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  \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}%
  \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany.
    Email: \texttt{s.sra@tum.de}.}}

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\hypersetup{pdftitle={The q-Foulkes problem III: On Foulkes' conjecture with a graded inner factor},pdfauthor={Suvrit Sra}}
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\keywords{$q$-Foulkes conjecture, Schur positivity, Foulkes--Howe map,
coinvariant algebra, current operator, Gaussian Newton expansion}
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\begin{document}
\title[The $q$-Foulkes problem III]{The $q$-Foulkes problem III:\\On Foulkes' conjecture with a graded inner factor}
\aiauthor
\begin{abstract}
  We show that surjectivity of the canonical map $\Sym^b(\Sym^aV)\to\Sym^a(\Sym^bV)$ survives grading of
  the inner factor. Consequently $h_b[H_a]-H_a[h_b]$ is Schur positive whenever the canonical map is
  surjective, where $H_a$ is the (modified) Hall--Littlewood polynomial; the same holds for (modified)
  Macdonald polynomials. We prove this surjectivity for $b\ge(a-1)^2$, improving the known bound
  $a(a-1)$. For Bergeron's $q$-analog of Foulkes' conjecture that lets both factors be graded, we prove
  nonnegativity of every Schur coefficient for certain special cases. %
  We also study the full conjecture under an expansion hypothesis on an explicit space of maps.
\end{abstract}
\preprintslink{qfoulkes3}{29 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}. Recall that \citeauthor{foulkes1950}'s
conjecture~(\citeyear{foulkes1950}) states that the difference of
plethysms $h_b[h_a]-h_a[h_b]$ is Schur positive for $a\le b$, i.e., it
expands with nonnegative integer coefficients in the Schur functions
$s_\lambda$, $\lambda\vdash ab$; here $h_n$ is the complete homogeneous
symmetric function. 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 $H_n(X;1)=h_1^n$, both plethysms 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 the conjecture at $q=1$ and showed that Foulkes' inequality also orders the
coefficients of $q^1$ \citep[Section~3]{bergeron2017foulkes}, so the known cases control only the two
lowest powers of $q$.

Recall that $h_b[h_a]$ and $h_a[h_b]$ are characters of the
domain and range of the \emph{Foulkes--Howe map}
\begin{equation}\label{spp:map}
 \Sym^b(\Sym^a V)\longrightarrow\Sym^a(\Sym^b V),
\end{equation}
so Foulkes' conjecture follows whenever this map is surjective.

Our first main result shows that this approach survives when the
inner factor is graded. The polynomial $H_a$ is the graded Frobenius
character of the coinvariant algebra $R_a$ of $S_a$, so replacing
$\Sym^aW$ by $\mathcal H_a(W)=(R_a\otimes W^{\otimes a})^{S_a}$ grades
the inner factor. Multiplication then maps $\Sym^b(\mathcal H_a(V))$
to $\mathcal H_a(\Sym^bV)$, and in coinvariant degree zero this map is
\eqref{spp:map}.
\begin{dashedbox}
\begin{theorem}[Graded lift]\label{spp:currenttheorem}
Suppose the canonical map \eqref{spp:map} is surjective for every
finite-dimensional complex $V$, for a fixed pair $a,b\ge1$.
Then multiplication gives a natural degree-preserving surjection
\begin{equation}\label{spp:halflift}
 \SPSym^b\bigl(\mathcal H_a(V)\bigr)
       \longrightarrow \mathcal H_a\bigl(\SPSym^bV\bigr)
\end{equation}
for every $V$. Consequently
\begin{equation}\label{spp:halfpositive}
 h_b[H_a]-H_a[h_b]\in\mathbb N[q]\{s_\lambda\}.
\end{equation}
\end{theorem}
\end{dashedbox}
The proof (Section~\ref{spp:currentlift}) uses derivations that
multiply the $i$th tensor row by a power of the $i$th coinvariant
variable. They preserve the image of multiplication and carry degree
zero to every coinvariant degree, so surjectivity propagates from
degree zero. The argument needs only that $R_a$ is generated by
variables that $S_a$ permutes, so it applies equally to the
Garsia--Haiman modules, whose characters are the modified Macdonald
polynomials $\widetilde H_\mu$, $\mu\vdash a$. Under the same
hypothesis, therefore,
$h_b[\widetilde H_\mu]-\widetilde H_\mu[h_b]\in\N[q,t]\{s_\lambda\}$
(Corollary~\ref{spp:macdonaldlift}).

Theorem~\ref{spp:currenttheorem} assumes that \eqref{spp:map} is
surjective, which the annihilator argument of
\citet{openai2026quadratic} proves for $b\ge a(a-1)$. Our second main
result slightly improves their bound.
\begin{dashedbox}
\begin{theorem}[Surjectivity]\label{spp:main}
For every $a\ge2$, every integer $b\ge(a-1)^2$, and every
finite-dimensional complex vector space $V$, the map
\eqref{spp:map} is surjective. Consequently
$h_b[h_a]-h_a[h_b]$ is Schur positive in this range.
\end{theorem}
\end{dashedbox}
The annihilator argument shows that every symmetric form killing the
image vanishes, by polarizing one variable at a time. We polarize by
whole powers of an operator instead. A whole power stays injective in
a wider range of degrees than its factors checked one by one
(Lemma~\ref{spp:power}), and this lowers the bound from $a(a-1)$ to
$(a-1)^2$. Together with Theorem~\ref{spp:currenttheorem},
Theorem~\ref{spp:main} makes the graded lift unconditional, so \eqref{spp:halfpositive} holds for every
$b\ge(a-1)^2$ (Corollary~\ref{spp:halfbound}).

Our third main result concerns Bergeron's $F_{a,b}$ itself. The lift
does not extend to a graded outer factor (Section~\ref{cur:section}),
so we compute instead. The depth of $\lambda$ is $r=ab-\lambda_1$, the
number of boxes below its first row. The coefficients of depth $r$
depend on $b$ only through a polynomial of degree at most $r$ in $q^b$
(Theorem~\ref{thm:depth}). We expand this polynomial in Gaussian
binomial coefficients in $b$, which have nonnegative coefficients and
increase with $b$. This Gaussian Newton expansion therefore proves
positivity and monotonicity for every $b$ at once whenever its
coefficients are nonnegative.
\begin{dashedbox}
\begin{theorem}[Positivity at bounded depth]\label{nw:layers}
For every $b\ge a\ge4$, every Schur coefficient of $F_{a,b}$ of depth
at most six belongs to $\mathbb N[q]$. The same conclusion holds
through depth seven if $a\ge5$, through depth eight if $a\ge6$, and
through depth ten if $a=7$.
In each of these ranges, the Gaussian Newton coefficients based at
$b=a$ are nonnegative, and for each fixed partition below the first
row the coefficient increases coefficientwise with $b$.
\end{theorem}
\end{dashedbox}
The inner size $a$ is unbounded as well, and a second Gaussian Newton
expansion in $q^a$ reduces each range to finitely many identities
that are exact over $\Q(q)$ (Appendix~\ref{app:certificates}). For
each fixed partition below the first row, the monotonicity is
Bergeron's $q$-stability conjecture
\citep[Conjecture~2]{bergeron2017foulkes}. Corollary~\ref{cor:depth-three-stability} proves it for every
$a\ge2$ at depth three, where the three coefficients are multiples of
one explicit polynomial.

