\documentclass[11pt]{amsart} \pdfoutput=1 \usepackage[usenames,dvipsnames]{xcolor} % \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[letterpaper,margin=1in]{geometry} \usepackage{lmodern} \usepackage{microtype} \usepackage{etoolbox} % \usepackage{lettrine} \usepackage{mathtools} \usepackage{amsmath,amssymb,amsfonts,amsxtra,bm} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{latexsym} \usepackage{nicefrac} % \usepackage{booktabs} \usepackage{makecell,multirow} % \usepackage{colortbl} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{float} \floatstyle{plaintop} \restylefloat{table} \usepackage{footnote, tablefootnote} \usepackage{enumitem} \usepackage{algorithm} \usepackage{algorithmic} \usepackage{comment} \usepackage{xspace} \usepackage{url} \usepackage[most]{tcolorbox} \usepackage[round]{natbib} \usepackage{hyperref} \numberwithin{equation}{section} \makeatletter \definecolor{darkblue}{rgb}{0.0,0.0,0.65} \definecolor{darkred}{rgb}{0.68,0.05,0.0} \definecolor{darkgreen}{rgb}{0.0,0.29,0.29} \definecolor{darkpurple}{rgb}{0.47,0.09,0.29} \renewcommand{\topfraction}{0.85} \renewcommand{\bottomfraction}{0.4} \renewcommand{\textfraction}{0.1} \renewcommand{\floatpagefraction}{0.9} \patchcmd{\section}{\scshape}{\fontfamily{ppl}\bfseries\scshape\selectfont}{}{% \PackageWarning{aiproofs}{Could not bold the section font; amsart may have changed}} \newcommand{\aiauthor}{% \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}% \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany. Email: \texttt{s.sra@tum.de}.}} \newcommand{\preprintslink}[2]{% \thanks{First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.}} \newcommand{\arxivlink}[1]{% \thanks{Also on arXiv: \url{https://arxiv.org/abs/#1}.}} \newcommand{\preprintsfooter}[2]{% \def\ps@firstpage{\ps@plain \def\@oddfoot{\hfil\parbox[t]{\textwidth}{\centering\normalfont\scriptsize First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.\\[3pt]\thepage}\hfil \global\topskip\normaltopskip}% \let\@evenfoot\@oddfoot \def\@oddhead{\@serieslogo\hss}% \let\@evenhead\@oddhead}} \makeatother \hypersetup{colorlinks=true,linkcolor=red!60!black,citecolor=green!48!black,urlcolor=blue!60!black} \setlist{nosep,leftmargin=*} \renewcommand{\bibnumfmt}[1]{{\color{darkblue}\bf\textsc{#1}}.\hskip 6pt} \newenvironment{aiusage}{\section*{AI usage}}{} \newcommand{\aiacknowledgements}[1][]{% \section*{Acknowledgements} \if\relax\detokenize{#1}\relax\else #1\ \fi The author (SS) thanks the Alexander von Humboldt (AvH) Foundation for their generous support via a Humboldt Professorship in AI.% } \newcommand{\reals}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\pp}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\bmOne}{\bm{1}} \newcommand{\C}{\mathbb{C}} \newcommand{\Cc}{\mathcal{C}} \newcommand{\Dc}{\mathcal{D}} \newcommand{\Nc}{\mathcal{N}} \newcommand{\Xc}{\mathcal{X}} \newcommand{\Fc}{\mathcal{F}} \newcommand{\symm}{\mathfrak{S}} \newcommand{\ma}{\bm{A}} \newcommand{\mx}{\bm{X}} \newcommand{\my}{\bm{Y}} \newcommand{\mz}{\bm{Z}} \newcommand{\vx}{\bm{x}} \newcommand{\vy}{\bm{y}} \newcommand{\vv}{\bm{v}} \newcommand{\vz}{\bm{z}} \newcommand{\vr}{\bm{r}} \newcommand{\mb}{\bm{B}} \newcommand{\dpp}{\textsc{Dpp}} \newcommand{\etal}{et~al.\@\xspace} \newcommand{\half}{\tfrac{1}{2}} \newcommand{\kron}{\otimes} \newcommand{\inv}[1]{{#1}^{-1}} \newcommand{\nlsum}{\sum\nolimits} \newcommand{\nlprod}{\prod\nolimits} \newcommand{\ip}[2]{\langle {#1},\, {#2} \rangle} \newcommand{\iplr}[2]{\left\langle {#1},\, {#2}\right\rangle} \newcommand{\set}[1]{\left\lbrace#1\right\rbrace} \newcommand{\norm}[1]{\|{#1}\|} \newcommand{\bignorm}[1]{\left\lVert #1\right\rVert} \newcommand{\pnorm}[2]{\|#1\|_{#2}} \newcommand{\enorm}[1]{\bigl\|#1\bigr\|_{2}} \newcommand{\frob}[1]{\|#1\|_{\text{F}}} \newcommand{\fdel}[2]{\frac{\partial {#1}}{\partial {#2}}} \newcommand{\what}{\widehat} \newcommand{\ld}{\ell{}d} \newcommand{\gm}{\#} \newcommand{\nf}[2]{\nicefrac{#1}{#2}} \newcommand{\pfrac}[2]{\left(\tfrac{#1}{#2}\right)} \def\sint{\begingroup\textstyle\int\endgroup} % \DeclareMathOperator{\trace}{tr} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Diag}{Diag} \DeclareMathOperator{\real}{Re} \DeclareMathOperator{\imag}{Im} \DeclareMathOperator{\per}{per} \DeclareMathOperator{\Vol}{Vol} \DeclareMathOperator{\Imm}{Imm} \DeclareMathOperator{\mdet}{det} \DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\qjsd}{QJSD} \DeclareMathOperator{\qjrd}{QJRD} \DeclareMathOperator*{\argmin}{argmin} \DeclareMathOperator*{\argmax}{argmax} \newtheorem{theorem}{Theorem}[section] \newtheorem{conj}[theorem]{Conjecture} \newtheorem{prop}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corr}[theorem]{Corollary} \newtheorem{rmk}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{defn}[theorem]{Definition} \newtheorem{prob}[theorem]{Problem} \definecolor{cdarkred}{RGB}{180,70,30} % \definecolor{cdarkgreen}{RGB}{0,130,0} % \definecolor{azure}{RGB}{240,255,255} \definecolor{cmidgreen}{RGB}{201,219,216} \definecolor{clightgreen}{RGB}{200,255,200} \definecolor{clightgray}{RGB}{240,240,240} \definecolor{cmygreen}{RGB}{10,120,0} \definecolor{cmyblue}{RGB}{50,50,255} \definecolor{sblue}{RGB}{100,100,255} \definecolor{cmypurple}{RGB}{150,50,180} \definecolor{cmyred}{RGB}{238, 64, 0} \definecolor{cdarkblue}{RGB}{30,75,170} \definecolor{csharpred}{RGB}{255,0,0} \newcommand{\sred}[1]{{\color{csharpred} #1}} \newcommand{\darkred}[1]{{\color{cdarkred} #1}} \newcommand{\bred}[1]{\textbf{\sred{#1}}} \newcommand{\darkblue}[1]{{\color{cdarkblue} #1}} \newcommand{\bblue}[1]{\textbf{\darkblue{#1}}} \newcommand{\darkbred}[1]{\textbf{\color{cdarkred} #1}} \newcommand{\darkgreen}[1]{{\color{cmygreen} #1}} \tcbset{dashedrounded/.style={no borderline,borderline={0.5pt}{0pt}{dashed},arc=4pt}} \tcbset{supporting/.style={colback=gray!8!white}} \newtcolorbox{dashedbox}[1][]{ before={\par\medskip\noindent}, after={\par\medskip}, colback=yellow!6!white, boxrule=0pt, enhanced jigsaw, top=2pt, left=2pt, right=2pt, bottom=0pt, dashedrounded, #1 } \newtcbox{\tcbmathbox}[1][]{ nobeforeafter, colback=yellow!30!white, boxrule=0pt, enhanced jigsaw, top=0pt, left=1pt, right=1pt, bottom=0pt, borderline horizontal={0.5pt}{0pt}{dashed}, borderline vertical={0.5pt}{0pt}{dashed}, #1 } \hypersetup{pdftitle={The q-Foulkes problem II: Schur positivity with high probability},pdfauthor={Suvrit Sra}} \allowdisplaybreaks[2] \setlength{\emergencystretch}{2em} \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\Htil}{\widetilde H} \newcommand{\ch}{\operatorname{ch}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\Ind}{\operatorname{Ind}} \newcommand{\Res}{\operatorname{Res}} \newcommand{\Cyc}{\operatorname{Cyc}} \newcommand{\im}{\operatorname{im}} \newcommand{\qbin}[2]{\begin{bmatrix}#1\\#2\end{bmatrix}_{q}} \newcommand{\Z}{\mathbb Z} \newcommand{\Prob}{\mathbb P} \newcommand{\Var}{\operatorname{Var}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Frob}{\operatorname{Frob}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\End}{\operatorname{End}} \newcommand{\maj}{\operatorname{maj}} \newcommand{\SYT}{\operatorname{SYT}} \newcommand{\gr}{\operatorname{gr}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\Span}{\operatorname{span}} \newcommand{\Dcal}{\mathcal D} \newcommand{\Kcal}{\mathcal K} \newcommand{\pos}{\succeq_s0} \newcommand{\cl}{c_{\lambda,k}} \theoremstyle{plain} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \newcommand{\recalledname}{} \newtheorem*{recalledthm}{\recalledname} \newenvironment{recalled}[1]{\renewcommand{\recalledname}{#1}\begin{recalledthm}}{\end{recalledthm}} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \subjclass[2020]{Primary 05E05, 20C30; Secondary 60E15, 60F10} \keywords{$q$-Foulkes conjecture, Schur positivity, Macdonald polynomial, Plancherel measure} \providecommand{\PFprob}{\mathbb P} \providecommand{\PFexp}{\mathbb E} \providecommand{\PFvar}{\operatorname{Var}} \providecommand{\PFPl}{\operatorname{Pl}} \providecommand{\PFch}{\operatorname{Frob}} \providecommand{\PFInd}{\operatorname{Ind}} \providecommand{\PFcodim}{\operatorname{codim}} \providecommand{\PFrank}{\operatorname{rank}} \providecommand{\PFsupp}{\operatorname{supp}} \providecommand{\PFreg}{\mathbb C} \providecommand{\PFH}{\widetilde H} \providecommand{\PFone}{\mathbf 1} \providecommand{\PFN}{\mathbb N} \providecommand{\PFZ}{\mathbb Z} \providecommand{\PFR}{\mathbb R} \providecommand{\PFQ}{\mathsf G} \begin{document} \title[The $q$-Foulkes problem II]{The $q$-Foulkes problem II:\\Schur positivity with high probability} \aiauthor \begin{abstract} Bergeron's $q$-analog of Foulkes' conjecture asserts that $(H_b[H_a]-H_a[H_b])/(1-q)$ is Schur positive for $a\le b$, where $H_n$ is the (modified) Hall--Littlewood