing $K_X$ and the marks, with only the tame $p$-level, omitting the dyadic level at $p=2$. Such a field exists by the local inertia description and the conductor bound in Corollary~\ref{cor:field-conductor}. On this field $L_\ell$ is normalized trace after multiplication by $(\ell-1)(1-\eta_\ell^{-1})/\ell$. The factor $\ell-1$ cancels the $p$-part of the degree of the $\ell$-primary cyclotomic field. At a complementary three the factor $2$ does this for $p=2$. The remaining local degrees are $p$-units. One may evaluate $F$ on any higher balancing level by normalized relative trace; the smaller field suffices for the coefficients of the embeddings. Galois covariance therefore gives a $p$-locally integral linear combination $B$ of the $[X^\sigma]$ with eigenvalues $b_Y=F(\chi_Y(X))$. We prove $|b_Y|0$ and $p\nmid m$, then $p\le B(m)$ and $a{\raggedright\arraybackslash}p{.24\linewidth}@{\hspace{.022\linewidth}}>{\raggedright\arraybackslash}p{.735\linewidth}@{}} \caption{Dimension families, grouped by the hypotheses on their prime divisors.} \label{tab:family-cones}\\ \toprule Dimension family & Conditions and references\\ \midrule \endfirsthead \multicolumn{2}{@{}l}{\emph{Table~\thetable\ continued}}\\ \toprule Dimension family & Conditions and references\\ \midrule \endhead \midrule \endfoot \bottomrule \endlastfoot \multicolumn{2}{@{}l}{\textbf{Uniform braided families}}\\* $p^Aq^B$ & At most two repeated primes \cite[Theorem~3.1]{DongWeakIntegral}.\\ $p^A q^B r^C$ & For $p\min(P,u,v)$. The caps $(a,B)=(7,5)$ hold in every order. The caps $(9,5)$ also hold unless $P$ is the least of four odd primes and $P\ge23$ (Theorems~\ref{fam:W:n:thm:unbounded}, \ref{thm:higher-two-squares} and~\ref{thm:cubic-lift}, Section~\ref{sec:finite-four-windows}).\\ $p^4(uvw)^3t^2$; $(puvwt)^3$ & Both five-prime families hold in every order (Theorems~\ref{fam:W:n:thm:main} and~\ref{fam:F:n:f14:five}).\\[2pt] \multicolumn{2}{@{}l}{\textbf{Three repeated primes with size, residue or word conditions}}\\* $t^a u^b s^C$ & Put $L=t^{\lfloor a/2\rfloor}u^{\lfloor b/2\rfloor}$. The size test is $stu\max(t,u)>L^2$; the residue and support alternatives are Proposition~\ref{prop:three-separation} and Theorem~\ref{fam:Q:4.1}.\\ $p^A q^B r^C$, $pp^h$ suffices. For $q>p^{h-1}$ the two companion caps may be rounded down to even; see Theorem~\ref{three:least-height} for the odd-prime refinement.\\ $p^A q^B r^4$, $pq$, then $C$ is free. For $A\ge6$, $q>p^{\lfloor A/2\rfloor-2}$ is another sufficient condition at $C=4$ (Proposition~\ref{fam:Q:5.5} and Theorem~\ref{fam:Q:5.6}).\\ $p^{2h}q^B r^4$; $p^{2h}q^{2k}r^4$; $p^{2h}q^B r^C$ & For odd $pp^{h-1}$ and $W(r,p^{h-2})$ suffice for unrestricted $B,C$ (Theorem~\ref{found:S:key:thm:twoarbitrary}).\\ $P^6Q^B r^C$ & At $2=PP^2$ and $W(r,1+\lfloor P^2/Q\rfloor)$ (Theorem~\ref{found:J:key:thm:middle}). For $P(uv-1)(uv)^2$ suffices (Corollary~\ref{ext:cor:five-two-free}).\\ $N r^B$ with covered $N$ & Write $N=\prod_{i=1}^k p_i^{a_i}$ with $a_i\ge2$, $p_1<\cdotsT^2$, or that $k=2$ and $rp_1p_2^2>T^2$ (Proposition~\ref{ext:gap:adjoin}).\\ $p^A m$, fixed $m$ & For $p\nmid m$ a finite list suffices even with $p$ variable, including the additional dyadic exponent for a reduced table (Corollary~\ref{found:D:cofactor-unified}).\\[2pt] \multicolumn{2}{@{}l}{\textbf{Conclusions without a braiding assumption}}\\* $D^2$ in a braided family & Apply the corresponding condition to $Q(D^2)=D^2$. This gives WGT for every source of dimension $D$ (\cite[Proposition~4.1]{ENOsolvable}).\\ $D\mid uvp^Aq^B$; $D$ square-free & The first family is solvable when $\min(u,v)<\min(p,q)$. Square-free $D$ on at most five primes gives solvability, and gives group-theoreticity in the integral case (Corollary~\ref{thm:sq30030}). Without an ordering condition, $D\mid p^Aq^Br$ gives WGT (Theorem~\ref{thm:three}).\\ \addlinespace \multicolumn{2}{@{}l}{\textbf{Rank families at the specified dimension}}\\* \multicolumn{2}{@{}p{.997\linewidth}@{}}{Here $R$ is the rank of a non-pointed simple integral modular category and $D=qM^2$ with $q$ square-free, except where a minimal obstruction is specified. Table~\ref{tab:intro-ranks} gives the uniform bounds.}\\* $q=1$ & Put $B=M/d_{\max}$. The bounds for $B\ge5$, $4\nmid M$, and $B\ge6$ are $36$, $39$, and $46$, respectively (Theorem~\ref{mrk:thm:modular-ranks}).\\ $q>1$ & Let $a$ be the least prime of $M$. For prime $q$, $R>qa^2$ since formal codegree $q$ is impossible (Lemma~\ref{small}). For composite $q$, $R>q+(a^2-1)(1+\sum_{p\mid q,\,p\text{ odd}}(p-1)/2)$ (Corollary~\ref{cor:squareclass-additive}, including $q=15$).\\ Prescribed prime support & For odd $s\mid M$, $s\nmid q$, collected in $S$, $R\ge B(q)(1+\sum_{s\in S}(s-1))$ with $B(q)$ as in Section~\ref{sec:joint-prime-capacity}. The joint-orbit, denominator and simultaneous-deficit bounds are Propositions~\ref{prop:crossed-packet-count}, \ref{prop:joint-prime-capacity}, \ref{mrk:prop:rank-envelope} and~\ref{prop:multiple-defect-one}.\\ $D=qU^2p^{2e}$ & For fixed $q,U$ and odd $p\nmid qU$, \eqref{eq:activity-rank} is uniform in $e\ge1$. For $q=3$ the ten-object refinement is Corollary~\ref{cor:saturated-twelve-prime}.\\ $q=10$, three primes & For $10M^2$ on three primes, $R\ge36$ in the ordinary case. At \emph{reduced} dimension $10M^2$ a minimal obstruction has full rank at least $68$ (Proposition~\ref{mrk:prop:ten-family}).\\ Two isolated primes & For the ordinary minimal root $P^hQ^kuv$ of the four-prime family above, $R\ge7+(u-1)j_u+(v-1)j_v$, where $j_u,j_v$ are its vacuum deficits. At odd dimension $R\ge13+uj_u+vj_v$ (Corollary~\ref{fam:rank:two-isolated}). Estimate~\eqref{mrk:eq:two-singletons} also uses the sizes of the two higher prime powers.\\ $D=M^2$, a prime denominator & A unique $M/d=p$ requires $v_p(M)=1$ and at least three prime factors of $M/p$ (Proposition~\ref{prop:rational-prime-square}). The further trace bounds are in Section~\ref{sec:unique-prime-denominator}.\\[2pt] \multicolumn{2}{@{}l}{\textbf{Minimal obstructions and decomposition}}\\* Five primes; three primes & A minimal obstruction on five repeated primes cannot have exact valuations $3,2,2$ at three positions, whatever the other two (Theorem~\ref{fam:F:n:f14:five}). Under the unordered caps $(6,6,4)$, a minimal obstruction must be ordinary with exact valuations $6,6,4$ and the quartic prime largest (Section~\ref{sec:boundary}). The only possible even dimension of such a minimal obstruction is $2^63^65^4$. At $5^67^623^4$ every non-WGT braided category has rank at least $553$ (Corollary~\ref{fam:rank:five-seven-twenty-three}).\\ $D\mid p^a q^b r^4$ & An integral modular or slightly degenerate category is WGT if the degrees in its principal block at odd $r$ are prime to $r$ (Theorem~\ref{fam:Q:n:quart:recognition}).\\ Prime supports of braided factors & For integral WGT modular categories use the components determined by $s\mid\ell^{2j}-1$, $1\le j\le\lfloor v_\ell(D)/2\rfloor$. Theorem~\ref{thm:uc-components} also treats integral categories with all repeated primes in one component and categories with pointed core.\\ $D=L$ or $L^2$ & For composite square-free $L$, Theorem~\ref{thm:uc-bipartite} gives the connected-graph existence test for a tensor-prime integral WGT category at $L$ (positive signs) and a prime integral WGT modular category at $L^2$ (one sign per target).\\ Modular or slightly degenerate dimensions & With $e=1$ or $2$, respectively, let $U$ be the square root of the largest square divisor of $D/e$. If $U<60$ the category is solvable. For a WGT category it suffices that no order of a nonabelian finite simple group divides $U$ (Proposition~\ref{prop:solv-square-divisor}).\\ \end{longtable} \endgroup \section{The four dimension bounds}\label{sec:numerical} \begin{theorem}[Uniform dimension bounds]\label{thm:uniform-new-bounds}\label{seq:end-bounds}\label{thm:unbraided-range} The four bounds in Table~\ref{tab:intro-bounds} hold. Every strictly weakly integral fusion category of dimension less than $10800$ is weakly group-theoretical. \end{theorem} \begin{proof} We compare the prime exponents of a divisor-minimal braided obstruction with those allowed by the family theorems. With a braiding, the square-full part of its dimension divides $Q(D)$. Without a braiding, the Drinfeld center has dimension $N^2$, so a valuation $e$ in a minimal obstruction requires valuation at least $\lceil e/2\rceil$ in $N$. At most two repeated primes are covered by Dong's theorem \cite[Theorem~3.1]{DongWeakIntegral}. The following lower bounds use only the families needed for this comparison. \[ \begin{array}{c|rrrr} \#\text{ repeated primes}&Q(D)&Q(D)\text{ odd}&N&N\text{ odd}\\\hline 3&2^63^65^4&3^65^67^4&5400&165375\\ 4&6300^2&121275^2&6300&121275\\ 5&2^3\,2310^2&3^3\,15015^2&9240&135135\\ \ge6&30030^2&255255^2&30030&255255 \end{array} \] At three primes $p1$ and $x^2+\operatorname{lcm}(x,y)^2\le\Delta$. By \eqref{mrk:e:orth} the gcd of all such candidates must be one. The ordinary searches apply the character, local-block, twist-mass and dimension-family tests after this degree check or earlier on a whole prefix when all completions violate them. A failed necessary condition discards the corresponding list. For the fermion quotient we use only identities established for that table, including its odd-primary twist identities. \begin{proposition}[Completeness of the degree search]\label{rank:search-completeness} Fix the table rank $r$, choose the ordinary or positive reduced table and choose the counting weight $w$ of \eqref{mrk:e:fraction}. Take the finitely many square-free $q$ and positive $B$ with $qB^20$, and each update decreases $h$. The terminal equations include the original pair of degrees for every completion; at the one-position boundary the same assertion uses $M=L$. Conversely the explicit final tests recover the dimension identity in the original units. A complete denominator list determines $M$, since its lcm is $M$ by Lemma~\ref{mrk:e:witness}. Grouping the last two divisors in a different order, or dispatching disjoint prefix subtrees, therefore leaves the enumeration exhaustive. The terminal arithmetic conditions are necessary; a retained list still requires the character and twist tests. \end{proof} \section{Counting labels by prime components of the twist}\label{mrk:packets} For an odd prime $p$ put $h_p=(p-1)/2$. A label with $p$-primary twist of exact order $p^j$ has $p^{j-1}h_p$ distinct square Galois translates of its phase, all at the same degree and with unchanged complementary twist. We call such a collection of labels a packet. Theorem~\ref{thm:twistdiv} requires $p^j\mid f_X$. In odd dimension each phase occurs in dual pairs. We first determine what happens within a fiber of the complementary twist, and then combine the degree and multiplicity requirements at several primes. For a set $T$ of primes, a complementary fiber $F$ fixes the specified twist components outside $T$; omitted components are summed over. On a nonempty fiber put $m(F)=\sum_{i\in F}d_i^2$ and $g(F)=\gcd_{i\in F}d_i$. In the positive fermion quotient the components are those of the twist modulo the fermion sign. Each odd component is well defined. The following estimates use the same fibers: at an odd prime of even valuation they equate primitive masses, at an odd prime of odd valuation they compare active and inactive masses, and at two they bound cardinality when $2\mid q$. The extra two-label test in Lemma~\ref{rank:ordinary-dyadic-test} uses the full ordinary action. \subsection{Local fibers and simultaneous prime bounds}\label{sec:crossed-fiber-count} \begin{proposition}[Even valuation]\label{prop:two-packets} Let $p$ be an odd prime with $v_p(\Delta)>0$ even. In either table, intersect a full $p$-block, or the whole label set, with prescribed complementary twist components. In this intersection the squared-degree masses at the $p-1$ primitive $p$th roots are equal. If any phase of exact order $p^j$ occurs, $j\ge1$, at least $p-1$ phases of that order occur there. In particular the whole table contains at least $p-1$ active labels. \end{proposition} \begin{proof} Take the vacuum at both entries of the local Bruhat calculation in Proposition~\ref{prop:odd-slice}. The block and complementary-phase projectors commute with the local factor. At even valuation the local torus fixes the vacuum vector with positive sign because $p$ is unramified in $\QQ(\sqrt\Delta)$. The odd local unipotent acts by the actual primary twists. Hence $\sum d_i^2\theta_{i,p}^{u}$ over the intersection is independent of the unit $u$ modulo the local level. These are all conjugates of the first moment, so the moment is rational. Its normalized trace to $\QQ(\zeta_p)$ kills the roots of higher $p$-power order. The unique rational relation among the $p$th roots then equates the masses at the nontrivial ones. The proof on fermion pairs uses the positive subspace of Section~\ref{sec:blocks}; the vacuum sign is unchanged. When $j\ge2$, a single square orbit has $p^{j-1}h_p\ge p-1$ phases. For the whole table, categorical Cauchy or its reduced form in Section~\ref{sec:blocks} supplies an active label. In ordinary data a related single-cycle obstruction appears in \cite[Lemma~4.7 and Theorem~4.8]{BGNPW}. \end{proof} \begin{lemma}\label{lem:crossed-fiber-count} Let $t$ be an odd prime with $v_t(\Delta)$ odd. For any nonempty complementary fiber $F$ for $\{t\}$, put $g=\gcd\{d_i:i\in F\}$ and $h_t=(t-1)/2$. Then \[ t\mid\sum_{i\in F}(d_i/g)^2. \] At least $h_t+1$ distinct $t$-primary phases occur in $F$, and $F$ contains at least $h_t+2$ labels. \end{lemma} \begin{proof} Apply Proposition~\ref{prop:odd-slice} with both test labels equal to the vacuum and with the whole label set as its block. Sum over all unspecified complementary components. The squared mass at exact $t$-order $t$ in $F$ is $t-1$ times the $t$-inactive squared mass. At any degree and exact $t$-primary order $t^j$, inertia at $t$ preserves the degree, the complementary components and the order. Its square action realizes all square-unit translates, of which there are $t^{j-1}h_t$. Multiplicities are equal within each such set of phases. Thus for $j\ge2$ the number of these labels at each fixed degree is a multiple of $t$. Dividing squared masses by $g^2$ gives the displayed divisibility. If a higher-order phase occurs it gives at least $th_t$ distinct phases and $th_t\ge h_t+2$. Otherwise the positive squared masses and the slice identity require both an inactive label and an orbit of at least $h_t$ distinct primitive phases. If precisely $h_t+1$ labels occur, they consist of one inactive label of degree $b$ and one label at each phase of a square orbit, all of a common degree $a$. The slice identity reads $h_ta^2=(t-1)b^2$, or $a^2=2b^2$, impossible for positive integers. This proves both assertions. \end{proof} For example, the congruence alone excludes a singleton fiber. At $t\equiv3\pmod4$ it also excludes any two-label fiber, since at least one of the two degrees divided by $g$ is prime to $t$ and $-1$ is not a square modulo $t$. For a set $T$ of odd primes of odd valuation put $\Gamma(\varnothing)=1$, and, for $t_0=\min T$ if $T$ is nonempty, put \[ \Gamma(T)=(h_{t_0}+2)\prod_{t\in T\setminus\{t_0\}}(h_t+1). \] Every nonempty complementary fiber for $T$ contains at least $\Gamma(T)$ labels. Indeed refine it successively by the $t$-components for $t\ne t_0$. Each nonempty intermediate fiber has at least $h_t+1$ nonempty children by Lemma~\ref{lem:crossed-fiber-count}. At the last prime each terminal parent has at least $h_{t_0}+2$ labels. The chosen order maximizes this bound because $(h+2)/(h+1)$ decreases with $h$. If $\Delta$ is odd, a complementary fiber that contains no vacuum consists of pairs of distinct duals of equal degree and twist, as in Lemma~\ref{lem:tables}. In this case the same argument counts pairs: the Galois action commutes with duality and each phase in a square orbit still occurs at the same degree, while all masses in the slice identity can be divided by two. Such a fiber therefore contains at least $2\Gamma(T)$ labels. The fiber containing the vacuum is only charged $\Gamma(T)$ labels below. \begin{proposition}\label{prop:crossed-packet-count} Let $u$ be any odd prime divisor of $\Delta$, $T=\{t:t\mid q,\ t \text{ odd},\ t\ne u\}$, and $\delta=2$ for odd $\Delta$, otherwise $\delta=1$. Set $s_u=u-1$ if $u\nmid q$ and $s_u=(u-1)/2$ if $u\mid q$. There are at least $\delta s_u\Gamma(T)$ $u$-active labels, and \[ \operatorname{rk}_{\rm table}\ge(1+\delta s_u)\Gamma(T),\qquad \sum_{\substack{d>1\\ u\mid\Delta/d^2}} \left\lfloor\frac{m_d}{\delta (u-1)/2}\right\rfloor \ge \frac{s_u}{(u-1)/2}\,\Gamma(T). \] \end{proposition} \begin{proof} Categorical Cauchy supplies a nontrivial $u$-primary phase. If higher $u$-order occurs, square covariance supplies at least $u(u-1)/2$ distinct nontrivial phases. If only order $u$ occurs, at odd valuation there are at least $(u-1)/2$ such phases. At even valuation every nontrivial $u$-phase has the same squared mass by Proposition~\ref{prop:two-packets}, hence all $u-1$ occur. Each resulting nontrivial $u$-fiber contains no vacuum, so its cardinality is at least $\delta\Gamma(T)$. The trivial $u$-fiber, containing the vacuum, contributes at least $\Gamma(T)$. At any fixed degree the $u$-active multiplicity is divisible by $\delta(u-1)/2$ by square covariance (and dual pairing in odd dimension). Its codegree is divisible by $u$ by Theorem~\ref{thm:twistdiv}. Dividing the active count into these degree packets proves the claimed bound. \end{proof} \begin{lemma}[Dyadic torus on a complementary fiber]\label{lem:dyadic-torus-fiber} If $2\mid q$, every nonempty fiber of any collection of odd-primary components has at least two labels in ordinary or positive reduced data. \end{lemma} \begin{proof} Work in the ordinary modular representation, using a minimal modular extension of dimension $4\Delta$ in the reduced case, with the congruence lift described in Section~\ref{sec:blocks}. Let $W,U(x),L(x),H_a$ denote the Weyl, upper and lower unipotents and diagonal torus in the dyadic local factor, with $H_a=\operatorname{diag}(a,a^{-1})$. The fiber space is preserved by $W$ and $U(2)$, since the odd twist operators belong to commuting local factors. In the reduced case these two operators also preserve the positive fermion subspace: even powers of the global twist preserve it, and the dyadic Weyl is the global Weyl times the inverses of the odd local Weyls, all of which preserve it (Section~\ref{sec:blocks}). The even local unipotents are powers of $U(2)$. For a one-dimensional character $\lambda$ of the subgroup generated by $W,U(2)$, Weyl conjugacy gives $\lambda(L(x))\lambda(U(x))=1$ for every even $x$. In the local coefficient ring $\mathbb Z/2^a\mathbb Z$ the matrix identity \[ U(2)L(2)=L(2/5)H_5U(2/5) \] therefore gives $\lambda(H_5)=1$. On the other hand, the local $H_5$ is the signed Galois permutation with pure dyadic multiplier five (\cite[Theorem~II]{DLNCongruence}). This automorphism negates $\sqrt2$, fixes $i$, and fixes all odd-conductor roots. It therefore negates $\sqrt\Delta$ for every even square class $q$. Every ordinary label here is integral, and in the extension the two representatives of each reduced label are integral. Applying the Galois formula to the positive vacuum coordinates $d_i/\sqrt\Delta$ (or $d_i/(2\sqrt\Delta)$ in the extension) shows that $H_5$ acts as minus a permutation on the specified ordinary basis, respectively on the basis of positive pair sums. For the latter, invariance of the positive subspace, already shown for the displayed product representing $H_5$, ensures that a pair sum maps to a pair sum with this common minus sign. On a singleton fiber it thus acts by $-1$, a contradiction. \end{proof} Refine any fixed odd-primary fiber successively by every $t\in T$ as in Lemma~\ref{lem:crossed-fiber-count}. This produces at least $\prod_{t\in T}(h_t+1)$ distinct tuples of odd phases. Each tuple has at least two labels by Lemma~\ref{lem:dyadic-torus-fiber}. Consequently for $2\mid q$ one may use \[ \Gamma_2(T)=2\prod_{t\in T}(h_t+1) \] in both bounds of Proposition~\ref{prop:crossed-packet-count}. At table rank at most 30 this leaves only $u=3$ as an additional odd prime for $q=10$ or $14$, and no such prime for $q=22$ or $26$. Since the dimension of a nontrivial simple perfect integral modular category contains at least three primes, for every odd prime $p\ge5$ the square class $q=2p$ forces rank at least $3(p+1)$: any third prime $u$ has even valuation and $u\ge3$. In particular the square classes 22 and 26 have rank at least 36 and 42 respectively, independently of the exponents. \begin{lemma}[Two labels in a local invariant subspace] \label{lem:two-label-local} Fix the linear congruence lift for an integral modular category. Let $p$ be a prime and suppose that the span of two coordinate vectors $v_X,v_Y$ is invariant under the $p$-local congruence factor. If its local Weyl element $W_p$ does not preserve $\mathbb C v_X$, then $p=2$, $d_X=d_Y$, and the ratio of their actual $p$-primary twists is a primitive fourth root. \end{lemma} \begin{proof} The local upper unipotent $U_p$ acts diagonally with ratio $\omega$ of $p$-power order: the anomaly normalization cancels in the ratio. Its two eigenvalues are distinct, since otherwise its conjugate lower unipotent is scalar too, and the two generate the local group. The two-dimensional unitary action is irreducible. Indeed, an invariant line would be a $U_p$ eigenline, whose orthogonal complement is a second invariant eigenline, contrary to the hypothesis on $W_p$. The group element $W_p^2=-I$ is central. After multiplying $W_p$ by a scalar of absolute value one, its restriction is a non-scalar Hermitian involution of the form \[ S'=\begin{pmatrix}a&b\\\bar b&-a\end{pmatrix},\qquad a\in\mathbb R,\quad b\ne0,\quad a^2+|b|^2=1 . \] The relation $(W_pU_p)^3=W_p^2$ implies that $S'U_p$ has projective order three: projective order one would make $S'$ diagonal. Its two eigenvalues therefore have ratio a primitive cube root, and their trace has absolute value one. Taking the trace in the displayed basis gives \[ a^2 |1-\omega|^2=1 . \] In particular $a\ne0$. Write $s=W_{p'}W_p$ with the complementary Weyl element commuting with $U_p$ and thus preserving every $U_p$-eigenspace. If the vacuum eigenvalue of $U_p$ were different from both eigenvalues on our subspace, the vacuum entries of both $s$-columns would vanish. It thus agrees with exactly one. On its eigenspace, the two columns are multiples of the same vector $W_{p'}v_X$ or $W_{p'}v_Y$. Their strictly positive vacuum entries give $d_Y/d_X=|a|/|b|$ or $|b|/|a|$. Consequently $a^2$ is rational. Thus $\omega+\omega^{-1}$ is a rational algebraic integer of absolute value at most two. A nontrivial prime-power root of unity with this property has order $2,3$ or $4$. At these orders $a^2$ is respectively $1/4,1/3,1/2$. The first two yield irrational vacuum ratios, and the last gives $d_X=d_Y$ as claimed. \end{proof} \begin{lemma}[Dyadic tests in an even square class]\label{rank:ordinary-dyadic-test} Let $D=qM^2$ in an ordinary non-pointed fusion-simple integral modular category, with $q$ square-free and $2\mid q$. A two-label odd-primary twist fiber in a union of full dyadic blocks has equal degrees. If $p\mid M$ is odd and $p\nmid q$, and $m_d$ counts nonunit labels of degree $d$ with $p\mid D/d^2$, then for $h=(p-1)/2$ one has \[ \sum_d \big(\lfloor m_d/h\rfloor+\lfloor m_d/(2h)\rfloor\big)\ge 6. \] An odd-length run at odd denominator needs a partner orbit at even denominator $c$ of length $4^{v_2(c)}$. \end{lemma} \begin{proof} The span of such a fiber is invariant under the dyadic congruence factor by Lemma~\ref{lem:local-block-invariance}, and $v_2(f_i)=1+2v_2(b_i)>0$. Lemma~\ref{lem:complement} excludes an invariant coordinate line, so Lemma~\ref{lem:two-label-local} applies. At an odd prime $p\mid D$ with $p\nmid q$, put $h=(p-1)/2$ and let $m_d$ count labels of each eligible nonunit degree ($p\mid D/d^2$). Put $P=\sum_d\lfloor m_d/h\rfloor$, $H=\sum_d\lfloor m_d/(2h)\rfloor$. We claim $P+H\ge6$. A phase of order $p^2$ or higher gives at least $ph$ distinct phases by square Galois covariance. Each nonempty fiber contains at least two labels, by invariance under the dyadic action and Lemma~\ref{lem:complement}. Translation by square units therefore gives $P\ge2p$. Otherwise both classes of nontrivial $p$-phases occur by Proposition~\ref{prop:two-packets}. In each class the fiber is either of size at least three, supplying three $h$-packets, or has two equal degrees, supplying two $h$-packets and one unit of $H$. The two classes are disjoint. At odd $b$, the dyadic defect is one and the character field is unramified at two by Theorem~\ref{por:th:exact-conductor}. Its block contains one other inertia orbit by Proposition~\ref{fam:M:2.10}. If that orbit is unramified, the two labels have equal degree by the two-label argument applied to the whole block. Otherwise the other orbit has even denominator $c$ and length $4^{v_2(c)}$ by the same proposition. An odd-length run at odd denominator needs at least one such partner orbit. A run of $n$ equal even denominators $c$ supplies at most $\lfloor n/4^{v_2(c)}\rfloor$. Both this bound and $P+H\ge6$ can be applied to a prefix by allowing the last run to grow and reserving the unused slots for any missing packets. \end{proof} \begin{proposition}[Degrees at the minimum fiber size]\label{prop:minimum-odd-fiber} In the setting of Lemma~\ref{lem:crossed-fiber-count}, if a complementary fiber has exactly $h_t+2$ labels, its degree multiset is $\{r,r,(2r)^{[h_t]}\}$ for some positive integer $r$, or, only at $t=3$, three copies of $r$. \end{proposition} \begin{proof} The fiber space is invariant under the odd local factor $G=\mathrm{SL}_2(\mathbb Z/t^a\mathbb Z)$, since its complementary primary twist operators commute with this factor. Use the ordinary congruence lift; in the reduced case restrict the odd factor to the positive fermion space as in Section~\ref{sec:blocks}. The determinant character of this $G$-representation cannot take the value $-1$: reduction from $\mathrm{SL}_2(\mathbb Z)$ is onto, its abelianization is generated by the upper unipotent, and the image of this generator in $G$ has odd order. Let $v$ generate the local units. Its quadratic character is negative. The torus $H_v$ acts on the fiber as minus a permutation, since its Galois sign on every integral vacuum coordinate $d_i/\sqrt\Delta$ is $(v/t)=-1$. In the reduced lift the two ambient representatives are integral, the normalization has the extra rational factor two, and their positive sums give the same signed action. Thus the number of cycles of this permutation must be even, by $\det(-P)=(-1)^{\#\text{cycles}(P)}$. Higher order is too large unless $t=3$ and the three labels form a single orbit of primitive ninth phases. Its single cycle is impossible. For $t\ge5$ there are exactly two inactive labels and $h_t$ active ones. The active set is one cycle under $v^2$ acting on twists, so the two inactive labels must be exchanged by $H_v$. They have the same degree $r$. The slice identity gives active degree $2r$. The same argument applies at three with two inactive labels; with one inactive and two active, the active labels must be exchanged, and the slice identity gives three equal degrees. \end{proof} For instance take an even-dimensional table, $u$ an odd divisor of positive even valuation, and another odd prime $t$ of odd valuation. If exactly $2(h_t+2)$ degree packets of size $h_u$ are available at $u$, all must be used. No higher $u$-order can occur: it would give at least $u h_u>2h_u$ distinct phases, each costing at least $h_t+2$ labels. Each nontrivial $u$-phase therefore has the minimum number $h_t+2$ of labels. Degrees are constant as multisets over each of the two $u$-square classes by inertia; the two class masses agree. The first pattern of Proposition~\ref{prop:minimum-odd-fiber} has mass $2tr^2$ and the exceptional ternary one has mass $3r^2$. When $t=3$ these masses cannot coincide across the two different patterns, since their ratio would be twice a rational square. Hence all nontrivial $u$-phases have the same multiset of degrees. The available packets themselves, counted once per $h_u$ labels, must number four at $r$ and $2h_t$ at $2r$, or six at $r$ for the equal-degree ternary alternative. \subsection{Simultaneous bounds on the missing denominators} \label{sec:joint-prime-capacity} We now combine several even-valuation primes. Work with ordinary integral modular data of dimension $D=qM^2$, where $q$ is square-free and the nonunit denominators $b_i=M/d_i$ are integers. Put $T=\{t:t\mid q,\ t\text{ odd}\}$, $a_t=(t+1)/2$, and $P=\prod_{t\in T}a_t$, with empty product one. Define \[ B(q)=\max\!\left(\{2^{\mathbf{1}_{2\mid q}}P\} \cup\{(a_t+1)P/a_t:t\in T\}\right). \] Every nonempty fiber obtained by prescribing odd-primary twist components outside $T$ contains at least $B(q)$ labels. Indeed, refine by the phases at $t\in T$. Each successive refinement has at least $a_t$ children by Lemma~\ref{lem:crossed-fiber-count}. Leaving $t$ to the last step gives at least $a_t+1$ labels in each of its parent fibers. Alternatively, after all refinements, each terminal fiber contains at least two labels if $2\mid q$, by Lemma~\ref{lem:dyadic-torus-fiber}. \begin{proposition}\label{prop:joint-prime-capacity} Let $S$ be any set of odd prime divisors of $D$ of even valuation. At least $B(q)\sum_{p\in S}(p-1)$ labels have twist order divisible by some $p\in S$. \end{proposition} \begin{proof} Project the twists to the components at $S$ and partition their values by the vector of exact orders. In a nonempty stratum let $A\subseteq S$ consist of the active coordinates. With all other components fixed, an active $p$-coordinate takes at least $p-1$ values of the prescribed order by Proposition~\ref{prop:two-packets}. This remains true if only the other $S$-coordinates are fixed: choose a full complementary twist occurring over them. Successively projecting away coordinates now bounds the number of distinct tuples in the stratum below by $\prod_{p\in A}(p-1)$. For nonempty $A$ this product is at least $\sum_{p\in A}(p-1)$, since each factor is at least two. Every $p\in S$ occurs in at least one active set by categorical Cauchy. Hence there are at least $\sum_{p\in S}(p-1)$ distinct nontrivial projected tuples, each supporting at least $B(q)$ labels by the fiber estimate. These tuples have disjoint fibers. \end{proof} For a nonempty prefix let $h$ be the unfilled slots, $L$ its lcm, $b_*$ its last denominator and $m_b$ its multiplicities. At an odd $p\mid L$ with $p\nmid q$, only runs with $p\mid b$ are eligible. Put $u_p=(p-1)/2$. Each closed run contributes whole $u_p$-packets, whereas every label already seen in the open run may count. The prefix capacity at $p$ and the total deficit are \[ C_p=u_p\!\sum_{b:\,p\mid b}\lfloor m_b/u_p\rfloor+ \mathbf{1}_{p\mid b_*}(m_{b_*}\bmod u_p),\qquad H=\!\sum_{\substack{p\mid L\\p\nmid 2q}}\max\{0,B(q)(p-1)-C_p\}. \] Apply Proposition~\ref{prop:joint-prime-capacity} to the set of primes with positive deficit. Its prefix union has at most the sum of their capacities, even if different primes use the same label. At most $h$ further labels occur in the suffix, so $H\le h$. An unseen odd $\ell\mid M$, $\ell\nmid qL$, has capacity zero. Adjoining it gives $H+B(q)(\ell-1)\le h$. This cutoff leaves two and all primes in $qL$ available (\RankFile{optimized/nonsquare/joint_capacity.hpp}). The trivial $S$-fiber adds $B(q)$ labels to the nontrivial ones, whence $r\ge B(q)(1+\sum_{p\in S}(p-1))$. At $q=6$ adding five and seven gives $4(1+4+6)=44$; at $q=15$ adding seven gives $9(1+6)=63$, with unrestricted positive even valuations. \subsection{Multiplicity at primes of defect one}\label{sec:defect} Here $D$ denotes the ordinary modular dimension unless specified otherwise. The defect of a label at $p$ is $v_p(f_X)$. \begin{proposition}[Simultaneous defect-one primes]\label{prop:multiple-defect-one}\label{prop:defect-one-cap} In a nontrivial ordinary or positive reduced simple perfect table, let $x>1$ have multiplicity $m$ and codegree $f=\Delta/x^2$. For $J=\{p\text{ odd prime}:v_p(f)=1\}$ one has \[ m\le f-1-\sum_{p\in J}\frac{p-1}{2}. \] \end{proposition} \begin{proof} At each $p\in J$ the $p$-part of the twist of a degree-$x$ label is trivial or has order $p$, by Theorem~\ref{thm:twistdiv}. For $S\subseteq J$, write $m_S$ for the number of such labels with order $p$ at every $p\in S$ and inactive at the other primes of $J$. Then \[ m_S\le C_S=\frac{f}{\prod_{p\in J}p}\prod_{p\in S}(p-1). \] In fact $v_p(\Delta)=2v_p(x)+1$ is odd. In a fixed prime-to-$p$ twist fiber, let $M_j$ be the squared mass of the labels of $p$-primary order $p^j$. By Proposition~\ref{prop:odd-slice}, $M_1=(p-1)M_0$. Thus $M_0$ and $M_1$ are at most $1/p$ and $(p-1)/p$ times the fiber mass, respectively, even in the presence of higher orders. Conditions already imposed at other primes give a union of such fibers. Imposing the conditions successively multiplies their upper bounds. Division by $x^2$ gives the displayed inequality. All $C_S$ are integers and their sum is $f$. In the empty pattern the vacuum gives strict inequality. For $S=\{p\}$ take the full principal $p$-block and the fiber where the complementary twist is trivial (up to fermion sign in the reduced table). The vacuum contributes to $M_0$, hence the same blockwise relation gives an order-$p$ label in this fiber. It lies outside the degree-$x$ class: square-fullness gives $v_p(\Delta)\ge3$, and principal labels have defect at least two by Corollary~\ref{fam:M:4.1}. This label lies in the singleton pattern, so $m_{\{p\}}1$ square-free and composite, and let $a$ be the least prime divisor of $M$. With no bounds on the exponents in $M$, \[ \operatorname{rk}(\mathcal C)> q+(a^2-1)\left(1+\sum_{\substack{p\mid q\\p\ \mathrm{odd}}} \frac{p-1}{2}\right). \] For $D=15M^2$, the following direct count gives rank at least $33$, and at least $55$ when $M$ is odd. In the latter case the enumeration of Theorem~\ref{mrk:thm:modular-ranks} strengthens the bound to $75$. \end{corollary} \begin{proof} Each degree divides $M$. If $m$ degrees equal $M$, all the others are at most $M/a$, and the unit is strictly smaller: the table is square-full and at least three distinct primes divide $M$. Thus $q0$. For $m=0$ the same upper bound on $m$ holds since $1+\sum_{p\mid q,\ p\ \mathrm{odd}}(p-1)/2\le q$. For $q=15$ this gives $r>27$ for even $M$. For odd $M$ it gives $r>47$, and $r\equiv D\equiv7\pmod8$ raises the lower bound to $55$. Take $M$ even and $r\le32$; then $30\mid M$. At least three nonunit degrees are five-free: two are required by Lemma~\ref{mrk:e:witness}, and two nonzero squares cannot sum to $-1$ modulo five. They are at most $M/5$. Write $b$ for the multiplicity at $M/2$. The following normalized square-mass maxima use $1/M^2\le1/900$. Nonunit tails outside the maxima and halves cost at most $1/9$ each. \[ \begin{array}{c|l} m\le10 & 10+18/4+3/25+1/900<15\\ m=11,\ b\le14 & 11+14/4+3/9+3/25+1/900<15. \end{array} \] The simultaneous defect-one bound gives $m\le11$; $m=11$ forces $b\le15$ by the vacuum. Thus $(m,b)=(11,15)$, leaving mass $1/4$. At $r\le31$ its upper bound is $1/9+3/25+1/900<1/4$. At $r=32$ there are five nonunit tails. If $4\mid M$, at least three are odd by reduction modulo four. Choose three odd and three five-free tails. Their overlap has size $j=1,2,3$ and they have denominators at least $4,5,20$ in the odd-only, five-free-only and common classes. The other $j-1$ tails cost at most $1/9$ each. But \[ \max_{1\le j\le3}\big(j/400+(3-j)/16+(3-j)/25+(j-1)/9\big)+1/900<1/4. \] If $2\parallel M$, the five tail squares sum to $(M/2)^2-1\equiv0\pmod8$. With $o$ odd tails this gives $o+4(5-o)\equiv0\pmod8$, hence $o=4$. Their denominators are $2\bmod4$ and larger than two, giving the final contradiction $4/36+1/9+1/900<1/4$. \end{proof} If $q$ is prime, degree $M$ is impossible by Lemma~\ref{small}. The same comparison, with $m=0$, therefore gives $r>qa^2$. \paragraph{Testing several defect-one degrees together.} For an odd prime $p\mid q$ put $w=(p-1)/2$. Select any set of \emph{closed} runs of nonunit denominators $b_j$ prime to $p$, with multiplicities $m_j$. A closed run is one whose multiplicity can no longer grow. There must be integers $0\le t_j\le m_j$ divisible by $w$ such that \[ \sum_j\frac{m_j-t_j}{b_j^2}<\frac qp,\qquad \sum_j\frac{t_j}{b_j^2}<\frac{(p-1)q}{p}. \] Indeed $t_j$ counts the labels of exact $p$-order in that run. No higher order is possible there. The slice identity of Proposition~\ref{prop:odd-slice} gives the two total mass upper bounds. Both inequalities are strict on the selected runs: the principal block lies outside all of them by Corollary~\ref{fam:M:4.1}, and contains the vacuum and positive exact-$p$ mass by the same slice identity. Scaling by $\lcm(b_j)^2$ gives a bounded integer sum with two strict endpoints. The program \RankFile{code/defect_mass.py} can apply it to a prefix or a complete profile. It uses at most the six smallest eligible runs, so discarding any further runs is conservative. \paragraph{Partition of the defect-one degrees.} \label{sec:defect-one-tiles} \begin{proposition}\label{prop:defect-one-tiles} In either integral categorical table, let $p$ be an odd prime with $v_p(\Delta)$ odd. The multiset of degrees $d$ with $v_p(\Delta/d^2)=1$ partitions into blocks of the following two forms, with exponents denoting multiplicities: \[ \{r^{[k^2]},(kr)^{[p-1]}\}\quad(k^2\mid p-1),\qquad \{r^{[2k^2]},(2kr)^{[(p-1)/2]}\}\quad(2k^2\mid p-1). \] For $p=3$ these become three copies of $r$, or two copies of $r$ and one of $2r$. \end{proposition} \begin{proof} By Proposition~\ref{lem:one-defect-block} a block containing a defect-one character consists of precisely two tame inertia orbits of defect one. The proof of Corollary~\ref{cor:defect-one-orbit-masses} applies in both tables: the odd-primary twist order divides the codegree and the two masses have ratio $1:(p-1)$ by Proposition~\ref{prop:odd-slice}. Thus one orbit has trivial primary twist and the other order $p$, with respective lengths $a,b\mid p-1$ and degrees $r,s$ satisfying $bs^2=(p-1)ar^2$. The square Galois action forces $(p-1)/2\mid b$. If $b=p-1$, then $a=(s/r)^2$ is an integer square. If $b=(p-1)/2$, then $2a=(s/r)^2$ is an even integer square. This gives the two lists. \end{proof} \begin{corollary}[Integer feasibility for defect-one degrees]\label{rank:defect-feasible} Let $m_d$ count the labels at each defect-one degree of Proposition~\ref{prop:defect-one-tiles}. Form the finite list $\mathcal T$ of its two allowed block multiplicity vectors whose support is among these degrees and whose entries do not exceed $m_d$. The required integer condition is \[ m_d=\sum_{T\in\mathcal T} n_T T_d\quad\text{for every such }d, \qquad n_T\in\mathbb Z_{\ge0}. \] For $p=3$ the two column types are $3[d]$ and $2[d]+[2d]$. At an incomplete prefix enforce the equalities on closed degrees and allow unused slots at compatible unprocessed degrees. \end{corollary} \begin{proof} Count the blocks of each permitted type. Conversely any integral solution partitions the multiplicity multiset into these types. This converse concerns only the multiset; the local action and the rest of the table impose further conditions. Projecting away unprocessed degrees only relaxes a necessary condition. \end{proof} The complement of the defect-one blocks has its own necessary signed-mass test when $v_p(D)\ge3$. For a nonunit degree $d$ occurring $m_d$ times there, let $t_d,u_d$ count its labels of primary twist order $p$ and of order at least $p^2$, respectively. In ordinary even dimension they are multiples of $h=(p-1)/2$ and $ph$, respectively. In odd dimension the multiples double because duality pairs labels of the same twist. Proposition~\ref{prop:odd-slice} on the whole set, less the same identity on the defect-one blocks, gives \[ \sum_d d^2\big[t_d-(p-1)(m_d-t_d-u_d)\big]=p-1. \] The right hand side is the omitted vacuum. On the principal block alone the same target must be attained by $\sum d^2[t_d-(p-1)i_d]$, where $i_d,t_d\ge0$, $i_d+t_d\le m_d$. Indeed Corollary~\ref{fam:M:4.1} removes all defect-one labels from this block, which is Galois- and duality-stable. The order-$p$ counts have the same divisibility, and the inactive counts are even in odd dimension. The implementation in \RankFile{code/odd_blocks.py} checks both integer conditions by suffix bounds and gcds, and the block partition by exact multiset subtraction. Each matching is a necessary condition. \paragraph{Exact prefix test for a family of odd primes.} Write $D=qM^2$ with $q$ square-free and put $z=M/d$ for a simple degree $d$. Let $p\mid q$ be an odd prime such that $8\nmid p-1$ and the odd part of $p-1$ is square-free. An individual label has $p$-defect one precisely when $p\nmid z$. Set $w=(p-1)/2$. The counts at these denominators must be a nonnegative sum of the two patterns \[ p\text{ labels at }z,\qquad w\text{ labels at }z\text{ and two at }2z . \] In the first block of Proposition~\ref{prop:defect-one-tiles}, $k=1$ or $2$. In the second block $k=1$. The case $k=2$ in the first list gives two copies of the second pattern, which proves the assertion. The system of Corollary~\ref{rank:defect-feasible} reduces to a recurrence on doubling chains. Process each completed run at $z$ in increasing order. Subtract from its multiplicity the counts already required by the run at $z/2$, leaving $v$. Negative $v$ fails. Set $y$ equal to the least nonnegative residue of $-2v$ modulo $p$. One requires $wy\le v$, and reserves $2y$ slots at $2z$. Indeed $v=px+wy$ and $w^{-1}\equiv-2\pmod p$. Choosing $y+kp$ in place of $y$ can be reversed by replacing $kp$ mixed tiles by $kw$ homogeneous tiles at $z$ and $2k$ at $2z$. Thus the least feasible choice suffices. A pending denominator skipped by the search, or more pending slots than remain, rules out that prefix. The primes through $31$ to which the test applies are $3,5,7,11,13,23,29,31$. When several primes are tested, a label can meet requirements at more than one prime. At a given denominator the maximum of their reservations therefore gives a valid joint lower bound. Summing these maxima over different denominators bounds the number of remaining slots. \paragraph{Principal witnesses on the degree list.} For distinct dimension primes $r,p$ with $r$ odd, and a dimension prime $t\ne r$ (odd in a reduced table), Lemma~\ref{ext:lem:two-free-crossing} and Corollary~\ref{fam:M:4.1} require nonunit degrees $x,y$ with $r\nmid x$, $t^2\mid\Delta/x^2$, $p\nmid y$ and $r^2\mid\Delta/y^2$. Their gcd $g$ satisfies $g\ge2$ and $r\le\max\{5,(g-1)g^2\}$. The test permits $p=t$ and $x=y$. It is implemented in \RankFile{code/spectrum_crossing.py}. \subsection{Mass and multiplicity of active degrees} For an odd prime $p$ of positive even valuation in the table dimension, Proposition~\ref{prop:two-packets} equates the squared masses of the two square classes at exact order $p$ in any full block and complementary fiber, including when a higher-order phase occurs. Set $h=(p-1)/2$, doubled when the table dimension is odd. At degree $d$ of multiplicity $m$, let $n=\lfloor m/h\rfloor$. If $p\nmid f_d$ there are no active labels. Otherwise let $k$ be the signed difference of the two numbers of exact-order-$p$ packets at this degree. Necessarily $|k|\le n$ and $\sum_d k_d d^2=0$. For fixed $k$, at most $h(n-\epsilon)$ labels are active, where $\epsilon=1$ if $n-|k|$ is odd and either $p^2\nmid f_d$ or $n-|k|1:\,p\mid f_d}h\lfloor m_d/h\rfloor d^2 \ \ge\ (p-1)D. \] \end{proposition} \begin{proof} Let $L_p$ be the total squared degree of the labels whose twist is $p$-active. Categorical Cauchy supplies such a label. A smallest nonunit simple generates the category, so Proposition~\ref{th:fs} gives $2pgL_p\ge(p-1)D$. Every active label satisfies $p\mid f_d$ by Lemma~\ref{lem:complement}. For odd $p$, fix a degree and an exact primary twist order $p^j$, $j\ge1$. The Galois square action on primitive roots has orbits of size $p^{j-1}(p-1)/2$ with constant multiplicity. In odd dimension every label in a nontrivial twist fiber pairs with a distinct dual of the same degree and twist. Thus the active count at each degree is a multiple of $h$. The vacuum is inactive. \end{proof} Here is a bound involving only a denominator threshold and a count. Take an odd $p\mid M$, $p\nmid q$. Suppose all nonunit degrees with $p\mid M/d$ have $M/d\ge B$. Suppose also that $K$ labels are known to be $p$-inactive. The number of $p$-active labels is a multiple of $h$, each of degree at most $M/B$. Therefore \begin{equation}\label{eq:activity-rank} R\ \ge\ K+h\left\lceil\frac{q(p-1)B^2}{2pgh}\right\rceil . \end{equation} Even before the type is known, $B=p$, $K=1$ and $g\le U=M/p^{v_p(M)}$ are valid. Indeed $p\mid D$, so reduction of $\sum d_i^2=D$ modulo $p$ forces a nonunit $p$-free degree, which divides $U$. In particular $R\ge1+h\lceil qp(p-1)/(2Uh)\rceil$ at $D=qU^2p^{2e}$ for every $e\ge1$ and $p\nmid qU$. This form bounds complete ranges of exponents at a fixed cofactor. \begin{corollary}\label{cor:saturated-twelve-prime} Let $D=3M^2$ and suppose ten labels have degree $M/2$. For every $p\ge5$ dividing $M$, equation~\eqref{eq:activity-rank} holds with $K=13$. In particular, throughout $D=2^{2a}3^{2b+1}5^{2c}$, $a,c\ge1$, if every nonunit $5$-eligible denominator $M/d$ is at least $B$, then \[ R\ \ge\ 13+2\left\lceil\frac{3B^2}{5g}\right\rceil . \] Here $5$-eligible means $5\mid M/d$. \end{corollary} \begin{proof} Degree $M$ is impossible by the ordinary small-codegree bound. In the three-defect-one partition (Proposition~\ref{prop:defect-one-tiles}), a copy of $M/2$ belongs either to an equal triple or to a block containing two copies of $M/4$ and one of $M/2$. Ten is not a multiple of three; the latter block must occur. The resulting twelve labels have codegrees $12$ or $48$, so their twists are $p$-inactive for $p\ge5$. The vacuum is a thirteenth such label. In the exceptional numerical case $M/4=1$ the mixed block would require two distinct invertibles, contrary to perfection. \end{proof} \paragraph{A divisor from the regular indicator.}\label{rank:regular-indicator-divisor} Suppose that $D=qM^2$ is the ordinary table dimension, $p$ is an odd prime dividing $M$ but not $q$, and there is no twist of $p$-primary order at least $p^2$. Let $C$ be the squared-degree mass on any one nontrivial $p$-primary phase. It is positive and independent of the chosen phase by Proposition~\ref{prop:two-packets}. Then \[ (qM/p)\mid C . \] Indeed, take the higher indicator of the regular object at the prime-to-$p$ part $k$ of the twist exponent. The ordinary indicator formula \cite[Theorem~7.5]{NSfs} and $\sum_X d_X N_{ij}^X=d_i d_j$ give the algebraic integer \[ \sum_X d_X\nu_k(X)=D^{-1}\left|\sum_i d_i^2\theta_i^k\right|^2 =(D-pC)^2/D . \] An integer $z$ with $D\mid z^2$ is divisible by $qM$. Since $qM\mid D$, the claim follows. The common phase mass is thereby constrained for every square-free kernel. In particular consider the family $D=qU^2p^{2e}$, $e\ge1$ and $p\nmid qU$. Put $\delta=2$ for odd $D$ and $1$ otherwise, $h=\delta(p-1)/2$. A degree $d>1$ can carry a nontrivial phase only if $p\mid f_d$ and $m_d\ge h$. If all such degrees with $p^2\mid f_d$ have multiplicity less than $ph$, the higher-order hypothesis follows from the inertia orbit bound. If just one eligible degree $d$ remains, the number of copies it requires is at least \[ \delta(p-1)\frac{qU p^{e-1}}{\gcd(qU p^{e-1},\delta d^2)} . \] Indeed $C=k'\delta d^2$ with integer $k'>0$, the same for all $p-1$ phases. With any number of eligible degrees, $C$ is a multiple of both $qM/p$ and $\delta\gcd_d(d^2)$. The total eligible mass, with multiplicities rounded down to multiples of $h$, must reach $p-1$ times the lcm of these integers. The implementation is \RankFile{code/regular_indicator.py}. \section{Character rows of small codegree}\label{sec:rows}\label{mrk:s:rows} Let $X$ be a nonunit in a nontrivial fusion-simple perfect integral modular category. Write $n=d_X$, $u_Y=S_{XY}/n$, $f=D/n^2$ and $c_Y=\operatorname{mean}|u_Y|^2$, where the mean is the normalized cyclotomic trace. The $(X,0)$ modular relation, its Galois conjugates and Cauchy--Schwarz give \begin{equation}\label{st} \sqrt f\le\frac1n\sum_Y d_Y\sqrt{c_Y},\qquad \sum_Y c_Y=f. \end{equation} On a nonzero entry $c_Y\ge(d_Y/\gcd(n,d_Y))^2$ by lcm integrality. We use $d_Y\ge6$ at nonunits (Lemma~\ref{lem:tables}) and $\sqrt D\ge5400$ (Corollary~\ref{cor:global-dimensions}). When $D=M^2$, write $b_Y=M/d_Y$ for the denominator, including $b_0=M$. \subsection{Trace and norm restrictions} \begin{lemma}[Prime codegrees]\label{small} No prime is a formal codegree of a nontrivial perfect integral modular category. \end{lemma} \begin{proof} By Lemma~\ref{lem:tables}, $D$ is square-full. If $D=pn^2$ and $f_X=p$, square-fullness implies $p\mid n$. All degrees divide $n$ and principal support and principal defect (Theorem~\ref{thm:principal-support} and Corollary~\ref{fam:M:4.1}) force the supported degrees to divide $n/p$. Now \eqref{st} gives $\sqrt p\le p^{-1}\sum c_Y=1$. \end{proof} \begin{lemma}[Integral trace vectors with a prescribed orbit]\label{mrk:l:trace-vector} Let $d_i$ be the degrees of a nontrivial simple perfect table with integral degrees, symmetric $S$ and ordinary or positive reduced orthogonality. Write $D=\sum_i d_i^2$ for its table dimension and $m_y$ for the multiplicity of the degree $y$. Fix a nonunit label $X$ of degree $x$ and put $f=D/x^2$. There is an integer $1\le h\le\min(m_x,f-1)$ and a family of integers $z_i$ satisfying \[ z_0=h,\qquad \sum_{i\ne0}d_i z_i=-h,\qquad \sum_{i\ne0}z_i^2=h(f-h). \] For $i\ne0$ with $\gcd(x,d_i)=1$ one has $z_i=0$. For every $i$, \[ \frac{d_i}{\gcd(x,d_i)} \frac{h}{\gcd(h,\operatorname{lcm}(1,\ldots,m_{d_i}))}\mid z_i . \] Among the $m_x$ labels of degree $x$, at least $h$ have a common coordinate $z_i$. In the ordinary case, if $h=1$ one also has $1+\sum_{i\ne0} d_i |z_i|\ge\lceil\sqrt D\rceil$. \end{lemma} \begin{proof} Let $K=\mathbb Q(\chi_X(i):i)$, with $\chi_X(i)=S_{Xi}/x$. This is an abelian extension, and its $h=[K:\mathbb Q]$ embeddings give $h$ distinct character rows. They have common degree $x$ and are mutually orthogonal with squared norm $f$. Set $z_i=\operatorname{Tr}_{K/\mathbb Q}\chi_X(i)$. These ordinary integers are the coordinates of the sum of the $h$ rows. Its squared norm is $hf$. It is orthogonal to the Perron character, and its vacuum coordinate is $h$. These facts give the equations and imply $h1/33$. We obtain $n=\operatorname{lcm}(k_1,k_2,k_3)\le4\cdot8\cdot32<2205$, a contradiction. \end{proof} \begin{proposition}[A rational row of prime-square codegree] \label{prop:rational-prime-square} In an ordinary fusion-simple perfect integral modular category, if the row of a nonunit simple $Z$ is rational and has codegree $p^2$ for a prime $p$, then $p\nmid d_Z$. In particular, if $D=M^2$ and the denominator $M/d=p$ occurs exactly once, then $v_p(M)=1$ and $M/p$ has at least three distinct prime divisors. \end{proposition} \begin{proof} Write $z=d_Z$ and $u_Y=\chi_Z(Y)\in\mathbb Z$. A nonzero $u_Y$ has nonzero trace. Thus if $p\mid z$, every nonunit degree hit by this row divides $z/p$, by Proposition~\ref{prop:quartic-prime-top-codegree}. The vacuum modular relation and the integer row norm give \[ p\le\frac1z+\frac1p\sum_{Y\ne0}|u_Y| \le\frac1z+\frac1p\sum_{Y\ne0}u_Y^2 =\frac1z+\frac{p^2-1}{p}

