\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}.}}

\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 I: A fusion inequality for Macdonald plethysms},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{\Ecal}{\mathcal E}
\newcommand{\Lcal}{\mathcal L}
\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 17B10, 14C05}
\keywords{q-Foulkes conjecture, fusion product, Macdonald polynomial,
Schur positivity}
\makeatletter
\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.}}
\makeatother
\begin{document}
\title[The $q$-Foulkes problem I]{The $q$-Foulkes problem I:\\A fusion inequality for Macdonald plethysms}
\aiauthor

\begin{abstract}
  We prove that the plethysm $H_m[\widetilde H_\mu]$ of a Macdonald polynomial by the Hall--Littlewood
  polynomial $H_m$ lies above the Macdonald polynomial $\widetilde H_{m\mu}$ obtained by fusing $m$
  copies of $\widetilde H_\mu$. More specifically, the quotient
  $(H_m[\widetilde H_\mu]-\widetilde H_{m\mu})/(1-q^m)$ is Schur positive with coefficients in
  $\mathbb N[q,t]$, and it is the character of an explicit module that we call the \emph{fusion
    defect}. For one-row shapes the inequality places both $q$-Foulkes plethysms $H_b[H_a]$ and
  $H_a[H_b]$ above the same polynomial $H_{ab}$, and comparing their two defects we obtain a distorted
  version of Bergeron's $q$-analog of Foulkes' conjecture, in which the subtracted plethysm is shifted
  by $q^{a(a-1)\lfloor
    b/2\rfloor}$. %
\end{abstract}
\preprintslink{qfoulkes1}{26 October 2026}
\maketitle
\addtocontents{toc}{\protect\setcounter{tocdepth}{-1}}
\raggedbottom

\vspace*{-5pt}
\section{Introduction}
We study two simple modifications of the $q$-analog of Foulkes' conjecture proposed by
\citet{bergeron2017foulkes}. 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}\label{eq:Hdef}
 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{conjecture}[\citealp{bergeron2017foulkes}]\label{conj:bergeron}
For all integers $a\le b$, the divided difference
\begin{equation}
  \label{eq:2}
 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}
Since $H_n(X;1)=h_1^n$, both plethysms in \eqref{eq:2} equal $h_1^{ab}$ at $q=1$, so the numerator of
$F_{a,b}$ is divisible by $1-q$; since $H_n(X;0)=h_n$, the conjecture at $q=0$ reduces
to Foulkes' conjecture. Bergeron proved that Conjecture~\ref{conj:bergeron} holds for $q=1$ \citep[Section~3]{bergeron2017foulkes}.

Conjecture~\ref{conj:bergeron} compares degrees. To see this, fix a partition $\lambda\vdash ab$, and
let $d_\lambda$ be the coefficient of $s_\lambda$ in $h_1^{ab}$. Since both plethysms equal $h_1^{ab}$
at $q=1$, the coefficient of $s_\lambda$ in $H_b[H_a]$ is a polynomial
$q^{\alpha_1}+\cdots+q^{\alpha_{d_\lambda}}$ with $\alpha_1\le\cdots\le\alpha_{d_\lambda}$; in
$H_a[H_b]$ it is $q^{\beta_1}+\cdots+q^{\beta_{d_\lambda}}$ with
$\beta_1\le\cdots\le\beta_{d_\lambda}$. Dividing by $1-q$ takes partial sums, so the coefficient of
$q^ks_\lambda$ in $F_{a,b}$ equals the difference of cardinalities $|\{i:\alpha_i\le k\}| - |\{i:\beta_i \le k\}|$. Thus Conjecture~\ref{conj:bergeron} asks that
$|\{i:\alpha_i\le k\}|\ge|\{i:\beta_i\le k\}|$ for every $k$, which, since both lists are sorted,
says exactly that $\alpha_i\le\beta_i$ for every $\lambda$ and $i$.

\vskip4pt Our first modification compares each of the two plethysms in~\eqref{eq:2} with the polynomial
obtained by fusing its copies of the inner factor. The second modification shifts the subtracted
plethysm by $q^s$, which amounts to asking only for $\alpha_i\le\beta_i+s$ ($s=0$ is Bergeron's
conjecture).

For the first modification, let $\widetilde H_\mu(X;q,t)$ be the
modified Macdonald polynomial of a partition $\mu$, the bigraded
character of its Garsia--Haiman module, so that
$\widetilde H_{(n)}(X;q,t)=H_n(X;q)$ (Section~\ref{sec:conventions}).
Placing $m$ copies of a current-algebra module with character
$\widetilde H_\mu$ at distinct points of a line and letting the points
collide gives the fusion product of \citet{feigin1999kostka}, named
after the fusion of fields in conformal field theory. Its character is
the fused polynomial $\widetilde H_{m\mu}$, the Macdonald polynomial of
$\mu$ with every row multiplied by $m$ \citep{KM}. We show that it lies
below the plethysm $H_m[\widetilde H_\mu]$, and call the difference the
\emph{fusion defect}.
\begin{dashedbox}
\begin{theorem}[Fusion defect]\label{thm:defect}
For every partition $\mu$ and $m\in\mathbb{N}$, with
$m\mu=(m\mu_1,m\mu_2,\ldots)$,
\[
 \frac{H_m[\widetilde H_\mu]-\widetilde H_{m\mu}}{1-q^m}
 \succeq_s0\qquad\text{over }\N[q,t],
\]
and the quotient is the bigraded character of an explicit module; $\succeq_s 0$ denotes Schur positivity.
\end{theorem}
\end{dashedbox}
The proof builds both characters on the tensor product of the $m$
copies. Fusion lets the current algebra act on it, and the plethysm
arises when every endomorphism of a single copy is allowed to act as
well \citep{Kedem}. More operators reach more vectors in each degree,
and the extra vectors, degree by degree, span a module whose character
is the fusion defect divided by $1-q$ (Section~\ref{sec:fusion}).

For the one-row shapes $\mu=(a)$, $m=b$ and $\mu=(b)$, $m=a$,
Theorem~\ref{thm:defect} places both $H_b[H_a]$ and $H_a[H_b]$ above
$H_{ab}$, and Conjecture~\ref{conj:bergeron} says that the fusion defect
of $H_b[H_a]$ is at least that of $H_a[H_b]$. To compare the two
plethysms with each other, we turn the inequality for $H_a[H_b]$
around, using a symmetry of the modified Macdonald polynomials that
reverses degrees \citep{haiman2001hilbert}. This costs a shift of
degrees, which we determine exactly (Theorem~\ref{thm:reversefusion}).
Against $H_{ab}$ the shift is quadratic in $b$, but it becomes linear in
$b$ if $H_a$ and $H_b$ are graded by separate variables $q$ and $t$, so
that both plethysms lie above the Macdonald polynomial of the rectangle
$(a^b)$ (Theorem~\ref{thm:shiftsharp}). Specializing $t=q^c$ afterwards,
and writing $H_n(z)=H_n(X;z)$, we obtain the following shifted form of
Conjecture~\ref{conj:bergeron}, which is our second main result.
\begin{dashedbox}
\begin{theorem}[$q$-distorted Foulkes comparison]\label{thm:qdistorted}
For all integers $a,b\ge2$ and $c\ge1$,
\[
 \frac{H_b(q^c)[H_a(q)]-q^{a(a-1)\lfloor b/2\rfloor}H_a(q)[H_b(q^c)]}{1-q}
 \succeq_s0\qquad\text{over }\N[q].
\]
In particular,
$(H_b[H_a]-q^{a(a-1)\lfloor b/2\rfloor}H_a[H_b])/(1-q)\succeq_s0$.
\end{theorem}
\end{dashedbox}

Bergeron's rectangular version \citep[Conjecture~4]{bergeron2017foulkes}
replaces the inner $H_a$ and $H_b$ by the modified Macdonald polynomials
$\widetilde H_{(a^k)}$ and $\widetilde H_{(b^k)}$ of rectangles with $k$
rows. Applying Theorem~\ref{thm:qdistorted} to the column character
$H_k(X;t)$ proves it up to a shift linear in $b$ for fixed $a$ and $k$
(Theorem~\ref{thm:rectlinear}). The
fusion defects of two inner shapes comparable in dominance are ordered
once $m$ is large, after the specialization $t=q^c$
(Theorem~\ref{thm:stretch} and Corollary~\ref{cor:defectdom}); the
proof pulls back along $x\mapsto x^m$ the family of \citet{GKX} that
joins the Garsia--Haiman modules of two shapes differing by one box.
Conjecture~\ref{conj:bergeron} itself would follow from a graded
embedding of the fusion defect module of $H_a[H_b]$ into that of $H_b[H_a]$, whose existence remains
open (Section~\ref{sec:todo}).

The companion papers show that negative coefficients of $F_{a,b}$ are
rare under Plancherel sampling \citep{foulkesII}, and prove that
$h_b[H_a]-H_a[h_b]$ is Schur positive for $b\ge(a-1)^2$, together with
positivity of families of Schur coefficients of $F_{a,b}$
\citep{foulkesIII}. Appendix~\ref{sec:matrix} polarizes each
Schur-positive quotient proved here into a positive semidefinite kernel
on tuples of vectors.

\subsection{Conventions}\label{sec:conventions}
We work over $\mathbb C$, and permuting tensor factors introduces no
signs. Plethysm acts on every grading variable through the Adams
operations $\psi_r(p_j)=p_{rj}$, $\psi_r(q)=q^r$, and $\psi_r(t)=t^r$,
and the involution $\omega$ sends $s_\lambda$ to $s_{\lambda'}$.
The coinvariant algebra
$R_m=\mathbb C[z_1,\ldots,z_m]/\langle\mathbb C[z_1,\ldots,z_m]^{S_m}_+\rangle$,
graded by $\deg z_i=1$, has graded Frobenius character $H_m$, and
$H_0=1$. We write $H_m[f;z]=H_m(z)[f]$ for the plethysm with outer
coinvariant grading $z$; grading variables in $f$ keep their meanings,
and $H_m[f]=H_m[f;q]$.
For a partition $\mu$ and an integer $n$, set
\[
 n(\mu)=\sum_i(i-1)\mu_i,\qquad
 c_n=\binom n2,\qquad \varepsilon_n=n\bmod2,
 \qquad [n]_q=\frac{1-q^n}{1-q},
\]
so that $[0]_q=0$, and write $\qbin nk=(q;q)_n/((q;q)_k(q;q)_{n-k})$
for $0\le k\le n$. For $\rho=1^{m_1}2^{m_2}\cdots$, put
$z_\rho=\prod_j j^{m_j}m_j!$; $d_\lambda$ is the dimension of the
Specht module $S^\lambda$, and $\succeq$ is the dominance order.

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}.
By Haiman's $n!$ theorem, for $\mu\vdash n$, $\widetilde H_\mu$ is the
bigraded Frobenius character of the Garsia--Haiman module of $\mu$
\citep{haiman2001hilbert}: the span of all partial derivatives of the
determinant $\det[x_k^jy_k^i]$, whose rows are indexed by
$k=1,\ldots,n$ and whose columns by the cells $(i,j)$ of $\mu$, with
zero-based row index $i$ and column index $j$; here $q$ counts the
degree in $x_1,\ldots,x_n$ and $t$ the degree in $y_1,\ldots,y_n$. For
$\mu=(n)$ the determinant is the Vandermonde determinant in
$x_1,\ldots,x_n$, whose derivatives span a copy of the coinvariant
algebra, so $\widetilde H_{(n)}(X;q,t)=H_n(X;q)$; similarly,
$\widetilde H_{(1^n)}(X;q,t)=H_n(X;t)$.