Finally, Section~\ref{cur:section} returns to maps. With both factors
graded,
the relations among the coefficients depend on how the
$ab$ positions are grouped into blocks, and extending the canonical
map polynomially does not respect them
(Proposition~\ref{gka:quadratic-descent}). Projecting onto harmonic
polynomials avoids this obstruction, giving a space of equivariant
maps that never lower degree. It remains open whether these maps
together expand every graded subspace; if they do, the conjecture
follows (Theorem~\ref{cur:criterion}).

\subsection{Conventions}\label{sec:conventions}
We work over $\C$. Plethysm acts on every grading variable through
the Adams operations $\psi_r(p_j)=p_{rj}$, $\psi_r(q)=q^r$, and
$\psi_r(t)=t^r$. Set
\[
 [n]_q=\frac{1-q^n}{1-q},\qquad
 \qbin nk=\frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}
 \quad(0\le k\le n),
\]
with $\qbin nk=0$ for $k>n$. Unless stated otherwise,
$a,b\ge2$ are integers and $N=ab$. For $\mu=1^{m_1}2^{m_2}\cdots$, put
$z_\mu=\nlprod_r r^{m_r}m_r!$. For $\lambda\vdash N$, $S^\lambda$ is the
irreducible $S_N$-module, $\chi^\lambda$ its character, and
$d_\lambda=\dim S^\lambda$; for a partition $\mu$, put
$n(\mu)=\nlsum_i(i-1)\mu_i$. For a semiring $K$,
$K\{s_\lambda\}$ is the set of symmetric functions whose Schur
coefficients lie in $K$.

\section{The canonical map and its graded lift}\label{spp:canonical}
Theorem~\ref{spp:main} supplies the surjectivity that
Theorem~\ref{spp:currenttheorem} assumes, so we prove it first. Both
proofs view the canonical map \eqref{spp:map} as arranging the tensor
positions in a rectangle and averaging over the symmetry groups of its
codomain, where we identify a symmetric power with invariant tensors by
\[
 \operatorname{Av}_a(w_1\otimes\cdots\otimes w_a)
 =\frac1{a!}\nlsum_{\sigma\in S_a}
 w_{\sigma(1)}\otimes\cdots\otimes w_{\sigma(a)}.
\]
In this picture the image of \eqref{spp:map} is the degree-$b$ part
of the subalgebra of $((\SPSym V)^{\otimes a})^{S_a}$ generated by
$\SPSym^aV$.

\subsection{Whole powers of polarization}
The proof of Theorem~\ref{spp:main} keeps the annihilator construction
of \citet{openai2026quadratic} and replaces its successive first
derivatives by whole powers of polarization operators. The injectivity
of these powers is the $\mathfrak{sl}_2$ argument in the proof of
\citet[Lemma~2.5]{OpenAI}; the following lemma records it with a norm
bound.

For independent vector variables $U=(U_i)$ and $Y=(Y_i)$, put
\[
 D=D_{Y,U}:=\nlsum_iY_i\frac{\partial}{\partial U_i},\qquad
 E=\nlsum_iU_i\frac{\partial}{\partial Y_i},\qquad
 H=N_U-N_Y,
\]
where $N_U$ and $N_Y$ count degrees in $U$ and $Y$.
We use the Fischer inner product, for which monomials are orthogonal
and $\|U^\alpha Y^\beta\|^2=\alpha!\,\beta!$, also for spectator
variables. Then $D^*=E$, $[E,D]=H$, and $[H,D]=-2D$, so each
application of $D$ lowers by two the weight $H$, which is the excess
of the degree in $U$ over the degree in $Y$. When the variables have at least
two coordinates, $D$ alone is injective on a space of fixed bidegree
only if the weight is positive. If we checked
$D^L$ one factor at a time, we would therefore need weight at least
$2L-1$, whereas on whole $\mathfrak{sl}_2$ strings weight $L$
suffices. Write $\SPfall{x}{L}=x(x-1)\cdots(x-L+1)$, with empty
product one.

\begin{lemma}[Power polarization]\label{spp:power}
Let $f$ have bidegree $(p,s)$ in $(U,Y)$, with arbitrary spectator
variables. If $L\ge0$ and $m=p-s\ge L$, then
\begin{equation}\label{spp:powernorm}
 \|D^L f\|^2\ge L!\,\SPfall{m}{L}\,\|f\|^2.
\end{equation}
In particular, $D^L$ is injective on this bidegree space, including
the boundary $p-s=L$.
\end{lemma}

\begin{proof}
Decompose each space of polynomials of fixed total degree into
orthogonal $\mathfrak{sl}_2$ strings. In a string of highest weight $\ell$, a
vector of weight $m$ is at position $j=(\ell-m)/2\ge0$. The squared
norm multiplier for $D^L$ is
\[
 \nlprod_{u=0}^{L-1}(j+u+1)(\ell-j-u)
 =\nlprod_{u=0}^{L-1}(j+u+1)(m+j-u).
\]
Since $m\ge L$, this multiplier is at least
$\nlprod_{u=0}^{L-1}(u+1)(m-u)=L!\,\SPfall{m}{L}>0$ when $L>0$.
For $L=0$ the assertion is the identity. Orthogonality gives the
estimate for an arbitrary sum of strings. Spectator variables do not
alter the calculation.
\end{proof}

For $L>0$, the threshold in this lemma is sharp on the full bidegree
space when $\dim V\ge2$. If $0\le p-s<L$, then $D^L$ kills the nonzero
polynomial
\[
 (U_1Y_2-U_2Y_1)^s U_1^{p-s}.
\]
If $p<s$, then $D$ already kills $(U_1Y_2-U_2Y_1)^pY_1^{s-p}$.