polynomial. We view this conjecture probabilistically: on each irreducible representation of $S_{ab}$, the two plethysms induce degree distributions, and positivity means that the first is stochastically smaller than the second. We prove that, for a random (Plancherel) irreducible representation and a degree sampled from its graded multiplicity, the corresponding coefficient is negative with probability at most $C\exp(-c(ab)^{1/3})$, uniformly in $a\le b$. Consequently, the $q$-Foulkes conjecture holds after deleting a fraction at most $C\exp(-c(ab)^{1/3})$ of each of the two graded modules. Additionally, we show that exact conditional means and variances yield complete positivity for explicit width and height families. \end{abstract} \preprintsfooter{qfoulkes2}{28 October 2026} \maketitle \addtocontents{toc}{\protect\setcounter{tocdepth}{-1}} \raggedbottom \vspace*{-6pt} \section{Introduction} We study the $q$-analog of Foulkes' conjecture proposed by \citet{bergeron2017foulkes} from a probabilistic point of view. Recall that \citeauthor{foulkes1950}'s conjecture~(\citeyear{foulkes1950}) states that, for $a\le b$, the difference of plethysms $h_b[h_a]-h_a[h_b]$ of complete homogeneous symmetric functions is Schur positive, i.e., it expands with nonnegative integer coefficients in the Schur functions $s_\lambda$, $\lambda\vdash ab$. This positivity is known for $a\le6$ \citep{thrall1942symmetrized,dent2000foulkes,mckay2008plethysm,cheung2017foulkes,OpenAI} and for $b$ large with respect to $a$ \citep{brion1993stable}, but it remains open in general. Bergeron's $q$-analog replaces $h_n$ by the (modified) Hall--Littlewood polynomial \begin{equation*} H_n(X;q) :=(q;q)_n h_n\left[\tfrac{X}{1-q}\right], \quad\text{where}\quad (q;q)_n :=\nlprod_{j=1}^n(1-q^j), \end{equation*} and his analog of Foulkes' conjecture is the following: \begin{dashedbox}[supporting] \begin{conjecture}[\citealp{bergeron2017foulkes}]\label{conj:bergeron} For all integers $a\le b$, the divided difference \begin{equation*} F_{a,b} :=\frac{H_b[H_a]-H_a[H_b]}{1-q} \end{equation*} is Schur positive with coefficients in $\N[q]$. \end{conjecture} \end{dashedbox} Since $H_n(X;0)=h_n$, at $q=0$ the conjecture reduces to Foulkes' conjecture. Bergeron proved the conjecture at $q=1$ \citep[Section~3]{bergeron2017foulkes}. Our starting point is a probabilistic view on Conjecture~\ref{conj:bergeron}. The plethysms $H_b[H_a]$ and $H_a[H_b]$ are the graded characters of two $S_N$-modules $M_{a,b}$ and $M_{b,a}$, where $N=ab$, and both are regular once the grading is forgotten (Lemma~\ref{lem:wreath}). Each irreducible $S^\lambda$ therefore occurs $d_\lambda=\dim S^\lambda$ times in each module, and the degree of a uniformly chosen copy defines a random variable $T_\lambda$ for $M_{a,b}$, the target, and $S_\lambda$ for $M_{b,a}$, the source. Dividing by $1-q$ takes cumulative sums, so $[q^ks_\lambda]F_{a,b}$ equals $d_\lambda(\Prob(T_\lambda\le k)-\Prob(S_\lambda\le k))$ (Proposition~\ref{prop:cdf}), and Conjecture~\ref{conj:bergeron} says that $T_\lambda$ is stochastically smaller than $S_\lambda$ for every $\lambda\vdash N$. The Plancherel-averaged degree laws admit an exact monotone coupling (Lemma~\ref{pfa:coupling}), but a small absolute error can reverse this order where the averaged gap is small. We control the character corrections relative to positive averaged gaps and tails. \begin{dashedbox} \begin{theorem}[Probabilistic $q$-Foulkes]\label{pfa:main} There are absolute constants $C,c>0$ with the following property. For integers $1\le a\le b$, put $N=ab$, sample $\Lambda\vdash N$ from Plancherel measure, and, conditionally on $\Lambda$, let $K$ have the law of $T_\Lambda$ or of $S_\Lambda$ according to a fair coin. Then \[ \Prob\bigl([q^Ks_\Lambda]F_{a,b}<0\bigr)\le C\exp(-cN^{1/3}). \] \end{theorem} \end{dashedbox} Theorem~\ref{pfa:main} says that Conjecture~\ref{conj:bergeron} holds with high probability, in that the probability of a $q$-Foulkes violation, a negative coefficient at a random shape and degree, is at most $Ce^{-cN^{1/3}}$. Concretely, suppose we weight each pair of a shape $\lambda$ and a degree $k$ by the dimension of the $\lambda$-isotypic part of the two modules in degree $k$. Then the pairs at which the conjectured inequality fails carry a fraction at most $Ce^{-cN^{1/3}}$ of the total weight. Moreover, outside a set of Plancherel mass at most $Ce^{-cN^{1/3}/2}$, the failing degrees carry at most $e^{-cN^{1/3}/2}$ of the shape's graded mass (Corollary~\ref{pfa:quenched}). When $a$ and $b$ are close, positivity holds throughout a central degree window. \vskip4pt \begin{dashedbox}[supporting] \begin{recalled}{Corollary}[\protect{informal Cor.~\ref{pfa:simultaneous-window}}] There are absolute constants $\kappa,C,c>0$ such that, if $2\le a0$ (Corollary~\ref{pfa:separated}). Because target degrees never exceed a fixed endpoint, both the failure probability in Theorem~\ref{pfa:main} and the deleted fraction are at most $e^{-cN}$ when $b\ge6a\ge18$, and at most $\exp[-(1/3+o(1))N\log N]$ when $b\asymp a^2$ and $a\to\infty$ (Corollary~\ref{pfa:support-rates}). Both extend to Bergeron's rectangular conjecture, where the inner factor $H_a$ becomes the modified Macdonald polynomial of a rectangle, with a second grading that the matching preserves (Theorem~\ref{pfa:qt-theorem}, Corollary~\ref{pfa:qt-support}). Theorem~\ref{pfa:main} and its consequences concern typical shapes. Our second main result proves the conjecture outright, at every degree, for explicit infinite families of shapes. Such families exist because a width bound separates the source support from the target mean by order $b^2$, while both conditional standard deviations are $O(b^{3/2})$. The exact conditional means and variances then order the two cumulative functions at every degree (see Theorem~\ref{thm:interval}). \begin{dashedbox} \begin{theorem}[Positivity at bounded width]\label{thm:width} Let $a\ge2$, and let $1\le w\le2a-1$, or $w=2a$ if $a$ is even. There is an explicit $B(a,w)$ such that $[s_\lambda]F_{a,b}\in\N[q]$ for every $b\ge B(a,w)$ and every partition $\lambda\vdash ab$ with $\lambda_1\le w$. \end{theorem} \end{dashedbox} The same criterion gives positivity under a height bound, even for widths growing linearly in $b$ (Corollary~\ref{cor:height}). For fixed $a\ge2$, let $\lambda_b\vdash ab$ have normalized transposition characters satisfying $\chi^{\lambda_b}(2,1^{ab-2})/d_{\lambda_b}\to x<1$. The moments certify a positive interval of length $\Theta(b^2)$ reaching within $O(b)$ of both conditional means (Corollary~\ref{cor:linear-margins}). In the macroscopic interior, the coefficients are asymptotic to their upper bound $d_{\lambda_b}$ (Theorem~\ref{thm:bulk}). Positivity at every degree of a typical shape remains open, since the rate $e^{-cN^{1/3}}$ does not rule out individual negative coefficients (Section~\ref{sec:whole-shape}). \subsection{Conventions}\label{sec:conventions} Throughout, $a,b\ge2$ are integers and $N=ab$ unless stated otherwise, $c_n=\binom n2$, and $[n]_q=(1-q^n)/(1-q)$. For a partition $\lambda$, $\lambda'$ is its conjugate, $\ell(\lambda)$ its length, $n(\lambda)=\sum_i(i-1)\lambda_i$, and its width, height, and depth are $\lambda_1$, $\ell(\lambda)$, and $|\lambda|-\lambda_1$. For $\rho=1^{m_1}2^{m_2}\cdots$, put $z_\rho=\prod_r r^{m_r}m_r!$. For $\lambda\vdash N$, $S^\lambda$ is the irreducible $S_N$-module over $\C$, $\chi^\lambda$ its character, and $d_\lambda=\dim S^\lambda$. Plethysm acts on every grading variable through the Adams operations $p_j\mapsto p_{rj}$, $q\mapsto q^r$, and $t\mapsto t^r$. \section{Regular modules and conditional degree laws}\label{sec:modules} This section turns Conjecture~\ref{conj:bergeron} into a comparison of two probability laws on each shape, which is the form every later result uses. The plethysms $P_{a,b}=H_b[H_a]$ and $Q_{a,b}=H_a[H_b]$ define the degree laws $T_\lambda$ and $S_\lambda$ of Theorem~\ref{pfa:main} because their underlying modules are regular. \subsection{The graded character of the coinvariant algebra} The coinvariant algebra is where $H_n$ comes from, and its regularity is what makes the degree laws exist. The coinvariant algebra $R_n$ is the quotient of the polynomial ring by invariant generators of degrees $1,\ldots,n$, so the graded trace of $\sigma\in S_n$ on $R_n$ is $(q;q)_n/\det(I-q\sigma)$. If $\mu$ is the cycle type of $\sigma$, this trace is \begin{equation}\label{eq:kappa} \kappa_n(\sigma;q):= \frac{(q;q)_n}{\prod_{r\in\mu}(1-q^r)}. \end{equation} Thus $\kappa_n$ is the graded character of $R_n$, and at $q=1$ it is the regular character. Expand $h_n=\sum_{\mu\vdash n}p_\mu/z_\mu$. The plethystic alphabet $1/(1-q)$ replaces $p_r$ by $p_r/(1-q^r)$, so $H_n=\sum_{\mu\vdash n}\kappa_n(\mu;q)p_\mu/z_\mu$ is the graded Frobenius character of $R_n$. \subsection{Wreath induction} Wreath induction realizes plethysm \citep[Chapter~I, Appendix~A, (6.2)]{macdonald1995}, and its graded form produces the modules behind $H_b[H_a]$. Partition $[N]$ into $b$ blocks of size $a$, with a fixed labeling $(i,j)$, $1\le i\le b$, $1\le j\le a$, and let $W_{a,b}=S_a\wr S_b$, of order $(a!)