\section{Fusion defects and their reversal}\label{sec:fusion}
Fusion places each plethysm above a fused Macdonald polynomial of the
same degree, and sign duality reverses that order after a shift. These
two comparisons are the inputs to every shifted comparison in
Section~\ref{sec:shifted}.

\subsection{Two fusion filtrations}\label{sec:fusion-filtrations}
To realize the fusion defect as a module, we filter one tensor power
of a module with character $\Htil_\mu$ in two ways. The first filtration
comes from the current algebra
$\mathfrak{gl}(V)[x,y]=\mathfrak{gl}(V)\otimes\C[x,y]$, spanned by the
currents $Yx^ry^\ell$ with $Y\in\mathfrak{gl}(V)$; the second comes from
all endomorphisms of the module. The first filtration is contained in
the second, and the quotient has the fusion defect as its character.
Placing the evaluation points at the $m$th roots of unity refines the
inclusion by degree modulo $m$, which gives the denominator $1-q^m$.

The fusion product of cyclic modules over a current algebra, evaluated
at distinct points $c_1,\ldots,c_m$, is the associated graded of their
tensor product for the filtration by current degree, where the variable
$x$ acts on the $i$th factor as $x+c_i$ \citep{feigin1999kostka}.
\citet[Corollary~6.16 and Theorem~6.19]{KM} construct cyclic bigraded
$\mathfrak{gl}(V)[x,y]$-modules $U_\mu(V)$ with character
$\Htil_\mu(X;q,t)$, where $q$ and $t$ count the $x$- and $y$-degrees;
the construction needs $\dim V\ge3N+1$, where $N$ is the total
polynomial degree of the module constructed. Their theorem identifies
fusion in $y$ with the union of rows, the partition whose rows are
those of all factors. Exchanging $x$ and $y$ and conjugating diagrams
\citep[Lemma~8.2]{KM} turns this into fusion in $x$: the fusion in $x$
of $m$ copies of $U_\mu$ has character $\Htil_{m\mu}$, where
$m\mu=(m\mu_1,m\mu_2,\ldots)$. These fusion products do not depend on
the evaluation points, and Haiman's $n!$ theorem is an input to the
identification of their characters.

Kedem's fusion theorem \cite[Theorem~4.3]{Kedem} identifies the
$m$-fold fusion product of the defining $\mathfrak{gl}(W)$-module with
$(R_m\otimes W^{\otimes m})^{S_m}$, whose graded character is
$H_m[W;q]$. The identification respects any grading of $W$, so that
grading survives in the character.
For the denominator $1-q^m$, we use the action of rotating the tensor
factors on fusion at roots of unity, described by
\citet[Proposition~8.1]{westbury2016cyclic}, on both filtrations at
once and compare its eigenspaces. The precise form
of Theorem~\ref{thm:defect} names the two quotient modules that the
proof constructs.

\begin{recalled}{Theorem~\ref{thm:defect}}[precise form]
For every partition $\mu$ and integer $m\ge1$,
\begin{equation}\label{eq:defect}
 \Dcal_{\mu,m}(X;q,t)
 :=\frac{H_m[\Htil_\mu(X;q,t);q]-\Htil_{m\mu}(X;q,t)}{1-q}
 \in\N[q,t]\{s_\lambda\}.
\end{equation}
For $\dim V\ge3m|\mu|+1$, $\Dcal_{\mu,m}$ is the character of the
finite-dimensional module $\Kcal_{\mu,m}(V)$ in \eqref{eq:Kdef}.
The quotient with denominator $1-q^m$ is also a character:
\begin{equation}\label{eq:defect-residue}
 \Ecal_{\mu,m}(X;q,t)
 :=\frac{H_m[\Htil_\mu(X;q,t);q]-\Htil_{m\mu}(X;q,t)}{1-q^m}
 \in\N[q,t]\{s_\lambda\},
 \qquad \Dcal_{\mu,m}=[m]_q\Ecal_{\mu,m}.
\end{equation}
The module $\Lcal_{\mu,m}(V)$ in \eqref{eq:Ldef} realizes this character.
\end{recalled}
\begin{proof}
For the empty partition, take $U_{\varnothing}=\mathbb C$ in degree
$(0,0)$ with trivial current action. Then $H_m[1;q]=1$, both
filtrations below are constant, and the defect is zero. Suppose
$|\mu|\ge1$.
Set $U=U_\mu(V)$, choose its cyclic vector $v$ in bidegree $(0,0)$,
and write $\rho$ for the current-algebra action. On $T=U^{\otimes m}$,
evaluate the $i$th factor at $c_i=\zeta^i$ in the $x$ variable, where
$\zeta$ is a primitive $m$th root of unity and indices are taken modulo
$m$. The first filtration is the fusion filtration by $x$-degree,
\[
 A_s=U(\mathfrak{gl}(V)[x,y])_{\deg_x\le s}v^{\otimes m}.
\]
It preserves the total $y$-degree,
and the fusion realization of \citet{KM} gives
$\ch_{q,t}\gr_A T=\Htil_{m\mu}$.

For the second filtration, regard $U$ as the defining representation
of $\mathfrak{gl}(U)$, and let $G_j$ be Kedem's fusion filtration of
$T$ at the same evaluation points. Each $G_j$ is
$\mathrm{GL}(U)$-invariant, so it decomposes by the internal $x$- and
$y$-degrees of $T$. Writing $T_d$ for the part of $T$ of internal
$x$-degree $d$, set
\[
 B_s=\bigoplus_{d\ge0}(G_{s-d}\cap T_d),\qquad G_j=0\ (j<0).
\]
The index $s$ adds the fusion degree and the internal $x$-degree, and
Kedem's theorem, with the internal grading kept, gives
$\ch_{q,t}\gr_B T=H_m[\Htil_\mu]$.

We claim $A_s\subseteq B_s$. A current $Yx^r y^\ell$ acts on $T$ as
\[
 \sum_{i=1}^m\sum_{h=0}^r
 \binom rh c_i^{r-h}\rho(Yx^h y^\ell)^{(i)},
\]
and we write $\rho_{\mathrm{sh}}(Yx^ry^\ell)$ for this operator. For
fixed $h$, the sum over $i$ is a multiple of the fusion current
$\sum_ic_i^{r-h}E^{(i)}$ of $E=\rho(Yx^hy^\ell)\in\End(U)$. This current
has fusion degree $r-h$, and $E$ raises the internal $x$-degree by $h$,
so the term maps $G_j\cap T_d$ into $G_{j+r-h}\cap T_{d+h}$, and hence
$B_s$ into $B_{s+r}$.
Since $v^{\otimes m}\in B_0$, every word in currents of total
$x$-degree at most $s$ maps $v^{\otimes m}$ into $B_s$, which proves
$A_s\subseteq B_s$.

The quotients of the two filtrations form the module
\begin{equation}\label{eq:Kdef}
 \Kcal_{\mu,m}(V)=\bigoplus_{s\ge0} B_s/A_s,
\end{equation}
in which the summand $B_s/A_s$ has $q$-degree $s$ and keeps its
$y$-degree. Both
filtrations eventually equal $T$, so $\Kcal_{\mu,m}(V)$ is finite-dimensional.
Taking differences of successive filtration spaces gives
\[
 \ch\gr_BT-\ch\gr_AT
 =(1-q)\sum_{s\ge0}q^s\ch_t(B_s/A_s).
\]
Since $\dim V\ge3m|\mu|+1$, the fusion realization applies and
$GL(V)$-characters detect every Schur coefficient of degree $m|\mu|$, so
this identity of characters proves \eqref{eq:defect}.

To keep track of degrees modulo $m$, let $R$ rotate the tensor factors
backward, so $R E^{(i)}R^{-1}=E^{(i-1)}$ for $E\in\End(U)$, let $Z$ act
on $T_d$ by $\zeta^d$, and set $\Gamma=ZR$. The operators $R$ and $Z$
commute, so $\Gamma^m=1$, and $\Gamma$ commutes with $GL(V)$ and
preserves the $y$-degree. In $\rho_{\mathrm{sh}}(Yx^ry^\ell)$,
conjugation by $R$ multiplies the term of index $h$ by $\zeta^{r-h}$,
and $Z$ multiplies it by $\zeta^h$, so
\[
 \Gamma\rho_{\mathrm{sh}}(Yx^ry^\ell)\Gamma^{-1}
 =\zeta^r\rho_{\mathrm{sh}}(Yx^ry^\ell).
\]
Since $\Gamma$ fixes $v^{\otimes m}$, it acts on $A_s/A_{s-1}$ as
$\zeta^s$. For the second filtration,
\[
 R\Bigl(\sum_i c_i^rE^{(i)}\Bigr)R^{-1}
 =\zeta^r\sum_i c_i^rE^{(i)},
\]
so $R$ acts on $G_j/G_{j-1}$ as $\zeta^j$, and $\Gamma$ acts on its part
of internal $x$-degree $d$ as $\zeta^{j+d}$. By the definition of
$B_s$, $\Gamma$ therefore acts on $B_s/B_{s-1}$ as $\zeta^s$.

Write $T^{[s]}=\ker(\Gamma-\zeta^s)$ and set filtration terms of
negative index equal to zero. On $T^{[s]}$, either filtration can grow
only at indices congruent to $s$ modulo $m$, so
\[
 A_{s-1}\cap T^{[s]}=A_{s-m}\cap T^{[s]},\qquad
 B_{s-1}\cap T^{[s]}=B_{s-m}\cap T^{[s]}.
\]
The inclusions $A_s\subseteq B_s$ define a finite bigraded module
\begin{equation}\label{eq:Ldef}
 \Lcal_{\mu,m}(V)
 =\bigoplus_{s\ge0}
 \frac{B_s\cap T^{[s]}}{A_s\cap T^{[s]}},
\end{equation}
with degree $q^s$ on the indicated summand and its original $y$-degree.
Taking differences at successive indices in each residue class gives
\[
 \ch_{q,t}\gr_BT-\ch_{q,t}\gr_AT
 =(1-q^m)\ch_{q,t}\Lcal_{\mu,m}(V).
\]
Both filtrations exhaust $T$, so the sum is finite, which proves
\eqref{eq:defect-residue}.
\end{proof}

Coefficientwise, Theorem~\ref{thm:defect} orders cumulative
multiplicities. For a Schur label $\nu\vdash m|\mu|$ and integers
$j,k\ge0$, expanding $1/(1-q)$ in \eqref{eq:defect} gives
\begin{equation*}
 \sum_{i=0}^k [q^it^js_\nu]
 \bigl(H_m[\Htil_\mu;q]-\Htil_{m\mu}\bigr)
 =[q^kt^js_\nu]\Dcal_{\mu,m}\ge0.
\end{equation*}
Summed over all $q$-degrees, the two multiplicities agree, because both
filtrations exhaust $T$; equivalently,
$\Htil_{m\mu}(X;1,t)=\Htil_\mu(X;1,t)^m$.

The refinement \eqref{eq:defect-residue} orders these multiplicities
within each residue class. For $0\le r<m$ and $j,k\ge0$,
\[
 \sum_{\ell=0}^k[q^{r+m\ell}t^js_\nu]
 \bigl(H_m[\Htil_\mu;q]-\Htil_{m\mu}\bigr)
 =[q^{r+mk}t^js_\nu]\Ecal_{\mu,m}\ge0.
\]
The totals in every residue class agree once $k$ exceeds the support.