\subsection{The annihilator argument}
Linear functionals on the codomain of \eqref{spp:map} correspond to
symmetric forms on $\SPSym^bV$. A functional vanishes on the image
exactly when its form vanishes on the diagonal of every product of $b$
vectors, so surjectivity means that every such form is zero. Polarization carries the vanishing from these special arguments
to arbitrary ones, one argument at a time. The obstacle is a degree
count. Polarization is injective only while the differentiated
variable has larger degree than the inserted one, and the decisive
step uses up this excess, polarizing $r=a-1$ arguments by blocks of
$r$ derivatives each. A block applied as a whole power needs an excess
of only $r$, rather than $2r-1$, so Lemma~\ref{spp:power} applies to
every block exactly when $b\ge r^2$.

\begin{proof}[Proof of Theorem~\ref{spp:main}]
The case $\dim V=0$ is immediate. We induct on $a$, allowing $a=1$
and every $b\ge0$ as the identity base case. The case $(a,b)=(2,1)$
is also an identity. Otherwise put
\[
 r=a-1,\qquad m=b-r.
\]
Thus $m\ge1$ and $b\ge r^2$.

\emph{\textbf{Forms vanishing on products.}}
Pure powers $v^a$ span $\SPSym^a V$. Therefore the image of
\eqref{spp:map} is spanned by $P^{\otimes a}$, where
$P=u_1\cdots u_b$ is a product of $b$ vectors: averaging sends $u^a$
to $u^{\otimes a}$, and multiplication sends
$u_1^{\otimes a}\cdots u_b^{\otimes a}$ to $P^{\otimes a}$.
A functional $\ell$ on the codomain corresponds to the symmetric form
$T(w_1,\ldots,w_a)=\ell(\operatorname{Av}_a(w_1\otimes\cdots\otimes w_a))$.
Thus the annihilator of the image consists exactly of the forms satisfying
\begin{equation}\label{spp:annihilator}
 T(P,\ldots,P)=0
 \quad\text{for every such product }P.
\end{equation}
We prove that every such $T$ is zero.

\emph{\textbf{Freeing the last argument.}}
To isolate one argument of $T$, we split a product into $r$ equal
factors and one different one. Fix $Q=y_1\cdots y_m$. The binary form $x^r+c t^r$ factors into $r$
linear factors over $\mathbb C$, so $(x^r+c t^r)Q$ is a product of
$b$ vectors. Taking the coefficient of $c$ in
\eqref{spp:annihilator} yields
\begin{equation*}
 T(x^rQ,\ldots,x^rQ,t^rQ)=0,
\end{equation*}
with $r$ copies of $x^rQ$; the coefficient is $a$ times the displayed
expression, and $a\ne0$ over $\mathbb C$.

To replace $t^rQ$ by $t^b$ in the last argument, define
\[
 F_x(y_1,\ldots,y_m,t)=T(x^rQ,\ldots,x^rQ,t^b).
\]
It has degree $b$ in $t$ and degree $r$ in each $y_i$. Moreover,
\[
 D_{y_m,t}\cdots D_{y_1,t}F_x
   =\SPfall{b}{m}\,T(x^rQ,\ldots,x^rQ,t^rQ)=0.
\]
Before applying the $i$th operator, the relevant degrees are
$p=b-i+1$ and $s=r$. Hence $p-s=m-i+1\ge1$.
Lemma~\ref{spp:power} with $L=1$ makes every step injective, so
\begin{equation*}
 T(x^rQ,\ldots,x^rQ,t^b)=0
 \quad\text{for all products }Q=y_1\cdots y_m.
\end{equation*}

\emph{\textbf{The other arguments, by induction.}}
For fixed $x,t$, the left side is the diagonal of a symmetric
$r$-linear form on $\SPSym^m V$. The induction hypothesis applies,
because
\[
 m=b-r\ge r^2-r\ge(r-1)^2.
\]
For $r=1$ the induction uses only the identity base case. It follows that
\begin{equation}\label{spp:induction}
 T(x^rQ_1,\ldots,x^rQ_r,t^b)=0
 \quad\text{for all }Q_i\in\SPSym^m V.
\end{equation}

\emph{\textbf{Whole powers of polarization.}}
It remains to replace each $x^rQ_i$ by an arbitrary pure power. Set
$f=T(v_1^b,\ldots,v_r^b,t^b)$, and introduce an
independent variable $x$ of initial degree zero. Apply, in order, the
$r$ operators $(D_{x,v_i})^r$. Before the $i$th block of $r$
derivatives, $v_i$ has degree $b$ and $x$ has degree $(i-1)r$.
Consequently
\[
 p-s=b-(i-1)r\ge r,
\]
so every whole block is injective by Lemma~\ref{spp:power}.
Their composition is
\[
 \left(\SPfall{b}{r}\right)^r
 T(x^rv_1^m,\ldots,x^rv_r^m,t^b)=0
\]
by \eqref{spp:induction}. Thus $f=0$. Pure powers span each copy of
$\SPSym^b V$, so $T=0$, and the map is surjective. Complete
reducibility gives the stated Schur positivity.
\end{proof}

\subsection{The graded lift}\label{spp:currentlift}
Theorem~\ref{spp:currenttheorem} transfers surjectivity from
coinvariant degree zero to every degree. Write the coinvariant algebra
as $R_a=\mathbb C[\xi_1,\ldots,\xi_a]/(e_1(\xi),\ldots,e_a(\xi))$, so
that $\mathcal H_a(W)=(R_a\otimes W^{\otimes a})^{S_a}$. The
difficulty is that the hypothesis says nothing in positive degree, and
a single variable $\xi_j$ cannot be multiplied in, because it is not
invariant under $S_a$. The transfer uses derivations that pair $\xi_i$
with the $i$th row instead. They commute with $S_a$, preserve the
image of multiplication, and act on the coefficient of a suitable
vector as multiplication by $\xi_j$.

\begin{proof}[Proof of Theorem~\ref{spp:currenttheorem}]
\emph{\textbf{One algebra for both sides.}}
Form the Segre algebra
\[
 C(V)=\bigoplus_{d\ge0}(\SPSym^d V)^{\otimes a},
 \qquad B(V)=(R_a\otimes C(V))^{S_a},
\]
with componentwise multiplication and simultaneous permutation of
the $a$ rows and coefficient variables. Write $B_d$ for Segre degree
$d$. Thus $B_1=\mathcal H_a(V)$ and
$B_b=\mathcal H_a(\SPSym^bV)$. Let $A\subseteq B$ be the subalgebra
generated by $B_1$. Multiplication gives \eqref{spp:halflift}, with
image $A_b$. In coinvariant degree zero it is \eqref{spp:map}.