^b b!$, preserve this partition. Write an element as $(\pi_1,\ldots,\pi_b;\sigma)$, acting by $(i,j)\mapsto(\sigma(i),\pi_i(j))$. For an outer cycle $C=(i_1,\ldots,i_r)$, define the monodromy $\pi_C=\pi_{i_r}\cdots\pi_{i_1}$, and put \begin{equation}\label{eq:theta} \Theta_{a,b}(\boldsymbol\pi,\sigma;q) =\kappa_b(\sigma;q) \prod_{C\in\Cyc(\sigma)}\kappa_a(\pi_C;q^{|C|}). \end{equation} \begin{lemma}[Graded wreath induction]\label{lem:wreath} The function $\Theta_{a,b}$ is the graded character of $R_a^{\otimes b}\otimes R_b$, with the natural wreath action, permuting tensor factors without signs, and \begin{equation*} \ch\Ind_{W_{a,b}}^{S_N}(R_a^{\otimes b}\otimes R_b)=H_b[H_a]. \end{equation*} \end{lemma} \begin{proof} The graded trace on an outer $r$-cycle is $\kappa_a(\pi_C;q^r)$: a homogeneous vector of inner degree $d$ is carried around $r$ tensor factors and contributes $q^{rd}$. Multiplication by the trace on $R_b$ proves \eqref{eq:theta}. The cycle-index formula for induction gives \[ \frac{1}{|W_{a,b}|}\sum_{w\in W_{a,b}} \Theta_{a,b}(w;q)p_{\mu(w)}=H_b[H_a]. \] Indeed, an outer $r$-cycle multiplies every cycle length of $\pi_C$ by $r$, exactly the Adams operation $p_j\mapsto p_{rj}$ together with $q\mapsto q^r$. \end{proof} Put \begin{equation}\label{eq:modules} M_{a,b}=\Ind_{W_{a,b}}^{S_N}(R_a^{\otimes b}\otimes R_b). \end{equation} By Lemma~\ref{lem:wreath}, $\ch_q M_{a,b}=P_{a,b}$. Ungraded, $R_a^{\otimes b}\otimes R_b$ is regular for $W_{a,b}$: a nonidentity outer permutation has zero trace on $R_b$, while for the identity outer permutation the inner traces vanish unless every inner permutation is the identity. Induction preserves regularity, so $M_{a,b}$ and $M_{b,a}$ are ungraded copies of $\C[S_N]$. Regularity turns the coefficient polynomials into probability laws. For $\lambda\vdash N$, write \[ u_\lambda(q)=[s_\lambda]P_{a,b},\qquad v_\lambda(q)=[s_\lambda]Q_{a,b}. \] These polynomials have nonnegative integer coefficients, and $u_\lambda(1)=v_\lambda(1)=d_\lambda$. Define integer-valued random variables by \begin{equation}\label{eq:laws} \E q^{T_\lambda}=u_\lambda(q)/d_\lambda,\qquad \E q^{S_\lambda}=v_\lambda(q)/d_\lambda. \end{equation} \begin{proposition}[The exact conditional comparison]\label{prop:cdf} For every integer $k\ge0$, \begin{equation}\label{eq:cdf} \frac{[q^ks_\lambda]F_{a,b}}{d_\lambda} =\Prob(T_\lambda\le k)-\Prob(S_\lambda\le k). \end{equation} Thus $F_{a,b}$ is Schur positive with coefficients in $\N[q]$ if and only if $T_\lambda$ is stochastically smaller than $S_\lambda$ for every $\lambda\vdash N$. \end{proposition} \begin{proof} Dividing by $1-q$ takes cumulative sums of coefficients, and the resulting series terminates because both totals equal $d_\lambda$. \end{proof} \begin{proposition}[Equivalent filtered formulations]\label{prop:matching} For fixed $a,b$, the following are equivalent: \begin{enumerate} \item $F_{a,b}$ is Schur positive with coefficients in $\N[q]$; \item there is a degree-nonincreasing $S_N$-equivariant linear bijection $M_{b,a}\to M_{a,b}$; \item there is a degree-preserving $S_N$-equivariant $\C[z]$-linear injection \[ M_{b,a}\otimes\C[z]\longrightarrow M_{a,b}\otimes\C[z],\qquad\deg z=1, \] with finite-dimensional cokernel of graded Frobenius character $F_{a,b}$. \end{enumerate} \end{proposition} \begin{proof} Fix an irreducible type and list its multiplicity degrees in increasing order as $s_1\le\cdots\le s_{d_\lambda}$ in the source and $t_1\le\cdots\le t_{d_\lambda}$ in the target. The cumulative inequalities are equivalent to $t_i\le s_i$ for every $i$. Matching these copies gives (2). Over $\C[z]$, send the source generator to $z^{s_i-t_i}$ times the target generator; the cokernel of this map has Hilbert polynomial $(q^{t_i}-q^{s_i})/(1-q)$. Conversely, either map gives the cumulative inequalities. The maps depend on the chosen matching. \end{proof} \section{Positivity with high probability}\label{pfa:overview} To measure how often the conjecture fails, we weight each coefficient $c_{\lambda,k}=[s_\lambda q^k]F_{a,b}$ by the dimension of its irreducible type and by its graded multiplicity in the two modules. The weights define the probability measure \begin{equation}\label{pfa:graded-measure} \boxed{\quad \PFQ_{a,b}(\lambda,k) =\frac{d_\lambda^2}{2N!} \bigl(\Prob(T_\lambda=k)+\Prob(S_\lambda=k)\bigr).\quad} \end{equation} Its shape marginal is $\PFPl_N(\lambda)=d_\lambda^2/N!$. Throughout, $\Lambda$ is a random shape with law $\PFPl_N$ and $\lambda$ is a fixed shape, so $T_\Lambda$ and $S_\Lambda$ are the conditional degree laws at $\Lambda$. Conditional on $\lambda$, it chooses either module with equal probability and samples the degree of a uniformly chosen element of a homogeneous basis of the multiplicity space. We call this law the \emph{graded Plancherel mixture}. Theorem~\ref{pfa:main} becomes, for all integers $1\le a\le b$, \[ \PFprob_{(\Lambda,K)\sim\PFQ_{a,b}} (c_{\Lambda,K}<0)\le C\exp(-cN^{1/3}). \] For $a=1$ or $a=b$ the divided difference is zero, so all nontrivial estimates below assume $2\le ae^{-cN^{1/3}/2}\right) \le C e^{-cN^{1/3}/2}. \end{equation*} \end{corollary} \begin{proof} Apply Markov's inequality to $V_\Lambda$. \end{proof} A shape's first negative cutoff always carries source mass. \begin{proposition}[The sampling law detects every violation] \label{pfa:detect} The following are equivalent: \[ \PFQ_{a,b}(c<0)=0;\qquad V_\lambda=0\text{ for every }\lambda;\qquad c_{\lambda,k}\ge0\text{ for every }\lambda,k. \] \end{proposition} \begin{proof} If a shape has a negative coefficient, take its first negative cutoff $k_0$, with $c_{\lambda,-1}=0$. Since $c_{\lambda,k_0}-c_{\lambda,k_0-1} =d_\lambda(\Prob(T_\lambda=k_0)-\Prob(S_\lambda=k_0))<0$, we have $\Prob(S_\lambda=k_0)>0$. Hence the negative coefficient has positive mass under \eqref{pfa:graded-measure}. \end{proof} \subsection{Positivity after a minimal deletion}\label{pfa:repair} Proposition~\ref{prop:matching} turns the conjecture into a degree-nonincreasing bijection $M_{b,a}\to M_{a,b}$. We seek the smallest deletion that repairs all degree cutoffs at once. Write $F_{T,\lambda}$ and $F_{S,\lambda}$ for the cumulative distribution functions of $T_\lambda$ and $S_\lambda$ in \eqref{eq:cdf}. The \emph{cumulative defect} of $\lambda$, \begin{equation*} D_\lambda=\sup_k(F_{S,\lambda}(k)-F_{T,\lambda}(k))_+, \end{equation*} measures the obstruction to matching each source copy to a target copy of no larger degree, and the following elementary lemma shows that the obstruction is exact. \begin{proposition}[Optimal ordered matching]\label{pfa:matching} Let $x_1,\ldots,x_f$ and $y_1,\ldots,y_f$ be two lists of integers. The largest partial matching with $x_i\le y_j$ has size \[ f-\max_k\bigl(\#\{j:y_j\le k\}-\#\{i:x_i\le k\}\bigr)_+. \] \end{proposition} \begin{proof} An initial segment of the $y$-list with $y_j\le k$ can only be matched to $x_i\le k$, so its excess is an unavoidable deficiency. Conversely, the neighbor sets of the $y$-vertices, namely $\{i:x_i\le y_j\}$, are nested. For any subset of $y$-vertices its Hall deficiency is bounded by the deficiency of the full initial segment ending at its largest degree. The maximum initial-segment deficiency is therefore the full Hall deficiency, and Hall's theorem, or greedy matching in increasing order, attains it. \end{proof} \begin{theorem}[Optimal filtered matching] \label{pfa:repair-theorem} For integers $1\le a\le b$, put $N=ab$ and $\Delta_{a,b}=\PFexp_{\PFPl_N}D_\Lambda$. There are graded submodules $M_{a,b}^\circ\subseteq M_{a,b}$ and $M_{b,a}^\circ\subseteq M_{b,a}$ of the same codimension, with an $S_N$-equivariant isomorphism $M_{b,a}^\circ\longrightarrow M_{a,b}^\circ$ preserving the increasing degree filtration, such that \begin{align} \frac{\PFcodim M_{a,b}^\circ}{N!} =\frac{\PFcodim M_{b,a}^\circ}{N!} &=\Delta_{a,b}\le C e^{-cN^{1/3}},\notag\\ \frac{\ch_q M_{a,b}^\circ-\ch_q M_{b,a}^\circ}{1-q} &\in\sum_{\lambda\vdash N}\PFN[q]s_\lambda.\label{pfa:exact-repair} \end{align} The normalized codimension is optimal. Equivalently, \begin{equation}\label{pfa:rank} \max_\phi\frac{\PFrank\phi}{N!