The cumulative order is weaker than an inequality in each degree. For
$\mu=(2)$ and $m=2$,
\[
 [s_{(1^4)}]\bigl(H_2[H_2]-H_4\bigr)=q^2-q^6,
 \qquad [s_{(1^4)}]\Dcal_{(2),2}=q^2[4]_q,
\]
so the difference has the negative coefficient $-q^6$ while the defect
is positive. The module of $H_2[H_2]$ is induced from the wreath
product $S_2\wr S_2$, the stabilizer in $S_4$ of a splitting of four
points into two pairs. The sign character of $S_4$ restricts to the
sign characters of the two inner factors and to the trivial character
of the outer one, so the sign constituent of $H_2[H_2]$ lies in degree
two, while that of $H_4$ lies in degree six. The refined
quotient has sign coefficient $[s_{(1^4)}]\Ecal_{(2),2}=q^2+q^4$.

\begin{samepage}
\subsection{The least shift for reversing fusion}\label{sec:reversefusion}
Theorem~\ref{thm:defect} places the plethysm $H_m[\Htil_\mu]$ above the
fused polynomial $\Htil_{m\mu}$. Comparing two plethysms also needs the
opposite order. Sign duality supplies it: reversing both gradings turns
the fusion comparison around at the cost of a degree shift. The least
shift is decided by the sign constituent, which occurs once in each
character. In the example of Section~\ref{sec:fusion-filtrations}, it
lies in degree $6$ of $H_4$ and in degree $2$ of $H_2[H_2]$, so the
reversed comparison of $H_4$ with $q^sH_2[H_2]$ needs $s\ge4$; by the
theorem below, $s=4$ suffices.

\begin{theorem}[Sharp reverse fusion]\label{thm:reversefusion}
Let $\mu\vdash n\ge1$ and $m\ge1$. Set
\begin{equation*}
 \Delta_{\mu,m}=c_m\left(\sum_i\mu_i^2-\varepsilon_n\right).
\end{equation*}
For every integer $s\ge0$,
\begin{equation}\label{eq:reversefusion}
 \frac{\widetilde H_{m\mu}(X;q,t)
       -q^sH_m[\widetilde H_\mu(X;q,t);q]}{1-q}
 \succeq_s0
 \quad\Longleftrightarrow\quad s\ge\Delta_{\mu,m}.
\end{equation}
Positivity is coefficientwise over $\mathbb N[q,t]$. With the
denominator $1-q^m$,
\begin{equation}\label{eq:reversefusion-residue}
 \frac{\widetilde H_{m\mu}(X;q,t)
       -q^sH_m[\widetilde H_\mu(X;q,t);q]}{1-q^m}
 \in\N[q,t]\{s_\lambda\}
 \quad\Longleftrightarrow\quad
 s\ge\Delta_{\mu,m}\ \text{and}\ m\mid s.
\end{equation}
\end{theorem}
\end{samepage}

\begin{proof}
For $m=1$, both quotients are $[s]_q\widetilde H_\mu$ and
$\Delta_{\mu,1}=0$. For $\mu=(1)$ the two numerator characters
coincide with $H_m$, and the assertions follow from the geometric sum
$(1-q^s)/(1-q^m)$ and its trivial Schur coefficient. Suppose
$m,n\ge2$, and abbreviate
\[
 A=\widetilde H_\mu,\qquad T=H_m[A;q],\qquad R=\widetilde H_{m\mu},
 \qquad u=n(\mu'),\quad v=n(\mu).
\]
Stretching $\mu$ horizontally to $m\mu$ gives
\[
 U:=n((m\mu)')=m^2u+c_mn,\qquad n(m\mu)=mv.
\]
Sign self-duality of the Garsia--Haiman module
\citep[Section~3.7, (45), and Section~3.9]{haiman2001hilbert},
with the sign twist explicit in \citep[(2.3)]{GKX}, gives
\[
 q^ut^v\omega A(q^{-1},t^{-1})=A(q,t),\qquad
 q^Ut^{mv}\omega R(q^{-1},t^{-1})=R(q,t).
\]
Since these characters have nonnegative degrees and coefficients,
$A$ has bidegrees bounded by $(u,v)$, $R$ by $(U,mv)$, and $T$ by
$(mu+c_m,mv)$. The bound for $R$ dominates the one for $T$, because
\[
 U-mu-c_m=c_m(2u+n-1)
         =c_m\left(\sum_i\mu_i^2-1\right)\ge0.
\]
Theorem~\ref{thm:defect} makes $E=(T-R)/(1-q^m)$ a polynomial.
Its numerator has $q$-degree at most $U$, so the bidegrees of a
nonzero $E$ are bounded by $(U-m,mv)$, where $U\ge m$ because
$m,n\ge2$.

To reverse the degrees of $T$, we use coinvariant reciprocity and the
plethystic parity rule,
\begin{equation}\label{eq:fusionreciprocity}
 q^{c_j}H_j(X;q^{-1})=\omega H_j(X;q),\qquad
 \omega(f[g])=(\omega^{\deg g}f)[\omega g],
\end{equation}
where $\omega$ acts only on the symmetric-function alphabet. The second
identity follows on power sums from $\omega p_j=(-1)^{j-1}p_j$, because
Adams operations fix an integer sign. Applied to $T$, the two
identities give
\begin{align*}
 \omega T(q^{-1},t^{-1})
 &=q^{-mu}t^{-mv}(\omega^nH_m[-;q^{-1}])[A]\\
 &=q^{-mu-c_m}t^{-mv}\overline T,
 \qquad
 \overline T=(\omega^{n+1}H_m[-;q])[A].
\end{align*}
Reversing the degrees of $E$ within its bounding box preserves
positivity, so, with $d_0=c_m(\sum_i\mu_i^2-1)$,
\begin{equation}\label{eq:reverse-general-twist}
 q^{U-m}t^{mv}\omega E(q^{-1},t^{-1})
 =\frac{R-q^{d_0}\overline T}{1-q^m}\succeq_s0.
\end{equation}
For odd $n$, $\overline T=T$ and $d_0=\Delta_{\mu,m}$, so
\eqref{eq:reverse-general-twist} is \eqref{eq:reversefusion-residue} at
$s=\Delta_{\mu,m}$.

For even $n$, $\overline T=(\omega H_m)[A]$, and
Lemma~\ref{lem:cyclicduality} supplies the positive character
\[
 J_m^{(m)}:=\frac{\omega H_m-q^{c_m}H_m}{1-q^m}\succeq_s0.
\]
Adding $q^{d_0}J_m^{(m)}[A]$ to
\eqref{eq:reverse-general-twist} proves \eqref{eq:reversefusion-residue}
at $s=d_0+c_m=\Delta_{\mu,m}$. In either parity $m$ divides
$\Delta_{\mu,m}$, because $m\mid c_m$ for odd $m$, and for even $m$ the
integer $\sum_i\mu_i^2-\varepsilon_n$ is even. For a larger multiple $s$
of $m$, adding $q^{\Delta_{\mu,m}}[ (s-\Delta_{\mu,m})/m ]_{q^m}T$
proves \eqref{eq:reversefusion-residue}. Multiplying by $[m]_q$ gives
\eqref{eq:reversefusion} at the threshold, and adding
$q^{\Delta_{\mu,m}}[s-\Delta_{\mu,m}]_qT$ gives it for every larger
integer $s$.

For necessity, we compute sign coefficients. The ungraded
Garsia--Haiman module is the regular $S_n$-module, so its sign
multiplicity is one, and sign self-duality pairs its constant trivial
constituent with $q^ut^v\sgn$. The sign coefficient of $A$ is therefore
$q^ut^v$.
The sign representation of $S_{mn}$ restricts to the sign
representation in each inner $S_n$ factor and to $\sgn^n$ on the
outer $S_m$, so
\[
 [s_{(1^{mn})}]R=q^Ut^{mv},\qquad
 [s_{(1^{mn})}]T=q^{mu+\varepsilon_nc_m}t^{mv}.
\]
If $L=s+mu+\varepsilon_nc_m<U$, the sign coefficient of the quotient
in \eqref{eq:reversefusion} is $-q^L[U-L]_q t^{mv}$, which is nonzero
and negative. Positivity therefore requires $L\ge U$, which is
$s\ge\Delta_{\mu,m}$. Positivity in \eqref{eq:reversefusion-residue}
implies \eqref{eq:reversefusion} after multiplication by $[m]_q$, so the
same bound is necessary there. Finally, $T$ and $R$ both have trivial
Schur coefficient $1$, since each is a graded regular representation
whose trivial constituent lies in degree zero (for $T$ by wreath
restriction). The trivial coefficient of the quotient in
\eqref{eq:reversefusion-residue} is therefore $(1-q^s)/(1-q^m)$, a
polynomial exactly when $m\mid s$.
\end{proof}

Since $\widetilde H_{\mu'}(X;q,t)=\widetilde H_\mu(X;t,q)$, conjugating
$\mu$ and exchanging $q,t$ turn Theorem~\ref{thm:reversefusion} into a
comparison in the variable $t$. Let $\mu^{\sqcup m}=(m\mu')'$ be the partition that repeats each row of
$\mu$ exactly $m$ times. The least shift in
\[
 \frac{\widetilde H_{\mu^{\sqcup m}}(X;q,t)
       -t^sH_m[\widetilde H_\mu(X;q,t);t]}{1-t}
\]
is $c_m(\sum_i(\mu_i')^2-\varepsilon_n)$. With the denominator
$1-t^m$, positivity holds exactly for the shifts of at least this size
that are divisible by $m$.

\section{Shifted Foulkes comparisons}\label{sec:shifted}
Adding a forward fusion comparison to a reversed one cancels the fused
Macdonald polynomial and leaves a shifted comparison of two plethysms.
This section proves Theorem~\ref{thm:qdistorted} in this way and
extends it to rectangular inner characters.

\subsection{A \texorpdfstring{$q$}{q}-distorted Foulkes comparison}\label{sec:shift}
To combine the forward comparison for $H_b[H_a]$ with the reversed
comparison for $H_a[H_b]$, we record the coinvariant grading of $H_a$
by $q$ and that of $H_b$ by $t$, in both plethysms:
\begin{equation}\label{eq:PQRtwo}
 \mathcal P=H_b[H_a[X;q];t],\quad
 \mathcal Q=H_a[H_b[X;t];q],\quad
 \mathcal R=\Htil_{(a^b)}(X;q,t).
\end{equation}
Both plethysms fuse to $\mathcal R$, one by fusion in $y$ and the other
by fusion in $x$; Macdonald transposition,
$\Htil_{(a^b)}(q,t)=\Htil_{(b^a)}(t,q)$, exchanges the two cases. Theorem~\ref{thm:defect} therefore places each
plethysm above $\mathcal R$.

\begin{proposition}[Two rectangular filtration quotients]\label{prop:UV}
For every $a,b\ge2$,
\begin{equation}\label{eq:UV}
 U=\frac{\mathcal P-\mathcal R}{1-t}\pos,
 \qquad
 V=\frac{\mathcal Q-\mathcal R}{1-q}\pos
\end{equation}
over $\N[q,t]$. As in Theorem~\ref{thm:defect}, the denominators
refine to $1-t^b$ and $1-q^a$:
\[
 \frac{\mathcal P-\mathcal R}{1-t^b}\pos,
 \qquad \frac{\mathcal Q-\mathcal R}{1-q^a}\pos.
\]
\end{proposition}
\begin{proof}
For $U$, fuse $b$ copies of the one-row module with character
$H_a[X;q]$ in the variable $y$. The currents give $\mathcal R$, and
all endomorphisms give $\mathcal P$. A current
$Yx^hy^r$ acts as $\sum_i c_i^r\rho(Yx^h)^{(i)}$, which is the fusion
current of the endomorphism $\rho(Yx^h)$ in the same $y$-degree $r$. The
filtration from the currents therefore lies inside the one from all
endomorphisms, and both keep the $q$-grading. The quotient argument of Theorem~\ref{thm:defect} gives
$U\succeq_s0$, and its cyclic part gives the denominator $1-t^b$.
Transposing the rectangle and exchanging $q$ and $t$ gives the same
statements for $V$, with the denominator $1-q^a$.
\end{proof}

To place the rectangle above a shifted $\mathcal Q$, the second
comparison must be reversed. For the rectangle, the least shift of
Theorem~\ref{thm:reversefusion} is linear in $b$.

\begin{theorem}[Sharp reverse fusion for a rectangle]\label{thm:shiftsharp}
For integers $a,b\ge2$, put $\varepsilon_b=b\bmod2$ and
\[
 \delta_{a,b}=(b-\varepsilon_b)c_a
 =a(a-1)\lfloor b/2\rfloor.
\]
For an integer $s\ge0$,
\begin{equation}\label{eq:shiftsharp}
 \frac{\mathcal R-q^s\mathcal Q}{1-q}\pos
 \quad\Longleftrightarrow\quad s\ge\delta_{a,b}.
\end{equation}
Positivity in \eqref{eq:shiftsharp} is over $\N[q,t]$.
\end{theorem}
\begin{proof}
Apply Theorem~\ref{thm:reversefusion} to $\mu=(1^b)$ and $m=a$.
Then $\widetilde H_\mu=H_b(X;t)$, $m\mu=(a^b)$, and
$\Delta_{\mu,a}=c_a(b-\varepsilon_b)$.
\end{proof}
The explicit specialization of the reversal identity is recorded in
Appendix~\ref{app:rectangle-reversal}.

