\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} \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 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} \newtheorem{ratioprop}{Proposition} \renewcommand{\theratioprop}{\thetheorem\alph{ratioprop}} \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 and decomposition} \author{Sebastian Burciu \and Jingcheng Dong \and S\'ebastien Palcoux} \date{} \newcommand{\cost}{\operatorname{cost}} \newcommand{\WGT}{\mathrm{WGT}} \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}} \@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 primes $p\ \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 its conductor divides $2\gcd(f_X,f_Y)$. 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}. \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} \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} \begin{proposition}[Multiplicity at codegree a prime power times three]\label{fam:M:2.8}\label{new:principal-gram} For the full principal class $P$ at $\ell\ne3$, suppose $t=\ell^e$, and a nonunit degree $z$ prime to $\ell$ has codegree $3t$. If $\sum_{Y\in P}d_Y\chi_Z(Y)\ne0$ for each $Z\in I_z$, then $n_z\le2t$, in either table. The trace hypothesis holds, for example, if all $d_Y$ with $Y\in P\setminus\{\one\}$ share a prime divisor. \end{proposition} \begin{proof} Put $I=I_z$, $k=|P|$ and let $B=sE_Ps^*$ be the conjugate of the coordinate projection onto $P$, $s=S/\sqrt\Delta$. The matrix $H=3B_{I,I}$ has integral entries: \[ H_{XZ}=t^{-1}\sum_{Y\in P}\chi_X(Y)\overline{\chi_Z(Y)} =3\tau(E_PXZ^*). \] Indeed the sum is a rational integer by Galois stability, and the last expression is integral at $\ell$ by \eqref{eq:local}. Because $B$ is a projection and $B_{X0}\ne0$ by hypothesis, $[H(3\operatorname{id}-H)]_{XX}=9\sum_{j\notin I}|B_{Xj}|^2>0$. Thus every diagonal is one or two and has at most one nonzero off-diagonal entry, necessarily $\pm1$. The resulting blocks have size at most two and positive rank. If all diagonals are two, $H$ is invertible. All entries on $I\times P$ are nonzero by principality and $\ell\nmid z$; Galois averaging gives $k\le tH_{XX}\le2t$. For $k\le t$ we obtain $n_z/2\le\operatorname{rank}H\le k$. Otherwise all diagonals are two and $n_z=\operatorname{rank}H\le k$. \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 $\widehat s$. 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{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} First consider ordinary data. Write $L_k^i=\nu_{k,1}^{(i,i^*)}$ for the indicator with insertion in $\Z(\C)\simeq\C\boxtimes\C^{\rev}$, extended linearly by \cite[Corollary~5.5]{NSind}. Its modular formula on a simple $c$ is $\sum_{a,b}s_{ia}s_{ib}(\theta_a/\theta_b)^kN_{ab}^c$ \cite[Proposition~6.1]{NSind}. Pairing with $f_ze_z$ replaces the fusion coefficients by $\overline{\chi_z(a)\chi_z(b)}$; duality in the second factor gives \begin{equation}\label{eq:indicator-entry} L_k^i(e_z)=|(sT^ks^{-1})_{iz}|^2 . \end{equation} \emph{A rotation congruence.} Put $m=N_{p'}$. Given a genuine object $A$, choose $b\ge\max(1,v_p(N))$ with $p^b\equiv1\bmod m$, and $n\ge2b$. Then \begin{equation}\label{eq:large-power} L_m^i(A^{p^n})\equiv L_{mp^b}^i(A^{p^{n-b}}) \pmod {p^{n-2b+1}\overline{\ZZ}_p}. \end{equation} Strand grouping identifies the two sides with traces of powers $p^n,p^{n-b}$ of the same rotation on $mp^n$ strands \cite[Lemmas~2.5, 2.7 and Proposition~2.8(ii)]{NSind}. Its eigenvalues are $mp^n$th roots, since the insertion has twist one, and all power traces lie in $\QQ(\zeta_N)$ by \cite[Proposition~5.7]{NSind}, since $N=\ord(T)=\operatorname{FSexp}(\C)$ by \cite[Theorem~7.7]{NSfs}. Thus Galois orbits over that field have constant multiplicity. The two powers agree on the prime-to-$p$ part and on roots of $p$-power order at most $p^{n-b}$. For order $p^t>p^{n-b}$, either power is constant on the subgroup orbit given by units $1\bmod p^b$ on the $p$-factor. Its size $p^{t-b}$ is divisible by $p^{n-2b+1}$, proving the congruence. \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 $L_k^0(y)=1$. Grouping $p$ strands gives \[ L_k^0(y)^p\equiv L_{k/p}^0(y^p) =f_z^{p-1}L_{k/p}^0(y)\pmod{p\overline{\ZZ}}. \] Indeed the $p$th power of a trace agrees modulo $p$ with the trace of the $p$th power. This applies first to each simple, then by additivity and the Frobenius congruence in the commutative fusion ring to the integral coefficients of $y$. 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 $P=s^*E_As$, $Q=TsE_Bs^*T^*$ 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} \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{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. At a zero residue, a nonzero $z$ must also ramify if $s\ge g$. \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 zero-residue case unramifiedness would make $z/s$ integral of house less than one. \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. It remains to justify this integral-word reduction. Write the level as $MN$ with $M$ supported on the specified primes, enlarged to include each, and $(M,N)=1$. The signed $M$th roots are a unimodular configuration \cite[Theorems~3, 13]{BH}; minimizing total weight therefore gives an integral optimum for an integral point. Use $\mathbb Z[\zeta_{MN}]=\mathbb Z[\zeta_M]\otimes_{\mathbb Z}\mathbb Z[\zeta_N]$, by coprimality and the cyclotomic degrees. Given an $N$th root $a\ne\pm1$, choose a basis in each primary factor of $N$ by omitting one root from each prime-order coset. Keep $1$ in every basis and, in a factor where the component of $a$ is not a sign, keep that component as well. This is possible also at two, where the two chosen elements then lie in different cosets. The coefficient of $1$ in the product basis defines $\lambda$ with $\lambda(1)=1$, $\lambda(a)=0$ and $\lambda(\mu_N)\subset\{0,\pm1\}$, since each omitted root is the negative sum of its remaining coset. The integral $\mathbb Z[\zeta_M]$-linear contraction $1\otimes\lambda$ preserves the congruence at each retained prime $s$: the radical of $(s)$ is generated by $1-\zeta_{s^{v_s(M)}}$ at both levels. The contracted word has rational weight at most $g$; unimodularity and the preceding alternatives give minimum weight exactly $g$. Any positive coefficient of an original summand with complementary component $a$ would give weight strictly below $g$ after contraction. Thus every complementary component is a sign, and unimodularity 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. To expose a primary support assign one to its $k$ marks in each $\mu_s$-coset and $-k/(s-k)$ elsewhere. The coset sums vanish, so this defines a functional on the primary field. Tensor with normalized trace on the complementary field. Nonidentity complementary roots have mean strictly below one, and in odd order their mean $\mu(n)/\varphi(n)$ is at least $-1/2$. Thus the product is strictly below one off the marks when $kb(g)$, 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

0$. Their special fibers are not isomorphic as algebras. \end{lemma} \begin{proof} Let $M=\OO^n$, $k=\OO/2\OO$. By Lemma~\ref{def:uniform-split-order}, $A_-=\OO\mathbf1+2H$, $H=\{y:\sum y_i\in2\OO\}$. Since $H$ is closed under coordinatewise squaring and $4M\subset A_-$, its special-fiber radical has \begin{equation}\label{eq:signed-radical-products} J_-^3=0,\qquad x^2=0\quad(x\in J_-). \end{equation} We show that the positive lift cannot have both properties. Write $A_+=\OO\mathbf1+2L$, $L=\{y:2y\in A_+\}$. The conductor and index give $4M\subset L$ and $\ell(M/L)=n/2+1$. Since $d_i^2\equiv1\pmod 8$, trace integrality on $2L$ and on $4LL$ gives $\sum y_i\in4\OO$ and $\sum y_iz_i\in2\OO$ for $y,z\in L$. Thus $C=(L+2M)/2M$ is self-orthogonal in $k^n$; the layer $(L\cap2M)/4M$ embeds by division by two in the sum-zero hyperplane. Their dimensions sum to $2n-\ell(M/L)=3n/2-1$, so $\dim C=n/2$ and $C=C^\perp$. If $J_+^3=0$, the trace of each $8yzw$, $y,z,w\in L$, is even. All triple dot products vanish in $C$, so $C=C^\perp$ is closed under coordinatewise products. This unital subalgebra consists of the functions constant on a partition. Self-orthogonality makes its $n/2$ blocks even, hence pairs. Lift a pair indicator to $y\in L$. If all radical squares vanished, $4y^2\in2A_+$ would give $y^2\in L$, contradicting $\sum y_i^2\equiv2\pmod4$. \end{proof} \subsection{Ramification and residue energy}\label{fam:M:sec:5.2} For a cyclotomic integer $x$ let $\mathsf M(x)$ be the mean of $|\sigma x|^2$ over its Galois conjugates and put $E(x)=[\QQ(x):\QQ]\mathsf M(x)$. At a fixed positive degree Galois permutes the columns of a row. We bound the squared norm spent by one orbit of entries, using their common residue and ramification at $p$. \begin{theorem}[Energy at a prescribed prime residue]\label{fam:M:5.3}\label{thm:cyclotomic-residue-energy} \label{prop:odd-defect-filtration}\label{lem:singleton-residue-energy}\label{lem:codegree-four-residue-energy} If a nonzero cyclotomic integer $x$ vanishes at all primes over $p$, then $E(x)\ge2p$. For odd $p$ write the ramification index of $\QQ(x)$ as $e=e_0p^s$, $e_0\mid p-1$. Then \[ \mathsf M(x)\ge W_p(e):= \begin{cases} p^2 & e=1,\\ 2p/e_0 &s=0,\ e_0>1,\\ 1+(p-1)/e_0 & s>0 . \end{cases} \] 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{lemma}[First tame shell and trace]\label{fam:M:5.4}\label{lem:fifth-residue-first-shell} Let $p$ be odd and $0\ne x\in\ZZ[\zeta_p]$ vanish above $p$. If $E(x)<3p$, then $E(x)=2p$. With $H$ its stabilizer in $\mathbb F_p^\times$, $x$ is either the difference of the sums on two nonzero $H$-cosets, or $\pm(1-\zeta_p^j)$ for $j\ne0$. The former has trace zero; the latter has trace $\pm p$ and sum of conjugate squares $p$. At $p=5$ the quadratic possibility in the first case is $\pm\sqrt5$. \end{lemma} \begin{proof} Take the unique coefficient vector of sum zero as in the theorem. For $h=|H|$, let $c_C$ be its integer coefficient on each nonzero index coset of $H$ and put $s=\sum_Cc_C$. The coefficient at zero is $-hs$. The same trace formula gives \[ E(x)=p\bigl(hs^2+\sum_C c_C^2\bigr),\qquad \Tr(x)=-ps. \] Below $3p$ the bracket must be two: for $s=0$ the nonzero entries are $1,-1$; for $s\ne0$ one has $h=1$ and a single entry $\pm1$. The sum of conjugate squares follows by summing $(1-\zeta_p^a)^2$. At five the difference of the two quadratic periods squares to five by the quadratic Gauss sum. \end{proof} \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. This holds on all conjugates and again gives $p\mid T_X$. 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} \subsection{The structure at defect two}\label{fam:M:sec:5.4} \begin{proposition}[Ramification of a principal defect-two row at two]\label{fam:M:5.8} \label{prop:principal-two-defect-two-ramification} Let a commutative symmetric integral order have cyclotomic characters, a perfect positive $*$-trace, rational positive codegrees, and the usual orthogonality of character rows. Let $\eta$ be inertia-fixed at two with $v_2(f_\eta)\ge3$. If a character $\chi$ of $v_2(f_\chi)=2$ has the same residue as $\eta$, then for a sufficiently large finite unramified extension $K/\QQ_2$ the local character field $K(\chi)$ is contained in $K(i)$. \end{proposition} \begin{proof} After unramified base change the conductor bound (Corollary~\ref{cor:field-conductor}) puts the local field inside $K(\zeta_8)$. The full extension has $(e,d)=(4,8)$ and the quadratic fields $K(\sqrt{\pm2})$ have $(2,3)$. Their branch conductor lengths $2e-d$ are respectively zero and one. By Lemma~\ref{lem:conductor-node} neither branch can coexist with the unramified $\eta$, whose conductor length is at least three. Only $K$ and $K(i)$ remain. \end{proof} At odd $p$ a defect-two field has at most one nontrivial wild layer by Corollary~\ref{cor:field-conductor}. \begin{lemma}[Wild coefficient vectors]\label{fam:M:5.9}\label{lem:wild-coefficient-vectors} Let $\chi$ be a character of rational codegree $f$, whose orbit under wild inertia at an odd prime $p$ has length $p$. Suppose its $p$-primary conductor is at most $p^2$. In a splitting field write $\zeta=\zeta_{p^2}$ and choose the wild generator $\tau$ with $\tau\zeta=\zeta_p\zeta$. There are integral tame vectors $A_j$, $0\le j0$ (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 $z_t=(W_t-I)e_0$. 