\emph{\textbf{Current operators.}}
Choose basis variables $x_{i,u}$ for the $i$th factor of $\SPSym V$.
For basis labels $u,v$ and $r\ge0$, define a derivation
\begin{equation*}
 \mathcal D_{uv,r}=\nlsum_{i=1}^a
       \xi_i^r x_{i,u}\frac{\partial}{\partial x_{i,v}}.
\end{equation*}
It descends modulo the coinvariant ideal, since it differentiates only
the $x$ variables. It preserves the degree in each row and commutes
with the simultaneous $S_a$ action. Hence it acts on every $B_d$,
preserves $B_1$, and raises the coinvariant degree by $r$. The Leibniz
rule shows that it preserves $A$ and every $A_d$. We call these
derivations current operators, because
$E_{uv}\otimes\xi^r\mapsto\mathcal D_{uv,r}$ is a representation of
the current algebra $\mathfrak{gl}(V)\otimes\mathbb C[\xi]$.

\emph{\textbf{Reduction to one weight space.}}
A diagonal current singles out one row only when the label it acts on
occurs once, so we reduce to the multilinear weight. For each fixed
coinvariant degree, the multiplication map is a
natural transformation of homogeneous polynomial functors of degree
$N=ab$. Its cokernel is therefore a direct sum of Schur functors
$S_\lambda$, with $\lambda\vdash N$, tensored with fixed multiplicity
spaces. Each $S_\lambda(\mathbb C^N)$ has a nonzero multilinear weight
space, isomorphic to the Specht module $S^\lambda$, and taking a
weight space is exact. It therefore suffices to show that the
cokernel vanishes on the multilinear weight space $(1,\ldots,1)$ of
$V=\mathbb C^N$; the cokernel functor then vanishes in every
dimension, both below and above $N$.

\emph{\textbf{Orbit vectors.}}
At this weight, the monomials of $C_b$ are indexed by ordered
partitions of the $N$ basis labels into $a$ blocks of size $b$.
Order each unordered partition $P$ as $(P_1,\ldots,P_a)$, and
put
\[
 m_P=\nlprod_{i=1}^a\nlprod_{\ell\in P_i}x_{i,\ell},\qquad
 w_{P,f}=\nlsum_{\sigma\in S_a}\sigma(f\otimes m_P),
 \quad f\in R_a.
\]
Since $b\ge1$, the orbit of the ordered monomial is free. Its
invariant coefficient space is therefore a copy of $R_a$: the
coefficient of $m_P$ in $w_{P,f}$ recovers $f$. The vector $w_{P,1}$
lies in $A_b$, because its coinvariant degree is zero, and in degree
zero $B_1=(V^{\otimes a})^{S_a}=\Sym^a V$ and multiplication is the
canonical map, which is surjective by hypothesis.

\emph{\textbf{Currents generate the coefficients.}}
Choose one label $\ell_j\in P_j$ for each block. In each orbit
summand, this label occurs exactly once, in the row carrying the
$j$th block. Consequently
\begin{equation*}
 \mathcal D_{\ell_j\ell_j,1}w_{P,f}=w_{P,\xi_j f},
\end{equation*}
so the diagonal current $\mathcal D_{\ell_j\ell_j,1}$ multiplies the
coefficient by $\xi_j$. Iterating these commuting currents generates
$w_{P,\xi_1^{n_1}\cdots\xi_a^{n_a}}$ from $w_{P,1}$, and the
monomials span $R_a$. Since $A_b$ is stable under the currents, it
contains the entire coefficient space over $P$. Summing over $P$
proves surjectivity in every coinvariant degree on the multilinear
weight space.

Finally the domain and codomain characters are $h_b[H_a]$ and
$H_a[h_b]$. Their difference is the graded character of the kernel
of \eqref{spp:halflift}, proving \eqref{spp:halfpositive}.
\end{proof}

Theorem~\ref{spp:main} removes the hypothesis of
Theorem~\ref{spp:currenttheorem} in its range.

\begin{corollary}\label{spp:halfbound}
For every $a\ge2$ and $b\ge(a-1)^2$,
\[
 h_b[H_a]-H_a[h_b]\in\mathbb N[q]\{s_\lambda\}.
\]
\end{corollary}

\begin{proof}
Combine Theorems~\ref{spp:main} and \ref{spp:currenttheorem}.
\end{proof}

The proof of Theorem~\ref{spp:currenttheorem} used only that $R_a$ is
generated by variables of one degree that $S_a$ permutes. Quotients
generated by several such lists give the same lift, and this brings
the modified Macdonald polynomials within reach.

\begin{proposition}[Other coefficient algebras]\label{spp:coefficientextension}
Let $C$ be a finite-dimensional nonnegatively multigraded quotient of
$\mathbb C[\xi_{i,r}:1\le i\le a,\ 1\le r\le d]$ by a homogeneous
$S_a$-stable ideal, where the generators in each list have a common
multidegree, $S_a$ permutes $i$, and $C_0=\mathbb C$.
Put $\mathcal C(W)=(C\otimes W^{\otimes a})^{S_a}$.
If the canonical map \eqref{spp:map} is surjective for every
finite-dimensional $V$, then
multiplication induces a natural multigrading-preserving surjection
\[
 \Sym^b(\mathcal C(V))\longrightarrow\mathcal C(\Sym^bV).
\]
\end{proposition}
\begin{proof}
Use $C$ in place of $R_a$ in the proof of
Theorem~\ref{spp:currenttheorem}. For every generator list, the current
$\nlsum_i\xi_{i,r}x_{i,u}\partial_{x_{i,v}}$ descends to the quotient,
commutes with $S_a$, and preserves the subalgebra generated in Segre
degree one. On the free orbit over an unordered block partition, its
diagonal version multiplies the coefficient by the corresponding
$\xi_{j,r}$. Monomials in these generators span every homogeneous
component of $C$. The degree-zero orbit vector therefore generates its
entire coefficient space inside the image. The reduction to one weight
space then gives the asserted surjection in every dimension.
\end{proof}

The Garsia--Haiman modules are quotients of this kind, and Haiman's
$n!$ theorem identifies their characters with the modified Macdonald
polynomials, which we normalize as
\[
 \widetilde H_\mu[X;q,t]
 =t^{n(\mu)}J_\mu\left[\frac{X}{1-t^{-1}};q,t^{-1}\right],
\]
where $J_\mu$ is the integral Macdonald form. This normalization is
the one of \citet{bergeron1999identities} and
\citet{haiman2001hilbert}; see also
\citet[Chapter~VI, (8.1') and (8.3)]{macdonald1995}. In particular
$\widetilde H_{(n)}(X;q,t)=H_n(X;q)$.