We combine Proposition~\ref{prop:UV} and Theorem~\ref{thm:shiftsharp}
while the two gradings are still separate. With
$G=(\mathcal R-q^{\delta_{a,b}}\mathcal Q)/(1-q)$,
\begin{equation}\label{eq:balance}
 \mathcal P-q^{\delta_{a,b}}\mathcal Q=(1-t)U+(1-q)G,
 \qquad U,G\pos.
\end{equation}
Specializing $t=q^c$ in \eqref{eq:balance} proves
Theorem~\ref{thm:qdistorted}.
\begin{recalled}{Theorem~\ref{thm:qdistorted}}
For all integers $a,b\ge2$ and $c\ge1$, with $H_n(z)=H_n(X;z)$ and
$\delta_{a,b}=a(a-1)\lfloor b/2\rfloor$,
\[
 \frac{H_b(q^c)[H_a(q)]-q^{\delta_{a,b}}H_a(q)[H_b(q^c)]}{1-q}
 \pos\qquad\text{over }\N[q].
\]
\end{recalled}
\begin{proof}
Substitute $t=q^c$ in \eqref{eq:balance}, so that $\mathcal P$ and
$\mathcal Q$ become the two plethysms above, and divide by $1-q$.
The resulting quotient is $[c]_qU(q,q^c)+G(q,q^c)$, and both summands
are positive.
\end{proof}

The identity \eqref{eq:balance} also allows both gradings to be
rescaled. For positive integers $r,s$, substituting $q=z^r$ and
$t=z^s$ and dividing by $1-z$ turns its right side into
$[s]_zU(z^r,z^s)+[r]_zG(z^r,z^s)$, which is Schur positive. This is a
comparison in the degree that gives $q$ weight $r$ and $t$ weight $s$.

\subsection{Rectangular modified-Macdonald comparisons}\label{sec:rectangular-comparisons}
Bergeron's rectangular conjecture replaces the inner $H_a$ by
$\widetilde H_{(a^k)}(X;q,t)$, whose two parameters both grade the inner
module, while the outer grading is again $q$. No variable is left for
the outer grading, as in Section~\ref{sec:shift}, so we compare the
rectangles through one-row plethysms instead.

For $a,b\ge2$ and $k\ge1$, set
\begin{equation*}
 \mathcal F^{(k)}_{a,b}
 =\frac{H_b[\Htil_{(a^k)}]-H_a[\Htil_{(b^k)}]}{1-q}.
\end{equation*}
Both plethysms fuse to $\Htil_{((ab)^k)}$, which cancels in the
difference:
\begin{equation}\label{eq:rectdefects}
 \begin{aligned}
 \mathcal F^{(k)}_{a,b}
 &=\Dcal_{(a^k),b}-\Dcal_{(b^k),a}\\
 &=[b]_q\Ecal_{(a^k),b}-[a]_q\Ecal_{(b^k),a}.
 \end{aligned}
\end{equation}
At $k=1$ the fused polynomial is $H_{ab}$, and the two defects are
\[
 \Dcal_{(a),b}=\frac{H_b[H_a]-H_{ab}}{1-q},\qquad
 \Dcal_{(b),a}=\frac{H_a[H_b]-H_{ab}}{1-q}.
\]
Both are positive by Theorem~\ref{thm:defect}, and
Conjecture~\ref{conj:bergeron} says that the first is at least the
second, coefficient by coefficient. For every $k$, the rectangular
conjecture says the same of $\Dcal_{(a^k),b}$ and $\Dcal_{(b^k),a}$.
For integers $c,d$ with $cd=ab$, which Bergeron's conjecture also
allows, the same cancellation gives
\begin{equation}\label{eq:rectangular}
 \frac{H_c[\Htil_{(d^k)}]-H_a[\Htil_{(b^k)}]}{1-q}
 =\Dcal_{(d^k),c}-\Dcal_{(b^k),a}.
\end{equation}
As a baseline, comparing $q$-supports alone gives a shift that works in
every $t$-degree separately.

\begin{proposition}[A rectangular support bound]\label{prop:rectbenchmark}
Let $a,b\ge2$ and $k\ge1$, and put
\[
 P^{(k)}_{a,b}=H_b[\widetilde H_{(a^k)};q],\qquad
 Q^{(k)}_{a,b}=H_a[\widetilde H_{(b^k)};q],\qquad
 M_{a,b,k}=bk c_a+c_b.
\]
Then the shift $M_{a,b,k}$ suffices:
\begin{equation*}
 \frac{P^{(k)}_{a,b}-q^{M_{a,b,k}}Q^{(k)}_{a,b}}{1-q}
 \in\mathbb N[q,t]\{s_\lambda\}.
\end{equation*}
More generally, if $ab=cd$, the same assertion with
$P^{(k)}_{a,b}$ replaced by $H_c[\widetilde H_{(d^k)};q]$ holds
with exponent $ckc_d+c_c$.
\end{proposition}
At $q=1$, fusion identifies both plethysms with $H_k(X;t)^{ab}$,
so their total multiplicities agree separately in every Schur label
and $t$-degree. Separating their $q$-supports gives the bound; the
proof is in Appendix~\ref{app:rect-support-proof}.

To beat the support bound, we pass the one-row comparisons of
Section~\ref{sec:shift} through the outer plethysm $H_m[-;q^c]$, which
preserves cumulative order in each $t$-degree.

\begin{lemma}[Preservation of cumulative order]\label{lem:filteredplethysm}
Let $f,g\in\mathbb N[q,t]\{s_\lambda\}$ be homogeneous of the same
positive symmetric-function degree. Suppose
$(f-g)/(1-q)\in\mathbb N[q,t]\{s_\lambda\}$.
For all positive integers $m,c$,
\begin{equation}\label{eq:filteredplethysm}
 \frac{H_m[f;q^c]-H_m[g;q^c]}{1-q}
 \in\mathbb N[q,t]\{s_\lambda\}.
\end{equation}
\end{lemma}
\begin{proof}
Choose a vector space $V$ large enough to detect all Schur types in
degree $m\deg f$, and realize $f,g$ as characters of bigraded polynomial
$GL(V)$-modules $M,N$. Fix an irreducible type and a $t$-degree.
The hypothesis says that the cumulative $q$-multiplicities in $M$
dominate those in $N$, and that their totals agree. List the degrees
of the multiplicity basis vectors in increasing order as
$\alpha_1\le\cdots\le\alpha_r$ for $M$ and
$\beta_1\le\cdots\le\beta_r$ for $N$.
The cumulative inequalities are equivalent to
$\alpha_i\le\beta_i$ for every $i$.
Matching these vectors gives a $GL(V)$-equivariant bijection
$\phi\colon N\to M$ that preserves $t$-degree and does not increase
$q$-degree.

Since $\phi^{\otimes m}$ commutes with permutations of the factors, we
may tensor it with the identity of $R_m$, graded with degrees multiplied
by $c$, and take $S_m$-invariants. This gives a bijection
\[
 (R_m\otimes N^{\otimes m})^{S_m}
 \longrightarrow (R_m\otimes M^{\otimes m})^{S_m}
\]
that preserves $t$-degree and does not increase total $q$-degree. Its
source and target have characters $H_m[g;q^c]$ and $H_m[f;q^c]$, so the
filtration inclusions give \eqref{eq:filteredplethysm}, and the
equality of total characters makes the quotient a polynomial.
\end{proof}

With Lemma~\ref{lem:filteredplethysm}, the rectangular comparison
reduces to the one-row one. The column character
$H_k(X;t)=\widetilde H_{(1^k)}$ fuses to both rectangles: $a$ copies
give $(a^k)$, and $b$ copies give $(b^k)$. Each rectangular plethysm
can therefore be compared with a plethysm of one-row plethysms of
$H_k(X;t)$, and Theorem~\ref{thm:qdistorted} compares those. The three shifts add up
as follows.

\begin{samepage}
\begin{theorem}[A distorted rectangular comparison]\label{thm:rectlinear}
Let $a,b\ge2$ and $k,c\ge1$ be integers. Put
\begin{equation*}
 S_{a,b,k}=c_a\bigl(b(k+1-\varepsilon_k)-\varepsilon_b\bigr).
\end{equation*}
Then the shift $S_{a,b,k}$ suffices:
\begin{equation}\label{eq:rectlinear}
 \frac{H_b[\widetilde H_{(a^k)}(X;q,t);q^c]
       -q^{S_{a,b,k}}
        H_a[\widetilde H_{(b^k)}(X;q^c,t);q]}{1-q}
 \in\mathbb N[q,t]\{s_\lambda\}.
\end{equation}
In particular, at $c=1$ the theorem compares $P^{(k)}_{a,b}$ and
$q^{S_{a,b,k}}Q^{(k)}_{a,b}$ with both Macdonald parameters retained.
At $k=1$, it is Theorem~\ref{thm:qdistorted}.
\end{theorem}
\end{samepage}
\begin{proof}
The plethysms of the column character $A=H_k(X;t)$ link the two
rectangles: after a shift, $\Htil_{(a^k)}$ lies above $H_a[A;q]$, and
$H_b[A]$ lies above $\Htil_{(b^k)}$, so Theorem~\ref{thm:qdistorted} for the
alphabet $A$ closes the chain. For the first step, set
$d_a=c_a(k-\varepsilon_k)$; Theorem~\ref{thm:reversefusion}, applied
to the column $(1^k)$, gives
\[
 \frac{\widetilde H_{(a^k)}-q^{d_a}H_a[A;q]}{1-q}\succeq_s0.
\]
Lemma~\ref{lem:filteredplethysm} therefore gives the first of the
following three positive quotients, where
\[
 X_0=H_b[H_a[A;q];q^c],\qquad
 Y_0=H_a[H_b[A;q^c];q],
\]
and $P_c,Q_c$ denote the first and second plethysms in
\eqref{eq:rectlinear} before shifting:
\begin{align*}
 E_1&=\frac{P_c-q^{bd_a}X_0}{1-q}\succeq_s0,\\
 E_2&=\frac{X_0-q^{\delta_{a,b}}Y_0}{1-q}\succeq_s0,\\
 E_3&=\frac{Y_0-Q_c}{1-q}\succeq_s0.
\end{align*}
The inner shift becomes $q^{bd_a}$ because the outer plethysm has
degree $b$. For $E_2$, substitute the positive alphabet $A$ into
Theorem~\ref{thm:qdistorted}; the coefficients of $A$ depend only on
$t$, so the substitution keeps the $q$-grading and the denominator
$1-q$.
For $E_3$, forward fusion gives
\[
 \frac{H_b[A;q^c]-\widetilde H_{(b^k)}(X;q^c,t)}{1-q}
 =[c]_q\Dcal_{(1^k),b}(X;q^c,t)\succeq_s0,
\]
and Lemma~\ref{lem:filteredplethysm} applies to the outer $H_a[-;q]$.
Finally, $S_{a,b,k}=bd_a+\delta_{a,b}$ and
\[
 \frac{P_c-q^{S_{a,b,k}}Q_c}{1-q}
 =E_1+q^{bd_a}E_2+q^{S_{a,b,k}}E_3\succeq_s0.
\]
\end{proof}

At $c=1$, Proposition~\ref{prop:rectbenchmark} and
Theorem~\ref{thm:rectlinear} both apply, so the smaller of $S_{a,b,k}$
and $M_{a,b,k}$ is a valid shift. Appendix~\ref{app:shift-meaning}
compares the two bounds, including the
parity cases and the larger shift obtained by direct reverse fusion.