p$ by exclusion of prime-power degrees. The contradiction proves the first statement. A singleton degree gives a rational row by Galois covariance. The three-prime requirement then follows from Proposition~\ref{prop:quartic-prime-top-codegree} with $z/\gcd(z,p)=z$. \end{proof} \subsection{A saturated prime-power degree class} Here $R$ denotes the rank. Since the vacuum contributes positive mass, the multiplicity at a nonunit degree of codegree $f$ is strictly less than $f$. \begin{proposition}\label{prop:saturation} Let $n$ be the largest nonunit degree, of codegree $f\in\{8,16,25,27\}$. If $f(R-f)\ge25$ at rank $R$, its multiplicity is at most $f-2$. \end{proposition} \begin{proof} Suppose the multiplicity is $f-1$ and let $A$ be the sum of twists of those labels. The twists have prime-power order supported at the prime dividing $f$ and belong to $K=\QQ(\zeta_{4f})$ by Theorem~\ref{thm:twistdiv}. Reduction at that prime gives $A\equiv-1$. The remaining squared degrees sum to $n^2$. Every conjugate of the Gauss sum has absolute value $\sqrt D$, so $|\sigma A|\le 1+f/\sqrt D$. The normalized trace of $A\bar A$ lies in $[K:\QQ]^{-1}\ZZ$, and $[K:\QQ]\le40$ for the four listed codegrees. Its nonzero norm puts the mean at least one, and $(1+27/5400)^2<1+1/40$ makes it exactly one. Equality in arithmetic--geometric mean makes $A$ a root of unity. Put $\eta=-A$. Multiplying the Gauss sum by $\bar A$ and taking real parts and normalized trace gives $\sum_{Y\notin I} d_Y^2(1-\operatorname{mean}(\theta_Y/\eta)) \le\sqrt D$, where $I$ is the largest degree class. A nontrivial root of order $k$ has mean $\mu(k)/\varphi(k)\le1/2$. Choose an odd dimension prime $\ell$ outside $4f$; there are at least three dimension primes. The $\ell$-active squared mass is at most $2\sqrt D$, since $\eta$ is $\ell$-inactive. Proposition~\ref{th:fs} bounds the least nonunit degree $g$ below by $(\ell-1)\sqrt D/(4\ell)$. Now $n^2\ge1+(R-f)g^2$ is impossible when $f(R-f)\ge25$ and $\ell\ge5$, or when $f(R-f)\ge36$ and $\ell=3$. We can take $\ell\ge5$ except possibly at $f=25$. The sole additional case is $f=25,R=26$, where the vacuum and one nonunit $Y$ would give $n^2=1+d_Y^2$. \end{proof} \subsection{Orbits and multiplicity at codegree eight} \begin{lemma}[Common projection of orthogonal rows]\label{rank:common-projection} Let $u_1,\ldots,u_k$ be mutually orthogonal vectors of squared norm $f$. If an orthogonal projection $P$ satisfies $Pu_i=v$ for every $i$, then $k\|v\|^2\le f$. \end{lemma} \begin{proof} $k^2\|v\|^2=\|P\sum_i u_i\|^2\le\|\sum_i u_i\|^2=kf$. \end{proof} \begin{theorem}[Ordinary rows of codegree eight]\label{prop:eight}\label{lem:even-eight}\label{cor:eight-triple}\label{cor:halfdegree}\label{cor:extra-primes}\label{cor:odd-eight} Let $x$ be a degree of formal codegree eight in a nontrivial fusion-simple perfect integral modular category. It occurs at most four times. If $x$ is even, its multiplicity is even and there are three nonunit labels with degrees $u\le v\le w$ dividing $x/2$ and satisfying $|u+v-w|=1$. If $x$ is a largest degree, at least three odd primes divide $x$ and it occurs at most twice, exactly twice when even. \end{theorem} \begin{proof} \emph{Rational rows at even degree.} Suppose the row at $X$ is rational, and put $z_Y=\chi_X(Y)\in\mathbb Z$. For $f_X=8$ the principal-support theorem places every hit in the principal dyadic block. Since $x$ is even, the nonzero integer $\chi_Y(X)=(x/d_Y)z_Y$ is even at every hit. Put $a=x/2$. Orthogonality gives $\sum_{Y\ne0} z_Y^2=7$ and $\sum_{Y\ne0}d_Yz_Y=-1$. If $|z_Y|=1$ then $d_Y\mid a$, and if $|z_Y|=2$ then $d_Y\mid x$. There are either seven entries of absolute value one, or three such entries and an entry of absolute value two. In the latter case the degree at the double entry cannot be $x$: the three single entries could cancel at most $3x/2$, whereas $2x-1>3x/2$. This degree is therefore at most $x/2$, giving $\sqrt8\,x\le1+5x/2$, impossible for $x\ge6$. In the former case all seven degrees divide $a$ and are at least six, so $a\ge6$. The positive and negative degree sums differ by one. The smaller is at least $2\sqrt2\,a-1>5a/2$, by the modular relation. There must be three labels of one sign and four of the other. All three degrees in the smaller group by cardinality must equal $a$, since a proper divisor of $a$ is at most $a/2$. The four degrees in the other group sum to $3a\pm1$. Exactly two must equal $a$: one or fewer gives at most $5a/2$, and three or more gives at least $3a+6$. The two proper divisors $u,v\le a/2$ left over therefore sum to $a-1$. They cannot both be $a/2$, so one is at most $a/3$, forcing $a\le6$. This is impossible since $u,v$ are proper divisors of $a$ of size at least six. Thus no row of this even degree is rational. All row fields at codegree eight have conductor dividing $8$ by the conductor bound of Theorem~\ref{por:th:exact-conductor}. Their Galois orbit lengths are powers of two, hence all are even. The multiplicity of $x$ is their sum. \emph{The three trace coordinates.} For a row at $X$ of degree $x$ write $h=[K_X:\mathbb Q]$ and $a_Y=h^{-1}\operatorname{Tr}_{K_X/\mathbb Q}\chi_X(Y)$. The conductor divides $8$. Normalized traces of the integers of $\mathbb Q(\zeta_{8})$ are integral since powers of $\zeta_{8}$ have normalized traces $0$ or $\pm1$. Since $h<8$, the preceding rational-row argument gives $h=2$ or $4$. For $h=4$, the norm identity $\sum_{Y\ne0}a_Y^2=8/h-1=1$ leaves one entry $\pm1$, contrary to $\sum_{Y\ne0}d_Ya_Y=-1$. Thus $h=2$, and there are exactly three nonunit entries $\pm1$. Each of their degrees divides $x$: the normalized trace of $S_{XY}/\lcm(x,d_Y)$ is integral, and multiplying by $d_Y/\gcd(x,d_Y)$ gives $a_Y$. In an imaginary quadratic field the orbit is $X,X^*$ with equal twists. Average their two ordinary modular identities to obtain $\sqrt8\,x\le1+\sum_{a_Y\ne0,\,Y\ne0}d_Y$. At least one of the three degrees is a proper divisor of $x$ by their signed sum, so the right hand side is at most $1+5x/2$, a contradiction. The real quadratic subfield of $\mathbb Q(\zeta_{8})$ is $\mathbb Q(\sqrt2)$. In this field write $S_{XY}/\lcm(x,d_Y)=A+B\sqrt2$ with $A,B\in\mathbb Z$. If $v_2(d_Y)\ge v_2(x)$, the factor $x/\gcd(x,d_Y)$ is odd. Multiplication by it gives $\chi_Y(X)$, which reduces to zero above two at every hit by principality. Thus $A$ is even. Then $a_Y=(d_Y/\gcd(x,d_Y))A$ cannot be $\pm1$. Each of the three degrees therefore divides $x/2$, and their signed sum gives the stated equation. \emph{The uniform bound of four.} Let $I$ consist of the $m$ labels of degree $x$. We use the mass inequality of Lemma~\ref{mass}: \[ \sum_{X,Y\in I}|S_{XY}|/x\ \ge\ \sqrt8\,m(2m/8-1). \] Suppose $m\ge5$. Some $I$-row meets $I$. Every $I$-row is supported in the principal dyadic block $P$, where every degree divides $x$ by square divisibility and the principal defect bound Theorem~\ref{thm:principal-support} and Corollary~\ref{fam:M:4.1}. If $x$ is odd, principality and symmetry give support exactly $P$ for every $I$-row. A hit within $I$ then gives $I\subset P$. By the row norm $|P|\le8$. Apart from the vacuum and $I$, it contains at least two distinct proper degrees, since a single nonunit divisor of $x$ would divide every nonunit support degree, contradicting \eqref{mrk:e:orth}. Thus $m=5$, and the two proper degrees $y,z$ are distinct coprime divisors of $x$. All eight entries on the support are roots of unity by equality in the Galois mean norm bound. The entries at $y,z$ are signs, since each occurs once in the support, and the sum over $I$ is integral by Galois symmetry. As $1+y+z13$, so $k\ge5$. At least two other nonunit hits are required: any single mixed-prime degree $y\mid x$ has $\gcd(y,x/2)>1$, whereas the nonunit support degrees have gcd one. Thus at least seven nonunit positions share norm seven. All have Galois mean squared norm one, so are roots of unity. This is impossible in the real character field $\QQ(\sqrt2)$ of the chosen row, since its real roots of unity are rational signs. Finally the parity just proved gives even multiplicity. \emph{Prime divisors of a largest degree.} At codegree eight every supported degree divides $x$. If $x$ is odd, their nonunit gcd is one and each has at least two prime divisors, so $x$ needs three odd primes. If $x$ is even, the preceding proof in fact excludes any $x$-row missing all columns of degree $x$ (the contradiction after choosing such a row used only its vanishing there). Its three odd nonunit hits therefore exist by the uniform bound argument, and their signed sum is $-1$. The same gcd argument applies to these three odd divisors of $x$. \emph{Odd largest degree.} It remains to take $x$ odd. A row of degree $x$ cannot vanish on the set $I$ of all positions of that degree: all its supported degrees divide $x$, and if they were proper then \eqref{st} would give $\sqrt8\le1/x+7/3$. Hence $I\subset P$ by the principal-support argument above. Its common support $P$ consists precisely of the odd-degree labels. Suppose $m=|I|$ is three or four; the larger values have been excluded. The unit, $I$, and at least two proper nonunit positions lie in $P$, and $|P|\le8$. Modulo four the dimension shows $4\mid |P|$. Hence exactly eight entries occur and all are roots of unity of two-power order. Their sum $A$ over $I$ is an integer congruent to $m$ modulo two. For $m=4$, there are three proper odd nonunit positions. The estimate $|xA|\le1+3x/3<2x$ forces $A=0$. The signs at the three proper positions are fixed by their signed degree sum $-1$. Indeed each singleton degree carries a rational root of unity; if two degrees coincide their two roots have integral sum in $\{-2,0,2\}$, because a sum $\pm1$ would require a third root in their ratio. A zero pair sum is impossible already at the remaining nonunit degree. If all three degrees coincided, their common degree would divide one. The four rows consequently agree on four signed entries including the vacuum, contradicting Lemma~\ref{rank:common-projection} since $4\cdot4>8$. For $m=3$, the four proper nonunit degrees $y_i$ have sum at least $x-1$ since $A$ is odd. Their gcd is one as they divide $x$. The integers $z_i=x/y_i$ in increasing order are odd, at least three, have lcm $x$, and satisfy $\sum_i 1/z_i\ge 1-1/x$. Here $x>105$ by the dimension bound. The first two denominators must be three since $1/3+3/5<1-1/105$. Three copies of three are impossible: the remaining mixed-prime divisor $y$ of $x$ has a prime in common with $x/3$. Thus $z_3=5$, since $2/3+2/7<1-1/105$, and $z_4\le7$, since $2/3+1/5+1/9<1-1/105$. The lcm is at most $105$, a contradiction. \end{proof} \subsection{The principal ternary block at codegree nine} \begin{theorem}[Exclusion of codegree nine]\label{thm:no-nine}\label{prop:nine}\label{prop:nine-single}\label{prop:nine-double}\label{cor:nine-two} A nontrivial fusion-simple perfect integral modular category has no label of formal codegree nine. \end{theorem} \begin{proof} Write $D=9n^2$, $I=\{X:d_X=n\}$, $m=|I|>0$, and let $P$ be the principal $3$-block. Every $I$-row is supported in $P$, whose degrees divide $n$, by Theorem~\ref{thm:principal-support} and Corollary~\ref{fam:M:4.1}. Here $n\ge1800$ by Corollary~\ref{cor:global-dimensions}. Suppose first that $3\mid n$. For $X\in I$ put $u_Y=\chi_X(Y)$. At a nonzero entry with $v_3(d_Y)=v_3(n)$, symmetry and principality give $u_Y=(d_Y/n)\chi_Y(X)\equiv0$ above three. Its field lies in $\QQ(\zeta_9)$ by Theorem~\ref{por:th:exact-conductor}. A wildly ramified entry would cost at least nine in the row norm over its conjugate columns by Theorem~\ref{thm:cyclotomic-residue-energy}; the vacuum already costs one. Hence $u_Y\in\QQ(\zeta_3)$. Its trace is zero by Proposition~\ref{prop:quartic-prime-top-codegree}, since $d_Y\nmid n/3$. The two terms for $Y,Y^*$ cancel in the twisted dimension sum, since $u_Y$ is purely imaginary and duality preserves twists. This holds at every embedding. Each remaining nonunit degree on the support divides $n/3$. Its mean squared entry $c_Y$ is at least one when nonzero. The vacuum modular relation gives $3\le 1/n+\sum_{Y\ne0}\!c_Y/3=1/n+8/3$, a contradiction. We now have $3\nmid n$. For any $X\in I$ and $Y\in P$, $\chi_Y(X)\equiv n\ne0$ above three, so the $X$-row has support exactly $P$. Thus $|P|\le9$. We claim $P$ meets $I$. Otherwise all nonunit degrees in $P$ are proper divisors of $n$ with gcd one. At most seven of the eight available positions have degree $n/2$; all other proper degrees are at most $n/4$, since $3\nmid n$. The mass of $P$ is consequently at most $1+7(n/2)^2+(n/4)^2<2n^2$, whereas Lemma~\ref{mass} makes it at least $mn^2$. It remains to exclude $m=1$ with $P\cap I=\varnothing$. Its unique row is rational, so the nonunit squared norm eight allows eight signs, one entry of absolute value two and four signs, or two entries of absolute value two. The last option is impossible: the two nonunit degrees would have gcd one by orthogonality, but every two in $P$ have nonzero crossing by principality and hence share a prime by Lemma~\ref{lem:tables}. In the middle case all five proper degrees must equal $n/2$ to meet the modular bound $3n-1\le \sum_{Y\ne0}d_Y|u_Y|$: the first smaller degree loses at least $n/4$. This contradicts their gcd. In the first case at least four degrees must equal $n/2$, since $3n/2+5n/4<3n-1$. Their signs agree by $b_Yu_Y\equiv n\pmod3$, where $b_Y=n/d_Y$. The other at most four degrees, each at most $n/4$, cannot cancel them to within one. An $I$-label in $P$ sees every $3$-free degree by principality. Conversely all degrees in $P$ divide $n$. Thus $P$ is the set of $3$-free labels and its twists are pure powers of three, by Lemma~\ref{lem:complement} applied to an $I$-row. Every $3$-active label has $3$-free degree since $v_3(D)=2$, so three is an isolated singleton. This contradicts Lemma~\ref{new:trace}, which excludes an isolated singleton at three. Its proof uses only coprime vanishing, the principal-block congruences and the singleton column identities, all available for the present fusion-simple perfect table. \end{proof} \subsection{Integral traces in the principal dyadic block}\label{mrk:pcover} Assume $D=M^2$, let $n=M/a$ be the largest simple degree, and suppose $a>1$ is a power of two. Write $f=a^2$ and $m$ for the multiplicity of $n$. Theorem~\ref{thm:principal-support} and Corollary~\ref{fam:M:4.1} put the support of each degree-$n$ row in the principal dyadic block $P$, where $4\mid f_Y$. Thus a hit of degree $d=M/b$ requires $2\mid b$. Its nonunit support has gcd one by orthogonality to the vacuum. Consequently the lcm of the even nonunit denominators is $M$. If $2m>f$, the degree-$n$ square submatrix contains a nonzero entry by Lemma~\ref{mass}. We can then choose its row $X$ in $P$, by symmetry and the support theorem. Otherwise choose any row $X$ of degree $n$. For each nonunit degree $d=M/b$ with even $b$, put \[ t=d/\gcd(n,d)=a/\gcd(a,b),\qquad \alpha_Y=\chi_X(Y)/t=S_{XY}/\operatorname{lcm}(n,d) \quad(d_Y=d). \] Lcm integrality and the character-field conductor bound (Corollary~\ref{cor:field-conductor}) place $\alpha_Y$ in the integers of a cyclotomic field $K=\mathbb Q(\zeta_{2^j})$. Enlarge $K$ to a common such field. The standard integral basis $1,\zeta,\ldots,\zeta^{[K:\mathbb Q]-1}$ is orthonormal for normalized trace of $u\bar v$. In particular the normalized trace of $\alpha_Y$ is its integer constant coefficient $k_Y$, and the mean $C_Y$ of $|\alpha_Y|^2$ is the sum of the squares of its integer coefficients. Reduction at the unique prime above two sends $\zeta$ to $1$. There are two circumstances prescribing the residue $\delta\in\{0,1\}$ of every $\alpha_Y$ at a given degree. If $n$ is even and $n/\gcd(n,d)$ is odd, then $\delta=0$. In fact $\alpha_Y=0$ off $P$, whereas on $P$ symmetry gives $(n/\gcd(n,d))\alpha_Y=\chi_Y(X)\equiv n\equiv0$. If the selected $X$ is known to be in $P$ and $t$ is odd, then $\delta=d\bmod2$ because $t\alpha_Y=\chi_X(Y)\equiv d$. In all other cases the residue may be left unspecified. Let $s$ be the multiplicity of $d$, $T=\sum_{d_Y=d}k_Y$, $B=\sum_{d_Y=d}C_Y$. The algebraic integer $\sum_{d_Y=d}\alpha_Y$ is rational by the Galois action on columns, so it equals $T$. If $\delta$ is prescribed then \[ T\equiv B\equiv s\delta\pmod2,\qquad C_Y\ge F(k_Y),\qquad F(k)=k^2+\mathbf{1}_{k\not\equiv\delta\pmod2}. \] Indeed the nonconstant coefficients must include an odd one when the constant coefficient has the wrong parity. Without prescribed residue put $F(k)=k^2$ and omit both congruences. Write $|T|=sj+r$ with $0\le rf/2$ in a nontrivial fusion-simple perfect integral modular category. Every label of degree prime to $p$ belongs to the principal $p$-block $P$ and has $p$-power twist order. Put $k=2m-f$. At least $k$ of the degree-$x$ labels belong to $P$. If $p\mid x$ and $p\nmid y$, then a label of degree $y$ satisfies \[ \sum_{Z:p\nmid d_Z}\!\big(d_Z/\gcd(d_Z,y)\big)^2 +k\big(x/\gcd(x,y)\big)^2\le D/y^2. \] For each other prime $\ell\mid D$, let $h$ have the value from Proposition~\ref{prop:activity-capacity}, let $g$ be the least nonunit degree and $m_d$ the multiplicity of $d$. Then \[ 2\ell g\!\sum_{\substack{p\mid d\\ \ell\mid D/d^2}} h\lfloor m_d/h\rfloor d^2\ \ge\ (\ell-1)D. \] \end{proposition} \begin{proof} Every row at $x$ is supported in $P$ by Theorem~\ref{thm:principal-support}. Its degree class $I$ has mass $m/f$ in $D$-normalized squared degrees, so $P$ has mass at least $m/f$ by Lemma~\ref{mass}. Thus $I\cap P$ has mass at least $2m/f-1$ and contains at least $k$ labels. Choose $X\in I\cap P$. For $p\nmid d_Y$, principality gives $\chi_X(Y)\equiv d_Y\ne0$ at the chosen prime over $p$. Thus $Y$ belongs to the support of the $X$-row and lies in $P$. For $\ell\ne p$ we have $\ell\nmid f_X$, so Theorem~\ref{thm:block} forces $S_{XY}=0$ if $Y$ is $\ell$-active. Consequently the twist of $Y$ has $p$-power order. Its row has nonzero entries on all labels of degree prime to $p$ by principality, and on each of the $k$ chosen degree-$x$ columns by symmetry. The sets are disjoint if $p\mid x$. Applying the Galois-mean bound to the nonzero lcm-scaled entries gives the norm inequality. Finally every $\ell$-active label has degree divisible by $p$. Apply the packet and mass argument of Proposition~\ref{prop:activity-capacity} on these degrees. \end{proof} \begin{theorem}\label{thm:odd-top-sixteen} An odd degree $x$ of formal codegree $16$ has multiplicity at most eight in a nontrivial fusion-simple perfect integral modular category. \end{theorem} \begin{proof} Suppose $m\ge9$. Every degree-$x$ row is supported exactly on $P$: for $Y\in P$, the reduction $\chi_Y(X)\equiv x$ is nonzero, and symmetry gives $S_{XY}\ne0$. By Proposition~\ref{prop:dyadic-concentrated} all odd labels lie in $P$. In fact $P$ consists entirely of odd labels. For an even degree $d_Y$ in $P$, every $X$-row at degree $x$ would have mean $|\chi_X(Y)|^2\ge(d_Y/\gcd(d_Y,x))^2\ge4$. The normalized column at $Y$ has norm one, whereas its $m$ entries on those rows would contribute a mean at least $4m/16>1$. Thus every member of $P$ has odd degree and divides $x$, since $D=16x^2$. Any two nonunit degrees there have a common prime: principality requires their entry to be nonzero and simplicity excludes a nonzero crossing at coprime degrees. The support degrees other than one have gcd one by dimension orthogonality. Thus at least three distinct proper nonunit divisors of $x$ occur. If $o=|P|$, the row norm gives $o\le16$, and $4\mid o$ since $4\mid D$ and all degrees outside $P$ are even. For $m\ge9$, the vacuum, the $m$ degree-$x$ positions and the three proper degrees require $o\ge13$. Thus $o=16$. Every entry of each $X$-row on $P$ has Galois-mean squared modulus one, so is a root of unity. In particular $m\le12$. Put $A=\sum_{Z:d_Z=x}\chi_X(Z)$. It is an integer, since the dimension class is Galois-stable and Galois action on entries can be taken on the column labels (the dimensions are rational and preserved). Each summand is a dyadic root of unity by the individual conductor bound. Its dyadic reduction is therefore one, so $A\equiv m\pmod2$. If $m=12$, the three proper degrees occur once each. They are odd proper divisors at most $x/3$. Dimension orthogonality gives $|A|\le1+1/x<2$, so $A=0$. At each singleton degree the root entry is rational, hence a sign. The three signs are fixed across all degree-$x$ rows by their weighted sum $-1$: two different choices would give a zero sum on a nonempty proper subset of the three distinct degrees, which is impossible. The rows therefore agree on the vacuum and three signs. Lemma~\ref{rank:common-projection} gives $4m\le16$, a contradiction. If $m=11$, the four proper degrees have sum at least $x-1$, since $A$ is an odd integer. Write them as $x/a_i$ where the $a_i$ are odd and at least three; their least common multiple is $x$ by the gcd-one property, and at least three are distinct. At least two $a_i$ are three since $1/3+3/5<1-1/x$. There are exactly two threes by distinctness. The two remaining distinct cofactors must be five and seven: $2/3+2/7<1-1/x$ and $2/3+1/5+1/9<1-1/x$. All these bounds hold since the dimension threshold gives $x\ge5400/4=1350$. The four cofactors have lcm $105$, a contradiction. If $m=10$, the five proper degrees are at most $x/3$, so $|A|\le 5/3+1/x<2$ and $A=0$ by parity. On each proper degree class, the sum of its entries is an integer of absolute value at most its multiplicity and congruent to that multiplicity modulo two. It can therefore be represented by a sum of formal signs, one for each position. Together they satisfy $\sum_{j=1}^5 \varepsilon_j y_j=-1$, where the $y_j$ are the five proper odd degrees. Any two sign assignments solving this equation have identical sign totals on each degree class: the subset on which they differ has weighted signed sum zero and even size. If its size is four, the remaining entry would have signed degree $-1$. If its size is two, the two flipped degrees are equal and their signs opposite. At least three distinct degrees occur among the five, so one of them is a singleton. Its coordinate is a fixed sign in all ten rows. The rows agree also at the vacuum. Lemma~\ref{rank:common-projection} on those two coordinates gives $2m\le16$. If $m=9$, the six proper degrees are at most $x/3$. The odd integer $A$ satisfies $|A|\le 2+1/x$, so $A=\pm1$. For a proper degree $d$ write $n_d$ for its multiplicity and $B_d$ for its class sum. As above the $B_d$ are realized by signs, with $1+Ax+\sum_d d B_d=0$. At fixed $A$ any two sign solutions differ on an even number of positions. Flipping two just exchanges signs at equal degrees. Flipping four leaves two proper degrees and $x$ with signed sum $-1$. This is impossible, since the two proper degrees share a prime dividing $x$. Flipping all six would give $Ax=-1$. Thus for each $A$ all $B_d$ are fixed across the corresponding rows. At least three distinct $B_d$ must be nonzero, by the same common-factor argument. If $\sum_d|B_d|\ge4$, Cauchy--Schwarz gives $\sum_d B_d^2/n_d\ge16/6$. Otherwise precisely three $B_d$ are $\pm1$. Their three multiplicities are odd and sum to at most five, giving $\sum_d B_d^2/n_d\ge1+1+1/3=7/3$. Including the vacuum, the common orthogonal projection of the rows with fixed $A$ onto the vacuum and the constant vectors on proper degree classes has squared norm at least $10/3$. Since the rows have squared norm sixteen, Lemma~\ref{rank:common-projection} allows at most four such rows for each sign of $A$. This excludes nine rows. \end{proof} \subsection{Fourteen labels of formal codegree sixteen}\label{sec:dyadic-zeros} Throughout this subsection $s=S/M$ denotes the normalized modular matrix. \paragraph{A zero when the denominator has exactly one factor two.} Suppose $8\mid M$. If $X$ has denominator four and $Y$ has denominator $b\equiv2\pmod4$, then $S_{XY}=0$. Indeed, otherwise $Y$ belongs to the principal dyadic block by the prime-power support theorem. Set $n=M/4$ and $\alpha=2s_{XY}$. This is an algebraic integer because $\mathrm{lcm}(n,M/b)=M/2$. The common conductor bound for the lcm-normalized entry puts $\alpha$ in $\mathbb Q(\zeta_8)$. Principality of $Y$ yields $\chi_Y(X)=(b/2)\alpha\equiv n\equiv0$ above two. The integer coefficients of $\alpha$ in the standard dyadic power basis thus have even sum. Their squared sum is the normalized trace of $|\alpha|^2$, at least two if $\alpha\ne0$. If $\alpha$ is rational it is even, so its entry would use at least one unit of squared norm. Otherwise its distinct Galois conjugates occur in the same row at distinct labels of degree $d_Y$ since $\sigma(s_{XY})=s_{X,\sigma Y}$ in integral square dimension. Their total squared norm is at least $[\mathbb Q(\alpha):\mathbb Q]\operatorname{mean}(|\alpha|^2)/4 \ge1$. Both bounds contradict the positive vacuum entry. Odd denominators already vanish on this row by the dyadic principal valuation bound. Since the nonunit degrees on a nonunit row have gcd one, the lcm of the nonunit denominators divisible by four must therefore equal $M$. In particular, if a prime power $p^a\Vert M$ is absent from the lcm of these denominators on a prefix, at least one later nonunit denominator is divisible by $\mathrm{lcm}(4,p^a)$. If $n$ is odd and the denominator four occurs at least four times, the same zero and lcm assertions hold. Indeed every such row has support exactly the principal dyadic block $P$: for $Y\in P$, $\chi_Y(X)\equiv n\not\equiv0$ above two. Consequently a nonzero column of denominator $b\equiv2\pmod4$ would meet every one of these rows. Each nonzero $2s_{XY}$ has mean squared modulus at least one, and the squared column norm would be exhausted before counting the vacuum. \begin{proposition}[Fourteen dyadic rows and their principal block] \label{prop:fourteen-principal} If the denominator four occurs exactly fourteen times, at least one of the corresponding labels is outside $P$. In particular \[ \sum_{\substack{4\mid b_Y\\ b_Y\ne4}}\frac1{b_Y^2}>\frac1{16}, \qquad \sum_{4\nmid b_Y}\frac1{b_Y^2}<\frac1{16}. \] The sums include the vacuum when eligible. Three denominators six are therefore impossible, in every rank. \end{proposition} \begin{proof} Let $I$ be the class of fourteen labels and suppose $I\subset P$. For a label $Y$ put $w_Y=\sum_{X\in I}\chi_X(Y)$. It is a rational integer by Galois stability of the degree class, and it is even since every summand has the principal residue $d_Y$ above two. In particular $w_0=14$, and row orthogonality against the vacuum gives \[ \sum_{Y\ne0} d_Y(w_Y/2)=-7 . \] We show that $w_Y=0$ at every nonunit of odd degree. For this $Y$ each $\chi_X(Y)$ is a dyadic integer with nonzero residue. In the standard power basis of a two-power cyclotomic field, its normalized mean squared modulus is an odd positive integer: the coefficients have odd sum, and their squares have the same parity. Unitarity, averaged over embeddings, gives \[ \sum_{X\in I}\operatorname{mean}|\chi_X(Y)|^2<16, \] since $s_{XY}=\chi_X(Y)/4$ and the additional vacuum coordinate is nonzero. Thus all fourteen means are one. In the orthonormal integral basis this means every $\chi_X(Y)$ is a signed basis root of unity, and so has absolute value one at the given embedding. The sections of columns $Y$ and $0$ on $I^c$ therefore both have squared norm $2/16$. Orthogonality of the columns gives $|w_Y|/16\le2/16$. If $|w_Y|=2$, the two restricted columns are proportional with scalar of absolute value one. Their vacuum entries have different absolute values, $d_Y/M$ and $1/M$. Thus $w_Y=0$, since it is even. The displayed integer identity is now impossible modulo two. This proves $I\not\subset P$. Choose $X\in I\setminus P$. It misses the whole of $I$ by symmetry and prime-power support, and its remaining hits have denominators divisible by four by the preceding zero. Write $H$ for these potential remaining columns, including the vacuum. The vacuum modular relation gives \[ M/4=\big|\sum_{Y\in H}s_{XY}d_Y\theta_Y\big| \le \big(\sum_{Y\in H}d_Y^2\big)^{1/2}. \] Equality would imply that $|s_{XY}|$ on $H$ is proportional to $d_Y$ with factor $4/M$, since the row has norm one. At the vacuum its entry is $1/4$, which would give $M=16$, excluded by Corollary~\ref{cor:global-dimensions}. Thus the first mass bound is strict. The fourteen labels use $14/16$ of the mass, so the complementary bound follows. In particular, either two denominators five or three denominators six exceed this complementary mass bound. \end{proof} \begin{corollary}[The exceptional dyadic row in rank at most 31] \label{cor:fourteen-nine-hits} At fourteen denominators four, $8\mid M$ and exactly one label $Z$ in the class lies outside $P$. Its row is rational. If the rank is at most $31$, the $Z$-row has nine nonunit hits, the primes of $M$ are $\{2,3,5\}$, at least five denominators eight occur, and at most one denominator six occurs. \end{corollary} \begin{proof} If $n=M/4$ is odd, all fourteen rows have exact support $P$ and $P$ meets their class by the mass-overlap inequality. Symmetry would put all fourteen in $P$. Thus $n$ is even. Let $E=I\setminus P$, which is nonempty by the proposition. All its rows miss $I$ and are supported on pure dyadic twists. The squared norm of the projection of $sT(d_Y)$ onto $E$ is $|E|n^2$ by the vacuum modular relation. Its input can be restricted to the pure dyadic labels outside $I$, of total squared mass strictly less than $2n^2$: some odd dimension prime has a nontrivial twist component by categorical Cauchy. Thus $|E|=1$. This singleton is Galois-stable since the dyadic principal block is stable, so $u_Y=\chi_Z(Y)\in\mathbb Z$. Its nonunit support consists of degrees $d_Y=n/h_Y$ for integers $h_Y\ge2$, only on pure dyadic twists. The row and modular identities give \[ \sum_{Y\ne0}u_Y^2=15,\quad \sum_{Y\ne0}u_Y/h_Y=-1/n,\quad \sum_{Y\ne0}|u_Y|/h_Y\ge4-1/n , \] where sums omit zero entries. For the last inequality the vacuum contribution has absolute value one. Since each hit $Y\in P$, $h_Yu_Y=\chi_Y(Z)\equiv n\equiv0$ above two: if $u_Y$ is odd then $h_Y$ is even. There are at least two odd dimension primes $p1$. \end{proof} \begin{theorem}\label{thm:no-fourteen} At rank at most $31$ the denominator four has multiplicity at most thirteen in a nontrivial fusion-simple perfect integral modular category of square dimension. \end{theorem} \begin{proof} Sixteen such labels would exhaust the dimension before the vacuum. Fifteen would leave the vacuum and at least six labels with a nontrivial odd-primary twist: there are at least two odd dimension primes by the two-prime WGT theorem, and Proposition~\ref{prop:joint-prime-capacity} gives the six labels outside the dyadic class. Hence $R\ge22$ and $16(R-16)\ge25$, so Proposition~\ref{prop:saturation} excludes fifteen at every rank. Suppose the multiplicity is fourteen. Let $Z$ be the unique label of that degree outside the principal dyadic block as in Corollary~\ref{cor:fourteen-nine-hits}, and let $B$ be its full dyadic block. The thirteen remaining labels of that degree, the nine nonunit hits of the $Z$-row, and the vacuum are distinct, lie in the principal block and have pure dyadic twists. At least six additional labels have nontrivial odd-primary twists by the two-odd-prime phase bound in the corollary. Thus \[ \#\{Y\in B:\theta_{Y,2'}=1\} \le31-13-9-1-6=2 . \] This set contains $Z$. Its coordinate span is invariant under the dyadic local factor, since the full block is invariant (Lemma~\ref{lem:local-block-invariance}), and the complementary upper unipotent commutes with that factor. The local Weyl cannot preserve the line of $Z$ because $2\mid f_Z$ (Lemma~\ref{lem:complement}). Hence the set has exactly two labels. Lemma~\ref{lem:two-label-local} makes their degrees equal, contrary to uniqueness of the degree-$M/4$ label outside the principal block. \end{proof} \subsection{Square dimension with denominators four, five and seven} \label{mrk:s:neighbouring-primes} Keep $D=M^2$ and $b_Y=M/d_Y$. Two restrictions will be used repeatedly. The field of a prime-power degree class is controlled by the square Galois action on twists, and every nonzero entry in a prime-square row is supported on a pure primary twist. \paragraph{Two prime-power degree classes in square dimension.}\label{rank:gauss-nearfull} Write $D=M^2$. Suppose $i$ labels of degree $M/4$ and $j$ of degree $M/5$ comprise their respective full degree classes. Put $t=400-25i-16j$, and let $A,B$ be their respective sums of twists. When $1\le t\le17$ both $A,B$ are rational integers. They satisfy \[ -i\le A\le i,\quad A\equiv i\pmod2,\qquad -j/4\le B\le j,\quad B\equiv j\pmod5,\qquad |25A+16B|1}m_d/(M/d)^2= 1-1/M^2$ by $3600$ is impossible since $0<3600/M^2<1$. This proves strictness. For example, immediate exclusions of exact multiplicity pairs are \[ (1,23),\ (2,21),\ (3,20),\ (5,17),\ (7,14),\ (10,9),\ (12,6),\ (14,3),\ (15,1). \] The pairs $(12,6)$ and $(10,9)$ cannot be increased in either coordinate without exceeding total dimension, so their two-prefix tests apply whenever at least the displayed multiplicities occur. For $(11,7)$ the only permitted pair is $(A,B)=(-1,2)$; for $(13,4)$ it is $(1,-1)$. \paragraph{Trace and isolation at a prime denominator.}\label{rank:prime-denominator-trace} Let $p$ be an odd prime, $y=M/p$, $J=\{X:b_X=p\}$, and $T=\{X\in J:\theta_X=1\}$. Set $t=|T|$ and $w_Z=\sum_{X\in T}\chi_X(Z)$. The twists in $J$ are pure $p$-power twists by Lemma~\ref{lem:complement}. Odd-primary twists transform by squares under the modular Galois permutation, so $T$ is Galois-stable, and all $w_Z$ are rational integers. Each $J$-row vanishes outside the principal $p$-block $P$ (Theorem~\ref{thm:principal-support}); this block is supported on $p\mid b_Z$ (Corollary~\ref{fam:M:4.1}). Every label hit by one of these rows has pure $p$-power twist, by Lemma~\ref{lem:complement}(1) at the other primes. For $t>0$, orthogonality and the vacuum modular relation yield \[ \sum d_Zw_Z=0,\qquad \sum d_Z\theta_Zw_Z=tM\xi, \qquad \sum|w_Z|^2=t p^2. \] Here $\xi$ is the multiplicative central charge. Since $M$ is an integer and the middle sum lies in a $p$-power cyclotomic field, $\xi\in\mu_8$ forces $\xi=\pm1$. The mean real part of a $p$-power root of unity lies in $[-1/(p-1),1]$. The positive and negative parts of $\sum d_Z w_Z$ have equal mass. Averaging the middle identity and applying Cauchy--Schwarz gives \begin{equation}\label{eq:septic-trace} \sum_Z d_Z |w_Z|\ \ge\ 2(p-1)tM/p,\qquad \sum_{w_Z\ne0} b_Z^{-2}\ \ge\ 4(p-1)^2t/p^4. \end{equation} If $p\mid y$, this support misses $J$: Proposition~\ref{prop:quartic-prime-top-codegree} gives zero absolute trace of $\chi_X(Z)$ for $X,Z\in J$, since $y\nmid y/p$. The sum over each complete Galois orbit in $T$ is therefore zero there. If $p\nmid y$, every $J$-row has support exactly $P$ by symmetry and the principal congruence Theorem~\ref{thm:principal-support}. If $P$ meets $J$, it contains $J$ by this same symmetry. At such a principal row the congruence ensures a hit at every $p$-free degree, all of whose twists must therefore be pure $p$-power. Since $v_p(D)=2$, every $p$-active label has $p$-free degree (Lemma~\ref{lem:complement}). Thus $p$ is isolated and Lemma~\ref{lem:complement}(2) puts every $p$-active label in $J$. In particular, if $j=|J|1$ labels. Its exact support $P$ is disjoint from $J$ by the preceding argument, and every nonunit in $P$ has degree a proper divisor of $y$ by Proposition~\ref{prop:quartic-prime-top-codegree}. Equation~\eqref{eq:septic-trace} gives $\sum_{P} b_Y^{-2}\ge4(p-1)^2/p^4$. At $p=5$, fourteen fours would leave only $1-14/16-1/25<64/625$ for $P$. For both $p=5$ and $7$ we can count its labels. Put $H=|P|-1$, $W=\sum_{P\setminus\{0\}}|w_Y|$. The proper hit degrees are at most $y/2$, at least one at most $y/3$ by their gcd one. Equation~\eqref{eq:septic-trace} gives $2(p-1)y\le1+Wy/2-y/6$. As $y\ge5400/p>6$, and $W\equiv\sum_{Y\ne0}w_Y^2=p^2-1\pmod2$, one has $W\ge4(p-1)+2$. Now $k^2\ge3k-2$ for positive integers $k$ gives $2H\ge3W-(p^2-1)\ge30$ for either prime. Choose a prime $\ell\mid y$ with $\ell\ge7$ for $p=5$, or $\ell\ge5$ for $p=7$: three distinct primes divide $y$, none equal to $p$. The $\ell$-active labels of Proposition~\ref{prop:two-packets} miss $P$, $Z$ and all denominator-four labels by twist purity. The four classes counted are disjoint, since $P$ requires $p\mid b$ and misses $Z$. Thus $R\ge (1+H)+1+m_4+(\ell-1)$, which gives $R\ge m_4+23$ at five and $R\ge m_4+21$ at seven. \paragraph{Eleven denominators four preclude six denominators five.}\label{rank:gauss-eleven-six} Suppose $D=M^2$ and eleven labels have largest degree $M/4$. Let $J$ be the labels of degree $y=M/5$ and suppose $j=|J|\ge6$. Their multiplicity is $6$ or $7$ by the dimension identity. Write $I$ for the eleven largest labels. The remaining squared mass is $\delta D$, where \[ \delta=1-\tfrac{11}{16}-\tfrac{j}{25}\le\tfrac{29}{400} <\tfrac{2}{25}. \] Every $J$-row is supported on the principal five-block $P$ (Theorem~\ref{thm:principal-support}); this block misses $I$ by Corollary~\ref{fam:M:4.1}. The actual twists in $J$ have five-power order (Lemma~\ref{lem:complement}). They have order at most five because a primitive higher phase would have a square orbit of at least ten labels at this degree. By Proposition~\ref{prop:two-packets}, the squared masses at the four primitive fifth phases agree even if other labels have twist order divisible by $25$. First suppose $5\mid y$. Equation~\eqref{eq:septic-trace} for all twist-one labels of $J$ charges only the other multiples of five. Even one such label would require mass $64D/625>\delta D$. Hence all $j$ twists in $J$ are primitive fifth phases. An odd $j=7$ is impossible since there are two elements in each square orbit. For $j=6$ the two quadratic classes have packet counts $c_+,c_-$ with $c_++c_-=3$. They contribute respective squared masses $2c_+y^2,2c_-y^2$, differing by at least $2y^2$. The class $I$ has no five-active labels. The remaining mass $\delta D<2y^2$ cannot restore the required equality. If $5\nmid y$, their exact support $P$ has normalized mass at least $j/25$ by Lemma~\ref{mass}. It misses $I$ and $\delta2/p^2$ by \eqref{eq:septic-trace}. Higher-order twists need at least $p(p-1)/2$ labels, so $J$ consists of one square packet of $(p-1)/2$ primitive phases. Their opposite class needs $(p-1)y^2/2\ge 2y^2$ outside $J$. If $p\nmid y$, the exact support $P$ has normalized mass at least $j/p^2$ by Lemma~\ref{mass}. If $\eta_pj$ active labels. This proves the stronger $\eta_p\ge j/p^2$ in this case. Thus $j=2$ at five requires $2/25$; $j=3$ requires $3/25$, since an orbit size divides $20$ and three labels must include a rational row, forcing $5\nmid y$ by Proposition~\ref{prop:rational-prime-square}. \begin{proposition}\label{prop:fourteen-seven} At rank at most $34$, fourteen labels of denominator four preclude any label of denominator seven. More precisely, if both denominators occur and there are at least fourteen fours, a unique seven forces rank at least $m_4+21$; multiplicities two, four, five and six of seven are impossible at every rank; and multiplicity three forces rank at least $35$. \end{proposition} \begin{proof} Write $p=7$, $j=|J|$, $m_4=|\{X:b_X=4\}|$, and $\eta=\sum_{7\mid b_Z,\ b_Z\ne7} b_Z^{-2}$, including the vacuum. The dimension equation gives $1\le j\le 6$ and $\eta\le1-14/16-j/49$. For $j=1$ the preceding single-denominator count at $p=7$ gives $R\ge m_4+21$. For $2\le j\le5$ there are no wild septic twists in $J$, and its active labels occur in packets of three under square translation. At $j=2$ all twists are trivial, so \eqref{eq:septic-trace} gives $\eta\ge288/2401>1-14/16-2/49$. For $j=4,5$ at least one label is inactive, giving $\eta\ge144/2401>1-14/16-j/49$. For $j=3$ either all are trivial, which gives $\eta\ge432/2401>1-14/16-3/49$, or they form one square class. In this case the opposite class needs $3/49$ compensating mass. Suppose $R\le34$. Each of the three opposite primitive seventh phases has squared mass at least $y^2$. The degree multisets on these opposite phases agree by the local square Galois action; this assertion concerns the seventh component, with the complementary component unrestricted. Let $s$ be their common label count. Every such label has $b=7k'$, $k'\ge2$, by Lemma~\ref{lem:complement} and the exact multiplicity of seven. It contributes at most $y^2/4$. Since the rank leaves at most $34-14-3-1=16$ other nonunit positions, $4\le s\le5$. For $s=5$ at least four labels on each phase have denominator $14$, since $3/4+2/9<1$. The fifth label gives strictly greater mass than $y^2$, so each of the three original phases also needs an additional label. This requires at least $15+3>16$ positions. Thus $s=4$ and all twelve labels on the three opposite phases have denominator $14$. Put $k=M/28$. The remaining $h\le4$ nonunit degrees satisfy \[ 1+\sum_{i=1}^{h}d_i^2=2k^2, \] because $14/16+3/49+12/196=1-1/392$. If an odd prime $\ell\ge5$ different from seven divides $M$, at least $\ell-1$ labels are $\ell$-active, all among the $h$ remaining positions by Lemma~\ref{lem:complement}. This forces $\ell=5$ and $h=4$. No wild quintic phase can occur in four labels. Equality of squared masses at the four primitive quintic phases forces all four degrees to coincide. Then $1+4d^2=2k^2$ contradicts parity. Otherwise $M$ has support contained in $\{2,3,7\}$. The two-prime and ordered three-prime divisor families (Theorems~\ref{fam:M:4.12}, \ref{seq:quartic-any} and~\ref{fam:Q:1.3}) dispose of all cases except $v_2(M)\ge3$, $v_3(M)\ge3$, $v_7(M)\ge2$: the corresponding covered ordered exponents of $D$ are $v_2(D)\le5$, $v_3(D)\le4$, or $v_7(D)\le3$. In particular $k$ is even. Reducing the residual equation modulo four forces exactly three odd degrees among the $h\le4$ terms. Each such degree is divisible by both three and seven, since no nonunit degree is a prime power. The degrees of denominators $4,7,14$ are likewise divisible by three. There remains at most one nonunit of degree prime to three, contradicting Lemma~\ref{mrk:e:witness}. Finally $j=6$ uses $6/49>1/16$ on denominators prime to four, contrary to Proposition~\ref{prop:fourteen-principal}. \end{proof} \section{Rank bounds for modular categories and fermion quotients}\label{mrk:certificates} Write $r$ for the number of labels of the table, $\Delta=qM^2$ with $q$ square-free, and $b_i=M/d_i$ for the denominators. In the modular case $r$ and $\Delta$ are the rank and dimension of the category. In the positive fermion quotient both quantities are halved. We use the row and twist conditions proved in Sections~\ref{mrk:packets}--\ref{mrk:dyadic} to conclude the finite searches defined by \eqref{mrk:e:fraction} and~\eqref{mrk:e:interval}. \subsection{The positive fermion quotient}\label{mrk:reduced} \begin{theorem}\label{mrk:thm:reduced-rank} A divisor-minimal slightly degenerate non-WGT obstruction has rank at least $34$. \end{theorem} \begin{proof} By Lemma~\ref{lem:proper-divisor-reduction} and Proposition~\ref{fam:M:4.7}, its positive table is simple and perfect, and $\Delta$ is even. We first describe a constraint specific to the two smallest codegrees in this search. For a row $X$ of prime-power codegree $f=p^a$, every hit lies in the principal $p$-block by Theorem~\ref{thm:principal-support}. Two nonunit hits cannot have coprime degrees. Indeed, one of their degrees, say $d_Y$, would be prime to $p$, and the column of the other hit $Z$ is principal. Then $\chi_Z(Y)\equiv d_Y$ at a prime above $p$, contradicting coprime vanishing. The gcd of all nonunit hit degrees is one by \eqref{mrk:e:orth}, so at least three different degrees occur. Each costs at least one in the mean squared row norm. Thus $f\ge4$ in the positive table. At $f=4$ all three hits have mean squared norm one, hence their character values are roots of unity. At each hit degree there is only one nonzero entry. The column Galois action preserves this degree and its nonzero entry, so the value is rational and equals $1$ or $-1$. Write $x=d_X$ and $a1$ one has $r\ge33$. If $D$ is odd then $r\ge57$, and if also $q>1$ then $r\ge75$. In square dimension put $B=M/d_{\max}$. Then $r\ge36$ if $B\ge5$, $r\ge39$ if $4\nmid M$, and $r\ge46$ if $B\ge6$. \end{theorem} In odd dimension count the nonunit denominators by dual pairs. Since all degrees are odd, $r\equiv D\equiv q\pmod8$. Let $f=qb^2$ and $s=\min_{p\mid f}v_2(p-1)$. Every run at $b$ contains a multiple of $2^{s-1}$ pairs. Indeed the character field $K$ has conductor dividing $f$. Complex conjugation acts nontrivially on $K$, since its label differs from its dual. On $\QQ(\zeta_f)$, complex conjugation is a $2^{s-1}$-st power, as one sees on each cyclic prime-power factor. Its root maps to an element of order divisible by $2^s$ in $\operatorname{Gal}(K/\QQ)$. The asserted divisibility follows, and it multiplies the odd factors supplied by the defect-one block partitions. For the nonsquare odd search this gives short bounds on each remaining denominator. If the support has precisely three primes in increasing order, their exponents in $D$ are at least $(6,5,4)$ by the dimension families. For each $p$ at least two $p$-free nonunit pairs are required; at three, at least four, since $1+2t\equiv0\pmod3$ for their number $t$. These sets of pairs are disjoint for three-prime support, since a nonunit degree contains at least two primes. Their denominators contain the full $p$-part of $M$. For additional primes outside $q$, the necessary number of eligible packets follows from the complementary-fiber bounds of Proposition~\ref{prop:joint-prime-capacity}. Sorting these mandatory lower bounds on the unfilled slots gives an upper bound on the available reciprocal-square mass. The vacuum term is at most $(15b_{\rm last})^{-2}$; the implementation rounds each summand upwards on a grid of mesh $2^{-64}$. \begin{proof}[Proof of Theorem~\ref{mrk:thm:modular-ranks}] A non-pointed fusion-simple category has trivial pointed subcategory. Put $f_{\min}=qB^2$. Lemma~\ref{lem:small-ordinary-codegrees} gives $f_{\min}\ge5$, whereas \eqref{mrk:e:fraction} gives $f_{\min}1$. This proves the first two bounds. For odd $D$, the \RankLink{code/odd_frontier.py}{paired recursion} visits $14\,520$ prefixes for $r\le56$ and leaves no type. With $q>1$ it visits $128$ prefixes for $r\le74$ and leaves no type. Both the \RankLink{data/revision/odd56.json}{unrestricted odd result} and the \RankLink{data/revision/nonsquare74.json}{odd nonsquare result} are regenerated by \RankLink{verify_revision.py}{the verifier}. Finally let $q=1$. Then $B^22$ is unique, hence its row is integral and has values $1,-1$ on $1,g$ and zero elsewhere by dimension orthogonality. These are the only invertibles, since each character is nonzero on every invertible. Every other character is one on $g$, so its permutation fixes every noninvertible: its sole eigenvalue $-1$ already occurs on $\{1,g\}$. Stabilizer divisibility \cite[Lemma~2.2(i)]{DNVFrobenius} makes their degrees even and $D\equiv2\pmod4$. The codegrees $2,3,7$ cannot all occur \cite[Corollary~2]{GreedyFusion}. The enumeration \RankFile{code/unbraided_small.py} covers every $D^2$ through rank five; at rank six these last two obstructions and the covered centers remove, respectively, $26,232,1310$ of $1568$ tuples (\RankFile{data/unbraided_small.txt}). This proves the unrestricted integral bound seven. For the braided bound take a counterexample of least rank. Its universal grading is trivial, since WGT passes to extensions. Its invertibles are therefore transparent: their double braiding defines a grading character trivial on the adjoint category. At codegree two, $g$ fixes a noninvertible and balancing gives $\theta_g=1$. Thus $\C=\mathcal B^{C_2}$ with $g$ the sign. A fixed simple of $\mathcal B$ has two lifts exchanged by $g$ \cite[Corollary~2.13]{BNEquiv}; only the vacuum is fixed. The other simples have half the integral degree of a free-orbit lift and hence integral degree; $\FPdim(\mathcal B)=D/2$ is odd. The involution of its commutative ring fixes just the dimension idempotent, by its trace in the simple basis. Each other pair $\chi,\chi'$ restricts to one character via the orbit-sum map from $K_0(\C)$; their distinct restrictions exhaust the rows other than $2,D$ because the image is the invariant algebra. The restricted norm is $2+\tfrac12\sum_{Y\ne1}|\chi(Y)+\chi'(Y)|^2 =f_\chi$ by orthogonality. All intermediate codegrees thus divide $D/2$. Using the divisor families and Theorem~\ref{thm:uniform-new-bounds}, \RankFile{code/formal_sieve.cpp} leaves no tuple through rank seven and $52$ tuples in $38$ dimensions at rank eight (\RankFile{data/formal_residual.txt}). At codegree two the cutoff is $2\cdot6\,087\,156\,075$, since the square-full part of $D$ is odd and at most $D/2$; otherwise use $29\,160\,000$. One tuple has all codegrees distinct, so the rows are integral, whereas the Gram determinant $\prod f_i$ is nonsquare. Each other tuple has two threes and all remaining codegrees distinct. Each row of norm three has two nonunit entries, both roots of unity: dimension orthogonality rules out only one and the conjugate mean square of a nonzero algebraic integer is at least