\begin{samepage}
\begin{corollary}[A bigraded Macdonald lift]\label{spp:macdonaldlift}
Let $\mu\vdash a$, $a\ge2$, and $b\ge(a-1)^2$. There is a natural
bigrading-preserving surjection between polynomial functors with
characters $h_b[\widetilde H_\mu]$ and $\widetilde H_\mu[h_b]$.
Consequently
\[
 h_b[\widetilde H_\mu(X;q,t)]
       -\widetilde H_\mu[h_b(X);q,t]
 \in\mathbb N[q,t]\{s_\lambda\}.
\]
Both grading variables undergo Adams substitution in the outer
plethysm, and $h_b$ is undeformed.
\end{corollary}
\end{samepage}
\begin{proof}
For the cells $(u_j,v_j)$ of $\mu$, with zero-based column and row
coordinates, let
$\Delta_\mu=\det(x_i^{u_j}y_i^{v_j})_{1\le i,j\le a}$ and use the
Garsia--Haiman quotient
\[
 C_\mu=\mathbb C[x_1,\ldots,x_a,y_1,\ldots,y_a]/I_\mu,
 \quad
 I_\mu=\{f:f(\partial_x,\partial_y)\Delta_\mu=0\}.
\]
Assign $\deg x_i=(1,0)$ and $\deg y_i=(0,1)$, recorded by $q$ and $t$,
respectively. The ideal is bigraded and $S_a$-stable, and it contains
sufficiently high powers of every variable, so $C_\mu$ is finite
dimensional. Since $\Delta_\mu\ne0$, its degree-zero part is $\mathbb C$.
With these gradings, Haiman's $n!$ theorem identifies its Frobenius character
with $\widetilde H_\mu(X;q,t)$
\citep[Section~3.9, equations~(55)--(56), and Theorem~3.2]{haiman2001hilbert}.
Haiman assigns $x$ to row coordinates and $y$ to column coordinates;
our interchange of these two lists makes $q$ record column degree.
The quotient is generated by the two equivariant lists $x_i,y_i$.
Apply Proposition~\ref{spp:coefficientextension} and
Theorem~\ref{spp:main}. Complete reducibility in each bidegree makes
the character of the kernel coefficientwise Schur positive.
\end{proof}

\section{Coefficients of bounded depth}\label{app:depth}
Theorem~\ref{nw:layers} asserts positivity for every $b$ at once. A
finite computation can prove it because, after restriction to a Young
subgroup, the coefficients of depth $r$ depend on $b$ only through
polynomials in $q^b$ of degree at most $r$.
\subsection{Restriction to a Young subgroup}\label{sec:depth}
Only the shapes of depth at most $r$ survive restriction to
$S_{N-r}\times S_r$ followed by invariants of the first factor, each
through a symmetric function of degree $r$. The size of the problem is then the
depth rather than $N=ab$.

We restrict the module whose character is $P_{a,b}=H_b[H_a]$. Partition
$[N]$ into $b$ blocks of size $a$, and let $W_{a,b}=S_a\wr S_b$ act on
$R_a^{\otimes b}\otimes R_b$, with $S_a^b$ acting factorwise and $S_b$
permuting the factors and acting on $R_b$. By graded wreath induction,
including the Adams operation on $q$
\citep[Lemma~2.1]{foulkesII}, the induced module and
its graded character are
\begin{equation}\label{eq:III-modules}
 M_{a,b}=\Ind_{W_{a,b}}^{S_N}(R_a^{\otimes b}\otimes R_b),
 \qquad \ch_q M_{a,b}=P_{a,b}.
\end{equation}
Put
\[
 \mathcal R_{a,b}^{(r)}(Y;q)=h_{N-r}^{\perp}P_{a,b}(Y;q).
\]
Here $h_k^\perp$ is adjoint to multiplication by $h_k$ under the Hall
scalar product. Representation-theoretically, $\mathcal R^{(r)}$ is
the $S_r$-character of the $S_{N-r}$-invariants in $M_{a,b}$.
The graded trace of the coinvariant algebra $R_n$ at a permutation of
cycle type $\mu$ is
\begin{equation}\label{eq:III-kappa}
 \kappa_n(\mu;q)=\frac{(q;q)_n}{\nlprod_{r\in\mu}(1-q^r)},
\end{equation}
and we write
\[
 \begin{bmatrix}b\\n_0,n_1,\ldots\end{bmatrix}_q
 =\frac{(q;q)_b}{\nlprod_j(q;q)_{n_j}}.
\]

\begin{theorem}[Fixed-depth formula]\label{thm:depth}
For $0\le r\le N$,
\begin{equation}\label{eq:depth}
 \mathcal R_{a,b}^{(r)}(Y;q)=
 \sum_{\substack{n_0+\cdots+n_a=b\\\nlsum_j jn_j=r}}
 \begin{bmatrix}b\\n_0,\ldots,n_a\end{bmatrix}_q
 \nlprod_{j=1}^{a}
 H_{n_j}\left[\qbin{a}{j}H_j(Y;q)\right].
\end{equation}
All plethysms on the right include the Adams action on $q$. The term
with $n_j=0$ is interpreted as $1$.
\end{theorem}
\begin{proof}
The proof is a Mackey decomposition, and the grading enters only
through the restriction of $R_a$ to a Young subgroup, which
\eqref{eq:III-kappa} settles. The double cosets for restriction to
$S_{N-r}\times S_r$ are indexed by
how the $r$ selected labels occupy the $b$ blocks: $n_j$ blocks contain
exactly $j$ selected labels. On a block of size $a$, the graded character of the coinvariant
algebra $R_a$ restricts to $S_j\times S_{a-j}$ as
\[
 \Res R_a=\qbin{a}{j}(R_j\boxtimes R_{a-j}).
\]
This identity follows from
\eqref{eq:III-kappa}, because the quotient of the trace factors is
$(q;q)_a/((q;q)_j(q;q)_{a-j})$. Taking the invariants on the unselected
labels leaves a graded multiplicity space of Hilbert series
$\qbin{a}{j}$ tensored with $R_j$.

Similarly, restricting the outer $R_b$ to
$\nlprod_jS_{n_j}$ produces the Gaussian multinomial factor and the
outer modules $R_{n_j}$. Blocks with equal occupancy are permuted, so
wreath induction gives
$H_{n_j}[\qbin{a}{j}H_j]$. The $j=0$ factor has only its degree-zero
invariant. Inducing the selected labels to $S_r$ multiplies the
Frobenius characters. Mackey decomposition over the occupancy
orbits gives \eqref{eq:depth}.
\end{proof}

The formula stays small when the depth is small. Only partitions of
$r$ occur in the occupancy sum, every symmetric function on the right
has degree at most $r$, and $a$ and $b$ enter only through Gaussian
coefficients. At $q=0$, \eqref{eq:depth} is the
decomposition of the restricted Foulkes module by
\citet[Proposition~2.11]{giannelli2013foulkes}.