\section{Open problems}\label{sec:todo}
Conjecture~\ref{conj:bergeron} and its rectangular version would follow
from a graded embedding of one fusion defect module into the other. By
\eqref{eq:rectdefects}, the question is whether there is an embedding
\begin{equation*}
 \Kcal_{(b^k),a}(V)\hookrightarrow\Kcal_{(a^k),b}(V),
 \qquad 2\le a\le b,\quad k\ge1,
\end{equation*}
that preserves both gradings and the $GL(V)$-action, for
$\dim V\ge3abk+1$, where the constructions of
Section~\ref{sec:fusion-filtrations} apply; their stable characters do
not depend on $\dim V$. For $c,d$ with $cd=ab$ and $a\le c,d\le b$, the
same question compares $\Kcal_{(b^k),a}$ with $\Kcal_{(d^k),c}$.

The least shift is also open. Reversing one fusion comparison costs
exactly the shift of Theorem~\ref{thm:reversefusion}, while
Theorem~\ref{thm:rectlinear} compares the two plethysms with a shift
linear in $b$, which beats the support bound of
Proposition~\ref{prop:rectbenchmark} for odd $k$, and for even $k$ once
$b$ is large (Appendix~\ref{app:shift-meaning}).

An intermediate shape does not help. For a non-row shape $\rho\vdash a$,
the defect of $\rho$ with $b$ fused copies has a smaller coefficient of
the standard character $s_{(ab-1,1)}$ and a smaller total dimension
than the defect of $H_a[H_b]$
(Proposition~\ref{prop:butler-bridge-obstruction}), so it cannot lie
between the two defects.

\begin{aiusage}
The results in this paper were developed through extensive use of AI
assistants in sustained mathematical discussion, including failed
attempts, counterexamples, audits, and repairs. AI assistants also helped
with exact computations, proof checks, and exposition. The external
representation-theoretic results are cited at their points of use.
\end{aiusage}
\aiacknowledgements
\begin{thebibliography}{18}
\providecommand{\natexlab}[1]{#1}
\providecommand{\url}[1]{\texttt{#1}}
\expandafter\ifx\csname urlstyle\endcsname\relax
  \providecommand{\doi}[1]{doi: #1}\else
  \providecommand{\doi}{doi: \begingroup \urlstyle{rm}\Url}\fi

\bibitem[Bergeron(2017)]{bergeron2017foulkes}
Fran{\c{c}}ois Bergeron.
\newblock A {$q$}-analog of {F}oulkes' conjecture.
\newblock \emph{The Electronic Journal of Combinatorics}, 24\penalty0
  (1):\penalty0 Paper 1.38, 2017.
\newblock \doi{10.37236/6004}.

\bibitem[Bergeron et~al.(1999)Bergeron, Garsia, Haiman, and
  Tesler]{bergeron1999identities}
Fran{\c{c}}ois Bergeron, Adriano~M. Garsia, Mark Haiman, and Glenn Tesler.
\newblock Identities and positivity conjectures for some remarkable operators
  in the theory of symmetric functions.
\newblock \emph{Methods and Applications of Analysis}, 6\penalty0 (3):\penalty0
  363--420, 1999.
\newblock \doi{10.4310/MAA.1999.v6.n3.a7}.

\bibitem[Brion(1993)]{brion1993stable}
Michel Brion.
\newblock Stable properties of plethysm: on two conjectures of {F}oulkes.
\newblock \emph{Manuscripta Mathematica}, 80:\penalty0 347--371, 1993.
\newblock \doi{10.1007/BF03026558}.

\bibitem[Cheung et~al.(2017)Cheung, Ikenmeyer, and
  Mkrtchyan]{cheung2017foulkes}
Man-Wai Cheung, Christian Ikenmeyer, and Sevak Mkrtchyan.
\newblock Symmetrizing tableaux and the 5th case of the {F}oulkes conjecture.
\newblock \emph{Journal of Symbolic Computation}, 80:\penalty0 833--843, 2017.
\newblock \doi{10.1016/j.jsc.2016.09.002}.

\bibitem[Dent and Siemons(2000)]{dent2000foulkes}
Suzie~C. Dent and Johannes Siemons.
\newblock On a conjecture of {F}oulkes.
\newblock \emph{Journal of Algebra}, 226\penalty0 (1):\penalty0 236--249, 2000.
\newblock \doi{10.1006/jabr.1999.8169}.

\bibitem[Feigin and Loktev(1999)]{feigin1999kostka}
Boris Feigin and Sergei Loktev.
\newblock On generalized {Kostka} polynomials and quantum {Verlinde} rule.
\newblock In \emph{Differential topology, infinite-dimensional {Lie} algebras,
  and applications}, volume 194 of \emph{American Mathematical Society
  Translations, Series 2}, pages 61--79. American Mathematical Society, 1999.
\newblock URL \url{https://arxiv.org/abs/math/9812093}.

\bibitem[Foulkes(1950)]{foulkes1950}
H.~O. Foulkes.
\newblock Concomitants of the quintic and sextic up to degree four in the
  coefficients of the ground form.
\newblock \emph{Journal of the London Mathematical Society}, s1-25\penalty0
  (3):\penalty0 205--209, 1950.
\newblock \doi{10.1112/jlms/s1-25.3.205}.

\bibitem[Guo et~al.(2026)Guo, Kang, and Xiong]{GKX}
Peter~L. Guo, Mingyang Kang, and Rui Xiong.
\newblock A solution to {Butler}'s positivity conjecture, 2026.
\newblock URL \url{https://arxiv.org/abs/2608.11543v1}.
\newblock Version 1, 12 August 2026.

\bibitem[Haiman(2001)]{haiman2001hilbert}
Mark Haiman.
\newblock Hilbert schemes, polygraphs and the {Macdonald} positivity
  conjecture.
\newblock \emph{Journal of the American Mathematical Society}, 14\penalty0
  (4):\penalty0 941--1006, 2001.
\newblock \doi{10.1090/S0894-0347-01-00373-3}.

\bibitem[Kedem(2004)]{Kedem}
Rinat Kedem.
\newblock Fusion products, cohomology of {$GL_N$} flag manifolds and {Kostka}
  polynomials.
\newblock \emph{International Mathematics Research Notices}, 2004\penalty0
  (25):\penalty0 1273--1298, 2004.
\newblock URL \url{https://arxiv.org/abs/math/0312478}.

\bibitem[Khoroshkin and Makedonskyi(2026)]{KM}
A.~Khoroshkin and I.~Makedonskyi.
\newblock On fusion product and {$N!/k$}-conjecture, 2026.
\newblock URL \url{https://arxiv.org/abs/2610.07461v1}.
\newblock Version 1, 5 October 2026.

\bibitem[Macdonald(1995)]{macdonald1995}
Ian~G. Macdonald.
\newblock \emph{Symmetric Functions and Hall Polynomials}.
\newblock Oxford University Press, Oxford, 2nd edition, 1995.

\bibitem[McKay(2008)]{mckay2008plethysm}
Tom McKay.
\newblock On plethysm conjectures of {S}tanley and {F}oulkes.
\newblock \emph{Journal of Algebra}, 319\penalty0 (5):\penalty0 2050--2071,
  2008.
\newblock \doi{10.1016/j.jalgebra.2007.12.003}.

\bibitem[{OpenAI}(2026)]{OpenAI}
{OpenAI}.
\newblock {Foulkes}' conjecture for the sixth symmetric power, 2026.
\newblock URL
  \url{https://www.emergentmind.com/openai-math-explorer/papers/373.pdf}.
\newblock OpenAI Math Explorer, manuscript 373, 25 September 2026.

\bibitem[Sra(2026{\natexlab{a}})]{foulkesII}
Suvrit Sra.
\newblock The {$q$}-{F}oulkes problem {II}: {Schur} positivity with high
  probability.
\newblock Preprint, 2026{\natexlab{a}}.

\bibitem[Sra(2026{\natexlab{b}})]{foulkesIII}
Suvrit Sra.
\newblock The {$q$}-{F}oulkes problem {III}: On {Foulkes'} conjecture with a
  graded inner factor.
\newblock Preprint, 2026{\natexlab{b}}.

\bibitem[Thrall(1942)]{thrall1942symmetrized}
R.~M. Thrall.
\newblock On symmetrized {K}ronecker powers and the structure of the free {L}ie
  ring.
\newblock \emph{American Journal of Mathematics}, 64:\penalty0 371--388, 1942.
\newblock \doi{10.2307/2371691}.

\bibitem[Westbury(2016)]{westbury2016cyclic}
Bruce~W. Westbury.
\newblock Invariant tensors and the cyclic sieving phenomenon.
\newblock \emph{The Electronic Journal of Combinatorics}, 23\penalty0
  (4):\penalty0 P4.25, 2016.
\newblock \doi{10.37236/4569}.
\newblock URL
  \url{https://www.combinatorics.org/ojs/index.php/eljc/article/download/v23i4p25/pdf/}.

\end{thebibliography}

\appendix
\addtocontents{toc}{\protect\setcounter{tocdepth}{1}}
\begingroup
\renewcommand{\contentsname}{Appendix contents}
\bigskip
\noindent\begin{minipage}{\linewidth}
\tableofcontents
\end{minipage}
\endgroup

\section{Shift bounds and exact checks}\label{app:shift-meaning}
This appendix proves the cyclic duality lemma and the support bound
used in Sections~\ref{sec:fusion} and~\ref{sec:shifted}, and compares
the admissible shifts with one another and with exact computations.

\subsection{Duality and its rectangular specialization}\label{app:rectangle-reversal}
For even $n$, the proof of Theorem~\ref{thm:reversefusion} corrects the
outer twist by the character $J_m^{(m)}$, which must be positive with
the denominator $1-q^m$ to keep the comparison in each residue class.
Its positivity comes from the rotation of tensor factors in the proof
of Theorem~\ref{thm:defect}.

\begin{lemma}\label{lem:cyclicduality}
For every $m\ge1$,
\[
 J_m^{(m)}=\frac{\omega H_m(X;q)-q^{c_m}H_m(X;q)}{1-q^m}
 \in\N[q]\{s_\lambda\}.
\]
\end{lemma}
\begin{proof}
On Kedem's fusion filtration of $W^{\otimes m}$, the backward rotation
$R$ acts on $G_j/G_{j-1}$ as $\zeta^j$, by the proof of
Theorem~\ref{thm:defect}, so $H_m(X;\zeta^k)$ is the trace of $R^k$ on
$W^{\otimes m}$, as a character of $GL(W)$. Since $R^k$ and $R^{-k}$
are conjugate permutations, $H_m(X;\zeta^k)=H_m(X;\zeta^{-k})$ for
every integer $k$, and coinvariant reciprocity
\eqref{eq:fusionreciprocity} then gives
\[
 \omega H_m(X;\zeta^k)=\zeta^{kc_m}H_m(X;\zeta^k).
\]
With $\dim W\ge m$, so that all Schur labels are detected, the
numerator of $J_m^{(m)}$ vanishes at every $m$th root of unity and is
divisible by $1-q^m$.