one. With two invertibles its values must be $1,1,-1$ on the two invertibles and one degree two, impossible for orthogonal rows. With one invertible, $1+a\alpha+b\beta=0$, $a,b>1$, forces consecutive degrees and opposite signs: $a=b$ would give the noninteger rational algebraic integer $-1/a$, and $|a-b|\le1$. Two orthogonal rows share one position of opposite sign; hence three consecutive degrees occur and the Gram determinant is square. With three invertibles the rows restrict to the nontrivial characters of $C_3$ and vanish elsewhere. Replacing them by their sum and their difference divided by $\sqrt{-3}$ gives a rational table, so $(\prod f_i)/3$ is square. The two nontrivial eigenvalues of a generator occur on the unit orbit, hence it fixes every other label; their degrees are divisible by three. The two square-class requirements leave respectively two and five tuples. In the one-invertible case subtract the unit and three consecutive squares from $D=576\,298\,800$, and test whether the remainder is four squares of nonunit divisors. In the three-invertible case, $D\equiv3\pmod9$ leaves only $(3,3,5,8,121,14600,2649920,351101150400)$. Its integral norm-five row is one on the three invertibles and has two more signs. Its signed dimension sum gives nonunit degrees $3a,3(a+1)$. The five nonunit degrees divided by three are divisors of $D/3$ with squared sum $(D-3)/9$. Neither integer test has a solution, by the pair sums in \RankFile{code/formal_check.py}. This proves the integral braided bound nine. \subsection{The odd unbraided boundary}\label{mrk:g:odd-unbraided} Ranks $3,5,7,9$ yield $1,3,13,115$ paired tuples, all covered at $D^2$. At rank eleven there are $2799$. We justify the additional conditions in \RankFile{code/odd_unbraided.py}, whose complete counts and last eight tuples are in \RankFile{data/odd_unbraided.json}. Write $f_1,\ldots,f_5$ for the paired denominators. When distinct, each pair is Galois-stable. Restrict the real parts of its characters and the dimension row to one representative per dual pair and the unit. The column metric is $\operatorname{diag}(1,2,\ldots,2)$ and the resulting rational Gram matrix has entries $D,f_1/2,\ldots,f_5/2$ on the diagonal. Hence $D\prod f_i$ is a square. We give the subcategory test used at $3,7$ and $5,11$. Suppose the earlier rows vanish outside a subcategory $\mathcal H_0$ of dimension $h$ and every other row equals dimension on it. At a single conjugate pair of prime codegree $f=2h+1$, write $a_i$ for its values on the exterior pairs and $t_i=a_i+\bar a_i$. The row field is imaginary quadratic because the row is nonreal and has one possible nontrivial conjugate. Orthogonality to its conjugate and its row norm give $\sum t_i^2=1$, $-\sum(a_i-\bar a_i)^2=f$. The unique nonzero trace is $-1$ at a pair of degree $h$ by dimension orthogonality. In $\mathbb Q(\sqrt{-m})$, $m$ square-free, the odd trace implies $m\equiv3\pmod4$; write $a_i-\bar a_i=k_i\sqrt{-m}$ with $k_i\equiv t_i\pmod2$. Then $m\sum k_i^2=f$ gives $m=f$. The pair has values $(-1\pm\sqrt{-f})/2$ and all other exterior values vanish. By orthogonality each subsequent character takes value $h$ on both new simples. Adjoining these simples gives a fusion subcategory $\mathcal H$. Indeed, the character formula for $N_{UV}^W$ vanishes for $U,V$ inside the enlarged set and $W$ outside, being $d_Ud_V\sum_\chi\overline{\chi(W)}/f_\chi=0$. It also gives $UY=d_UY$ for every exterior $Y$ and every $U$ inside, by evaluation on the characters. The $\mathcal H$-module component containing $Y$ has only this simple, hence $\FPdim(\mathcal H)\mid d_Y^2$ \cite[Proposition~8.15 and Remark~8.17]{ENOannals}. Starting at the unit gives the pointed $C_3$ when $f_1=3$ and then a subcategory of dimension $21$ at a single pair of sevens. A subsequent pair at $43$ would give a subcategory with degrees $1,1,1,3,3,21,21$, contrary to the odd rank-seven classification \cite[Theorem~1.7]{ABDPIntegral}. In this greedy sequence uniqueness of each pair follows from the remaining reciprocal sum. At dimensions three and twenty-one all exterior degrees are multiples of $3$ and $21$ respectively, giving $D\equiv3\pmod9$ and $D\equiv21\pmod{882}$ in the latter case. For use here and in the perfect case, consider any row $\chi$ of codegree five. If the unit were the only invertible, orthogonality to dimension would require at least two dual pairs of hits: their degrees have gcd one. Equality of the mean-square bound then gives two pairs of root-of-unity values $u,\bar u,v,\bar v$. Their traces $s=u+\bar u,t=v+\bar v$ and degrees $d,e>1$ satisfy $s^2+t^2=3$ and $ds+et=-1$ by orthogonality to $\bar\chi$ and to dimension. Rational traces are integers and cannot square to three in this way. Otherwise elimination gives the quadratic trace $-2d/(d^2+e^2)$ of $s$, strictly between $-1$ and zero. Thus there are at least three invertibles. The order of the group of invertibles is odd and at most five, since the row is nonzero on the group. At order three there would be one extra dual pair of roots $a,\bar a$. A trivial restriction to $C_3$ gives $3+a^2+\bar a^2=0$ by orthogonality to $\bar\chi$, while a nontrivial restriction gives both $a+\bar a=0$ by dimension orthogonality and $a^2+\bar a^2=0$. Consequently there are five invertibles, and the Galois translates of the restriction give four codegree-five rows. They restrict to the four nontrivial characters of $C_5$ and vanish elsewhere. A single pair of elevens adds two objects of degree five. In rank eleven the remaining two dual pairs have degrees divisible by $55$, so $(D-55)/6050$ must be two positive squares. We recall the indicator identity $\operatorname{Tr}(\theta^a_{I(X)})=D\nu_a(X)$ of Ng--Schauenburg \cite[Theorem~4.1]{NSfs}. If $g$ is an invertible of prime order $p$ fixing some simple under left tensor product, its pointed category admits a rank-one module category. Its associator class is therefore trivial, so $\nu_p(g)=1$ and $\nu_1(g)=0$. Positivity of dimensions forces every summand of $I(g)$ to have $p$th-root twist, and the relation at $a=1$ forces the dimension at each of the $p$ twists to be $D/p$, by the minimal polynomial of a primitive $p$th root. Write $r=\rank(\C)$ and, for a simple central object $A$, put $m_X(A)=[F(A):X]$. Frobenius reciprocity gives $[I(X):A]=m_X(A)$ and $F(I(X))=\bigoplus_Y YXY^*$. In particular when the Grothendieck ring is commutative and $g$ invertible, \[ \langle I(g),I(g)\rangle=r,\qquad \langle I(1),I(g)\rangle=\#\{Y:gY=Y\}. \] \begin{lemma}\label{lem:rank-eleven-cubic} Suppose $r=11$ and $f_1=3$. There are three invertibles, with generator $g$ fixing every noninvertible, and $D\equiv3\pmod9$. For $u_j$ the least nonnegative residue of $D/f_j-1$ modulo three, one has $\sum u_j=8$. If \[ \frac{D}{3}-\sum_j u_j\frac{D}{f_j}=1 \] and the dimension character is the only character of codegree $D$, then $\mathcal C$ de-equivariantizes by a central $\operatorname{Rep}(C_3)$ to a fusion category of dimension $D/3$. \end{lemma} \begin{proof} The subcategory argument with $h=1$ gave the three invertibles, the fixed noninvertibles and the dimension congruence. Pair the nonunit summands of $I(1)$ into dual pairs, choosing $A_j$ in each; their degrees are $D/f_j$. Put $v_j=m_g(A_j)+m_{g^2}(A_j)$. Since the central category has odd dimension and duality interchanges $g$ and $g^2$, and since the external dimensions are multiples of three, $v_j\equiv u_j\pmod3$, $\sum v_j=8$. The known summands contribute at least $\sum\lceil v_j^2/2\rceil$ to $\langle I(g),I(g)\rangle$. The two nontrivial twists each require at least one more summand. If some $v_j\ge3$, the five nonnegative integers of sum eight give $\sum\lceil v_j^2/2\rceil\ge10$ (minimum attained at $3,2,1,1,1$). This would exceed the available norm eleven. Thus $v_j=u_j$. At twist one the remaining mass in $I(g)$ is the displayed gap. A gap of one supplies an invertible central lift of $g$ with twist one. Its cube lies above $1$ and is the unit because $I(1)$ has no other invertible summand. This gives the asserted central Tannakian subcategory. Its embedding is fully faithful after forgetting since its generator maps to $g$. \end{proof} \begin{lemma}[Schur elements of cyclic invariants]\label{rank:cyclic-trace} Let $B$ be a finite-dimensional semisimple complex algebra with a cyclic $C_\ell$-action, $\ell$ prime, preserving a symmetrizing trace $\tau$. Write $\tau|_{\operatorname{End}(V)}=f_V^{-1}\operatorname{Tr}_V$, with $\operatorname{Tr}_V$ the ordinary matrix trace. Suppose $B^{C_\ell}$ is commutative, and let $e$ be one of its primitive idempotents. On a free block orbit $e$ is the sum of the identities of $\ell$ scalar blocks, so $\tau(e)=\ell/f_V$. On a fixed block it has matrix rank one and $\tau(e)=1/f_V$. \end{lemma} \begin{proof} On a free orbit the invariant algebra projects isomorphically to any constituent matrix block. Its commutativity makes that block scalar, and invariance makes the $\ell$ trace weights equal. On a fixed block a generator acts by conjugation by a matrix $T$ with $T^\ell$ scalar. Rescaling $T$ makes $T^\ell=1$. Its centralizer is a product of endomorphism algebras of its eigenspaces. Since the centralizer is commutative, each nonzero eigenspace has dimension one. The trace formula follows by evaluating the corresponding projector. \end{proof} \begin{lemma}\label{lem:double-central-odd} The tuple $(3,7,55,179,24323;5028658635)$ cannot occur in a non-weakly-group-theoretical rank-eleven category. \end{lemma} \begin{proof} For this tuple $\sum u_j=8$, the twist-one deficit is one, and $D$ occurs only once. Lemma~\ref{lem:rank-eleven-cubic} therefore supplies a de-equivariantization $\mathcal E$ of dimension $M=D/3$. The canonical action of $C_3$ on $\mathcal E$ recovers $\mathcal C$ by equivariantization. In the equivariantization a fixed simple contributes three simple equivariant structures (the stabilizer is cyclic); a free orbit contributes one simple. The dual action by the three invertibles of $\mathcal C$ permutes the three structures in the first case and fixes the resulting simple in the second. Since the three invertibles are the only nonfixed labels in $\mathcal C$, the unit is the only fixed label of $\mathcal E$. Thus a noninvertible of $\mathcal C$ maps to a sum of three distinct simples of equal dimension in $\mathcal E$. The subcategory of dimension $21$ yields seven invertibles in $\mathcal E$. The other three dual pairs in $\mathcal C$ have dimensions $21e_i$. Here $\sum e_i^2=(D-21)/882=5701427$. The three $e_i$ are distinct. The calculation in \RankFile{code/odd_unbraided.py} tests $5701427-2t^2$ for a positive square for $1\le t\le\lfloor\sqrt{(5701427-1)/2}\rfloor$ and finds none. Hence $\mathcal E$ has rank 25 and its 18 noninvertibles occur in three sets of six, each degree divisible by seven. Its group $C_7$ fixes every one under both left and right tensor product. Put $B=K_0(\mathcal E)\otimes\mathbb C$. The forgetful map $F$ from $K_0(\mathcal C)\otimes\mathbb C$ has image $B^{C_3}$, since the orbit sums and the unit span it. This image has dimension nine (one fixed label and eight free label orbits). As $F(g)=1$, the two scalar blocks on which $g$ acts nontrivially lie in the kernel; dimension now gives equality. For each remaining primitive source idempotent $e_\chi$, its coefficients at $1,g,g^2$ all equal $1/f_\chi$ by the character-idempotent formula, and no noninvertible label contributes to the unit after forgetting. Thus $\tau_B(F(e_\chi))=3/f_\chi$. By Lemma~\ref{rank:cyclic-trace} this equals $3/f_V$ on a free block orbit and $1/f_V$ on a fixed block. The latter alternative would give $f_V=f_\chi/3$, impossible for any of the eight characters of codegrees $7,55,179,24323$ by algebraic integrality. The eight free orbits already account for 24 scalar summands of the 25-dimensional algebra $B$. The last summand is scalar and carries the dimension character. In particular $B$ is commutative. Apply the indicator identity and the induction norm count in $\mathcal E$ for each $h\in C_7\setminus\{1\}$. For any canonical $A\subset I(1)$ let $x_A=\sum_{h\ne1} m_h(A)$. There are 25 canonical summands, six of each of the four listed codegrees and the unit. From the eighteen fixed simples, $\sum_A x_A=6\cdot18=108$. The congruence $x_A\equiv d_A-1\pmod7$ follows from divisibility of external degrees. At the eighteen summands of codegree $55,179,24323$ it requires $x_A\ge6$. Thus $x_A=6$ there and vanishes on the rest. Summing the twist-one mass over the six nontrivial $h$, the deficit from the canonical objects is \[ 6M/7-6\!\sum_{A:\,f_A\in\{55,179,24323\}}d_A =6M/7-6(M-1-6M/7)=6. \] For any noncanonical summand $Z$ contributing to that deficit, put $s_Z=\sum_{h\ne1}m_h(Z)$. Its contribution is $s_Zd_Z$. As $m_1(Z)=0$ we have $d_Z\equiv s_Z\pmod7$ and $1\le s_Z\le d_Z\le6$, whence $d_Z=s_Z$. Each $d_Z$ divides $M$ by modular divisibility in the center, so is odd. Thus all $d_Z=1$ and six central invertibles of twist one lift the six nontrivial elements of $C_7$. Indeed, two lifts of the same element would produce a nontrivial invertible above $1$, whereas the dimension character alone has codegree $M$. They yield a central $\operatorname{Rep}(C_7)$ with faithful forgetful restriction. All powers of a twist-one invertible have twist one. The second de-equivariantization has dimension $D/21=5\cdot11\cdot13\cdot179\cdot1871$ and is therefore solvable by Corollary~\ref{thm:sq30030}. Weak group-theoreticity goes back up both equivariantizations. \end{proof} For clarity, the successive counts in \RankFile{data/odd_unbraided.json} are as follows. Of the $2799$ paired tuples, $2134$ have a covered center. The stabilizer congruence removes $533$, the cubic induction norm $17$, the central $C_3$ descent $88$, and the real determinant $19$. The dimension-$55$ and dimension-$21$ tests remove six and one, respectively. Lemma~\ref{lem:double-central-odd} excludes the last. This proves the general odd bound $13$. If the Grothendieck ring is noncommutative, the published bound for a nonpointed odd-dimensional integral category is $23$ \cite[Theorem~1.8]{ABDPIntegral}. \begin{corollary}\label{cor:all-dim-rank} A nontrivial perfect integral fusion category of odd dimension has rank at least $15$. With a braiding its rank is at least $57$, and at least $75$ in nonsquare dimension. \end{corollary} \begin{proof} An odd-dimensional WGT category is solvable by the group-extension argument of Corollary~\ref{thm:sq30030} and the odd-order theorem. A nontrivial solvable category contains a nonunit invertible \cite[Proposition~4.5(iv)]{ENOsolvable}. Thus it suffices to exclude rank thirteen in the first assertion. At that rank the ring is commutative by \cite[Theorem~1.8]{ABDPIntegral}. Perfection rules out a row of codegree three (its support would be a single dual pair of nonunit degree, contradicting orthogonality), and the norm-five calculation above excludes five. Hence the reciprocal denominators are odd and at least seven. In a commutative perfect odd-dimensional integral fusion ring arising from a category, a character $\chi$ of codegree seven has nonreal field contained in $\QQ(\zeta_7)$ by Corollary~\ref{cor:field-conductor}. Its support consists of the unit and two or three dual pairs, by the preceding norm and gcd estimates. With three pairs all six nonunit values are roots in $\mu_{14}$. Orthogonality to $\bar\chi$ says that the seven squares of the entries sum to zero. The cyclotomic relation forces these squares to run through $\mu_7$. Orthogonality to the dimension character is then a relation on all seventh roots: after writing each value as a signed seventh root, the powers are distinct and the signed coefficients are its simple degrees. The vacuum coefficient is one, so every coefficient equals one, contrary to perfection. With two pairs, all Galois conjugates of $\chi$ have the same five-label support and are mutually orthogonal, so the field has degree at most five. The only nonreal proper subfield of $\QQ(\zeta_7)$ is $\QQ(\sqrt{-7})$. The two nonzero squared absolute values are positive integer norms summing to three. Thus on a pair $A$ the character is $\epsilon=\pm1$ and on a pair $B$ its values have norm two and sum $\delta=\pm1$ by the norm form of that field. Their degrees satisfy $1+2\epsilon d_A+\delta d_B=0$, so $d_B>d_A$. Associate the ordered pair $A\to B$ to $\{\chi,\bar\chi\}$. Two such pairs have distinct heads, because $\chi-\bar\chi$ is supported just at the head, and two differences with that same support would be proportional and orthogonal. Their tails too are distinct: at a common tail the inner product of the two character sums $\chi+\bar\chi$ would be $4\pm8$. For distinct heads and tails this inner product is four from the unit and $\pm4$ per head--tail coincidence. Its vanishing therefore requires exactly one such coincidence. Three pairwise consecutive directed edges with distinct heads and distinct tails would form a cycle, which strict degree increase precludes. There are at most two complex-conjugate pairs of codegree seven. At rank thirteen the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/verify_odd13.py}{recursion} with six paired denominators at least seven and one final $D$ gives 520 tuples. Of these, 17 have $D\ge121275$ and at most two paired denominators equal to seven. Testing their squares by the exponent comparisons of Theorem~\ref{thm:uniform-new-bounds} leaves the paired denominators and dimensions \[ (7,7,11,11,31,4785;1038345),\qquad (7,7,9,11,95,575;7571025). \] In both, $v_7(D)=1$ and no other paired denominator is divisible by seven. In $D\tau=\sum_\chi (D/f_\chi)\chi$, reduce at a prime above seven. The two codegree-seven pairs give two distinct reduced characters: the two quadratic-conjugate values agree at the unique prime of $\QQ(\sqrt{-7})$ over seven, and the values on each pair $A,B$ are nonzero modulo seven. The two supports differ, since their edges are consecutive and directed by strictly increasing degrees. The surviving relation contains only the dimension character and these two distinct characters, each of the latter with nonzero coefficient $2D/7$. This contradicts independence of distinct characters over the residue field, even if one is the dimension character. Hence rank thirteen is impossible. For a perfect braided category of odd dimension every fusion subcategory is perfect. Its symmetric center is Tannakian, and a nontrivial group of odd order is solvable, hence has a nontrivial linear character. It follows that each subcategory is nondegenerate. Successive M\"uger decompositions express the category as a product of nontrivial simple perfect modular categories. Ranks multiply, and Theorem~\ref{mrk:thm:modular-ranks} gives $57$ for every factor. If the total dimension is nonsquare, at least one factor has nonsquare dimension and hence rank at least $75$. \end{proof} There is a consequence for simple integral categories in which no braiding exists. Their formal codegrees exclude $3$ provided the category is nonpointed. Indeed, their Drinfeld centers are simple by \cite[Theorem~5.1]{LPRsimple} and integral by the forgetful functor. They are nonpointed since their global dimension is a nontrivial square, whereas a simple pointed modular category has prime dimension. A codegree $f$ of the original category gives a simple of dimension $D/f$ in the center \cite[Theorem~2.13]{OstrikPivotalRankThree}. Its modular codegree is $f^2$, so the assertion follows from Theorem~\ref{thm:no-nine}. \subsection{Counting twists in odd braided dimension}\label{mrk:g:odd-braided} Suppose that $\C$ is an odd-dimensional braided non-WGT category of least rank $R$. Its universal grading is trivial. If $\C$ is modular, $(\C_{\ad})'=\C_{\pt}$ makes it perfect, and Corollary~\ref{cor:all-dim-rank} already gives $R\ge57$. Otherwise its symmetric center is $\Rep(G)$ for a nontrivial group $G$ of odd order. Write $k$ for its number of conjugacy classes. The central de-equivariantization $\mathcal M$ is modular, $\C=\mathcal M^G$, and $\FPdim(\C)=|G|D$ for $D=\FPdim(\mathcal M)$ \cite[Section~4]{DGNO}. The category $\mathcal M$ is non-WGT since WGT passes to equivariantizations, so at least three primes divide $D$ by \cite[Theorem~1.6]{ENOsolvable}. Braided equivalences preserve the canonical positive spherical structure. Thus every twist of $\mathcal M$ occurs in $\C$ under the dominant forgetful functor. Write $t$ for the number of its distinct nontrivial twists. We have \[ R\ge k+2t . \] Indeed, all $k$ central simples have twist one, and a nontrivial twist accounts for at least one pair of distinct duals in odd dimension. \paragraph{A lower bound on the number of twists.} Fix a set $S$ of primes of even valuation in $D$. Partition the projected $S$-tuples of twists by the exact orders of their coordinates. In a nonempty part let $A\subset S$ be the primes where the coordinate is nontrivial. Fixing the other coordinates still leaves at least $p-1$ choices at $p\in A$ by Proposition~\ref{prop:two-packets}, first with the full complementary twist fixed and then after projection. Successively forgetting coordinates gives at least $\prod_{p\in A}(p-1)$ distinct tuples in this part. For nonempty $A$ this is at least $\sum_{p\in A}(p-1)$. Every prime occurs by categorical Cauchy, and the vacuum contributes the trivial tuple. There are therefore at least $1+\sum_{p\in S}(p-1)$ projected twists. At a prime $p$ of odd valuation, every nonempty fiber over complementary twist coordinates has at least $(p+1)/2$ different $p$-phases. A phase of order at least $p^2$ supplies enough square Galois conjugates. Otherwise Proposition~\ref{prop:odd-slice} at the vacuum gives positive inactive mass and order-$p$ mass equal to $(p-1)$ times that mass. An active phase has $(p-1)/2$ square conjugates. If $D$ has at least two primes of odd valuation, use these two fiber bounds over the projection to any third prime $q$. The latter has at least $(q+1)/2$ values by categorical Cauchy, square Galois conjugation and the vacuum. There are at least $2\cdot3\cdot4=24$ twists. If precisely $p$ has odd valuation, choose two other primes $up-1$ square conjugates. We will use the following equality argument twice. Suppose all nontrivial phases are pure of prime order and only the unit of $\mathcal M$ has twist one. At each $p$, let $m_p>0$ be the common mass at each primitive $p$th root (Proposition~\ref{prop:two-packets}). The dimension and the real Gauss sum satisfy \[ n^2=1+\sum_{p\mid n}(p-1)m_p,\qquad \tau^+=1-\sum_{p\mid n}m_p=-n . \] Writing $P$ for the greatest prime of $n$ gives $n^2\le 1+(P-1)(n+1)$, hence $n\le P$. This contradicts the presence of three distinct prime divisors. If $R-k=24$, then $t=12$ and the primes are $3,5,7$. Every nontrivial twist uses exactly two simples of $\C$, so its twist-one simples are exactly those in $\Rep(G)$. Dominance and the fact that their images are multiples of the unit imply that only the unit of $\mathcal M$ has twist one. This is the equality argument above. Consequently $R-k\ge32$. Since $k\ge3$ for a nontrivial odd-order group, a counterexample of rank below $41$ must have $R-k=32$ and $k\in\{3,5,7\}$. \paragraph{The orbits at the boundary.} No nonunit $G$-orbit of simples in $\mathcal M$ is its own dual: duality would fix a point of that odd orbit. The odd rank-seven classification \cite[Theorem~1.7]{ABDPIntegral} applied to $\Rep(G)$ allows orders $3,5,7,21,39,55$. For a nonabelian group $F_{pq}=C_q\rtimes C_p$ of order $pq$ ($p0$, as $q\ge3$ and $r\ge5$. \end{proof} The unsigned alternatives of Theorem~\ref{first:balance} also determine all monodromy phases and masses. At one residue with $r>2g$, at most $g0$ then $2b\le2+2m\le g+1\le r$. In both cases $bq\ge5$ be odd primes, $g=s+t$, $1\le tg,\quad w(g)>t,\quad (s-1)w(a)+w(g+a)>g\quad(a\ne0). \] Then every positive weight-$g$ expression for $z$ has the same phase multiplicities. They are nonnegative integers. The $q$-parts of the terms consist of $t$ copies of $1$ and $s$ copies of one $q$-primary root $\eta$. All remaining parts are $s$-primary. At every $s$-position $x$ its multiplicity $n_x$ satisfies $n_x\equiv g\,1_{x=1}\pmod q$. \end{proposition} \begin{proof} Give positive roots cost one and negative roots cost $c$. First consider an integral signed word in $q$- and $s$-primary roots with the two residues and cost at most $g$. Erase the $q$-parts and let $m_x$ be the signed multiplicity at $s$-position $x$. Modulo $q$, the numbers $m_x-g1_{x=1}$ are constant on each coset of $\mu_s$ by the cyclotomic relations. The identity coset cannot have a nonzero constant $a$ by the last cost inequality. A nonidentity coset with nonzero constant would cost at least $s$, besides the identity position's cost $w(g)>t$. Thus all constants vanish and the signed augmentation is $g$ modulo $q$. Erase instead the $s$-parts, obtaining a signed $q$-primary word of value $v$. Its residue over $s$ gives $\alpha=(v-t)/s$ integral, as $s$ is unramified in the $q$-power field. Each conjugate of $\alpha$ has absolute value at most $1+2t/s$. A nonzero cyclotomic integer in a $q$-power field of mean square less than $2(q-2)/(q-1)$ is a signed $q$-power root. Indeed, write it as an integer sum of $q$-power roots. Cassels's trace identities \cite[(3.4) and (3.16), pp.~114--115]{Cassels} make different $\mu_q$-cosets orthogonal for normalized trace and give the contribution $\sum_{ig$. For the plus sign, mark $1$ and $\eta$, assigning value $1$ at each marked position and $-k/(q-k)$ at the other positions of a coset containing $k$ marks. The resulting functional is strictly below cost on every signed letter other than the marked positive letters. Its value at $t+s\eta$ is $g$. Thus cost at most $g$ forces an unsigned word of length exactly $g$ with precisely the stated $q$-parts, by linear independence of the one or two marked roots. Apply Lemma~\ref{lem:cyclotomic-optimum} with retained primes $q,s$ and the cost $c<\ell-1$. The integral lower bound $g$ just proved holds for every target with the two residues. Thus every phase of the original expression has only $q$- and $s$-parts, and its coefficients are a convex combination of integral optima. For uniqueness it suffices that the union of the $s$-positions of two optima have fewer than $s$ points in each $\mu_s$-coset. Each optimum has at most $1+\lfloor g/q\rfloor$ positions, including $1$ since $w(g)>0$. Hence the union has size at most $1+2\lfloor g/q\rfloor25$, $w(25)=4>2$. For $a=1$ one has $22+w(26)=26$, and for other nonzero residues $22w(a)\ge44$. \begin{corollary}\label{found:C:3.2}\label{found:C:key:cor:orbit22} At any nonunit crossing of the two-residue type just described, tame $23$-inertia has orbit size $22$ on the normalized entry $z$. In particular both fields of the two labels ramify at $23$. This applies to all opposite complementary rows of the principal seven- and $23$-blocks with common $5$-part $25$, for arbitrary exponents of the other degree factors. \end{corollary} \begin{proof} The $23$-projection contains $1$ and at most three further positions. If it is just $1$, then $z=2+23\eta$. If $\eta=1$ this violates strictness; otherwise the full-mass two-position character has relative order at least seven, excluded by Theorem~\ref{found:S:4.1}. By uniqueness every stabilizer in the order-$22$ tame group preserves the nonzero positions (in additive $23$-exponents). Its order divides their number, since that group acts freely there. Only order two could survive. In that case the projection is $A=\{0,a,-a\}$. The full positive adjoint word on the same row has $23$-components in $A-A=\{0,\pm a,\pm2a\}$ and $7$-components in $\{0,\pm b\}$, writing $b$ for the exponent of $\eta$. This entire product of sets is exposed. In each primary factor mark the prescribed positions, give them value one, and use $-k/(v-k)$ elsewhere in a $\mu_v$-coset with $k$ marks. Each density is less than one half, so the product functional (with normalized trace on any complementary factor) exposes the prescribed set. An independent $7$-unit moves the nonzero pair off itself, and a tame $23$-unit moves the two nonzero opposite pairs off themselves (at most four of eleven multipliers can give a match modulo sign). The original and conjugate adjoints would then have a common nonunit whose balancing phases are all $1$, contradicting strictness (Lemma~\ref{fam:Q:2.2}). This rules out the last stabilizer. \end{proof} \begin{proposition}\label{found:C:3.3}\label{found:C:key:prop:fourmarks} In the $g=25$, $q=7$, $s=23$ crossing of Proposition~\ref{found:C:key:prop:above-prime} in an ordinary minimal table on $5,7,23$, the four $23$-positions belong to $\mu_{23}$ and, in additive exponents, are affinely equivalent to $\{0,1,3,14\}\subset\mathbb F_{23}$. Their multiplicities are four at zero and seven at each of the other three. At most two letters have $7$-part different from $\eta$. The joint inertia orbit of the entry at $7,23$ has size $22\varphi(\operatorname{ord}\eta)$. With $v_{23}(23)=1$, \[ v_{23}(z-2-23\eta)=1/22 \] at every prime above $23$. At the prime-weight crossing $z=\eta(2+7\sum_{i=1}^3\xi_i)$ of Proposition~\ref{fam:Q:3.1} at $7,23$, the same affine classification gives $v_{23}(z)=1/22$. \end{proposition} \begin{proof} Write $A$ for the $23$-projection of the weight-$25$ word, $D=A-A$ and $B=\{0,\pm b\}$ for the differences of the two possible $7$-parts. The rectangle $B\times D$ exposes the adjoint word even when the two components are correlated. Use the product of the two coset functionals in Corollary~\ref{found:C:key:cor:orbit22}, and the normalized trace on the remaining primary factor. Indeed the negative values in the $7$- and $23$-factors have magnitudes bounded by $3/4$ and $13/10$, respectively, since $|A|\le4$. Their product is at most $39/40<1$. In the remaining factor the trace is at least $-1/4$. Thus the functional attains one precisely on this rectangle. An independent tame $7$-unit moves the nonzero pair in $B$ off itself. For every $23$-unit $u$, comparison with the conjugate rectangle consequently gives a common nonunit whose entire word on the fixed row is pure $23$-primary on $D\cap uD$. Likewise comparison of the rectangle with any subsequently forced pure mask uses precisely their $23$-coordinate intersection. Every pure word on this row has at least five opposite pairs of nonzero differences. Otherwise a minimizing one with $h\le4$ has exposed differences and violates $10\min\{j:j(j+1)\ge h\}\le h(h-1)$ (Lemma~\ref{fam:Q:2.3}). Thus each of the intersections just obtained contains at least two opposite pairs. If $A$ meets two order-$23$ cosets, at most three of the at most six opposite pairs in $D$ have order $23$. With $l$ such pairs the average overlap over all units is at most $(l^2+(6-l)^2/23)/11<2$, a contradiction. Hence $A\subset\mu_{23}$ since it contains the identity. The remaining finite calculation takes place in $\mathbb F_{23}$. Normalize two marked positions to $0,1$ and take the other two in increasing order. For each of the $\binom{21}2=210$ supports, start with the set $B$ of its nonzero differences modulo sign. Store one representative per dilation class. Given two stored sets $B,C$ and $u\in\mathbb F_{23}^{\times}$, put $H=B\cap uC$. Reject if $|H|<2$ or if the nonzero differences of $\{0\}\cup(\pm H)$ comprise fewer than five opposite pairs. Otherwise add these differences if the resulting support $L$ is exposed, i.e.\ $3(1+2|L|)<46$. Repeat until rejection or no new dilation class appears. The preceding rectangle argument justifies each use of the initial set, and later sets are pure. Of the $18$ distinct initial dilation classes, the surviving supports among the $210$ are exactly the $12$ affine translates and dilates of $\{0,1,3,14\}$ normalized in this way. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/word_closure.py}{closure calculation} with \texttt{check 23 4 --ell 3 --classes} reproduces these counts (the least conductor prime in this exposure test may be taken to be three). A starting mask with fewer than five opposite pairs is likewise impossible, since its mean overlap over nonidentity tame units is at most $4\cdot3/10<2$. Thus the same four-position proof gives precisely the indicated affine class. Fewer than four positions give at most three difference pairs. The congruences in Proposition~\ref{found:C:key:prop:above-prime} now fix the four multiplicities. The full $23$-unit group has trivial stabilizer of these positions: a stabilizer acts freely on the three nonzero ones and has order dividing $22$. Under a joint action the $23$-projection is still preserved by uniqueness, so the same argument applies. With the $23$-action then trivial, the $7$-component must fix $\eta$ since it has multiplicity $23$. This gives the orbit size. Write $0,a_1,a_2,a_3$ for the projected positions. Their affine centroid is outside the support, since the centroid of the displayed representative is $16$ modulo $23$. In particular $U=7\sum_{j=1}^3a_j\ne0\pmod{23}$. Color two letters with $7$-part one and the others with $7$-part $\eta$. This convention works also when $\eta=1$. Let $V$ be the sum of the exponents of the two colored letters. For $\pi=\zeta_{23}-1$, the linear coefficient in $z-(2+23\eta)$ is $V+(U-V)\eta$ modulo $\pi$. It is nonzero: if $\eta=1$ it equals $U$, and otherwise the reduction of $\eta$ has $7$-power order greater than one and hence lies outside $\mathbb F_{23}$. This proves the first valuation. For the pure prime-weight word the linear coefficient after division by $\eta$ is $U$, and its constant term is $23$. The same argument applies. \end{proof} \begin{lemma}\label{found:C:3.4}\label{found:C:key:lem:jointfourier} Let $X,Y$ be integral labels in a normalized minimal table of dimension $D$, let $L=\operatorname{lcm}(d_X,d_Y)$ and $z=S_{XY}/L$. Let $G$ be the Galois group of the abelian field generated by $z$. For every character $\psi$ of $G$, \[ L\left|\sum_{u\in G}\psi(u)u(z)\right|\le\sqrt D . \] \end{lemma} \begin{proof} Choose a Galois lift $\sigma_u$ of every $u\in G$. The labels $X^{\sigma_u}$ are distinct since their entries at $Y$ are distinct. The same holds for $Y^{\sigma_u}$. The submatrix of $S/L$ on these rows and columns has entries $(uv)(z)$ because degrees are integral. Testing on a unit vector with coordinates $\psi(v)/\sqrt{|G|}$ gives the claimed inequality by whole-table orthogonality. \end{proof} In either crossing of Proposition~\ref{found:C:key:prop:fourmarks}, if $\eta\ne1$ and $Q=\operatorname{ord}(\eta)$, then \[ \frac{\sqrt D}{L}\ \ge 19\cos(2\pi/11)\sqrt{23Q}. \] Indeed $G=(\mathbb Z/Q)^\times\times\mathbb F_{23}^{\times}$ by the proven orbit size and conductor, also for the pure prime-weight word by the same stabilizer argument. Use a primitive character at $Q$ and a nontrivial character $\chi$ at $23$. Summing first at $Q$ kills each term with $7$-part one and gives a Gauss sum of modulus $\sqrt Q$ for every remaining term. Summing at $23$ kills the zero position and yields a common Gauss sum of modulus $\sqrt{23}$ times $\sum_{j=1}^3 m_j\overline{\chi(a_j)}$, where $m_j$ counts the letters with $7$-part $\eta$ at $a_j$. Thus $\sum_jm_j\ge19$. For each origin $c=0,1,3,14$ in the affine representative found by the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/word_closure.py}{four-position enumeration}, use the character sending the primitive root $5$ to $\exp(2\pi i k/22)$ with $k=18,7,11,9$, respectively. At the three other points shifted by $-c$, its exponents modulo $22$ are respectively $\{0,2,4\},\{10,11,14\},\{11\},\{2,4,5\}$. An overall dilation rotates all three values together. Projection onto the bisector of the resulting arc gives the indicated bound. In particular if $D=5^6 7^{2e}23^4$ and $5^3 7^{e-1}23\mid L$, then $\eta=1$ without a separation hypothesis: otherwise the upper bound is $161$ while $19\cos(2\pi/11)\sqrt{23\cdot7}> 19\sqrt{161/2}>161$. \subsection{Crossings of weight five at the primes three and five}\label{sec:weight-five-crossing} Consider an ordinary minimal table with dimension primes $2,3,5$. For a jointly $3$- and $5$-active column, suppose $v_5(D)=4$. Proposition~\ref{prop:principal-support} gives a two-free nonunit $Y\in P_3$ meeting the column with $d_Y=3^j5$; the height at five cannot be two by defect-zero vanishing. Take any three-free nonunit $X\in P_5$ of degree $2^i5$. The normalized crossing $z=S_{XY}/\ell$, $\ell=\operatorname{lcm}(d_X,d_Y)$, is congruent to $5$ above three and to $0$ above five by principality. Apply Proposition~\ref{found:C:5.1} with $(q,r)=(3,5)$. Its actual phase masses are either $2\ell,3\ell$ at $\eta,\eta\xi$, for $\xi$ primitive of order five by Theorem~\ref{found:S:4.1}, or precisely $\ell$ at each of $\eta,-\eta\eta_t\zeta_5^t$, $t=1,2,3,4$. In particular, tame five-inertia has trace $5\eta$ in either case. In the second case there are four phases at which the dyadic monodromy equals $-1$ and a fifth at which it is $1$. Both witnesses may be chosen in the fibers of trivial prime-to-principal twist, by Proposition~\ref{prop:principal-support}, so their actual dyadic twists are trivial. In the signed case at most one of the $r-1$ negative letters cancels the $r$-part of $\theta_X\theta_Y$, and at most $k+1$ cancel its $q$-part, where $k=(r-2)/q$. Thus at least $(r-3-k)\ell$ of the total dimension belongs to summands active at all three primes $2,q,r$. In particular, at the sextic frontier, absence of a twist of order divisible by $30$ forces the first spectrum for every such crossing. There is also an exact local valuation at the sextic frontier: at every prime above five, $v_5(z)=1/4$, with $v_5(5)=1$. This is immediate in the first case from $2+3\xi=5+3(\xi-1)$. In the second case expand in $\pi=\zeta_5-1$. The constant coefficient vanishes and, up to a unit, the linear coefficient is $\sum_{t=1}^4 t\eta_t$. If $\theta\mu_3$ is the ternary coset of the three nondistinguished letters, this reduces to $a+\theta B$, $1\le a\le4$, $B\in\mathbb F_{25}$. When $\theta^3\ne1$ its residue is outside $\mathbb F_{25}$, since $3\Vert25-1$. Hence the coefficient is nonzero. If $\theta^3=1$, there are indices $i,j$ with ternary part one and $k,l$ with parts $\omega,\omega^2$ respectively, with $\omega$ primitive of order three. The linear coefficient is $(i+j-l)+(k-l)\omega$, nonzero at five as $k\ne l$ and $\omega\notin\mathbb F_5$. Tame five-inertia consequently gives at least four different conjugates of this crossing and of each witness. Indeed it fixes the ternary coefficients, and the leading residue of $\pi$ is multiplied by each fourth root of unity in the residue field. \subsection{The conductor of a cyclic character subfield}\label{found:C:7} \begin{jointideallemma}[Local ideals on a joint twist fiber]\label{lem:joint-twist-ideals} Let $V$ have a basis $(v_i)$ and commuting actions of $G_p=\mathrm{SL}_2(\ZZ/p^{b_p}\ZZ)$ for a set $S$ of odd primes. Write $U_p(x),L_p(x)$ for the upper and lower unipotents, and $H_{p,a}=\operatorname{diag}(a,a^{-1})$ for the diagonal torus element or its action. Suppose $U_p(1)v_i=\theta_{i,p}v_i$. For $0\le k_pv_p(\ord\theta_X)$, and put $k_p=c_p-1$. Take $V$ spanned by the labels with the same actual twist components as $X$ outside $J$. Each local group at $p\in J$ preserves $V$ because it commutes with the complementary upper unipotents. On the orbit of $X$ under the product of the local tori in $J$, the unsigned action is the permutation action on a quotient of this abelian product. Its stabilizer fixes every $\chi_X(Y)$ and hence $F$. Thus $\psi$ restricted to the product occurs on the orbit, and the represented tori have a common eigenline of character $\epsilon_D\psi$ in $V$. For each $p\in J$ choose a local unit $a_p\equiv1\pmod{p^{k_p}}$ with $\psi(a_p)\ne1$, extending it by one in the other factors. Such a unit exists by the local conductor of $\psi$. Among the odd primes in $J$ set aside those for which $c_p=1$ and $v_p(D)$ is odd. Call this set $R$, and write $S$ for the other odd primes in $J$. For $p\in S$ the local quadratic sign is trivial at $a_p$. Indeed it is trivial on the entire local factor when $v_p(D)$ is even and on the principal units at every odd prime. Hence the eigenvalue of $A=\prod_{p\in S}(H_{p,a_p}-I)$ on the chosen line is $\prod_{p\in S}(\psi(a_p)-1)\ne0$. The space $E$ of Lemma~\ref{lem:joint-twist-ideals} is therefore nonzero. Its labels satisfy $p^{c_p}\mid\ord\theta_i$ for every $p\in S$. If $2\in J$, the unsigned permutation $P_{a_2}=\epsilon_D(a_2)H_{2,a_2}$ has eigenvalue $\psi(a_2)$ on the same line. Thus $(P_{a_2}-I)A\ne0$. The dyadic assertion of the lemma supplies a label of $E$ whose dyadic twist order is at least $2^{c_2}$. Fix the actual twist components of the resulting label outside $R$. Proposition~\ref{prop:blockwise-odd-slice} gives a proportion at least $\prod_{p\in R}(1-1/p)>0$ of squared-dimension mass active at every $p\in R$ in this fiber. Since $c_p=1$ for these primes, any such label has all required orders and retains the prescribed twists outside $J$. \end{proof} \begin{corollary}\label{found:C:7.2} If $\operatorname{ord}\theta_X$ is maximal under divisibility among the orders of all simple twists, then $K_X\subseteq\QQ(\theta_X)$. \end{corollary} \begin{proof} For every cyclic subfield of $K_X$, the theorem and maximality force its conductor to divide $\operatorname{ord}\theta_X$, so it lies in $\QQ(\theta_X)$. Their compositum is $K_X$, since the characters of the finite abelian group $\operatorname{Gal}(K_X/\QQ)$ separate its elements. \end{proof} \begin{corollary}\label{found:C:7.3}\label{found:C:key:new:klein} If three primes ramify in the field of a single row, and at least one of their inertia images has order greater than two, some simple has twist order divisible by their product. More precisely, absence of such a twist forces the three inertia images to be the distinct subgroups of order two in a Klein four group. \end{corollary} \begin{proof} Let $G=\operatorname{Gal}(K_X/\QQ)$ and let $I_p,I_q,I_r$ be the three nontrivial inertia images in $G$. By Theorem~\ref{found:C:key:new:single-twist} every character of $G$ is trivial on at least one of these subgroups if the stated twist is absent. Thus the three proper annihilators cover $G^\vee$. Recall the elementary three-subgroup argument. None of these subgroups is contained in the union of the other two: a group covered by two subgroups equals one of them. Each two of them therefore have the same intersection $N$ as all three; for if $x$ were common to the first two but outside the third, adding $x$ to an element only in the third gives one in none. Modulo $N$, the pairwise intersections are zero. Choose $a$ only in the first and $b$ only in the second. For any nonzero $a'$ in the first, both $a+b$ and $a'+b$ are in the third, whence $a-a'=0$. By symmetry all three images have size two. Thus the quotient is a Klein four group; dualizing gives the claimed inertia images. \end{proof} A second consequence uses the actual twist of the given row. If $q$ and $r$ ramify in $K_X$ and $\theta_X$ is $p$-active for a third prime $p$, some twist is active at all three. Indeed the two proper annihilators of the inertia images cannot cover the dual of $\operatorname{Gal}(K_X/\QQ)$. Choose a character outside both: its cyclic field has conductor divisible by $qr$, and the theorem retains the $p$-activity of $X$. For example, in the absence of a twist of order divisible by $30$, each complementary witness whose tame five-inertia orbit has length four is unramified at two or at three. This applies to the weight-five crossing of Proposition~\ref{found:C:key:new:zero} as well as the complementary weight-four crossing of Section~\ref{found:S:5}. More directly, if the twist of $J$ is active at two of the three dimension primes and its row field ramifies at the remaining prime, choose a cyclic subfield ramified there. The theorem produces a twist active at all three. Hence, at the ordinary sextic core in the absence of a twist of order divisible by $30$, every jointly active row is unramified at the third prime. In particular every jointly two- and three-active column has a five-unramified field. The principal hit chosen above at two in the trivial prime-to-three twist fiber of $P_3$ therefore has odd degree $75$ or $225$, and its twist has three-power order (possibly one). This applies even when both two and three mix separately with five. \subsection{Changing the primary components of a twist}\label{found:J:4} Work first in ordinary integral modular data with positive spherical structure. Write $s=S/\sqrt D$, $v_X$ for the coordinate vector of a simple, and $W_I$ for the product of the local Weyl operators at the primes of a subset $I$ of the congruence level. We use the linear lift of Dong--Lin--Ng \cite[Theorem~II]{DLNCongruence}, with the normalization of Section~\ref{subsec:sqvac}. Thus $s=W_IW_J$ when $I,J$ partition the level primes. A common scalar convention for $s$ does not affect any conclusion below. Different local groups commute. \begin{theorem}\label{found:J:4.1}\label{found:J:key:pr:route} Let $A,B$ be sets of labels such that their coordinate subspaces are invariant under $W_I,W_J$, respectively. Then the submatrix $s_{A,B}$ has exactly $|A\cap B|$ nonzero singular values, all equal to one. In particular \[ \sum_{X\in A,Y\in B}|S_{XY}|^2=D\,|A\cap B|. \] If $S_{XY}\ne0$, any choice of the $I$-components of $\theta_Y$ and the complementary components of $\theta_X$ occurs as a simple twist. If $I=\{p\}$, the realizing simple can be chosen in the full $p$-block of $X$. \end{theorem} \begin{proof} Write $P_A,P_B$ for the coordinate projectors. The two Weyl operators are unitary, so the invariance gives \[ P_A s P_B=W_I P_{A\cap B} W_J . \] For the second statement take $A$ to prescribe the $J$-components of $X$, and $B$ the $I$-components of $Y$. The nonzero entry forces a nonempty intersection. At a singleton $I$ we may intersect $A$ with the full $p$-block of $X$ by Lemma~\ref{lem:local-block-invariance}. \end{proof} \begin{proposition}\label{found:J:4.2}\label{found:J:key:pr:fiberdef} Fix a simple $X$, a prime $p\mid D$, its full $p$-block and all of its prime-to-$p$ twist components. In this set there is a simple $Z$ with trivial $p$-twist and $(W_p)_{XZ}\ne0$. There is a simple $Y$ in the same set with nontrivial $p$-twist and $(W_p)_{YX}\ne0$ if and only if $p\mid f_X$. If $X$ is itself $p$-active, the inactive $Z$ has $p\mid f_Z$. \end{proposition} \begin{proof} Use the proof of Theorem~\ref{found:J:key:pr:route} with $Y$ there the tensor unit. Since $s_{X0}=d_X/\sqrt D\ne0$, a term $(W_p)_{XZ}(W_{p'})_{Z0}$ is nonzero, giving the first assertion. If $p\nmid f_X$, its locally integral primitive idempotent makes its full $p$-block a singleton, and it is inactive by Theorem~\ref{thm:block} and Lemma~\ref{lem:complement}. Conversely suppose $W_pv_X$ has only inactive coordinates. The local upper unipotent is a scalar multiple of the actual $p$-twist, so both it and its conjugate by $W_p^{-1}$ preserve the $X$-line. They generate the local group; hence $W_p$ preserves the line. By Lemma~\ref{lem:complement} this implies $p\nmid f_X$. Every coordinate reached by $W_p$ is in the stated block and fiber. An inactive $Z$ with $p\nmid f_Z$ would lie in a singleton block and could not be joined by $W_p$ to an active $X$. \end{proof} A simple with twist order maximal under divisibility satisfies $\operatorname{rad}(f_X)=\operatorname{rad}(\operatorname{ord}\theta_X)$. Indeed Proposition~\ref{found:J:key:pr:fiberdef} would enlarge the twist order at any missing prime of $f_X$; the converse inclusion follows from Lemma~\ref{lem:complement}. More generally, if no twist is divisible by $p\prod_{t\in T}t$, then every label active at all primes of $T$ has $p$-unit codegree and its row vanishes on $p$-active labels. For square $D=n^2$, this is precisely $v_p(d_X)=v_p(n)$. The conductor-by-conductor bounds for inactive labels in the same fiber are proved in Proposition~\ref{found:F:3.1}. \subsection{A congruence for primes missing from joint twists}\label{found:J:5} \begin{proposition}\label{found:J:5.1}\label{found:J:key:pr:missing} Let $p$ be any prime and let $T$ be a nonempty set of other odd primes in a square-dimensional integral modular category. Prescribe some twist components outside $\{p\}\cup T$. Suppose no label in that fiber is active at every prime of $\{p\}\cup T$. If $X$ is in the fiber and $p\mid f_X$, then for every label $U$ and arbitrary independent local $t$-inertia automorphisms $\sigma_t$, \[ \prod_{t\in T}(1-\sigma_t)\chi_X(U)\in p\mathcal O , \] where $\mathcal O$ is the ring of integers of a common cyclotomic field. The ideal here is generated by $p$, including any ramification. \end{proposition} \begin{proof} All indicated local groups preserve the fiber. By Lemma~\ref{lem:joint-twist-ideals} at $k_t=0$, the image of $P=\prod_{t\in T}(I-H_t)$ lies in the span of $T$-local translates of its simultaneous $T$-active coordinate space: the ideals generated by $I-U_t(1)$ contain the respective torus differences and the factors commute. The $p$-local modular factor preserves the joint space, on which its upper unipotent acts by the normalization scalar, hence the Weyl $W_p$ acts by $\gamma_p^3$ there and on its translates (Proposition~\ref{deep:cluster}). Thus $W_pPv_X=\gamma_p^3Pv_X$. Write $B=\prod_{u\ne p}W_u$ over the complementary local level factors, so $s=BW_p$. If $k=v_p(\sqrt{f_X})\ge1$, every entry of the $X$-column of $B$ has $p$-valuation at least $1-k$. This is the proof of Theorem~\ref{fam:W:2.1} for the product of the complementary factors: choose an exponent divisible by the full $p$-level and equal to one on its complement. The Bruhat identity identifies $sT^m s^{-1}$ with $B^{-1}$ up to diagonal phases and a scalar unit. The mixed indicator and the $p$-strand congruence give $2v_p(B_{iX})+2k>0$, using the local torus at $-1$ on the other insertion as in that theorem. The entries of $B$ are unramified at $p$ by Galois covariance, so the valuation bound follows. It applies to every Galois conjugate of $X$. Multiply $sPv_X=\gamma_p^3 BPv_X$ by $\sqrt{f_X}$ and use Galois covariance of characters. \end{proof} \begin{corollary}\label{found:J:5.2}\label{found:J:key:pr:unsignedall} Suppose there is no twist of order divisible by $30$ in an ordinary minimal table on $2,3,5$. At every complementary $P_3,P_5$ crossing of weight four with five-free and three-free degrees, respectively, the lcm-normalized word is $1+3\xi$, $\operatorname{ord}\xi=5$. At every crossing of an odd $Y\in P_3$ of degree $3^j5$ with a three-free $X\in P_5$ of degree $2^i5$, it is $\eta(2+3\xi)$ with $\eta$ ternary. If $X$ is dyadically active in the latter case, $\eta=1$. \end{corollary} \begin{proof} Section~\ref{found:S:5} and Proposition~\ref{found:C:5.1} classify both words without bounds on the other exponents. In their signed alternatives the four nonidentity five-positions have coefficients $-\eta\eta_a$, $a=1,2,3,4$, with the multiset of the $\eta_a$ one copy of one plus one coset of the cubic roots in the ternary roots. At weight four $\eta=1$ and the constant coefficient $c_0=0$; at weight five $c_0=\eta$. If both witnesses are dyadically inactive, all four negative packets are dyadically active. At most one cancels the five-twist of the product of the factors, and at most two cancel its three-twist. This forces a twist divisible by $30$. If the $P_3$ witness is dyadically active, apply the preceding congruence with $p=3,t=5$ in its dyadic twist fiber. Its $3$-defect is positive by principality, and the opposite degree divided by the weight is a $3$-unit. Thus all five-inertia conjugates of the normalized word agree modulo $3\mathcal O$. A tame generator permutes the four nonzero five-positions transitively and fixes the coefficient at zero. Over the ternary cyclotomic integers the only relation on these five roots modulo three is the constant vector, by linear disjointness and the monic cyclotomic basis. The four nonzero coefficients would agree modulo $3\mathcal O$. Distinct three-power roots differ by an element of $3$-valuation at most $1/2$. This rules out the signed alternative. If the $P_5$ witness is dyadically active, use $p=5,t=3$. The opposite degree divided by the weight is a five-unit, hence $c_0-c_a$ is invariant modulo five under ternary inertia for every $a$, by the same integral-basis argument now reduced modulo five. At weight four this would fix all the distinct $\eta_a$ under ternary inversion. At weight five one $\eta_a$ equals one, so this first fixes $2\eta$, whence $\eta=1$, and then every $\eta_a$, again impossible. Reduction modulo a prime over five is injective on three-power roots. In the unsigned weight-five case the difference $ (\eta-\sigma\eta)(2+3\xi)$ is divisible by five; the second factor has five-valuation $1/4$, forcing $\eta=1$ as asserted. \end{proof} The congruence extends to whole residue blocks. In the absence of a joint $p,t,r$ twist, let $B$ be a full $p$-block containing a label of positive $p$-defect. If local $t$-inertia moves its residue character, every label in $B$ is $r$-inactive. Indeed each member has positive $p$-defect, for one of defect zero is a singleton block. An $r$-active member would have $t$-invariant residue character by the singleton congruence proved above. Galois covariance in the \emph{argument} shows that this applies to the entire $t$-orbit of $B$ even at a fixed prime above $p$. \subsection{Character fields at twist one}\label{found:J:6} Let $S$ be ordinary symmetric integral modular data of dimension $D$, with $SS^*=DI$. Let $G$ be the Galois group over $\mathbb Q$ of a cyclotomic field containing all the character fields $K_X=\mathbb Q(\chi_X)$. For a character $\psi$ of $G$, use the same symbol for its descent to a quotient whenever it is trivial on the kernel. \begin{proposition}\label{found:J:6.1}\label{found:J:key:new:fourierblock} Choose representatives $X$ and $Y$ of any two lists of distinct Galois orbits (the two lists may intersect). Form a matrix indexed by the representatives for which $\psi$ descends to their character fields. Put $L_{XY}=\operatorname{lcm}(d_X,d_Y)$, $z_{XY}=S_{XY}/L_{XY}$ and $F_{XY}=\mathbb Q(z_{XY})$. The following matrix has operator norm at most $\sqrt D$: \[ A_{XY}= L_{XY}\sqrt{[K_X:F_{XY}][K_Y:F_{XY}]} \sum_{u\in\operatorname{Gal}(F_{XY}/\mathbb Q)} \psi(u)u(z_{XY}) \qquad\text{if $\psi$ descends to $F_{XY}$}. \] The entry is zero otherwise. In particular every row and every column has squared norm at most $D$. \end{proposition} \begin{proof} On the orbit of $Y$, take the unit vector whose coordinates are $\psi(v)/\sqrt{[K_Y:\mathbb Q]}$, with $v$ ranging over $\operatorname{Gal}(K_Y/\mathbb Q)$. On the orbit of $X$ take the analogous vector with $\overline\psi$ in place of $\psi$. These are orthonormal families on each side. The matrix coefficients of $S$ between them are the double averages of $L_{XY}\psi(u)\psi(v)(uv)(z_{XY})$. Here lifts from the two fields give the same answer by symmetry and the covariance of characters, since the degrees are rational. Summing on the fibers of restriction to $F_{XY}$ gives zero if $\psi$ is nontrivial on a kernel; otherwise it gives the displayed coefficient. Thus $A$ is a compression of $S$ between two isometries. \end{proof} \begin{proposition}\label{found:J:6.2}\label{found:J:key:pr:carry} In ordinary integral modular data with positive spherical structure let $u,v,w$ be three distinct primes. Suppose the other local groups act by scalars. Let $X\in P_u$ have $u$-power twist and $Y\in P_v$ have $v$-power twist, allowing twist one in either case. Write $G$ for the Galois group of the cyclotomic splitting field. Let $\psi$ be a character of $G$. For $j=u,v$ suppose that inverting just the $j$-primary component of $\psi$ leaves its kernel unchanged. In particular this holds for a character of prime-power conductor. If \[ \sum_{a\in G}\psi(a)\,a(S_{XY})\ne0 , \] then $P_u$ and $P_v$ have twist-one members in a common full $w$-block. Their character fields both contain the cyclic field cut out by $\psi$. \end{proposition} \begin{proof} By Proposition~\ref{found:J:key:new:fourierblock} the orbit vectors affording $\overline\psi$ at $X$ and $\psi$ at $Y$ have a nonzero matrix coefficient for $s$. Now $s$ is a scalar multiple of $W_uW_wW_v$. Move $W_u^*$ onto the first vector and $W_v$ onto the second. For completeness, when $D$ is nonsquare the diagonal torus is $H_a=\epsilon_D(a)P_a$, with $P_a$ the permutation matrix and $\epsilon_D(a)=a(\sqrt D)/\sqrt D$ (Theorem~\ref{found:C:7.1}). This quadratic character is unchanged by inversion in any subset of the local factors. Thus the conjugation rule for a Weyl and a torus also holds with $P_a$. A local Weyl inverts the torus on its own factor, so both new characters have the original kernel. The new row vector is supported on pure $u$-power twists in $P_u$, and the new column vector on pure $v$-power twists in $P_v$. A nonzero $W_w$-entry between their supports preserves the complementary twists and the full $w$-block (Lemma~\ref{lem:local-block-invariance}). Hence the two twists are one. The stabilizer of any nonzero coordinate of an eigenvector is in the kernel of its character, giving the two field inclusions. \end{proof} \subsection{Multiplicities of quadratic inertia characters}\label{found:J:7} In square-dimensional ordinary integral modular data let $H_p$ be the group of local diagonal tori at an odd prime $p$, acting on the simple labels by the Galois permutations. Use the linear lift of Section~\ref{subsec:sqvac}. For a set $T$ of odd primes dividing the level, consider a coordinate subspace $E$ stable under the local groups at every $p\in T$. For example, one may prescribe the complementary primary twist components, and for a single $p$ one may also prescribe its full $p$-block. Write $H=\prod_{p\in T}H_p$ and let $\lambda=\prod_{p\in T}(\frac{\cdot}{p})$ be its product of quadratic characters. Retain the $H$-orbits in $E$ on which $\lambda$ is trivial on the stabilizer; write $m_1,\ldots,m_b$ for their sizes. \begin{proposition}\label{found:J:7.1}\label{found:J:key:pr:qmult} One has $2^{|T|}\mid b$. Moreover, putting $K=\QQ(\sqrt p:p\in T)$, \[ \prod_{i=1}^b m_i\ \in \operatorname{disc}(K)^{\,b/[K:\QQ]}\, N_{K/\QQ}(K^\times)\,(\QQ^\times)^2 . \] In particular, in every full five-block and specified prime-to-five twist fiber, the number of five-inertia orbits of tame order four is even, and so is the number of tame order two. \end{proposition} \begin{proof} The rational $\lambda$-eigenspace $V$ has the orthogonal basis of signed orbit sums on the retained orbits. Its positive form thus has Gram matrix $\operatorname{diag}(m_1,\ldots,m_b)$. Write $p^*= (-1)^{(p-1)/2}p$. On $V$ the Weyl operator $W_p$ preserves the character, since it inverts the local torus and the latter acts by signs. For a cyclotomic automorphism $\sigma_a$ the congruence representation gives $\sigma_a(W_p)=H_{a,p}W_p$, with the opposite torus convention producing the same signs. This follows by conjugation by $\operatorname{diag}(a,1)$ in Lemma~\ref{lem:local-vacuum-column}. Consequently $A_p=\sqrt{p^*}\,W_p$ has rational matrix on $V$. Here $W_p^2=H_{-1,p}=(\frac{-1}{p})I$. Unitarity gives $A_p^*=A_p$ and $A_p^2=pI$. These commuting operators make $V$ a vector space over the stated multiquadratic field $K$, of degree $2^{|T|}$. The zero-dimensional case is immediate. The rational symmetric form $B$ is balanced over $K$ since all $A_p$ are self-adjoint. Nondegeneracy of the field trace defines a symmetric $K$-bilinear form $B_K$ by $\operatorname{Tr}_{K/\QQ}(uB_K(x,y))=B(ux,y)$ for all $u\in K$. Diagonalize it over $K$. Taking the trace of a one-dimensional summand with coefficient $c$ gives determinant $\operatorname{disc}(K)N_{K/\QQ}(c)$ up to rational squares, proving the formula. For $p=5$, the retained orbit sizes have the form $2\cdot5^a$ or $4\cdot5^a$. The prime-to-five unit of every nonzero norm from $\QQ(\sqrt5)$ is a square modulo five: in $x^2-5y^2$ the two summands have valuations of opposite parity, and $-1$ is a square modulo five. The discriminant is $5$ and the unit part of a rational square is a square. Since two is a nonsquare, the number of tame-order-two orbits is even. Subtraction from the even dimension proves the assertion at tame order four. \end{proof} \subsection{The conductor filtration inside a twist fiber}\label{found:F:3} Let $D=n^2$ in ordinary integral modular data. Fix an odd prime $p$, one full $p$-block (or all blocks), and the exact prime-to-$p$ twist components. Write $E$ for the resulting nonempty fiber and $P_E$ for its coordinate projector. Let $\Gamma_E$ be the subgroup of all diagonal tori preserving $E$, and let $b$ count its orbits on the $p$-inactive labels of $E$. Write \[ m(\alpha)=\sum_{X\in E:\,\theta_{X,p}=\alpha}d_X^2,\qquad J=\{j\ge2:\ m(\alpha)>0\text{ for some $\alpha$ of order }p^j\}. \] \begin{proposition}\label{found:F:3.1}\label{found:F:key:pr:low-conductor}\label{found:J:4.3}\label{found:J:key:prop:inactive-multiplicity} All nonidentity $p$th roots have the same mass $m_*$. One has \[ m_*\le(p+1)m(1),\qquad b\ge |J|+\mathbf1_{m_*>0} +\mathbf1_{m_*<(p+1)m(1)}\ge |J|+1 . \] For the number of inactive labels, replace $|J|$ by $\sum_{j\in J}|G_j^2|$, where $G_j=(\mathbb Z/p^{\lfloor j/2\rfloor})^\times$. At odd $p$ one has $|G_j^2|=\varphi(p^{\lfloor j/2\rfloor})/2$. If all inactive labels lie in one orbit and the fiber is active, its only twists at $p$ are the $p$th roots, with $m_*=(p+1)m(1)$ and total mass $p^2 m(1)$. In particular $m(1)=d_Y^2$ for a singleton $Y$. \end{proposition} \begin{proof} Write $U,L,H,W$ for the local upper, lower, torus and Weyl operators. Put $v=nP_E W_{p'}v_0$. The upper unipotents and diagonal torus at $p$ fix $v$, as does $\Gamma_E$, since conjugation across $W_{p'}$ only inverts the complementary tori and all tori fix the vacuum. The vector $Wv$ has entries $d_X$ on $E$ and zero elsewhere. At local level $p^a$ let $P_j$ average over the kernel of reduction modulo $p^j$, for $1\le j\le a$. Its lower--diagonal--upper factorization is unique, the diagonal entries being $1\bmod p^j$. Averaging the factors in this order therefore gives $P_jv=\operatorname{avg}_{p^j\mid t}L(t)v$. Conjugation by $W$ shows that $W P_jv$ is exactly the projection of $Wv$ onto $p$-twists of order dividing $p^j$. (Take $j=1$ to mean the identity projection when $a=1$.) The mutually orthogonal vectors $(P_j-P_{j-1})v$, $2\le j\le a$, are thus nonzero precisely at $j\in J$. Each is upper-fixed and $\Gamma_E$-fixed, because its projectors commute with the local group and the other local factors. They lie in the space of invariant vectors on the inactive labels, of dimension $b$. Fix such a nonzero layer $v_j=(P_j-P_{j-1})v$, on which the action factors through level $p^j$. Put $l=\lceil j/2\rceil$ and $d=j-l$. The kernel of reduction modulo $p^l$ is abelian, of elements $I+p^l A$ with $A\in\mathfrak{sl}_2(\mathbb Z/p^d)$. As $v_j$ is upper- and torus-fixed, the characters in its decomposition on this kernel have the form $A\mapsto\exp(2\pi i cA_{21}/p^d)$. Some $c$ is a unit: otherwise $L(p^{j-1})$ fixes $v_j$, although $Wv_j$ has exact twist order $p^j$. Every square-unit multiple of this $c$ occurs by torus invariance. Their character components are orthogonal, and are upper-fixed since upper conjugation fixes $A_{21}$. They give the indicated number of independent vectors on the inactive labels within this layer; different layers are orthogonal. It remains to calculate in the quotient $G=\mathrm{SL}_2(\mathbb F_p)$ on $u=P_1v$; its Weyl image gives all masses at orders one and $p$. Let $B$ be its upper Borel, $P_B$ its average, and $A=P_B W P_B$ on $B$-fixed vectors. Then \[ A=P_G-p^{-1}(I-P_G). \] Indeed the two double cosets are $B$ and $BWB$, of relative sizes $1:p$. Averaging over $G$ therefore gives $P_G=(I+pA)/(p+1)$ on this space. Write $u=u_0+u_1$ with $u_0=P_Gu$ and $u_1=u-u_0$. The mass at one equals $\|Au\|^2$, since $Wu$ is torus-fixed. For $t\in\mathbb F_p^\times$, the element $W^{-1}U(t)W$ belongs to $BWB$, so the $t$th moment of the masses is $\langle u,Au\rangle$. Finite Fourier inversion gives \[ m(1)=\|u_0\|^2+p^{-2}\|u_1\|^2,\qquad m(\alpha)=(p+1)p^{-2}\|u_1\|^2 \quad(\alpha\in\mu_p\setminus\{1\}). \] Each nonzero $u_i$ supplies a further orthogonal invariant vector on the inactive labels. There is at least one such vector: $v\ne0$ is upper-fixed, hence there are inactive labels, so $m(1)>0$. \end{proof} The gap between the two masses also gives a signed integer identity at every local level: \[ m(1)-m_*= n\!\sum_{X\in E:\,\theta_{X,p}=1} d_X\,(W_{p'}v_0)_X . \] Indeed, for a unit $t$ in the local ring, $W^{-1}U(t)W\in BWB$ for its upper triangular subgroup $B$. Its coefficient at $v$ equals $\langle v,Wv\rangle$. Averaging such coefficients over the units kills roots of order at least $p^2$ and averages a primitive $p$th root to $-1/(p-1)$. The entries $n(W_{p'}v_0)_X$ are rational integers divisible by $p$ by Lemma~\ref{lem:local-vacuum-column}. The same result holds at $p=2$ whenever the dyadic anomaly scalar $\gamma$ is $1$ or $-1$. If $\gamma=-1$, tensor the local lift with the sign character of $\mathrm{SL}_2(\mathbb F_2)\cong S_3$: the normalized upper generator then becomes the actual dyadic twist, the diagonal tori are unchanged, and the Weyl changes only by a sign. In particular this applies if a nonempty exact odd-twist fiber has only dyadically inactive labels. The dyadic group preserves this fiber, so its upper acts there by $\gamma^{-1}$, and Bruhat gives $W_2=\gamma^3 I$ there. Its square is an unsigned torus permutation, hence $\gamma^6=1$; the lift has dyadic $\gamma$, giving the assertion. At $D=2^63^65^4$ a joint three--five column whose five-twist has order $25$ consequently requires two Galois-inequivalent five-inactive labels in its full five-block with the same dyadic and ternary twist components (inequivalence is under $\Gamma_E$). This strictly strengthens a count of two labels when those labels are conjugate. There is a joint three--five column at this dimension, so absence of a twist of order divisible by $30$ supplies the odd-twist fiber just used and gives the dyadic assertion on every block and complementary fiber. In particular, if $b=1$ there and the fiber is dyadically active, its actual dyadic twists are $1,-1$ and the mass at $-1$ is precisely three times the mass at $1$. \begin{proposition}[Joint conductor-profile transform]\label{lem:conductor-profile-transform} In ordinary integral modular data of square dimension $D=n^2$, assume that the dyadic anomaly is $1$ or $-1$ whenever two occurs, with the sign correction above in the latter case. Write $M_{\boldsymbol i}$ for the squared-degree mass at twists of exact order $\prod_p p^{i_p}$. On an array indexed by these exponents apply the following operation in each prime coordinate, with the other indices fixed: \[ (M_0,M_1,M_2,\ldots)\longmapsto \left(M_0-\frac{M_1}{p^2-1},\, \frac{p^2M_1}{p^2-1},\,M_2,\ldots\right). \] All entries of the resulting array $V$ are nonnegative, and the number of its positive entries is at most the number of distinct Galois orbits of twist-one labels. \end{proposition} \begin{proof} Put $Q_{p,0}=P_{G_p}$, $Q_{p,1}=P_1-P_{G_p}$, and $Q_{p,j}=P_j-P_{j-1}$ for $j\ge2$. Then $V_{\boldsymbol j}=n^2\|\prod_p Q_{p,j_p}v_0\|^2$. To see this, compress each local twist-order projector, conjugated by its Weyl, to the space of upper-Borel fixed vectors. At order one its compression is $Q_{p,0}+p^{-2}Q_{p,1}$, at order $p$ it is $(1-p^{-2})Q_{p,1}$, and at order $p^j$, $j\ge2$, it is $Q_{p,j}$. These are the squared-norm formulas in the proof above, applied to an arbitrary upper-Borel fixed vector and polarized. Distinct local factors commute, so their compressed projectors can be multiplied inside the vacuum coefficient. Inverting the resulting two by two tame relation gives the stated array. Its nonzero vectors are orthogonal, fixed by every torus and every upper unipotent. They are consequently supported on twist-one labels and constant on their Galois orbits. \end{proof} In the forward relation of Proposition~\ref{lem:conductor-profile-transform}, a profile $n'$ can contribute to order $m$ only if $n'=md$ with $d$ square-free and coprime to $m$. Every positive $V$-profile also occurs as an actual twist order. Thus every occurring $m$ for which no such strict extension occurs gives its own positive entry. At the even core without a twist order divisible by $30$, this applies to \emph{every distinct mixed order}, even when one mixed order divides another. Hence their number is bounded by the number of Galois orbits at twist one. If all three pairs occur there are at least three such orbits, of which at least two are inequivalent nonunit orbits. There is an unconditional form on the branch admitting an order divisible by $30$. Fix $a\ge0$ and take $E$ to be the union of all dyadic phases of exact order $2^a$. Work only with the local groups at three and five, with initial vector $n P_E W_2v_0$. It is fixed by both upper Borels and by all tori, and its two local Weyls give the degree vector on $E$. Applying the proof of Proposition~\ref{lem:conductor-profile-transform} only at the two odd primes gives \[ \#\{(b,c)\in\Z_{>0}^2:2^a3^b5^c\in\mathcal T\} \ \le\ \#\bigl(\{X:\ord(\theta_X)=2^a\}/\operatorname{Gal}\bigr). \] Here $\mathcal T$ is the set of actual twist orders. Indeed each profile with both odd exponents positive already specifies every odd prime, so no other profile can contribute to it. This holds for each $a$, with arbitrary dyadic central charge, and with the same proof for any prescribed set of odd primes in a Galois-stable union of complementary twist fibers. \paragraph{Primitive odd components at a fixed dyadic phase.} The filtration can equally be applied at just three and five, fixing the dyadic twist. In the next formulas the total masses sum over both possible signs at three and all four possible primitive phases at five, as appropriate. Fix an individual nontrivial dyadic phase $\delta$ at $D=2^63^65^4$. Write $M_{ij}$ for the mass with $\theta_{X,2}=\delta$ and odd primary orders $3^i,5^j$, $i,j\in\{0,1\}$. Then \[ M_{11}\le 8M_{01},\qquad M_{11}\le24M_{10},\qquad M_{10}\le8M_{00},\qquad M_{01}\le24M_{00}, \] and \[ 192M_{00}-24M_{10}-8M_{01}+M_{11}\ge0. \] Every one of the eight individual phase pairs $(\zeta_3^u,\zeta_5^v)$, $u=1,2$ and $1\le v\le4$, has mass $M_{11}/8$. This follows by applying the individual balance successively, fixing the other primary twist each time. If this mass is positive, the degree transfers and activity height bounds give the triple degrees $2^i3^j5$, $0\le i\le2$, $1\le j\le2$. The full quartic tame five-inertia result for jointly three- and five-active labels shows that the unique involution of tame five-inertia moves every such label. It fixes each of these eight phases, by the square action on the fifth roots, and preserves the degree. Thus each degree occurs with even multiplicity at each phase. Since every such degree is divisible by $15$, \[ 3600\mid M_{11}. \] There are at least sixteen labels across the eight phases, and at least four distinct full Galois orbits: the square classes of the two odd components are unchanged under the square Galois action. All these conclusions concern exact odd primary orders $3$ and $5$. \paragraph{Opposite dyadic phases for arbitrary central charge.} Write $v_E=\sum_{X\in E}d_X v_X$ for the degree vector restricted to a fiber. Let $p=2$, fix the odd twist components and optionally one full two-block, and write $m(\delta)$ for the mass of an individual dyadic phase in the resulting set $E$. For the multiplicative central charge $\xi$ one has \[ m(\delta)=m(-\delta)\qquad(\delta\ne\pm\xi). \] In fact the integral lift has $\gamma^3=\xi$, $\gamma^8=1$, and the actual dyadic twist operator is $\gamma U(1)$. Conjugating the upper unipotents on the vacuum shows that $L(y)v_E=\gamma^y v_E$ for the standard lower unipotents. The dyadic diagonal tori $H(a)$ fix $v_E$. For odd $u$ the identity \[ U(u)=L(u^{-1})H(-u)W L(u^{-1}),\qquad W=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \] therefore gives \[ \sum_{X\in E}d_X^2\theta_{X,2}^{\,u} =\gamma^{u+2u^{-1}}v_E^*Wv_E=\xi^u v_E^*Wv_E . \] The exponent calculation uses $u^{-1}\equiv u\pmod8$ (enlarge the local modulus to include $8$ if necessary). The coefficient $v_E^*Wv_E$ is an integer by the local Weyl vacuum column formula. Thus $\sum_{X\in E}d_X^2\theta_{X,2}/\xi$ is rational. In the power basis of a dyadic cyclotomic field, opposite roots cancel and one representative of each opposite pair gives a basis. Rationality forces equality of the two coefficients except in the pair $\{1,-1\}$. If $\xi$ is nonreal, every nonempty odd-twist fiber in a full two-block has the same positive mass at its dyadic phases $1$ and $-1$, since it contains a dyadically inactive label by Weyl routing. Consequently any joint ternary and quintic phases, including arbitrary higher orders, then occur with dyadic phase exactly $-1$ in that block. At any central charge, when a nonexceptional $\delta$ admits a triple with odd primary orders $3$ and $5$, the opposite phase $-\delta$ contains the same mass at each of the eight individual odd-phase pairs. \paragraph{Wild inertia at an inactive prime.} Work in ordinary integral modular data of square dimension. At a prime $p$ let $U(t)$, $L(t)$, $H(a)$ be the upper, lower and diagonal subgroups of the local modular representation at level $p^e$, and $W_p$ its Weyl element. Torus actions are the unsigned Galois permutations. At two a common normalization scalar shifts the phase labels; we formulate the local statement for the actual representation and use its categorical consequences only at odd $p$. \begin{proposition}\label{por:prop:wild} Suppose $v$ is fixed by $U(t)$ and has torus character $\psi$. Let $p^b$ be the conductor of $\psi^2$, $b\ge1$. The squared masses of $W_pv$ on the distinct upper-unipotent phases are constant on each coset of $\mu_{p^b}$ in $\mu_{p^e}$. In particular the squared mass on any single phase is at most $p^{-b}\|v\|^2$. The assertion also applies to the global normalized $S$-image of a vector with these local properties, since the complementary Weyl is a commuting unitary. \end{proposition} \begin{proof} If $j=v_p(t)U,\qquad W(s,j+1)\text{ holds for every }1\le j\le m \text{ and every prime }s>jq. \] For every prime $r>q$ and every $B\ge0$, each braided fusion category of dimension dividing $p^{2h}q^B r^4d$ is weakly group-theoretical. \end{theorem} \begin{proof} In a remaining divisor-minimal table the $r$-valuation is four, while square divisibility bounds every $p$-height by $h$. Lemma~\ref{three:complementary} gives an $r$-free nonunit in the principal $q$-block and a $q$-free nonunit in the principal $r$-block. For every such pair the degrees are $p^i q^j$ and $p^t r$, and the weight of the lcm-normalized entry is $g=p^{\min(i,t)}\ge q$. Thus $m=0$ is impossible. If $r\le U$, then $j=\lfloor r/q\rfloor$ satisfies $1\le j\le m$, $r>jq$ and $j+1U$. If any of the above pairs has $g\le U$, then $r\ge g+2$ since both are odd. Theorem~\ref{first:balance}, applied to the two principal residues, expresses the normalized entry as a positive $r$-word of weight $g$ on at most $1+\lfloor g/q\rfloor$ positions. Multiplying by $\operatorname{lcm}(d_X,d_Y)/d_Y$ gives a character of full weight $d_X$. Since $r>g\ge q\lfloor g/q\rfloor$, the word hypothesis excludes this pair. Consequently every opposite pair has $g=G$. All its witnesses have $p$-height exactly $h$, forcing the dimension's $p$-valuation to be exactly $2h$ and their codegrees to be $p$-units. Every $q$-active column has a nonzero entry at an $r$-free nonunit of the principal $q$-block by Proposition~\ref{prop:principal-support}, applied modulo $r$. This row has $p$-unit codegree, hence vanishes on every $p$-active column by Theorem~\ref{thm:block}. Thus $p$ and $q$ never occur together in a twist. The same argument on the principal $r$-block separates $p$ from $r$. By Lemma~\ref{first:isolate}, the quartic $r$ cannot be isolated, so some column is both $q$- and $r$-active. At this column apply Proposition~\ref{prop:principal-support} on the principal $q$-block modulo $p$. The resulting $p$-free nonunit has degree $q^j r^v$ for $v\in\{1,2\}$ by Lemma~\ref{three:complementary} and square divisibility. At $v=2$ its row would vanish on this column by its $r$-unit codegree. Thus $v=1$, exactly the hypothesis that gives the first word in Proposition~\ref{fam:Q:3.1}. Its support has at most $1+\lfloor r/q\rfloor$ positions, with no restriction on $p$ in this construction. If $rG$, apply Theorem~\ref{first:balance} to an opposite pair of weight $G$: the resulting word contradicts $W(r,1+m)$. Equality $r=G$ is impossible. \end{proof} \begin{corollary}\label{found:S:2.2}\label{found:S:key:cor:topheight} The word hypothesis in Theorem~\ref{found:S:key:thm:topheight} holds whenever $q\ge(m+1)^2$. In particular the theorem applies whenever $h\ge3$ and $q>p^{h-1}$. For $h\ge6$ it applies throughout $(p-1)q>p^{h-1}$. \end{corollary} \begin{proof} With $E=j(j+1)/2$, put $g(v)=\min\{t\ge1:t(t+1)\ge v\}$ and $C(E)=1+\max_{1\le v\le E}v(v-1)/g(v)$ as in Lemma~\ref{fam:Q:2.3}. At $j=1$ the two-position theorem applies. For $j\ge2$, the scalar estimates in the proof of Theorem~\ref{lem:principal-crossing-prime-bound} bound both $3(1+2E)/2$ and $2C(E)+1$ by $j(j+1)^2$. This is strictly less than $s$ since $s>jq$ and $q\ge(m+1)^2$. If $h\ge3$ and $q>p^{h-1}$, then $m+1\le p$ and $q>p^{h-1}\ge p^2$. Finally, if $h\ge6$ and $(p-1)q>p^{h-1}$, then $m+1\le p(p-1)$ and $q>p^{h-2}\ge p^4>(m+1)^2$. \end{proof} The all-$r$ criterion in Corollary~\ref{fam:Q:4.3} required $m+111$. This includes $r=23$, where the ratio test of Theorem~\ref{fam:Q:1.2} also has $1+\lfloor r/q\rfloor=3$. Direct use of the strict scalar inequalities gives examples below $q=p^{h-1}$: the same theorem covers $5^6 23^B r^4d$ and $7^6 47^B r^4d$ for all $B$ and all $r$ above the middle prime. Indeed their respective values of $m$ are five and seven, and for $j=1,\ldots,7$ the exact values of $2C(j(j+1)/2)+1$ are $3,9,33,63,108,193,1527/5$. Each required inequality follows from $s>jq$. These examples illustrate the criterion; no prime value is used in its proof. \begin{proposition}\label{found:S:2.3}\label{found:S:key:prop:twoisolated} Let $h,k\ge2$, $U=p^{h-1}$, $V=q^{k-1}$, and $m=\max\{\lfloor U/q\rfloor,\lfloor V/p\rfloor\}$. For any prime $r>\max(U,V)$ for which $W(r,1+m)$ holds, every braided fusion category of dimension dividing $p^{2h}q^{2k}r^4d$ is weakly group-theoretical. \end{proposition} \begin{proof} Repeat the opposite-pair argument of Theorem~\ref{found:S:key:thm:topheight} first with $p$ as the common degree prime. Every common weight $g\le U$ would force a word on at most $1+\lfloor U/q\rfloor$ positions, so all complementary degrees have top $p$-height. The codegree and prime-avoidance argument isolates $p$. Now use the principal blocks at $p$ and $r$ instead, so the common degree prime is $q$. Lemma~\ref{three:complementary} gives for every such pair a common weight $g=q^a\ge p$. At $g\le V$ we have $r\ge g+2$ and Theorem~\ref{first:balance} gives a word on at most $1+\lfloor V/p\rfloor$ positions. Thus all those witnesses have upper height $k$, forcing valuation $2k$ and $q$-unit codegrees. Applied on the principal blocks at $p,r$, prime avoidance now isolates $q$. No twist can then contain $r$ and either other prime, contrary to Lemma~\ref{first:isolate} at the quartic $r$. \end{proof} \subsection{An isolated prime with unramified principal witnesses}\label{found:S:3} \label{found:S:key:sec:oddunram} This obstruction depends only on the exponent at the least prime. It does not require a quartic at either companion. \begin{lemma}[Trace and mass at an isolated prime]\label{found:S:3.1}\label{found:S:key:lem:unramisol} Let $p$ be isolated in an odd ordinary minimal table with $v_p(D)=2h$, $h\ge3$. Write $D=p^{2h}n^2$. Each $p$-active $Z$ has degree $p^jn$ with $0\le j0$, no $X\in P_p$ of height $h-1$ can have nonzero absolute trace $T_X$ in the $Z$-row of \eqref{eq:r8-tracenorm}. \item If all active labels have degree $n$, their number $N$ satisfies $N\ge2p^{h-1}(p-1)$. In that case $P_p$ has no nonunit of height $h-1$. \end{enumerate} In particular some active column has no hit on a principal row of height $h-1$ unramified at $p$. \end{lemma} \begin{proof} The descriptions of $d_Z$ and $K_Z$ follow from isolation, Lemma~\ref{lem:complement} and Corollary~\ref{cor:field-conductor}. In (1), $x=\chi_Z(X)$ satisfies $x/p^{h-1-j}\in\mathcal O_{K_Z}$ by lcm divisibility. By symmetry this quotient equals $\chi_X(Z)d_X/(p^{h-1-j}d_Z)$, whose second factor is a $p$-unit. Principality gives zero residue above $p$. Since $K_Z$ is totally ramified at $p$, the rational trace of the quotient from $F_X=\QQ(x)\subseteq K_X\cap K_Z$ is divisible by $p$. A nonzero $T_X$ would thus have square at least $p^{2(h-j)}=f_Z$, leaving no room for the unit. For (2), choose a CRT exponent $m$ equal to one on the $p$-level and zero on the complementary level, including the dyadic anomaly. Lemma~\ref{lem:local-vacuum-column} and Proposition~\ref{deep:cluster} give \[ p^{-h}\sum_{Z\ \text{$p$-active}}(1-\theta_{Z,p}) =p^h(1-(W_p)_{00})\in\mathbb Z . \] The integer is positive by taking real parts and even since distinct dual labels have the same twist and $p$ is odd. Each summand on the left before division has normalized trace at most $p/(p-1)$, giving the lower bound. Any $X\in P_p$ hits every one of these labels by principality. If $v_p(d_X)=h-1$, then $S_{XZ}/(p^{h-1}n)$ is a nonzero algebraic integer by lcm divisibility. Its Galois mean squared modulus is at least one, by the norm and the arithmetic--geometric mean inequality. Averaging the row norm and counting the unit gives $Np^t,\qquad (s-p^t)(\ell-1)>p^t . \] At an eligible height put \[ b_s(t)=\frac{s-1}{\max_{a\mid s-1,\,a\le p^t} a/\gcd(a,2)}, \qquad c_s(t)=(p-1)b_s(t). \] At every other height set $b_s(t)=c_s(t)=1$. Let \[ A_s=\min_t p^{2(t-1)}c_s(t), \] omitting eligible $t$ for which $W(s,p^t)$ holds, and interpreting an empty minimum as infinity. Then $A_q+A_r0$. For each companion $s$, the unit term of \eqref{eq:r8-tracedim} is one, so reduction modulo $s$ supplies an $s$-free nonunit $X$ with nonzero trace $T_X$ on the $Z$-row. It belongs to $P_p$ by Theorem~\ref{thm:principal-support}. Write $d_X=p^t u^i$, where $u$ is the other companion and $t,i>0$ by Lemma~\ref{three:complementary}. Defect-zero vanishing excludes $t=h$, and Lemma~\ref{found:S:3.1} excludes $t=h-1$. If $j\le t$, the proof of that lemma gives $p^{t+1-j}\mid T_X$: divide $\chi_Z(X)$ by $p^{t-j}$, then use its zero residue and total ramification of its own field at $p$. If $j>t$, integrality alone suffices. In either case $d_Z^2 T_X^2\ge p^{2(t+1)}n^2$. At an eligible height cross $X$ with a $u$-free nonunit of $P_s$. The gcd is some $w=p^k$ with $1\le k\le t$, and the normalized entry has residue $w$ over $s$. Lemma~\ref{fam:Q:2.5} gives a positive integral $s$-word of weight $w$, since $(s-w)(\ell-1)>w$. It is unique: a word shorter than $s$ misses a position in every $\mu_s$-coset, so its coefficients are the unique nonnegative coset representatives with minimum zero. Strictness gives a nonidentity term. The tame stabilizer acts freely on these terms and preserves multiplicities, hence has order $a\mid s-1$ with $a\le w$. If $a$ is even, its involution acts as conjugation on the pure word. The entry field is then real whereas $K_X$ is nonreal by odd duality. Thus $[K_X:\QQ]\ge b_s(t)$. Since $s-1>p^t$ (both $s,p^t$ are odd), the entry ramifies at $s$. Separation gives $K_X\cap K_Z=\QQ$, so the full orbit of $X$ agrees on the $p$-active columns. The square action at $p$ produces at least $(p-1)/2$ phases in the orbit of $Z$, each with a distinct dual pair, hence $[K_Z:\QQ]\ge p-1$. Finally $W(s,p^t)$ would forbid the word just obtained upon rescaling to full weight. The $X$-orbit therefore costs at least $p^{2(t+1)}n^2 c_s(t)$ in \eqref{eq:r8-tracenorm} multiplied by $d_Z^2$: at eligible heights use the two field degrees over their rational intersection, and otherwise just the trace square. The $q$-free and $r$-free orbits are disjoint by degree. Counting also the unit yields $p^4n^2(A_q+A_r)U$, $W(r,1+\lfloor U/q\rfloor)$, and for every $1\le t\le h-2$ either \[ W(r,p^t)\qquad\text{or}\qquad (p-1)(r-1)\ge p^{2(h-t-1)}H_t . \] Then every braided fusion category of dimension dividing $p^{2h}q^Br^Cd$ is weakly group-theoretical for all $B,C\ge0$. In particular $r>U$ and $W(r,p^{h-2})$ suffice. \end{theorem} \begin{proof} In a remaining divisor-minimal table take complementary nonunits of the principal $q$- and $r$-blocks, of degrees $p^i q^b,p^k r^c$. If their common weight $g$ is at most $U$, then $r\ge g+2$ and the two residues in Theorem~\ref{first:balance} give a full $r$-word on at most $1+\lfloor U/q\rfloor$ positions. Thus every such pair has weight $p^h$ and their codegrees are $p$-units. Prime avoidance and defect-zero vanishing isolate $p$ and force $D=p^{2h}n^2$, as in Theorem~\ref{found:S:key:thm:topheight}. This argument only uses $qU$ and works in either order of $p,q$. In Theorem~\ref{lem:isolated-trace-budget}, all heights at $r$ are eligible since $r>U\ge p^{t+1}$. Every retained height gives $p^{2(t-1)}c_r(t)\ge p^{2(t-1)}(p-1)(r-1)/H_t\ge p^{2h-4}$, a contradiction. The specialization uses $1+\lfloor U/q\rfloor\le p^{h-2}$. \end{proof} \begin{corollary}\label{found:S:3.3}\label{found:S:key:cor:allthree} For $h\ge3$, the conclusion of Theorem~\ref{found:S:key:thm:twoarbitrary} holds whenever \[ r>\max\bigl\{p^{h-1},\,33,\,(p^{h-2}-1)^3\bigr\}. \] In particular, $3^6q^B r^C d$ is covered for all $B,C$ and all odd primes $3\max(33,j^3)$ and $j\ge1$. For $j=1,2,3$ substitute directly. For $j\ge4$, write $E=j(j+1)/2$ and use the $g(v)$ of Corollary~\ref{found:S:key:cor:topheight}. If $g(v)\le j-2$ then $v\le g(v)(g(v)+1)$; if $g(v)\ge j-1$, then $v-1\le E-1 \le(j-1)^2$. Thus $(v-1)/g(v)\le j-1$ and $2C(E)+1\le3+j(j+1)(j-1)\le j^3$. Also $3(1+2E)/29$ throughout that range. The quartic assertion follows also for the smaller primes from Corollary~\ref{found:C:2.3} below. \end{proof} \subsection{The trace budget at height three}\label{found:C:2} \label{found:C:key:sec:oddrectangle} For the two companions $s=q,r$ put \[ e_s=\frac{s-1}{\max_{a\mid s-1,\,a\le p}a},\qquad e_s^+=\frac{s-1}{\max_{a\mid s-1,\,a\le p}(a/\gcd(a,2))}. \] \begin{proposition}[Height-three substitutions]\label{found:C:2.1}\label{found:C:key:prop:sexticrect}\label{found:S:3.4}\label{found:S:key:prop:sexticorb} Suppose $p$ is isolated in an odd ordinary minimal table of dimension $p^6n^2$ with $n$ supported on precisely $q,r$, $qp,\quad (s-p)(\min(p,q)-1)>p &(p-1)e_s^+\\ \text{otherwise}&1 \end{array} \] If their sum is at least $p^2$, isolation is impossible. When $pp^2$ and $W(r,1+\lfloor p^2/q\rfloor)$, every divisor of $p^6q^Br^Cd$ is a WGT dimension if $(p-1)e_r\ge p^2$. \end{proposition} \begin{proof} The table substitutes $h=3$, $t=1$ into the trace budget (an omitted height has infinite cost). When $pp^2$, every divisor of $p^6q^Br^Cd$ is a WGT dimension for all $B,C\ge0$, in either order of $p,q$. If $pp^2$. The first alternative of Proposition~\ref{fam:Q:3.1} gives the forbidden word, and the proof of Theorem~\ref{fam:Q:1.2} excludes the second at this bound. For $r>p^2$, a divisor-minimal obstruction has ordinary data by Proposition~\ref{three:odd-reduced}. The complementary-pair argument of Theorem~\ref{found:S:3.2} at $h=3$ isolates $p$ at valuation six, in either order of $p,q$. In Proposition~\ref{found:C:key:prop:sexticrect} the companion $r>2p$ is eligible and $e_r^+\ge e_r\ge p+1$ since $r-1>p^2$. The other companion costs at least one, giving the contradiction $A_q+A_r\ge1+(p-1)(p+1)=p^2$. \end{proof} In particular all $r$ are covered for the sextic--quartic conclusion if $p\ge5$ and $q^3\ge p^4$. Indeed, with $j=\lfloor\min(r,p^2)/q\rfloor$ we have $j^3p^2$ use $(p^2/q)^3\le p^233$, use the scalar obstruction in Corollary~\ref{found:S:3.3}. At $j=0,1,2$ strictness and the small scalar bounds suffice (at $j=2$, $r>2q$). At $j=3$ and $r\le33$ we have $r>3q$, so the only possible $q$ is $7$ and thus $p=5$, incompatible with $q^3\ge p^4$. For $p=5,q=7$ and $r=29,31$, the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/exposed_support.py}{four-position calculation} in Corollary~\ref{fam:A:strips} gives $W(r,4)$, so both companion exponents are unrestricted. For $p=5,q=7,r=23$ the word bound $W(23,4)$ remains undecided. Nevertheless $e_7=2$ and $e_{23}=11$, so Proposition~\ref{found:C:key:prop:sexticrect} excludes isolation at five for arbitrary companion exponents whenever the five-valuation is six. \begin{corollary}\label{found:C:2.3}\label{found:C:key:cor:concrete} Let $p3&r\ge11\\ 5&q>5&r\ge29\\ 7&q>7&r\ge53 . \end{array} \] For $C=4$ the conclusion holds at $p=3,7$ for every $q,r$. At $p=5$ the sole assignment left by these tests is $(q,r)=(7,23)$. \end{corollary} \begin{proof} In the three lines $r>p^2$, and $1+\lfloor p^2/q\rfloor$ is at most $2,4,5$, respectively. At two positions use Theorem~\ref{found:S:4.1}. For four and five positions use the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/exposed_support.py}{exposed-support calculations} of Corollary~\ref{fam:A:strips}. At five positions these give the cases $53,59,61$; the scalar bound applies from $67$ onward. For the quartic the only smaller triple at $p=3$ is $(3,5,7)$, where $r<2q$. At $p=5$, $r<29$ gives $r<3q$ except for $(q,r)=(7,23)$. Both ratio tests follow from Theorem~\ref{fam:Q:1.2}. When $p=7$ and $r<53$ we have $r<5q$, so Corollary~\ref{fam:A:strips} applies. \end{proof} For $r>33$ the scalar bound alone has a useful consequence for the location of any surviving odd triple: \[ q^3p^2$. Independently, if $m=\lfloor r/q\rfloor$ works and $m+13m+2$. Mark $1$ and every $q$-root in the chosen word for $\alpha$. On a $\mu_q$-coset containing $k$ marks prescribe a functional equal to $1$ there and $-k/(q-k)$ on the unmarked positions. Its coset sums are zero, so these prescriptions give a rational-linear functional. There are at most $m+2 Bs\left(1-\frac{1}{s}-\frac{m+1}{q-b}\right)>\frac47 Bs , \] using $g<(m+1)s$, $b\le B$, $1/s<1/15$ and $(m+1)/(q-b)<2/9$. This contradicts $N\le 4sd/7$. Thus every $a_C\ge0$ in both ranges. Their sum is $-d\le0$, so all vanish, $d=0$ and $N=0$. The word for $\alpha$ has signed sum $m$ and length at most $m+1$; by parity it consists of $m$ positive letters. The functional now takes value $g$ on $v$, so $P=g$ and every $q$-part is marked. The marks are linearly independent, since they miss a position in each $\mu_q$-coset. This gives the stated $q$-multiplicities, and $a_C=0$ gives the congruence on the $s$-multiplicities. Lemma~\ref{lem:cyclotomic-optimum} with $c=2$ eliminates all spectator components (their primes are at least five) and puts the physical coefficient vector in the convex hull of integral signed optima. In each optimum the nonidentity $s$-positions have multiplicity at least $q$ and the identity occurs since $q\nmid g$. Two optima together use at most $1+2\lfloor g/q\rfloorh(h-1)\quad(1\le h\le E). \end{equation} Indeed, at the fixed row choose a full word minimizing the sum of its numbers $h_s$ of opposite nonzero differences, subject to $h_s\le E_s$ and to the initially constant coordinates staying constant. Its adjoint support lies in the product of the coordinate difference sets. The coset functionals in Lemma~\ref{fam:Q:2.2} have absolute value strictly less than one off the marks under the first inequality. Their product, with normalized trace on the complement, therefore exposes this box. In each nonconstant coordinate the nonidentity tame multipliers modulo sign have average overlap at most $h_s(h_s-1)/(H_s-1)$ by freeness. Choose independently one with overlap $j_sp$, the sole alternative to full $p$-height is $i=2$. Their normalized crossing would then have the two residues and weight $g=p^2$ of Proposition~\ref{found:J:key:lem:near-updated}. Its $r$-projection has at most $k$ positions, excluded by the mixed support obstruction. Thus every such witness has full $p$-height three. The opposite principal hits separate $p$ from both $q$ and $r$ as in Theorem~\ref{found:C:2.2}, without a hypothesis on their exponents. Here both companion primes are eligible. Proposition~\ref{found:C:2.1} excludes this isolated sextic by the stated orbit inequality. \end{proof} Here are the small-prime substitutions into the preceding theorem. Each entry in the third column is $e_r^+$ in the order displayed. \[ \begin{array}{c|l|l|l} p&r&e_r^+&q\\\hline 5&17,19,23&8,6,22&11\le qp^2$, all divisors of $p^6q^Br^Cd$ are WGT dimensions for all $B,C$. \end{corollary} \begin{proof} Here $k\le2$. For $r>p^2$ apply Theorem~\ref{found:C:2.2}. Otherwise $2r-2>p^2$ since $r\ge q+2$, and \eqref{found:J:eq:exposure} holds. Put $h=(p+1)/2$. For either $s=q,r$ we have $s-1\ge(p-1)h>p(h-1)$, so $e_s^+\ge h$. Equality would require an effective stabilizer $a/\gcd(a,2)\ge p-1$ for a divisor $a\le p$ of $s-1$. Its only possible value is $p$, so equality forces $s=ph+1$. The two distinct primes cannot both have this value; hence $e_q^++e_r^+\ge2h+1=p+2$, as required. \end{proof} For $g=p^2$ and $s=r9$: the height-one auxiliary pure words are available since $(s-3)(3-1)>3$ for $s=5,7$. At weight nine use the proof of Proposition~\ref{found:J:key:lem:near-updated} with negative cost $7/4$, permitting contraction at the spectator prime three. The coset cost is at most $9/(7-1)<7/4$, so each $a_C$ is nonnegative and their sum is $0$ or $1$. Here $-3\le L-1\le1$, hence the congruence $\sum a_C\equiv L-1\pmod5$ is equality. The functional on at most three marks in a five-coset takes values in $[-3/2,1]$. Consequently $2+7L\le P+3N/2\le9-N/4$, forcing $L=1$, $N=0$ and all $a_C=0$. As in the proposition the two five-parts have masses $2,7$ and the two seven-parts have masses $4,5$, with $1$ first in each coordinate (combining the masses in a singleton projection). A singleton projection would give an excluded one- or two-phase word (Theorem~\ref{found:S:4.1}). Write the nonzero exponents of the two projections as $a,b$, respectively. If either relative order is wild, its adjoint projection $\{0,\pm a\}$ or $\{0,\pm b\}$ has at most one mark per prime-order coset. In the product coset functional the largest possible single negative magnitude is $3/2$ and the product of the two negative magnitudes is at most $(3/2)/4<1$. Tensoring with the normalized spectator trace (at least $-1/2$) exposes the adjoint rectangle. Independent units moving the nonzero pair off itself in each coordinate then contradict the common-nonunit argument. Thus both relative orders are tame. If an initial corner is missing, after scaling the exponents and choosing signs, the adjoint marks are the $3\times3$ square of positions $0,\pm1$ in each coordinate with $(1,1),(-1,-1)$ omitted. The following functional exposes these seven marks. In that square put $1$ on the marks and $0$ on the omitted corners. In the other four columns put $-3/4$ on row zero and $-1/2$ on rows $\pm1$. In each of the other two rows put $-3/2,-1,-1$ on columns $0,1,-1$ and $7/8$ on the four remaining columns. Every row and every column sums to zero, so the array is a linear functional on the cyclotomic configuration. Extend by zero on other product cosets and tensor with the normalized spectator trace. As all its values are greater than $-2$, equality at one is possible only on the seven marks. The independent dilations again yield a contradiction. With all corners occupied the margins force exactly the four stated coefficients. In particular the value generates $\QQ(\zeta_{35})$ by uniqueness of the two projections, and its tame trace at five is $4(1+\xi)-(3+4\xi)=1$. These are necessary conditions. The $(3,5,7)$ assignment with two unrestricted companion exponents stays open; its quartic case follows from Section~\ref{found:S:3}. \paragraph{Twist-one orbits at the least prime three.} For a minimal table in the unbounded-exponent case $(3,5,7)$ of Section~\ref{found:J:2}, the weight-nine pair may be chosen in the pure five- and seven-twist fibers. Indeed Proposition~\ref{prop:principal-support} gives each hit in those fibers; if all their complementary pairs had full three-height, the vacuum-block argument of Section~\ref{found:J:2} would again isolate three. Take faithful characters at five and seven of orders four and six. At $1+\xi+3\eta+4\eta\xi$ their mixed Fourier coefficient is four times a product of two nonzero Gauss sums. On $(\ZZ/35)^\times$ their product has kernel generated by simultaneous negation: the two images intersect in $\{1,-1\}$ and the characters are faithful. Inverting either factor preserves this kernel. Therefore Proposition~\ref{found:J:key:pr:carry} supplies twist-one labels in $P_5,P_7$ in the same full three-block with fields containing $\QQ(\zeta_{35})^+$. Each orbit has at least $24$ members, since a nonunit in odd dimension has nonreal field and this real subfield has degree $12$. More precisely, use characters with local orders $(4,6),(4,3),(2,6)$. All three mixed coefficients are nonzero by the same product of Gauss sums. With fixed generators $a,b$ of the order-four and order-six groups, their kernels on $(\ZZ/35)^\times$ are respectively $\langle(a^2,b^3)\rangle$, $\langle(1,b^3)\rangle$, and $\langle(a,b^3)\rangle$. Inverting either component preserves each kernel. The kernel intersections are pairwise trivial, so any two of the three fields generate $\QQ(\zeta_{35})$. In each of $P_5,P_7$ therefore either a twist-one row contains $\QQ(\zeta_{35})$, or at least $24+12+6=42$ nonunit twist-one labels occur in three distinct orbits. The common three-block can depend on the character. \subsection{The sextic exponent at either smaller prime}\label{found:J:3} For any fixed distinct odd primes $P,Q$, Theorem~\ref{found:J:3.1} gives $P^6 Q^B R^C$ for all sufficiently large primes $R$, independently of both $B$ and $C$. An explicit sufficient threshold is $R>\max\{P^2,33,\lfloor P^2/Q\rfloor^3\}$ by the scalar word bound in Section~\ref{found:C:2} and Corollary~\ref{found:S:3.3}. \paragraph{A cubic scalar bound.} For integer $j\ge1$ put $E=j(j+1)/2$. For $1\le h\le E$ one has \[ \frac{h}{k(h)}\le \sqrt{h+1/4}+1/2\le(j+1)/\sqrt2+1/2, \qquad k(h)=\min\{v:v(v+1)\ge h\}. \] Consequently $3+2h(h-1)/k(h)<(j+1)^3/\sqrt2$. To verify the strict estimate, replace $2(h-1)$ by $j(j+1)-2$. The difference between the proposed upper bound and the resulting expression is $(j+1)(j+3)/\sqrt2-(j^2+j)/2-2>0$. Also $3(1+2E)/2<(j+1)^3/\sqrt2$ for $j\ge1$. Thus $r\ge(j+1)^3/\sqrt2$ forces $W(r,j+1)$ by Lemma~\ref{fam:Q:2.3}. \paragraph{A uniform floor criterion.} For $5\le pq$, all divisors of $p^6 q^B r^4d$ are WGT dimensions if \[ J\le1\quad\text{or}\quad (J+1)^3\le\sqrt{2}\,Jq . \] For $r>p^2$ both companion exponents can be arbitrary. Indeed put $j=\lfloor\min(r,p^2)/q\rfloor$. At $j=0,1$ use strictness and Theorem~\ref{found:S:4.1}, respectively. At $2\le j\le J$, the function $(x+1)^3/x$ increases on $x\ge1$, so $(j+1)^3\le\sqrt2\,jq<\sqrt2\,r$. The preceding cubic bound and Theorem~\ref{found:C:2.2} finish the proof. Two convenient sufficient tests are $(J+2)^2\le\sqrt2\,q$ and $(p^2/q+1)^3\le\sqrt2\,p^2$. For the first use $(J+1)^3/J<(J+2)^2$ for $J\ge2$; for the second use the same monotonicity at $x=p^2/q\ge J$. In particular either $\sqrt2\,q^3\ge(p^2+2q)^2$ or $q^3\ge\alpha p^4$ for fixed $\alpha>1/\sqrt2$ and sufficiently large $p$ suffices. For $\alpha=4/5$, $p\ge125$ suffices since $(5/4)^{1/3}+p^{-2/3}<27/25+1/25<2^{1/6}$. For $(p,q)=(31,97)$ the floor test applies although $q^30$, then summing the overlaps over $u\ne1$ gives $\sum_a h_a(h_a-1)$. Two or more nonempty orbits decrease this sum from $h(h-1)$ by at least $2(h-1)$. This proves $\delta\ge0$ and, under the strict bound on $\delta$, identifies $A$ with an $h$-element subset of a single cyclic orbit. Identify that orbit with $\mathbb Z/H$. Under reduction modulo $b$ let $x_i$ count the members of $A$ in residue class $i$. The left side of the displayed equation counts all ordered pairs with difference $j$ modulo $b$. Each of the $H/b$ group differences in that class occurs at least $k$ times; difference zero in the full group occurs $h$ times. Summing gives the asserted equations and total excess. At $b=2$ subtraction of the odd equation from the even gives $(x_0-x_1)^2=h-k+e_0-e_1$. The two populations sum to $h$. \end{proof} For arbitrary odd primes $pp^2$ the exponent at $r$ may also be arbitrary. Here are instances where the scalar inequality alone leaves a feasible $h$ but the order-two quotient excludes it. Every column uses the same rule. The last column brackets the interval in the preceding column by consecutive squares of the required parity. \[ \begin{array}{c|r|r|r|r|c|c} m&r&h&k&\delta&[h-k-\delta,h-k+\delta]&\text{bracketing squares}\\\hline 17&3877&153&12&12&[129,153]&11^2,13^2\\ 29&17981&435&21&21&[393,435]&19^2,21^2\\ 33&26449&552&23&23&[506,552]&22^2,24^2\\ 34&29453&595&24&30&[541,601]&23^2,25^2\\ 45&66889&1035&32&14&[989,1017]&31^2,33^2 \end{array} \] To check uniqueness of $h$ without enumerating supports, on the interval $k(k-1)193$ is impossible. \emph{The primes from $173$ to $193$.