}=1-\Delta_{a,b}, \end{equation} where $\phi\colon M_{b,a}\to M_{a,b}$ ranges over equivariant filtration-preserving linear maps. The minimum probability of $T_\Lambda>S_\Lambda$ in a coupling of the two conditional degree laws that preserves $\Lambda$ is also exactly $\Delta_{a,b}$. \end{theorem} \begin{proof} For each $\lambda$, list the $d_\lambda$ homogeneous multiplicity degrees of the two modules. The unmatched number in Proposition~\ref{pfa:matching} is the integer $\max_k(-c_{\lambda,k})_+=d_\lambda D_\lambda$. Delete this many copies of $S^\lambda$ on each side, and match the rest. The common dimension loss is $\sum_\lambda d_\lambda^2D_\lambda=N!\Delta_{a,b}$. Every matched pair $x\le y$ contributes \[ s_\lambda\frac{q^x-q^y}{1-q} =s_\lambda(q^x+\cdots+q^{y-1}), \] with the contribution zero when $x=y$; summing over matched pairs proves \eqref{pfa:exact-repair}. The same counting of initial segments bounds the nullity of any filtered equivariant map from below by $\sum_\lambda d_\lambda^2D_\lambda$, proving optimality and the rank identity \eqref{pfa:rank}. A coupling of two uniform degree lists is a doubly stochastic matrix. The linear matching cost attains its minimum at a permutation matrix. Greedy ordered matching gives failure probability exactly $D_\lambda$; the same counting of initial segments gives the reverse inequality. Average over $\lambda$. Finally, $D_\lambda\le2V_\lambda$ (Lemma~\ref{pfa:elementary}) and Theorem~\ref{pfa:main} give the rate. \end{proof} \subsection{The averaged degree laws}\label{pfa:annealed} The proof of Theorem~\ref{pfa:main} measures each conditional law against its Plancherel average, so we first show that the averages are explicit and already ordered. They are sums of independent uniform coordinates, which can be coupled monotonically. Let $U_j$ be uniform on $\{0,\ldots,j\}$ and define $\mathsf M_n=\sum_{j=1}^{n-1}U_j$, with all summands independent. For $M_{u,v}$, the Plancherel-averaged generating function is its graded dimension divided by $N!$: \begin{equation}\label{pfa:pgf} g_{u,v}(q)=\left(\frac{[u]_q!}{u!}\right)^v \frac{[v]_q!}{v!}. \end{equation} For $P_{a,b}$ and $Q_{a,b}$ these laws are those of \begin{equation}\label{pfa:XY} X=\sum_{i=1}^{b}\mathsf M_a^{(i)}+\mathsf M_b,\qquad Y=\sum_{i=1}^{a}\mathsf M_b^{(i)}+\mathsf M_a. \end{equation} Repeated terms denote independent copies. Write $p_X(k)=\PFprob(X=k)$ and $p_Y(k)=\PFprob(Y=k)$, and put $\mu_X=\PFexp X$ and $\mu_Y=\PFexp Y$. Their mean difference is \begin{equation*} d=\mu_Y-\mu_X=\frac{(a-1)(b-1)(b-a)}4. \end{equation*} Also \begin{equation*} \PFexp\mathsf M_n=\frac{n(n-1)}4,\qquad \PFvar\mathsf M_n=\frac{n(n-1)(2n+5)}{72}. \end{equation*} \begin{lemma}[Positive coupling and separation]\label{pfa:coupling} There is a coupling of $X,Y$ with $X\le Y$ almost surely. In this coupling, if $a0$. There is also a deterministic real cutoff $t$ such that \begin{equation}\label{pfa:separation} \PFprob(X>t)+\PFprob(Y\le t) \le 2\exp\left(-\frac{N(b-a)^2}{256b^2}\right). \end{equation} \end{lemma} \begin{proof} Remove the common summands $\mathsf M_a+\mathsf M_b$. There remain $m_0=(a-1)(b-1)$ uniform coordinates on each side. For $1\le i\le m_0$, put \[ A_i=\left\lceil\frac{i}{b-1}\right\rceil,\qquad B_i=\left\lceil\frac{i}{a-1}\right\rceil. \] With independent uniform $\xi_i\in[0,1)$, couple the remaining coordinates as $\lfloor(A_i+1)\xi_i\rfloor$ and $\lfloor(B_i+1)\xi_i\rfloor$. Their laws are the required multiset of uniform coordinates, and $A_i\le B_i$. The total expected displacement is $m_0(b-a)/4=d$. Each coordinate displacement is between zero and $b-a$. Hoeffding's inequality \cite{ProbHoeffding} gives \eqref{pfa:displacement}, for example with $c_0=1/32$. For the last assertion, let $Q_X,Q_Y$ be the sums of squared uniform widths in \eqref{pfa:XY}. Then \[ Q_X=b\frac{a(a-1)(2a-1)}6+\frac{b(b-1)(2b-1)}6, \] and $Q_Y$ is obtained by interchanging $a,b$. Both are at most $Nb^2/2$, and $d\ge N(b-a)/16$. At $t=\mu_X+d\sqrt{Q_X}/(\sqrt{Q_X}+\sqrt{Q_Y})$, Hoeffding bounds the sum of the two tails by $2\exp\{-2d^2/(\sqrt{Q_X}+\sqrt{Q_Y})^2\}$. The displayed estimates imply \eqref{pfa:separation}. \end{proof} The cutoff of Lemma~\ref{pfa:coupling} bounds the optimal loss $\Delta_{a,b}$ and the failure probability when the degree is sampled from the target. \begin{corollary}[Separated parameters]\label{pfa:separated} For $2\le a0$, both quantities are at most $2\exp[-\delta^2N/(256(1+\delta)^2)]$. \end{corollary} \begin{proof} For every real $t$ and every shape, \[ D_\lambda,\ \PFprob(c_{\lambda,T_\lambda}<0) \le1-F_{T,\lambda}(t)+F_{S,\lambda}(t). \] For the defect, cutoffs at most $t$ have defect at most $F_{S,\lambda}(t)$, and larger cutoffs have defect at most $1-F_{T,\lambda}(t)$. The bound on the sign when the degree is sampled from the target is \eqref{pfa:target-sign}. Average and use the cutoff of Lemma~\ref{pfa:coupling}. The last assertion follows from $(b-a)/b\ge\delta/(1+\delta)$. \end{proof} Writing $G(k)=\PFprob(X\le k)-\PFprob(Y\le k)$, the same coupling gives \begin{equation*} \PFexp_{\PFPl_N}\frac{c_{\Lambda,k}}{d_\Lambda} =G(k)=\PFprob(X\le kk_+). \] Lemma~\ref{pfa:tilt} gives constants $B,c_1>0$ with \begin{equation}\label{pfa:delta} e^{-Bz^2}\le\delta_B\le C e^{-c_1z^2},\qquad \delta_B\ge c\frac{\sigma}{1+z}p_Y(k_+). \end{equation} After fixing $\Gamma$, choose $\varepsilon$ sufficiently small as below. Proposition~\ref{pfa:retained} bounds the sum of the retained tail corrections by $\delta_B/4$ outside an event of probability $4\cdot16^{-p}$. Lemma~\ref{pfa:fpr}, with $m=\lceil\Gamma z^2\rceil$, gives \begin{equation*} \gamma_m(W_{b,a}) \le\exp(-\Gamma(\log2)z^2+o(1))+C e^{-c_2\sqrt N}. \end{equation*} For the tail projector, use $\PFexp|\mathcal R_m|^2\le\gamma_m$. Chebyshev and \eqref{pfa:delta} show that its residual exceeds $\delta_B/4$ with probability at most $16\gamma_m e^{2Bz^2}$. Choosing $\Gamma$ so that $\Gamma\log2>2B+10$ gives \begin{equation}\label{pfa:conditional-survival} \PFprob_{\PFPl_N} \left(1-F_{S,\Lambda}(k_+)<\delta_B/2\right) \le C e^{-c_3z^2}. \end{equation} \subsubsection{Separated means} Suppose $d\ge\Omega\sigma z$, where $\Omega$ is a sufficiently large absolute constant chosen after $B$. The optimized cutoff from Lemma~\ref{pfa:coupling} gives $t\in(\mu_X,\mu_Y)$ such that \begin{equation}\label{pfa:very-small-tails} \PFprob(X>t)+\PFprob(Y\le t) \le C e^{-(B+10)z^2} \end{equation} after increasing $\Omega$. Markov's inequality and $\delta_B\ge e^{-Bz^2}$ imply \[ \PFprob_{\PFPl_N} \left(1-F_{T,\Lambda}(t)>\delta_B/4\right) \le C e^{-10z^2}. \] On this good event and the one in \eqref{pfa:conditional-survival}, monotonicity gives, simultaneously for all integers $t0. \end{equation*} When the coin selects the source, a violation can therefore occur only below $t$, above $k_+$, or on an exceptional shape, with total probability at most $Ce^{-cz^2}$. When it selects the target, averaging the distribution-free bound \eqref{pfa:target-sign} and using \eqref{pfa:very-small-tails} gives the same conclusion, and so does the mixture. \subsubsection{Close means} Suppose instead $d<\Omega b\sqrt N z$. Since $d\ge N(b-a)/16$, we have $(b-a)/b\le16\Omega z/\sqrt N=o(1)$. Consequently $a,b\asymp\sqrt N$ and $b-a\le C_\Omega z$, and the variances of $X$ and $Y$ are both comparable to $N^2$. Take the integer window \begin{equation*} I_N=[\mu_X-L_0Nz,\ \mu_Y+L_0Nz]\cap\PFZ \end{equation*} with a fixed $L_0\ge1$. The two averaged degree laws have mass at most $Ce^{-cz^2}$ outside $I_N$. To compare corrections with $G(k)$, we bound $G(k)$ below relative to the point masses throughout this window. On this window, both $X$ and $Y$ satisfy the estimates of Appendix~\ref{pfa:smoothing}, with multiplier depending only on $\Omega,L_0$. Take an integer $\ell_0\asymp N/(1+z)$, small enough that $\ell_0\le d/2$; such an $\ell_0$ exists because $b-a\ge1$ gives $d\ge cN$. In the monotone coupling of Lemma~\ref{pfa:coupling}, \[ G(k)=\PFprob(X\le k0$ such that, whenever $2\le au\bigr) &\le\eta_{a,b}/u\qquad(u>0).