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 $e_X-b_X e_0$ and its inverse Weyl image vanish on the active sector. Both local unipotents, and hence the local torus, fix it. The torus fixes $e_0$ 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 $X$, \[ (W_t-I)e_X=[d]_t(W_t-I)e_0,\qquad a_t(X)=-\epsilon j_t[d]_t . \] 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}[Small-degree cutoff]\label{fam:M:6.7}\label{lem:twist-cut} At an ordinary minimal modular table of dimension \(n^2\), partition the primes of \(n\) into clusters such that no twist meets two clusters. For a cluster \(J\) write \(n_J=\prod_{v\in J}v^{v_v(n)}\), and put \(B=\min_J n/n_J\). Every nonunit degree \(d\) satisfies \(d^2\ge B\). \end{lemma} \begin{proof} A label active in \(J\) has degree divisible by \(n/n_J\), by Lemma~\ref{lem:complement} at every complementary prime. Thus all labels of degree below \(B\) have twist one. If \(d_X^20$. 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} Let $\mathcal C$ be an integral modular category of dimension $n^2$ with positive spherical structure. Write $s=S/n$, $f_z=n^2/d_z^2$, and $E_z$ for the primitive character idempotent and $e_z$ for the coordinate vector of label $z$. Fix distinct primes $q,p$. For the congruence lift of Lemma~\ref{fam:M:6.1}, let $U_q,W_q$ be the upper unipotent and Weyl element of the $q$-primary factor. The diagonal matrix $T_q$ of actual $q$-primary twists differs from $U_q$ by a scalar root of unity. Call $z$ active at $q$ if $(T_q)_{zz}\ne1$. \begin{theorem}\label{fam:W:2.1}\label{fam:W:n:thm:foreign-column} Put $a=v_p(f_z)$. At every prime above $p$, normalized by $v(p)=1$, each nonzero entry $w_{iz}$ of $W_q^{\pm1}$ 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 $q$-active label $z$, all entries of $W_q^{\pm1}$ are integral at every prime above $p$. \end{theorem} \begin{proof} Choose $m>0$ congruent to one at the $q$-primary level and to zero at every other primary level, with $p\mid m$. Put $A=sT^m s^{-1}$ and $w=W_q^{-1}$. The identity $U_q(W_qU_q W_q^{-1})U_q=W_q^{-1}$ makes $A$ equal to $w$ up to diagonal root-of-unity factors. For $H=W_q^2$, the local torus at $-1$, Galois covariance of the linear lift gives $\overline w=w^{-1}=Hw$. The matrix $H$ permutes the labels, since the torus signs in integral square dimension are positive. Moreover every cyclotomic automorphism with exponent one modulo the $q$-primary level fixes $w$: its action conjugates the group element by $\operatorname{diag}(k,1)$, which fixes the $q$-component for these exponents; all other components of the group element are trivial. Thus $w_{iz}$ lies in a $q$-power cyclotomic field, unramified at $p$ (Lemma~\ref{fam:M:6.1} and its proof). For labels $i,j$ let $L_m^{ij}=\nu_{m,1}^{(i,j^*)}$ under $\mathcal Z(\mathcal C)\simeq\mathcal C\boxtimes\mathcal C^{\rm rev}$. The modular formula \cite[Proposition~6.1]{NSind} reads $L_m^{ij}(c)=\sum_{a,b}s_{ia}s_{jb} (\theta_a/\theta_b)^m N_{ab}^c$. Contracting with the coefficients of $y=f_z E_z=\sum_c\overline{\chi_z(c)}[c]$ replaces $N_{ab}^c$ by $\overline{\chi_z(a)\chi_z(b)}$. Changing $b$ to $b^*$ in the second sum and using $s_{0z}^{-2}=f_z$ gives the mixed identity \[ L_m^{ij}(y)=f_z A_{iz}\overline{A_{jz}} . \] For any central simple insertion $X$, abbreviate $L_m=\nu_{m,1}^X$. The rotation is of finite order since its $m$th power is the scalar $\theta_X^{-1}$. For each fusion basis element $c$, the $p$th power of its trace is congruent at every prime above $p$ to the trace of its $p$th power. Composition of rotations and grouping $p$ strands identify the latter with $L_{m/p}(c^p)$ \cite[Lemmas~2.5, 2.7 and Proposition~2.8(ii),(iii)]{NSind}. The functionals are additive by \cite[Corollary~5.5]{NSind}. Thus for all integral cyclotomic coefficients $a_c$, with $y'=\sum_c a_c[c]$, the commutative fusion-ring congruence gives \[ L_m(y')^p\equiv\sum_c a_c^p L_{m/p}(c^p) \equiv L_{m/p}((y')^p)\pmod{\mathfrak p}. \] In particular $L_m^{ij}(y)$ is an algebraic integer, and if $p\mid f_z$ its valuation is positive since $y^p=f_z^{p-1}y$. Take $j=Hi$. Then $\overline{w_{jz}}=w_{iz}$. The mixed identity gives $a+2v(w_{iz})\ge0$, strictly if $a>0$. The valuation $v(w_{iz})$ is integral by unramifiedness, proving the bound also for $W_q=Hw$. Work in a cyclotomic splitting field $K$ and put $R=\bigcap_{\mathfrak p\mid p}\OO_{K,\mathfrak p}$. The column bound puts $w_z=W_q e_z$ in $R^{\Irr(\mathcal C)}$ for every $q$-active $z$ when $v_p(f_z)\le2$. Complex conjugation permutes the primes above $p$, so the entries of $w_z^*$ also lie in $R$. By unitarity, \[ W_q T_q W_q^{-1}=I+\sum_{z\text{ active at }q} (\theta_{z,q}-1)w_z w_z^* . \] Thus $W_q U_q W_q^{-1}$ has entries in $R$, since $U_q$ differs from $T_q$ by a root of unity. The unipotent identity above gives the assertion for $W_q^{-1}$, and taking the adjoint gives it for $W_q$. \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{Conductor, local Weyl coefficients and twist masses} \label{por:tools} For an integral modular category put $d_Z=\FPdim Z$, $f_Z=D/d_Z^2$, $s=S/\sqrt D$ and $\chi_Z(Y)=S_{ZY}/d_Z$. We use the positive spherical structure, and the lift of Lemma~\ref{lem:local-vacuum-column} with normalization scalar $\gamma$ of dyadic order and $\gamma^3=\xi$, the multiplicative central charge. All uses of Frobenius--Schur indicators below concern actual categories. Let $\mathcal C$ be an integral modular category of dimension $D=n^2$. For any integer $m\ge1$ put \[ M_m(Z)=\sum_Y d_Y\chi_Z(Y)\theta_Y^m. \] For a set $J$ of primes in the level write $W_J$ for the product of its local Weyl operators in the normalized congruence representation, and $\theta_{Y,J}$ for the $J$-part of $\theta_Y$. \begin{theorem}\label{por:th:weyl-div} For all $m,Z$, the number $|M_m(Z)|^2/D$ is an algebraic integer. For every $J,Z$, \[ k_{J,Z}:=\frac n{d_Z}(W_J e_0)_Z\in\mathbb Z . \] More precisely, if $m$ is one on the local levels in $J$ and zero on the complementary levels, then $M_m(Z)=n\gamma_J^3\theta_{Z,J}^{-1}k_{J,Z}$, with $\gamma_J$ the $J$-part of the normalization scalar. Every prime $q$ outside $J$ dividing $f_Z$ also divides $k_{J,Z}$. \end{theorem} \begin{proof} Extend $\nu_m$ linearly to virtual objects with cyclotomic integer coefficients; its values are integral because each simple indicator is a finite-order trace. For every $B$, inserting Verlinde into the Ng--Schauenburg formula \cite[Theorem~7.5 and Remark~7.6]{NSfs} gives $\nu_m(B)=\sum_Y\chi_Y(B^*)|(sT^ms e_0)_Y|^2$, where $*$ extends duality linearly. The two factors are conjugate since $X,X^*$ have the same twist. For $B=\sum_X\chi_Z(X)[X]$ the coefficient at $Y$ is \[ \sum_X\chi_Z(X)\chi_Y(X^*)=f_Z\mathbf1_{Y=Z}. \] Thus $\nu_m(B)=f_Z|(sT^ms e_0)_Z|^2=|M_m(Z)|^2/D$. For the indicated $m$, $sT^m s e_0=sT^m s^{-1}e_0$ since charge conjugation fixes the vacuum. Only the $J$-local upper unipotent occurs in the middle of the latter word, apart from the scalar $\gamma_J$. The matrix identity $W U W^{-1}=U^{-1}W^{-1}U^{-1}$ in each selected factor shows that its $Z$-coordinate is $\gamma_J^3\theta_{Z,J}^{-1}(W_J^{-1}e_0)_Z$. The vectors $W_J^{-1}e_0=W_J e_0$ are rational in square dimension (Lemma~\ref{lem:local-vacuum-column}). Since the same coordinate equals $d_Z M_m(Z)/D$, the claimed expression for $M_m(Z)$ follows with rational $k_{J,Z}$. Its square is an algebraic integer by the first assertion, proving integrality. For a prime $q$ the rotation identity \cite[(5.4) and proof of Theorem~5.1, pp.~75--76]{NSpivot} gives $\nu_{qv}(X)^q\equiv\nu_v(X^{\otimes q})$ at a prime above $q$: trace to the $q$th power agrees in the residue field with trace of the $q$th power, and rotation by $q$ places is rotation by one block of length $q$. This extends to every cyclotomic-integral virtual $B$ by additivity and the Frobenius homomorphism in the commutative fusion ring modulo $q$. If $q\notin J$, choose the CRT parameter in the form $m=qv$. The above $B$ equals $f_Z$ times a primitive idempotent (at $Z^*$), hence $B^q=f_Z^{q-1}B$. For $q\mid f_Z$ the congruence now gives $k_{J,Z}^{2q}\equiv0$ at that prime, proving $q\mid k_{J,Z}$. \end{proof} \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 $W_pe_0$ gives \[ \sum_Z p^{2v_p(d_Z)}0$. 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 $t^i z_A=\chi_A(V)\equiv0$ above $h$, while $z_A\ne0$ 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, $w=W_oe_0$, 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 $Pw=w$, $Ww=se_0$, 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. In all cases the missed phase has a nonempty fiber on which no $q$- or $r$-active column can occur. The two companion groups act trivially there, since their generating upper unipotents and their conjugates do. Thus Lemma~\ref{lem:complement} gives the asserted codegree for any $Z$ of this fiber. 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} gives a unique unsigned $r$-word. Its tame stabilizer acts freely on its nonidentity roots and has order at most $H_{p^h}(r)$; hence the $r$-inertia orbit of $X$ has 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 $\mathcal B=\langle g\rangle'$. Write $R_+=\K(\mathcal B)/(g-1)$. An even pair $[Y]=\{Y,gY\}$ has opposite actual dyadic twists and hence a well-defined squared twist $\epsilon_Y=\theta_{Y,2}^2$. Odd mates have the same actual dyadic twist. The positive space $A$ has orthonormal basis $(v_Y+v_{gY})/\sqrt2$ on even pairs. Let $B$ be the antisymmetric even space and $C$ the symmetric odd space. In the integral congruence lift, $U=\lambda T_2$, $\lambda^8=1$, exchanges $A,B$ and preserves $C$. The dyadic Weyl $W$ interchanges $B,C$ and preserves $A$, because it interchanges fermion permutation and monodromy sign. Put $K=W|_A$ and write $W_o=W_{2'}$ for the complementary Weyl. \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. The two corresponding signed 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 $w=W_o e_{[1]}$ 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 $\widehat s e_{[1]}=Kw$, $|w_{[j]}|^2=2d_j^2/\Delta$ on $\epsilon_j=1$ and vanishes otherwise. Lift back to the full vacuum $w^{\rm full}=W_o e_1$. 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 $tr$. For each complementary prime-power level $u^a$ contract the cyclotomic lattice by a random integral functional. On $\mu_u$ give $1$ the value one, a uniformly selected nonidentity root the value $-1$, and all the others zero. On every other $\mu_u$-coset give all positions zero. The coset relations make this well defined. Tensor all the functionals, leaving the $t,r$-level alone. The two prescribed residues are preserved since the radicals are generated by a primary cyclotomic uniformizer in the retained factor. When an original positive summand has a nontrivial complementary part its expected signed cost is at most $3/(2(u-1))<1$ for any nontrivial component $u$. The trivial complementary part always costs one. If any complementary part occurred, some contraction would thus cost less than $r$. Its value is integral, and unimodularity of the signed two-prime configuration \cite[Theorems~3 and~13]{BH} gives an integral point attaining the minimum of the same asymmetric linear cost. This contradicts the preceding paragraph. 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 $z$ be a cyclotomic integer expressed as a positive rational combination of roots of total weight $g\in\mathbb Z_{>0}$. Suppose every prime of their orders 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. We justify the integral-word reduction. Apply the contraction used in Theorem~\ref{first:balance} onto the $q,r$-primary field. A rational integral-basis functional on the complementary field fixes $1$ and takes values $0,\pm1$ on roots. It preserves both residue conditions. The signed primary configuration is unimodular, so an algebraic integer with a rational expression of mass at most $a$ there has a signed integral expression of length at most $a$. The preceding lower bound applies also without strict house (only $m>0$ used strictness). For each specified complementary root different from a sign the functional can be chosen to kill it, as in the same proof. Any such letter in the original positive expression would then leave mass strictly below $r$. Hence all complementary parts are signs, and unimodularity supplies the asserted integral word at $z$ itself. 