For each Schur label, the first numerator term is supported in
$[0,c_m]$, and the second in $[c_m,2c_m]$. By divisibility, their
total multiplicities agree in every residue class modulo $m$.
Before degree $c_m$, each residue-wise cumulative difference is
nonnegative because only the first term contributes. At and after
$c_m$, the first term has reached the shared total, while the second
has contributed at most that total. Every coefficient of the quotient
is therefore a nonnegative integer.

With the ordinary denominator, writing
$\omega H_m=\sum_{\nu,j}a_{\nu,j}q^js_\nu$ and using reciprocity gives
\[
 [m]_qJ_m^{(m)}
 =\sum_{\nu,j}a_{\nu,j}q^j[2(c_m-j)]_q s_\nu,
 \qquad a_{\nu,j}\ge0,\quad 0\le j\le c_m.
\]
\end{proof}

For the rectangle of Section~\ref{sec:shift}, with $a,b\ge2$, the
characters \eqref{eq:PQRtwo}, the quotient $V$ of \eqref{eq:UV}, and
$\overline{\mathcal Q}=(\omega^{b+1}H_a[-;q])[H_b(X;t)]$, the reversal
identity \eqref{eq:reverse-general-twist} reads
\begin{equation*}
 q^{bc_a-1}t^{ac_b}\omega V(q^{-1},t^{-1})
 =\frac{\mathcal R-q^{(b-1)c_a}\overline{\mathcal Q}}{1-q}\succeq_s0.
\end{equation*}
The parity correction and the sign obstruction are those of the proof
of Theorem~\ref{thm:reversefusion}.

\subsection{Support separation in each grading}\label{app:rect-support-proof}
The support bound uses only that the two plethysms agree at $q=1$ and
have bounded $q$-degree.
\begin{proof}[Proof of Proposition~\ref{prop:rectbenchmark}]
Theorem~\ref{thm:defect}, at $q=1$ with inner partition $(1^k)$,
gives $\widetilde H_{(a^k)}(X;1,t)=H_k(X;t)^a$.
Since $H_b[-;1]$ is the ordinary $b$th power,
\begin{equation}\label{eq:rectslice}
 P^{(k)}_{a,b}(X;1,t)=Q^{(k)}_{a,b}(X;1,t)=H_k(X;t)^{ab}.
\end{equation}
The $q$-degree of $\widetilde H_{(a^k)}$ is at most $kc_a$, so
$\deg_qP^{(k)}_{a,b}\le M_{a,b,k}$. Fix a Schur label and a $t$-degree.
Below degree $M_{a,b,k}$ the shifted source
$q^{M_{a,b,k}}Q^{(k)}_{a,b}$ contributes nothing, and from degree
$M_{a,b,k}$ on the target has reached its total multiplicity, which
equals that of the source by \eqref{eq:rectslice}. Every cumulative
difference is therefore nonnegative, and the equal totals make the
quotient a polynomial. For $cd=ab$ the same argument uses
$\deg_qH_c[\widetilde H_{(d^k)};q]\le ckc_d+c_c$.
\end{proof}

\subsection{Comparing admissible shifts}
A shift is admissible when the corresponding shifted quotient is Schur
positive. Evaluated at $c=1$ and $0\le q<1$, Theorem~\ref{thm:qdistorted}
compares real Schur coefficients,
\begin{equation*}
 H_b[H_a](q)\succeq_s q^{\delta_{a,b}}H_a[H_b](q),
\end{equation*}
so, for any $0<\eta<1$,
\[
 q\ge(1-\eta)^{1/\delta_{a,b}}
 \quad\Longrightarrow\quad
 H_b[H_a](q)\succeq_s(1-\eta)H_a[H_b](q),\qquad 0\le q<1.
\]

At $k=1$, Proposition~\ref{prop:rectbenchmark} gives the shift
$M=bc_a+c_b$, quadratic in $b$ for fixed $a$, while
Theorem~\ref{thm:qdistorted} gives $\delta_{a,b}=a(a-1)\lfloor b/2\rfloor$,
linear in $b$; by Theorem~\ref{thm:shiftsharp}, $\delta_{a,b}$ is the
least shift for the comparison of $\mathcal R$ with $\mathcal Q$.

For $c=1$, both $S_{a,b,k}$ and $M_{a,b,k}$ are admissible, and their
difference is
\[
 M_{a,b,k}-S_{a,b,k}
 =\begin{cases}
 c_b+\varepsilon_bc_a,&k\text{ odd},\\
 c_b-c_a(b-\varepsilon_b),&k\text{ even}.
 \end{cases}
\]
For odd $k$, $S_{a,b,k}$ is always the smaller; for even $k$, it is
smaller exactly when $c_b>c_a(b-\varepsilon_b)$. For general $c$, the
support argument gives the exponent $bk c_a+c c_b$.

Reversing fusion directly is costlier. Theorem~\ref{thm:reversefusion}
with $\mu=(b^k)$ and $m=a$ gives the least shift
$\Delta_{(b^k),a}=c_a(kb^2-\varepsilon_{bk})$ for the fused character
$\widetilde H_{((ab)^k)}$ to dominate the shifted $Q^{(k)}_{a,b}$, and
adding the forward defect of $P^{(k)}_{a,b}$ turns this into a full
comparison, but with an exponent larger than $M_{a,b,k}$:
\[
 \Delta_{(b^k),a}-M_{a,b,k}
 =c_b(2kc_a-1)-\varepsilon_{bk}c_a>0.
\]
Theorem~\ref{thm:rectlinear} avoids this cost by passing through
one-row plethysms of the column character, where the shift is the
smaller $\delta_{a,b}$.

Since $\widetilde H_{\mu'}(X;q,t)=\widetilde H_\mu(X;t,q)$,
Theorem~\ref{thm:rectlinear} also compares the conjugate rectangles,
with the same shift: for $a,b\ge2$ and $k,c\ge1$,
\begin{equation*}
 \frac{H_b[\widetilde H_{(k^a)}(X;q,t);t^c]
       -t^{S_{a,b,k}}H_a[\widetilde H_{(k^b)}(X;q,t^c);t]}{1-t}
 \in\mathbb N[q,t]\{s_\lambda\},
\end{equation*}
and at $c=1$ the exponent $\min\{S_{a,b,k},M_{a,b,k}\}$ is also
admissible.

\subsection{Exact checks}\label{app:bivariate-checks}
Exact Schur expansions over $\mathbb Q[q,t]$ check forward and reverse
fusion for 25 pairs $(\mu,m)$ with $|\mu|\le4$, $m\le3$, and
$m|\mu|\le8$, including $m=1$ and both parities of $|\mu|$. At every
positive threshold $\Delta_{\mu,m}$, decreasing the shift by one
produces a negative coefficient, as the sign obstruction requires.

\begin{samepage}
Six rectangular comparisons through total degree twelve check both the
shift $S_{a,b,k}$ of Theorem~\ref{thm:rectlinear} and the support
bound, including $c=2$, the case $a=b$, and both orders of $a,b$, with
Adams operations acting on both grading variables. Representative
parameters are
\[
\begin{array}{c|c|c}
 (a,b,k,c)&S_{a,b,k}&bk c_a+c c_b\\\hline
 (2,3,1,2)&2&9\\
 (2,3,2,1)&8&9\\
 (2,2,3,2)&6&8
\end{array}
\]
\end{samepage}
Another 36 cases check the denominator $1-q^m$ over $\mathbb Q[q,t]$,
for forward positivity and for reverse positivity at the sharp shift:
all nontrivial pairs with $m|\mu|\le8$, all partitions of size at most
four at $m=1$, and four cases with empty shape. A bivariate example is
\[
 \Ecal_{(1,1),2}=s_{(2,2)}+t\,s_{(2,1,1)}+t^2s_{(1^4)}.
\]
The residue condition in \eqref{eq:reversefusion-residue} is a
polynomiality requirement; increasing the sharp shift by one when
$m>1$ violates it, as the trivial coefficient already shows.

\section{The stretched Butler curve and defect comparisons}\label{app:stretchproof}
To compare fusion defects with different inner shapes, we deform one
Garsia--Haiman module into another. Haiman's proof of the $n!$ theorem
realizes the Garsia--Haiman module of $\mu$ as the fiber, at the
monomial ideal of $\mu$ (spanned by the monomials $x^jy^i$ with
$(i,j)\notin\mu$), of a vector bundle on the Hilbert scheme
$\operatorname{Hilb}^n(\C^2)$ of $n$ points in the plane, the Procesi
bundle \citep{haiman2001hilbert}. When two shapes differ by moving one
box, \citet[Lemmas~2.1--2.2 and Theorem~2.3]{GKX} join their monomial
ideals by a family of ideals over a projective line. A vector bundle
on a projective line splits into line bundles, and for the restricted
Procesi bundle the splitting writes both Macdonald polynomials in
terms of the same positive characters.

Let $\lambda,\mu\vdash n$ differ by moving one box up and to the right,
\begin{equation}\label{eq:boxmove}
 \lambda=\eta\cup\{(i,j+d)\},\qquad
 \mu=\eta\cup\{(i+e,j)\},\qquad d,e>0,
\end{equation}
where a cell $(i,j)$ has zero-based row index $i$, counted by $t$, and
column index $j$, counted by $q$. For example, $\lambda=(2)$ and
$\mu=(1,1)$ differ by such a move, with $\eta=(1)$ and $d=e=1$. The
union of $\lambda$ and $\mu$ is a partition of $n+1$, and their
intersection is $\eta$.

Along the projective line joining the monomial
ideals of $\lambda$ and $\mu$, the tautological bundle, whose fiber at
an ideal is the quotient ring by that ideal, has a determinant of
degree one, which confines the splitting degrees of the Procesi bundle
to $0$ and $1$. At $m=1$, the theorem below writes $\Htil_\lambda=I+q^dJ$
and $\Htil_\mu=I+t^eJ$ with $I,J\succeq_s0$; in the example,
$\Htil_{(2)}=s_{(2)}+qs_{(1,1)}$ and $\Htil_{(1,1)}=s_{(2)}+ts_{(1,1)}$,
so $I=s_{(2)}$ and $J=s_{(1,1)}$. The positivity of $I$ is
Butler's conjecture of 1994, proved by \citet{GKX}. Pulling the family
back along $x\mapsto x^m$ stretches $\lambda$ and $\mu$ to $m\lambda$
and $m\mu$, and the same splitting then has $m+1$ pieces, with
$\mathcal J^{(m)}_{\lambda,\mu}$ the $m$-fold version of $J$.

\begin{theorem}[Positive stretched divided difference]\label{thm:stretch}
For every $m\ge1$, there are characters
$C_0,\ldots,C_m\in\N[q,t]\{s_\nu\}$ such that
\begin{equation}\label{eq:commonC}
 \Htil_{m\lambda}=\sum_{l=0}^m q^{mdl}C_l,
 \qquad
 \Htil_{m\mu}=\sum_{l=0}^m t^{el}C_l.
\end{equation}
Consequently
\begin{equation}\label{eq:Jstretch}
 \mathcal J^{(m)}_{\lambda,\mu}
 :=\frac{\Htil_{m\lambda}-\Htil_{m\mu}}{q^{md}-t^e}
 =\sum_{l=1}^m C_l\sum_{r=0}^{l-1}q^{md(l-1-r)}t^{er}
 \pos.
\end{equation}
\end{theorem}

\subsection{The geometric construction}
The characters $C_l$ of Theorem~\ref{thm:stretch} are the quotients of
the splitting filtration of an equivariant vector bundle on
$\mathbb P^1$; we construct the bundle and record its weights at both
endpoints.