\label{pfa:support-quenched} \end{align} There is an explicit filtered comparison with normalized dimension loss exactly $\eta_{a,b}$ from each module; the optimal loss $\Delta_{a,b}$ of Theorem~\ref{pfa:repair-theorem} may be smaller. \end{theorem} Hoeffding's inequality and integer-tuple counting bound this scalar tail. The proofs and boundary cases are in Appendix~\ref{pfa:support-estimates}. \begin{corollary}[Exponential and entropy rates]\label{pfa:support-rates} For $a\ge3$ and $b\ge6a$, \begin{equation*} \max\{\PFQ_{a,b}(c<0),\Delta_{a,b}\} \le\eta_{a,b}\le e^{-8N/2187}. \end{equation*} For $b\ge a\ge3$, \begin{equation}\label{pfa:support-entropy-simple} \eta_{a,b}\le \min\left\{1,\left[e^2\left(\frac ab+\frac1a\right)\right]^{N-1}\right\}. \end{equation} Consequently, if $m_*:=\min(a,b/a)\to\infty$, then \[ \log\eta_{a,b}\le-(N-1)\log m_*+O(N). \] In particular, for $b\asymp a^2$ with $a\to\infty$, \begin{equation}\label{pfa:support-square} \max\{\PFQ_{a,b}(c<0),\Delta_{a,b}\} \le\exp\bigl[-(1/3+o(1))N\log N\bigr]. \end{equation} \end{corollary} \section{Positivity on explicit shape families}\label{app:moments} Theorem~\ref{pfa:main} is a statement about typical shapes. This section proves the conjecture outright on explicit families, using the exact conditional means and variances. The means of $T_\lambda$ and $S_\lambda$ are explicit character values, and for fixed $a$ their variances are $O(b^3)$ (Theorem~\ref{thm:variance}). Whenever a shape family separates each mean from the opposite support endpoint by order $b^2$, a one-sided Chebyshev bound therefore compares the cumulative functions at every degree. Width and height families have this separation; for other families the same moments locate positive intervals. Write $r_\lambda(\rho)=\chi^\lambda(\rho1^{N-|\rho|})/d_\lambda$ when the specified nontrivial cycle type fits in $S_N$. In particular, $x_\lambda=r_\lambda(2)$ and $r_j^\lambda=r_\lambda(2^j)$. We omit the index $\lambda$ when it is fixed. \begin{proposition}[Exact conditional means]\label{prop:means} For every $\lambda\vdash ab$, \begin{align} \mu_T:=\E T_\lambda &=\frac{bc_a}{2}(1-x_\lambda)+\frac{c_b}{2}(1-r_a^\lambda),\label{eq:meanT}\\ \mu_S:=\E S_\lambda &=\frac{ac_b}{2}(1-x_\lambda)+\frac{c_a}{2}(1-r_b^\lambda).\label{eq:meanS} \end{align} Also \begin{equation}\label{eq:contentratio} x_\lambda=\frac{n(\lambda')-n(\lambda)}{c_N}. \end{equation} \end{proposition} The proof is in Appendix~\ref{prob:mean-derivation}. Write $\sigma_T^2$ and $\sigma_S^2$ for the variances of $T_\lambda$ and $S_\lambda$. They have explicit character expressions (Theorem~\ref{thm:exactvariance}), and the uniform estimate \eqref{eq:coarsevariance} gives $\sigma_T\sigma_S\le4a^3b^3$ for $b\ge a\ge2$. \subsection{Support and moment criteria}\label{sec:support} Support bounds settle thresholds outside the overlap of the two degree distributions. For a partition $\lambda$, define \begin{equation}\label{eq:eta} \eta_m(\lambda)=\sum_{i\ge1}\left\lfloor\frac{i-1}{m}\right\rfloor\lambda_i. \end{equation} Recall $D_+=bc_a+c_b$. \begin{theorem}[Support bounds]\label{thm:support} For every $\lambda\vdash ab$, \begin{align*} \eta_b(\lambda)&\le T_\lambda\le D_+-\eta_b(\lambda'),\\ \eta_a(\lambda)&\le S_\lambda\le ac_b+c_a-\eta_a(\lambda'). \end{align*} \end{theorem} The proof is in Appendix~\ref{prob:support-calculations}. Fusion and a one-variable expansion sharpen the upper bounds when the first row is long. \begin{corollary}[Upper support from the shape]\label{cor:depth-support} For $\lambda\vdash N=ab$ of depth $r=N-\lambda_1$, \[ T_\lambda,S_\lambda \le\min\{c_N-n(\lambda'),\ r(a+b-2)\}. \] If $r\le a\le b$, the second bound can be decreased to $r(a+b-2)-c_r$. At every integer cutoff at or above any of these common upper bounds, $[q^ks_\lambda]F_{a,b}=0$. \end{corollary} The proof is in Appendix~\ref{prob:support-calculations}. For $\lambda=(N-r,\nu)$, $c_N-n(\lambda')=Nr-r(r+1)/2-n(\nu')$. At fixed $a,r$, the depth bound confines nonzero coefficients to degrees of order $b$. Set $L_\lambda=\eta_a(\lambda)$ and \[ U_\lambda=\min\{D_+-\eta_b(\lambda'),\ c_N-n(\lambda'),\ r(a+b-2)\}, \qquad r=N-\lambda_1. \] If $L_\lambda\ge U_\lambda$, every coefficient of $[s_\lambda]F_{a,b}$ is nonnegative by support separation. Otherwise only the integer thresholds $L_\lambda\le k0$, Markov's inequality applied to $(\xi+\sigma^2/s)^2$ on $\{\xi\ge s\}$ gives $\Prob(\xi\ge s)\le\sigma^2/(\sigma^2+s^2)$. Apply this inequality to the upper tail of $T$ and the lower tail of $S$. Subtracting both bounds from one gives the stated expression. For integer-valued $T$, $\{T>k\}=\{T\ge k+1\}$, giving the stated lattice improvement. \end{proof} The exact variances in Theorem~\ref{thm:exactvariance} give a rational criterion. For an integer $k$ with $k+1>\mu_T$ and $k<\mu_S$, \begin{equation}\label{eq:exactcantelli} \bigl((k+1-\mu_T)(\mu_S-k)\bigr)^2 \ge\sigma_T^2\sigma_S^2 \quad\Longrightarrow\quad [q^ks_\lambda]F_{a,b}\ge0. \end{equation} A strict inequality gives a strictly positive coefficient; Example~\ref{ex:exactmoment} certifies one coefficient inside a support overlap in this way. The concavity of $(k-\mu_T)(\mu_S-k)$ reduces the comparison at all degrees to the two endpoints of the support overlap. \begin{theorem}[Complete support--moment comparison]\label{thm:interval} Suppose $S\ge L$, $T\le U$, and $\mu_T0$, define \begin{equation}\label{eq:explicitB} B(a,w)=\left\lceil\max\left\{ a,4,\frac{2(c_a+2a+w+1)}g, \frac{2(c_a+a+w+1)}h,\frac{16a^3}{gh} \right\}\right\rceil. \end{equation} For every $b\ge B(a,w)$ and every $\lambda\vdash ab$ of width at most $w$, $[s_\lambda]F_{a,b}\in\N[q]$. In particular, $g,h>0$ for $w\le2a-1$, and for $w\le2a$ if $a$ is even, so the conclusion holds in these cases. \end{recalled} The proof is in Appendix~\ref{prob:width-proof}. For $a=6$, sharper variance bounds give the smaller thresholds of Theorem~\ref{thm:explicit}. Height bounds give the same separation for families whose first row can grow linearly in $b$. \begin{corollary}[Complete positivity at large height]\label{cor:height} Fix $a\ge2$ and a real $h\le a$. Suppose \[ h>\max\{1,\sqrt{a/2}\}\quad\text{if $a$ is even}, \qquad h>\sqrt a\quad\text{if $a$ is odd}. \] For all sufficiently large $b$, depending only on $a,h$, every $\lambda\vdash ab$ with $\ell(\lambda)\ge hb$ satisfies $[s_\lambda]F_{a,b}\in\N[q]$. \end{corollary} The proof is in Appendix~\ref{prob:width-proof}. For $a=6$, Theorem~\ref{thm:explicit} gives the explicit range $b\ge10240$ and $\ell(\lambda)\ge\lceil7b/4\rceil$. \subsection{Positive intervals on general shapes}\label{sec:bulk} Overlapping supports can still leave a middle interval where the target cumulative probability is near one and the source cumulative probability is near zero. \begin{theorem}[A conditional macroscopic interval]\label{thm:bulk} Fix $a\ge2$ and let $\lambda_b\vdash ab$ have $x_{\lambda_b}\to x<1$. If \[ \frac{1-x^a}{4}K_a/G_a(x)$ and all sufficiently large $b$, \[ [q^ks_{\lambda_b}]F_{a,b}>0 \quad\text{whenever}\quad \lceil\mu_T+cb\rceil\le k\le\lfloor\mu_S-cb\rfloor. \] \end{corollary} \begin{proof} The mean calculation in Theorem~\ref{thm:bulk} gives $(\mu_S-\mu_T)/b^2\to G_a(x)>0$. In \eqref{eq:mastervariance}, $I(a,b)=O_a(b)$, $O(a,b)/b^3\to(9a+5)/18$, and $I(b,a)/b^3\to7a/9$; the source outer term is bounded by $c_a/2$. Hence \[ \limsup_{b\to\infty}\frac{\sigma_T\sigma_S}{b^3}\le K_a. \] At each endpoint of the asserted interval, $(k+1-\mu_T)(\mu_S-k)/b^3\to cG_a(x)>K_a$. By concavity, the margin product is minimized at an endpoint. The strict form of \eqref{eq:exactcantelli} proves the claim. \end{proof} For hooks of growing depth, the exact variances of Theorem~\ref{thm:exactvariance} replace $K_a$ by the limit of $\sigma_T\sigma_S/b^3$ and so sharpen the constant $c$ (Theorem~\ref{thm:hookinterval}); for $a=6$ and the hooks $(5b,1^b)$ the margin $b/5$ suffices for all sufficiently large $b$ (Corollary~\ref{cor:growinghookinterval}). \subsection{The whole-shape question}\label{sec:whole-shape} Theorems~\ref{pfa:main} and~\ref{thm:width} approach the conjecture from two sides, typical shapes at sampled degrees and every degree on special families. This subsection records what each side says about an arbitrary shape at every degree. By integrality, a negative coefficient of $[s_\lambda]F_{a,b}$ forces $V_\lambda\ge1/(2d_\lambda)$. The dimension bound $d_\lambda\le\sqrt{N!}$ is too large to convert the rate $e^{-cN^{1/3}}$ into a vanishing bound on the Plancherel probability that a shape has a negative coefficient. Theorem~\ref{pfa:main} weights each shape by $d_\lambda^2$. If instead every shape and degree is counted once, the support bounds and the exact means still make negative coefficients rare, because every degree between the target endpoint $D_+$ and the top of the source support carries a strictly positive coefficient. \begin{corollary}[Unweighted count of negative coefficients]\label{cor:unweighted-count} For integers $2\le a0$ and \[ \frac{B}{B+P}\le\min\Bigl\{1,\ \frac2a+\frac{2(a-1)}{b-1}\Bigr\}. \] \end{corollary} \begin{proof} Let $p(N)$ be the number of partitions of $N$. By Theorem~\ref{thm:support}, $T_\lambda\le D_+$, so \eqref{eq:cdf} gives $[q^ks_\lambda]F_{a,b}=d_\lambda(1-\Prob(S_\lambda\le k))\ge0$ for $k\ge D_+$. Hence every negative coefficient has $k0$, and since $B\le p(N)D_+$ and $B/(B+P)$ increases with $B$, \[ \frac{B}{B+P}\le\frac{D_+}{D_+ + (ac_b/2-D_+)}=\frac{2D_+}{ac_b} =\frac2a+\frac{2(a-1)}{b-1}. \] Finally, summing \eqref{eq:cdf} over $k$ and over $\lambda$ with weights $d_\lambda$ gives $\sum_\lambda d_\lambda(\mu_{S,\lambda}-\mu_{T,\lambda})$, which equals $N!