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. On every such fiber both complementary local groups act by scalars. By Lemma~\ref{fam:M:3.3} every label $Z$ in it has 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 $p$. It is unique and has a nonidentity term by strictness. As the tame units act freely on nonidentity $r$-power roots, its stabilizer has order at most $pq$, 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 $q3$, 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 $w=-G Pse_0$ 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 $(e_Y+e_{gY})/\sqrt2$ for even fermion pairs. Write $W_o$ for the product of the local Weyls away from three and $w=W_o e_0$. It interchanges the fermion permutation and its monodromy sign. Therefore the coordinates of $W_o(e_0+e_g)/\sqrt2$ 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. It has neither $2$- nor $r$-activity and is invariant under both corresponding local groups. Their upper unipotents act by scalars there, hence their conjugates and local Weyl operators do too. By the complement lemma every row $Z$ in that fiber has $2$- 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 word of length three in $r$-power roots. It is unique at a common level, and it has at least one nonidentity term. The stabilizer in tame $r$-inertia acts freely on its nonidentity roots, so its order is at most three. Its orbit therefore has 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 $3g+1$. All complementary witnesses therefore have height $k$ and singleton positive two-blocks. The singleton rows vanish on all nontrivial squared phases. Every squared phase contains a $q$-active pair, which hits a witness by principal-block orthogonality modulo $r$. Thus the even twists are signs. By Lemma~\ref{lem:fermion-singleton-lift}, opposite principal witnesses see the same odd phase, and an odd $q$-active column hits such a witness by the same orthogonality on the full block (the opposite $q$-unit odd mate excludes $g$). Every odd dyadic fiber has a $q$-active label by Lemma~\ref{lem:odd-valuation-joint-twist}, contrary to Lemma~\ref{seq:fermion-one-spin}. \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} Write $\mathcal B$ for the category, $\mathcal M$ for an integral minimal modular extension, and $g$ for its fermion. An \emph{even label} belongs to $\mathcal B$, the neutral fermion-monodromy component. An \emph{odd degree} refers to the integer $d_X$. The only invertibles of $\mathcal M$ are $1,g$. For each noninvertible $X$ of $\mathcal M$, the constituents of $XX^*$ generate $\mathcal B$. Projective centralization would put the other simple in $\mathcal B'=\langle g\rangle$. Thus strict house and coprime vanishing apply to all pairs of full noninvertibles (Lemma~\ref{fam:M:4.4} and its proof). The case of a missing repeated prime follows from the two-prime reduction. At each odd prime $s$, a positive principal pair has a full principal representative by the sign argument in the proof of Theorem~\ref{fam:M:7.3}. It follows that no full noninvertible can have prime-power degree $t^i$. Choose $s\in\{3,r\}$ distinct from $t$. Prime avoidance on positive $P_s$ supplies a $t$-free nonunit pair by Propositions~\ref{fam:M:3.4} and~\ref{fam:M:4.2}. Its full $s$-principal lift must hit the degree $t^i$, contradicting coprime vanishing. 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{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 $|z|2g$ the single-residue case of Theorem~\ref{three:word} gives a unique unsigned word of $g$ roots of $t$-power order. The tame stabilizer permutes its terms, freely on every nontrivial root; at least one is present by strictness. Its order divides $t-1$ and is at most $g$; both character fields contain $z$. \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 $x_0-(L/e)e_0$ 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 Lemma~\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} 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. \begin{proposition}[The degree $p^2q$ cannot occur alone]\label{fam:M:12.5} \label{new:sole-A-global} $A$ cannot occur as the sole common degree; in particular $uv$ is absent. \end{proposition} \begin{proof} 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. \end{proof} \begin{proposition}[Consequences of a common degree $p^2q$]\label{fam:M:12.6}\label{new:common-A} Let $n=p^3q^2uv$ be an odd ordinary minimal root, $p\ge5$ least, $ubv . \] All labels of degree $p^iu$ or $p^iv$ ramify at $q$ and are unramified at both singletons. If $L=p^3q$ also occurs, every $A$- and $L$-label is unramified at both singletons, and some $B$-label is singleton-ramified. \end{proposition} \begin{proof} 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. \end{proof} 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 $p<3s/4$. \begin{proposition}[The degree $p^2q^2$ cannot occur alone]\label{fam:M:12.7}\label{new:sole-B-global} The sole common degree cannot be $B=p^2q^2$. \end{proposition} \begin{proof} 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 $0bv$ (Propositions~\ref{nov:no-L-ram} and~\ref{new:common-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 Proposition~\ref{new:common-A}. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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. The recurrence for the nonempty axis sets minimizes the sum of the squared costs at each residue modulo the omitted repeated prime. For the common degrees, enumerate their shared multiplicities in the remaining two norm budgets and impose their mixed Gram sum $-j_uj_v$. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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/main/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 trace and Gram test 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 all pairs of roots by CRT and retain prime representatives within the proven window. At each endpoint $t$, with $w=uv/t$, both a $P$-axis $P^iw$ and a $Q$-axis $Q^lw$ must occur. Each has at least two labels by odd duality. From the norm budget $j_t(t+1-j_t)$ subtract twice the least possible $A_t(P^iw)^2$ and twice the least possible $A_t(Q^lw)^2$, using admissible degrees. The remainder must be positive, since the mixed Gram sum is $-j_u j_v<0$. Some common degree $y=P^iQ^l$ must contribute negatively to that sum. Its two signed costs $2A_u(y)^2,2A_v(y)^2$ must fit the two budgets. It must be admissible at both endpoints and pass the common-degree unramifiedness test of Lemma~\ref{fam:M:8.13} at one endpoint. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/code/four_primes/prime_projection.py}{first} enumerates residues by multiplication until repetition. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 $u,v$. It follows that, for arbitrary $A\ge0$, every divisor of $2^{2H+1}q^A u^3v^2 e$ is covered if $q\notin S_H$ and $e$ is square-free and coprime to the four primes. Indeed, for the two squared companions a divisor-minimal failure retains all four repeated primes. Its reduced case is excluded by Lemma~\ref{fam:M:8.4}, and its ordinary table is square by Proposition~\ref{fam:M:8.3}. Lower dyadic heights are one, covered by Theorem~\ref{thm:cubic-lift}, two, covered by Theorem~\ref{fam:W:1.2}, or, at $H=4$, three, where $S_3\subseteq S_4$. A single opposite root height is covered by the same one-high-prime theorem. On loss of a repeated prime, there are at most two, or three with a squared companion. Finally Theorem~\ref{fam:M:10.3} raises either designated square to a cube. \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. \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 $z_H=(W_H-I)e_0$. 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 Lemma~\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} A divisor-minimal failure of the five-prime theorem must retain all five repeated primes. On losing one, Theorems~\ref{fam:M:4.12} and~\ref{fam:M:10.3} apply. If the quartic drops, use Theorem~\ref{fam:F:2.1}. Section~\ref{fam:F:sec:4} excludes the reduced table when all exponents survive. Thus for the square case the ordinary table has dimension $(p^2abcd)^2$. The quartic prime two was excluded in Section~\ref{fam:F:sec:5.4}. In the odd case, Corollary~\ref{fam:C:2.3} rules out mixing two squared primes and Propositions~\ref{fam:C:3.1}--\ref{fam:C:3.3} rule out full isolation. \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 $h1$. 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 $ad_A$. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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. \par In the nonpointed odd-dimensional case with noncommutative fusion ring the lower bound is $23$ by \cite[Theorem~1.8]{ABDPIntegral}. \end{proof} For comparison, the Anyon Atlas \cite[Section~3]{AnyonAtlas} uses a numerical search to construct modular data through rank $22$. The proof above uses the exact rank-$15$ list. \section{Cyclotomic support and local ramification}\label{found:group:support} Throughout these sections a minimal table is one of the two tables in Lemma~\ref{lem:minimal}. The dimension of a positive fermionic quotient is half the dimension of its category. Write $P_t$ for the principal block at a chosen prime above $t$, and call $X$ \emph{$t$-active} when the $t$-part of $\theta_X$ is nontrivial. The twist graph has the dimension primes as vertices and an edge $t,u$ when a simple twist is divisible by $tu$. An edge need not have the same witness as another edge. We call $t,u$ \emph{separated} when this edge is absent. In the ordinary square-dimensional case put $n=\sqrt D$, $s=S/n$, $f_X=D/d_X^2$ and $K_X=\QQ(\chi_X)$. A nonzero $S$-entry will be called a hit. A full word at nonunits $X,Y$ is a positive rational combination of phases, of total mass $d_X$, equal to $\chi_Y(X)$ up to a common root of unity. Write $W(t,k)$ for the impossibility of such a word on at most $k$ positions of $t$-power order in odd ordinary minimal data. All assertions about divisor families allow a coprime square-free factor $d$. \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 $|E\cap C|3$, the sets $\{0,\pm d\}$ and $\{0,\pm2d\}$ intersect only at zero. If $a>1$, the three positions of each are in distinct order-$s$ cosets, and every capacity is at least one. If $a=1$ and $s\ge7$, a coset with three marked positions has at least four unmarked ones. Both cases force exposure, contrary to the proposition. At $s=5$ the only coset which can fail has three marked positions and two unmarked. Formula~\eqref{mass:witness} at $h=2$ forces an unmarked position to admit a complementary root of order two. The primary part at that position is a nonidentity fifth root. The order of the actual phase is therefore ten. Odd ordinary data admit only odd-order phases (Section~\ref{fam:Q:sec:2}). \end{proof} \paragraph{A test from the unresolved family.