By Haiman's theory \citep{haiman2001hilbert}, the Procesi bundle
$\mathcal P_n$ on $\operatorname{Hilb}^n(\C^2)$ is globally generated,
has fiber character $\Htil_\mu$ at the monomial ideal of $\mu$, and,
with $\mathcal B_n$ the tautological bundle, satisfies
\citep[(2.3)]{GKX}
\[
 \mathcal P_n\cong
 \mathcal P_n^\vee\otimes\det\mathcal B_n\otimes\sgn.
\]
A vector bundle on $\mathbb P^1$ has a canonical filtration by its
splitting degrees, which is equivariant because it is intrinsic.

\begin{proof}[Proof of Theorem~\ref{thm:stretch}]
Put $\nu=\lambda\cup\mu$ and use coordinates $z,y$ on the
original affine plane. For a partition $\rho$, let $I_\rho$ denote
the monomial ideal whose standard monomials are indexed by $\rho$.
In $\C[z,y]/I_\nu$, let
\[
 \alpha=z^{j+d}y^i,\qquad \beta=z^jy^{i+e},\qquad
 W=\operatorname{span}\{\alpha,\beta\}.
\]
Both classes are corners of $\nu$, so $z$ and $y$ annihilate them, and
for a line $\ell\subset W$ the ideal $I_\ell=I_\nu+(\ell)$ has
colength $n$. These ideals form a flat family
$f\colon\mathbb P(W)\to\operatorname{Hilb}^n(\C^2)$, with endpoints
$I_\lambda$ at $\ell=\C\beta$ and $I_\mu$ at $\ell=\C\alpha$, and with
tautological sequence
\[
 0\longrightarrow\mathcal O_{\mathbb P(W)}(-1)
 \longrightarrow\mathcal O_{\mathbb P(W)}\otimes(\C[z,y]/I_\nu)
 \longrightarrow f^*\mathcal B_n\longrightarrow0.
\]
By \citet[Lemma~2.1]{GKX}, the sequence gives
$\deg\det f^*\mathcal B_n=1$.

To stretch, pull the family back under $(z,y)=(x^m,y)$. The extension
$\C[z,y]\subset\C[x,y]$ is free with basis $1,x,\ldots,x^{m-1}$, so the
family stays flat, has length $mn$, and has endpoints $m\lambda$ and
$m\mu$. With $z$ of weight $q^m$, the quotient line bundle
$(\mathcal O\otimes W)/\mathcal O(-1)$ has endpoint weights
$q^{m(j+d)}t^i$ and $q^{mj}t^{i+e}$, and twisting by the character
$q^{-mj}t^{-i}$ gives a degree-one line bundle $\mathcal L$ with
endpoint weights $q^{md}$ and $t^e$, the weights in \eqref{eq:commonC}.

Pulling $\mathcal P_{mn}$ back gives a vector bundle $E$ on
$\mathbb P^1$ whose fibers at the endpoints have the two Macdonald
characters in \eqref{eq:commonC}. The tautological bundle of the
pulled-back family is a direct sum of $m$ copies of the original one,
twisted by characters, so its determinant has degree $m$. If
$\mathcal O(k)$ occurs in the splitting of $E$, self-duality makes
$\mathcal O(m-k)$ occur as well, and global generation gives
$k,m-k\ge0$, so every splitting degree lies in $[0,m]$.

The canonical filtration of $E$ by splitting degree is equivariant for
the torus and for $S_{mn}$, and its degree-$l$ quotient is
$\mathcal L^l\otimes C_l$ for a representation $C_l$ whose torus
character is a priori a Laurent polynomial. Taking fibers proves
\eqref{eq:commonC} as identities of Laurent characters. A weight
$q^\sigma t^\tau$ of $C_l$ contributes $q^{\sigma+mdl}t^\tau$ at one
endpoint and $q^\sigma t^{\tau+el}$ at the other; neither endpoint has
negative degrees, and all multiplicities are nonnegative, so
$\sigma,\tau\ge0$. Each $C_l$ is therefore a positive polynomial
character, and subtraction proves \eqref{eq:Jstretch}.
\end{proof}

The quotient $\mathcal J^{(m)}_{\lambda,\mu}$ is the character of the
module $H^0(\mathbb P^1,E\otimes\mathcal L^{-1})$: after the twist
every splitting degree is at least $-1$, so $H^1$ vanishes, and the
character of $H^0(\mathcal L^{l-1})$ is the geometric sum in
\eqref{eq:Jstretch}.

\subsection{Defect comparisons under a box move}
\label{app:butler-defect-comparisons}
A fusion defect compares a plethysm with a fused polynomial, so across
a box move both differences are needed. The fused side is
Theorem~\ref{thm:stretch}; at $m=1$, with $I=C_0$ and $J=C_1$, it reads
\begin{equation*}
 \Htil_\lambda=I+q^dJ,\qquad
 \Htil_\mu=I+t^eJ,\qquad I,J\pos.
\end{equation*}
For the plethysms, the one-row Cauchy identity
$H_m[I+zJ]=\sum_r\qbin mr z^rH_{m-r}[I]H_r[J]$, at $z=q^d$ and at
$z=t^e$, gives a second positive quotient, in which every term is a
graded character:
\begin{align*}
 \Xi_{\lambda,\mu,m}
 &:=\frac{H_m[\Htil_\lambda]-H_m[\Htil_\mu]}{q^d-t^e}\\
 &=\sum_{r=1}^m\qbin mr H_{m-r}[I]H_r[J]
       \sum_{l=0}^{r-1}q^{d(r-1-l)}t^{el}\pos.
\end{align*}
The difference of the two defects combines $\Xi_{\lambda,\mu,m}$ and
$\mathcal J^{(m)}_{\lambda,\mu}$, with coefficients that are positive
once $t=q^c$ for suitable $c$.

\begin{theorem}[Fusion defects under a box move]\label{thm:costmove}
For the move \eqref{eq:boxmove}, let $c\in\Z_{>0}$ satisfy $d\le ce\le md$.
Then
\begin{align*}
 &\Dcal_{\lambda,m}(q,q^c)-\Dcal_{\mu,m}(q,q^c)\\
 &\quad=q^d[ce-d]_q\,\Xi_{\lambda,\mu,m}(q,q^c)
       +q^{ce}[md-ce]_q\,\mathcal J^{(m)}_{\lambda,\mu}(q,q^c)\pos.
\end{align*}
\end{theorem}
\begin{proof}
Before specializing,
\[
 (1-q)(\Dcal_{\lambda,m}-\Dcal_{\mu,m})
 =(q^d-t^e)\Xi_{\lambda,\mu,m}
 -(q^{md}-t^e)\mathcal J^{(m)}_{\lambda,\mu}.
\]
Setting $t=q^c$ and dividing each monomial difference by $1-q$ gives
the stated formula, whose $q$-integers $[ce-d]_q$ and $[md-ce]_q$ are
nonnegative because $d\le ce\le md$.
\end{proof}

Dominance is generated by box moves, so applying
Theorem~\ref{thm:costmove} along a chain orders the defects of all
comparable shapes once $m$ is large.

\begin{corollary}[Uniform dominance comparison of defects]\label{cor:defectdom}
If $\lambda\succeq\mu$ are partitions of $n\ge2$ and $m\ge(n-1)^2$, then
\[
 \Dcal_{\lambda,m}(q,q^{n-1})-\Dcal_{\mu,m}(q,q^{n-1})\pos.
\]
In particular, for $\rho\vdash6$ and $b\ge25$,
\[
 \Dcal_{(6),b}(q)-\Dcal_{\rho,b}(q,q^5)\pos.
\]
\end{corollary}
\begin{proof}
Each box move \eqref{eq:boxmove} in a chain from $\lambda$ to $\mu$ has
$1\le d,e\le n-1$, so $c=n-1$ gives $d\le ce\le(n-1)^2\le m\le md$.
Theorem~\ref{thm:costmove} therefore applies to every move, and adding
the comparisons along the chain proves the corollary.
\end{proof}

\subsection{Why a non-row intermediate cannot bridge the two defects}
\label{app:butler-bridge-obstruction}
The unshifted comparison of $\Dcal_{(a),b}$ with $\Dcal_{(b),a}$ could
pass through the defect $\Dcal_{\rho,b}$ of a non-row partition
$\rho\vdash a$: Corollary~\ref{cor:defectdom} compares $\Dcal_{(a),b}$
with $\Dcal_{\rho,b}$ for the same $m=b$, so it would remain to show
$\Dcal_{\rho,b}\succeq_s\Dcal_{(b),a}$. Proposition~\ref{prop:butler-bridge-obstruction}
shows that this comparison fails, already for total dimensions. Both
obstructions use two invariants of a defect, its coefficient of the
standard character and its total dimension, which we compute first.

\begin{lemma}[Standard multiplicity and total dimension]
\label{lem:defect-standard-dimension}
Let $\mu\vdash n\ge1$, $m\ge2$, and $N=mn$, and define
\[
 C_{r,m}(q)=\frac{[r]_q[m]_q-[rm]_q}{1-q}.
\]
Then the coefficient of the standard character $s_{(N-1,1)}$ in the
defect is
\begin{equation}\label{eq:defect-standard}
 [s_{(N-1,1)}]\Dcal_{\mu,m}
 =\sum_{i\ge0}t^i C_{\mu_{i+1},m}(q).
\end{equation}
For $r\ge2$,
\[
 C_{r,m}=[r]_q\sum_{j=1}^{m-1}q^j[j(r-1)]_q
\]
has degree $rm-2$ and leading coefficient one; $C_{1,m}=0$.
Writing $\operatorname{Dim} f=\sum_{\lambda\vdash N}
d_\lambda[s_\lambda]f(1,1)$, we also have
\begin{equation}\label{eq:defect-total-dimension}
 \operatorname{Dim}\Dcal_{\mu,m}
 =\frac{N!}{4}\,m(m-1)\left(\sum_i\mu_i^2-1\right).
\end{equation}
\end{lemma}
\begin{proof}
With $B_\mu(q,t)=\sum_{i\ge0}t^i[\mu_{i+1}]_q$, the modified Macdonald
evaluation formula \citep{macdonald1995,haiman2001hilbert}, in
zero-based cell coordinates, reads
\[
 \Htil_\mu[1-u;q,t]
 =\prod_{(i,j)\in\mu}(1-uq^jt^i)
 =1-uB_\mu(q,t)+O(u^2).
\]
To expand the plethysm, we use the coinvariant Cauchy identity, which
follows from \eqref{eq:Hdef},
\[
 H_m[1+Y;q]=\sum_{j=0}^m\qbin mj H_j[Y;q].
\]
Adams operations send $u$ to $u^\ell$, so an alphabet $Y=O(u)$ gives
$H_j[Y;q]=O(u^j)$, and substituting the Macdonald expansion gives
\[
 H_m[\Htil_\mu;q][1-u]
 =1-u[m]_qB_\mu(q,t)+O(u^2).
\]
Schur coefficients are read off from such evaluations, since each
homogeneous $f$ of degree $N$ satisfies
\[
 f[1-u]=(1-u)\sum_{r=0}^{N-1}(-u)^r
 [s_{(N-r,1^r)}]f;
\]
by Jacobi--Trudi, non-hook Schur functions vanish at $1-u$, and a hook
evaluates to $(1-u)(-u)^r$. Both numerator characters in
\eqref{eq:defect} have trivial coefficient one, so comparing the
coefficients of $u$ gives \eqref{eq:defect-standard}. Finally,
$[rm]_q=[r]_q\sum_{j=0}^{m-1}q^{rj}$ gives the positive formula for
$C_{r,m}$ and its leading term.