The restriction $\mathcal R^{(r)}$ mixes the layers of depth at most
$r$, where the layer of depth $d$ collects the Schur coefficients of
the shapes $(N-d,\nu)$; the following triangular relation separates
them.

\begin{prop}[Triangular recovery of the layers]\label{prop:triangular}
For $N\ge2r$, let
\[
 C_d(Y;q)=\nlsum_{\nu\vdash d}
 [s_{(N-d,\nu)}]P_{a,b}\;s_\nu(Y),\qquad 0\le d\le r.
\]
Then the restrictions and the layers are related by
\[
 \mathcal R_{a,b}^{(r)}=\nlsum_{d=0}^r h_{r-d}C_d,
 \qquad
 C_r=\nlsum_{j=0}^r(-1)^j e_j\mathcal R_{a,b}^{(r-j)}.
\]
The same identities hold for differences and for division by $1-q$.
\end{prop}
\begin{proof}
By the Pieri rule, $s_\nu$ occurs in
$h_{N-r}^\perp s_\lambda$ precisely when $\lambda/\nu$ is a horizontal
strip of size $N-r$. Such a $\lambda$ has $\lambda_1\ge N-r$.
Write $\lambda=(N-d,\tau)$ with $d\le r$. Since $N\ge2r$, the
horizontal-strip condition is equivalent to $\nu/\tau$ being a
horizontal strip of size $r-d$. This equivalence proves the first identity.
The second uses
$(\nlsum_{j\ge0}h_jz^j)^{-1}=\nlsum_{j\ge0}(-1)^je_jz^j$.
\end{proof}

\subsection{Gaussian Newton expansions}\label{nw:section}
Theorem~\ref{thm:depth} bounds how the coefficients of depth $r$
depend on $b$. We now expand them in a basis that is itself positive
and increasing in $b$, so that positive expansion coefficients give
positivity for every $b$.

For integers $a\ge1$ and $r\ge0$, choose a baseline $B\ge a$ with
$aB\ge2r$. Writing a shape of depth $r$ as $(ab-r,\nu)$, we call
$\nu\vdash r$ its tail and collect the coefficients of depth $r$ in
\[
 \mathcal G_{a,r}(b;Y)=
 \nlsum_{\nu\vdash r}[s_{(ab-r,\nu)}]F_{a,b}\,s_\nu(Y),\qquad b\ge B.
\]
The stable-range condition $aB\ge2r$ ensures that every displayed shape is a
partition. Theorem~\ref{thm:depth} and Proposition~\ref{prop:triangular}
make $\mathcal G_{a,r}$ a polynomial of degree at most $r$ in $z=q^b$,
by counting occupied blocks. A term of $\mathcal R^{(r)}_{a,b}$ with
$m$ occupied blocks contains $\qbin bm$ and has $z$-degree $m\le r$.
In $\mathcal R^{(r)}_{b,a}$, an inner term of degree $j$ contains
$\qbin bj$, of $z$-degree $j$. An Adams operation of order $t$ replaces
$(q,z)$ by $(q^t,z^t)$, so a total occupied degree $r$ still gives
$z$-degree at most $r$, and the lower restrictions in
Proposition~\ref{prop:triangular} have smaller degree. Hence
$\mathcal G_{a,r}$ has a unique expansion
\begin{equation}\label{nw:expansion}
 \mathcal G_{a,r}(b;Y)
 =\nlsum_{j=0}^r A^{(B)}_{a,r,j}(Y;q)\qbin{b-B}{j}.
\end{equation}
Evaluation at $b=B,B+1,\ldots,B+r$ gives the triangular system
\begin{equation}\label{nw:inversion}
 A^{(B)}_{a,r,j}
 =\mathcal G_{a,r}(B+j;Y)
  -\nlsum_{i<j}\qbin ji A^{(B)}_{a,r,i},
\end{equation}
so the coefficients are integer polynomials in $q$.

Through depth three the expansion is explicit. For every $2\le a<b$,
the coefficients of depth at most two vanish, and the three of depth
three are $q$, $1+q^2$, and $q$ times one polynomial with nonnegative
coefficients (Theorem~\ref{thm:main}). Beyond depth three,
Theorem~\ref{nw:layers} needs positivity for unbounded $a$ as well as
$b$. A second Gaussian Newton expansion in the inner parameter reduces
each of its ranges to finitely many exact coefficient arrays
(Appendix~\ref{app:newton-double}), which are positive;
Appendix~\ref{app:layers} reconstructs them and proves
Theorem~\ref{nw:layers}.

A positive Gaussian expansion also proves monotonicity, because
\begin{equation}\label{nw:monotonicity}
 \qbin{n+1}j-\qbin nj=q^{n+1-j}\qbin n{j-1}
 \qquad(1\le j\le n+1),
\end{equation}
and terms with $j>n+1$ vanish. Every positive Newton expansion is
therefore coefficientwise increasing in $n$, which gives the
monotonicity assertion of Theorem~\ref{nw:layers}.

\section{Grading both factors}
\label{cur:section}
With both factors graded, positivity becomes a comparison of degrees.
The modules $\mathcal P=M_{a,b}$ and $\mathcal Q=M_{b,a}$ of
\eqref{eq:III-modules} have characters $P_{a,b}=H_b[H_a]$
and $Q_{a,b}=H_a[H_b]$, and both are ungraded copies of
$\mathbb C[S_N]$. Each irreducible $S^\lambda$ therefore occurs equally
often in the two modules, and $[s_\lambda q^k]F_{a,b}$ counts how many
more of its copies $\mathcal P$ places in degrees at most $k$. In the
language of Paper~II \citep{foulkesII}, the conjecture says that on
every isotype the degree distribution of $\mathcal P$ is
stochastically smaller than that of $\mathcal Q$. It would
suffice to have an injective equivariant map from $\mathcal P$ to
$\mathcal Q$ that never lowers degree, because such a map embeds the
copies above each degree $k$ in $\mathcal P$ into those above $k$ in
$\mathcal Q$.

The lift of Theorem~\ref{spp:currenttheorem} does not provide such a
map. It uses one coefficient algebra for every position, whereas here
the relations among the coefficients depend on how the positions are
grouped into blocks, and extending the canonical map polynomially
does not respect them (Proposition~\ref{gka:quadratic-descent}).
Projecting onto harmonic representatives repairs the relations and
gives an explicit space of maps that never lower degree, built in
Appendix~\ref{gka:free-currents}. No single map in this space needs to
be injective; it suffices that the maps together expand every graded
subspace.