(a-1)(b-1)(b-a)/4$ by \eqref{eq:meanT}, \eqref{eq:meanS}, and $\sum_\lambda d_\lambda\chi^\lambda(g)=0$ for $g\ne1$. This is positive for $a0$. \end{proof} For $b=a^2$ the bound is $2/a+2/(a+1)<4N^{-1/3}$, so the negative coefficients are a vanishing fraction of the nonzero ones as $a\to\infty$ and $b/a\to\infty$. The positive coefficients counted here lie above the target support, so the sign pattern below $D_+$ is untouched. For general shapes, the moment criteria certify the intervals of Corollary~\ref{cor:linear-margins}, and the thresholds inside the support overlap remain open. \section{The rectangular conjecture} \label{pfa:qt} Both arguments of Section~\ref{pfa:overview} extend to Bergeron's rectangular conjecture \citep[Conjecture~4]{bergeron2017foulkes}, which replaces the inner factor $H_a$ by the modified Macdonald polynomial of a rectangle and so carries a second grading. An equivariant bijection preserving this grading requires equal multiplicities in each shape and $t$-degree after forgetting the $q$-degree. The rectangular plethysms agree at $q=1$, so the matching argument of Section~\ref{pfa:overview} applies within each such fiber, and the averaged degree laws bound the dimension loss. Our normalization of the modified Macdonald polynomial is \[ \widetilde H_\mu[X;q,t] =t^{n(\mu)}J_\mu\left[\frac{X}{1-t^{-1}};q,t^{-1}\right], \] where $J_\mu$ is the integral Macdonald form. It is the normalization of \citet{bergeron1999identities} and \citet{haiman2001hilbert}; see also \citet[Chapter~VI, (8.1') and (8.3)]{macdonald1995}. In particular $\widetilde H_{(n)}(X;q,t)=H_n(X;q)$. Let $r\ge1$, $2\le ax_0)+\PFprob(Y_r\le x_0). \] The optimized Hoeffding cutoff yields the sharper formula \begin{equation*} \Delta^{(r)}_{a,b} \le2\exp\left\{-\frac{2d_r^2} {(\sqrt{Q_{A,r}}+\sqrt{Q_{B,r}})^2}\right\}. \end{equation*} Since $N-a-b+1\ge N/4$ and $Q_{A,r},Q_{B,r}\le Nb^2/2$, we obtain \eqref{pfa:qt-repair}. \end{proof} The exponent in \eqref{pfa:qt-repair} is $N(b-a)^2/b^2=r(a/b)(b-a)^2$; it grows when $b-a$ or $r$ does. The target support also controls sampled signs over the full $q$-degree range. Sample $(\Lambda,J)$ with law $w$, and then sample a $q$-degree $K$ from its target or source multiplicity with equal probabilities. This law is the bigraded Plancherel mixture. \begin{corollary}[A support bound for bigraded signs]\label{pfa:qt-support} Put $D_r=rb c_a+c_b$, $\eta_r=\PFprob(Y_r1. \end{equation} \end{lemma} \begin{proof} The part of $P(E)$ above $t$ is at most $1-F_P(t)$. On the remaining part, $F_P(K)0$, and take $k_*$ with $g(k_*)=-D$. Let $j0$, \begin{equation}\label{pfa:tail-lower} e^{-B_Lz^2}\le\PFprob(Z>k_+) \le C_L e^{-c_Lz^2},\qquad \PFprob(Z>k_+)\ge c_L\frac{\sigma}{1+z}p_Z(k_+). \end{equation} \end{lemma} \begin{proof} Put $\psi(\theta)=\log\PFexp e^{\theta Z}$. A uniform coordinate of width $w\le2u$ has tilted point probabilities between $e^{-w|\theta|}$ and $e^{w|\theta|}$ times its original probabilities. The identity $\Var U=\tfrac12\PFexp(U-U')^2$, with $U'$ an independent copy, therefore bounds its tilted variance between $e^{-2w|\theta|}\Var U$ and $e^{2w|\theta|}\Var U$. For fixed $A$, the condition $z=o(\sqrt N)$ gives $2uAz/\sigma=o(1)$, so $c\sigma^2\le\psi''(\theta)\le C\sigma^2$ uniformly on $|\theta|\le Az/\sigma$. Choose $A=A_L$ so that $cA>L$. Integrating this lower bound shows that $\psi'$ at the two endpoints brackets every $k$ in the stated window. Its strict monotonicity gives the mean-setting tilt and \eqref{pfa:theta}. Since $I'(x)=\theta_x$, integration from $\PFexp Z$ proves the bounds on $I$. Each tilted uniform is discrete log-concave, and convolution preserves this property \cite[Corollary 4.3]{ProbBMM}. At an integer mean $k$, any such law $p$ satisfies \begin{equation}\label{pfa:entropy-at-mean} p(k)\ge\{2\pi e(\PFvar(p)+1/12)\}^{-1/2}. \end{equation} Indeed, the piecewise linear interpolation of $\log p$ is concave, so Jensen gives $\log p(k)\ge-H(p)$. Adding an independent uniform variable on $[-1/2,1/2]$ gives differential entropy $H(p)$ and variance $\PFvar(p)+1/12$; the Gaussian entropy upper bound proves \eqref{pfa:entropy-at-mean}. The upper bound $\max p\le2/\sqrt{1+4\PFvar(p)}$ is the discrete log-concave maximal-atom bound \cite[Theorem 1.1]{ProbBMM}. Undoing the tilt proves \eqref{pfa:tilt-density}. The derivative bound on $I$ proves comparison over intervals of length $c\sigma/(1+z)$. Sum over an interval immediately to the right of $k_+$ to obtain the last inequality in \eqref{pfa:tail-lower} and the lower exponential bound. Polynomial factors are absorbed using $\log N=o(z^2)$. The upper bound is Hoeffding's inequality, since the sum of squared uniform widths is $O(\sigma^2)$. \end{proof} \begin{lemma}[Growing finite differences, relative to the density] \label{pfa:relative-difference} Suppose $r0$. Suppose a nonnegative reference function $J(k)$ satisfies $J(k)\ge c_L\sigma p_Z(k)/(1+z)$ on a moderate window of multiplier $L$. On the character event \eqref{pfa:good-characters}, the sum of the absolute class corrections of indices $1\le r2B+10$; finally an absolute $\varepsilon>0$ so small that each geometric series in \eqref{pfa:geometric-correction}, for both uses, has the required sum, which is possible because the constants in these series depend only on $\Gamma,\Omega,L_0$. The close residual has an extra $\log N$ in its exponent and absorbs every fixed $B_{\Omega,L_0}$. With these fixed choices, all necessary growth conditions hold: \[ \begin{gathered} ps=O(N^{2/3})0$, Hoeffding's inequality \cite{ProbHoeffding} gives \eqref{pfa:support-hoeffding}; otherwise the bound is one. There are exactly $a(b-1)+(a-1)=N-1$ nontrivial uniform coordinates. Each point in their product sample space has probability $1/((b!)^a a!)$. Dropping the upper coordinate bounds leaves $\binom{D_++N-2}{N-1}$ nonnegative integer tuples whose sum is at most $D_+-1$, proving \eqref{pfa:support-counting}. \end{proof} \begin{proof}[Proof of Corollary~\ref{pfa:support-rates}] For $a\le b$, the inequality $(a-1)(2a-1)\le2a^2\le3b^2-b$ gives $Q_Y\le Nb^2/3$. For $a\ge3$, the normalized gap is \[ \frac{\delta}{Nb} =\frac14\left(1-\frac2a-\frac{2a-1-2/a}{b} +\frac{a-1}{b^2}\right). \] Its derivative in $b\ge a$ is positive. At $b=6a$ it equals $1/6-65/(144a)+11/(144a^2)$, which increases for $a\ge3$ and takes the value $2/81$ at $a=3$. Thus $\delta\ge2Nb/81$ in the stated range, and \eqref{pfa:support-hoeffding} gives $2\delta^2/Q_Y\ge8N/2187$. For the entropy estimate put $n=D_++N-2$ and $k=N-1$. The bounds $\binom nk\le(en/k)^k$ and $j!\ge(j/e)^j$ imply \[ \frac{\binom nk}{(b!)^a a!} \le\left(\frac{e^2n}{bk}\right)^k\frac{e^{a+1}}{ba^a} \le\left(\frac{e^2n}{bk}\right)^k, \] where the last step uses $b\ge a\ge3>e$. Also \[ \frac{D_++N-2}{b(N-1)} \le\frac{2D_+}{bN}+\frac1b =\frac ab+\frac1a-\frac1{ab}. \] For the first inequality, clearing the positive denominators leaves $D_+(N-2)+N\ge0$. The last two displays prove \eqref{pfa:support-entropy-simple}. Since $a/b+1/a\le2/m_*$, $\log\eta_{a,b}\le-(N-1)\log m_*+(N-1)\log(2e^2)$. If $b\asymp a^2$, then $N\asymp a^3$ and $\log m_*=(1/3)\log N+O(1)$, proving \eqref{pfa:support-square}. \end{proof} The restriction $a\ge3$ is necessary: for $a=2$, $\delta=1/2-b$ and \eqref{pfa:support-hoeffding} gives no decay, while for $a=3$, $\delta=(b^2-13b+6)/4$ and the strict-cutoff Hoeffding bound is less than one precisely for integers $b\ge13$. Nonnegative multiplicities make the support bound at $t=1$ valid in each $t$-degree. \begin{proof}[Proof of Corollary~\ref{pfa:qt-support}] By \eqref{pfa:qt-specializations}, $\mathcal T^{(r)}(X;q,1)=H_b[H_a(X;q)^r]$ has $q$-degree at most $D_r=rb c_a+c_b$. Since all bigraded Schur multiplicities are nonnegative, each target $(\lambda,j)$ fiber has the same support bound. Apply the proof of Theorem~\ref{pfa:support-tail} in that fiber: both its conditional sign-failure mass and its cumulative defect are bounded by its source probability of degree below $D_r$. Averaging with $w$ gives the bound by $\eta_r$. The independent-uniform representation of $Y_r$ has $M_r$ nontrivial coordinates and $(b!)^{ra}a!$ equiprobable outcomes. Dropping the upper coordinate bounds leaves $\binom{D_r+M_r-1}{M_r}$ nonnegative integer tuples with sum at most $D_r-1$, proving the counting bound. Write $M=M_r$. The binomial and factorial bounds give \[ \frac{\binom{D_r+M-1}{M}}{(b!)^{ra}a!