} At $s=173$ the inequality allows $h=19,20$, and at $179,181,191,193$ only $h=20$. In all cases $g(h)=4$. Let $l$ be the number of nonzero opposite pairs in $D$ of exact order $s$. For a uniform random unit at the chosen $s$-power level, the expected overlap on an orbit of $n$ opposite pairs containing $b$ pairs of $D$ is $b^2/n$. The order-$s$ orbit has $H$ pairs, and every higher-order orbit has at least $sH$. Summing gives \[ 4H\le l^2+(h-l)^2/s . \] If $l\le18$, the right side is at most $18^2+20^2/173<344\le4H$. Thus at most one pair has higher order. No nonidentity tame multiplier modulo sign can match a higher-order pair with itself, so every overlap for such a multiplier lies in $\mathbb F_s$. The mean nonidentity overlap there is $l(l-1)/(H-1)\le380/85<5$. Some intersection therefore equals $T=\{0,\pm x_1,\ldots,\pm x_4\}$. A common nonunit has support contained in $T$. Its differences lie in $T-T$, which has at most $4\cdot5=20$ pairs and at least $h$. These actual differences are exposed. Comparing their dilates as above shows that $T-T$ meets each of its dilates in at least four pairs. The following short finite check is on differences of four residues and is independent of the dimension. Scale one pair to $\pm1$ and choose the other three from the $H-1$ remaining pairs. Form the mask of $x_i,2x_i,x_i\pm x_j$ modulo sign. Retain masks with at least nineteen pairs at $s=173$ and twenty elsewhere, and test their intersections with every nonidentity multiple. The columns count normalized choices of $T$, including choices with the same mask. The loops in the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/short_words.py}{residue enumeration} reproduce the counts using only integer arithmetic. \[ \begin{array}{r|rrr} s&\text{choices}&\text{at least the required size}&\text{all overlaps at least four}\\\hline 173&98770&52012&0\\ 179&109736&50376&0\\ 181&113564&52658&0\\ 191&134044&65108&0\\ 193&138415&67813&0 \end{array} \] This excludes every possible $T$ in this range. \emph{The primes from $127$ to $167$.} Let $h_0$ be the least $h\le21$ satisfying $(H-1)g(h)\le h(h-1)$. Every word on this row has at least $h_0$ differences modulo sign, by the minimizing argument above. Here $g(h_0)=4$. The initial seven positions lie in one coset of $\mu_s$. Indeed, if split across cosets, at most $\binom62=15$ of their nonzero opposite differences have exact order $s$, by summing the binomial coefficients of the coset populations. All difference spectra in this initial comparison are exposed since $432$, hence $4\mid L$. There are only three odd nonunit labels, a contradiction. All active labels now have degree $2n$, so the saturation of the row norm just proved gives $S_{ZZ'}=0$ for any two of them. Let $A$ be their coordinate span. The odd local factors act trivially, up to scalars, on $A$ by isolation (as in the proof of Proposition~\ref{deep:cluster}). Thus $W_2 A=sA\perp A$. The actual dyadic twist is the identity on $A^\perp$, and its $W_2$-conjugate is the identity on $(W_2 A)^\perp$. In particular both preserve $A$. The corresponding upper and lower unipotents differ from these operators by scalars and generate the dyadic group. This would make $W_2 A=A$, contradicting orthogonality. It remains that every active degree is $n$. Let $N$ count them. The same highest-order indicator calculation, now without its last sum, gives $N\ge4$. Take an $r$-free even $X\in P_2$ by Proposition~\ref{prop:principal-support}. Its degree $2^i q^j$ has $i=1$ or $2$ by Corollary~\ref{cor:principal-valuation-bound}. It hits all $N$ active labels. On each of them the $r$-inertia conjugates of the $X$-row agree, since all active fields have dyadic conductor. Averaging those $t$ orthogonal $S$-rows gives squared norm $D/t$. On the active columns the lcm scale is $2^i n$; averaging also over all Galois automorphisms gives squared mass at least $N4^i n^2$, in addition to the vacuum. Thus $4^itN<64$. First $i=1$, then $t\ge(r-1)/2\ge2$ by the orbit estimate, giving $N\le7$. By Lemma~\ref{lem:local-vacuum-column} the modulus of the vacuum coordinate of $u$ is \[ |u_0|=\tfrac18\bigl|64-\sum_{Z\text{ active}} (1-\theta_{Z,2})\bigr| \ge (64-2N)/8>6 . \] Hence $u_0^2\ge49$. The $N$ odd active coordinates cost at least four in total. The $t\ge2$ even $r$-free conjugates cost at least eight, and a $q$-free even member of $P_2$ costs at least four, by its height one or two. These disjoint contributions exceed $64$. \end{proof} \begin{theorem}\label{aux:dyadic-six} Let $q5$, Theorem~\ref{first:balance} gives $1+3\xi$ at the normalized crossing, where $\xi$ has nontrivial $r$-power order. Theorem~\ref{found:S:key:thm:two-phases} excludes the resulting two-position character. Hence $r=5$. All complementary witnesses have degree $4\cdot3^i$ or $4\cdot5^j$ and positive dyadic defect one. Take the integral minimal modular extension of Lemma~\ref{lem:integral-minimal-extension}, with fermion $g$ and matrices $S,s=S/\sqrt{4\Delta}$. Put $\chi_X(Y)=S_{XY}/d_X$. Even and odd refer to monodromy with $g$. Let $A,B$ be the symmetric and antisymmetric even spaces under tensoring by $g$, and $C,D$ the corresponding odd spaces. The full Weyl interchanges this permutation and the monodromy sign, while odd local factors commute with both (Section~\ref{sec:blocks}). Thus the dyadic Weyl $W$ preserves $A,D$ and interchanges $B,C$. The dyadic upper unipotent $U=\lambda T_2$ interchanges $A,B$ and preserves $C,D$ since even mates have opposite twists and odd mates equal twists. Write $K=W|_A$, $W_o$ for the product of odd Weyls and $\widehat s=s|_A=K W_o|_A$. The genuine congruence representation of Section~\ref{sec:blocks} has $\lambda^8=1$ and trivial odd normalizing scalars. The dyadic group preserves full two-blocks and $K$ preserves positive two-blocks by lifting their projectors (Lemmas~\ref{lem:local-block-invariance} and~\ref{lem:positive-defect-one-phases}). \smallskip\noindent\emph{Positive twists.} The first assertion of Lemma~\ref{lem:positive-defect-one-phases} applies to each complementary witness, since $v_2(4\Delta)=7$. Thus its row sees only the two squared dyadic phases $+1,-1$. A pair of any other squared phase cannot be active at three or five, as Proposition~\ref{prop:principal-support} modulo the opposite prime would force a complementary hit. On each such extra phase fiber the odd local groups act trivially: they preserve the fiber, their upper unipotents act identically there, and conjugate unipotents generate. Their Weyls in particular fix both full lines of its even pairs (the actual dyadic twists are opposite). By Lemma~\ref{lem:complement} such a pair $V$ has pure dyadic positive codegree; hence both $b,c$ are even. Set $n=3^{b/2}5^{c/2}$ in this case. The column $K v_V=\widehat s v_V$ has nonzero vacuum coefficient, so $V\in P_2$. It vanishes modulo two on even degrees. The estimate $f_V\ge[\QQ(\chi_V):\QQ]+8$ of Corollary~\ref{cor:defect-two-energy-stratum} leaves only $f_V=32$ among $32,8,2$, giving $d_V=n$. Dimension orthogonality on such a row modulo five gives a five-free nonunit hit $X$, necessarily of degree $2\cdot3^h$ since $X\in P_2$ by Theorem~\ref{thm:principal-support}, Corollary~\ref{cor:principal-valuation-bound}, Lemma~\ref{lem:tables}. Crossing with a complementary $4\cdot5^j$ witness gives a weight-two entry of strict house below two and residue two at five. Its unique unsigned five-primary word is a nonscalar sum of two roots (Theorem~\ref{first:balance}), so the tame stabilizer has order at most two. The five-inertia orbit of $X$ has size $t\ge2$. Every extra pair, being of degree $n$ odd, is hit by every member of this orbit by principality. All extra pairs are five-unramified. The averaged norm of Lemma~\ref{fam:Q:2.1}, retaining the vacuum and all $L$ extra pairs and using $\lcm(n,d_X)=2n$, gives $4tL<32$. Thus $L\le3$. If these phases occurred, take the nonempty set $H$ of pairs whose squared phase has highest order $2^m$, $m\ge2$. The operator $J=(T_2^2)^{2^{m-1}}$ on $A$ is $1-2P_H$ with $P_H$ their coordinate projection. The trace of $JKJ^{-1}K^{-1}$ is an algebraic integer since the scalar of the powered unipotent cancels and the resulting group element has finite order. Its expansion is \[ \dim A-4|H|+\tfrac18\sum_{V,Y\in H}|\chi_V(Y)|^2 , \] using $K v_V=\widehat s v_V$. The sum is rational by Galois covariance and stability of $H$, hence belongs to $8\ZZ$. Its terms are congruent to one above two by principality; thus $|H|=2$. If both labels are individually fixed, the four entries are odd-integer squares. Otherwise their fields are the same quadratic dyadic field. The squared moduli form a matrix with diagonal $y$ and off-diagonal $\sigma y$, where $\sigma$ interchanges the labels: use symmetry and Galois covariance in either argument. For rational $y$ the sum is four modulo eight by oddness. Otherwise $y$ belongs to the real quadratic dyadic field and is $a+b'\sqrt2$ with $a$ odd; the sum is $4a$, again impossible. Hence all even dyadic twists have order dividing four. Lemma~\ref{lem:positive-defect-one-phases} now gives full dyadic level dividing eight and two equal-degree positive labels of opposite squared twists in each complementary block. It also gives Gaussian integer values on even objects for every purely dyadic odd character. \smallskip\noindent\emph{Odd spins.} Put $Z=T_2^2|_A$, $F=KZK^{-1}$. These involutions anticommute, and $U^4$ is scalar on $C$, by Lemma~\ref{lem:signed-dyadic-pair}. Over each complementary block the two odd orbits are free, and their actual dyadic twists have ratio $\pm i$. Their degrees satisfy $u^2+v^2=2d^2$ and $v_2(u)=v_2(v)=2$. For $d=4\cdot3^i$, descent modulo three gives $u=v=d$; for $d=4\cdot5^j$, reduction modulo three gives $3\nmid uv$. At its odd prime $p$ each complementary witness has a full principal lift. Indeed, if $Y$ is an odd label of $p$-unit degree, its square is even, so its character value is congruent to $\pm d_Y$. Tensoring the witness by $g$ chooses the positive sign. For every odd $V$, evaluating the even product $YV$ then gives the dimension residue on $V$. Such a $Y$ is supplied by the opposite odd mates just found. This lift hits both odd mates of every opposite block by principality. As $W$ preserves its full two-block and $W_o$ preserves odd dyadic spin fibers, its row sees only the two odd spins of its own block. Thus the same pair $\delta,i\delta$ occurs in every complementary block. At each odd prime the full principal block has only even labels (evaluate at $g$) and excludes $g$ (evaluate at an odd mate of unit degree). Therefore every odd label active at three or five hits a complementary witness by Proposition~\ref{prop:principal-support} modulo the other prime. All such labels have one of these two spins. Some other odd spin occurs, since the two-spin conclusion of Lemma~\ref{lem:signed-dyadic-pair} would contradict perfectness of $R_+$. By scalar $U^4$ all other odd orbits have spin $-\delta$ or $-i\delta$. Call their set $E$. The odd local groups fix their full lines, by the same fiber argument as above, so their full codegrees are dyadic by Lemma~\ref{lem:complement}. This forces $b,c$ even. Write $n=3^{b/2}5^{c/2}$. For $V\in E$, $Wv_V=sv_V$ has nonzero vacuum coefficient, hence $V$ is full two-principal. Its degree is $2^j n$, $j=0,1,2$, by Corollary~\ref{cor:principal-valuation-bound} at valuation seven. \smallskip\noindent\emph{The signed norm.} At least eight odd-degree positive pairs lie in $P_2$. Its rank is even since $Z,F$ restrict to anticommuting involutions. Its sum of squared degrees is divisible by 32 by Theorem~\ref{thm:principal-support} on the unit row. Write $k_0,k_1$ for the numbers of odd and even degrees. The even degrees have height one by Corollary~\ref{cor:principal-valuation-bound}. Modulo four, $4\mid k_0$, so $k_1$ is even; modulo eight $k_0+4k_1\equiv0$. The vacuum gives $k_0\ge8$. Each of these $k_0$ values at $V\in E$ is nonzero by principality and has squared modulus at least one since it is Gaussian. Compare the signed character of $V$ with that of an odd mate of a $4\cdot5^j$ witness. These rows differ, since they have different spins. In their restricted orthogonality (Lemma~\ref{lem:fermion-quotients}), the vacuum survives modulo three and all three-bearing degrees contribute zero by lcm-integrality on the three-free mate. This gives a nonunit three-free hit $X_3$. A $4\cdot3^i$ mate gives a five-free hit $X_5$. Both belong to the full and positive principal two-block by Theorem~\ref{thm:principal-support}, Lemma~\ref{lem:fermion-quotients}, with degrees $2\cdot5^l$ and $2\cdot3^h$, respectively, $l,h>0$. Each $\chi_V(X)$ is a nonzero Gaussian integer vanishing above two. When its squared modulus is at least four, its Galois orbit costs at least four per pair. When it is two, $\chi_V(X)^2=\pm2i$ and $\chi_X(V^2)=(d_V/d_X)^2\chi_V(X)^2$. Since $V^2$ is even, $i$ belongs to the positive character field of $[X]$. Five-inertia fixes $i$, so another coset of its orbit is disjoint from the first. Every conjugate there also has signed absolute square two, by Galois covariance, using full lifts and ignoring the fermion sign. The five-inertia orbit of $[X_5]$ has size at least two by the weight-two crossing above; for $[X_3]$ use the trivial lower bound one. The disjoint supports cost at least eight and four, respectively. Thus the signed codegree is at least $8+8+4=20$. By \eqref{eq:fermion-restriction-codegree} it equals $32/4^j$ for a free orbit and $64/4^j$ for a fixed one. Hence $j=0$. A fixed label would have even $d_V^2$, since $VV^*$ would be $g$-invariant and all even simples occur in free pairs. Every $V\in E$ is therefore free of odd degree $n$. \smallskip\noindent\emph{Strict multiplicity.} Take $X=X_5$ as above. Every $V\in E$ hits $X$, since $X$ is two-principal and $d_V$ is odd. Lcm-integrality makes $\chi_V(X)$ a nonzero multiple of two in $\ZZ[i]$, of squared modulus at least four. The five-inertia orbit of $[X]$ has size $t\ge2$. Its full orbit also has size $t$: the entries at the $V$ are fixed by five-inertia by symmetry, equal degrees along the orbit, and pure dyadic codegree at $V$, whereas passage to $gX$ negates them. All chosen representatives have the same dyadic twist. Put $a_X=(v_X+v_{gX})/\sqrt2$, $b_X=(v_X-v_{gX})/\sqrt2$, and similarly $c_V,d_V^-$ for symmetric and antisymmetric odd pair vectors. For $v=\sum_X b_X$ over the $t$ representatives, $\|v\|^2=t$. The coefficient of $s b_X$ at $c_V$ is $2S_{XV}/(\sqrt{128}\,n)$, independent of the representative and of squared modulus $|\chi_V(X)|^2/32\ge1/8$. Hence $t|E|\le8$. If $|E|=4$, then $t=2$ and equality gives $sv\in\langle c_V:V\in E\rangle$. The odd Weyls fix this space pointwise and commute with $s$, so $W_o v=v$. Applying $U$ and using equal dyadic twist of the representatives gives $W_o a=a$ for $a=\sum_X a_X$. Thus $sa=Ka$ is supported on the positive $P_2$. A $4\cdot5^j$ witness $Y$ lies outside $P_2$ by the dyadic height bound. Yet the $Y$-coordinate of $sa$ is a nonzero common factor times $\sum_X\chi_Y(X)\equiv2d_X\ne0$ modulo a prime over five by five-principality. We have proved $1\le |E|\le3$. \smallskip\noindent\emph{Trace on the odd antisymmetric space.} On $D$ put $x=\lambda\delta$, $T=U/x$. Its eigenvalues lie in $\{1,i,-1,-i\}$. For the projection $P_E$ onto $\langle d_V^-:V\in E\rangle$ one has \[ J=1-2P_E=\tfrac12\big((1-i)T+(1+i)T^{-1}\big). \] We claim $\Tr_D(JWJW^{-1})\in\ZZ$. Unitarity and cyclicity give \[ \Tr_D(JWJW^{-1}) =\Im\Tr_D(TWTW^{-1})+\Re\Tr_D(TWT^{-1}W^{-1}). \] The corresponding words in $U,W$ in $\mathrm{SL}_2(\ZZ/8)$ have matrices \[ h=\begin{pmatrix}0&1\\-1&1\end{pmatrix}, \qquad k=\begin{pmatrix}2&1\\1&1\end{pmatrix}. \] Both have order dividing six: $h^3=-I$ and $k^3=5I$ modulo eight. Both are conjugate to their inverses in this group: the Weyl inverts $k$, and the coordinate swap inverts $h$ while its product with $I+2h$ has determinant one modulo eight. Their traces on $D$ are consequently real algebraic integers in $\QQ(\zeta_6)$, hence integers. Replacing $U$ by $T$ multiplies the first trace by $x^{-2}\in\mu_4$ and leaves the second unchanged, proving the claim. Expanding the projector gives \[ \Tr_D(JWJW^{-1})=\dim D-4|E|+4\sum_{V,Y\in E}|W_D(V,Y)|^2 . \] The odd Weyls fix each of these columns, so $W_D(V,Y)$ is the full $s$-entry in antisymmetric pair bases, $2S_{VY}/(\sqrt{128}\,n)$. It follows that \begin{equation}\label{found:S:key:eq:ferm-eight} \sum_{V,Y\in E}|\chi_V(Y)|^2\in8\ZZ . \end{equation} We prove an orbit congruence valid for any size of this packet. Its entries $\chi_V(Y)$ are dyadic units by principality and belong to $L=\QQ(\zeta_8)$ by the pure codegrees and the level. The group $G=\Gal(L/\QQ)$ acts on the pairs: odd inertia fixes each full character, and the Galois square fixes its dyadic spin at level eight. Equal degrees and symmetry imply that conjugation in either index conjugates the entry, up to the sign from changing a fermion representative. Thus $y_{VY}=|\chi_V(Y)|^2$ is well defined on pairs and satisfies $y_{\sigma V,\tau Y}=\sigma\tau(y_{VY})$. Partition $E$ into $G$-orbits of sizes $n_i\in\{1,2,4\}$. On a rectangle $O_i\times O_j$ the sum of the $y_{VY}$ is $n_i n_j m_{ij}$, where $m_{ij}$ is their normalized absolute trace. Indeed averaging independently over two group elements averages their product uniformly over $G$. For an entry written $\alpha=\sum_{k=0}^3 a_k\zeta_8^k$ with $a_k\in\ZZ$, the normalized trace of $|\alpha|^2$ is $\sum_k a_k^2$. It is odd since $\alpha$ is a dyadic unit and $\zeta_8$ reduces to one. When either orbit is a singleton, conjugating that index acts on $\alpha$ only by sign. The rational character lines in $L$ are $\QQ,i\QQ,\sqrt2\QQ,\sqrt{-2}\QQ$; an integral dyadic unit on these lines is an odd integer times one or $i$. Then $m_{ij}\equiv1\bmod8$. In every other diagonal rectangle $4\mid n_i^2$, so oddness of $m_{ii}$ gives $n_i^2m_{ii}\equiv n_i^2\bmod8$. The remaining off-diagonal rectangles occur in transpose pairs of equal sum with $8\mid2n_i n_j$. Altogether \[ \sum_{V,Y\in E}|\chi_V(Y)|^2\equiv \Big(\sum_i n_i\Big)^2=|E|^2\pmod8 . \] Consequently \eqref{found:S:key:eq:ferm-eight} forces $4\mid |E|$, contradicting $1\le |E|\le3$. For the unordered box, an odd-dimensional obstruction is ordinary. When the dyadic valuation is at most five, Theorem~\ref{fam:Q:1.3} covers the reduced source, with the lower faces in Section~\ref{sec:three}. Every remaining even source has dyadic valuation six, just excluded. \end{proof} \subsection{Restrictions on an ordinary dyadic obstruction}\label{found:S:6} Write $d_i=\dim i$, $f_i=D/d_i^2$, $\chi_i(j)=S_{ij}/d_i$, $P_u$ for the full principal $u$-block of ordinary integral modular data, and $K_i$ for the cyclotomic field generated by $\chi_i$. A label is $u$-active when its twist has nontrivial $u$-part. We work with a perfect fusion-simple category and the positive spherical structure. \begin{lemma}[Trace transfer across principal blocks]\label{rev:trace-transfer} Let $s,t$ be distinct primes. Choose $E=P_t$ or its intersection with the fiber of trivial prime-to-$t$ twist. Suppose $X\in P_s$ is $t$-inactive, $H$ is a finite group of cyclotomic automorphisms preserving its $t$-inactivity, and $g$ is a positive integer. For each $s$-free nonunit $Y\in E$, suppose that $\gcd(d_X,d_Y)=g$ and that \[ \sum_{\sigma\in H}\sigma\!\left(S_{XY}/\operatorname{lcm}(d_X,d_Y)\right)=\tau \] for a common rational integer $\tau$. If $v_s(g|H|-\tau)=1$, every $t$-active $J$ has $s\mid d_J$. More precisely, at every chosen prime over $s$ there is an $s$-bearing $Y\in E$ with $v_s(\chi_J(Y))<1$, normalizing $v_s(s)=1$. Consequently $1\le v_s(d_Y)\le v_s(d_J)$. If $J$ also lies in $P_s$, then $K_X$ and $K_J$ both ramify at $s$. \end{lemma} \begin{proof} Indeed set $F(Y)=g\sum_{\sigma\in H}\chi_{X^\sigma}(Y)-\tau d_Y$. The characters transform by ordinary Galois action, with $d_{X^\sigma}=d_X$. Then $F(1)=g|H|-\tau$ and $F(Y)=0$ at every $s$-free nonunit of $E$. Principality gives $F(Y)\equiv(g|H|-\tau)d_Y\equiv0$ at every prime over $s$. Indeed $\sigma\chi_X(Y)=\chi_X(Y^\sigma)$ by symmetry, Galois covariance and invariance of integral dimensions. Thus the defining principal congruences are stable at a fixed prime under every cyclotomic automorphism (see also the proof of Theorem~\ref{thm:principal-support}). This also gives the congruence at every conjugate prime. Restricted orthogonality gives \[ 0=\sum_{Y\in E}\chi_J(Y)\overline{F(Y)}. \] For the full block this is Theorem~\ref{thm:block}. For its indicated fiber, both projectors commute with the $t$-local group and their product still does; conjugation by the global Weyl yields an operator commuting with the $t$-twist, exactly as in Proposition~\ref{prop:principal-support}. The vacuum term has valuation one. Therefore, at some $s$-bearing $Y$, the product has valuation at most one. Since $\overline{F(Y)}$ has positive valuation, this forces the first assertion. The lcm divisibility of $S_{JY}$ implies the height inequality. If $J\in P_s$, its character also has positive valuation at such $Y$. Unramifiedness of either field would make that factor divisible by $s$, since $\overline{F(Y)}$ lies in the abelian field $K_X$. This is impossible. \end{proof} \paragraph{The ordinary sixth dyadic valuation.} Assume $D=2^6 3^b5^c$. If two mixes with three, every five-active label has degree divisible by three. If two mixes with five, every three-active label has degree divisible by five. More precisely, Lemma~\ref{rev:trace-transfer} applies with $(s,t)=(3,5)$ and $(5,3)$, respectively. For a column $J$ mixing two and $s$, apply Proposition~\ref{prop:principal-support} at $s$, inside its trivial complementary fiber and avoiding $t$. It produces a nonunit $t$-free $X\in P_s$ with $S_{XJ}\ne0$ and $\theta_{X,t}=1$. By Lemmas~\ref{lem:complement} and~\ref{three:complementary}, $v_2(d_X)=2$: it is at least two, and at height three the row misses $J$. Every $s$-free nonunit $Y\in P_t$ has dyadic height at least two by the same height bound, hence the gcd is four. For the normalized crossing, principality gives residue four above three and five and strict house less than four. Theorem~\ref{first:balance} leaves $1+3\xi$, for a nonidentity five-power root, or $-\sum_{j=1}^4\eta_j\zeta_5^j$, in which the $\eta_j$ are one copy of one plus one coset of the cube roots of unity. In the first case $\xi$ has order five by the two-position relative-order result (Theorem~\ref{found:S:4.1}), applied to the full-mass expression $(d_Y/4)(1+3\xi)$ for $\chi_X(Y)$. Take $H$ to be the order-four tame subgroup of five-inertia in a common cyclotomic level, fixing the other primary factors. In the first case its trace is $4-3=1$; in the second its trace is $\sum_j\eta_j=1$. Thus $g|H|-\tau=15$ in both orientations, using \emph{the same} five-inertia group in each. \begin{lemma}[Pure dyadic phase of a fusion product]\label{rev:pure-product} In the ordinary modular table, if all constituents of $XY$ have the same dyadic twist $\delta$ and no dyadically active row sees both $X$ and $Y$, then $\delta=1$. In particular the conclusion holds when either factor has odd codegree. \end{lemma} \begin{proof} Indeed the vector $v=[XY]$, written in the fusion basis, is an eigenvector of the dyadic upper unipotent with eigenvalue $a\delta$, where $a$ is its common normalization scalar. On the other hand, $(Sv)_J=d_J\chi_J(X)\chi_J(Y)$ for the unnormalized $S$. Its support has trivial dyadic twist, so the conjugate generator $S^{-1}T_2 S$ has eigenvalue $a$ on $v$. These two conjugate unipotents generate the dyadic modular group, so that their actions on the common eigenline coincide. The odd-codegree specialization is Lemma~\ref{lem:complement}. \end{proof} Applying Lemma~\ref{rev:pure-product} to the two ordinary weight-four phase profiles gives the following consequence. When both factors belong to their trivial complementary fibers and one has full dyadic height, the negative profile is excluded. Without the full-height hypothesis that profile forces a dyadically active row meeting both factors. \paragraph{Complementary height under separation.} Assume further $c=4$ and that three and five never mix. Since an isolated quartic prime is excluded (Lemma~\ref{first:isolate}), two mixes with five. Choose the resulting $Y\in P_5$ of dyadic height two and three-free degree as above. For every five-free nonunit $X\in P_3$ the gcd is four. All five-active labels have full three-height by Lemma~\ref{lem:complement}; thus $b$ is even, $Y$ and $X$ are five-inactive (Lemma~\ref{three:complementary} bounds $X$ below full three-height). The negative profile is impossible: at most one of its four distinct five-parts permits a five-inactive constituent, and at most two of its four three-parts permit a three-inactive constituent. The positive profile puts total dimension $3\operatorname{lcm}(d_X,d_Y)$ on its nontrivial five-phase. It follows that $3^{b/2}\mid3\operatorname{lcm}(d_X,d_Y)$ and hence $v_3(d_X)=b/2-1$, again by Lemma~\ref{three:complementary}. In particular the height is exactly two at $(6,6,4)$. \begin{lemma}[Separated twists and inertia orbits]\label{u2:separation} Let an ordinary perfect fusion-simple integral modular table have square dimension $n^2$, where $n=A s^e r^c$, $e,c\ge1$, $s,r$ are distinct primes, $s$ is odd, and $\gcd(A,sr)=1$. Suppose $s$ and $r$ never divide the same twist order. Each $s$-active $Z$ has degree $z s^h r^c$, with $z\mid A$ and $h0$ has at least $b$ Galois conjugates and $mbM^2\ge A^2$, then $h=0$ for all $s$-active labels and \[ \sum_{Z\ {\rm active\ at}\ s} z^2 < A^2s^2/m,\qquad m\operatorname{rad}(A)s^e < 2As^2 . \] \end{lemma} \begin{proof} Since $r\nmid f_Z$, inertia at $r$ fixes $K_Z$ by Corollary~\ref{cor:field-conductor}. Average $m$ distinct $I_r$-conjugate unnormalized $S$-rows of $X$. Their average has squared norm $n^2/m$ by orthogonality. At every conjugate $Z'$ of $Z$ its entry is $S_{XZ'}$, since the $m$ entries coincide there. The vacuum contributes positively. At a nonzero entry divisible by an integer $L$ in the algebraic integers, the mean squared modulus over conjugations is at least $L^2$, by the norm. These estimates still hold after conjugation since the Galois action permutes the rows and preserves degrees. If $h>0$, principality makes $\chi_X(Z)$ vanish at every prime above $s$; unramifiedness gives $s\mid\chi_X(Z)$. This and lcm divisibility make $S_{XZ}$ a nonzero algebraic-integer multiple of $\operatorname{lcm}(x,z)s^e r^c$. Counting the $Z$-orbit and the vacuum therefore gives $mbM^20$. For any deficient rows $A$ and columns $B$, the residue rank at $p$ of $(\chi_x(y))_{A,B}$ bounds below the number of intermediate labels in either factor order $W_t V_p=V_p W_t$ between $A$ and $B$, where $V_j=\gamma_j^{-3}W_j$. Only the full blocks and complementary twist fibers preserved by each factor need be counted. In particular, every deficient $Z$ has a $Y\in P_p$ in its full $t$-block with the same prime-to-$t$ twist and $(W_t)_{ZY}\ne0$. If $t\nmid f_Z$, one may take $Y=Z$. Hence a $t$-full label deficient at both $p$ and $q$ belongs to both principal blocks. \end{proposition} \begin{proof} The $p$-local translates of the nontrivial $p$-phase spaces span an invariant subspace on which the $q$-upper unipotent acts by its normalization scalar: separation gives this on each phase space, and the $q$-action commutes with its translates. The generating conjugates also act by scalars there, so $V_q=I$ there and $V_p=I$ on the orthogonal complement, as in Proposition~\ref{deep:cluster}. Thus $(V_p-I)(V_q-I)=0$ and, up to a scalar root of unity, $S/n=W_t(V_p+V_q-I)$; a dyadic anomaly outside $n$ acts by scalars. The $q$-active columns have $p$-unit codegrees, so $V_q$ is $p$-integral by the lower-unipotent expansion in the preceding Gauss congruence. The single-column estimate in Theorem~\ref{fam:W:2.1} bounds the valuation of column $y$ of $W_t$ below by $1-k_y$ and, by unitarity applied to $W_t^{-1}$, that of row $x$ by $1-k_x$. Commutativity gives \[ v_p((W_t(V_q-I))_{xy})\ge 1-\min(k_x,k_y). \] After scaling row $x$ by $n/d_x$, the character minor therefore equals the scaled $W_t V_p$ minor in the residue field, up to a common unit. A nonzero residue minor is nonzero in characteristic zero; the factorization in either order gives the claimed count. For $A=\{Z\}$, $B=\{0\}$ the entry $\chi_Z(0)=1$ is a unit. The intermediate label for $W_t V_p$ lies in $P_p$ by the vacuum-column lemma and in the required block and fiber by $W_t$. The $t$-block is a singleton if $t\nmid f_Z$ by Theorem~\ref{thm:block}. \end{proof} \paragraph{A trace bound for several principal residues.} Let $Z$ be a nonunit in a perfect fusion-simple symmetric integral table with exactly three dimension primes $a,b,t$. Write $H$ for any subset of these primes at which $Z$ is principal. Then \[ f_Z\ \ge\ [K_Z:\mathbb Q]+ \sum_{q\in\{a,b,t\}}\left(\prod_{p\in H\setminus\{q\}}p\right)^2. \] Indeed for each $q$, the dimension identity \eqref{eq:r8-tracedim} in Lemma~\ref{deep:field-energy} has vacuum term one; reduction modulo $q$ forces a nonzero absolute trace $T_X$ at some nonunit $q$-free degree. That degree has both other prime factors by Lemma~\ref{lem:tables}, so the three column orbits are distinct. Every conjugate of $\chi_Z(X)$ vanishes above each $p\in H$ dividing $d_X$, by principality and Galois covariance. Their absolute trace is a rational integer therefore divisible by the product of these primes. The orbit norm identity (Lemma~\ref{deep:field-energy}), \eqref{eq:r8-tracenorm} counts at least $T_X^2$ for each of these three orbits in addition to the vacuum contribution. Consequently when $a,b$ are separated, every $t$-full degree misses at least two powers of $a$ or at least two powers of $b$ relative to $n$. Otherwise let $H$ be the set of deficient primes. The preceding Weyl argument puts $Z$ in $P_p$ for every $p\in H$, while $f_Z=(\prod_{p\in H}p)^2$. The term $q=t$ alone equals this value, a contradiction after counting the vacuum. Codegree one is impossible by the ordinary row norm as soon as a nonunit is present. This excludes in particular $4\cdot3^e\cdot5$ when two and five are separated and $4\cdot3^{e-1}\cdot25$ when two and three are separated. \subsection{Congruences and quartic fields at the dyadic sextic}\label{sec:dyadic-fields} Proposition~\ref{found:J:4.2} determines the full height at a missing prime. At $D=2^63^{2e}5^4$, in the absence of a twist whose order is divisible by $30$, the joint $3,5$ degrees are divisible by eight, the joint $2,5$ degrees by $3^e$, and the joint $2,3$ degrees by $25$. If all three pairs occur, Lemma~\ref{rev:trace-transfer} and its application at dyadic height six make every joint $3,5$ degree equal to $8\cdot3^j5$ with $1\le jt$, respectively. Indeed the raw value $\sum_Xd_X^2\prod_{t\in T}(1-\theta_{X,t})$ is rational by the vacuum Weyl formula and inclusion--exclusion: all the CRT exponents kill the dyadic normalization. It is therefore equal to its normalized trace $L_T$. Every surviving term is divisible by $u^{2v_u(n)}$ in the cyclotomic integer ring because its label has full $u$-height (Proposition~\ref{found:J:4.2}). At the even core the combined divisors give $43200\mid L_{\{3,5\}}$. Equivalently, writing $d_J=8\cdot3^{j}5$, $a_3=2c_3$, $a_5=4c_5$ on the joint sector, \[ 24\mid \sum_J 9^{j-1}a_3(J)a_5(J). \] When two is included, the factors $c_2,c_3$ have denominators prime to five and $c_2,c_5$ have denominators prime to three. Therefore the omitted prime's full degree square divides the rational simultaneous sum directly in these two cases. Together with the outside divisor in Section~\ref{found:C:6}, the three divisibilities in this triangle are \[ 135000\mid L_{\{2,3\}},\qquad 145800\mid L_{\{2,5\}},\qquad 43200\mid L_{\{3,5\}} . \] If an order divisible by $30$ occurs, both dyadic edges occur. One can then take $X$ and $U$ at the two dyadic edges so that $d_X=12$ or $36$ and $d_U=20$. Both have twist one by the same degree transfers. Their fields already contain $\QQ(z)$, which in the signed case is $\QQ(\zeta_{15})^+$ or $\QQ(\zeta_{15})$ according as the positions carrying the repeated ternary phase are opposite or nonopposite (Section~\ref{found:S:5}). The two distinct degrees give two distinct nonunit orbits. In particular the triangle, with or without a triple twist, has at least nine twist-one simples, rising to seventeen for a nonopposite signed crossing. \paragraph{Thirteen twist-one simples without a triple.} Suppose all three edges occur and no twist order is divisible by $30$. There are nonunit twist-one labels of each of the following degree types: \[ 12\ \text{or}\ 36,\qquad 20,\qquad 15\ \text{or}\ 45 . \] Each can be chosen with character field containing $\QQ(\zeta_5)$. For the first two use $X,U$ from Lemma~\ref{rev:weight4-transfer}. Their unsigned crossing $1+3\xi$ gives the field on both sides. For the third, use Section~\ref{found:C:6} at a jointly three- and five-active column to obtain an odd $Y\in P_3$ in the pure three-twist fiber, of degree $3^j5$ with $1\le j\le2$. By Section~\ref{sec:dyadic-fields} its crossing with $U$ has normalized word $2+3\xi'$ for a primitive fifth root. Its constituent packet at $\xi'$ has dimension $3\lcm(20,d_Y)$. If $Y$ had nontrivial ternary twist, every simple in this packet would be three- and five-active, since $\theta_U=1$ and $Y$ is pure ternary. All those degrees are divisible by eight in the absence of the triple (Proposition~\ref{found:J:4.2}). But the packet dimension has dyadic valuation two. Thus $\theta_Y=1$, and $\QQ(\zeta_5)\subseteq K_Y$. The three degree types distinguish three Galois orbits, giving at least twelve nonunit twist-one simples in addition to the unit. \paragraph{The quartic field throughout a dyadic residue block.} There is also a block restriction valid when a twist is divisible by $30$. More generally, take labels $J,Y$ in ordinary integral data, put $x=\chi_J(Y)$ and $F=\QQ(x)$, and suppose that $\ell\nmid\operatorname{Tr}_{F/\QQ}(x)$. Let $F_{(\ell)}$ be the maximal subfield of $F$ of $\ell$-power degree. Every label in the full $\ell$-block of $J$ has character field containing $F_{(\ell)}$, and the orbit of that block under the cyclotomic Galois group has size divisible by $[F_{(\ell)}:\QQ]$. Indeed this Galois group acts on the full blocks at a fixed prime over $\ell$: equivalence of two characters can be tested after permuting their arguments. If $h$ stabilizes the block of $J$, then $gx\equiv ghx$ at that prime for every Galois element $g$, by using $Y^g$ as the test argument. If the image of the block stabilizer on $F$ had order divisible by $\ell$, it would contain an element of order $\ell$. Grouping the absolute trace into the orbits of this element would make it zero in the residue field. Thus that image has order prime to $\ell$. It fixes $F_{(\ell)}$, since $F/\QQ$ is abelian. The stabilizer of a label in the block is contained in the block stabilizer, proving both assertions. For a joint three--five label $J$, use the $Y,x,F,K$ chosen above. Since $K\subseteq F_{(2)}$, every label of its full two-block has field containing $K$. Its four tame five-conjugate blocks are distinct: five-inertia has image of order four on $K$. None is principal, and every label in them has five-height at most one by Corollary~\ref{cor:field-conductor}. \paragraph{Retaining the quartic field on removing the five-twist.} Let $J$ be any jointly three- and five-active column at the ordinary even sextic core. Its full five-block contains a five-inactive label $Z$ with exactly its dyadic and ternary twists and with \[ \mathbb Q(\zeta_5)\subseteq K_Z\qquad\text{or}\qquad \mathbb Q(\zeta_{15})^+\subseteq K_Z. \] Without an order divisible by $30$, the first inclusion holds. For the same $Y,x,F,K$, the integral trace $\operatorname{Tr}_{F/K}(x)$ lies outside the quadratic subfield of $K$ since its absolute trace is odd. Consequently $x$ has nonzero Galois Fourier coefficient for a faithful character of $\operatorname{Gal}(K/\mathbb Q)$ inflated to the splitting field. Equivalently, for the corresponding character orbit vector $v$ on the conjugates of $Y$ we have $(sv)_J\ne0$ (replacing the character by its inverse if required). All its coordinates are five-inactive. Let $E$ project onto the full five-block and complementary twists of $J$. Since $s=W_5W_{5'}$ and $E$ commutes with $W_5$, the vector $W_{5'}v$ has a nonzero coordinate in $E$. It is still five-inactive and still affords the same character: the local Weyls at two and three invert only a trivial or quadratic component. The stabilizer of a nonzero coordinate therefore fixes $K$, proving the claim. In the absence of a $30$-multiple, the same $Z$ can moreover be chosen in $P_3$. The full dyadic height of $J$ makes its two-block a singleton, so its row of $W_2$ is scalar on the $J$-line. From $(W_2W_5W_3v)_J\ne0$ we now get $E W_3v\ne0$. The initial orbit lies in $P_3$, which $W_3$ preserves. As before, $W_3v$ is five-inactive and affords the faithful conductor-five character. In the no-$30$ triangle the resulting $Z$ is even by the exclusion of ternary-active odd degrees $15,45$ in $P_3$ proved by the packet argument above (which also applies if the odd label has nontrivial five-twist: dilating the primitive fifth root on the opposite witness makes the three-coefficient packet five-active): three-activity forces five-height at least one, ramification at five bounds it by one, and an odd principal degree would be $15$ or $45$. Thus each such joint label requires in its own full five-block four even non-self-dual, five-inactive principal ternary conjugates at its exact ternary phase. \subsection{Residue ranks on a separated path}\label{found:F:4} We apply Proposition~\ref{u2:deficient-minors} with $(p,q,t)=(2,3,5)$; write $z_{xy}=S_{xy}/\operatorname{lcm}(d_x,d_y)$. Now take $D=2^63^{2e}5^4$, $e\ge3$, and suppose two and three are separated. Choose a pure five-primary $U\in P_5$ of degree $20$ from a joint two-five column. For a joint three-five column choose an odd $V\in P_3$ of degree $3^j5$ in its pure ternary fiber. In fact $V$ has twist one, since a three-active degree is divisible by eight. Let $H$ be tame five-inertia, taken of order four. The weight-five calculation in Section~\ref{sec:dyadic-fields} gives $z_{UV}=2+3\xi$ for a primitive fifth root $\xi$. Index the four $H$-conjugates of either label by the nonzero elements of $\mathbb F_5$, and append the vacuum with index zero. The residue character matrix between these two sets of five labels at a fixed prime above two is \[ (\chi_{U_a}(V_b)\bmod\mathfrak p)_{a,b\in\mathbb F_5} =(\overline{\xi}^{\,ab})_{a,b\in\mathbb F_5}. \] For nonzero $a,b$ this follows from $\chi_{U_a}(V_b)=3^j(2+3\xi^{ab})$; the vacuum entries are odd. The matrix is invertible by the Vandermonde determinant. All five rows belong to $P_5$ and have pure five-primary twists. For every full two-block $C$ containing $b_C$ of the five columns, the forward factor order thus gives at least $b_C$ twist-one labels in $P_5\cap C$. In particular if $V\in P_2$ there are at least five twist-one labels in $P_2\cap P_5$, since $P_2$ is Galois-stable. For the reverse factor order let $B$ be the five-block of $V$. Five-inertia fixes $B$. The four blocks $C_a$ of the $U_a$, $a\ne0$, are distinct and nonprincipal by their residues at $V$. The reverse factor order at column $V$ places a label in each $C_a\cap B$ with pure five-primary twist $\theta_{U_a,5}$: $V_p$ preserves the complementary phases of the row $U_a$, and $W_t$ preserves those of the column $V$. The vacuum row similarly gives a twist-one label in $P_2\cap B$. Thus each such five-block meets all five of these specified two-blocks. Every label $Z$ of $C_a$ has $\mathbb Q(\xi)\subseteq K_Z$. For any Galois automorphism $h$ fixing $Z$, the two values $\chi_{U_a}(V^h)$ and $\chi_{U_a}(V)$ agree modulo the same prime above two, since one may substitute $Z$ for $U_a$. The displayed crossing and prime-to-two injectivity of reduction give $h(\xi)=\xi$. In particular all four intersections $C_a\cap B$ contain non-self-dual labels with this precise subfield. This applies to each distinct five-block of an odd pure-ternary $P_3$ witness of five-height one. There is also a second hit for the same joint $J$. In $E=P_3\cap\{\theta_{3'}=1\}$ set $F(Y)=\sum_{h\in H}\chi_{U^h}(Y)-d_Y$. All four rows and the unit are three-inactive, whereas $J$ is three-active. Restricted orthogonality gives $\sum_{Y\in E} F(Y)\chi_J(Y)=0$, using the dual of $J$ in the conjugated row. Here $F(1)=3$. At every odd nonunit of five-height one in $E$ the crossing above gives $F(Y)=0$. Other odd nonunits have full five-height and miss $J$. At $8\mid d_Y$, lcm divisibility gives $F(Y)\in2\mathcal O$. It follows, at every prescribed prime above two, that some even $Y\in E$ has both $F(Y)$ and $\chi_J(Y)$ nonzero in the residue field. Its two-height is one or two, its five-height is at most one, and $Y$ itself has twist one by separation. Its tame five-inertia orbit has size four: otherwise the trace defining $F(Y)$ counts each value an even number of times. This supplies distinct even and odd twist-one $P_3$ orbits of full tame five-inertia meeting every joint $J$. The rank conclusions apply to this second orbit as well. Put $r_h=\overline{\chi_{U^h}(Y)}$ in the residue field at two, where here the bar denotes reduction. Its sum is nonzero by the choice of $Y$. The four-by-four matrix $(r_{ab})_{a,b\in H}$ is invertible: after inverting the column indices it is the regular multiplication matrix of $\sum r_h[h]$ in the group algebra of the cyclic group $H$ of order four. Its augmentation ideal in characteristic two is nilpotent. Thus the forward count in each two-block holds separately for this even orbit. Its five-block also meets each $C_a$ at the pure five-phase of $U_a$, using any nonzero entry in the corresponding row, and meets $P_2$ at twist one: the vacuum crossing with $Y$ is nonzero by the two-sided valuation estimate after scaling by $n/d_Y$. If $v_2(d_Y)=2$, the four conjugates of $Y$ occupy four different full two-blocks, since $\chi_{Y^b}(U_a)=(20/d_Y)\chi_{U_a}(Y^b)$ differs by a two-unit from the residue matrix just shown invertible. \paragraph{A height bound from the exceptional fifth-root adjoint.} Suppose at nonunits $X,Y$ the two monodromy packets have primitive fifth-root ratio, and their larger dimension is $\lambda d_Xd_Y$, $1/2\le\lambda<1$. Suppose also $\theta_{Y,2}=1$. Among the adjoints of $X$ and of a tame five-conjugate $X'$, there is a nonunit constituent $Z$ and there is a constituent $T$ of $YZ$ for which the dyadic monodromy is exactly $-1$. Furthermore \[ d_Z\le d_X^2-1,\qquad d_T\le\lambda d_X^2d_Y . \] Indeed the adjoint character word at row $Y$ has the full mass on $A=\{1,\xi,\xi^{-1}\}$, where $\xi$ is the relative fifth root. On the identity coset of the five-primary positions put $1$ on $A$ and $-3/2$ on its two vacant positions, and put zero on every other coset. Tensor with normalized trace on the prime-to-five factor. This rational functional has value one only on $A$, and is less than one everywhere else except at the negatives of the two vacant roots, where its value is $3/2$. Indeed a complementary root other than $-1$ has normalized trace at least $-1/2$. Apply also the dilated functional to the conjugate with multiplier two at five, for which $A\cap A^2=\{1\}$. If the two balancing expansions of the adjoints avoided their respective exceptional negative phases, equality at the full mass would force all phases of their common nonunit constituent into $\{1\}$. Strictness and the common-nonunit lemma in the proof of Theorem~\ref{found:S:4.1} prohibit this. The unit has no negative phase. For either adjoint every such $T$ occurs in $(YX)X^*$ or its counterpart with $X'$. The first product $YX$ splits into the two indicated packets, of which each simple has degree at most $\lambda d_Xd_Y$. The same holds after moving just the $X$-label by tame inertia: the resulting two positive fifth positions are exposed by the two-mark functional in Section~\ref{found:S:5}, so their weights persist. Apply this to $X=U$ of degree $20$ and any odd $Y\in P_3$ of degree $3^j5$ in the pure ternary fiber. Here $\lambda=3/5$ and $Y$ has twist one. At least one of $Z,T$ is dyadically active and therefore has degree at least $2\cdot3^e$, since it has full three-height and no nonunit has prime-power degree. As $399<1200\cdot3^j$, the two bounds imply $3^{e-j}\le600$, hence $j\ge e-5$. If $e\ge5$, every $Z$ in this argument is dyadically inactive since $2\cdot3^e>399$. The actual dyadic component of $T$ is then $-1$. Both the lower height bound and this prescribed phase hold uniformly without an upper bound on $e$. The five-free $P_3$ degrees satisfy a similar bound. Write one of them $d_Y=2^i3^j$ with $i=2$ or $3$ and $j0$, a residue $c\in\{0,\ldots,\ell-1\}$, and $0\le t(24/2^i)^2$. For $i=0$, lcm integrality makes the character even at even degrees. Its squared norm $576$ modulo four requires at least three odd nonunit values besides the vacuum; each is divisible by three. The five-free hit has even degree and value divisible by six. These hits, the vacuum and $A$ already cost $3\cdot9+36+1+8\cdot64=576$. At a two-active column three-avoidance gives a further three-free nonunit in $P_2$ outside $A$ (its dyadic height is at most two); its entry is nonzero by three-principality. This contradiction gives $|A|=4$. It is a single Galois orbit. Proposition~\ref{prop:principal-support} provides a member with pure five-twist by avoiding three, and all three-free labels have pure three-twist by Section~\ref{found:C:6}; thus all four have twist one. Put $a=\frac12\sum_{X\in A} v_X$ and $w=sa$. The orbit is stable under duality, so $w$ is real. On $B$ every coordinate has absolute value $2/3$. The modular relation $sTs=\xi T^{-1}sT^{-1}$ and $Ta=a$ give \[ \left|\sum_Z w_Z^2\theta_Z\right|=|\langle a,w\rangle|. \] If there were two labels in $B$, their phases would have sum of modulus at least one since both lie in $\mu_3$. Their combined contribution on the left has modulus at least $4/9$. The remaining mass is $1/9$, giving $|\langle a,w\rangle|\ge1/3$. But $w_0=80/n>0$ lies outside $A\cup B$, and hence $|\langle a,w\rangle|\le\|w|_A\|<1/3$. Thus $B=\{Y\}$ and Galois invariance gives $K_Y=\mathbb Q$. In particular $Y$ is self-dual. Let $b_U=\chi_Y(U)\in\ZZ$, $i=v_2(d_Y)$, and $q=n/d_Y$. Thus $i=0,1,2$, $q=24/2^i$, and $|b_U|=8/2^i$ on $A$. Let $U$ be a five-free nonunit in $P_3$. Its degree is $2^k3^j$ with $k=2$ or $3$, $j\ge1$ (Lemma~\ref{three:complementary}). If $b_U\ne0$, then $3\cdot2^{k-i}\mid b_U$ by three-principality and lcm divisibility. For $k=2$ its crossing with any member of $A$ is $1+3\zeta$ after lcm normalization, with $\zeta$ a primitive fifth root: the signed weight-four crossing would produce a joint $2,5$ twist (Section~\ref{found:S:6}). Thus the Galois orbit of $U$ has at least four members, all with the same $b_U$ since $Y$ is rational. For $k=3$ a single label suffices. Either orbit costs at least $(24/2^i)^2=q^2$ in $\sum_U b_U^2=q^2$. This leaves no room for the unit. Hence $b_U=0$ for every five-free nonunit $U$ in $P_3$. All members of $P_3$ have purely ternary twist (Section~\ref{found:C:6}). If $Y$ were three-active, prime avoidance at $Y$ with avoided prime five (Proposition~\ref{prop:principal-support}) would contradict the vanishing just proved. Thus $\theta_Y=1$. Moreover every three-active $J$ must have five-bearing degree. Indeed block orthogonality for distinct ternary twists gives \[ \sum_{U\in P_3}\overline{\chi_Y(U)}\chi_J(U)=0. \] The only five-free contribution is the unit. If $5\nmid d_J$, every five-bearing $U$ contributes zero modulo a prime over five by lcm divisibility, a contradiction. The diagonal modular relation and $\theta_Y=1$ give $|\sum_Z b_Z^2\theta_Z|=q|b_Y|$. Since the row norm is $q^2$, its squared mass on twist one is at most $(q^2+q|b_Y|)/2$. At a dyadically active column use Proposition~\ref{prop:principal-support} on the principal two-block in the pure dyadic twist fiber, avoiding three. The resulting nonunit $W$ is three-free, hence also has purely ternary twist and therefore twist one. It is outside $A$ since its dyadic height is below three. Its degree is even, and $b_W\ne0$ by three-principality of $Y$. By symmetry $\chi_W(Y)=d_Yb_W/d_W$ is an integer, congruent to $d_Y$ modulo two by two-principality. Thus $|b_W|\ge2$ for $i=0,1$, and $|b_W|\ge1$ for $i=2$. By dimension orthogonality reduced modulo five there is at least one five-free nonunit $Z$ with $b_Z\ne0$. Every such $Z$ has pure five-twist because it can be neither two-active (that requires full five-height) nor three-active (as just proved), and is five-inactive by defect-zero vanishing in the row of $Y$, since five does not divide $q^2$. Hence $\theta_Z=1$. Write $d_Z=2^k3^j$ with $k,j\ge1$, using the exclusion of nonunit prime-power degrees. Lcm divisibility and three-principality give \[ 3\cdot 2^{\max(k-i,0)}\mid b_Z . \] The unit, $A$, $Y$ and $W$ give distinct twist-one contributions outside these positions. Subtracting them from the twist-one budget leaves at most \[ \frac{q^2+q|b_Y|}{2}-1-4(8/2^i)^2-b_Y^2-b_W^2 =\frac{68}{4^i}-1-b_W^2-(|b_Y|-6/2^i)^2 . \] For $i=0,1,2$ this is at most $63,12,9/4$, respectively. Every five-free hit costs at least $36,9,9$, respectively. This excludes $i=2$ outright. For $i=0,1$ exactly one hit remains and it has $k=1$, since $k\ge2$ costs at least $144,36$, respectively. Uniqueness and the rationality of $Y$ make this $Z$ Galois-fixed, hence rational. Now cross $Z$ with any degree-$40$ label $X\in A\subset P_5$. The lcm-normalized entry $z=S_{XZ}/\operatorname{lcm}(d_X,d_Z)$ is rational since $Z$ is rational, and integral by lcm divisibility. Since $\gcd(d_X,d_Z)=2$ and $5\nmid d_Z$, the congruence $\chi_X(Z)=(d_Z/2)z\equiv d_Z\pmod5$ gives $z\equiv2\pmod5$. Strictness gives $|z|<2$, an impossibility. At the first core the surviving degrees satisfy a common residue equation. More generally, take any jointly two- and three-active $J$ at $D=2^6 3^{2e}5^4$ on the path separating two and five. For every prime above two, \[ \sum_{\substack{U\in P_3,\ d_U=25\cdot3^j\\1\le j\le e-2}}\chi_J(U) \equiv1\pmod{\mathfrak p},\qquad \mathfrak p\mid2. \] Indeed block orthogonality gives $\sum_{U\in P_3}d_U\chi_J(U)=0$. Only odd degrees survive reduction. Every odd nonunit degree is $3^j5^t$, with $1\le j\le e-1$ and $t=1,2$, by Corollary~\ref{cor:principal-valuation-bound} and the exclusion of prime-power degrees. At $t=1$ the weight-five crossing gives an orbit of length four under the order-four subgroup of tame five-inertia (Section~\ref{found:F:5}). That subgroup fixes $J$, since two-activity requires full five-height. The character is constant on the four arguments, so the contribution vanishes modulo two. At $t=2$ the upper ternary height has just been excluded. This proves the congruence. At $e=3$ only degree $75$ occurs in the sum; in particular every such $J$ meets at least one degree-$75$ row. If $d_J$ is odd, the same argument works modulo four after retaining the factors $d_U$: lcm divisibility makes $d_U\chi_J(U)$ a multiple of four at every even degree. Thus at $e=3$ one has $\sum_{d_U=75,\ U\in P_3}\chi_J(U)\equiv1\pmod4$ for such $J$, since $75\equiv-1\pmod4$. The degree-$75$ witness at $e=3$ can be chosen with character field contained in $\QQ(\zeta_{27})$. Indeed group the last residue sum by dyadic inertia. It is a two-group, and conjugate summands in an inertia orbit have the same residue. A nonzero contribution therefore comes from a fixed label. Its field is unramified at two, and at five by its codegree. It is consequently a ternary cyclotomic subfield. The weighted inequality for the sets $B_j$ in Proposition~\ref{found:F:5.1} gives $4\cdot9|B_1|<729$, so its entire orbit has size at most twenty. A ternary field of conductor at least $81$ has inertia order divisible by $27$. This proves the claim. \subsection{The local Weyl bound for a nontrivial tame character}\label{found:P:2} \label{found:P:key:sect:weyl} \begin{lemma}[Weyl compression in a tame torus character]\label{rev:tame-compression} Let $s=S/n$ in an ordinary integral modular category of square dimension $D=n^2$. For an odd prime $p$, take a character $\chi$ of the group of cyclotomic tori whose local factor $\psi$ at $p$ is inflated from $\mathbb F_p^\times$ and is nontrivial. In particular the diagonal torus is unsigned. Write $P_0$ for the projector onto $p$-inactive labels. On the $\chi$-isotypic space, \[ \|P_0s P_0\|\le p^{-1/2}. \] More precisely, let $v$ be a $p$-inactive vector of torus type $\chi$ and $E$ fix the complementary twist components, optionally inside one full $p$-block. When $\psi^2\ne1$, in $Esv$ the sums of squared absolute coordinates at the $p$ individual phases $1,\zeta_p,\dots,\zeta_p^{p-1}$ are all equal. Higher $p$-power phases may also occur. \end{lemma} \begin{proof} Indeed put $w=EW_{p'}v$, using $s=W_pW_{p'}$. The projector $E$ commutes with the local group, so $w$ is upper-unipotent-fixed and has local torus type $\psi$. At local level $p^a$, let $Q_1$ average over the kernel of reduction mod $p$. The lower--diagonal--upper factorization of this kernel as in Section~\ref{found:F:3} gives $Q_1w=\operatorname{avg}_{p\mid t}L(t)w$: the torus character is trivial on units congruent to one modulo $p$. Hence $W_pQ_1w$ is precisely the restriction of $W_pw=Esv$ to phases in $\mu_p$. Write $u=Q_1w$ in the level-one quotient and $B$ for its upper Borel. If $t\ne0$ in $\mathbb F_p$, $W_p^{-1}U(t)W_p\in B W_p B$. In the explicit Bruhat decomposition the two outer upper unipotents fix $u$, and the intervening torus factor has argument $t$ or its inverse up to fixed sign. Thus the coefficient at $u$ is a constant times $\psi(t)$ or $\psi(t)^{-1}$. Its sum over $t\ne0$ is zero. Finite additive Fourier inversion at the zero frequency gives squared mass $\|u\|^2/p$ at the trivial phase. If $\psi^2\ne1$ the coefficient vanishes individually since $W_p$ sends the torus type $\psi$ to the distinct type $\psi^{-1}$. Each of the $p$ masses is then $\|u\|^2/p$. Taking $E=1$ proves the norm bound. The same proof in commuting local factors gives the norm bound $\prod_{p\in T}p^{-1/2}$ if input and output are inactive at all these primes and each local factor of $\chi$ is tame and nontrivial: after moving the other Weyls past the projectors, the local compressed operators preserve the other local character types. \end{proof} In particular, for two lists of $p$-inactive orbits the Fourier matrix of Proposition~\ref{found:J:6.1} in this mode has norm at most $n/\sqrt p$. Its $(X,Y)$ entry remains the usual product of $\operatorname{lcm}(d_X,d_Y)$, the square root of $[K_X:F][K_Y:F]$, and $\sum_{g\in\operatorname{Gal}(F/\mathbb Q)}\chi(g)g(z)$, where $z=S_{XY}/\operatorname{lcm}(d_X,d_Y)$ and $F=\mathbb Q(z)$, provided the character descends to $F$. \paragraph{Two missing principal degrees on the separated path.