Fix $b\ge a\ge2$ and $N=ab$, and put
\[
 U=\mathcal P_0=\Ind_{S_a\wr S_b}^{S_N}\mathbf1,\qquad
 D_P=b\binom a2+\binom b2,\quad
 D_Q=a\binom b2+\binom a2.
\]
The integers $D_P,D_Q$ are the top degrees of $\mathcal P$ and
$\mathcal Q$. The maps run from $\mathcal P$ to $\mathcal Q$, and their
adjoints run in the direction of the filtered matchings of Paper~II
\citep[Proposition~2.3]{foulkesII}. We realize
$\mathcal P$ and $\mathcal Q$ as spaces of harmonic polynomials with
coefficients in permutation modules
(Appendix~\ref{gka:free-currents}), write $P_{\mathcal Q}$ for the
orthogonal projection onto $\mathcal Q$, and use the operators
$A_k^U(d)$ of Proposition~\ref{gka:free-factor}, which are positive
and monic of degree $k$ in a parameter $d$. These modules and maps
have rational forms; subspaces below are taken after extending scalars
to $\mathbb C$.

\subsection{The current space}
We call an equivariant map from $U=\mathcal P_0$ into $\mathcal Q_r$
a seed. Appendix~\ref{gka:free-currents} extends each seed to a map
$T_d$ on all of $\mathcal P$ that never lowers degree and depends
polynomially on the parameter $d$. The parameter is auxiliary, so we
keep every coefficient of this dependence.
Put $\overline A_k^U(\varepsilon)=\varepsilon^k A_k^U(1/\varepsilon)$.
This reversed factor is a polynomial of degree at most $k$ in
$\varepsilon$, including at $\varepsilon=0$, because $A_k^U(d)$ is
monic of degree $k$ in $d$ (Proposition~\ref{gka:free-factor}).
For an equivariant homogeneous seed
$\sigma_r\colon U\to\mathcal Q_r$, let
$\widehat\sigma_r(f\otimes u)=f\,\sigma_r(u)$. On the harmonic modules,
define the maps $\mathcal A_j(\sigma_r)$ by their degree blocks
\begin{equation}\label{cur:coefficients}
 \mathcal A_j(\sigma_r)|_{\mathcal P_k}
   =P_{\mathcal Q,k+r}\widehat\sigma_r
      [\varepsilon^j]\overline A_k^U(\varepsilon)|_{\mathcal P_k}.
\end{equation}
The block is zero when $j>k$ or $k+r>D_Q$.
By \eqref{gka:harmonic-block}, these maps are the coefficients of
$T_d$ after removing the factor $(d/N)^k$ and setting
$\varepsilon=1/d$. Each map is equivariant and has degree shift
$r\ge0$. The operators $A_k^U$ and the projections in
\eqref{cur:coefficients} are evaluated before selecting an isotypic
component.

Taking all rational equivariant seeds gives the current space
\begin{equation*}
 \mathscr M_{a,b}=\Span_{\mathbb Q}
 \bigl\{\mathcal A_j(\sigma_r):
   \sigma_r\in\Hom_{\mathbb Q[S_N]}(U,\mathcal Q_r),\
   0\le r\le D_Q,\ 0\le j\le D_P\bigr\}.
\end{equation*}
The module $\mathcal Q$ is an ungraded regular module, so
$\dim\Hom_{\mathbb Q[S_N]}(U,\mathcal Q)=\dim U$.
Linearity in the seed therefore gives
\begin{equation*}
 \dim\mathscr M_{a,b}\le(D_P+1)\dim U
       =(D_P+1)\frac{(ab)!}{(a!)^b b!}.
\end{equation*}
In particular, bases of the finite seed Hom spaces give a homogeneous
spanning family for $\mathscr M_{a,b}$, so for each pair $a,b$ the
condition below is a finite problem in linear algebra.

Write the isotypic decompositions as
\[
 \mathcal P=\bigoplus_{\lambda\vdash N}
             S^\lambda\otimes\mathcal E_\lambda^P,\qquad
 \mathcal Q=\bigoplus_{\lambda\vdash N}
             S^\lambda\otimes\mathcal E_\lambda^Q.
\]
Both multiplicity spaces have total dimension $d_\lambda$. By Schur's
lemma, each map in $\mathscr M_{a,b}$ acts on multiplicity spaces; let
$\mathscr M_\lambda$ be the resulting space of complex matrices from
$\mathcal E_\lambda^P$ to $\mathcal E_\lambda^Q$, and define
\[
 \mathscr M_\lambda(W)
       =\Span\{Mv:M\in\mathscr M_\lambda,\ v\in W\}.
\]
Call $\lambda$ \emph{neutral} when
$\dim\mathcal E_{\lambda,k}^P=\dim\mathcal E_{\lambda,k}^Q$
for every degree $k$. Such an isotype contributes zero to $F_{a,b}$.

\subsection{A sufficient condition}
We ask that the matrices in $\mathscr M_\lambda$ together expand every
graded subspace of $\mathcal E_\lambda^P$,
\begin{equation}\label{cur:expansion}
 \dim\mathscr M_\lambda(W)\ge\dim W
 \quad\text{for every nonneutral $\lambda$ and every graded }
 W\subseteq\mathcal E_\lambda^P.
\end{equation}
Here graded means $W=\bigoplus_k(W\cap\mathcal E_{\lambda,k}^P)$.
A subspace violating this inequality is called \emph{shrunk}. On the
degree tails of $\mathcal E_\lambda^P$, hypothesis
\eqref{cur:expansion} gives the inequalities that positivity of
$F_{a,b}$ requires.

\begin{theorem}[A sufficient condition]\label{cur:criterion}
For fixed $b\ge a\ge2$, hypothesis~\eqref{cur:expansion} implies
\[
 F_{a,b}(X;q)=\frac{H_b[H_a]-H_a[H_b]}{1-q}
       \in\mathbb N[q]\{s_\lambda\}.
\]
\end{theorem}
\begin{proof}
Fix a nonneutral $\lambda$ and an integer $k\ge0$. Every generator
of $\mathscr M_\lambda$ has nonnegative degree shift. For the graded
tail $W=\mathcal E_{\lambda,\ge k+1}^P$, hypothesis
\eqref{cur:expansion} therefore gives
\[
 \dim\mathcal E_{\lambda,\ge k+1}^P
 \le\dim\mathscr M_\lambda(W)
 \le\dim\mathcal E_{\lambda,\ge k+1}^Q.
\]
Subtracting from the common total dimension $d_\lambda$ yields
\[
 [s_\lambda q^k]F_{a,b}
   =\nlsum_{j\le k}\bigl(\dim\mathcal E_{\lambda,j}^P
                         -\dim\mathcal E_{\lambda,j}^Q\bigr)
   \ge0.
\]
Neutral isotypes contribute zero. Equality of the ungraded characters
makes the divided difference a polynomial, proving the assertion.
\end{proof}

The proof uses hypothesis~\eqref{cur:expansion} only on degree tails.
We state it for every graded subspace because it then excludes
shrunk subspaces of every kind (by homogeneity), which is the
condition that
operator scaling tests through capacity; the supplement
\citep{foulkesIIIsupp} uses this to certify \eqref{cur:expansion} on
individual blocks. It remains open whether hypothesis~\eqref{cur:expansion} holds in
general, and Proposition~\ref{gka:quadratic-descent} shows why it
does not follow from the one-layer surjection.