} At $2^6 3^6 5^4$ one surviving complementary crossing can have the normalized word $z=1+3\zeta_5$. This word has strict house below four, residue four over three and five, and tame trace one at five. On the alphabet $\mu_{30}$ the pure coefficient vector for $|z|^2$ is $(10,3,0,0,3)$. The minimum in Theorem~\ref{mass:thm:mass} occurs at $t=3$ and is thirteen: \[ |z|^2=7+3(-\zeta_5^2)+3(-\zeta_5^{-2}). \] The required adjoint mass is sixteen. Given \emph{any} $\omega\in\mu_{30}$, add $\omega(1+\zeta_3+\zeta_3^2)=0$ to the right. This gives a positive integral expression of mass sixteen containing $\omega$. Thus even the sharpest one-entry support bound at this alphabet retains every phase. Nevertheless Corollary~\ref{mass:cor:two} forces the indicated order-ten event once this two-position alternative is selected: after scaling, the crossing gives a full character word. It is a disjunction about specified products. The joint restrictions on the surviving modular data are developed in Section~\ref{por:frontier}. \subsection{Two phases of arbitrary relative order}\label{found:S:4} In this subsection exposure is taken among all roots at any common multiple of the conductors of the expressions compared. \begin{theorem}\label{found:S:4.1}\label{found:S:key:thm:two-phases} If a full-weight word has exactly two distinct phases at a pair of nonunits in a minimal ordinary or positive reduced table, their ratio has order $2,3,4,5$ or $6$. In odd ordinary data only order $3$ is possible. \end{theorem} \begin{proof} Write the ratio as $\rho$ and its order as $n$. We prove that for $n\ge7$ the set $D=\{1,\rho,\rho^{-1}\}$ is exposed at every common conductor. We may choose any primitive $n$th root for $\rho$ by Galois symmetry. We first give a construction that handles all but six orders at once. Split off a prime-power factor $s^b$ of the common conductor, writing each root as $\alpha y$, with $\alpha$ of $s$-power order and $y$ of order coprime to $s$. At a marked primary position write $y_i$ for the complementary component of its marked root. The three primary positions will always be distinct in this construction. Prescribe there the functional $f_i(y)=\Tr_{\rm norm}(y/y_i)$, normalized absolute trace on the complementary field. It takes the value $1$ on a root only when $y=y_i$, and lies in $[-1,1]$ at all roots. On each $\mu_s$-coset of primary positions with $k$ marks, put $-\sum f_i/(s-k)$ at each vacant position. These assignments define a rational functional by linear disjointness and the primary cyclotomic relation that the values sum to zero on each coset. If $s\mid n$ and $s\ge7$, then $k\le30$. At every root other than the former maximizer the old functional is at most $1/2$: for a root of order $m>1$ the normalized trace is $\mu(m)/\varphi(m)$. Since $|L|\le2$ on roots, the new maximum one occurs precisely at the prescribed phases. The nonidentity $r$th roots are linearly independent over the coprime cyclotomic field, so the prescribed phases are rationally independent. Applying the functional to the balancing expression forces every phase into the displayed set; independence fixes each mass. At most one phase cancels the $r$-part of $\theta_X\theta_Y$, and at most $k+1$ cancel its $q$-part, since each coset uses each position at most once. The remaining number is at least $r-3-k>0$, 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. It remains to pass from integral signed words to arbitrary positive expressions and possibly additional primes. We use the contraction and two-prime unimodularity of Theorem~\ref{first:balance} and Lemma~\ref{fam:Q:2.4}. In each complementary primary factor, the random integral functional fixes $1$, sends a uniformly chosen nonidentity root of prime order to $-1$, and sends the remaining positions to zero. These maps preserve both residues. A nontrivial complementary factor in a positive term has expected signed cost at most $c/(\ell-1)<1$. Its presence would give a contracted expression of cost below $g$. Linear optimization on the signed two-prime cyclotomic configuration has an integral minimum, contradicting the first part. Thus all original phases have only $q$- and $s$-parts. Their expression is a convex combination of integral optima, since the optimal face is bounded and its vertices are integral by the same unimodularity. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 $(e_i)$ 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 $U_p(1)e_i=\theta_{i,p}e_i$. 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 $s=S/\sqrt D$, $e_X$ 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 $E_A,E_B$ for the coordinate projectors. The two Weyl operators are unitary, so the invariance gives \[ E_A s E_B=W_I E_{A\cap B} W_J . \] 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 $W_pe_X$ 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 $W_pPe_X=\gamma_p^3Pe_X$. 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 $sPe_X=\gamma_p^3 BPe_X$ 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. Denote the resulting nonempty fiber and its coordinate projector by $E$. 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 $v=nE W_{p'}e_0$. 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: \[ m(1)-m_*= n\!\sum_{X\in E:\,\theta_{X,p}=1} d_X\,(W_{p'}e_0)_X . \] 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 $n(W_{p'}e_0)_X$ 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 $1$. \paragraph{Simultaneous orders and distinct twist-one orbits.} Apply the filtration simultaneously at all primes, assuming the dyadic normalization above when two occurs. 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. In fact 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 $V_{\boldsymbol j}=n^2\|\prod_p Q_{p,j_p}e_0\|^2$. 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. In the forward relation 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 $n E W_2e_0$. It is fixed by both upper Borels and by all tori, and its two local Weyls give the degree vector on $E$. The same array argument shows that \[ \#\{(b,c)\in\Z_{>0}^2:2^a3^b5^c\in\mathcal T\} \ \le\ \#\bigl(\{X:\ord(\theta_X)=2^a\}/\operatorname{Gal}\bigr). \] 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 $v_E=\sum_{X\in E}d_X e_X$ 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} \begin{proposition}\label{found:S:3.4}\label{found:S:key:prop:sexticorb} For $h=3$, the numerical alternative in Theorem~\ref{found:S:3.2} says that all divisors of $p^6q^B r^C d$ are WGT dimensions whenever $r>p^2$, $W(r,1+\lfloor p^2/q\rfloor)$ and $(r-1)(p-1)\ge p^2\max\{a:a\mid r-1,\ a\le p\}$. \end{proposition} \subsection{Two traces at an isolated sextic prime}\label{found:C:2} \label{found:C:key:sec:oddrectangle} Let $p,q,r$ be distinct odd primes, with $qp$ and $(s-p)(\min(p,q)-1)>p$, and $c_s=1$ otherwise. \begin{proposition}\label{found:C:2.1}\label{found:C:key:prop:sexticrect} An isolated sextic as above is impossible if $c_q+c_r\ge p^2$. In particular, for $pp^2$, the exponent at $r$ can also be arbitrary. \end{theorem} \begin{proof} For $rp^2$. The first alternative of Proposition~\ref{fam:Q:3.1} gives the forbidden word. The proof of Theorem~\ref{fam:Q:1.2} excludes its second alternative when $kp^2$, the two-residue argument of Theorem~\ref{found:S:3.2} excludes every complementary common $p$-weight $g\le p^2$: since $r\ge g+2$ it gives a full word in at most $1+\lfloor g/q\rfloor$ positions. Thus all opposite witnesses have full $p$-height three, and their defect-zero rows isolate $p$ at valuation six. This step works for arbitrary companion exponents. Now $e_r\ge p+1$ because $r-1>p^2$, and $e_q\ge1$, so $(p-1)(e_q+e_r)\ge(p-1)(p+2)\ge p^2$. Proposition~\ref{found:C:key:prop:sexticrect} now excludes the isolated table. \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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 $m+1q$ be primes and $sg$ occupied positions. Erase the $s$-parts instead and write $v$ for the resulting integer in the $q$-power field. Set $\alpha=(v-t)/s$, which is integral because this field is unramified at $s$. Its signed rational word length is at most $(g+t)/s<3$: subtract $t$ from the word for $v$ and divide all coefficients by $s$. On a single prime-power configuration the minimal signed length of an integral value is attained integrally. Indeed its coefficient vectors differ by constants on the prime-order cosets, and subtracting an integral median on each coset minimizes the sum of absolute values. Thus $\alpha$ is the sum of at most two signed $q$-power roots, the empty word being permitted. Let $M\in\{-2,-1,0,1,2\}$ be the sum of their signs. Reducing $v=t+s\alpha$ above $q$ shows that the signed augmentation is $t+sM$ modulo $q$. Since $g=t+s$, the sum of the coset constants of the preceding paragraph is $M-1$ modulo $q$. For $M=-2,-1,0$ the sole nonzero constant would be $-3,-2,-1$, respectively. At every one of at least $s-1$ positions this requires signed cost at least two, impossible. For $M=2$ the element $v$ is a positive expression of mass $t+2s$ on at most three $q$-positions. These are exposed among signed roots by the coset functional that is one at the $m\le3$ marked positions and $-m/(q-m)$ elsewhere; its absolute value on unmarked positions is strictly below one as $q\ge7$. The cost of the expression for $v$ would thus exceed $g$. Consequently $M=1$ and $\alpha$ is a single positive $q$-root $\eta$. Exposing $1,\eta$ forces precisely the announced unsigned word of length $g$. Since $M=1$, all coset constants vanish. For clarity, the passage to rational physical expressions is unchanged from Proposition~\ref{found:C:3.1}. Independent integral functionals on the complementary factors fix $1$, send a uniformly chosen nonidentity prime-order root to $-1$, and vanish at the other nontrivial roots. Any term with complementary support has expected signed cost at most $2/(5-1)<1$. The loss contradicts the signed integral minimum just proved, by two-prime unimodularity. All optimal integral expressions have at most $1+\lfloor g/q\rfloor$ primary $s$-positions including $1$. Two such supports together contain 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 Lemma~\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} For $p=5$, this covers every $r\in\{17,19,23\}$ with $11\le q7$. The numerical criterion holds in every case. \begin{corollary}\label{found:J:2.3}\label{found:J:key:cor:halfsquare} For odd primes $pp^2$, 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} \begin{ratioprop}[Two positive residues at arbitrary ratio]\label{prop:arbitrary-ratio-rigidity} Let $q3m+2$. 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. Its values lie in $[-1/4,1]$ because $k\le m+2\le q/5$. Applied to the two expressions for $v$ it gives \[ t+sL\le P+N/4\le g-7N/4 . \] Put $d=m-L\ge0$. If some $a_C$ is negative, write $a_C=-b_C$ on these cosets, put $B=\sum b_C$ and $b=\max b_C$. Then $B\ge d$ and $2b\le m+1$. At each of at least $s-1$ positions of such a coset there are at least $b_C$ negative letters, or there are at least $q-b_C$ positive letters. The number of positions where the first bound fails is at most $g/(q-b)$. Consequently \[ N\ge B(s-1)-\frac{bg}{q-b} > Bs\left(1-\frac{1}{s}-\frac{m+1}{q-b}\right)>\frac47 Bs , \] using $g<(m+1)s$, $b\le B$, $1/s<1/15$ and $(m+1)/(q-b)<2/9$. This contradicts $N\le 4sd/7$. Hence every $a_C\ge0$. 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. To pass to the original rational expression, split off all primary factors other than $q,s$ as in Proposition~\ref{found:C:3.1}. In each such $u$-power factor use the integral functional which fixes $1$, takes value $-1$ on one uniformly chosen nonidentity $u$th root, and vanishes on the remaining roots. Tensor these independent functionals with the identity on the retained factors. The values on each $\mu_u$-coset sum to zero. The contraction fixes $g$ and preserves the two radical congruences: the radical at either retained prime is generated by its primary cyclotomic uniformizer. A term with a nontrivial complementary $u$-part has expected signed cost at most $2/(u-1)\le1/2$. Its presence would give some integral contraction target with a rational signed expression of cost less than $g$. The two-prime cyclotomic configuration is unimodular \cite[Theorems~3 and~13]{BH}. In an integral basis of its column lattice, every nonsingular maximal minor has determinant $\pm1$; this property persists on adjoining the negative columns. A basic optimum for an integral target therefore has integral coefficients, contrary to the preceding calculation. Only the $q,s$-factors remain. The cost-$g$ face of the polyhedron of signed representations of $z$ is a bounded optimal face. By the same unimodularity its vertices are integral, so the rational expression is a convex combination of integral 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 $1+2\lfloor g/q\rfloor9$: the height-one auxiliary pure words are available since $(s-3)(3-1)>3$ for $s=5,7$. At weight nine repeat the argument of Lemma~\ref{found:J:key:lem:near-updated} with negative cost $7/4$. This keeps the contraction at the spectator prime three strict. The signed residue $-3,-2$ or $-1$ modulo five costs at least $7/4$ at each of six positions, already more than nine. For $M=2$ the functional on at most three marked positions in a five-coset has negative magnitude at most $3/2<7/4$, so this case also fails. The rest of that proof gives two $5$-parts of mass $2,7$ and two $7$-parts of mass $4,5$, with $1$ the first position in each coordinate. 