} Take $D=2^63^{2e}5^4$, $e\ge3$, with two and five separated and both joined to three. Every member of $P_3$ and every three-free simple is five-inactive by Section~\ref{found:C:6}. Choose a three-free nonunit $U\in P_5$ tame at three by Section~\ref{found:C:6} at a five-active column, avoiding three. Its degree is $40$ by Section~\ref{found:S:6}. Let $R$ range through Galois orbit representatives of members of $P_3$ of degrees $4\cdot3^{j_R}$ or $5\cdot3^{j_R}$. Each type occurs: apply prime avoidance at a column mixing two and three to omit five, and at a column mixing three and five to omit two, respectively (Section~\ref{found:S:6} and Section~\ref{found:C:6}). Here the odd hit has five-height one because a full-five label misses a five-active column. In the even case the hit has dyadic height two because it sees a two-active column. At the even type the normalized entry is $1+3\xi$, with $\xi$ a primitive fifth root. At the odd type it is $\eta(2+3\xi)$ for a ternary root $\eta$ (Corollary~\ref{found:J:5.2}). Its tame five-inertia trace is $5\eta$, hence $\eta\in K_U$. Tameness at three gives $\eta\in\mu_3$. The entry field $F_R$ is $\mathbb Q(\zeta_5)$ for $\eta=1$ and $\mathbb Q(\zeta_{15})$ otherwise, interpreting $\eta=1$ for the even type. For a faithful quartic character at five alone the absolute Fourier coefficient on $F_R$ is $3\sqrt5$ in every case: when $\eta$ is primitive cubic its two conjugates sum to $-1$. The lcm is always $40\cdot3^{j_R}$. Dividing the improved column bound by $5\cdot9\cdot40^2$ gives \[ \sum_R 9^{j_R}[K_U:F_R][K_R:F_R]\le 9^{e-1}. \] A single exponent $e-1$ would exhaust this budget, although the opposite degree type is compulsory. Thus both types satisfy $j_R\le e-2$, with no hypothesis on the high full-five degrees. At $e=3$ each of the two types exists, all such degrees are respectively $12,15$, and the displayed sum of relative index products without the weights is at most nine. In particular the chosen degree-$40$ witness has field $\QQ(\zeta_5)$ or $\QQ(\zeta_{15})$ at the first dimension. Wild five-inertia would contribute a factor five to each of the two compulsory index products, exceeding nine. This field is already unramified at two and at most tame at three. \paragraph{Pure fifth phases on the triangle.} At the core with all three edges, take the twist-one $X\in P_3$, $U\in P_5$ of degrees $12$ or $36$, and $20$ from Section~\ref{found:F:2}. Their crossing has a nonzero Fourier coefficient for a character cutting out $K=\mathbb Q(\zeta_5)$ in the unsigned case and $K=\mathbb Q(\zeta_{15})^+$ in the signed case. Both characters restrict at five to a faithful quartic. Apply the equal-norm identity to the $\chi$-orbit vector on $X$ in the full principal five-block with trivial complementary phases. Its coordinate at $U$ after applying $s$ is nonzero. By equality every primitive fifth phase has a nonzero coordinate there. This fiber is globally Galois-stable, and $s$ inverts the torus character, so the field of each such label contains $K$. Tame five-inertia inversion acts nontrivially on $K$ and preserves each fifth phase by the square action. Hence at least two such labels occur at each of the four phases in $P_5$. For the nonopposite signed word, both quartic characters have a nonzero coefficient. For the conductor-five character, the trace over three assigns coefficient $2$ to the repeated positions and $-1$ to the others, up to global sign. The resulting nonzero coefficient has magnitude $3\sqrt{10}$ because the two repeated positions are nonopposite; the character twisted by the quadratic character at three was treated in Section~\ref{found:F:2}. Each phase therefore has witnesses for both quartic fields. If the witnesses are in distinct Galois orbits each contributes at least two. If they are in the same orbit its field contains their compositum $\mathbb Q(\zeta_{15})$, and the orbit has at least eight labels at precisely two fifth phases with equal multiplicities. Thus each primitive fifth phase occurs at least four times in this case. If the triangle has no triple, let $Z\in P_3$ be the third twist-one witness of Section~\ref{found:F:2}. Write $d_X=4\cdot3^j$, $d_Z=5\cdot3^k$ with $j,k\in\{1,2\}$, and $m_T=[K_T:\mathbb Q(\zeta_5)]$ for these three labels. Their two unsigned crossings against $U$ have faithful-quartic coefficient of magnitude $3\sqrt5$. The compressed column norm gives \[ m_U(9^{j-1}m_X+9^{k-1}m_Z)\le 36. \] Consequently a degree-$36$ or degree-$45$ witness has relative-index product with $U$ at most three. If both occur, $m_U(m_X+m_Z)\le4$. \subsection{Which principal rows meet a joint three--five column?}\label{found:P:4} Consider $D=2^63^{2e}5^4$, $e\ge3$, on the ordinary path where two and three are separated. Let $P_3$ be the full principal three-block. A joint three--five column is a label whose twist order is divisible by $15$. Every such column meets an odd member of $P_3$ of five-height one and an even member of five-height at most one (Section~\ref{found:F:4}). We determine which even hits have five-height one and bound the number of distinct character values there. Choose a joint two--five column. Apply the wild-inertia refinement of Section~\ref{found:C:6} to the principal five-block at this column, avoiding three. The resulting three-free label $U$ has degree $20$ by Section~\ref{found:S:6} and its field $K_U$ is tame at three. Write $h=[K_U:\mathbb Q]$ and $\tau$ for normalized field trace to $\mathbb Q$. Its normalized orbit sum and eigenvalues are \[ B=\frac8h\sum_{U'\in\operatorname{Gal}\cdot U}[U'], \qquad b_Y=8\tau(\chi_Y(U)),\qquad b_1=160. \] The normalized-trace argument gives $B$ in the three-local fusion ring and $b_Y\in\mathbb Z$. Indeed $3\nmid h$ by tameness and the prime support $\{2,3,5\}$, and normalized traces of cyclotomic integers in this support lie in $\tfrac18\mathbb Z$. For a nonunit $Y$ one has $-1605$, one can require ramification at five. In particular the first assertion applies at $e=3$, and the second applies to every such $J$ at $e\ge7$. \end{proposition} \begin{proof} For every even-degree $Y\in P_3$ with $v_5(d_Y)=1$ and $S_{JY}\ne0$, put $x=\chi_Y(U)$ and $b=8\tau(x)$, where $\tau$ denotes normalized field trace to $\QQ$. By Proposition~\ref{found:P:4.1}, the distinct $b$ other than $10,40$ satisfy \[ \sum_b v_3(160-b)\ \ge\ 2e-j-2. \] All conjugates of $x$ have modulus strictly below $20$ and are congruent to $20$ above three by $Y\in P_3$. They are also congruent to $20$ above five, since $U\in P_5$ gives $\chi_U(Y)=(d_Y/20)x\equiv d_Y$ and $d_Y/20$ is a five-adic unit. Galois covariance gives these congruences at every conjugate prime. Suppose $F_Y$ is unramified at five. Then $z=x/5$ is an algebraic integer of conductor supported on two and three, by the character-field conductor bound. Hence $k=2\tau(z)\in\ZZ$. Indeed the normalized trace of a dyadic root of unity is integral, that of a root of three-power order is half-integral, and the traces multiply at coprime orders. Roots of unity span the cyclotomic integer ring. Moreover $[F_Y:\QQ]$ is prime to three since $F_Y\subseteq K_U$, the latter is tame at three, and the cyclotomic degrees at two and five are prime to three. Thus the congruences for the conjugates of $z$ give $k\equiv8\pmod3$. The bound $|k|<8$ leaves $k=-7,-4,-1,2,5$, with $b=20k$. The value $k=2$ is discarded. The others contribute at most $1+1+2+1=5$ to the sum, proving the five-ramification assertion. If $F_Y$ is unramified also at three, its conductor is dyadic. Then $\tau(z)$ is integral, so $k$ is even. Alternatively, if $v_2(d_Y)=1$, lcm divisibility gives $x\in2\mathcal O_{F_Y}$, hence $z\in2\mathcal O_{F_Y}$ and again $k$ is even. In both cases only $k=-4,2$ remain. Their values $b=-80,40$ contribute at most $v_3(240)=1$. The three-activity of $J$ gives $j\le e-1$, so the required sum is at least $e-1\ge2$. This proves the first assertion. \end{proof} \section{The surviving odd twist path}\label{found:group:odd} \subsection{Compulsory edges and orientation} Throughout this section a pair of primes occurs when their product divides some twist order. The larger prime is $r=23$. The first result locates the two compulsory edges; the second fixes a common orientation for all complementary principal witnesses on the remaining separated path. \begin{theorem}\label{found:J:8.1}\label{found:J:key:thm:twenty-three}\label{found:C:4.1}\label{found:C:key:prop:septmix} Suppose that an ordinary perfect fusion-simple minimal table has dimension $D=5^6 7^B23^4$. For every $B$ some twist order is divisible by $7\cdot23$. At $B=6$ some twist order is also divisible by $5\cdot23$. \end{theorem} \begin{proposition}\label{found:D:6.3}\label{found:D:key:prop:odd-oriented-path} Consider an ordinary minimal table of dimension $5^6 7^6 23^4$ with separated twist primes five and seven. Let $U$ range over all seven-free nonunits of $P_5$ and $V$ over all five-free nonunits of $P_7$. A single $t\in\{5,7\}$ ramifies in every one of these row fields; the other prime $s$ is unramified in all of them. If $t=5$ every such $V$ has seven-height two, and if $t=7$ every such $U$ has five-height two. Every $s$-active degree has $23$-height at most one. There is a twist-one orbit of size at least $44$ among the $V$; when $t=5$ the same holds among the $U$. \end{proposition} The following notation applies throughout the separated path at $B=6$. \begin{center}\small \begin{tabular}{@{}llp{.54\linewidth}@{}}\toprule Orientation & $(s,t,r)$ & Complementary rows\\\midrule Seven unramified & $(7,5,23)$ & $s$-active rows have $23$-height $\le1$; excluded by Theorem~\ref{por:th:odd-seven-elim}\\ Five unramified & $(5,7,23)$ & $s$-active rows have $23$-height $\le1$; the remaining orientation\\\bottomrule \end{tabular} \end{center} Here $t$ ramifies in the complementary principal row fields and $s$ is the other endpoint. The twist-one orbit from Proposition~\ref{found:D:6.3} enters Corollary~\ref{fam:rank:five-seven-twenty-three}. \subsection{The remaining ordered triple at least prime five}\label{found:C:4} Write $P_t$ for the principal $t$-block. We use the following consequence of Proposition~\ref{found:C:key:prop:above-prime}. At the ordinary $5,7,23$ table, a complementary $P_7,P_{23}$ crossing of common weight $25$ has four distinct $23$-positions and trivial $7$-parts if its entry is unramified at seven. Its actual masses are $4\ell,7\ell,7\ell,7\ell$, where $\ell$ is the lcm of the degrees. Indeed Proposition~\ref{found:C:key:prop:above-prime} gives a unique expression in which the seven-parts are two copies of $1$ and $23$ copies of $\eta$. If the entry is unramified at seven, its unique seven-projection is invariant under seven-inertia, so $\eta=1$. The word obstruction $W(23,3)$ then forces four positions, and the congruence of the multiplicities modulo seven gives the stated masses. The tame $23$-inertia orbit has length $22$ by Corollary~\ref{found:C:key:cor:orbit22}. If a column is active at both five and $23$, principal support (Proposition~\ref{prop:principal-support}) at $23$ modulo seven supplies a seven-free nonunit $C\in P_{23}$ hitting it. At $5^6 7^{2e}23^4$ its degree is $25\cdot23$: its $23$-height is one by quartic principality, its five-height is at least two by the complementary bound in Section~\ref{found:C:6}, and it cannot be three by defect-zero vanishing on the chosen column. \begin{proposition}\label{found:C:4.2}\label{found:C:key:prop:remaining-edges} In the frontier dimension $5^6 7^6 23^4$ the following constraints hold on the two possible separations involving five. \begin{enumerate} \item If five and $23$ are separated, every seven-free nonunit $C\in P_{23}$ has degree $125\cdot23$ and a tame $23$-inertia orbit of length $22$. \item If five and seven are separated, there is a seven-free $C\in P_{23}$ of degree $25\cdot23$. Its crossings with each five-free $B\in P_7$ of $23$-height one and with each $23$-free $X\in P_7$ of five-height two have only $23$-primary monodromy phases. Both resulting $23$-words have four positions. \end{enumerate} \end{proposition} \begin{proof} Five cannot be isolated, by Proposition~\ref{found:C:key:prop:sexticrect}. If five and $23$ are separated, a five- and seven-active column therefore exists. A $23$-free $X\in P_7$ hitting it has five-height exactly two: the complementary bound gives at least two and defect-zero vanishing rules out three. Each seven-free nonunit $C\in P_{23}$ has $23$-height one and five-height at least two, as above. If its five-height were two, neither $X$ nor $C$ could be $23$-active, since $23$-activity forces full five-height. At their weight-$25$ crossing, Proposition~\ref{found:C:key:prop:above-prime} and Corollary~\ref{found:C:key:cor:orbit22} give a nonidentity $23$-position with multiplicity $m\in\{7,14,21\}$. Its summands are $23$-active, so $125\mid m\operatorname{lcm}(d_X,d_C)$, impossible at common five-height two. Thus $d_C=125\cdot23$, and the same crossing still has common weight $25$ and a tame orbit of length $22$. Suppose instead that five and seven are separated. Five must mix with $23$. Use the degree-$25\cdot23$ witness $C$ above for that mixed column. Seven-active degrees have full five-height, so $C$ and both indicated partners are seven-inactive. The weight-$23$ crossing with $B$ has one common seven-part $\eta$ by Proposition~\ref{fam:Q:3.1}. If $\eta\ne1$, all constituents would be seven-active, contrary to $125\nmid d_Bd_C$. At the weight-$25$ crossing with $X$, a nontrivial $\eta$ in Proposition~\ref{found:C:key:prop:above-prime} would put total mass $23\operatorname{lcm}(d_X,d_C)$ on seven-active constituents. This again is not divisible by $125$. The words at both crossings are therefore pure $23$-words on at most four positions, and $W(23,3)$ forces all four. \end{proof} \paragraph{The middle-prime height of an outer witness.} For any odd $pm+L$ and $z$ is unramified at $s$, then \[ z=d+s\sum_{j=1}^m\eta_j \] for $p$-power roots $\eta_j$. Indeed $(z-d)/s$ is integral in the complementary two-prime field; normalized $s$-trace and two-prime unimodularity give a signed integral word of length at most $L$. Erase its $p$-parts. The resulting $q$-primary word equals $m$ modulo every prime above $p$, and $p$ is unramified there. Its difference from $m$ has house at most $m+L2m+1$. When $g=2s+d$, $049$. All joints therefore have $h=0$ and every such $E$ hits each one by principality. Fix $E$ and set $L=125\cdot49\cdot23$. Write $b=S_{EJ}/L$ and let $c=7$ or $3$ according as $k=1$ or $2$. Since $\chi_E(J)=125b$, principality gives $b\equiv c\pmod{23}$ in the unramified field. At least two five-inertia conjugates of $E$ agree on the $J$-orbit. Taking their average gives $M(b)<161^2/(2\cdot66)<197$, where $M$ is the Galois mean square. If $b\ne c$, the integral element $(b-c)/23$ has mean square at least one; the triangle inequality then gives $M(b)\ge(23-c)^2 \ge256$. Therefore $b=c$. At $k=1$ this makes $\chi_E(J)=d_J$, contrary to strictness. All joints have degree $125\cdot49$ and $b=3$. If their number is $N$, their total cost gives $18N<161^2$. More precisely, if the primary twist orders of a joint are $7^u,23^v$, its orbit length is a multiple of $66\cdot7^{u-1}23^{v-1}$ by the same square action and duality. Our bound thus gives $v=1$ and $u=1$ or $2$. Write the counts in these two sectors $66A$ and $462B'$; then $1\le A+7B'\le21$. The simultaneous Gauss congruence of Proposition~\ref{rev:simul-gauss} for $T=\{7,23\}$ makes $5\sqrt D$ divide the positive integer \[ (125\cdot49)^2 \left(66A\frac{7}{6}\frac{23}{22} +462B'\frac{23}{22}\right) =(125\cdot49)^2\frac{7\cdot23}{2}(A+6B'). \] In particular $23\mid A+6B'$, a contradiction. This eliminates the five-ramified alternative. In the surviving alternative $z$ is unramified at five. Apply the converse test at $49=9\cdot5+4$ to get the unique word $z=4+5\sum_{j=1}^9\xi_j$ in $23$-power roots. The $23$-tame orbit length of every such $U,E$ is at least eleven. In particular any five- and seven-active $Z$ whose degree is divisible by five has a $23$-free hit of degree exactly $5\cdot49$ in $P_5$. Indeed take a hit $U$ by prime avoidance. Its seven-height is two: it is at least two as above, and full height would miss $Z$. Write $d_Z=5^h7^k23^2$ and $d_U=5^i49$. As $K_U$ is unramified at five, $5\mid\chi_U(Z)$, and $S_{UZ}$ is a nonzero integral multiple of $5^{\max(h,i+1)}49\cdot23^2$. The $Z$-orbit has at least twelve members, by the square twist action and odd duality. At least eleven inertia-conjugate $U$ rows agree there. If $i=2$, their norm bound would give $11\cdot12<49$. Hence $i=1$ as claimed. We exclude the second alternative. The preceding unsigned word on at most ten $23$-positions is exposed among all phases: on a $\mu_{23}$-coset with $b\le10$ marks use value one on the marks and $-b/(23-b)$ at unmarked positions, tensoring with normalized trace at complementary primes (Lemma~\ref{fam:Q:2.2}). It is also linearly independent on its support by the cyclotomic coset relations. Thus every actual nonidentity balancing packet of $U,E$ has dimension $5m\operatorname{lcm}(d_U,d_E)$ for a positive $m$, with the sum of these $m$ at most nine. Both factors are $23$-inactive since neither has full five-height, and every $23$-active constituent has dimension divisible by $125$. If $i=1$, then $5\mid m$ for every such packet. There would be at most one nonidentity position, contradicting strictness or Theorem~\ref{found:S:4.1}. Hence all $23$-free nonunits of $P_5$ have $i=2$. The degree-$5\cdot49$ conclusion just proved rules out every five-bearing joint five- and seven-active label. The remaining joint labels have degree $7^k23^2$, $k=1,2$, by the seven-divisibility transfer and activity. Their set is nonempty since five is not isolated. Fix a $23$-free $U\in P_5$ hitting one of them. Its degree must now be $25\cdot49$, and it hits every joint by principality. At such a joint $J$, put \[ L=25\cdot49\cdot23^2,\qquad b=S_{UJ}/L. \] Since $K_U$ is unramified at five and $K_J$ at $23$, $b$ lies in a seven-power field. The congruence $\chi_U(J)=23^2b\equiv 7^k23^2$ above five therefore gives $b\equiv7^k\pmod5$. Counting at least eleven $U$-rows agreeing on the entire $J$-orbit as before, we get \[ M(b)<\frac{35^2}{11\cdot12}<9.29 . \] For $k=2$ this forces $b=-1$, since $(b+1)/5$ is integral and a nonzero value would give $M(b)\ge(5-1)^2$. For $k=1$ let $w=(b-2)/5$. A nonzero integer in a seven-power field has mean square at least $5/3$ unless it is a signed seven-power root (Proposition~\ref{found:C:3.1}). The reverse triangle inequality would then give $M(b)\ge(5\sqrt{5/3}-2)^2>9.29$. For a signed seven-power root $M(b)=29+20\operatorname{Tr}_{\mathrm{mean}}(w)$, and this is at most $9.29$ only at $w=-1$. Thus at $k=1$ we have $b=2$ or $-3$. All these entries are rational. The whole Galois orbit of $U$ now agrees on every joint column. The $23$-tame inertia orbit has size $11$ or $22$, so eleven divides the full orbit length. Duality has no fixed points on that orbit in odd dimension. Its full length is therefore divisible by $22$. If $N_k$ counts the joint labels of each degree, the averaged row gives $4N_1+N_2<35^2/22<56$. Each joint orbit length is divisible by $12\cdot5^{u-1}7^{v-1}$, where its actual primary twist orders are $5^u,7^v$. Thus $u=v=1$ throughout and, writing $N_1=12A$, $N_2=12B'$, we have $4A+B'\le4$ and $A+B'>0$. The joint Gauss sum is now \[ L_{5,7}=\frac{5}{4}\frac{7}{6}\,23^4 \bigl(7^2 N_1+7^4N_2\bigr). \] It is divisible by $23\sqrt D$, hence $25\mid A+49B'$. Consequently $A\equiv B'\pmod{25}$, impossible in the indicated range. We have proved that a twist divisible by $5\cdot23$ is compulsory. Together with the first result of this section both edges to the quartic prime are compulsory; only separation of five and seven, or a complete three-edge graph, remains. \paragraph{A bound on two crossing fields.} In the ordinary $5^6 7^{2e}23^4$ family, apply this to either of the two crossings of Proposition~\ref{found:C:3.3}, at weights $23$ and $25$. Suppose $5^37^{e-1}23\mid L_{XY}$. By Lemma~\ref{found:C:3.4}, the seven-part is trivial. In particular the entry field is $F=\mathbb Q(\zeta_{23})$. Thus for $n_X=[K_X:F]$ and $n_Y=[K_Y:F]$ we obtain \[ n_Xn_Y\le3. \] Indeed the three nonzero positions all have weight seven. At each of the four origins in the affine support classified by the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/word_closure.py}{enumeration in Proposition~\ref*{found:C:3.3}}, Section~\ref{found:C:3} provides a nontrivial character at $23$ sending those positions, after rotation, into an arc spanning at most four steps modulo $22$, of length $4\pi/11$. A multiplicative Gauss sum of modulus $\sqrt{23}$ consequently gives a Fourier coefficient of modulus at least $21\sqrt{23}\cos(2\pi/11)$. On the other hand $\sqrt D/L_{XY}\le161$. Since $\cos(2\pi/11)>5/6$ and $35\sqrt{23}>161$, the single-entry bound in Proposition~\ref{found:J:key:new:fourierblock} excludes $n_Xn_Y\ge4$. In particular one of the two character fields equals $\mathbb Q(\zeta_{23})$ and the other has relative degree at most three. Neither label admits wild inertia at $5,7$ or $23$. At most one of the labels can have a twist divisible by seven, since a label with seven-active twist has at least three distinct seven-inertia conjugates by the square action, all fixing $F$. If such a label occurs, its seven-part has order exactly seven, its field has relative degree three, and the other label has field $F$. More explicitly, the relative field of nontrivial degree can only be obtained by adjoining to $F$ one of \[ \sqrt5,\quad\sqrt{-7},\quad\sqrt{-35},\quad \zeta_7+\zeta_7^{-1}. \] Indeed after excluding wild inertia the ambient field is $\mathbb Q(\zeta_{5},\zeta_{7},\zeta_{23})$. A subfield containing the last factor splits as its compositum with a subfield of $\mathbb Q(\zeta_5,\zeta_7)$, by the product description of the Galois group. The trivial, quadratic and cubic subfields of this latter field give the list. The square Galois action now shows that both original twists divide $5\cdot7\cdot23$, that at least one has order dividing $23$, and that neither of them has twist order divisible by $35$. In fact any five- or seven-active twist forces an extra inertia image of order at least two or three, respectively, and activity at both forces their relatively prime orbit sizes to divide the same relative degree. \paragraph{If five and seven are separated.} Each of $P_5$ and $P_7$ contains a twist-one orbit of length at least $22$ outside the vacuum. In $P_5$ its row field contains $\mathbb Q(\zeta_{23})^+$; in $P_7$ it contains $\mathbb Q(\zeta_{23})$. In fact take a seven-free $U\in P_5$ in the pure five-twist fiber hitting a joint five- and $23$-active column, and a five-free $V\in P_7$ in the pure seven-twist fiber hitting a joint seven- and $23$-active column, by Proposition~\ref{prop:principal-support}. Their degrees are $5^i23,7^j23$, respectively, $i,j\in\{1,2\}$, by Lemma~\ref{three:complementary} and defect-zero vanishing at $23$. Since $U$ lacks full seven-height it cannot be five-active; likewise $V$ cannot be seven-active. Thus both twists equal one. Take $E\in P_{23}$ of degree $49\cdot23$ and $C\in P_{23}$ of degree $25\cdot23$ meeting the corresponding joint columns as above. The normalized crossing of $U,E$ has weight $23$, positive residue $23$ at five by $U$-principality ($\chi_U(E)=49 z$), and residue zero at $23$ by $E$-principality ($\chi_E(U)=5^i z$). The odd prime-weight result in Lemma~\ref{fam:Q:2.4} gives a common five-part $\eta$ on the actual monodromy, and an $r$-word $h+5\sum_{j=1}^m\xi_j$ after its removal, where $r=23$, $h+5m=23$, and $1\le m\le4$ counts nonidentity positions with multiplicity in the sum. Both inputs are five-inactive by separation (their seven-heights are zero and two). If $\eta\ne1$ then every constituent is five-active and has degree divisible by $7^3$, whereas $7^3\nmid d_Ud_E$. Thus $\eta=1$. Uniqueness in the prime-weight normal form shows that the order of the tame stabilizer divides $m$, so the tame $23$-inertia orbit of the entry has size $11$ or $22$. Since the value is in a pure $23$-power field, it generates a field containing the unique degree-eleven subfield $\mathbb Q(\zeta_{23})^+$. The orbit of $U$ has even size by duality. At the crossing of $V,C$, the same calculation uses $\chi_V(C)=25z$, $\chi_C(V)=7^jz$ with primes seven and $23$. Both inputs are seven-inactive by separation, and $125\nmid d_Vd_C$ excludes any nontrivial common seven-part in Lemma~\ref{fam:Q:2.4}. The resulting pure $23$-word is $2+7(\xi_1+\xi_2+\xi_3)$: at most three nonidentity terms occur and $W(23,3)$ requires four distinct positions including the identity. Its tame stabilizer order divides three and $22$, so is one. This pure cyclotomic value therefore generates a field containing $\mathbb Q(\zeta_{23})$. In fact all three principal blocks have large twist-one orbits here. At $V,C$, Proposition~\ref{found:C:3.3} puts the three distinct nonidentity positions in $\mu_{23}$. The row $C$ has pure $23$-twist by separation and its deficient five- and seven-heights. For a faithful character at $23$, the Fourier coefficient of this crossing is a nonzero Gauss sum times the sum of three $22$nd roots of unity. The latter cannot vanish: three unit vectors with sum zero differ by cube roots, and $3\nmid22$. Proposition~\ref{found:J:key:pr:carry} gives twist-one members of $P_7,P_{23}$ in a common full five-block, both with character field containing $\QQ(\zeta_{23})$. Any additional dyadic factor acts by scalars since the actual twists have odd order. Similarly the pure $23$-word at $U,E$ is moved by the order-eleven part of tame inertia. At least one of its nonzero Fourier components has order divisible by eleven, and its character has $23$-power conductor. Here $E$ has pure $23$-twist by its deficient five- and seven-heights. The same proposition gives members of $P_5,P_{23}$ in a common full seven-block with twist one and fields containing $\QQ(\zeta_{23})^+$. The two choices of member in $P_{23}$ may differ. \subsection{Residues of mixed words at five, seven and twenty-three}\label{sec:odd-mixed-residues} Apply Proposition~\ref{found:J:5.1} at the odd assignment $(5,7,23)$ without assuming that a particular edge is missing. Suppose no twist is active at all three primes. At a weight-$25$ crossing of a degree $25\cdot 23$ label $C\in P_{23}$ and a degree $25\cdot7^j$ label $X\in P_7$, both five-heights are two. At a weight-$23$ crossing of the same $C$ with a five-free $B\in P_7$ of degree $7^j23$, the argument is identical. In both cases take $p=5,T=\{7,23\}$ at $C$ and note that the opposite degree divided by the common weight is a five-unit. Write the $23$-positions as the identity and three nonidentity positions as in Proposition~\ref{found:C:3.3}. If the seven-phase $\eta$ is nontrivial, apply seven-inversion and any $23$-dilation carrying one occupied nonzero position to a vacant one. At weight $23$ the coefficient of $\eta$ at the occupied position is seven. At weight $25$, at most two of the letters have seven-phase one, so at least one of the three nonidentity positions has all seven letters of phase $\eta$. Use that position. The mixed difference has zero constant coefficient, and at the image position its coefficient is, up to sign, $7(\eta-\eta^{-1})$, a unit modulo five. It cannot be zero modulo $5\mathcal O$: the only relation on the $23$ positions over the seven-primary integers, also after reduction modulo five, is the constant coefficient vector. Thus $\eta=1$ in both cases. At the weight-$25$ crossing $X$ is seven-inactive as well: if it were seven-active the singleton test with $p=5,T=\{23\}$ at $X$ would make all $23$-dilations of that pure word congruent modulo five, which the same occupied and vacant positions disprove. At $(5,7,23)$ the row $C$ of degree $25\cdot23$ in either crossing just considered forces $22$ distinct full five-blocks, each entirely seven-inactive in the absence of a joint twist. Its reduced character at the opposite label is a nonzero scalar modulo five times a pure four-position $23$-word. The three marked nonidentity exponents have common coefficient $7\not\equiv0\pmod5$. The polynomial of a primitive $23$rd root is irreducible modulo five since $5$ has multiplicative order $22$ modulo $23$. A dilation stabilizing the reduction must preserve the three nonzero marked positions, by the constant-vector relation and the unchanged identity coefficient. Its order divides both three and $22$, hence it is the identity. All $22$ reduced values are distinct. \subsection{An orientation for two separated odd primes}\label{found:D:6} \label{found:D:key:sec:odd-orientation} Write $P_s$ for the principal $s$-block. A positive expression of weight $g$ is a sum of roots of unity with nonnegative rational coefficients adding to $g$. The balancing expression for $z=S_{UV}/\operatorname{lcm}(d_U,d_V)$ has weight $g=\gcd(d_U,d_V)$, and all conjugates satisfy $|\sigma z|g\ell/(\ell-1)$, then $\QQ(z)$ ramifies at $s$ or $t$. \end{lemma} \begin{proof} If both primes are unramified then $(z-g)/(st)$ is integral. Since $st>g$, the number $1+(z-g)/(st)$ is a convex combination of $1$ and $z/g$ with both coefficients positive. Its house is strictly below one, so its norm forces it to vanish. This gives $z=g-st<-g/(\ell-1)$. The normalized trace of a root of order $n$ is $\mu(n)/\varphi(n)\ge-1/(\ell-1)$ under the stated prime restriction. Averaging the positive expression gives a contradiction. \end{proof} \Needspace{4\baselineskip} \begin{lemma}\label{found:D:6.2}\label{found:D:key:lem:unramified-orientation} Let $s,t,r$ be distinct odd primes and suppose $z$ is a cyclotomic integer with a positive expression of weight $r$ in roots supported on these primes. Assume $z\equiv r$ above $s,t$ and $\QQ(z)$ is unramified at $s$. Write $r=sm+c$, $023\cdot5/4$. Each row field contains the entry field, and separation prevents a row field from ramifying at both endpoints (Proposition~\ref{deep:cluster}). Pair the two height-one witnesses, then pair arbitrary members of either set with the opposite height-one witness. This forces the common $t$, which ramifies also in every entry field with $\min(a,b)=1$. For $(s,t)=(7,5)$ the parameters $(m,c,L)$ of Lemma~\ref{found:D:key:lem:unramified-orientation} are $(3,2,3)$, and for $(s,t)=(5,7)$ they are $(4,3,5)$. Both satisfy its hypotheses, including exposure since $4<6$ and $5/2<4$, respectively. Strictness forces some $\eta_j\ne1$. At a nonidentity phase the total dimension of the corresponding summands of $U\otimes V$ is $s m_x\operatorname{lcm}(d_U,d_V)$ for $1\le m_x\le m0$, choose a $t$-free principal $s$-row $Y$ hitting $H$ by Proposition~\ref{prop:principal-support}. Its field is unramified at $s$, so $S_{YH}$, lying in both fields, is rational. Principality gives $s\mid\chi_Y(H)$, and $Y$ has $s$-height two. By lcm divisibility, $S_{YH}$ is then a nonzero integer multiple of $5^3 7^3 23^2=\sqrt D$, impossible in its row norm. If $k=0$, choose the canonical $23$-height-one $Y$ in the same block. Principality makes it hit every conjugate of $H$, with rational entries divisible by $\sqrt D/s$. The orbit of $Y$ has size at least $22$ (Section~\ref{found:J:8}). The orbit of $H$ has size at least $s-1$ by the square action on its $s$-twist and free duality on each twist fiber. The averaged $Y$-row has squared norm $D/|\operatorname{Gal}\cdot Y|$ and agrees with $Y$ on this entire orbit of $H$. This would give $22(s-1)1$. The sum of $c_s(H)c_r(H)$ on that orbit, with $c_p$ from the simultaneous Gauss congruence in Proposition~\ref{rev:simul-gauss}, is $Asr k_s(u)k_r(v)/2$. Write $B_0$ for the sum of $A k_s(u)k_r(v)$ on the $s$-free joint orbits. The congruence $s^3\mid L_{s,r}=\sum_H d_H^2c_s(H)c_r(H)$ applies since $s^3\mid\sqrt D$. After multiplication by $2$, each positive $s$-height orbit contributes a multiple of $s^3$. The remaining sum is $(t^3r)^2srB_0$, hence $s^2\mid B_0$. Comparing the orbit size with $A k_s(u)k_r(v)$ gives $N_0\ge (s-1)(r-1)B_0/2$. A nonzero $N_0$ would therefore be at least $1100$ at $s=5$ or $3234$ at $s=7$; \eqref{found:T:eq:squared} gives respectively $N_0<886$ and $N_0<814$. \end{proof} \begin{proposition}\label{found:T:10.2}\label{found:T:key:pr:joint-exact} Every label jointly active at $s,r$ has twist of order exactly $sr$. \end{proposition} \begin{proof} By Proposition~\ref{found:T:key:pr:joint-positive} write $d_J=t^3s^h r^k$ for such a joint label, with $h=1,2$ and $k=0,1$. The degrees follow from separation and activity. By Proposition~\ref{prop:principal-support} choose a complementary principal $s$-row $Y$ hitting $J$ with pure $s$-primary twist. Its twist is one, because $s$-activity would require full $t$-height. It has $d_Y=s^2r$: the $s$-height is two by Section~\ref{found:D:6}, and full $r$-height would miss $J$. Put $n=\sqrt D$ and $S_{YJ}=(n/r)b$. Then $b$ is a nonzero algebraic integer. Indeed, $\chi_Y(J)$ is divisible by $s$ because $Y\in P_s$, $s\mid d_J$ and $K_Y$ is unramified at $s$; lcm divisibility supplies the other factors. The field of $J$ is unramified at $t$ and that of $Y$ at $s$, so $b$ belongs to an $r$-power cyclotomic field. Suppose the primary twists of $J$ have orders $s^u,r^v$. Its orbit has at least $(s-1)(r-1)s^{u-1}r^{v-1}/2$ labels: use independent square actions on the two primary twists and the distinct duals in each fiber. At least two $t$-inertia conjugates of $Y$ agree on this whole orbit. Their average has squared row norm at most $D/2$, with a positive vacuum contribution. Writing $M(b)$ for the mean of $|b|^2$ over embeddings, this gives \[ (s-1)(r-1)s^{u-1}r^{v-1}M(b)11\cdot22^2$, all the indicated sums vanish by the algebraic norm. Then $c_a$ is constant on squares and on nonsquares, so $b\in\mathbb Q(\sqrt{-23})$. A nonrational integer of that field has mean square at least six, whereas $M(b)<529/132$. A rational nonzero $b$ would give $m_Ym_J<529$ by averaging the entire $Y$-orbit, also impossible. Thus $m_Ym_J\le5324$, forcing $m_J=66$ and $m_Y=44$: the only further multiple of $22$ in range is $66$, and $3$ cannot divide $m_Y$ since $K_Y$ is supported on $5,23$. Its two required subfields now give the asserted compositum. At these orbit sizes the right hand side of \eqref{found:T:eq:planes} divided by $m_Ym_J$ is $11/6<2$. Any nonzero inner sum $w$ is therefore a root of unity: $1-w\overline w$ has all conjugates of modulus less than one. To read this condition, write $\mathbb F_{23}^{\times}= \{\pm a^j:0\le j<11\}$ with $a$ of order eleven, and put $A_j=c_{a^j}+c_{-a^j}$, $B_j=c_{a^j}-c_{-a^j}$. The order-eleven Fourier coefficient of each vector is zero or a root of unity in $\mathbb Q(\zeta_{11})$. The only coefficient relation on the eleven positions is the constant relation. Thus each vector is constant apart from at most one change by $\pm1$. Since $A_j\equiv B_j\pmod2$, any changes occur at the same index. Both constant would again place $b$ in the quadratic field. Removing one signed term $\zeta_{23}^{\pm a^j}$ therefore leaves $\beta\in\mathcal O_{\mathbb Q(\sqrt{-23})}$. Write $\beta=(x+y\sqrt{-23})/2$, $x\equiv y\pmod2$. For a primitive $23$rd root $\zeta$, the quadratic Gauss sum gives \[ M(\beta\pm\zeta)\ge (x^2+23y^2)/4+1-(|x|+23|y|)/22. \] For $y\ne0$ this exceeds $529/132$ (use $|x|\ge1$ when $|y|=1$). Hence $\beta$ is rational. The constant Fourier coefficient, by the two orbit norms, has absolute value at most $23/\sqrt{44\cdot66}<1/2$. Since the mean of $\zeta$ is $-1/22$, this forces $\beta=0$. Choose the same faithful $\psi$ for every such hit orbit. The resulting unit vectors are orthogonal, while their Weyl projections onto the $J$-orbit are collinear: that transitive orbit has at most one copy of their common full Galois character. Each projection has squared norm $44\cdot66/(22^2\cdot23)=6/23$. For $k'$ hit orbits a normalized linear combination therefore gives squared norm $6k'/23$ there, at most $11/23$ by the same phase estimate. Thus $k'=1$. Lemma~\ref{found:T:key:lem:separated-double-principal} puts a member of the unique hit orbit in $P_5$ and hence puts all of it there. In its congruence the signed primitive roots over this orbit sum to $\pm2$. Since $d_Y=7^2 23$ and $\chi_J(Y)=\pm7^{3-h}23^{1-k}\zeta$, it gives $2\cdot7^{5-h}23^{2-k}\equiv\pm1\pmod5$. Among $h=1,2$ and $k=0,1$ this leaves exactly the two asserted profiles. Finally the pure five-word at $Y,U$ in Section~\ref{found:D:6} is $2+7\sum_{j=1}^3\eta_j$. Its nontrivial field is contained in $\mathbb Q(\sqrt5)$. The multiset of three roots is unique, since it has fewer than five positions in each defining coset. Every nonidentity orbit under the subgroup fixing $\sqrt5$ has size $2\cdot5^{e-1}$ at order $5^e$. The three roots therefore consist of one identity and one inverse pair of order five. \end{proof} \begin{corollary}\label{found:T:10.4}\label{found:T:key:cor:odd-one-degree} If $s=7$ all jointly seven- and $23$-active labels have the single degree $125\cdot7\cdot23$. Every seven-free nonunit of $P_5$ has degree $25\cdot23$, and every label of full $23$-height has twist one. \end{corollary} \begin{proof} A seven-free nonunit $U\in P_5$ has degree $5^i23^a$, $i,a\in\{1,2\}$. Pair it with a $23$-free nonunit $X\in P_7$ from Proposition~\ref{prop:principal-support}. If $i=1$, their lcm-normalized crossing $z$ has weight five. Since $d_U/5$ is a seven-unit, $X$-principality gives $z\equiv5$ above seven; $K_U$ is unramified there by Section~\ref{found:D:6}. The integral $1+(z-5)/7$ has strict house below one, so $z=-2$. The normalized trace of every root on $5,7,23$ is at least $-1/4$, giving a contradiction. Thus $i=2$. At least one joint exists by Section~\ref{found:J:8}, hence so does its pure hit $Y$ from Proposition~\ref{found:T:key:pr:odd-pinned}. For every $U$ above, the normalized entry there is $9+7(\eta+\eta^{-1})$ of Galois mean $11/2$. If $a=2$, then $n/\operatorname{lcm}(d_U,d_Y)=35$. Averaging the $44$ conjugate $Y$-rows and retaining just $U$ would give $44(11/2)^2\le35^2$, impossible. Thus every such $U$ has degree $25\cdot23$. Fix the particular twist-one $U$ of Section~\ref{found:J:8}. Its orbit has at least $44$ members by Section~\ref{found:D:6}. On all the pure $Y$ hit orbits described above, the averaged $U$-row has the same normalized entry $11/2$. Now $n/\operatorname{lcm}(d_U,d_Y)=805$, so the number $b$ of these $Y$-orbits satisfies $44^2 b(11/2)^2<805^2$, hence $b\le11$. Each such $Y$ meets at most one joint orbit of each $23$-twist square class: its faithful orbit vector puts $6/23$ squared mass on each joint orbit it meets, whereas the eleven possible phases of one square class carry at most $11/23$. Thus there are at most $22$ joint orbits. All have size $66$ and exact twist order $7\cdot23$. Let $A_{hk}$ count those of $7,23$-degree heights $(h,k)$. The only possibilities are $(1,1)$ and $(2,0)$. In $2L_{7,23}$ the two contributions together equal \[ 7\cdot23\cdot5^6 \bigl(7^2 23^2 A_{11}+7^4 A_{20}\bigr). \] Since $23^2\mid L_{7,23}$ by Proposition~\ref{rev:simul-gauss}, we have $23\mid A_{20}$. The orbit bound forces $A_{20}=0$. Finally suppose a five-active $H$ has full $23$-height. Its degree is $7^3 5^h23^2$, $h\le2$, and its field is unramified at $23$. Choose a seven-free hit $U\in P_5$ of the now determined degree. Its $23$-inertia orbit contains at least eleven labels by the pure word against degree $49\cdot23$ (Section~\ref{found:J:8}). They agree on at least four conjugates of $H$, by its five-twist square action and duality. Since $\operatorname{lcm}(d_H,d_U)=n/5$, their averaged row would give $44<25$. Thus no such $H$ exists; Section~\ref{found:D:6} excludes seven-activity at full $23$-height, and full height itself excludes $23$-activity. \end{proof} \subsection{The unramified endpoint at seven in dimension \texorpdfstring{$5^6 7^6 23^4$}{5\string^6 7\string^6 23\string^4}} \label{por:sect:odd-seven} Put $r=23$, $n=\sqrt D$, and $m_Z=[K_Z:\QQ]$. We first record the two estimates on pure seven-twist columns used at both possible $r$-heights. Throughout, a pure hit means a five-free $Y\in P_7$ with pure seven-primary twist and $S_{YH}\ne0$. Such a $Y$ has twist one by separation. \paragraph{Fourier estimate on inactive orbits.