} \le\left[\frac{e^2(D_r+M)}{bM}\right]^M \frac{e^{ra+1}}{b^{ra-a+1}a^a}. \] For $a\ge3$, the last factor is at most $(e/a)^{ra+1}\le1$. Also $M\ge N/2$ and $D_r=N(a-1)/2+b(b-1)/2\le(a-1)M+bM/(ra)$. Thus $(D_r+M)/(bM)\le a/b+1/(ra)$, proving the entropy bound. For the exponential bound, let $\delta_r=\PFexp Y_r-D_r$. Directly, \[ \frac{\delta_r}{Nb} =\frac{b-2a+1}{4b} -\frac{2b(b-1)-a(a-1)}{4rab^2}. \] This expression increases with $r\ge1$. The $r=1$ calculation in Corollary~\ref{pfa:support-rates} therefore gives $\delta_r\ge2Nb/81$ when $b\ge6a\ge18$. The squared-width sum satisfies $Q_{B,r}\le rQ_{B,1}\le Nb^2/3$. Hoeffding's inequality at the strict cutoff $D_r-1$ yields \[ \eta_r\le\exp\left(-\frac{2(\delta_r+1)^2}{Q_{B,r}}\right) \le e^{-8N/2187}. \] Finally, if $b\asymp ra^2$ and $ra\to\infty$, then $a/b+1/(ra)\asymp1/(ra)$ and $(N-M_r)\log(ra)=O(N)$. The entropy bound gives $\log\eta_r\le-N\log(ra)+O(N)$. At fixed $a$, $N=rab\asymp r^2a^3$, so $\log(ra)=\tfrac12\log N+O(1)$. \end{proof} \section{Spectral realization and conditional moments}\label{app:variance-proofs} Section~\ref{app:moments} uses the exact conditional means and variances of $T_\lambda$ and $S_\lambda$, and its width and height families need variances of order $b^3$ uniformly in the shape. Expanding the plethysms gives neither. Instead, a Hermitian element of the group algebra realizes each grading on the irreducible spaces, and conjugation averages turn its moments into character values, which can be bounded uniformly. \subsection{Operators on irreducible spaces}\label{sec:operators} The fake-degree polynomial of $S^\nu$ in $R_n$ is $f^\nu(q)=\sum_{T\in\SYT(\nu)}q^{\maj(T)}$. In a unitary Wedderburn decomposition of $\C[S_n]$, choose a Hermitian element $Z_n\in\C[S_n]$ whose matrix on $S^\nu$ is diagonal in the Young basis with entries $\maj(T)$. It exists because $\C[S_n]\cong\bigoplus_\nu\End(S^\nu)$. Right multiplication by $Z_n$ on the regular module has graded Frobenius character $H_n$, and $0\le Z_n\le c_nI$ in every unitary representation. Embed identical copies of $Z_a$ in the $b$ blocks, and embed $Z_b$ in the subgroup permuting whole blocks. Put \begin{equation}\label{eq:A} A_{a,b}=K_{a,b}+L_{a,b},\qquad K_{a,b}=\sum_{j=1}^bZ_a^{(j)},\quad L_{a,b}=Z_b^{\mathrm{blocks}}. \end{equation} The inner sum is invariant under block permutations, so $K_{a,b}$ and $L_{a,b}$ commute. \begin{proposition}[Spectral realization]\label{prop:spectral} For every $\lambda\vdash N$, \begin{equation}\label{eq:spectral} u_\lambda(q)=\Tr_{S^\lambda}q^{A_{a,b}},\qquad v_\lambda(q)=\Tr_{S^\lambda}q^{A_{b,a}}. \end{equation} Thus $T_\lambda,S_\lambda$ are normalized eigenvalue distributions of explicit Hermitian elements on the same irreducible space. \end{proposition} \begin{proof} Use the same graded left-module isomorphism $R_a\cong\C[S_a]$ in every factor, and $R_b\cong\C[S_b]$. The resulting tensor identification with $\C[W_{a,b}]$ respects outer permutations. Right multiplication by the sum of the inner $Z_a$'s acts independently of the outer coordinate, because that sum is invariant under its permutations. Right multiplication by the outer $Z_b$ changes only the outer coordinate. Hence their sum gives the product grading in \eqref{eq:modules}. Induction identifies it with right multiplication by $A_{a,b}$ on $\C[S_N]$. In the bimodule decomposition of the regular representation, the left $S^\lambda$-multiplicity space is its dual. The right action on that dual has the same eigenvalue multiset as the chosen Hermitian matrix, proving \eqref{eq:spectral}. \end{proof} \subsection{Classical fixed-shape moments and conditional means}\label{prob:mean-derivation} Conjugation averaging reduces the conditional means to the mean and variance of the major index on a random tableau, which are classical. Let $\operatorname{ct}(\nu)=\sum_{(i,j)\in\nu}(j-i)$ be the content sum of $\nu\vdash n$, put $v_n=n(n-1)(2n+5)/72$, and let $\E_\nu$ and $\Var_\nu$ refer to a uniformly random standard Young tableau of shape $\nu$. The hook formula gives \begin{equation}\label{eq:majmoments} \E_\nu\maj=\frac{c_n-\operatorname{ct}(\nu)}{2},\qquad \Var_\nu\maj=\frac1{12}\left(\sum_{j=1}^n j^2- \sum_{c\in\nu}h_c^2\right)\le v_n. \end{equation} These moment identities are due to Adin--Roichman \citep[Propositions~3.1--3.2]{AR}. They follow by taking first and second cumulant derivatives of $q^{n(\nu)}[n]_q!/\prod_{c\in\nu}[h_c]_q$ at $q=1$. The last bound uses $\sum_ch_c^2\ge n$. Averaging $Z_n$ over conjugation by $S_n$ (twirling) therefore gives \begin{equation}\label{eq:twirlZ} \mathsf E_nZ_n=\frac12\left(c_nI-\sum_{i2$ contributes zero to this derivative. For cycle types $1^j$, $2,1^{j-2}$, $3,1^{j-3}$, and $2^2,1^{j-4}$, the second Euler derivatives over $j!$ are $c_j^2/4+v_j$, $-c_j/2$, $2/3$, and $1/2$, respectively. For the identity, $Q_j/j!$ is the probability generating function of a sum of independent uniforms on $\{0,\ldots,i-1\}$, $1\le i\le j$. Their means sum to $c_j/2$ and their variances sum to $v_j$. For a transposition, differentiate $Q_j(q)(1-q)/(1+q)$. For the last two classes differentiate $Q_j(q)(1-q)^2/[3]_q$ and $Q_j(q)(1-q)^2/[2]_q^2$. Thus, with $C_\rho$ denoting the sum of all permutations of the indicated nontrivial cycle type in $S_j$, \[ \mathsf E_jZ_j=\frac12(c_jI-C_2), \] \[ \mathsf E_jZ_j^2 =\left(\frac{c_j^2}{4}+v_j\right)I -\frac{c_j}{2}C_2+\frac23C_3+\frac12C_{2^2}. \] There are $c_j$ transpositions, $2\binom j3$ three-cycles, and $3\binom j4$ double transpositions. Taking normalized traces gives \[ \tau(Z_n^2)=\frac{c_n^2}{4}+v_n-\frac{c_n^2}{2}x +\alpha_n r(3)+\beta_n r(2^2). \] Independent averaging over distinct inner blocks $i,j$, whose supports are disjoint, gives \[ \tau(Z_n^{(i)}Z_n^{(j)}) =\frac{c_n^2}{4}\bigl(1-2x+r(2^2)\bigr). \] Summing the $m$ diagonal terms and $m(m-1)$ ordered off-diagonal terms, then subtracting $\tau(K)^2=m^2c_n^2(1-x)^2/4$, proves the formula for $V_K$. Applying the same one-block second-moment identity to the outer $S_m$ multiplies each cycle length by $n$, proving the formula for $V_L$. For the covariance, $K$ is invariant under the block permutation group. Consequently $\tau(KL)=\tau(K\mathsf E_mL)$. Let $B_{uv}$ exchange blocks $u,v$. To average $Z_n^{(i)}$ while preserving $B_{uv}$, use arbitrary conjugation in block $i$ if $i\notin\{u,v\}$, and simultaneous identical conjugation in blocks $u,v$ otherwise. An inner transposition multiplied by $B_{uv}$ has cycle type \[ 2^{n+1}\quad(i\notin\{u,v\}),\qquad (4,2^{n-2})\quad(i\in\{u,v\}). \] It follows that \[ \tau(KB_{uv}) =\frac{mc_n}{2}y -\frac{c_n}{2}\bigl((m-2)r(2^{n+1})+2r(4,2^{n-2})\bigr). \] Use $\mathsf E_mL=(c_mI-\sum_{u0$ when $w>1$; at $w=1$ it is $2a-2>0$. For even $a$, the boundary $w=2a$ has $4g=(2a)^{-a}>0$. The second coefficient $h$ is positive in all these cases. \end{proof} Lemma~\ref{lem:shape} bounds the source support and mean in terms of height, without restricting width. \begin{proof}[Proof of Corollary~\ref{cor:height}] Put $l=\lceil hb\rceil$ and suppose $\ell(\lambda)\ge l$. Lemma~\ref{lem:shape} gives \[ L_\lambda\ge\frac{l(l-1)}{2a}-\frac{(a-1)l}{a} =\frac{h^2}{2a}b^2-O_{a,h}(b). \] If $a$ is even, \eqref{eq:transpower} gives $r_a\ge x_\lambda^a-4(a-1)/b\ge-4(a-1)/b$. The target mean therefore satisfies $\mu_T\le b^2/4+O_a(b)$. For odd $a$, the bound $r_a\ge-1$ instead gives $\mu_T\le b^2/2+O_a(b)$. For the source mean, $r_b\le1$ and the height estimate for $x_\lambda$ give \[ \mu_S\ge\frac{ab(b-1)(l-1)}{2(ab-1)} \ge\frac h2b^2-O_{a,h}(b). \] Also $U_\lambda\le D_+=b^2/2+O_a(b)$. Thus $\mu_S-U_\lambda\ge\tfrac{h-1}{2}b^2-O_{a,h}(b)$, while the leading coefficient of $L_\lambda-\mu_T$ is at least $h^2/(2a)-1/4$ for even $a$ and $h^2/(2a)-1/2$ for odd $a$. Both gaps are positive of order $b^2$ under the stated hypotheses. If $L_\lambda\ge U_\lambda$, use support separation. Otherwise each endpoint product in \eqref{eq:intervaltest} is at least $(L_\lambda-\mu_T)(\mu_S-U_\lambda)$, of order $b^4$ uniformly over the shapes. Since $\sigma_T\sigma_S\le4a^3b^3$, Theorem~\ref{thm:interval} applies for all sufficiently large $b$. \end{proof} The same argument can use the column sizes directly. For fixed $a$, let $\lambda_b\vdash ab$ and $\gamma_b=(ab)^{-2}\sum_j(\lambda_b')_j^2$. The shape and mean bounds give \[ \frac{L_{\lambda_b}-\mu_T}{b^2} \ge\frac{2a\gamma_b-1+x_{\lambda_b}^a}{4}-O_a(b^{-1}), \qquad \frac{\mu_S-U_{\lambda_b}}{b^2} \ge\frac{a(1-x_{\lambda_b})-2}{4}-O_a(b^{-1}). \] Consequently, positive lower limits of both displayed leading terms suffice for complete positivity, without a width or height cutoff. \subsection{Sixth-power families}\label{sec:explicit} At $a=6$ the uniform variance bound of Theorem~\ref{thm:variance} is too crude for the thresholds in Theorem~\ref{thm:explicit}, so we first sharpen it. \begin{corollary}[Variance bounds for the sixth power]\label{prob:sixth-variance} For $a=6$, the sharper bounds \begin{align} \sqrt{\Var T_\lambda} &\le B_T(b):=\sqrt{\frac{695}{6}b}+\sqrt{\frac{59}{18}b^3},\label{eq:BT}\\ \sqrt{\Var S_\lambda} &\le B_S(b):=\sqrt{\frac{14}{3}b^3}+\frac{15}{2}\label{eq:BS} \end{align} hold for every $\lambda\vdash6b$. For $b\ge32$, each variance is at most $5b^3$. \end{corollary} \begin{proof} For $n=6$, direct simplification gives $I(6,b)=\frac{695}{6}b$, $O(6,b)\le\frac{59}{18}b^3$, and $I(b,6)\le(14/3)b^3$. Use the range bound $c_6/2=15/2$ for the source outer term. Then \eqref{eq:mastervariance} gives \eqref{eq:BT}--\eqref{eq:BS}. For $b\ge32$, $B_T(b)/b^{3/2}<11/32+11/6=209/96<\sqrt5$, while $B_S(b)/b^{3/2}<13/6+1/24=53/24<\sqrt5$. The inequality $15/(2b^{3/2})<1/24$ follows from $32^3>180^2$. \end{proof} For $a=6$, define $M_b=(b^2+79b-20)/4$ and $U_b=(b^2+29b)/2$. The exact means and \eqref{eq:transpower} imply $\mu_T\le M_b$ and $T_\lambda\le U_b$. Indeed $r_6\ge x^6-20/b\ge-20/b$ and $1-x\le2$. With Corollary~\ref{prob:sixth-variance}, these bounds give smaller thresholds than \eqref{eq:explicitB}. \begin{theorem}[The explicit ranges]\label{thm:explicit} For $\lambda\vdash6b$, each condition below implies $[s_\lambda]F_{6,b}\in\N[q]$: \[ \begin{array}{c|c} \text{shape condition}&\text{sufficient range}\\\hline \lambda_1\le6&b\ge120\\ \lambda_1\le11&b\ge1200\\ \ell(\lambda)\ge2b+1&b\ge350\\ \ell(\lambda)\ge\lceil7b/4\rceil&b\ge10240. \end{array} \] \end{theorem} \begin{proof} For width at most $w$, Lemma~\ref{lem:shape} gives $S_\lambda\ge L_b:=3b^2/w-3b$. When $6b\ge w^2$, the same lemma gives $x\le0$, so $\mu_S\ge m_b:=3b(b-1)/2$. For an overlapping support interval, the endpoint products in Theorem~\ref{thm:interval} are at least $(L_b-M_b)(m_b-L_b)$ and $(U_b-M_b)(m_b-U_b)$. If $w=6$ and $b\ge120$, then $L_b-M_b=(b^2-91b+20)/4\ge29b/4$ and $m_b-L_b=b^2+3b/2$, whose product exceeds $5b^3$. Also $U_b-M_b\ge b^2/8$ and $m_b-U_b\ge b^2/2$, so the other product is at least $b^4/16>5b^3$. If $w=11$ and $b\ge1200$, then $L_b-M_b=(b^2-1001b+220)/44\ge199b/44$ and $m_b-L_b\ge27b^2/22$, whose product is at least $(5373/968)b^3>5b^3$. The same upper-endpoint estimate applies. The support--moment theorem and \eqref{eq:BT}--\eqref{eq:BS} prove both width assertions. Now suppose $\ell(\lambda)\ge2b+1$. Then $S_\lambda\ge L_b:=(4b^2-8b-5)/12$ by summing $\lfloor i/6\rfloor\ge(i-5)/6$ for $0\le i\le2b$. Also $x\le1-4b/(6b-1)<1/3$, hence $\mu_S\ge m_b:=b(b-1)$. For $b\ge350$, $L_b-M_b=(b^2-245b+55)/12\ge35b/4$ and $m_b-L_b\ge3b^2/5$, whose product is at least $(21/4)b^3>5b^3$. At the other endpoint, $U_b-M_b\ge b^2/8$ and $m_b-U_b\ge b^2/4$. Their product is at least $b^4/32>5b^3$. Theorem~\ref{thm:interval} applies again, and disjoint relaxed supports give the result directly. Finally suppose $\ell(\lambda)\ge\lceil7b/4\rceil$. The height estimate in Lemma~\ref{lem:shape} and $\mu_T\le M_b\le b^2/4+20b$ give $L_\lambda-\mu_T\ge b^2/192-1037b/48\ge b^2/384$ for $b\ge8296$. The source mean is at least $(b-1)(7b/4-1)/2$, so $\mu_S-U_\lambda\ge3b^2/8-127b/8\ge3b^2/16$ for $b\ge85$. If the supports overlap, both endpoint products are at least $b^4/2048\ge5b^3$ for $b\ge10240$. Corollary~\ref{prob:sixth-variance} and Theorem~\ref{thm:interval} prove the last assertion; disjoint supports again give it directly. \end{proof} For $a=6$, Chebyshev's inequality and the exact means give an effective form of Theorem~\ref{thm:bulk}. \begin{corollary}[A uniform sixth-power interval]\label{cor:bulkfinite} Let $0<\varepsilon\le1/2$, $b\ge1280\varepsilon^{-4}$, and $\lambda\vdash6b$ satisfy $\lambda_1\le(1-\varepsilon)6b$. Every integer $k$ with $\mu_T+\varepsilon^2b^2/8\le k\le\mu_T+\varepsilon^2b^2/4$ satisfies $[q^ks_\lambda]F_{6,b}\ge d_\lambda/2$. \end{corollary} \begin{proof} The elementary estimate $x\le(\lambda_1-1)/(N-1)$ implies $1-x\ge\varepsilon$. From the exact means and $|r_6-x^6|\le20/b$, $\mu_S-\mu_T\ge\frac34b(b-1)(1-x)^2-20b\ge\frac12\varepsilon^2b^2$ in the stated range. Here $x^6-6x+5=(1-x)^2[3+(1+x)^2(x^2+2)]$. Chebyshev's inequality in \eqref{eq:cdf}, with the variance bound $8b^3$, gives a total error at most $640/(\varepsilon^4b)\le1/2$, which proves the claim. \end{proof} \subsection{Hooks of growing depth} Hook characters have the closed form below, so for hooks the exact variances of Theorem~\ref{thm:exactvariance} can be evaluated and sharpen the constant in the order-$b$ margins of Corollary~\ref{cor:linear-margins}. The exterior powers of the standard representation give the hook characters: \begin{equation}\label{eq:hookchar} \sum_{r=0}^{N-1}\chi^{(N-r,1^r)}(g)z^r =\frac{\prod_{C\in\Cyc(g)}(1-(-z)^{|C|})}{1+z}. \end{equation} \begin{theorem}[Hook intervals from exact variances]\label{thm:hookinterval} Fix $a\ge2$, let $r_b/(ab)\to\rho\in(0,1)$, and set $\lambda_b=(ab-r_b,1^{r_b})$. Write \[ x=1-2\rho,\qquad z=\rho(1-\rho),\qquad G=\frac{a(1-x)-1+x^a}{4}>0, \] \[ C_a(\rho)=\frac1{36}+\frac29(1-3z)^a -\frac{x^{2a}}4-\frac{azx^{2a-2}}4. \] Then the exact conditional variances satisfy \[ \frac{\operatorname{Var}T_{\lambda_b}}{b^3}\longrightarrow C_a(\rho), \qquad \frac{\operatorname{Var}S_{\lambda_b}}{b^3}\longrightarrow\frac{az}{12}. \] Consequently, for every constant $c>\sqrt{C_a(\rho)az/12}/G$, the degree $k_b=\lceil\mu_T(\lambda_b)+cb\rceil$ has $[q^{k_b}s_{\lambda_b}]F_{a,b}>0$ for all sufficiently large $b$. The same conclusion holds for every integer degree between this lower endpoint and $\lfloor\mu_S(\lambda_b)-cb\rfloor$. \end{theorem} \begin{proof} Put $M=ab-1$, $\rho_M=r_b/M$, $x_M=1-2\rho_M$, and $z_M=\rho_M(1-\rho_M)$. The hook character at $j$ disjoint transpositions is \[ r(2^j)=\sum_{h=0}^j(-2)^h\binom jh \frac{(r_b)_{\underline h}}{(M)_{\underline h}} =x_M^j-\frac{2j(j-1)z_Mx_M^{j-2}}M+O(M^{-2}) \] for fixed $j\ge2$; for $j=1$ it is exactly $x_M$. The error is uniform when $\rho_M$ stays in a compact subinterval of $(0,1)$. Expand the falling-factorial quotients to first order in $1/M$ and differentiate $(1-2\rho_M)^j$ twice. The hook determinant formula also gives $r(3^a)\to(1-3z)^a$, by selecting the finitely many nontrivial eigenvalues of the $a$ three-cycles. The expansion of $r(2^j)$ at $j=a$ and $j=2a$ gives \[ r(2^{2a})-r(2^a)^2 =-\frac{4a^2z_Mx_M^{2a-2}}M+O(M^{-2}). \] In the exact outer variance, use $\beta_b-c_b^2/4=-b(b-1)(2b-3)/8$ and group the last two terms as \[ V_L=v_b+\alpha_b r(3^a) +(\beta_b-c_b^2/4)r(2^{2a}) +\frac{c_b^2}{4}\bigl(r(2^{2a})-r(2^a)^2\bigr). \] Dividing by $b^3$ gives $C_a(\rho)$. For the target, $V_K=O_a(b)$ by the formula for $V_K$ and $r(2^2)-x_M^2=O(b^{-1})$. Cauchy--Schwarz gives $C_{KL}=O_a(b^2)$, so these two terms do not affect the limit. For the source, the formula for $V_K$ can be regrouped as \[ V_K=av_b+a\alpha_b r(3) +a(\beta_b-c_b^2/4)r(2^2) +\frac{a^2c_b^2}{4}\bigl(r(2^2)-x_M^2\bigr). \] Here $r(3)\to1-3z$ and the last difference is exactly $-4z_M/(M-1)$. The coefficient of $b^3$ is \[ a\left(\frac1{36}+\frac29(1-3z)-\frac14x^2\right)-\frac{az}{4} =\frac{az}{12}. \] The source outer operator has range $[0,c_a]$, hence bounded variance; its covariance with the inner sum is $O_a(b^{3/2})$, proving the second variance limit. Proposition~\ref{prop:means} gives $(\mu_S-\mu_T)/b^2\to G$. At the stated two endpoints the lattice margin product divided by $b^3$ tends to $cG$, whereas the exact standard-deviation product divided by $b^3$ tends to $\sqrt{C_a(\rho)az/12}$. The strict inequality in the hypothesis proves eventual coefficient positivity. Concavity of the lattice margin product extends the endpoint estimates to the full integer interval. \end{proof} \begin{corollary}\label{cor:growinghookinterval} For all sufficiently large $b$, every integer $k$ satisfying \[ \left\lceil\mu_T+b/5\right\rceil\le k\le \left\lfloor\mu_S-b/5\right\rfloor \] has $[q^ks_{(5b,1^b)}]F_{6,b}>0$. \end{corollary} \begin{proof} At $a=6$ and $\rho=1/6$, the constants in Theorem~\ref{thm:hookinterval} are $C_6(1/6)=33733345/1088391168$, $a\rho(1-\rho)/12=5/72$, and $G=793/2916$. The choice $c=1/5$ satisfies the strict hypothesis because $1/25-C_6(1/6)(5/72)/G^2=1578804259/144886809600>0$. \end{proof} Exact variances can certify coefficients when the supports overlap. \begin{example}\label{ex:exactmoment} For $b=1200$, $\lambda=(6000,1^{1200})$, and $k=331423$, \eqref{eq:hookchar} and Theorem~\ref{thm:exactvariance} give the rational bounds \[ \mu_T<331217.24,\qquad \mu_S>719507.43,\qquad \sigma^2_{6,b}<53592619,\qquad \sigma^2_{b,6}<119979996. \] These bounds imply the strict inequality in \eqref{eq:exactcantelli}; hence $[q^{331423}s_{(6000,1^{1200})}]F_{6,1200}>0$. The support overlap is $[119600,725400)$, so support separation alone does not settle this coefficient. \end{example} \end{document}