For the dimension, both numerator characters in \eqref{eq:defect} have
total $S_N$-dimension $N!$, so the divided difference at $q=1$ is $N!$
times the mean $x$-degree of the fused character minus that of the
plethysm. Sign reciprocity gives
the inner Garsia--Haiman module mean $x$-degree $n(\mu')/2$, and the
wreath construction adds the degrees of $m$ inner factors and an outer
coinvariant factor, so the mean degrees of the plethysm and of the
fused character are, respectively,
\[
 \frac{m n(\mu')+c_m}{2},
 \qquad \frac{n((m\mu)')}{2}.
\]
Since $n((m\mu)')-m n(\mu')=c_m\sum_i\mu_i^2$, the difference of the
two means gives \eqref{eq:defect-total-dimension}.
\end{proof}

For a non-row inner shape, both invariants fall short of those of
$\Dcal_{(b),a}$.

\begin{samepage}
\begin{proposition}[Obstruction to a non-row intermediate]
\label{prop:butler-bridge-obstruction}
Let $a\ge2$, $b\ge a$, and $\rho\vdash a$ with $\rho\ne(a)$.
Then
\begin{equation}\label{eq:butler-bridge-top}
 [q^{ab-2}s_{(ab-1,1)}]
 \bigl(\Dcal_{\rho,b}(q,q^{a-1})-\Dcal_{(b),a}(q)\bigr)=-1.
\end{equation}
Moreover, independently of the specialization $t=q^c$,
\begin{equation*}
 \operatorname{Dim}\Dcal_{\rho,b}
 -\operatorname{Dim}\Dcal_{(b),a}
 \le-\frac{(ab)!}{4}(a-1)(b-1)(a+b)<0.
\end{equation*}
Equality in the displayed bound occurs for $\rho=(a-1,1)$.
\end{proposition}
\end{samepage}
\begin{proof}
By \eqref{eq:defect-standard}, a nonzero summand of the standard
coefficient of $\Dcal_{\rho,b}(q,q^{a-1})$ has $r=\rho_{i+1}\ge2$ and
degree $br-2+(a-1)i$. If $i=0$, then $r<a$ because $\rho\ne(a)$; if
$i\ge1$, the rows above give $a-r\ge i$, hence
\[
 ab-br-(a-1)i\ge(b-a+1)i>0.
\]
Every such degree is therefore below $ab-2$, while the standard
coefficient of $\Dcal_{(b),a}$ is $C_{b,a}$, of degree $ab-2$ with
leading coefficient one, which proves \eqref{eq:butler-bridge-top}.

For every non-row partition $\rho$ of $a$,
$\sum_i\rho_i^2-1\le(a-1)^2$, with equality for $(a-1,1)$, and
\eqref{eq:defect-total-dimension} applied to both defects gives
\begin{align*}
 \operatorname{Dim}\Dcal_{\rho,b}
 -\operatorname{Dim}\Dcal_{(b),a}
 &\le\frac{(ab)!}{4}
 \bigl(b(b-1)(a-1)^2-a(a-1)(b^2-1)\bigr)\\
 &=-\frac{(ab)!}{4}(a-1)(b-1)(a+b).
\end{align*}
Evaluation at $(q,t)=(1,1)$ does not see the specialization $t=q^c$.
\end{proof}

The standard coefficient alone also gives an obstruction independent of
the specialization, since
\[
 C_{r,m}(1)=m(m-1)\binom r2,\qquad
 [s_{(N-1,1)}]\Dcal_{\mu,m}(1,1)=m(m-1)n(\mu').
\]
For a non-row $\rho\vdash a$, this value is strictly smaller than the
corresponding total for $\Dcal_{(b),a}$.
In contrast, $C_{a,b}=C_{b,a}$, so the standard coefficient of the
full $F_{a,b}$ vanishes identically, while
\[
 \operatorname{Dim}F_{a,b}
 =\frac{(ab)!}{4}(a-1)(b-1)(b-a)\ge0.
\]
Thus $\Dcal_{(a),b}$ has sufficient total dimension,
but every non-row intermediate is too small. Constant convex
combinations of such intermediates have the same obstruction.

\section{Matrix polarization and positive kernels}\label{sec:matrix}
Polarization turns each positive character of this paper into a kernel
on tuples of vectors that is positive semidefinite coefficient by
coefficient, which is stronger than nonnegativity of the corresponding
matrix function on Gram matrices.

For a permutation $\sigma\in S_n$ of cycle type $\rho$, the graded
trace of $\sigma$ on the coinvariant algebra $R_n$ is
\[
 \kappa_n(\sigma;q)=\frac{(q;q)_n}{\prod_{j\in\rho}(1-q^j)},
\]
because quotienting the polynomial ring by invariant generators of
degrees $1,\ldots,n$ turns the graded trace $1/\det(I-q\sigma)$ into
$(q;q)_n/\det(I-q\sigma)$. Hence
$H_n=\sum_{\rho\vdash n}\kappa_n(\rho;q)p_\rho/z_\rho$.

\subsection{Polarization and positive kernels}\label{sec:polar}
For a homogeneous symmetric function $f$ of degree $n$ and an
$n\times n$ matrix $A$, with ordinary commuting entries, put
\[
 \mathfrak M_n(f)(A)=[x_1\cdots x_n]f(XA),\qquad X=\diag(x_1,\ldots,x_n),
\]
where $f(XA)$ is evaluated through $p_r(XA)=\tr((XA)^r)$, so that no
diagonalization is needed. If $f=\sum_{\mu\vdash n}a_\mu p_\mu$, define
the central function $\chi_f(\sigma)=z_{\mu(\sigma)}a_{\mu(\sigma)}$.
Then
\[
 \mathfrak M_n(f)(A)=\sum_{\sigma\in S_n}\chi_f(\sigma)
 \prod_i a_{i,\sigma(i)},
\]
and in particular $\mathfrak M_n(s_\lambda)$ is the immanant
$\Imm_\lambda(A)=\sum_\sigma\chi^\lambda(\sigma)\prod_ia_{i,\sigma(i)}$
of the irreducible character $\chi^\lambda$ of $S_n$.

To see the formula, expand each trace
$p_r(XA)=\operatorname{tr}((XA)^r)$. A squarefree term in
$p_\rho(XA)$ is a permutation with cycle type $\rho$; choosing the
starting point of each cycle and ordering cycles of equal length gives
exactly $z_\rho$ copies of its matrix monomial. Hence
$\mathfrak M_n(p_\rho/z_\rho)$ is the sum of those monomials.

Since $H_n=\sum_\mu\kappa_n(\mu;q)p_\mu/z_\mu$, the formula gives
$\mathfrak M_n(H_n)(A;q)=\sum_\sigma\kappa_n(\sigma;q)\prod_i a_{i,\sigma(i)}$.
In particular,
\[
 \mathfrak M_n(H_n)(A;0)=\per A,\qquad
 \mathfrak M_n(H_n)(A;1)=n!\prod_i a_{ii}.
\]

For tuples $\mathbf u=(u_1,\ldots,u_n)$ and $\mathbf v=(v_1,\ldots,v_n)$
in $\C^d$, the cross-Gram kernel of $f$ is
\[
 K_f(\mathbf u,\mathbf v)=
 \mathfrak M_n(f)\bigl([\langle u_i,v_j\rangle]_{i,j=1}^n\bigr).
\]
The kernel $K_f$ detects the Schur coefficients of $f$.

\begin{prop}[Schur coefficients are kernel eigenvalues]\label{thm:kernel}
Suppose $f=\sum_{\lambda\vdash n}c_\lambda s_\lambda$ has real coefficients.
The kernel $K_f$ on $(\C^d)^n$ is positive semidefinite if and only if
$c_\lambda\ge0$ for every $\lambda$ with $\ell(\lambda)\le d$.
For $d\ge n$, this condition is equivalent to Schur positivity of $f$.
The same equivalence holds coefficient by coefficient for polynomial
parameters.
\end{prop}
\begin{proof}
Let $T_f=\sum_\sigma\chi_f(\sigma)U_\sigma$, where $U_\sigma$ permutes
the factors of $(\C^d)^{\otimes n}$. For the pure tensors associated
with $\mathbf u,\mathbf v$, the kernel is
$K_f(\mathbf u,\mathbf v)=\langle u_1\otimes\cdots\otimes u_n,
T_f(v_1\otimes\cdots\otimes v_n)\rangle$. Pure tensors span the
tensor space, so positivity of the kernel is equivalent to $T_f\succeq0$.
Character orthogonality gives $T_f$ the eigenvalue
$(n!/d_\lambda)c_\lambda$ on the $S_n$-isotypic component of type
$\lambda$, and by Schur--Weyl duality exactly the types with
$\ell(\lambda)\le d$ occur in $(\C^d)^{\otimes n}$. Applying the
argument to each coefficient of the parameters gives the last
assertion.
\end{proof}

By Proposition~\ref{thm:kernel}, Bergeron's conjecture says exactly
that the kernel $K_{F_{a,b}}$ is positive semidefinite coefficient by
coefficient. Nonnegativity of $\mathfrak M_{ab}(F_{a,b})(A)$ on positive
semidefinite $A$ is weaker, because it tests $T_{F_{a,b}}$ only on pure
tensors.

\subsection{Kernels from the positive characters}
By Proposition~\ref{thm:kernel}, each positive character of this paper
defines a kernel that is positive semidefinite coefficient by
coefficient. In degree $m|\mu|$, the defects $\Ecal_{\mu,m}$ and
$\Dcal_{\mu,m}$ give such kernels in $q,t$, and the shifted quotient of
Theorem~\ref{thm:qdistorted} gives one in $q$ for all $a,b\ge2$ and
$c\ge1$. The same holds, in their stated parameter ranges, for the
reversed quotients of Theorem~\ref{thm:reversefusion}, the rectangular
quotients of Proposition~\ref{prop:rectbenchmark} and
Theorem~\ref{thm:rectlinear}, the quotient
$\mathcal J^{(m)}_{\lambda,\mu}$ of Theorem~\ref{thm:stretch}, and the
dominance differences of Corollary~\ref{cor:defectdom}.
All these kernels act on arbitrary tuples of vectors; consequences of
the unshifted results on restricted tuples are in the companion paper
\citep{foulkesIII}.

\subsection{Rectangular plethysms}\label{sec:rectangular}
Bergeron's rectangular conjecture replaces the inner one-row character
by a modified Macdonald polynomial. By Haiman's positivity theorem and
Proposition~\ref{thm:kernel}, the polarizations
$\mathfrak M_n(\Htil_\mu)(A;q,t)$ define positive semidefinite
cross-Gram kernels for $q,t\ge0$; for example,
\[
 \mathfrak M_3(\Htil_{(2,1)})(A;q,t)=
 \per A+(q+t)\Imm_{(2,1)}(A)+qt\det A.
\]

For $a\le c,d\le b$, $ab=cd$, and $k\ge1$,
\citet[Conjecture~4]{bergeron2017foulkes} asserts Schur positivity of
\eqref{eq:rectangular}, and by Proposition~\ref{thm:kernel} this is
equivalent to the positivity of the cross-Gram kernel of the difference
in \eqref{eq:rectangular}, with the parameters read in the same way.
\citet{foulkesIII} realizes these rectangular plethysms through wreath
products.

\end{document}