}\label{rev:seven-inactive-fourier} Let $Y,H$ be $r$-inactive, $22\mid m_Y$, $m_Y\ge44$, $6\mid m_H$. If $S_{YH}/n=a+b/r$, with $a\in\QQ$ and $b$ an algebraic integer in an $r$-power cyclotomic field, then $b\in\ZZ$. Indeed, if $b$ lies outside the quadratic field and has conductor $r^c$, Fourier inversion detects a character $\psi$ of this conductor with $\psi^2$ of the same conductor (at $c=1$ omit the trivial and quadratic characters). It factors through $\QQ(b)$ and defines orbit vectors on both inputs. Writing $b=\sum_j d_j\zeta_{r^c}^j$ with integral $d_j$, the primitive Gauss formula of Proposition~\ref{found:J:6.1} gives their coefficient in absolute value as \[ \frac{\sqrt{m_Ym_H}\,r^{c/2}}{r\varphi(r^c)} \left|\sum_{(j,r)=1}d_j\overline{\psi(j)}\right|. \] Proposition~\ref{por:prop:wild} bounds it by $r^{-c/2}$. Thus $m_Ym_H|\sum d_j\overline{\psi(j)}|^2\le484$ for every conjugate. The sum is a nonzero algebraic integer, whereas $m_H$ has the additional divisor eleven if $c=1$ or $23$ if $c>1$. This is impossible. In the quadratic case $b=(x+y\sqrt{-23})/2$, the orbit coefficient has modulus $\sqrt{m_Ym_H}|y|/(2\sqrt{23})$. Lemma~\ref{rev:tame-compression} bounds it by $1/\sqrt{23}$, forcing $y=0$. \begin{lemma}[Pure principal hits at seven]\label{rev:pure-seven-hit} Suppose $(s,t,r)=(7,5,23)$. If $H$ has pure seven-active twist and $7\mid d_H$, each pure hit $Y$ satisfies \[ d_Y=49r,\qquad S_{YH}/n=\pm1/r,\qquad (m_Y,m_H)=(44,6). \] Thus $K_Y=\QQ(\zeta_r,\sqrt5)$ and the twist of $H$ has order seven. The pure hits and the witnesses for joint columns in Proposition~\ref{found:T:10.3} together occupy at most eleven $Y$-orbits. \end{lemma} \begin{proof} Separation, activity and Proposition~\ref{found:D:6.3} give $d_H=125\cdot7^hr^k$, $h=1,2$, $k=0,1$, and $d_Y=49r^a$, $a=1,2$. The lcm-normalized crossing $z$ is divisible by seven: $\chi_Y(H)=125z\equiv0$ above seven and $K_Y$ is unramified there. At $a=2$ this would make $S_{YH}/n$ a nonzero algebraic integer, exhausting the Galois-averaged norm. Thus $a=1$, and $b=rS_{YH}/n$ is a nonzero cyclotomic integer, supported only at $r$ since $K_H$ is unramified at five. The crossing with degree $25r$ in $P_r$ gives $\QQ(\zeta_r)\subset K_Y$ (Section~\ref{found:J:8}); ramification at five gives $m_Y\ge44$. Seven-activity and duality give $6\mid m_H$. The preceding Fourier estimate makes $b$ rational. The constant unit vector on the $Y$-orbit has squared mass $m_Ym_H b^2/529<1$ on the $H$-orbit after applying $S/n$, since the vacuum also contributes. As $3\nmid m_Y$ (its conductor is supported at five and $23$), this gives $b=\pm1$ and $(m_Y,m_H)=(44,6),(44,12)$ or $(88,6)$. The last two pairs cost $528/529$. Retain also the twist-one $U\in P_5$ of degree $25r$ and orbit size at least $44$ from Corollary~\ref{found:T:10.4}. At $U,Y$ the word is $2+7\sum_{j=1}^3\eta_j$ in five-power roots (Lemma~\ref{found:D:key:lem:unramified-orientation}). By uniqueness wild five-inertia permutes the three letters, since $5\nmid m_Y$, and hence fixes each. With $t$ trivial letters the mean $(35t-13)/4$ has absolute value at least $13/4$. The additional squared mass is at least $(44\cdot13/(4\cdot805))^2>1/529$, since $n/\lcm(d_Y,d_U)=805$. Only $(44,6)$ remains. The square action on the seven-primary twist now gives exact order seven. The degree-$44$ field containing $\QQ(\zeta_r)$ and ramified at five is $\QQ(\zeta_r,\sqrt5)$. The last argument of Proposition~\ref{found:T:10.3} thus identifies the three letters as one identity and an inverse pair. This gives mean $11/2$ against the fixed $U$ both for every pure hit and for every joint-column witness of that proposition. By the averaged $U$-row their total number $b'$ of orbits satisfies $44^2b'(11/2)^2<805^2$, so $b'\le11$. \end{proof} \begin{lemma}[Height of a seven-active label]\label{rev:seven-active-height} In the orientation $(s,t,r)=(7,5,23)$ every seven-active $H$ has $v_{23}(d_H)=1$. At each such $H$ there is a twist-one $23$-free hit in $P_7\cap P_{23}$, and $W_7$ has integral entries at every prime over $23$. \end{lemma} \begin{proof} Full $r$-height is excluded by Proposition~\ref{found:D:6.3}. If $r\nmid d_H$, Corollary~\ref{found:T:10.4} excludes a joint twist, so $H$ is pure and $d_H=125\cdot7^h$, $h=1,2$, by prime-power exclusion. A pure hit exists by Proposition~\ref{prop:principal-support}. Its normalized crossing $z$ is divisible by seven, whereas at $h=1$ its house is strictly below seven. Thus $h=2$. By Lemma~\ref{rev:pure-seven-hit}, each pure hit orbit costs $44\cdot6/529$ in the averaged $H$-row. At most two occur. Each one in $P_5$ contributes $\pm44(49r)(7r)\equiv\pm3\pmod5$ to the double-principal residue (Lemma~\ref{found:T:key:lem:separated-double-principal}). One cannot give $-1$ modulo five, so exactly two occur, both in $P_5$. Conversely each $Y$-orbit meets at most two such $H$-orbits by the same norm. The eleven-orbit bound gives at most eleven $H$-orbits. In $L_7$ each contributes $7(125\cdot49)^2$, and every $r$-bearing label contributes zero modulo $r$. Proposition~\ref{rev:simul-gauss} gives $r\mid L_7$, so the orbit count is divisible by $23$ and must be zero. Now use the double-principal lemma with $(p,q)=(7,23)$. Since $v_{23}(f_H)=4-2=2$, Theorem~\ref{fam:W:n:thm:foreign-column} also gives the asserted integrality at every prime above $23$. \end{proof} \begin{theorem}\label{por:th:odd-seven-elim} The separated five--seven path at $5^67^623^4$ cannot have the unramified endpoint $s=7$. In any remaining separated table the complementary principal rows ramify at seven and are unramified at five. \end{theorem} \begin{proof} A seven-free seven-active $H$ would have degree $125r$ by Lemma~\ref{rev:seven-active-height} and pure twist by Corollary~\ref{found:T:10.4}. The orbit of $44$ pure witnesses $Y$ from a joint column meets every such $H$ by seven-principality. Write $S_{YH}/n=z/(7r)$. As $K_Y=\QQ(\zeta_r,\sqrt5)$ and $K_H$ is unramified at five, $z\in\mathcal O_{\QQ(\zeta_r)}$. Principality gives $z\equiv r\equiv2\bmod7$. Apply the \hyperref[rev:seven-inactive-fourier]{inactive Fourier estimate} with $a=2/(7r)$, $b=(z-2)/7$ to get $z\in\ZZ$, hence $|z|\ge2$. The averaged $Y$-row bounds the total such labels by $N_0<(7r)^2/(44\cdot4)<148$. For an orbit of exact seven-primary order $7^u$ write its size as $6\cdot7^{u-1}A$, $A\ge1$, using the independent square action and duality. The orbit sum of $c_7$ is $7A k(u)$ where $k(1)=1$ and $k(u)=6\cdot7^{u-2}$ for $u>1$. In the singleton congruence $7^3\mid L_7$ all seven-bearing orbits disappear modulo $7^3$. The remaining degrees equal $125r$, so $49$ divides $\sum A k(u)$ over the seven-free orbits. If nonempty this gives $N_0\ge6\cdot49=294$, a contradiction. Every remaining seven-active degree is $125\cdot7^hr$, $h=1,2$. For pure twists Lemma~\ref{rev:pure-seven-hit} gives the double-principal contribution of each $Y$-orbit as $\pm44\cdot49r\cdot7^{3-h}$, congruent to $\pm2$ at $h=1$ and $\pm1$ at $h=2$ modulo five. At $h=1$ at least two contributing orbits are needed and at $h=2$ at least one. Let $A_h$ count the pure seven-active orbits of height $h$, and $B$ the joint orbits. Each joint has height one and a pure witness by Proposition~\ref{found:T:10.3} and Corollary~\ref{found:T:10.4}. For each of the at most eleven witness orbits, at most two pure orbits meet it by the row norm, and at most two joints by the primitive $r$-phase bound. Two pure orbits cost $528/529$ on its constant unit vector; a joint would cost another $44\cdot66/(22^2\cdot529)=6/529$, since its signed primitive $r$-root has mean $\mp1/22$. Thus the number of incident pure orbits plus half the incident joint count is at most two, giving \[ 2A_1+A_2+B/2\le22. \] All seven-active twists have seven-component of exact order seven. Each of the two quadratic classes of primitive seventh phases occurs, by the quadratic Gauss balance of Lemma~\ref{rev:simul-sign}. At a fixed primitive phase $\alpha$ let $P_\alpha$ be its projector. Its squared-degree mass $M_\alpha=\sum_{\theta_{Y,7}=\alpha}d_Y^2$ is a rational integer. Since $sv_0=(d_Y/n)_Y$ and the complementary Weyl is unitary and commutes with $P_\alpha$, we have \[ \frac{M_\alpha}{D}=\|P_\alpha W_7 v_0\|^2 =\sum_{\theta_{Y,7}=\alpha}(W_7)_{Y0}\,\overline{(W_7)_{Y0}} . \] Every summand is integral at all primes above $r=23$ by the preceding all-primes bound and conjugation. The rational number $M_\alpha/D$ therefore belongs to $\ZZ_{(23)}$. Since $v_{23}(D)=4$, we obtain $23^4\mid M_\alpha$. Write $A_{h,\pm},B_\pm$ for the orbit counts in the corresponding square class at seven. A pure orbit gives two labels over each of its three seventh phases, and a joint gives twenty-two over each of its three phases. Dividing the squared degree sum by $2(125\cdot7r)^2$ gives \[ 529\mid A_{1,\pm}+49 A_{2,\pm}+11 B_\pm . \] Both quantities are positive, whence $A_1+49 A_2+11 B\ge1058$. The joint quadratic balance at $7,23$ from Lemma~\ref{rev:simul-sign} gives equal mass at both values of the sign product. Every joint orbit has the same size and degree and one sign product, so the nonempty joint count $B$ is even. But the incidence inequality gives \[ A_1+49 A_2+11B\le1078-97A_1-\tfrac{27}{2}B\le1051 . \] This contradiction proves the assertion. \end{proof} The opposite ramification orientation at this odd dimension, and the ordinary modular case at $5400^2$, remain open. \begin{corollary}\label{fam:rank:five-seven-twenty-three} Every non-WGT braided fusion category of dimension $D=5^6 7^6 23^4$ has rank at least $553$. The bound is $1121$ if a twist order is divisible by $5\cdot7\cdot23$. In particular these bounds hold for every non-pointed simple integral modular category of dimension $D$. \end{corollary} \begin{proof} Every proper divisor of $D$ divides one of $5^5 7^6 23^4$, $5^6 7^5 23^4$ or $5^6 7^6 23^3$. These three products are covered, respectively, by Theorems~\ref{fam:Q:1.3}, \ref{fam:F:3.2} and~\ref{thm:three}. Lemma~\ref{lem:minimal} therefore gives an ordinary simple perfect table, since $D$ is odd. Conversely, a non-pointed simple integral modular category is non-WGT by the proof of Corollary~\ref{cor:global-dimensions}. Write $S$ for the set of odd prime divisors of a twist order. Every singleton $S=\{p\}$ occurs. Indeed, the space on which all complementary primary twists are trivial contains the vacuum and is invariant under the local $p$-factor. If $p$ were inactive there, its upper unipotent would act trivially (the normalization is dyadic), as would the conjugate lower unipotent. They generate the local factor, so $W_p$ would fix the vacuum, contrary to Lemma~\ref{lem:complement} since $p\mid f_{\one}=D$. All three dimension valuations are even. In any nonempty stratum of fixed primary orders, each active $p$-coordinate admits at least $p-1$ phases with the other coordinates fixed (Proposition~\ref{prop:two-packets}). Successively projecting away the coordinates gives at least $\prod_{p\in S}(p-1)$ distinct tuples. For $S\ne\varnothing$ each tuple occurs on a pair of distinct duals (Lemma~\ref{lem:odd-codegrees-duality}). The three singleton supports therefore account for at least $2(4+6+22)=64$ nonunits. Theorem~\ref{found:J:key:thm:twenty-three} requires orders divisible by $5\cdot23$ and $7\cdot23$. Without a triple these give disjoint supports and at least $2(4\cdot22+6\cdot22)=440$ further labels. A triple itself gives at least $2\cdot4\cdot6\cdot22=1056$. Adding the vacuum gives the asserted triple bound. In the remaining case the singleton and mixed supports account for $504$ nonunits. If five and seven occur in a common twist, the support $\{5,7\}$ gives $2\cdot4\cdot6=48$ more labels. If they are separated, Proposition~\ref{found:D:key:prop:odd-oriented-path} gives an orbit of at least $44$ nonunit labels of twist one. The rank is thus at least $553$ or $549$, respectively. Since every degree is odd, $R\equiv D\equiv1\pmod8$, so both cases give $R\ge553$. \end{proof} \section{Degree and support restrictions at the first open dimension}\label{found:group:degrees} \subsection{Four disjoint orbits of witnesses}\label{found:R:2} \label{found:R:key:sec:quartets} Say that a pair $p,q$ occurs if $pq$ divides some twist order. The pair $3,5$ occurs and at least one of $2,3$ and $2,5$ occurs. We recall the following four disjoint sets, each of four labels of one degree. \begin{itemize} \item Complementary witnesses in $P_3,P_5$, of pure three- and five-primary twist, respectively, have degrees $a\in\{12,24,36,72\}$ and $b\in\{20,40\}$. Their weight-four crossing gives four distinct five-inertia conjugates of each (Theorem~\ref{found:F:2.1}). \item At a jointly $3,5$-active column, an odd witness in $P_3$ of pure ternary twist has degree $c=15$ or $45$. Its weight-five crossing gives four distinct five-inertia conjugates. The joint column itself has four such conjugates by Sections~\ref{sec:weight-five-crossing} and~\ref{found:C:6}. Its degree $d$ divides $360=2^3 3^2 5$, since both active primes are deficient in its degree (Lemma~\ref{lem:complement}). The first three sets have pure twists and distinct degree types, so all four sets are disjoint. \end{itemize} \subsection{Formal codegrees and the largest simple degree}\label{found:R:1} Write $n=5400$, $d_U=\operatorname{FPdim}(U)$, $b_U=S_{ZU}/d_Z$ for the character row at $Z$, and $f=f_Z=D/d_Z^2$. Recall that $d_U\mid n$, that every nonunit degree has at least two distinct prime factors, and that $S_{ZU}/\lcm(d_Z,d_U)$ is an algebraic integer (Section~\ref{sec:wgt-preliminaries}). Each nonunit row satisfies \[ b_0=1,\qquad \sum_U d_U b_U=0,\qquad \sum_U |b_U|^2=f,\qquad n=\left|\sum_U d_U b_U\theta_U\right|. \] The last equality is the vacuum--$Z$ entry of the modular relation $STS=p^+T^{-1}ST^{-1}$, using $|p^+|=n$. Reducing the first orthogonality sum at each of $2,3,5$ gives a nonzero nonunit entry of degree prime to that prime. These three kinds of degrees are disjoint. We give two details of the estimates on a row. Cyclotomic Galois conjugation of its entries permutes the argument $U$ and preserves $d_U$. In an orbit $O$ with $b_U\ne0$, the mean of $|b_U|^2$ is at least one by the arithmetic--geometric mean inequality applied to its absolute norm. Thus $\sum_O|b_U|\le\sum_O|b_U|^2$ by Cauchy--Schwarz, and the support has at most $f$ labels. If a nonzero algebraic-integer value $c$ in an orbit reduces to zero at a prime over $p$, its rational norm is divisible by $p$. Writing $t=[\QQ(c):\QQ]$, its distinct conjugates have squared mass at least $t p^{2/t}$. For $p=2$ this is at least $4$, and for $p=3$ at least $6$, as follows by checking $t=1,2,3$ and monotonicity thereafter. For $f$ a power of $p$, the row is supported in the principal $p$-block $P_p$ (Theorem~\ref{thm:principal-support}). On this block $v_p(d_U)\le v_p(n)-1$ (Corollary~\ref{cor:principal-valuation-bound}) and $S_{UZ}/d_U\equiv d_Z$ at the chosen prime above $p$. Codegree four is excluded by Lemma~\ref{lem:small-ordinary-codegrees}. Theorem~\ref{thm:no-nine} excludes codegree nine in this ordinary fusion-simple perfect case. \paragraph{Codegree sixteen.} Here $d_Z=1350$, and the support lies in $P_2$. For an even supported degree of two-height $k$, the number $b_U/2^{k-1}$ is integral by lcm divisibility. It has zero residue at two by two-principality of $U$: up to a dyadic unit it equals $\chi_U(Z)$. Its orbit costs at least $4\cdot4^{k-1}$. Consequently $k=1$ at every even hit. In particular the three-free and five-free nonunit hits have degrees at most $50$ and $54$, respectively, and cost at least four each. No two rows of degree $1350$ have a nonzero mutual entry, including a row paired with itself. Otherwise symmetry and prime-power support put both in $P_2$. A row in $P_2$ meets every odd degree by principality. There are at least eight odd-degree labels: the unit and four odd conjugates described in Section~\ref{found:R:key:sec:quartets} already give five, and their total number is divisible by four by reduction of $\sum d_U^2=D$ modulo four. All odd positions therefore cost at least eight in total. The compulsory three-free and five-free nonunit orbits are even. Their additional cost exhausts sixteen before the proposed degree-$1350$ hit. All remaining support degrees are at most $675$. Indeed $k\le1$, and the divisors of $2\cdot3^3\cdot5^2$ bigger than $675$ consist only of $1350$. In $\sum d_U|b_U|$, replace the sum of absolute values on each orbit by its squared mass, then reserve the unit and the two compulsory even orbits. This gives \[ n\le 675\cdot16-(675-1)-4(675-50)-4(675-54)=5142. \] This contradiction excludes codegree sixteen. Codegree $25$ is excluded by Proposition~\ref{prop:quartic-two-codegree4} at the quartic prime five. The degree $5400$ would alone exhaust $D$. The only divisors of $5400$ above $900$ are $1080,1350,1800,2700,5400$, completing the degree bound. \subsection{Excluding a simple object of degree six}\label{found:D:1} \begin{theorem}\label{found:D:1.1}\label{found:D:key:thm:spectral-six} Neither an ordinary divisor-minimal table nor its positive reduced counterpart has a nonunit label of degree six. In particular every nonunit degree in either table is at least ten. \end{theorem} \begin{proof} Suppose $d_X=6$ and write $\Delta$ for the table dimension, equal to the category dimension in the ordinary case and half of it in the reduced case. By Corollary~\ref{lem:sixseven} its prime support is $\{2,3,5\}$. By Lemma~\ref{three:complementary} a three-free nonunit in $P_5$ has five-height $j\ge1$ with $2j+2\le v_5(\Delta)$. Thus in both tables $v_5(\Delta)\ge4$. Work in a common cyclotomic splitting field with Galois group $G$. Use prime avoidance at any five-active column in the full principal five-block, avoiding three. The wild-inertia argument of Section~\ref{found:C:6} gives a three-free nonunit $U$ with $K_U$ tame at three. Indeed at a five-active $J$ the orthogonality relation $\sum_{Y\in P_5}d_Y\chi_J(Y)=0$ holds in either table (Proposition~\ref{prop:principal-support}). Wild ternary inertia preserves $P_5$ and permutes its arguments without changing their residues at three. Nonsingleton orbits contribute zero there. The vacuum term forces a singleton nonunit orbit with nonzero contribution. Its degree is $2^i5^j$ with $i,j\ge1$. The entry $z=S_{XU}/\lcm(6,d_U)$ has weight two and residue two above five, since $\chi_U(X)=3z$. By Theorem~\ref{first:balance}, it is a sum of two five-power roots. They are distinct by strictness, and their ratio has order five by Theorem~\ref{found:S:4.1}, applied to the full-weight word $3z$. Consequently $|\chi_U(X)|^2=9(2+t)$, where $t=\zeta+\zeta^{-1}$ for a primitive fifth root $\zeta$. The unit occurs exactly once in $XX^*$, also in the positive quotient: for a lift $X$ the fermion $f$ does not occur in $XX^*$ since $fX\ne X$. The other constituents have total degree $35$. Each of these degrees is supported on $2,3,5$ and has at least two prime divisors. The only possible odd degree at most $35$ is $15$, occurring once. The remaining mass $20$ can only be one degree $20$ or two degrees $10$. Write $XX^*=1+V+B$, with $d_V=15$ and $d_B=20$. Neither $v=\chi_U(V)$ nor $b=\chi_U(B)$ is rational. Indeed, if $v$ is rational, $v/3$ is a rational integer of absolute value below five by lcm integrality. Five-principality gives $v\equiv0$ above five, whence $v=0$. Now $b=17+9t$ has a conjugate greater than $20$. If $b$ is rational, then $b\in\ZZ$, $|b|<20$ and $b\equiv0\pmod5$. Since $v=17+9t-b\in3\mathcal O$, where $\mathcal O$ is the ring of algebraic integers, $b\equiv2\pmod3$. The possibilities $b=-10,5$ both give a conjugate of $v$ greater than $15$. For every $\sigma\in G$, the adjoints of $X$ and $X^\sigma$ share a nonunit (Theorem~\ref{found:S:4.1}). If $B$ is one simple or twice one simple, every $\sigma$ fixes $V$ or $B$. Their stabilizers are proper because $v,b$ are irrational, and a group is never the union of two proper subgroups. Hence $B=W+W'$ with $W,W'$ distinct of degree ten. No degree-ten simple can be fixed by $G$. Otherwise its character eigenvalues would be integers, the vacuum value $10$ and all others strictly between $-10$ and $10$. Applying $\prod_{k=-9}^9(T-k)$ to this simple in the fusion ring gives $19!$ times the vacuum idempotent, whose unit coefficient is $19!/\Delta$. This is impossible since $v_5(19!)=312$, then $d_X\ge18$ and $|A|\le3599$. Since all degrees in $A$ are at least $15$, the degree-$45$ witness gives \[ \sum_{Y\in A}(900-d_Y^2)\le675\cdot8091-1800. \] In $P_3$ a five-free witness has degree at least $12$, and gives scale at least five on $B$, so $|B|\le8099$. All its degrees are at least $10$, and the complementary $P_5$ quartet has degree $20$ or $40$. Thus the analogous sum over $B$ is at most $800\cdot8099-4\cdot300$. Use both the degree-$45$ row and a row of the complementary quartet of common degree $b=20$ or $40$ in $P_5$ (Section~\ref{found:R:2}). Put $s_m(d)=d/\gcd(d,m)$. For every five-free nonunit degree $d$, \[ 35(900-d^2)\le1557s_{45}(d)^2+172s_b(d)^2. \] Indeed, the only such degrees below $30$ are $12,18,24$. The pairs of squared scales are $(16,9),(4,81),(64,s_b(24)^2)$, respectively. The coefficients give equality in the first two cases, and the first term alone suffices in the third. Summing the two row-norm bounds gives \[ \sum_{Y\in C}(900-d_Y^2)\le \frac{1557(14400-1)+172(72900-1)}{35} =\frac{34957871}{35}. \] All remaining nonunits have degree at least $30$. Hence \[ \operatorname{rank}\le1+\left\lfloor \frac{D-1+5459625+6478000+34957871/35}{900} \right\rfloor=46774. \] The degree-$45$ witness is supplied by Section~\ref{found:C:6}. \subsection{The four primitive fifth phases on the path separating two and three} \begin{lemma}[Fourier test for the pure fifth phases]\label{rev:boundary-fourier} In the ordinary minimal table of dimension $5400^2$, suppose no twist order is divisible by six. If every degree-$20$ label of $P_5$ has primitive fifth twist, every jointly two--five-active column $K$ meets a degree-$20$ label at each of the four primitive fifth phases. \end{lemma} \begin{proof} Write $\alpha=\theta_{K,5}$. Let $E$ be the intersection of $P_5$ with the fiber with trivial complementary twists. In the local modular representation write $s=W_5W'$, $v=W'v_0$, and $w=W'v_K$. At five, $v$ is fixed by the upper unipotents and the torus, and $w$ has upper-unipotent phase $\alpha$. The coordinate projector $P_E$ commutes with this local group. For every unit $u$ modulo the five-part of the level we have \[ H(u):=\sum_{Y\in E}d_Y\chi_K(Y)\theta_{Y,5}^{\,u} =C\alpha^{-u^{-1}}. \] In fact $H(u)$ is $D/d_K$ times $\langle v,P_E W_5^{-1}U(u)W_5w\rangle$. In the lower-unipotent Bruhat decomposition the left upper-triangular factor fixes $v$, and the right upper unipotent has parameter $-u^{-1}$. This gives the formula with a constant $C$ independent of $u$; in particular $C=\alpha H(1)$ is integral. Also $H(0)=0$ by restricted block orthogonality. Reduce at a prime above three. Every three-free nonunit in $P_5$ has degree $20$ or $40$ by Section~\ref{found:S:6}; degree $40$ misses $K$ by full dyadic height. By hypothesis the degree-$20$ labels of $P_5$ have primitive fifth twist. Hence $\overline H$ is the discrete Fourier transform of masses on $\mu_5$, with mass one at the trivial phase (only the vacuum contributes there). It is periodic modulo five and nonzero by Fourier inversion. Since $H(0)=0$, the constant $C$ has nonzero residue. Comparing $u=1$ and $u=6$ gives $\alpha^5=1$, since reduction is injective on five-power roots. Thus $\alpha$ is primitive of order five. Fourier inversion at phase one gives $\overline C=-5$. At every other fifth phase $\delta$, the mass is consequently \[ -\sum_{u\in\mathbb F_5^\times} \overline{\alpha^{-u^{-1}}\delta^{-u}} . \] This sum is nonzero: $\Phi_5$ is irreducible over $\mathbb F_3$, so a vanishing sum of four fifth roots would have all five multiplicities congruent modulo three. Their sum is four, so each would be at least two. Thus $K$ meets degree-$20$ labels at each primitive phase. \end{proof} \subsection{Counting active twists by a local torus character}\label{found:T:1} \label{found:T:key:sec:geometry} Work in ordinary integral modular data of square dimension with unsigned cyclotomic tori. Let $E$ be a coordinate subspace invariant under the local modular group $G_p$, and let $H_p$ be its diagonal torus. Examples include the whole table and a full $p$-block restricted to specified complementary twist phases: the upper unipotent preserves the coordinates and the local Weyl preserves both conditions (Lemma~\ref{lem:local-block-invariance} and Section~\ref{found:F:2}). \begin{proposition}\label{found:T:1.1}\label{found:T:key:prop:torus-active-count} Let $p\ge5$ and let $L\subset E$ be an $m$-dimensional space supported on $p$-inactive labels, of torus type obtained by inflating a character $\psi$ of $\mathbb F_p^\times$ (hence trivial on the wild subgroup $1+p\mathbb Z/p^a\mathbb Z$ at level $p^a$), with $\psi^2\ne1$, $\psi(-1)=-1$. If the projector to local level at most $p$ has rank $r$ on $L$, then $E$ contains at least $2(p-1)r$ labels of twist order exactly $p$ at $p$, and at least $p(p-1)(m-r)$ of higher $p$-power order. They may be selected from contributing torus orbits, each of size divisible by the order of $\psi$. In particular, at five, $m$ disjoint inactive orbits of size divisible by four force $8m$ active labels in equal-degree quartets within $E$. \end{proposition} \begin{proof} Write $Q_1$ for averaging over the kernel of reduction modulo $p$. For $v\in L$ the squared norm of the $\alpha$-phase projection of $W_pv$ is $\|Q_1v\|^2/p$ for every $\alpha\in\mu_p$ (Lemma~\ref{rev:tame-compression}). Each such projection has rank $r$. The unsigned involution $H_p(-1)$ fixes each phase and acts by minus one on the image, requiring two labels per independent vector. Sum over the $p-1$ primitive phases. The Weyl image of the $(m-r)$-dimensional kernel is supported at higher orders. Torus type $\psi^{-1}$ restricts to a space of dimension at most one on each orbit, with stabilizer in $\ker\psi$ whenever the restriction is nonzero. Its support thus meets at least $m-r$ orbits. At order $p^a$, the square action gives $p^{a-1}(p-1)/2$ phases, each containing at least two positions by oddness. This proves the second bound. At five each specified inactive orbit affords the same faithful tame quartic character. The contributing active orbits have constant degree and size divisible by four, giving the stated selection. \end{proof} \paragraph{Compulsory active orbits.}\label{found:T:key:cor:core-lower-rank} Consider the four disjoint quartets in Section~\ref{found:R:2}. The complementary rows $X\in P_3,U\in P_5$ have degrees $a\in\{12,24,36,72\}$ and $b\in\{20,40\}$, respectively. An odd $V\in P_3$ has degree $c\in\{15,45\}$, and a joint three--five label $J$ has degree dividing $360$. Each has a local five-inertia orbit of size divisible by four, by Section~\ref{found:F:2} and Sections~\ref{sec:weight-five-crossing} and~\ref{found:C:6}. The selected $X,V$ have purely ternary twist and $U$ purely quinary twist. For each five-inactive orbit of this kind the permutation representation of the cyclic group $H_5$ contains the same tame quartic character, since its orbit size is divisible by four. Distinct orbits thus count independently in Proposition~\ref{found:T:key:prop:torus-active-count}. Every five-active degree divides $1080$ and is at most $540$ by Section~\ref{found:R:1}. We also use the five-inactive label $Z$ of Section~\ref{found:F:2} in the full five-block of $J$, with the same dyadic and ternary twists. Its field contains $\mathbb Q(\zeta_5)$ or $\mathbb Q(\zeta_{15})^+$, giving a fourth-order five-inertia action and five-height at most one. If no triple occurs, one can take $Z\in P_3$; then its ternary height is at most two. If a triple occurs, take $J$ triple-active and its ternary activity again bounds that height by two. In all cases $d_Z\mid360$. In the triangle $X,U$ have twist one (Section~\ref{found:F:2}). In the absence of a triple, that argument also supplies the odd $V$ of twist one; $Z$ is three-active and so distinct from all three quartets. With a triple, $Z$ is both two- and three-active, while $V$ has pure ternary twist. Thus $X,U,V,Z$ give four distinct inactive quartets and at least $32$ five-active labels. Their $J$ quartet has degree at most $360$. If ten is missing, $(a,b,c)=(12,40,15)$ and the chosen $U$ has twist one (Sections~\ref{found:P:1} and~\ref{found:P:2}). Again there are at least $24$ five-active labels. All have full two-height and three-bearing degree (Section~\ref{found:C:6}); their degrees divide $1080$, and the largest such degree below $1080$ with full two-height is $360$. There are also five-inactive odd labels of degrees $45,75$ by Section~\ref{found:C:6} and Theorem~\ref{found:P:1.1}. Counting the unit, these two and the four degree-$15$ labels gives seven odd labels; their total number is divisible by four, so there is another, of degree at most $675$. At a two-active column obtain a three-free nonunit $W\in P_2$ by prime avoidance; its degree divides $100$ and is even, and it is five-inactive since it has deficient two-height. It is distinct from the degree-$40$ quartet by that same deficiency. It remains that six is missing. Here $b=20$. Every three-active label has full two-height by separation and is five-bearing because two and five occur together (Section~\ref{found:S:6}). Hence the pure ternary $X,V$ have twist one. The three-active $Z$ is distinct from them and from $U$, and has full two-height and five-height one. In fact every degree-$20$ row of $P_5$ has pure quinary twist (its two- and three-heights are deficient) and full quartic five-inertia by its crossing with $V$. If even one is five-inactive, there are four inactive quartets and hence at least $32$ active labels. Otherwise there are at least eight active degree-$20$ labels. Indeed a five-phase of order at least $25$ gives an orbit of size divisible by $20$, by the square action and the quartic inertia. If all their phases have order exactly five, apply Lemma~\ref{rev:boundary-fourier}: at a jointly two--five-active column the degree-$40$ rows miss by full two-height, so the only three-free terms are the vacuum and degree-$20$ rows at primitive fifth phases. That argument gives a hit at each primitive phase; its inertia involution supplies a second label of the same degree. Thus the three inactive quartets of $X,V,Z$ force at least twenty-four five-active labels. A three-free witness $W\in P_2$ in the pure dyadic fiber has twist one: every two-active degree has full three-height by separation. Its degree distinguishes it from $X,V$, and its twist distinguishes it from $Z$. Take a joint two--five label $K$. Its full five-block contains a label $T$ with the same dyadic twist and trivial five-twist, by removal of the primary twist (Proposition~\ref{found:J:4.2}). Its ternary twist is trivial by separation. It has full three-height, deficient dyadic height, and five-height at most one: a full five-height label would be a singleton five-block and could not share that block with $K$. Thus $d_T\le540$. Its nontrivial pure dyadic twist distinguishes it from the four witness orbits. Now consider the full five-block of $J$ at its exact ternary phase, denoted by $E$. Each label there has degree $40\cdot3^j$, $j=0,1,2$: three-activity gives full two-height and a factor of five, and full five-height would make its block a singleton. If $n_j$ count these three degrees then \[ n_0+9n_1+6n_2\equiv0\pmod{25}. \] Indeed the coordinate projector $P_E$ commutes with $W_5$, so $\sum_{Y\in E}d_Y^2=D\|P_E W_2W_3 v_0\|^2$. On this nontrivial ternary phase $W_2$ is scalar: the dyadic upper and its conjugates are scalar there by separation. The operator $W_3$ is integral at every prime above five, by the enhanced Weyl estimate (Section~\ref{found:C:6}), since every three-active degree has five-height at least one. Hence the sum is divisible by $5^4$ and division by $40^2$ gives the congruence. Section~\ref{por:sec:jointfibers} uses this congruence to give the size of each such fiber. \paragraph{Intersections of principal blocks on the path missing ten.}\label{por:double-missing-ten} In the ordinary minimal table at $D=5400^2$, suppose two and five are separated. Each five-active column has a degree-$40$ twist-one hit in $P_5\cap P_3$, and each two-active column has a three-free twist-one hit in $P_2\cap P_3$ of degree $10,20,50$ or $100$. Indeed, in this graph three divides every two-active and every five-active degree (Section~\ref{found:C:6}). Apply Lemma~\ref{found:T:key:lem:separated-double-principal} with $q=3$ and $p=5,2$, respectively. The three-free nonunits in $P_5$ have degree $40$ by Section~\ref{found:S:6}; the positive two-height in $P_2$ is at most two (Corollary~\ref{cor:principal-valuation-bound}), and a nonunit degree has at least two prime divisors. Both hits can be chosen tame at three: in the lemma's principal-block congruence group the terms by wild three-inertia. Every nonsingleton orbit contributes zero to the residue (Section~\ref{found:C:6}). In particular the chosen degree-$40$ label has field $\mathbb Q(\zeta_5)$ or $\mathbb Q(\zeta_{15})$ (Section~\ref{found:P:2}). For each jointly three- and five-active column $J$, the same lemma at $(p,q)=(5,2)$ gives a degree-$45$ twist-one hit in $P_5\cap P_2$ whose field is unramified at two. In fact any odd nonunit $X\in P_5$ meeting this $J$ must have degree $3^j5$: it has positive ternary and five-height by the exclusion of prime-power degrees, and the five-height in $P_5$ is at most one (Corollary~\ref{cor:principal-valuation-bound}). The odd-prime-weight bound in Section~\ref{found:C:6} gives $j\ge2$, while three-activity of $J$ excludes $j=3$ by defect-zero vanishing. Hence $j=2$. Wild two-inertia fixes some contributing row in the resulting intersection by the same orbit argument (and the tame quotient at two is trivial). For a jointly two- and three-active column the five-free hit from $(p,q)=(2,5)$ lies in $P_2\cap P_5$, has twist one, and has degree $12,18$ or $36$: the two-height and three-height are each at most two, and degree six is excluded by Section~\ref{found:D:1}. \paragraph{Four further principal labels when six is missing.}\label{por:four-neutral} At $D=5400^2$ suppose two and three are separated. At any jointly three--five-active $J$, Lemma~\ref{found:T:key:lem:separated-double-principal} gives an odd twist-one hit $V\in P_2\cap P_3$. Its degree is $5\cdot3^j$, $j=1,2$: it is deficient at three by principality and at five since it meets $J$. At any jointly two--five-active column the reversed lemma gives a three-free twist-one $W\in P_2\cap P_3$ of degree $2^i5$, $i=1,2$, by the corresponding dyadic and quinary height bounds. In fact $V\notin P_5$. Otherwise $z=S_{WV}/\operatorname{lcm}(d_W,d_V)$ has weight five and residue five above each of $2,3,5$. At two use $\chi_W(V)=3^jz$ and $W\in P_2$; at three and five use $\chi_V(W)=2^iz$ and $V\in P_3\cap P_5$. The congruences hold at every conjugate prime. By Proposition~\ref{found:C:5.1}, the residues at three and five and strictness leave just \[ z=\eta(2+3\xi) \quad\hbox{or}\quad z=\eta\left(1-\sum_{t=1}^{4}\eta_t\zeta^t\right), \] where $\eta$ is ternary, $\xi$ is a nonidentity five-power root, $\zeta$ is a primitive fifth root and $\{\eta_t\}=\{1,1,\omega,\omega^2\}$ for a primitive cubic root $\omega$. The first expression gives $\eta\xi=1$ on reduction at two, impossible by injectivity on odd-order roots. For the second expression work at a common odd conductor. Two is unramified, so the congruences at every prime over two are equality modulo $2$. The four powers $\zeta^t$, $1\le t\le4$, are an integral basis over the ternary cyclotomic integer ring (coprime conductors), also after reduction. Substituting $1=-\sum_{t=1}^4\zeta^t$ in the congruence shows that all $\eta_t$ are equal modulo two, again impossible. Consequently $P_2\cap P_5$ contains at least four nonunit twist-one labels distinct from the $V$ quartet. Indeed, Section~\ref{found:F:4} uses the four tame five-conjugates of $V$ together with the vacuum as columns, and the analogous four degree-$20$ labels $U\in P_5$ together with the vacuum as rows. Their residue matrix at two is the invertible five-by-five Fourier matrix. All five columns now belong to $P_2$. The forward factorization in that section therefore requires five distinct twist-one labels in $P_2\cap P_5$, including the vacuum. The $X$ quartet is outside $P_2$ since $\chi_X(U)=5(1+3\xi')\not\equiv d_U=20$ above two, with $\xi'$ primitive of order five. No three-free nonunit in $P_5$ belongs to $P_2$: its degree is $20$ or $40$, the latter has full dyadic height, and in degree $20$ its value at $V$ reduces to a primitive fifth root instead of the odd dimension residue. This separates the four new labels from $W$ and $U$ as well. \section{Twist fibers and rank at the first open core}\label{sec:portrait-proofs} \label{por:frontier} At $D=5400^2=29160000$, every nonunit degree divides $5400$, lies between $10$ and $900$ and has at least two distinct prime factors (Section~\ref{found:D:1}). A joint label is active at the indicated primes. Write $m_z$ for the number of simples of degree $z$, $R$ for rank, $\tau$ for normalized field trace and $\mathsf M(b)=\tau(b\overline b)$. \Needspace{9\baselineskip}\begin{theorem}\label{por:ranktheorem} A perfect fusion-simple integral modular category of dimension $5400^2$ has rank $101\le R\le46774$. The last column gives the pages where the bounds for each twist configuration are proved. \begin{center} \begin{tabular}{@{}lrrrr@{}}\toprule Twist orders & Min. rank & Min. five-active & Rank without $900$ & Proof, p.\\\midrule None divisible by $6$ & $141$ & $60$ & $\ge141$ & \pageref{por:count-no-six}\\ None divisible by $10$ & $101$ & $40$ & $\ge127$ & \pageref{por:count-no-ten}, \pageref{por:certificate}\\ All pairs, none divisible by $30$ & $106$ & $56$ & $\ge130$ & \pageref{por:count-triangle}\\ Some divisible by $30$ & $112$ & $56$ & $\ge142$ & \pageref{por:count-triangle}\\\bottomrule \end{tabular} \end{center} At $R=101$ at least $21$ labels have degree $900$, and at least $18$ of their rows meet a label jointly active at two and three. At every rank $m_{675}\le7$. \end{theorem} \subsection{Degree caps used in the rank proof}\label{found:T:2}\label{found:T:key:sec:large} For the remainder of this section let $\mathcal C$ be a perfect fusion-simple integral modular category with $D=5400^2$. By Section~\ref{found:R:1} every simple degree divides $5400$ and every nonunit degree is at most $900$. \begin{corollary}\label{por:prop:small-ninehundred} If no twist order is divisible by six at $5400^2$, every simple degree is at most $675$. \end{corollary} \begin{proof} A degree-$900$ row in Proposition~\ref{por:prop:separated-high-row} has $p=2$, $q=3$, $a=2$, $b=3$ and $f=36$. A two-active degree is divisible by $27$ and at most $900$. Among divisors of $5400$ with these constraints, the largest is $675$. But $2\cdot675\cdot35=47250<48600=5400\cdot3^2$. The next possible degree after $900$ is $675$. \end{proof} \begin{corollary}\label{por:th:joint-at-ninehundred} At $5400^2$, with at most three exceptions among all degree-$900$ simples, every such row has a nonzero entry on a label whose twist is jointly two- and three-active and five-inactive. Any exception has twist one and belongs to $P_2\cap P_3$. It requires $\xi=\pm1$, and the possible pairs $(k_{2,Z},k_{3,Z})$ are $(3,-2\xi)$ and $(-3,4\xi)$. In particular every orbit of at least four degree-$900$ simples meets the indicated joint sector at every member. \end{corollary} \begin{proof} A degree-$900$ row misses all five-active labels. If it also misses the indicated joint sector, the separated-sum equation $H_2+H_3=5400\xi\theta_Z^{-1}$ holds for this row even without global separation. By Theorem~\ref{por:th:weyl-div}, $3\mid k_{2,Z}$ and $2\mid k_{3,Z}$ since $f_Z=36$. Neither integer can vanish, as the other would then have modulus one. Local Weyl vacuum columns belong to their respective principal blocks and have trivial twist at all complementary primes, by local commutation. This puts $Z$ in both principal blocks at twist one. Each such label accounts for at least $9/36$ of the norm squared of $W_2v_0$. Four labels would exhaust this norm on nonunit twist-one labels, impossible by the strict norm estimate following Proposition~\ref{por:prop:separated-high-row}. Likewise $|k_{2,Z}|=6$ is impossible as it would exhaust that norm on $Z$ alone. Thus $k_{2,Z}=\pm3$. Since $\gamma_3=1$, $\gamma_2^3=\xi$ and $\theta_Z=1$, the separated-sum equality becomes $k_{3,Z}=\xi(1-k_{2,Z})$. This proves the assertions; the exceptional property is preserved by Galois. \end{proof} \begin{proposition}\label{por:prop:cap675} At $5400^2$ write $k$ for the total number of degree-$675$ labels. Then $k\le7$. If any member of the full principal two-block has degree divisible by four, then $k\le3$. Consequently, when $k\ge4$ every even member of that block has dyadic height one. \end{proposition} \begin{proof} For $W\in P_2$ put $i=v_2(d_W)$. Since every row of degree $675$ hits $W$ by two-principality and $d_W/\gcd(d_W,675)=2^i$, Lemma~\ref{por:orbitbound} gives $ek4^i<64$ for its five-inertia orbit size $e$. There exists a five-free nonunit $W\in P_2$ by Proposition~\ref{prop:principal-support}. Its degree involves two and three. If $i\ge2$ the bound gives $k\le3$. Otherwise $d_W=18$ or $54$, since degree six is excluded. A three-free $U\in P_5$ has degree $20$ or $40$ (Section~\ref{found:R:2}). Their crossing, of weight two, is a sum of two distinct five-power roots up to a scalar (Theorem~\ref{three:word}). Its squared absolute value generates a field containing $\QQ(\sqrt5)$, so $e\ge2$ and $k\le7$. \end{proof} \subsection{Local Weyl transitions in joint twist fibers} \label{por:sec:jointfibers} Write $E$ for a full five-block restricted to fixed dyadic and ternary twists. Proposition~\ref{found:T:1.1} applies inside this fiber. \paragraph{Two distinct ternary phases.}\label{por:two-ternary-fibers} Let $J$ be jointly three- and five-active at $5400^2$. There are two different nontrivial ternary phases with the \emph{same} dyadic and quinary components as $J$ at which a joint label occurs. Indeed at local order three the two primitive cubic phases have equal positive squared mass when the complementary phases are fixed (Section~\ref{found:F:3}). At order $3^a\ge9$, the ternary square action alone supplies $3^{a-1}$ phases. In each such phase choose a joint label, and take its full five-block and exact complementary phases. The resulting fibers $E$ are disjoint. Each contains a five-inactive quartic orbit by Section~\ref{found:F:2}. Consequently each contains at least eight five-active labels. Every active label there is jointly three- and five-active, so the local five-inertia involution moves it (Sections~\ref{found:C:6} and~\ref{found:F:2}); in particular all active labels occur in quartets. \paragraph{Equal mass at tame fifth phases.}\label{por:tame-fiber-masses} In a tame fifth fiber $E$ as above suppose the active degrees are $d_0b$ with $b$ odd. Section~\ref{found:F:3} gives equal squared mass at each of the four primitive phases. After division by $d_0^2$, reduction modulo eight shows that the four label counts agree modulo eight. If all active orbits are quartets, each contributes twice its squared degree at each phase of its square class. Cancelling this factor two before reduction shows that the numbers of quartets in the two classes agree modulo eight. In particular, five quartets are impossible and four split two to each class. If in addition the two-term sums of the available $b^2$ are distinct up to order, the two classes have the same multiset of quartet degrees. \paragraph{At least twenty-four labels in each joint fiber.}\label{por:24-joint} Every such $E$ contains at least $24$ labels, including at least $16$ active at five. All degrees in it are at most $360$, since the two odd heights are deficient: the ternary phase is nontrivial, and full five-height makes a five-block a singleton. When the dyadic phase is nontrivial too, the upper degree bound drops to $180$. Write $A$ for the number of five-active labels in $E$, all of them jointly three- and five-active. These labels occur in local five-inertia orbits of size divisible by four. If a phase has five-primary order $5^a\ge25$, its orbit contains at least $2\cdot 5^{a-1}$ distinct phases by the square action, with at least two labels over every phase by the inertia involution. Then $A\ge20$. When $A<20$ all active twists are tame and each active orbit is one quartet: the local group action factors modulo five because $U(5)$ acts trivially, and its conjugates generate the kernel of reduction modulo five. At any successive depth the conjugates of an upper elementary matrix span the trace-zero quotient over $\mathbb F_5$, which proves the normal generation by induction. The parity theorem (Section~\ref{found:J:7}) says that the number of local five-orbits of tame order four in $E$ is even. If $A=8$, its two active quartets therefore require an even number of inactive quartets. There is at least one, so at least two, forcing $A\ge16$ by the local Weyl count. If $A=12$, the three active quartets cover both square classes since the masses at all four individual fifth phases are equal by Section~\ref{found:F:3}. Each quartet contributes twice its squared degree to each of the two individual phases in its class. The two quartets on one side and the single quartet on the other would therefore have degrees satisfying $d_1^2+d_2^2=d_3^2$. When two and three are separated or when two and five are separated, these degrees all have full dyadic height by separation. This also holds in the triangle without a triple (Proposition~\ref{found:J:4.2}). After removing their common dyadic part the equation is impossible modulo four. For a triangle containing a triple, we choose the starting $J$ triple-active, so all active labels in these fibers have five-height exactly one: positivity of the five-height is the ternary degree transfer and the upper bound is activity (Section~\ref{found:S:6}). After dividing by five, every degree is supported on two and three. In a primitive Pythagorean triple the odd hypotenuse must have a prime divisor equal to one modulo four, since each of its prime divisors admits a square root of $-1$ modulo that prime. Thus the required equality is again impossible. The same height-one argument works for a dyadically trivial fiber whenever the ternary degree transfer is available. It follows that $A\ge16$. If $A=16$, its four tame active orbits force an even number of inactive quartic orbits by the same parity theorem. Hence there are at least eight inactive labels in this case; if $A\ge20$ we use the four already proved. This gives $|E|\ge24$ as asserted. On the path missing six every degree in $E$ is $40,120$ or $360$ (Section~\ref{found:T:1}). In fact $|E|\ge25$. If equality held in the bound $24$ with twenty active labels, the four inactive labels would form a single orbit. The orbit filtration (Section~\ref{found:F:3}) then gives only orders dividing five on $E$. With sixteen active labels the same tameness is automatic. The normalized squares $1,9,81$ have distinct two-term sums by their base-nine digits. The tame count above excludes twenty active labels; at sixteen the degree multisets in the two square classes agree. The eight inactive labels are two quartets. The fiber mass divided by $40^2$ is divisible by $25$ (Section~\ref{found:T:1}). Removing its common factor four, its residue would be $s+2t$ with $s,t\in\{2,7,10,12,15,18\}$, the sums of two residues from $1,9,6$. None is the negative of twice another. This excludes size twenty-four. Two nontrivial ternary phases therefore supply at least $50$ labels of these three degrees, including at least $32$ jointly active at three and five. The twist-one quartets of $X,V$ supply another $16$ purely five-active labels by the local estimate. In addition, a jointly two- and five-active phase gives at least four jointly two- and five-active labels: at quintic order five use equal mass at the four primitive phases with the dyadic twist fixed, and a higher order already has at least ten square phases. With a ternary phase of order at least nine at the joint $J$, the corresponding three--five counts rise to $75$ and $48$. \paragraph{When all three pairs occur.}\label{por:triangle-fibers} In the absence of a triple the other three inactive quartets $X,U,V$ have twist one, giving at least $24$ pure five-active labels outside the two fibers. In the presence of a triple, choose $J$ triple-active, so the fibers share a nontrivial dyadic phase. The three inactive quartets again give $24$ active labels outside those fibers: all three have trivial dyadic phase, even when the pure ternary $V$ is active at three. There are consequently at least $56$ five-active labels in the triangle. If a joint label at the chosen dyadic phase has ternary order at least nine, the corresponding argument gives at least $72$. \paragraph{Five inactive orbits on the path separating two and five.}\label{por:five-inactive-orbits} In an ordinary minimal table of dimension $2^63^{2e}5^4$, $e\ge3$, suppose the ternary prime occurs jointly with each of two and five, which are separated. There are at least five distinct five-inactive local five-inertia orbits of size divisible by four. Consequently there are at least forty five-active labels in equal-degree quartets. At $e=3$ one can choose all five inactive orbits with ternary defect. Let $H_5$ be local five-inertia. Inflate a character of order four of $\mathbb F_5^\times$ to the local torus and denote the result by $\psi$. For each five-inactive orbit $O_Y=H_5Y$ whose size is divisible by four, put $\ell_Y=|O_Y|=4m_Y$. Its stabilizer lies in $\ker\psi$ since the group of units at a power of five is cyclic. The input and output unit vectors are \[ v_Y=\frac1{\sqrt{\ell_Y}}\sum_{g\in H_5/\mathrm{Stab}(Y)} \overline{\psi(g)}\,v_{gY},\qquad v_Y^-=\frac1{\sqrt{\ell_Y}}\sum_{g\in H_5/\mathrm{Stab}(Y)} \psi(g)\,v_{gY}. \] We use the Hermitian inner product linear in its second variable. They form orthonormal bases of the five-inactive spaces $L_\psi$ and $L_{\psi^{-1}}$ of the indicated torus types. Each space has one line per orbit. Every such orbit is dyadically inactive, since two-activity forces full five-height. The dyadic group acts scalarly: its diagonal upper generator is scalar on this invariant space and its conjugates generate the group. The ternary group acts on $L_\psi$. Section~\ref{found:P:2} gives three compulsory orbits $U,X,V$ with $d_U=40$, $d_X=4\cdot3^j$, $d_V=5\cdot3^k$, $1\le j,k\le e-2$. Their fields contain $\QQ(\zeta_5)$; $U$ is neutral and tame at three. For $Y=X,V$ the full-weight normalized entries are $1+3\zeta$ and $\eta(2+3\zeta')$, respectively, with primitive fifth roots $\zeta,\zeta'$ and a ternary scalar $\eta$. The local five-orbit fixes $\eta$. In the double sum for $\langle v_Y^-,sv_U\rangle$, the entry at $(gY,tU)$ depends on the tame product $gt\bmod 5$. Each of its four values occurs $4m_Ym_U$ times with the same character weight. Division by $\sqrt{\ell_Y\ell_U}$ leaves the factor $\sqrt{m_Ym_U}$. The constant term has zero quartic Fourier coefficient and the remaining Gauss sum has squared modulus $3^2\cdot5$. Since $n=200\cdot3^e$ and $\lcm(d_U,d_Y)=40\cdot3^{v_3(d_Y)}$, this gives \[ 5|\langle v_Y^-,s v_U\rangle|^2 =5^a/9^h,\qquad 5^a=m_Um_Y,\quad h=e-v_3(d_Y)-1\ge1. \] The local five-block at each of two distinct nontrivial ternary phases contains an inactive quartic orbit. For $e=3$ this was proved above. For varying $e$, the avoided-two trace selection of Section~\ref{found:C:6} in the full principal three-block of a joint label gives an odd hit of degree $5\cdot3^l$ with five-height one by defect-zero vanishing. Its crossing with $U$ puts $\QQ(\zeta_5)$ in its field. The selected value has odd relative field index and odd absolute trace, so a faithful quartic Fourier coefficient is nonzero (Section~\ref{found:F:2}). Applying the complementary Weyls gives the required inactive orbit at the specified phase. At $e=3$ first set aside the coordinate orbits of full ternary height in each mode space. Each label then has ternary-unit codegree, hence a singleton three-block and trivial ternary twist. Its coordinate line affords the trivial ternary representation since the upper unipotent acts trivially and its conjugates generate the local group. These lines contribute no transitions between different ternary phases. For $e>3$ retain all orbits. Suppose the retained space has dimension at most four. It contains the three compulsory orbits at distinct degrees $d_U,d_X,d_V$. A ternary phase of order at least nine has at least three square conjugates at the same degree, on different five-orbits, forcing at least five orbits in total. Thus the local upper unipotent has cube one. Its representation factors through $\mathrm{SL}_2(\mathbb F_3)$. Indeed, in each congruence quotient of depths $j,j+1$ with $j\ge1$, the image of the representation kernel contains the conjugates of $E_{12}$, from the upper unipotent at depth $j$. These span $\mathfrak{sl}_2(\mathbb F_3)$: interchanging the coordinate axes gives $-E_{21}$, and conjugation by a lower unipotent gives a nonzero diagonal component. Descending induction kills the entire first congruence subgroup. Write $z=-I$ in this quotient. As an unsigned torus permutation it preserves degree and permutes the five-orbit lines. If a compulsory orbit is fixed setwise, the action on its labels commutes with $H_5$ and therefore is translation by an element of $H_5/\mathrm{Stab}(Y)$. Because $K_Y$ contains $\QQ(\zeta_5)$ and $z$ fixes the fifth roots, this translation is wild, hence of odd order. The involution fixes that orbit pointwise. At most one additional orbit is available. If $z$ exchanges it with a compulsory orbit the exchanged pair has a single negative eigenline, and otherwise only the extra line can be negative. On the negative eigenspace $z$ has determinant one. Indeed the Weyl matrix and $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)$ anticommute over $\mathbb F_3$, so their commutator is $z$; the eigenspace is stable under both. Its negative multiplicity is therefore even, hence zero. This works for either quartic mode. Both retained representations descend to $A_4$, with upper generator $u$ a $3$-cycle and Weyl element $w$ a nonidentity element of the Klein four-group. Let $E_0$ project onto the five-inactive coordinates, let $B=E_0W_5|_{L_\psi}:L_\psi\to L_{\psi^{-1}}$, and let $Q$ project onto local level at most five. Both operators here are defined on the full mode spaces, including any trivial lines set aside at $e=3$. The five-phase estimate of Section~\ref{found:P:2}, polarized, gives $5B^*B=Q$ on $L_\psi$. Both $B$ and $Q$ intertwine the ternary action. The compression of $s$ between the two mode spaces equals $BW_3$ up to the dyadic unit scalar. At least one of $X,V$ is nonneutral since both primitive cubic phases occur and at most four orbit lines are available. There is thus a nonzero transition from $U$ to a different $u$-phase. Character summands cannot give this transition: $w$ acts trivially on them and $B$ intertwines $u$. Lemma~\ref{lem:tetrahedral-weyl} therefore gives a unique tetrahedral summand $T$ on each side. The projection $Q$ acts identically on the input $T$, since it commutes with the group and $B$ is nonzero on $T$. Thus $\sqrt5 B$ identifies the two tetrahedral summands unitarily. Any coordinate lines set aside at $e=3$ were trivial constituents and do not affect this identification. A transition from a simple $u$-phase to a different simple phase has squared modulus $4/9$ for $\sqrt5 BW_3$, by the lemma. On either compulsory transition the actual value $5^a/9^h$ is one modulo four, where all denominators are odd. If the neutral phase were simple, $X,V$ would therefore have to be the two orbit lines at one doubled nonneutral phase. The sum of their two squared transitions would be $4/9$, the squared norm of the projection onto that phase from the neutral tetrahedral line. The sum is two modulo four. This forces the neutral phase to be doubled, with basis $v_U,v_{U'}$ in the input; the two nonneutral phases are simple and the fourth constituent is trivial. For a nonneutral $Y\in\{X,V\}$ we now have \[ \tfrac49 x=5|\langle v_Y^-,s v_U\rangle|^2=5^a/9^h, \qquad x=\|P_Tv_U\|^2 . \] Applying the last assertion of Lemma~\ref{lem:tetrahedral-weyl} in the input space gives \[ |\langle v_{U'},W_3v_U\rangle|^2 =\tfrac{16}{9}x(1-x) =\frac{5^a(4\cdot9^{h-1}-5^a)}{9^{2h-1}}. \] The ternary torus fixes every label of $O_U$. Its wild part fixes $U$ by tameness and commutes with $H_5$. Its involution preserves $O_U$ because it acts identically on the retained mode space, and the translation argument above applies. The Galois formula of Lemma~\ref{lem:local-vacuum-column} sends $W_3$ to $W_3$ times a ternary torus element on the right. The columns of $W_3$ indexed by $O_U$ are consequently rational. In particular, writing $G_Z=H_5/\mathrm{Stab}(Z)$, the two normalized input modes give \[ \sqrt{\ell_{U'}\ell_U}\,\langle v_{U'},W_3v_U\rangle =\sum_{g\in G_{U'},\ t\in G_U}\psi(g)\overline{\psi(t)}(W_3)_{gU',tU}\ \in\QQ(i). \] Its squared modulus is a Gaussian norm. So is $\ell_{U'}\ell_U=16m_{U'}m_U$, since $5=|2+i|^2$. Hence the squared transition has even valuation at every prime congruent to three modulo four. But $4\cdot9^{h-1}-5^a$ is positive (it is nonnegative since $x\le1$, and odd) and is three modulo four, so it has such a prime to odd valuation. Neither $5^a$ nor the square denominator changes this parity. This contradiction proves the five-orbit assertion. Proposition~\ref{found:T:key:prop:torus-active-count} gives the forty five-active labels. \subsection{Rank from compulsory simple degrees} \label{por:sec:rank-two-branches} For rank $R$, let $m_d$ count simples of degree $d$. Suppose distinct positions have upper degree bounds $t_1,\dots,t_j$. For any cutoff $b$, \begin{equation}\label{por:eq:budget600} b^2R+\sum_{d>b}(d^2-b^2)m_d \ \ge\ D+\sum_{i=1}^j\bigl(b^2-\min\{t_i,b\}^2\bigr). \end{equation} Indeed the left side minus $D$ counts $b^2-\min\{d,b\}^2$ at every position. We use this with $b=600$ or $675$. The following disjoint positions enter this budget. Here $E_1,E_2$ have distinct nontrivial ternary phases. A quartet comprises four equal-degree labels in one five-inertia orbit. \begin{center}\small \begin{tabular}{@{}p{.18\linewidth}p{.72\linewidth}@{}}\toprule Twist graph & Positions counted before the remaining degree deficits\\\midrule Missing $6$ & $|E_1|,|E_2|\ge25$, degrees $40,120,360$; neutral $X,V,Q$ quartets and the three-free $W$; at least $24$ pure five-active labels; four joint two--five labels and one pure two-active label (\hyperref[por:count-no-six]{first case}).