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} Let $P,Q,R$ be distinct odd primes with $QP^2$. \begin{theorem}\label{found:J:3.1}\label{found:J:key:thm:middle} If $W(R,1+\lfloor P^2/Q\rfloor)$ holds, then every divisor of $P^6 Q^B R^C d$ is a WGT dimension for arbitrary $B,C$ and coprime square-free $d$. The exponent six can thus occur at the middle prime: there is no ordering hypothesis on $P$ and $Q$. \end{theorem} \begin{proof} For the two principal blocks at $Q,R$, the complementary-pair argument in Theorem~\ref{found:S:3.2} at $h=3$ isolates $P$ at valuation six, since $R>P^2$ and $W(R,1+\lfloor P^2/Q\rfloor)$ holds. The repeated odd primes permit ordinary data by Proposition~\ref{three:odd-reduced}. In Proposition~\ref{found:C:2.1}, $R>2P$ is eligible and $e_R^+\ge e_R\ge P+1$, since $R-1>P^2$. The other companion costs at least one. Thus $c_R+c_Q\ge(P-1)(P+1)+1=P^2$. \end{proof} For example, for any fixed distinct odd $P,Q$, the theorem 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{An all-$r$ inequality.} Let $pp^2$ both companion exponents are arbitrary. In particular the all-$r$ statement holds whenever $\sqrt2\,q^3\ge(p^2+2q)^2$; for every fixed $\alpha>1/\sqrt2$ it also holds throughout $q^3\ge\alpha p^4$ once $q$ is sufficiently large in terms of $\alpha$. Indeed put $j=\lfloor\min(r,p^2)/q\rfloor\le J$. The case $j=0$ is strictness and $j=1$ follows from the two-position theorem since $r\ge7$. For $j\ge2$, $(j+1)^3/j<(j+2)^2\le(J+2)^2\le\sqrt2\,q$, and $jq0$, 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 $k(k-1)q$ and every $B$. In particular, for any $c>1/\sqrt2$ and all sufficiently large $p$, it is enough that $q^3\ge c p^4$; the choice $c=4/5$ works for $p\ge125$. This uniform estimate is already implicit in the exact scalar test, whereas the preceding quotient test eliminates witnesses that satisfy that test. If $q\ge p^2$ then $j=\lfloor\min(r,p^2)/q\rfloor=0$ and strictness suffices. Otherwise set $x=\min(r,p^2)>q$. The function $(x/q+1)^3/x$ is increasing for $x\ge q$. The displayed prime condition gives $(j+1)^3\le (x/q+1)^3\le\sqrt2\, x\le\sqrt2\,r$. Now apply Theorem~\ref{found:C:2.2}. The asserted asymptotic form follows by dividing the cube root of the prime condition by $p^{2/3}$. For the explicit constant, $(5/4)^{1/3}<27/25$ and $28/25<2^{1/6}$. \subsection{A small intersection at the largest quartic prime}\label{found:S:8} \label{found:S:key:sec:minword} \begin{proposition}\label{found:S:8.1}\label{found:S:key:prop:sevenword} At every prime $s\ge173$, a full positive $s$-primary word of a nonunit label at a nonunit row cannot have at most seven positions, in either type of minimal table. \end{proposition} \begin{proof} Write $H=(s-1)/2$, $g(v)=\min\{t\ge1:t(t+1)\ge v\}$. Among all full positive $s$-primary words of nonunit labels on the given row, take one minimizing the number $h$ of nonzero opposite pairs in its difference set $D$. We allow overall root-of-unity factors. The initial word gives $h\le21$. Since $43193$ is impossible. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 $T$. \end{proof} \begin{corollary}\label{found:S:8.2}\label{found:S:key:cor:sevenstrip} For arbitrary nonnegative $A,B$, all divisors of $p^Aq^Br^4d$ are WGT dimensions whenever $11\le p6q\ge174$ gives $r\ge179$, and $1+\lfloor r/q\rfloor=72$, 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 $K e_V=\widehat s e_V$ 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 $K e_V=\widehat s e_V$. 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 $V\in E$, $We_V=se_V$ 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 $a_X=(e_X+e_{gX})/\sqrt2$, $b_X=(e_X-e_{gX})/\sqrt2$, 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. \paragraph{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 $s$. 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. \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 the preceding 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. \paragraph{A spectrally pure product.} 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. 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}. The two ordinary weight-four phase profiles therefore have 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 $30$-twist, 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, Section~\ref{found:S:6} makes 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 \[ 135000\mid L_{\{2,3\}},\qquad 145800\mid L_{\{2,5\}},\qquad 43200\mid L_{\{3,5\}} . \] \subsection{A quartic field at twist one without a separation hypothesis}\label{found:F:2} \begin{theorem}\label{found:F:2.1}\label{found:F:key:thm:fields} At $D=29160000$ the ordinary minimal table has twist-one labels $V_3\in P_3$ and $V_5\in P_5$ in the same full two-block whose fields both contain one of \[ \QQ(\zeta_5),\qquad \QQ(\zeta_{15})^+ . \] If no twist has order exactly $30$, the witnesses can be chosen with the first field. In particular each principal block contains at least four nonunit simples of twist one. The two chosen labels may coincide. \end{theorem} \begin{proof} Choose complementary pure-twist principal witnesses $X\in P_3$, $U\in P_5$, of degrees $2^i3^j$, $2^k5$. Here pure allows twist one. Use Proposition~\ref{prop:principal-support} at the odd endpoint of a dyadic edge and avoid the other odd prime. This gives dyadic height two at that witness by Section~\ref{found:S:6}. Use the same proposition at an active column for the other odd prime. Both dyadic heights are at least two, so the common degree factor is four. Put $z=S_{XU}/\lcm(d_X,d_U)$. By Section~\ref{found:S:5} its unsigned form is $1+3\xi$ with $\xi$ a primitive fifth root. Its signed form is \[ -\sum_{a=1}^4\eta_a\zeta_5^a,\qquad \{\eta_a\}=\{1,1,\omega,\omega^2\},\quad \ord(\omega)=3. \] The exclusion of a wild cubic coset in this formula is precisely Proposition~\ref{found:S:5.2}. A nonzero Fourier coefficient for a Galois character whose kernel is preserved upon inversion of either the three- or five-component puts the corresponding cyclic field into twist-one witnesses in the same full two-block (Proposition~\ref{found:J:6.2}). In the unsigned case use a faithful quartic character $\psi$ at five. Its coefficient is a nonzero multiple of the primitive Gauss sum of $\psi$. In the signed case let $i,j$ be the two exponents carrying $\omega,\omega^2$. Take the product of $\psi$ and the nontrivial character at three. Antisymmetrizing at three leaves a nonzero scalar times $\zeta_5^i-\zeta_5^j$. The five Fourier coefficient is a nonzero Gauss sum times $\overline\psi(i)-\overline\psi(j)\ne0$. On $(\ZZ/15)^\times$ the product character has kernel generated by simultaneous complex conjugation, unchanged by either local inversion. Its fixed field is $\QQ(\zeta_{15})^+$. It remains to see which orders the signed case produces. The four signed terms are actual monodromy phases with equal total dimension $\lcm(d_X,d_U)$ at each phase (Proposition~\ref{found:S:5.1}). Both inputs have trivial dyadic twist. At most one of the four terms cancels the five-twist product, and at most two cancel the three-twist product. Thus some constituent is active at all three primes. Both dyadic edges now occur. By the two degree transfers (Section~\ref{found:S:6}), every three-active degree is five-bearing and every five-active degree three-bearing. Hence the five-free pure ternary $X$ and the three-free pure quinary $U$ both have twist one. The two marked phases with $\eta_a\ne1$ are therefore actual twists of order exactly $30$, and the other two have order $10$. More generally replace one input by each of its Galois conjugates; it still has twist one, and the normalized entry is conjugated by the same automorphism. The uniqueness of the physical monodromy packets now realizes all eight primitive thirtieth roots and all four primitive tenth roots. Finally the orbit of a label whose field contains a quartic field has size at least four, and the rational vacuum is outside it. \end{proof} 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 $X,U$ in the preceding proof. 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 jointly three- and five-active $J$ take $Y\in P_3$ of odd degree $15$ or $45$ by the refined prime avoidance of Section~\ref{found:C:6}. It ensures that $[K_Y:F]$ and $\operatorname{Tr}_{F/\QQ}(x)$ are odd. A weight-five crossing puts one of $\QQ(\zeta_5)$, $\QQ(\zeta_{15})^+$ in $K_Y$ (Section~\ref{sec:dyadic-fields}), so that field lies in $F_{(2)}$. The whole two-block of $J$ has that field and the four tame five-conjugates of the block are distinct: five-inertia has image of order four on either quartic field. In particular none of these blocks is principal, and every label in them has five-height at most one, since a five-unit codegree would have five-unramified character field (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. Take the odd $Y\in P_3$ in its pure ternary fiber supplied by the avoided-two trace selection in Section~\ref{found:C:6}. For $x=\chi_J(Y)$ and $F=\mathbb Q(x)$, both $[K_Y:F]$ and $\operatorname{Tr}_{F/\mathbb Q}(x)$ are odd. The degree of $Y$ is $15$ or $45$, and the quartic-field calculation of Section~\ref{sec:dyadic-fields} puts one of the displayed cyclic quartic fields $K$ in $K_Y$, the first in the absence of a $30$-multiple. Then $K\subset F$, by the odd index. The algebraic integer $\operatorname{Tr}_{F/K}(x)$ cannot lie in 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 $a=\frac12\sum_{X\in A} e_X$ 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} Let $s=S/n$ be the integral square-dimensional normalization, $D=n^2$. 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 $E_0$ for the projector onto $p$-inactive labels. On the $\chi$-isotypic space, \[ \|E_0s E_0\|\le p^{-1/2}. \] 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. 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. 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{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 Section~\ref{found:C:6} 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. Its length is divisible by eleven by the tame $23$-bound (the only possibilities are $11,22$), and even by odd-dimensional duality. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 orbit size is $11$ 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 Section~\ref{found:C:6}, 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 Section~\ref{found:C:6}, 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} Consider the branch of Section~\ref{found:T:10} at $D=5^67^623^4$, with five and seven separated and the complementary principal rows unramified at seven. Put $r=23$, $n=\sqrt D$. Every seven-active label $H$ then has \[ v_{23}(d_H)=1 . \] Thus the $23$-free possibility is excluded even for a pure seven-twist. At each such $H$, Lemma~\ref{found:T:key:lem:separated-double-principal} supplies a twist-one $23$-free hit in $P_7\cap P_{23}$. Moreover $W_7$ has integral entries at every prime above $23$. \begin{proof} Full $23$-height was excluded in Section~\ref{found:T:10}. Suppose $r\nmid d_H$. Separation and the absence of prime-power degrees give $d_H=125\cdot7^j$, $j=1,2$. Take a five-free nonunit $Y\in P_7$ in the pure seven fiber meeting $H$; its twist is one, since seven-activity would require full five-height. Write $d_Y=49r^a$, $a=1,2$, by the orientation result (Section~\ref{found:D:6}). The lcm-normalized crossing $z$ is unramified at seven in $K_Y$, and $\chi_Y(H)=125z\equiv0$ at every prime over seven, so $z\in7\mathcal O$. At $j=1$ its house is strictly below seven, impossible. At $a=2$ the entry $S_{YH}=n(z/7)$ already costs at least $n^2$ in the Galois-averaged row norm. Thus $j=2,a=1$. Write $S_{YH}=(n/r)b$. Then $b$ is a nonzero integer of an $r$-power cyclotomic field since $K_H$ is unramified at five and $K_Y$ at seven. Any such pure $Y$ of degree $49r$ has $K_Y\supseteq\mathbb Q(\zeta_r)$ by its crossing with the seven-free degree-$25r$ row of $P_r$ (Sections~\ref{found:J:8} and~\ref{found:T:10}). Its field ramifies at five by the orientation in Section~\ref{found:D:6}. Thus $m_Y=[K_Y:\mathbb Q]$ is a multiple of $22$ and at least $44$. Also $6\mid m_H$ by the seven-twist square action and odd duality. We claim $b\in\ZZ$. Otherwise let $r^c$ be its conductor at $r$, and express $b=\sum_j d_j\zeta_{r^c}^j$, $d_j\in\ZZ$. If $b$ lies outside the quadratic subfield there is a character $\psi$ of exact conductor $r^c$, with $\psi^2$ of the same conductor, such that its Fourier coefficient is nonzero: this follows by Fourier inversion and minimality of $c$ (at $c=1$ omit the two characters of order at most two). This character descends to $\mathbb Q(b)$ and hence gives orbit vectors on both inputs. Their normalized $s$-matrix coefficient has absolute value \[ \frac{\sqrt{m_Ym_H}}{r\varphi(r^c)}\, r^{c/2} \left|\sum_{(j,r)=1}d_j\overline{\psi(j)}\right|. \] This is the primitive Gauss-sum formula applied to the orbit matrix as in Proposition~\ref{found:J:6.1}. Both inputs are $r$-inactive, by Section~\ref{found:T:10} for the $r$-free $H$ (all jointly seven--$r$-active degrees there are $125\cdot7r$). Proposition~\ref{por:prop:wild} bounds the coefficient by $r^{-c/2}$. Thus $m_Ym_H|\sum d_j\overline{\psi(j)}|^2\le (r-1)^2=484$, also for every conjugate character. The nonzero sum is an algebraic integer. But $m_H$ is divisible by eleven if $c=1$, and by $23$ if $c>1$, in addition to six, a contradiction. If $b=(x+y\sqrt{-23})/2$ instead, the quadratic orbit coefficient has absolute value $\sqrt{m_Ym_H}\,|y|/(2\sqrt{23})$. The tame inactive compression (Section~\ref{found:P:2}) applies also to the quadratic character, giving $m_Ym_H y^2\le4$. Thus $b$ is rational. Let $v_Y$ be the unit vector constant on the orbit of $Y$ and zero off it. The squared mass of $s v_Y$ on the $H$-orbit is $m_Ym_H b^2/529<1$, since the vacuum also contributes. Thus $b=\pm1$. We have $22\mid m_Y$, $m_Y\ge44$ and $6\mid m_H$; moreover $3\nmid m_Y$, since $K_Y$ has conductor supported at five and $23$. The possible pairs $(m_Y,m_H)$ are $(44,6),(44,12),(88,6)$. To exclude the two pairs with product $528$, retain also the disjoint orbit of the twist-one $U\in P_5$ from Corollary~\ref{found:T:key:cor:odd-one-degree} and Proposition~\ref{found:D:key:prop:odd-oriented-path}, of degree $25r$ and size $m_U\ge44$. By Lemma~\ref{found:D:key:lem:unramified-orientation} the lcm-normalized entry at $U,Y$ has the unique weight-$23$ expression $z=2+7\sum_{j=1}^3\eta_j$, with $\eta_j$ of five-power order. Wild five-inertia fixes $K_Y$ since $5\nmid m_Y$. It permutes the three letters by uniqueness, so fixes each: its orbits on roots of order $5^a>5$ have size $5^{a-1}$. If $t$ letters are trivial, the normalized Galois trace of $z$ is $(35t-13)/4$, of absolute value at least $13/4$. Since $n/\lcm(d_Y,d_U)=805$, averaging the $Y$-orbit on the $U$-orbit gives squared mass at least $(44\cdot13/(4\cdot805))^2>1/529$ (use $44\cdot13/4=143>805/23$). This and $528/529$ exceed one. Consequently $m_H=6$ and $m_Y=44$. The seven-twist of $H$ has order exactly seven. The two required subfields determine $K_Y=\mathbb Q(\zeta_{23},\sqrt5)$. Such a $Y$ crosses each seven-free nonunit $U\in P_5$ in the same $9+7(\eta+\eta^{-1})$, up to Galois conjugation, as in Section~\ref{found:T:10}: the proof uses the $5$-primary word on three letters and the displayed quadratic five-part of $K_Y$. In particular for the fixed $U$-orbit of size at least $44$ the mean normalized crossing is $11/2$. Here $d_U=25r$ and $n/\operatorname{lcm}(d_U,d_Y)=805$. The averaged $U$-row bounds the number $B$ of these $Y$-orbits by $44^2 B(11/2)^2<805^2$, so $B\le11$. For each $H$-orbit at most two of these $Y$-orbits meet it, since three would cost $3\cdot44\cdot6>529$ in the averaged $H$-row. The double-principal residue at five in Section~\ref{found:T:10} counts only those lying in $P_5$. Each contributes $\pm44\cdot(49r)\cdot(7r)\equiv\pm3\pmod5$, since $\chi_H(Y)=\pm7r$. One orbit cannot give $-1$ modulo five. Thus precisely two meet $H$, both double-principal. Conversely, a fixed $Y$-orbit can meet at most two such $H$-orbits by the same norm bound. Hence there are at most eleven $H$-orbits. Finally the singleton Gauss sum $L_7=\sum_Z d_Z^2c_7(Z)$ of Section~\ref{found:C:6} is divisible by $23$. The terms at $23$-bearing labels are zero modulo $23$ (the denominator of $c_7$ divides six). Each orbit of $23$-free seven-active labels would contribute exactly $7(125\cdot49)^2$ because its size is six and its seven-twist has order seven. Their number is therefore a multiple of $23$, hence zero. We have proved $v_{23}(d_H)=1$ for every seven-active $H$. This gives the divisibility hypothesis of Lemma~\ref{found:T:key:lem:separated-double-principal} with $(p,q)=(7,23)$. It also gives $v_{23}(f_H)=v_{23}(D)-2v_{23}(d_H)=4-2=2$. Apply the all-primes conclusion of Theorem~\ref{fam:W:n:thm:foreign-column} with $(q,p)=(7,23)$. \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} We first exclude seven-free seven-active labels. They have degree $125r$ by the preceding result and are pure seven-active by Section~\ref{found:T:10}, since all joints have degree $125\cdot7r$. Fix the orbit of $44$ labels $Y$ of degree $49r$ from a joint column as in Section~\ref{found:T:10}. Each hits every such $H$ by seven-principality. Write $S_{YH}=n z/(7r)$. The lcm-normalized crossing $z$ lies in $\QQ(\zeta_r)$: $K_Y=\QQ(\zeta_r,\sqrt5)$ and $K_H$ is unramified at five. It satisfies $z\equiv r\equiv2$ modulo seven by the seven-principality of $Y$. Thus $z=2+7y$ for a cyclotomic integer $y$. Every nontrivial $r$-Fourier coefficient of $z/(7r)$ equals the corresponding coefficient of $y/r$. Both labels are $r$-inactive. The same estimate for inactive orbit vectors gives $y\in\ZZ$: a nonquadratic $y$ forces $66\mid m_H$, making $m_Ym_H>484$, and the quadratic coefficient would require $m_Ym_H\le4$. Therefore $z$ is a rational integer congruent to two modulo seven. Averaging the $44$ rows on this set bounds its size 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. Now every seven-active degree is $125\cdot7^hr$, $h=1,2$. At each pure seven-active $H$ repeat the five-free pure principal-hit analysis above. A hit $Y$ of $r$-height two would again give $S_{YH}/n$ nonzero integral at full lcm scale: the common crossing is divisible by seven because $Y\in P_7$ and $\chi_Y(H)=125z$. Thus $d_Y=49r$, and $b=S_{YH}/(n/r)$ is nonzero integral in the $r$-primary field. The same inactive compression makes $b$ rational, and the two disjoint orbit estimates above again give $b=\pm1$, $m_H=6$ and $m_Y=44$. Indeed $6\mid m_H$ by seven-activity and duality, and the $Y$-orbit and the fixed $U$-orbit have the same degree and field constraints used there. Thus $K_Y=\QQ(\zeta_r,\sqrt5)$ and the seven-twist has exact order seven. In the double-principal residue modulo five each contributing $Y$-orbit now gives \[ \pm44\cdot49r\cdot7^{3-h} \equiv\begin{cases}\pm2,&h=1,\\ \pm1,&h=2.\end{cases} \] At $h=1$ at least two contributing orbits are needed and at $h=2$ at least one. Let $A_1,A_2$ count the pure seven-active orbits at the respective heights and $B$ the joint orbits. Every joint has height one, orbit size $66$, and a double-principal pure $Y$-orbit hitting it, with $S_{YJ}/(n/r)$ a signed primitive $r$th root by Section~\ref{found:T:10}. All $Y$-orbits used for the two kinds of columns together number at most eleven: every one has the field and degree giving the common mean $11/2$ against the fixed $U$-orbit, so the same $805^2$ bound applies to their union. For each $Y$ there are at most two incident pure orbits by the row norm and at most two joint orbits by the primitive $r$-phase bound (Section~\ref{found:T:10}). If two pure orbits are incident, the unit orbit vector constant on $Y$ has, after $s$, squared mass $2\cdot44\cdot6/529=528/529$ on them. Each incident joint would use an additional $44\cdot66/(22^2\cdot529)=6/529$, since the signed primitive $23$rd root has Galois mean $\mp1/22$. Therefore the number of incident pure orbits plus half the incident joint count is at most two for each $Y$. Counting these incidences gives \[ 2 A_1+A_2+B/2\le22. \] 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 Section~\ref{found:C:6}. 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 $se_0=(d_Y/n)_Y$ and the complementary Weyl is unitary and commutes with $P_\alpha$, we have \[ \frac{M_\alpha}{D}=\|P_\alpha W_7 e_0\|^2 =\sum_{\theta_{Y,7}=\alpha}(W_7)_{Y0}\,\overline{(W_7)_{Y0}} . \] 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 Section~\ref{found:C:6} 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. \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}. \paragraph{Codegree nine.} Here $d_Z=1800$ and the support lies in $P_3$. Any supported degree exceeding $600$ must have three-height exactly two and contain all three primes: the other possible heights give degrees at most $8\cdot3\cdot25=600$, and the prime-omitting bounds are at most $225,200,72$ in $P_3$. At such a position, $b_U=(d_U/d_Z)\chi_U(Z)$ vanishes modulo the prime over three, since $d_U/d_Z$ is a local unit there and $U\in P_3$. Its orbit costs at least six in squared mass in addition to the unit and the three disjoint prime-omitting orbits. This exceeds nine. All supported degrees are therefore at most $600$. Since the unit has degree one, the modular relation gives $5400<600\cdot9$ if the row exists, a contradiction. \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 largest five-free principal degree on the path missing six}\label{found:T:4} Suppose $D=n^2=2^6 3^{2e}5^4$, $e\ge3$, in an ordinary minimal table, and no twist order is divisible by six. The two--five and three--five pairs both occur. Every three-active degree is divisible by $40$, and every two-active degree has full three-height. These are the separation and degree-transfer results of Section~\ref{found:S:6}. \begin{proposition}\label{found:T:4.1}\label{found:T:key:prop:missing-six-upper} Suppose $P_3$ contains a five-inactive $X$ of degree $8\cdot3^{e-1}$ ($72$ at the first core). All such labels form a single Galois orbit, with $K_X=\QQ(\zeta_5)$. Every degree-$20$ label of $P_5$ has the same field and twist of order exactly five. There are between two and four such orbits. Every jointly two--five-active $K$ has five-component of order exactly five and meets a degree-$20$ label of $P_5$ at each of the four primitive fifth phases. \end{proposition} \begin{proof} A jointly two--five-active column supplies $U\in P_5$ of degree $20$; fix any such degree-$20$ label. Its twist is purely quinary by the two degree transfers above. At a joint three--five column choose an odd $V\in P_3$ in the pure ternary fiber of degree $5\cdot3^k$, $1\le k0$, outside the support of $a$. This excludes the second label. \end{proof} \subsection{Character fields inside a signed fusion product}\label{found:T:6} Suppose an ordinary minimal table at $n=5400$ has all three twist pairs. Take the twist-one principal witnesses $X\in P_3$, $U\in P_5$ of degrees $4\cdot3^j$, $20$, with $j=1,2$, as in Section~\ref{found:F:2}. Consider a signed crossing \[ z=\frac{S_{XU}}{L} =-\sum_{a=1}^4\eta_a\zeta^a,\qquad L=20\cdot3^j,\qquad \{\eta_a\}=\{1,1,\omega,\omega^2\}, \] where $\zeta,\omega$ have orders five and three. Write $k,l$ for the two positions with coefficients $\omega,\omega^2$ respectively. For a label $A$ let $K_A$ be its character field. Put $F=\mathbb Q(z)$ and $h_A=[K_A:F]$ for $A=X,U$. \begin{proposition}\label{found:T:6.1} If $k=-l\pmod5$, then $9^{j-1}h_Xh_U\le36$; otherwise $9^{j-1}h_Xh_U\le18$. In particular, for $d_X=36$ both fields are tame at five. In the nonopposite case their compositum is at most quadratic over $\mathbb Q(\zeta_{15})$ and tame at three as well. \end{proposition} \begin{proof} Take the product $\chi$ of a faithful quartic character $\psi$ at five and the quadratic character at three. The sum of $\chi(g)g(z)$ over $g\in\operatorname{Gal}(\mathbb Q(\zeta_{15})/\mathbb Q)$ has squared modulus \[ 3\cdot5\,|\overline\psi(k)-\overline\psi(l)|^2 . \] Indeed ternary antisymmetrization gives $-(\omega-\omega^2)(\zeta^k-\zeta^l)$, and the remaining sum is a quartic Gauss sum times the difference of the character values. This gives $60$ at opposite positions and $30$ at nonopposite positions. In the former case $F=\mathbb Q(\zeta_{15})^+$ and in the latter $F=\mathbb Q(\zeta_{15})$ (Section~\ref{found:F:2}). The character is trivial on simultaneous complex conjugation, so the sum over $\operatorname{Gal}(F/\mathbb Q)$ has squared modulus $15$ and $30$, respectively. Since $X,U$ are both inactive at three and five, the two-prime local Weyl bound in Section~\ref{found:P:2} bounds $L^2 h_Xh_U$ times this squared modulus by $n^2/15$. Substituting $L$ yields the inequalities. At $j=2$ the relative-index products are at most four and two respectively, giving the field assertions. \end{proof} Here the Fourier estimate can also control the constituents of the tensor product. We give the underlying orbit identity for arbitrary primes. For simples $A,B$ in ordinary square-dimensional integral modular data, let $E$ be an abelian field in the splitting field containing $K_A,K_B$ and a twist value $t$. If the total degree of the constituents of $AB$ of twist $t$ is $M_t$, then \[ M_t=\sum_V N_{AB}^V d_V [EK_V:E], \] taking one representative per orbit under $\operatorname{Gal}(\overline{\mathbb Q}/E)$ among those constituents. In fact in integral square dimension the Galois permutation is a based ring automorphism: applying $\sigma$ to each character identity permutes the arguments without signs and preserves degrees. Automorphisms fixing the inputs therefore preserve all multiplicities. The square Galois action preserves the indicated twist when $t$ belongs to $E$ (in the normalized lift the anomaly correction is trivial, since the two-primary scalar has order dividing eight and every odd square equals one modulo eight). The orbit length is the stated field index. In particular, if all degrees in the packet are multiples of $g$ and a prime $p$ does not divide $M_t/g$, some constituent has $p\nmid (d_V/g)[EK_V:E]$. For the signed crossing, each of its four phases has $M_t=L$ by uniqueness of the balancing packets, and conjugating one input realizes every primitive thirtieth root (Section~\ref{found:F:2}). These conjugate products have the same input fields. At order $30$ all constituents have degrees $2^i3^a5$ with $0\le i\le2$, $1\le a\le2$, by the two principal degree transfers. Consequently: \begin{itemize} \item If $d_X=12$, each of the eight primitive thirtieth roots occurs on a simple of degree $15$, $30$ or $60$. Indeed its packet degree divided by $15$ is four, and at least one quotient degree is prime to three. The tame fifth-inertia involution moves each jointly three- and five-active label and preserves its twist and degree (Section~\ref{found:C:6}). Thus each phase has at least two labels of one of these three degrees. \item If $d_X=36$, each primitive thirtieth root occurs on a simple whose character field is tame at five. Take $E=K_XK_U\mathbb Q(\zeta_{15})$, which is tame at five by the proposition, and divide the orbit identity by $15$. The resulting packet mass is $12$, a five-unit. Moreover any constituent of degree at least $45$ is tame at five since $[EK_V:E]\le180/d_V<5$. In the nonopposite case all constituents of degree at least $90$ are likewise tame at three, as their indices over $E$ are at most two. A degree-$180$ constituent has its entire field inside $E$. \end{itemize} \subsection{Five-free hits and degree-forty-five traces}\label{found:T:7} At $n=5400$ assume that every pair of dimension primes occurs in a twist. At a jointly two- and three-active column pick $X\in P_3$ by five-avoidance in the pure ternary fiber, with $K_X$ tame at five (Section~\ref{found:C:6}). It has twist one and degree $4\cdot3^j$, $j=1,2$ (Section~\ref{found:S:6} and Section~\ref{found:F:2}). Write $\tau$ for the normalized trace to $\mathbb Q$. The five-local fusion-ring element \[ B=\frac8{[K_X:\mathbb Q]}\sum_{X'\sim X}[X'] \] has integer eigenvalues $b_Y=8\tau(\chi_Y(X))$ with $b_0=32\cdot3^j$. Indeed the conductor primes are two, three, five, and tameness at five makes the orbit size a five-unit. The normalized traces of primary roots have denominators dividing $1,2,4$ respectively. Principality gives $b_Y\equiv b_0\pmod5$ on $P_5$. Put $u_Y=(b_0-b_Y)/5$ there. \paragraph{Five-free witnesses.} Let $J$ be jointly three- and five-active. For every prime above five, either some five-free nonunit $Y\in P_5$ with $\chi_J(Y)$ a unit there has $b_Y\equiv b_0\pmod{25}$, or such hits attain at least three distinct nonzero residues $u_Y\bmod5$. The possible degrees of these hits are $12,18,24,36,72$: perfectness rules out prime-power degrees, three-activity rules out full three-height and Theorem~\ref{found:D:1.1} excludes degree six. For the proof, the vacuum coefficient of $E_{P_5}[J](b_0-B)^a$ is five-integral. Dividing the resulting numerator by $d_J5^a$ and using $d_Y^2\chi_Y(J)=d_Jd_Y\chi_J(Y)$ gives \[ \sum_{Y\in P_5} d_Y\chi_J(Y)u_Y^a\equiv0\pmod{\mathfrak p}, \qquad 0\le a\le2,\quad\mathfrak p\mid5, \] since $v_5(d_J)=1$ and $v_5(n^2)=4$. A polynomial of degree at most two therefore cannot vanish on all contributing nonvacuum residues and be nonzero at zero. Five-bearing degrees drop out after reduction. More generally the same argument gives at least $v_p(D)-v_p(d_J)$ distinct nonzero residues, unless zero is realized at a nonunit $p$-free hit, for any $p$-local element with rational integral eigenvalues congruent to the vacuum on $P_p$. In particular, if all these five-free unit hits have rational $x=\chi_Y(X)$, at least two distinct ones have ternary height two. Indeed write $t=3^{j-1}$, so $d_X=12t$. For $j=2$ the integer $x$ is divisible by three by lcm integrality and three-principality at $X$: if $v_3(d_Y)=1$, the identity $\chi_X(Y)=(d_Y/d_X)x\equiv d_Y$ even gives $9\mid x$, and if $v_3(d_Y)=2$ it gives $3\mid x$. Always $x/t$ is an integer, is congruent to $12$ modulo five and has absolute value below $12$. Hence $x/t\in\{-8,-3,2,7\}$, giving four nonzero residues for $u_Y=8(12t-x)/5$ modulo five. At $v_3(d_Y)=1$ three-principality gives $3\mid x/t$ also when $j=1$, leaving only $x/t=-3$. At least two of the three prescribed residues must therefore occur at the higher ternary height. The rest of this section determines the possible rational traces on the known odd degree-$45$ rows of $P_5$ when no twist is divisible by $30$. This is a constraint on the spectrum even when many five-free rows meet $J$. \paragraph{The first prime below twice the weight.} Suppose ordinary square modular data have prime support $\{2,q,r\}$ with $q,r$ distinct odd primes and $r\ge5$. For two nonunits $A,B$ in a perfect fusion-simple table set $L=\operatorname{lcm}(d_A,d_B)$ and $z=S_{AB}/L$. Assume \[ g=\gcd(d_A,d_B)=(r+1)/2,\qquad z\equiv g\pmod {\mathfrak r}\quad(\mathfrak r\mid r), \] and that $z$ is tame at $r$. If $q\ge5$, then $v_r(g-\tau(z))=1$. The same holds for $q=3$ when no twist is divisible by $6r$, $v_2(L)0$. Their $q$-traces vanish. The identity-coset component $A$ is fixed by the tame action, hence belongs to $\mathbb Q(\zeta_r)$. Averaging its signed expression at the tame $q$-factor and using the one-prime integer-word minimum gives a pure $r$-word for $A$, of length at most $g-(q-1)m$. Consider one nonidentity coset. There is nothing to prove if $m\ge r$. Otherwise write $h_a$ for the signed sum of its $q$-letters in position $a\in\mathbb F_r$ of a minimum word. On erasing $r$-parts the coset has zero residue modulo $r$: the rational congruence holds above every prime, and erasure puts the congruence in the $q$-cyclotomic ring, unramified at $r$, where the different cosets remain independent. The $q$ integer coefficients of $\sum_a h_a$ on this coset are equal modulo $r$. Their pairwise differences have absolute value at most $m7035$. \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} \paragraph{Where all odd degrees must lie if a quadratic field occurs at $675$.} Suppose a degree-$675$ label $Z$ has quadratic field $\QQ(\sqrt\delta)$, where $\delta=-1,2$ or $-2$. Then every odd-degree label belongs to $P_2$. In particular every jointly three- and five-active label is even by the nonprincipal two-block result (Section~\ref{found:F:2}). There are at least four degree-$675$ labels and every even degree in $P_2$ has dyadic height one. Indeed write $b_Y=x_Y+\sqrt\delta\,y_Y$ with rational integers $x_Y,y_Y$. Each is divisible by $2^{v_2(d_Y)}$ by lcm integrality. Orthogonality of the two conjugate rows gives \[ \sum x_Y^2=32,\qquad \sum y_Y^2=32/|\delta|. \] At twist order at most eight the twists of $Z$ and its conjugate coincide. Adding their vacuum modular identities gives $5400\le\sum d_Y|x_Y|$. For nonzero $x_Y$ the ratio $d_Y/|x_Y|$ is at most $225$ unless $d_Y=675$ and $|x_Y|\le2$: the odd part divides $675$ and any proper divisor is at most $225$. Multiples $2\cdot675$ are excluded by the degree bound. Let $m_1,m_2$ count degrees $675$ with $|x_Y|=1,2$. Their excess over $225x_Y^2$ is $450$ in either case. The vacuum, three-free nonunits and five-free nonunits have respective ratios at most $1,25,27$ and compulsory squared masses at least $1,4,8$. For the last bound, a five-free position has dyadic height at least one, and height one entails degree $18$ or $54$ (degree six excluded), with at least a conjugate pair by the weight-two crossing with $U\in P_5$ recalled above. Thus \[ 5400\le4592+450(m_1+m_2). \] In particular some $675$ hits a $675$. At twist order sixteen we have $\delta=2$ and the two conjugate twists are opposite. Subtracting the two modular identities gives $5400/\sqrt2\le\sum d_Y|y_Y|$. We have $\sum y_Y^2=16$, so if all $y_Y$ vanished on degree $675$ the right side would be at most $225\cdot16=3600$. This again gives a mutual hit. In fact multiplying the difference identity by $\sqrt2\,\xi^{-1}\theta_Z$ and taking normalized trace gives \[ 5400=\sum_Y d_Y y_Y\,\tau(\sqrt2\,\xi^{-1}\theta_Z\theta_Y). \] Here the difference was taken with the factor $2\sqrt2$ removed, so that its right side before multiplication is $5400\xi\theta_Z^{-1}/\sqrt2$. Each displayed trace has modulus at most one: $\sqrt2=\zeta_8+\zeta_8^{-1}$, and for any dyadic root at most one of the two resulting terms has nonzero normalized dyadic trace. Traces factor across the odd conductor. The trace vanishes unless $\theta_Y$ has dyadic order exactly sixteen, as $\xi$ has order dividing eight. Each degree $675$ with $|y_Y|=1,2$ can improve the allowance $225y_Y^2$ by $450$. Thus there must already be at least $(5400-225\cdot16)/450=4$ degree-$675$ labels in this case. Every degree-$675$ row has support exactly $P_2$: its codegree is a two-power, and each principal position gives $\chi_Y(Z)\equiv675\not\equiv0$ above two. By symmetry a mutual hit puts a degree-$675$ label in $P_2$, then every one of them in $P_2$. Testing any odd degree on such a principal row shows that all odd labels are in its support. At a joint three--five column pick the odd $V\in P_3$ of degree $15$ or $45$ with field unramified at two (Section~\ref{found:C:6}); its five-inertia orbit has at least four labels. On these four $b_Y$ is a rational odd integer by principality and the intersection of the two fields. Their contribution saves at least $4(225-45)=720$ against $225x_Y^2$, disjoint from the reserved positions. The bound $4592$ at order at most eight can consequently be replaced by $3872$, proving $m_1+m_2\ge4$. If $W\in P_2$ has dyadic height $k$, it hits every degree $675$ with lcm-scale $675/\gcd(675,d_W)=675\cdot 2^k/d_W$. Averaging its row norm over the Galois group and reserving the vacuum gives $4(675\cdot2^k/d_W)^26$, the only possibility is $h=9$, so the field is the real subfield of $\QQ(\zeta_{27})$ and $a_Y=\tau(b_Y)$. Now $\sum a_Y^2=81/9=9$. In particular no nonzero coefficient occurs at ternary height two or higher, since lcm divisibility would make it a multiple of three. The five-free entries thus have degree $12$ or $24$ (degree six was excluded). Their allowance is at most four in squared norm after reserving the vacuum, the odd type and the three-free type. A nonzero signed sum in a single one of these degrees must therefore come from exactly three equal signs at that degree. A fourth nonzero entry could neither preserve the multiple of three at that degree nor give a multiple of three at the other degree. Modulo five the equation is $1\pm3d\equiv0$, so $d=12$ and all three signs are negative. However every degree-$12$ member of $P_3$ has tame five-inertia orbit of size divisible by four. To see this, cross it with a three-free $U\in P_5$ of degree $20$ or $40$ (Section~\ref{found:R:2}). The common weight is four. In the unsigned case of Section~\ref{found:S:5} the crossing is $1+3\xi$ with $\xi$ primitive of order five. In the signed case its four nonidentity fifth positions have coefficients given, up to a common sign, by one copy of one and a coset of the three cube roots. The square of a tame generator pairs the four positions and cannot preserve these coefficients since some multiplicity is odd. The sole relation on the fifth roots is the constant vector over the ternary field, and the position zero stays fixed. This proves the orbit assertion in both cases. Since $a$ is an ordinary normalized trace, it is constant on that orbit. This contradicts the three labels and gives $h\le6$. The cyclic Galois group at $81$ then puts $K_Z$ inside $\QQ(\zeta_9)$. Any higher ternary twist would require at least nine square conjugates, and the codegree excludes every other prime. \end{proof} \paragraph{The cubic conductor at degree $600$ on the path missing six.}\label{por:field600-sep} In fact the degree-$600$ character field has degree at most three on that path. To exclude $h=6$, let $K_Z=\QQ(\zeta_9)$ and use the two conjugate selectors $a_Y=\tau(\beta b_Y)$, $a_{Y^*}=\tau(\bar\beta b_Y)$ defined in the proof for degree $600$. Then $\sum a_Y^2=18$. A three-free $U\in P_5$ of degree $20$ exists on this path. Each five-free $Y$ with nonzero $a_Y$ belongs to $P_3$. Its ternary height is at most two by lcm divisibility. At height two the possibilities $18,36,72$ have $3\mid a_Y$; their five-inertia orbit sizes are multiples of two, four and four respectively, by their weight-two or weight-four crossings with $U$ by the weight-two and weight-four classifications. Even two such positions would exhaust the norm before the vacuum. The remaining five-free degrees are $12,24$. Their crossings with $U$ have weight four and hence five-inertia orbit sizes divisible by four. Both signed degree sums $\sum_{d_Y=d}a_Y$ are therefore multiples of twelve by the degree-sum obstruction at three. Reserving the odd type, the three-free type and the unit leaves at most $13$ in squared mass. Modulo five the only possibility is twelve positive degree-$12$ coefficients: $1+12\cdot12\equiv0$, whereas neither choice of sign at $24$ works. A thirteenth position of that type cannot preserve the signed sum. There are five further units of norm after the vacuum and the twelve. A three-free position is needed. The odd hits are of degree $15$ or $75$ as a higher ternary height would cost nine. Their signed sums are multiples of three, and at $15$ also of four by the weight-five crossing with $U$ (Section~\ref{found:J:5}). Thus precisely three odd positions of degree $75$ have equal sign. The other two positions are three-free: a three-bearing signed sum at a new degree cannot be obtained from a single spare unit, and neither known type admits one additional position. The possible three-free degrees are $10,20,40,50,100,200$. The signed dimension equation $1+144\pm225+e_1+e_2=0$ now gives negative sign and $(e_1,e_2)=(40,40)$ or $(100,-20)$. No hit can be active at two or five, since the degree of $Z$ has full height at both. A three-active degree has full dyadic height on this path, so the only possible three-active trace positions are the two of degree $40$. If the twist of $Z$ has order at most three, the sum of $a_Y d_Y\theta_Y$ has modulus $5400$, impossible on the listed support. Thus $\theta_Z$ has order nine. Averaging over wild ternary inertia makes this sum zero, since $\beta$ is fixed and the three conjugate twists are equally spaced. In the two-$40$ alternative this gives $40(\theta_1+\theta_2-2)=0$. Consequently every trace position has twist one. Let $G=\operatorname{Gal}(K_Z/\QQ)$ and put $u_\chi(gZ)=\chi(g)$ on the orbit and zero elsewhere. Put $v(g)=g(\beta)$ on the orbit. The $s$-image of $v$ is supported at twist one. As in the proof at $675$, $sT^{-1}v$ is supported on the orbit. Here $v$ has both trivial and quadratic components, so $T^{-1}v$ has the two components $\lambda,\lambda\epsilon$, where $\lambda(g)=\theta_Z/\theta_{gZ}$ is cubic and $\epsilon$ is the quadratic character. Torus equivariance and $s^2$ equal to duality now put the $s$-images of these two modes and their two inverses on this orbit. Each of the four internal Fourier coefficients $\sum_g\chi(g)g(b_Z)$ has modulus $5400/600=9$. Write $c_Y$ for the mean of $b_Y$ over the three wild ternary conjugates; then $\sum|c_Y|^2=81/3=27$ by row orthogonality, and its orthogonal complement has norm squared $54$. The four characters are all wild-nontrivial. Parseval shows that their contribution on the orbit of $Z$ alone is $4\cdot9^2/6=54$. Thus $b_Y=c_Y\in\QQ(\omega)$ off the orbit. At a positive degree-$12$ position write $b_Y=x+y\omega$ with $x,y\in\ZZ$. The two selectors are $x-y$ and $x$. Their sum $2x-y$ is zero modulo three since $b_Y$ has zero residue at three (principality and $v_3(12)=v_3(600)$). Yet both selectors at degree $12$ are zero or one, and at this position one is positive. This final contradiction excludes the sextic field. \paragraph{A restriction on the central charge.} On every graph, the existence of a degree-$600$ simple forces $\xi=\pm1$. Indeed a hit of this simple has trivial dyadic and quinary twist components by full height of $600$ at two and five. Its character field is ternary and $\theta_Z$ is a ninth root. Thus $5400\xi\theta_Z^{-1}=\sum_Yd_Y\chi_Z(Y)\theta_Y$ belongs to a cyclotomic field of odd ternary conductor. The dyadic root of unity $\xi$ in this field is real. \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. Fix a tame character $\psi$ of order four at five. 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 \[ v_Y=\frac1{\sqrt{\ell_Y}}\sum_{g\in H_5/\mathrm{Stab}(Y)} \overline{\psi(g)}\,e_{gY},\qquad v_Y^-=\frac1{\sqrt{\ell_Y}}\sum_{g\in H_5/\mathrm{Stab}(Y)} \psi(g)\,e_{gY}. \] 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 $675$. \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}. Five-ramification supplies 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 $W_{\{2,3\}}e_0$ side. Theorem~\ref{por:th:weyl-div} gives the factor $25$ because every degree in the fiber has positive five-defect. The normalized coordinate is constant on each selected five-inertia group. 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. The character-field assertions follow from Theorem~\ref{por:th:no-twelve}, Proposition~\ref{por:field600} and its \hyperref[por:field600-sep]{separated-case refinement}; Proposition~\ref{por:prop:cap675} bounds $m_{675}$. 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/f7f9f9a02f3116e61eb9dca4eebfed77001494e8/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 $m_{900}\ge21$; Theorem~\ref{por:th:no-twelve} bounds each orbit by six, Corollary~\ref{por:cor:thirteen} bounds the multiplicity at one twist, and Corollary~\ref{por:th:joint-at-ninehundred} allows at most three rows missing the joint sector. \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 $n=5400$. Its center is a counterexample in dimension $n^2$. The minimal reduction at that dimension therefore applies to the center itself. In particular it is integral and fusion-simple, its only invertible simple is the unit, and its nonunit degrees belong to \[ \mathcal D=\{a:10\le a\le900,\ a\mid n,\ a\text{ has at least two prime divisors}\}. \] 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$ one also has $H^n(G,\CC^\times)=\bigoplus_iH^n(G_i,\CC^\times)$: positive integral homology is killed by the group order, so the mixed tensor and Tor terms in K\"unneth vanish, and $\CC^\times$ is divisible. In the Picard classifying space the next two homotopy groups are $A_1\oplus A_2$ and $\CC^\times$; external product includes the individual groups $A_i$ and multiplies the scalar coordinates \cite[Theorem~5.2, Proposition~7.3 and Theorem~7.12]{ENOH}. By naturality the two factor $H^3(G_i,A_i)$ obstructions are the components of the vanishing obstruction for the given extension. Correct the two lower lifts by the components of their $H^2(G,A_1\oplus A_2)$ difference from the prescribed lift. Their product now has vanishing scalar fourth obstruction. Upon restriction to $G_i$ the other lift is constant and the inclusion of the $i$th factor is the identity on scalars, so each factor fourth obstruction vanishes. The remaining difference of the two extensions is in $H^3(G,\CC^\times)$; its two inflated components adjust the factor associators. Equivariantization by the product group then gives the assertion. \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 it holds whenever $2\mid D$ or $3\mid D$. 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 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} \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 $5$--$31$. \subsection{Multiplicities and prime-power codegrees}\label{fam:M:sec:13.2} 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}. \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 \[ s\longrightarrow t\quad (s\in S,\ t\in T,\ s\mid t-\epsilon_t). \] Connectivity 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. \section*{Code, data and certificates} \addcontentsline{toc}{section}{Code, data and certificates} \label{sec:code-data-certificates} The verification programs, their inputs and the certificates are available at\\ \url{https://github.com/sebastienpalcoux/WGT-fusion-categories}. To reproduce a calculation, look up its theorem or subsection in the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/main/certificates/manifest.json}{verification manifest}. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/48d2813db714e314684fd79f5ba565b81c16a2be/code/README.md#dimension-comparison}{dimension-check instructions} give the command and expected output for the four bounds in Theorem~\ref{thm:uniform-new-bounds}. The printed locations in the manifest refer to the repository edition. The source labels identify the corresponding results in the present text. Each entry gives the hypotheses, the command to run from the \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/tree/main/code}{\texttt{code/}} directory, and the expected output. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/tree/main/data}{\texttt{data/}} directory contains the rank-$15$ data of Dong and Palcoux. The \href{https://github.com/sebastienpalcoux/WGT-fusion-categories/blob/main/certificates/certificates.json}{witness file} records the seven terminal trace obstructions of Theorem~\ref{fam:M:8.15} and the five pairs of disjoint fibers at rank $100$ in Section~\ref{por:certificate}. \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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