\\ Triangle & Unit, three inactive quartets ($d_X\le72,d_U\le40,d_V\le45$), $24$ five-active labels outside $E_1,E_2$ ($d\le540$), and at least $48$ positions in those fibers ($d\le360$, or $180$ on a triple); see the \hyperref[por:triangle-fibers]{triangle fibers}.\\ Missing $10$ & With at least two neutral quartets, count eight neutral positions, sixteen pure five-active labels and $48$ in $E_1,E_2$. With only one neutral quartet reserve all five inactive quartets, ten active quartets and the unit, then the five outside labels in the \hyperref[por:count-no-ten]{separated case}.\\\bottomrule \end{tabular} \end{center} Different twist fibers are disjoint. Within a fiber the proofs below use distinct inertia orbits, blocks or degrees to separate the labels. An inactive quartet inside $E_i$ contributes to its size only once. \paragraph{The path separating two and three.}\label{por:count-no-six} Here $m_{900}=0$ by Corollary~\ref{por:prop:small-ninehundred}. Put $g=600^2$, $C(d)=g-d^2$ and $k=m_{675}$. By the \hyperref[found:T:key:cor:core-lower-rank]{compulsory active orbits} and the \hyperref[por:four-neutral]{double-principal-block argument}, the following labels are distinct: the vacuum, neutral $X,V$ quartets with degrees at most $72,45$, a neutral three-free $W\in P_2\cap P_3$ of degree $10$ or $20$, and four further neutral members of $P_2\cap P_5$. There is also a pure two-active member of the five-block of a joint two--five label, of degree at most $540$, and at least four jointly two--five-active labels with the same upper bound. The \hyperref[por:two-ternary-fibers]{two joint ternary fibers} have different ternary twists. The \hyperref[por:24-joint]{fiber count} in this separated case gives at least $25$ labels of degrees $40,120,360$ in each. The four neutral members of $P_2\cap P_5$ can be taken in a single tame five-inertia orbit $Q$. Indeed, let $H$ be the order-four tame subgroup and adjoin the unit both to $HU$, for the pure five-primary $U\in P_5$ of degree $20$, and to $HV$ for the neutral $V\in P_2$. Section~\ref{found:F:4} gives a minor of $W_5V_2$ between their five-dimensional coordinate spaces. After multiplying row $Y$ by $n/d_Y$, its reduction at two is the Fourier matrix $(\bar\zeta^{ij})_{i,j\in\mathbb F_5}$, up to a scalar unit, for a primitive fifth root $\zeta$. The complex minor is invertible. Its factorization through $V_2$ and $W_5$ only uses intermediate labels of twist one in $P_2\cap P_5$, by block and complementary phase preservation. Both coordinate projections commute with $H$; $V_2$ commutes with $H$ and $W_5$ intertwines its action with inversion. A faithful quartic eigenline in the domain survives in the intermediate space, giving the asserted orbit. The \hyperref[por:four-neutral]{block distinctions} separate $Q$ from $X,V,U$ and $W$. The three neutral quartets $X,V,Q$ supply at least $24$ pure five-active positions, selectable in equal-degree quartets by the local Weyl count. The same conclusion holds if wild inertia enlarges one of these orbits: take its trivial-wild, faithful-tame character. The central involution acts negatively on the relevant Weyl images at both tame and wild phases. If a degree-$20$ row in $P_5$ is neutral, it gives a fourth neutral quartet, and we get $32$ pure five-active positions. In the other case at least eight pure five-active labels have degree $20$, in full quartets (Section~\ref{found:T:1}); choose them among the $24$. Since the joint ternary fibers supply $32$ five-active labels and the two--five edge supplies four others, there are at least $60$ five-active labels in either case. For $Y\in Q$ let $i=v_2(d_Y)$. The inertia-orbit estimate of Lemma~\ref{por:orbitbound}, with $z=675$ and $p=5$, gives $k4^i<16$: every degree-$675$ row hits $Y$ by two-principality. Hence the common degree of $Q$ is bounded by $v=540,270,135$ for $k=0$, $1\le k\le3$, $4\le k\le7$, respectively. In the last case it is odd, of degree $15,45$ or $135$. To count the two other five-free labels required by the dimension identity, put $x=d_X$. For $x=12,24,36,72$ the residues $1+4x^2$ modulo $25$ lie in $\{2,5,10,12\}$, none zero or the negative of a unit square. A five-free position outside $Q$ and the $X$ quartet contributes at least $540^2-216^2=244944$ beyond its previous allowance, whether filling a $540$ slot or newly reserved. The joint ternary fibers, $V,W$ and all degree-$20$ labels are five-bearing. If $Q$ itself is five-free, its four degrees are at most $108$ at $k=0$ and at most $54$ at $k>0$. Accordingly the additional five-free contribution is at least $2\cdot244944$, $4(270^2-54^2)$, $2\cdot244944$ in the three ranges of $k$. In the middle range this is the cheaper of the two possibilities for $Q$. For the branch with a neutral degree-$20$ quartet, the resulting lower bound on $gR+95625k$ is the sum of $D$, these five-free contributions, and \[ C(1)+4C(72)+4C(45)+C(20)+4C(v)+5C(540) +50C(360)+4C(20)+32C(540). \] In the other branch replace the last two terms by $8C(20)+16C(540)$, increasing the bound by $344000$. For the first branch the baseline and its remaining allowance at $R=140$ are \[ \begin{array}{c|r|r} k & \text{baseline} & 140g+95625k-\text{baseline}\\\hline 0&48983451&1416549\\ 1,2,3&49648299&847326,\ 942951,\ 1038576\\ 4,5,6,7&50076951&705549,\ 801174,\ 896799,\ 992424 . \end{array} \] In any such fiber, if no inactive label has the minimum ternary height, at least $36$ labels attain it. Indeed $\sum_Ed_Y^2=\sum_{Y\ {\rm inactive}}d_Y^2k_{\{2,3\},Y}^2$. Divide by the square of the smallest degree in $E$ and reduce modulo nine: the number attaining it is a multiple of nine, and it is also a positive multiple of four since all such labels occur in equal-degree quartets. The two fibers $E$ may have more than fifty labels or degrees below $360$. Let $t=\sum_E(|E|-25)$, and let $a,b$ count their labels of degree $40,120$ respectively. Since each fiber mass is divisible by $25\cdot40^2$ (Section~\ref{found:T:1}), their extra contribution obeys \[ 230400t+128000a+115200b,\qquad 6t-5a+3b\equiv0\pmod{25}. \] Outside the fibers the previously reserved three-free positions are the unit, $W$ of degree $w=10,20$, and four, respectively eight, positions of degree $20$. Denote the number of further three-free positions outside by $c$. Each contributes at least $290000$: a pure five-active three-free degree is at most $40$ instead of its cap $540$; an unreserved three-free degree is at most $200$. No dyadically active label here is three-free, and none of the neutral nonunits in $P_2\cap P_5$ is three-free (see the \hyperref[por:four-neutral]{four further principal labels on the path missing six}). None of these three-free savings coincide with a five-free one, by prime-power exclusion. In the first branch $a+c$ is a positive multiple of three: divisibility follows modulo three; if zero, modulo nine gives $1+4\cdot20^2+w^2=0\bmod9$, hence $w=10$. The unique degree-ten label would be rational, contrary to Section~\ref{found:T:3}. In the second branch $a+c\equiv2\bmod3$. The extra cost is at least $870000$, including the $344000$ branch correction where applicable. If $a=0$ use $c\ge3,2$, respectively. If $a>0$, the only fiber cost below $870000$ has $(t,a,b)=(0,5,0)$ and costs $640000$. Indeed $t\le3$, $b\equiv-2t\bmod5$ and $5a\equiv6t+3b\bmod25$. At $t=0$ test $b=0,5$; $b=5$ forces $a\ge3$. At $t=1,2$ the least $b$ is $3,1$, respectively, forcing $a\ge3$ at these minima; at $t=3$, already $b\ge4$. In the surviving case the first branch needs $c\ge1$; the second incurs the branch correction. This excludes $k=1,4,5$. The odd-degree count is divisible by four. At $k=0$ three more odd labels are needed beyond the unit and $V$ quartet. Each saves at least $540^2-135^2=273375$ in a capped slot or $C(225)$ outside. An odd degree distinct from $675$ properly divides $675$ and is at most $225$; it is three- and five-bearing. At $k=2$ at least one more odd is required. If $Q$ is odd, its improved cap and the external five-free allowance already add $4(270^2-135^2)+(489888-279936)>273375$; otherwise use the same odd saving outside $Q$. At $k=6$ the unit, $V,Q$ and the six degree-$675$ labels give fifteen odd labels, so one more is needed. These savings exclude $k=0,2,6$. At $k=3$, five-bearing $Q$ increases the five-free allowance by $489888-279936$. Five-free $Q$ has degree $d=18$ or $54$ and $1+4x^2+4d^2\ne0\bmod25$ for the four values of $x$, costing another $244944$ outside the two quartets. Both exceed the allowance. Finally $k=7$ forces $w=10$ by Proposition~\ref{por:prop:cap675}. Theorem~\ref{found:T:key:local:ten} forces five-ramification on this separated path, supplying another neutral degree-ten conjugate outside the reserved positions. Its cost $359900$ leaves less than $640000$, so $a=0$. We have $c\le3$; as the selected pure five-active positions are grouped in full quartets, none of these $c$ positions lies there. Each besides the new degree-ten label therefore costs at least $C(200)=320000$. This gives at least $359900+2\cdot320000$ in the first branch or $344000+359900+320000$ in the second. Both exceed $992424$. This establishes $R\ge141$. \paragraph{All three pairs occur.}\label{por:count-triangle} Select $85$ distinct positions: the vacuum, the three inactive quartets $X,U,V$ of degrees at most $72,40,45$, twenty-four five-active positions outside the joint fibers capped at $540$, and forty-eight labels in the \hyperref[por:24-joint]{two joint fibers}. The \hyperref[por:triangle-fibers]{complementary twist conditions} separates these fibers from the other five-active positions. These last degrees are at most $u=360$ if there is no triple, or $u=180$ if a triple is used to define the fibers. At least two additional five-free labels besides the $X$ quartet are required by the same residue argument. All degrees in the two chosen fibers are five-bearing, by the ternary degree transfer, and so are $U,V$. Each additional five-free label gains $540^2-216^2$ even if already counted among the twenty-four five-active labels. There is also a further three-free label: the vacuum and the $U$ quartet give five, whereas their total number is a multiple of three. It differs from every five-free nonunit. If in a joint fiber, its degree divides $40$ by the five-defect; the generic five-active positions outside have three-bearing degrees by the degree transfer, and an unreserved three-free degree divides $200$. This adds at least $u^2-40^2$. With a baseline $G$ equal to $900^2$ or $600^2$, put \[ B(G,u)=D+(G-1)+4\!\sum_{t\in\{72,40,45\}}(G-t^2) +24(G-540^2)+48(G-u^2)+2(540^2-216^2)+(u^2-40^2). \] Thus $900^2 R\ge B(900^2,u)$, giving $R\ge106,112$, respectively. In the absence of degree $900$, \eqref{por:eq:budget600} gives $gR+95625k\ge B(g,u)$, whose right hand sides are $47123451,51691851$. With $k\le7$ this proves $R\ge130,142$, respectively. \paragraph{The path separating two and five.}\label{por:count-no-ten} At $n=5400$, a five-inertia quartet of degree $12,15,20,36$ or $45$ meets at most eight degree-$900$ labels. A quartet of degree $24,40$ or $72$ meets at most two. Thus the three mandatory quartets of degrees $12,15,40$ on the path missing ten collectively meet at most eighteen such labels. Every other degree-$900$ row vanishes on all twelve columns. There is a stronger bound on that path. Let $4a$ count all degree-$12$ labels in $P_3$. We have $1\le a\le8$: in the weighted sum of Section~\ref{found:P:2} a degree-$12$ orbit costs at least one quarter of its size, and the opposite degree-$15$ type costs at least one of the available nine units. At any jointly two- and three-active $J$, orthogonality on $P_3$ gives \[ 1+12\sum_{d_X=12,\ X\in P_3}\chi_J(X)\equiv0\pmod{\mathfrak p}, \qquad \mathfrak p\mid5. \] Indeed a five-free nonunit of $P_3$ has degree $2^i3^j$ with $i=2,3$ (Lemma~\ref{three:complementary}); full dyadic height misses $J$, and the other type has degree $12$ by Section~\ref{found:P:2}. The degree of $J$ has full five-height by separation, so its row agrees on the four tame five-conjugates in each quartet. Write $b_J$ for one quarter of the sum in this congruence. It is a rational integer: the full set of $X$ is Galois-stable, and the sum is four times an algebraic integer. Thus $b_J\equiv-2\pmod5$. Applying $s$ to the normalized indicator of the $4a$ labels gives \[ \sum_{J\text{ jointly two-, three-active}}d_J^2 b_J^245$ five-active labels. Thus both fibers contain sixteen tame active labels. The normalized squares $1,9,25,225$ available in a joint fiber (activity excludes $216$) have distinct two-term sums. Thus its quartet degrees are $(u,u,v,v)$ by the same tame count. For a fixed fifth phase the two primitive cubic phases also have equal squared mass, so the very same $u,v$ occur in both fibers. The ten selected quartet degrees altogether consist of four $u$, four $v$, and a pair $h,i\in A$ at $U$. We initially allow this last pair to be unequal. Here is the exact second computation underlying the last line of the table. Take every $u\le v$ in $J$ and $h\le i$ in $A$ with \[ 576 S=4u^2+4v^2+h^2+i^2 . \] Form a list of possible five-inactive positions in the two fibers. Include the four quartets $12,15,b,c$ as groups of multiplicity four. From $L\cup\{o\}$ keep the degrees dividing $360$. If $y=18$ first remove its twist-one conjugate of degree $18$ from $L$. Group the remaining $18$'s and $54$'s in pairs, and use multiplicity one for the others. This pairing is allowed because each five-inertia orbit at these two degrees has even size and the complementary phases and full five-block are invariant. The already reserved quartets and conjugate pair are themselves stable under tame five-inertia. Take disjoint sublists for the two fibers. For each sublist $I$ of pairs $(m,d)$ require $\sum_I m\ge8$, and put \begin{equation}\label{por:eq:two-fiber-norms} F=8(u^2+v^2)+\sum_I md^2,\qquad F=25\sum_I md^2 x_{m,d}^{\,2},\quad x_{m,d}\in\ZZ , \end{equation} with a separate variable for each group, including repeated pairs. This identity follows by commuting the fiber projector through $W_5$ in the squared norm of the vacuum column: only five-inactive positions contribute on the $W_{\{2,3\}}v_0$ side. Theorem~\ref{por:th:weyl-div} gives the factor $25$ because every degree in the fiber has positive five-defect. Indeed the unsigned five-torus fixes $v_0$ and commutes with $W_{\{2,3\}}$. Thus $W_{\{2,3\}}v_0$ and the degree are constant on each five-inertia orbit. This justifies one integer coordinate per selected quartet or pair, even when a larger orbit is split into such pieces: giving the pieces independent integers enlarges the feasible set. Equal degrees on different orbits impose no equality of coordinates. To test the equality, start with the single remainder $F/25$, discarding nonintegers. At each $(m,d)$ replace each remainder $t$ by the set of $t-md^2 x^2$ for integers $0\le x\le\sqrt{t/(md^2)}$. Keep the sublist if zero occurs after its last group. Two successful sublists must be disjoint, and unused positions are permitted. These recursions give the following complete output, all at $R=100$, $w=50$, and $u=v=360$. \[ \begin{array}{c|c|c|l|c|c} (b,c)&o&y&L&S&(h,i)\\\hline (15,18)&135&12&360,72,12,12&2034&72,360\\ (15,36)&135&12&360,36,12,12&2034&72,360\\ (18,180)&135&12&72,30,12,12&2034&72,360\\ (24,360)&15&36&600,600,270,24,12&1882&24,216\\ (36,180)&135&12&36,30,12,12&2034&72,360 \end{array} \] \paragraph{Eliminating the five remaining profiles.}\label{por:five-survivors} In each row at most two entries of $L$ could be five-active. Hence the pure five-fiber at $U$ contains at most ten active labels. The two selected quartets cannot come from a wild orbit, which would need twenty labels if it contained the faithful quartic mode. An additional wild orbit would itself require at least ten more. By the \hyperref[por:tame-fiber-masses]{tame-phase count} the four phase counts agree modulo eight. All are positive with sum at most ten, hence equal. There are just eight active labels, two quartets in opposite square classes, of equal degree. This contradicts every displayed pair $(h,i)$ and proves the bound. \begin{proof}[Proof of Theorem~\ref{por:ranktheorem}] The \hyperref[por:rank-upper]{three row-norm estimates} on p.~\pageref{por:rank-upper} give the upper bound. Proposition~\ref{por:prop:cap675} gives the multiplicity bound at $675$. For the lower bounds, the \hyperref[por:two-ternary-fibers]{two disjoint ternary fibers} each contain \hyperref[por:24-joint]{at least $24$ labels}. If $10$ is missing, the \hyperref[por:five-inactive-orbits]{five inactive orbits} give $40$ five-active labels. The degree budget and congruences in the second row's proof give $R\ge95$ and $R\ge127$ if $m_{900}=0$. In Section~\ref{por:certificate}, the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/check_frontier_rank.py}{rank enumeration} tests the degree multisets and the two disjoint fiber norms \eqref{por:eq:two-fiber-norms}. It leaves five profiles, all of rank~$100$. Their \hyperref[por:five-survivors]{incompatible tame phases} exclude all five. At $R=101$ the same budget gives $m_{900}\ge21$. At most three of those rows miss the joint sector by Corollary~\ref{por:th:joint-at-ninehundred}. \end{proof} \section{The unbraided boundary in dimension \texorpdfstring{$5400$}{5400}} \label{sec:unbraided-5400} Let $\mathcal A$ be a non-weakly-group-theoretical fusion category of dimension $n=5400$. If $\Z(\mathcal A)$ were WGT, then $\mathcal A$ would be WGT by \cite[Proposition~4.1]{ENOsolvable}. The center is modular of dimension $5400^2$, and every proper divisor of $5400^2$ is below the braided threshold in Theorem~\ref{thm:uniform-new-bounds}. Thus Lemma~\ref{lem:minimal} applies to the center itself. Its modular alternative gives integrality, fusion-simplicity and perfection; Section~\ref{found:R:1} and Theorem~\ref{found:D:1.1} put its nonunit degrees in \[ \mathcal D=\{a:10\le a\le900,\ a\mid n,\ a\text{ has at least two prime divisors}\}. \] Dominance of the forgetful functor $F:\Z(\mathcal A)\to\mathcal A$ implies that $\mathcal A$ is integral as well: a positive sum of square roots of integers is integral only when every summand is integral, by independence of distinct squarefree radicals. \begin{theorem}\label{th:unbraided-5400} Every proper fusion subcategory of $\mathcal A$ has dimension at most $216$. Every formal codegree of $K_0(\mathcal A)$ other than that of the dimension character lies between $6$ and $540$. The rank satisfies the following bounds. \[ \begin{array}{l|cc} & K_0(\mathcal A)\text{ commutative}&K_0(\mathcal A)\text{ noncommutative}\\\hline \text{arbitrary }\mathcal A&10\le \rank\mathcal A\le1512&12\le \rank\mathcal A\le1512\\ \mathcal A\text{ simple and perfect}&11\le \rank\mathcal A\le462&13\le \rank\mathcal A\le490 \end{array} \] \end{theorem} \subsection{The canonical induction algebra and character moments}\label{sec:5400-proof} Write $I$ for the right adjoint of $F$. For each irreducible representation $E$ of $K_0(\mathcal A)_\CC$, let $m_E=\dim E$ and $f_E$ be its formal codegree. Ostrik's induction theorem identifies $E$ with a distinct simple summand of $I(\one)$, of degree $a_E=n/f_E$ and multiplicity $m_E$ \cite[Theorem~2.13]{OstrikPivotalRankThree}. In particular \begin{equation}\label{eq:5400-induction} \sum_E m_Ea_E=n,\qquad \sum_E m_E^2=\rank\mathcal A,\qquad n\tau=\sum_E a_E\chi_E , \end{equation} where $\tau$ takes the coefficient of the unit. The dimension character has $(m_E,a_E)=(1,1)$, and all other $a_E$ belong to $\mathcal D$. This proves the codegree assertion. \paragraph{Indices and the length of induction.} A subcategory of index $e$ corresponds to a connected \'etale subalgebra of $I(\one)$ of degree $e$ \cite[Theorem~4.10]{DMNOWitt}. Hence $e\mid n$ and $e-1$ is a sum from $\mathcal D$. Below $24$ the only odd such sum is $15$, whereas $16\nmid n$. For the odd divisors $12K$. A singleton $(8)$ cannot occur since every other label is incompatible with it. Every remaining partition is compatible with itself, and any two occurring partitions must have a common compatible partner. A $(3,5)$ partition is compatible only with itself; if one occurs, all $P_X$ equal $U+V$ with $d_U=3,d_V=5$. Here $U=U^*$ and $U^2=\one+U+V$. Left multiplication by $U$ preserves the nonempty set of degree-three labels other than $U$: for $Y\ne U$, the product $UY$ has only degree-three constituents and $[UY:U]=[U^2:Y]=0$. This is impossible, since $U$ generates the category and $Y^*Y$ contains the unit. For $(4,4)$ the only compatible pairs of different types are $aa,ab$. Two distinct repeated pairs, or two distinct pairs with unequal entries, are incompatible. A repeated pair $aa$ and a pair $bc$ with $a\notin\{b,c\}$ have no common compatible partner. Thus there is a pair compatible with all, and unique among occurring pairs of its kind (repeated or distinct): two repeated pairs $aa,bb$ force the distinct type to be $ab$, and two different distinct pairs force a unique repeated type. If only one kind occurs, all pairs coincide. Choose $X$ with this pair. Since $\|XX^*\|^2=\|X^*X\|^2$, the type at $X^*$ is the same and $P_{X^*}=P_X$. For $Y\ne X$ of degree three, $X^*Y$ has only degree-three constituents and omits $X^*$ since $[XX^*:Y]=0$. Multiplying these constituents by $X$ again gives only degree-three labels; it gives no $X$ since $[XX^*Y:X]=\langle X^*Y,X^*X\rangle=0$. Thus the nonunit constituents of $XX^*$ and their duals preserve the degree-three labels other than $X$, contradicting generation as before. This proves \eqref{eq:5400-three}. \paragraph{Upper ranks.} Set $H=\sum_{j\ge4}j^2n_j$ and $E=\sum_{j\ge4}(j^2-9)n_j$. If $n_3\le1$ the degree-square sum gives rank at most $338$. Otherwise \[ 5399=9n_3+H\le18K+H\le60E/11 . \] The last inequality holds coefficientwise except at $4,5$ where it is $198\min(4n_4,5n_5)\le244n_4+685n_5$. Since $E=5399-9(\rank\mathcal A-1)\equiv8\pmod9$, this yields $E\ge998$ and rank at most $490$. If the fusion ring is commutative, reduce the last identity of \eqref{eq:5400-induction} at a prime above two. Distinct algebra characters over the residue field are linearly independent, so some $\chi\ne\FPdim$ with $a_\chi$ odd reduces to $\FPdim$. In particular $a_\chi\ge15$, $f_\chi\le360$. At each odd-degree label the character is nonzero. Its squared norm, and that of each Galois conjugate, is $f_\chi$; a nonzero algebraic integer has mean squared conjugate modulus at least one. Thus $O=\sum_{j\ge3,\ j\ {\rm odd}}n_j\le359$. From \eqref{eq:5400-three} we have $n_3\le10n_5+12n_6+18n_9$. If $n_3\ge2$, comparison of the coefficients at $3,4,5,6,9$ gives \[ \rank\mathcal A-1\le\frac{5399}{16}+\frac{61O}{176} +\frac{n_3-10n_5-12n_6-18n_9}{11} \le\frac{5399}{16}+\frac{61\cdot359}{176}<462 . \] The two upper rank bounds are attained by degree multisets satisfying these numerical inequalities. In the general and commutative cases, respectively, they are \[ \begin{aligned} 5400 &= 1+401\cdot3^2+48\cdot4^2+38\cdot5^2+2\cdot6^2,\\ 5400 &= 1+328\cdot3^2+100\cdot4^2+31\cdot5^2+2\cdot6^2. \end{aligned} \] Their ranks are $490$ and $462$, and the second has $328+31=359$ odd-degree nonunits. These are arithmetic profiles. Any improvement of the upper bounds must use information beyond the inequalities they satisfy. \paragraph{The two endpoint ranks.} For rank ten, or for noncommutative rank twelve, the lower bound $L\ge10$ and \eqref{eq:5400-induction} force $L=10$, respectively with all $m_E=1$ or with exactly one $m_E=2$. In fact for commutative rank ten the assertion is immediate; in the other case $\sum m_E(m_E-1)\ge2$ and equality is necessary. There are nine nonunit summands counted with multiplicity. Group the non-dimension representations by their common $a=a_E$, and put $k_a=\sum m_E$, $b_a=\sum m_E^2$, $f_a=n/a$ within each group. For a nonunit simple $X$ put $d=d_X$, $s_a=\sum\chi_E(X)$ and $t_a=\sum\chi_E(XX^*)$. These are integers by Galois invariance; $t_a$ is a sum of squared Hilbert--Schmidt norms in unitary realizations of the based-ring representations. The necessary row conditions are \begin{equation}\label{eq:5400-charrows} \sum a s_a=-d,\quad \sum a t_a=n-d^2,\quad t_a\ge0,\quad s_a^2\le k_at_a,\quad |s_a|0$, restriction gives \[ H^n(G,\CC^\times)\simeq H^n(G_1,\CC^\times)\oplus H^n(G_2,\CC^\times). \] Indeed the K\"unneth tensor and Tor terms involving positive homology of both factors vanish by coprimality, and $\CC^\times$ is divisible. Products of inflated scalar classes give the inverse. The Picard-space classification identifies the two higher homotopy groups with $A$ and $\CC^\times$ \cite[Theorem~5.2, Proposition~7.3 and Theorem~7.12]{ENOH}. External product includes each $A_i$ as its summand and multiplies scalars. Apply the two cohomology splittings in the following order. \begin{center}\small \begin{tabular}{@{}p{.25\linewidth}p{.21\linewidth}p{.46\linewidth}@{}}\toprule Stage & Group for $G$ & Comparison with the two factors\\\midrule First obstruction & $H^3(G,A)$ & Restrict the zero obstruction of the given extension and project to $A_i$. Each factor admits a lower lift.\\ Choice of lower lift & $H^2(G,A)$ & Choose lower lifts on the factors. Decompose the difference between their product and the prescribed lift, and add its two components to the respective choices.\\ Scalar obstruction & $H^4(G,\CC^\times)$ & At the corrected product lift the obstruction vanishes. Its restriction to $G_i$ is the obstruction for the $i$th lift, since the opposite lift restricts to the constant map.\\ Choice of associator & $H^3(G,\CC^\times)$ & The difference from the prescribed associator is the product of two inflated classes. Twisting each factor by its component realizes the given extension.\\\bottomrule \end{tabular} \end{center} The two choice groups act on torsors, so differences are taken only after fixing a lift at the preceding stage. Equivariantizing the external product by $G_1\times G_2$ gives the product of the two equivariantizations. \end{proof} We use the Hall-factor criterion from the order graph of Erd\H{o}s and P\'alfy \cite{ErdosPalfy}. Partition $\pi(|G|)$ into two parts. If across the cut $s\nmid\ell^j-1$ for $1\le j\le v_\ell(|G|)$ in either direction, $G$ is the product of its two Hall factors. Indeed, for even order every odd prime joins two. For odd order use solvability \cite[p.~775]{FT} and induct on $|G|$. Choose a minimal normal subgroup $M\cong\mathbb F_\ell^e$. In $G/M$ the opposite Hall factor is normal; its preimage is $M\rtimes H$ by Schur--Zassenhaus. The cut makes $|H|$ coprime to $|\mathrm{GL}_e(\ell)|$, so this preimage is $M\times H$ and its unique opposite Hall subgroup $H$ is normal in $G$. A complementary Hall subgroup $P$ commutes with $H$ modulo $M$ and $[P,H]\subseteq M\cap H=1$. \begin{theorem}[Arithmetic and action components]\label{fam:M:13.3}\label{thm:uc-components} For $D=\prod_\ell\ell^{a_\ell}$ form a graph $\Gamma(D)$ on the primes of $D$. Join distinct $s,\ell$ if in either direction \[ s\mid\ell^{2j}-1\qquad (1\le j\le\lfloor a_\ell/2\rfloor). \] An integral WGT modular category of dimension $D$ is a braided product over the connected components $\Pi$ of $\Gamma(D)$; the factor for $\Pi$ has dimension $\prod_{\ell\in\Pi}\ell^{a_\ell}$. This conclusion holds for every integral modular category when all repeated primes lie in one component, including the square-free case. In particular this component formula applies whenever $2\mid D$ or $3\mid D$. It gives a nontrivial splitting precisely when the graph has at least two components. For any modular category with pointed core, the same decomposition holds after adjoining the edges $2$--$\ell$ for the odd primes with $a_\ell=1$ and $v_\ell(d_X^2)$ odd for some simple $X$. When the action is known, one may instead use the following finer action graph on $\pi(|G||A|)$. Join $s,\ell$ if in either direction $s\mid |\rho_\ell(G)|$, or if $s,\ell\mid |G|$ and $s\mid\ell^j-1$ for some $1\le j\le v_\ell(|G|)$. \end{theorem} \begin{proof} Across a cut in the finer graph, $G$ splits by the Hall-factor criterion. Each factor acts trivially on the opposite primary summands of $A$. Lemma~\ref{lem:uc-descent} thus splits the gauging into factors of dimensions $|G_i|^2|A_i|$. It remains to bound the edges of this graph. The group edges are in $\Gamma(D)$ since $2v_\ell(|G|)+v_\ell(|A|)=a_\ell$. For an action edge choose an isometry of $A_\ell$ of prime order $s\ne\ell$. If $s$ is odd, a primitive $s$th root occurs among its eigenvalues on $A_\ell/\ell A_\ell$: the kernel of reduction is an $\ell$-group. Put $f=\operatorname{ord}_s(\ell)$ and $m=v_\ell(|A|)$. The composition factors of $A_\ell$ as a $\mathbb Z[C_s]$-module have total $\mathbb F_\ell$-dimension $m$. Their eigenvalues over $\overline{\mathbb F}_\ell$ are stable under Frobenius, and under inversion because the polar pairing identifies $A_\ell$ with its character dual as a $C_s$-module. If $f$ is even the Frobenius orbit already has size $f$; if $f$ is odd its inverse orbit is disjoint and adds another $f$. Thus $s\mid\ell^{2j}-1$ with $j=f/\gcd(2,f)\le\lfloor m/2\rfloor$. This works also at $\ell=2$ and retains the rank bound $f\le\dim_{\mathbb F_\ell}(A_\ell/\ell A_\ell)$. If $s=2$ and $a_\ell\ge2$, the edge belongs to $\Gamma(D)$. If $a_\ell=1$, then $A_\ell=C_\ell$ has isometry group $\{1,-1\}$. The action is nontrivial exactly when some $(1-\rho(g))A$ has $\ell$-valuation one; this is exactly the odd-valuation test on $d_X^2$ by Proposition~\ref{prop:gauged-degrees}. In an integral gauging $A_\ell$ is therefore fixed whenever $|A_\ell|=\ell$ is odd, also when $\ell\mid |G|$, by the same moved-subgroup formula. More generally a pointwise fixed $A_\ell$ with $\ell\nmid|G|$ splits off by Lemma~\ref{lem:uc-descent}. Finally, even without WGT all exponent-one primes in an integral modular category split off as pointed modular primary factors \cite[Corollary~1.22]{BP}. If the repeated primes are in one component, every other component is an isolated exponent-one prime, giving the required decomposition. If $2$ or $3$ divides $D$, it joins each other repeated prime through $\ell^2-1$. \end{proof} \begin{lemma}[Factors of a cyclic gauging]\label{fam:M:13.6}\label{lem:cyclic-gauging} Let a cyclic group $G$ act faithfully on a finite metric group $A$ of coprime order. Join $s\mid |G|$ to $t\mid |A|$ when the Sylow subgroup $G_s$ acts nontrivially on $A_t$. Suppose that each target has a neighbor and $A_t^{G_s}=0$ on each edge. The action admits a modular gauging, and the connected components give the prime braided factors of every gauging. \end{lemma} \begin{proof} Coprimality gives $H^3(G,A)=0$ and cyclicity gives $H^4(G,\CC^\times)=0$, so the two obstructions to a braided crossed extension vanish \cite[Proposition~7.3 and Theorem~7.12]{ENOH}. Its gauging is modular \cite[Proposition~4.56(ii)]{DGNO}, and Lemma~\ref{lem:uc-descent} splits it over the components. Conversely, its entire pointed subcategory is $\Rep(G)$: Proposition~\ref{prop:gauged-degrees} excludes invertibles over $g\ne1$ by faithfulness, and in the neutral component only $0\in A$ is fixed by $G$. Any braided factorization therefore splits $\Rep(G)$ as the product of the pointed parts of its factors. This splits $G$ as $G_1\times G_2$ of coprime orders, since $G$ is cyclic. De-equivariantizing factor by factor splits $A=A_1\perp A_2$, with $G_i$ acting trivially on $A_{3-i}$. If $G_s\le G_i$ acts on $A_t$, the opposite summand of $A_t$ lies in $A_t^{G_s}=0$. Thus both ends of every edge belong to one factor. Since the factor dimensions are $|G_i|^2|A_i|$, a connected component admits no further splitting. \end{proof} \subsection{Two exact criteria on a square-free set of primes}\label{fam:M:sec:13.3} For a bipartition $S\sqcup T$ of a set of primes and a sign $\epsilon_t\in\{1,-1\}$ at each $t\in T$, draw the allowed edges \[ s\longrightarrow t\quad (s\in S,\ t\in T,\ s\mid t-\epsilon_t). \] Here the signed graph concerns the \emph{existence} of a prime category. By contrast, $\Gamma(D)$ bounds the interactions that the dimension permits, and the action graph of Theorem~\ref{thm:uc-components} uses a specified gauging. Connectivity in the following test refers to the underlying undirected graph. The signs belong to targets, not to individual edges. \begin{theorem}\label{fam:M:13.7}\label{thm:uc-bipartite} Let $L$ be composite and square-free. \begin{enumerate} \item A tensor-prime integral WGT fusion category of dimension $L$ exists if and only if some bipartition has connected allowed graph with all signs $+1$. Moreover for any integral WGT category of this dimension, tensor-primeness is equivalent to primeness of its center. \item A prime integral WGT modular category of dimension $L^2$ exists if and only if some bipartition and choice of target signs has connected allowed graph. \end{enumerate} \end{theorem} \begin{proof} \emph{Groups of square-free order.} Every group $H$ of square-free order has the form $F\rtimes E$, where $F=[H,H]$ and $E$ are cyclic of coprime orders \cite[Theorem~10.1.10]{Robinson}. Draw the actual nontrivial actions from the prime factors of $E$ to those of $F$. Every edge $s\to t$ has $s\mid t-1$. Subgroups belonging to distinct components commute and have coprime orders, so $H$ is their direct product. Conversely a direct product must separate the prime factors without cutting an edge. Every connected all-positive allowed graph is realizable: let each $C_s$ act nontrivially on every incident $C_t$ through its cyclic automorphism group. By \cite[Proposition~9.1]{ENOsolvable}, an integral WGT category of dimension $L$ has a group-theoretical presentation $\mathcal C(H,\omega,K,\alpha)$ with $|H|=L$. A coprime product $H=H_1\times H_2$ splits this presentation. Indeed $K=K_1\times K_2$ by taking powers to isolate each coordinate, and both $\omega$ and its trivialization on $K$ can be chosen as products by the scalar cohomology calculation in Lemma~\ref{lem:uc-descent}. The twisted algebra and its category of bimodules then split. If $H$ is directly indecomposable, its center is trivial: a nontrivial center is a central Hall subgroup and splits off by Schur--Zassenhaus, while a cyclic group of composite square-free order is decomposable. Its twisted double, the center of every presentation above, is prime by \cite[Proposition~6.7(ii), p.~4215]{NNW}: a decomposition would give commuting normal subgroups with product $H$ and intersection in $Z(H)=1$. The center of a tensor product is a braided product. This proves (i), using $\Vec_H$ for existence. \emph{Modular categories of square-free root.} For even $L$ the all-positive star with source two is connected, and (i) supplies a prime center. Now let $L$ be odd, and write $L^2=|G|^2|A|$ as above. The order of $G$ is square-free and coprime to $|A|$. Each primary summand $A_t$ is weakly anisotropic, since an isotropic subgroup invariant under all its isometries would be $G$-invariant. As $t$ is odd and $|A_t|=t^2$, $A_t$ is a metric plane over $\mathbb F_t$ by \cite[Theorem~A.24(i),(iv)]{DGNO}. The odd image of $G$ acts on it through the cyclic rotation group. The plane is anisotropic, since rotations of a split plane preserve its isotropic lines, contrary to maximality of $\Rep(G)$. Thus for $G=F\rtimes E$ as above, $F$ acts trivially on the planes. The prime factors of $E$ act on the cyclic factors $F_t$ through groups of order $t-1$ and on $A_t$ through rotations of order dividing $t+1$. These are the positive and negative targets of a single signed graph. Across any cut $G$ splits by its actual group actions, and its factors fix the opposite planes. Lemma~\ref{lem:uc-descent} splits the category, so primeness implies connectivity. Conversely, given a connected signed graph, take $G=\prod_{s\in S}C_s$ and one metric plane $A_t$ for each target, split for sign $+1$ and anisotropic for sign $-1$. Its rotation group is cyclic of order $t-\epsilon_t$; choose the image of $G$ there to have order the product of the incident primes. The action is faithful and every nonidentity rotation has no nonzero fixed vector. Lemma~\ref{lem:cyclic-gauging} gives a prime modular gauging; it is WGT, integral since each moved plane contributes $t^2$ to $|(1-\rho(g))A|$, and has dimension $L^2$. \end{proof} Both tests apply without a WGT hypothesis when $L$ has at most five prime divisors by Theorem~\ref{seq:five-squares}, using the center in (i). For three odd primes $p1$ gives the metaplectic categories \cite[Theorem~3.2]{ACRW}. At $M=35$ the two extra edges of Theorem~\ref{thm:uc-components} connect the isolated vertices of $\Gamma(140)$ through two. In dimension $4qr$ with $q,r$ distinct odd primes, integral modular categories are pointed \cite[Theorem~3.1]{BGNPW}. Every fusion category of this dimension is WGT by applying Theorem~\ref{thm:three} to its center. \subsection{Prime divisors of degrees and codegrees}\label{fam:M:sec:13.2} \paragraph{Primes seen by simple dimensions.} For integral modular $\C$, let $S_0=\{p:p\mid d_X\text{ for some }X\}$. If $p\mid D$ and $p\notin S_0$, Burciu's categorical It\^o--Michler theorem gives $p^{v_p(D)}\mid |U(\C)|$ and splits off a pointed modular factor of dimension $p^{v_p(D)}$ \cite[Proposition~1.3]{BurciuIM}. The divisibility also implies $p\notin S_0$ \cite[Corollary~1.7]{BurciuIsaacs}. Since $D=|U(\C)|\FPdim(\C_{\ad})$, it follows that $S_0=\pi(\FPdim(\C_{\ad}))$. Burciu extends this splitting to weakly integral braided categories using conjugacy-class dimensions \cite[Theorem~1.1]{BurciuClassGraph}. For WGT categories the graph of the remaining centralizer is $\Gamma(D)[S_0]$; removing only the exponent-one factors gives $\Gamma(Q(D))$. An exponent-one vertex may have joined two repeated primes. In particular, if all $d_X$ are odd and $D=2^\alpha M$ with $M$ odd, the remaining odd-dimensional centralizer has dimension $M$. WGT for it gives solvability of $\C$, for instance when $M=p^a q^b c$ with $c$ square-free and coprime to $pq$ \cite[Theorem~7.4]{NataleWGT}. In the Frobenius--Perron prime graph \cite[Definitions~2.1 and~4.1]{NP}, which we call the simple-dimension graph, one joins distinct primes $p,q$ if $pq\mid d_X$ for some $X$. In a product over arithmetic components $\Pi$, it is the join of the induced graphs on the nonempty $S_0\cap\Pi$: witnesses from different factors can be tensored. Thus each connected component of its complement lies in one arithmetic factor, and with two nonempty factor supports the simple-dimension graph has diameter at most two. In the integral modular case, the class dimensions are the $d_X^2$, so the simple-dimension graph is also Burciu's class graph; if it is disconnected then the category is solvable \cite[Theorem~1.2 and Lemma~6.2(1)]{BurciuClassGraph}. A nonpointed prime integral modular category has $S_0=\pi(D)$. Natale and Pacheco Rodr\'iguez treat Frobenius groups with abelian kernel and complement in \cite[Example~8.6]{NP}, illustrating the calculation with $C_{341}\rtimes C_{10}$. Here take $G=C_{341}\rtimes C_5$, with faithful actions on $C_{11}$ and $C_{31}$. The nonidentity fluxes have abelian centralizers of index $5$ or $341$, and the pure charges have degrees $1$ or $5$. Thus the simple dimensions of $\mathcal Z(\Vec_G)$ are $1,5,341$. The simple-dimension graph has edge $11$--$31$; the arithmetic graph has edges $5$--$11$ and $5$--$31$. \paragraph{Multiplicities and prime-power codegrees.} For a weakly integral modular category let $n_\lambda$ count the simples of degree $\lambda$, $N=\operatorname{ord}(T)$ and $K=\QQ(S)$ for the unnormalized matrix. Then \[ \prod_{\substack{\ell\mid D\\\ell\ {\rm odd}}}\frac{\ell-1}{2} \ \biggm|\ \frac{\varphi(N)}{|(\ZZ/N\ZZ)^\times[2]|} \ \biggm|\ [K:\QQ]\ \biggm|\ \prod_\lambda n_\lambda! . \] By \cite[Theorems~2.3 and~2.5(a), Lemma~2.6]{BNRWClass}, $K\subseteq\QQ(\zeta_N)\subseteq\QQ(\zeta_n)$, where $n$ is the order of a normalized twist matrix, and $\operatorname{Gal}(\QQ(\zeta_n)/K)$ has exponent at most two. The same holds for its quotient $\operatorname{Gal}(\QQ(\zeta_N)/K)$. Galois permutes the characters $S_{ij}/d_j$ faithfully, since their values generate $K$ (the unit column contains the $d_i$), and preserves the positive simple dimensions \cite[Lemma~3.9]{BNRWClass}. The categorical Cauchy theorem \cite[Theorem~3.9]{BNRWFinite} identifies the primes of $N$ and $D$. For instance $7\mid D$ forces a repeated dimension of multiplicity at least three. If $K$ is multiquadratic, then $N\mid240$ by \cite[Corollary~3.14]{BNRWClass}. Pointed core alone does not imply this: a nondegenerate cyclic metric group of order seven has $K=\QQ(\zeta_7)$. \begin{proposition}[The smallest codegrees at prime-power dimension]\label{fam:M:13.4} \label{prop:prime-power-no-defect-two} An integral modular category of dimension $p^n$, $n\ge4$, has no simple of formal codegree $p^2$. \end{proposition} \begin{proof} Such a category is nilpotent by successive cyclic gradings \cite[Theorem~8.28]{ENOannals}, hence WGT, so use the pointed-core description above. For a simple above $Y\in\mathcal F_g$ with projective stabilizer representation $\pi$, the dimension formula gives \[ f=|A^g|\,|G_Y|^2/(\dim\pi)^2\ \ge\ |A^g|\,|G_Y|. \] Here $(\dim\pi)^2\le |G_Y|$ by the dimension of the twisted group algebra, and $g\in G_Y$: the crossed braiding identifies $Y\otimes Y^*$ with ${}^g(Y^*)\otimes Y$, which therefore contains the unit. Suppose $f=p^2$. Then $p^n=f d_X^2$ is square, hence so is $|A|=p^n/|G|^2$. If $A\ne0$, a $p$-group element fixes at least $p$ elements of $A$ by the orbit equation. Since $|(1-g)A|$ is square by integrality, $|A^g|$ is square as well and is at least $p^2$. The inequality forces $G_Y=1$, hence $g=1$ and $|A|=p^2$. Every $h\in G$ then fixes all of $A$ by the same argument. The neutral label $Y\in A$ has stabilizer $G$, giving $G=1$, contrary to $n\ge4$. If $A=0$ the crossed category has one invertible in each component, so the gauging is the twisted double of a $p$-group $G$ of order at least $p^2$. Now $G_Y=C_G(g)$ has order at least $p^2$: for noncentral $g$ it contains $Z(G)\langle g\rangle\supsetneq Z(G)$, and for central $g$ it is $G$. Equality $f=p^2$ would force the twisted group algebra of $C_G(g)$ to be a single matrix algebra. For $g=1$ its cocycle is trivial. For $g\ne1$ its basis element at $g$ is central: the centralizer cocycle is $\alpha_g(x,y)=\omega(g,x,y)\omega(x,y,g)/\omega(x,g,y)$ for $x,y\in C_G(g)$ \cite[(8.1)]{NP}, so $\alpha_g(g,x)=\alpha_g(x,g)=\omega(g,x,g)$. Either way the center has dimension greater than one. \end{proof} \begin{corollary}[Nilpotent modular categories and prime-power doubles]\label{fam:M:13.5} \label{cor:nilpotent-no-defect-two} In an integral nilpotent modular category of dimension $D$, $v_p(D)\ge4$ implies $v_p(f_X)\ne2$. In particular every simple of $\Z(\Vec_P^\omega)$ has codegree divisible by $p^4$ when $P$ is a $p$-group of order at least $p^2$. \end{corollary} \begin{proof} Use the prime-power decomposition \cite[Theorem~1.1]{DGNOnil} and multiplicativity of codegrees. In a double of dimension $|P|^2$ all codegrees are even powers of $p$ greater than one. \end{proof} The nilpotence hypothesis is necessary. The center of a Tambara--Yamagami category on an abelian group of order $p^2$, $p$ odd, is integral modular of dimension $4p^4$. Its degree-$p$ simples $Z_{\rho,\Delta}$ have codegree $4p^2$ and lie outside the principal $p$-block: their character values on $Y_{a,b}$, $a\ne b$, are zero rather than two modulo $p$ \cite[Proposition~4.1 and the $S$-matrix on p.~977]{GNNcenters}. Gelaki, Naidu and Nikshych credit Izumi with an earlier computation of these modular data \cite[Remark~4.3]{GNNcenters}. \section*{Code, data and certificates} \addcontentsline{toc}{section}{Code, data and certificates} \label{sec:code-data-certificates} The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/tree/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c}{verification release} fixes every program, input and receipt cited here. Its \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/certificates/manifest.json}{manifest} identifies the hypotheses, command, expected output and source labels for each calculation. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/certificates/README.md}{verification guide} gives the commands and the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/certificates/statement-index.tsv}{statement index} gives their printed numbers and pages. The file digests permit the same checks on an offline copy. For the dimension bounds, the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/README.md#dimension-comparison}{exponent comparison} recovers the four thresholds in Theorem~\ref{thm:uniform-new-bounds}. For the boundary calculations, the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/code/verify_witnesses.py}{terminal verifier} checks seven trace obstructions and the two disjoint fiber assignments for each of the five numerical rank-100 survivors. The theoretical exclusion of those survivors is in Section~\ref{por:certificate}. For the ordinary rank bounds the \RankLink{ordinary_release.py}{ordinary verifier} checks the degree lists, dimensions and exclusions. Its replay option regenerates the lists from the rank and compares them individually. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/acb9ac6438697f2ca083b9ca9ff0a1ed1b07f66c/rank/README.md}{rank guide} gives separate commands for the positive quotient, nonsquare, odd, and conditional square-dimension searches, and records the earlier implementations for comparison. The nonbraided argument uses \RankFile{code/unbraided_small.py} and the \RankLink{code/odd_unbraided.py}{odd rank-eleven verifier}. The remaining square case at rank $31$ is open. \section*{Acknowledgments} Sebastian Burciu was supported by a grant of the Ministry of Research, Innovation and Digitization, CNCS--UEFISCDI, project number PN-IV-P1-PCE-2023-0264, within PNCDI IV. S\'ebastien Palcoux is supported by the National Natural Science Foundation of China (NSFC), Grant no.~12471031. GPT-6 Astra served as a research assistant in developing and drafting the arguments, identifying references, and checking finite calculations. The authors are continuing their review and have yet to reach a complete understanding of all the proofs. \begin{thebibliography}{99}\small\phantomsection\addcontentsline{toc}{section}{References} \bibitem{ABDPIntegral} M.~A. 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