\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
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\end{thebibliography}

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

\section{Positivity at bounded depth}\label{app:certificates}
Theorem~\ref{nw:layers} concerns unbounded $a$ and $b$, so its finite
certificates must determine the coefficients as polynomials in both
$q^a$ and $q^b$. The degree bounds from the restriction formula make
this reconstruction finite. At the first nonzero layer, the same
formula gives a closed expression valid for every inner size.

\subsection{The first nonzero layer}\label{sec:layer}

The three coefficients at depth three share one positive factor, so a
single polynomial governs their positivity, their Newton expansion,
and their stability. The only delicate point is the division by
$[3]_q$ in this factor, since a quotient of polynomials with
nonnegative coefficients can have negative ones; the residues of $a$
and $b$ modulo $3$ settle it.

\begin{samepage}
\begin{theorem}[The first nonzero layer]\label{thm:main}
Let $2\le a<b$, and put
\begin{equation}\label{eq:Gamma}
 \Gamma_{a,b}(q)=
 \frac{q^{a-1}(1+q)^2}{[3]_q}
 \qbin{a}{2}\qbin{b}{2}[b-a]_q.
\end{equation}
Then $\Gamma_{a,b}\in\N[q]$. All Schur coefficients of $F_{a,b}$ with
$\lambda_1>N-3$ vanish. The three coefficients at depth three are
\begin{align*}
 [s_{(N-3,3)}]F_{a,b}&=q\Gamma_{a,b}(q),\\
 [s_{(N-3,2,1)}]F_{a,b}&=(1+q^2)\Gamma_{a,b}(q),\\
 [s_{(N-3,1,1,1)}]F_{a,b}&=q\Gamma_{a,b}(q).
\end{align*}
\end{theorem}
\end{samepage}

The vanishing concerns the four shapes $(N)$, $(N-1,1)$, $(N-2,2)$,
and $(N-2,1,1)$.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Theorem~\ref{thm:depth} at $r\le3$ gives the three restrictions
explicitly, and Proposition~\ref{prop:triangular} converts their
differences into the Schur coefficients of $F_{a,b}$.
Abbreviate
\[
 A=[a]_q,\quad B=[b]_q,\quad
 C_a=\qbin{a}{2},\ C_b=\qbin{b}{2},\quad
 E_a=\qbin{a}{3},\ E_b=\qbin{b}{3}.
\]
The first three occupancy formulas are
\begin{align}
 \mathcal R^{(1)}_{a,b}&=AB\,s_1,\label{eq:R1}\\
 \mathcal R^{(2)}_{a,b}
 &=B C_a H_2(Y;q)+C_b H_2[A Y],\label{eq:R2}\\
 \mathcal R^{(3)}_{a,b}
 &=B E_a H_3(Y;q)+(1+q)A C_aC_b\,H_2(Y;q)h_1(Y)
       +E_bH_3[A Y].\label{eq:R3}
\end{align}
The middle term uses the identity $[b]_q[b-1]_q=(1+q)C_b$.

\emph{\textbf{Cancellation through depth two.}}
We substitute $p_1[[n]_q]=[n]_q$ and $p_2[[n]_q]=[n]_{q^2}$ in
$H_2=s_2+qs_{11}$ and obtain the two Schur coefficients
\begin{align*}
 [s_2]H_2[[n]_qY]&=[n]_q+2q^2\qbin{n}{2},\\
 [s_{11}]H_2[[n]_qY]&=q[n]_q+q(1+q^2)\qbin{n}{2}.
\end{align*}
Consequently the coefficients of \eqref{eq:R2} are
\[
 B C_a+A C_b+2q^2C_aC_b,
 \qquad
 q\bigl(B C_a+A C_b+(1+q^2)C_aC_b\bigr).
\]
Both are symmetric under $a\leftrightarrow b$. The invariants at depth
zero are $1$ in either order, and \eqref{eq:R1} is symmetric as well.
Proposition~\ref{prop:triangular} proves the stated cancellation of the
four top Schur coefficients.

\emph{\textbf{The three depth-three coefficients.}}
The three terms of \eqref{eq:R3} need three families of Schur
coefficients. For $\nu\vdash3$, put
\[
 K_\nu(q)=[s_\nu]H_3,\qquad
 \ell_\nu(q)=[s_\nu](H_2h_1),\qquad
 L_\nu(n;q)=[s_\nu(Y)]H_3[[n]_qY].
\]
The coefficient $L_\nu(n;q)$ is a polynomial in $q^n$ of
degree at most three that vanishes at $n=0$. The Gaussian coefficients
$[n]_q$, $\qbin n2$, and $\qbin n3$ form a basis for these polynomials.
Therefore
\begin{equation}\label{eq:Newton}
 L_\nu(n;q)=K_\nu(q)[n]_q+u_\nu(q)\qbin n2+v_\nu(q)\qbin n3.
\end{equation}
The coefficient of $[n]_q$ follows by taking $n=1$, and $u_\nu$ follows
by taking $n=2$, in
\[
 L_\nu(n;q)=\nlsum_{\mu\vdash3}
  \frac{\kappa_3(\mu;q)\chi^\nu(\mu)}{z_\mu}
  \nlprod_{r\in\mu}[n]_{q^r},
\]
and the three $S_3$ characters give
\begin{center}
\begin{tabular}{@{}llll@{}}
\toprule
$\nu$&$K_\nu$&$\ell_\nu$&$u_\nu$\\\midrule
$(3)$&$1$&$1$&$q^2(1+q)(q+2)$\\
$(2,1)$&$q+q^2$&$1+q$&$q(1+q)^2(2q^2+1)$\\
$(1,1,1)$&$q^3$&$q$&$q^2(1+q)(q^3+q+1)$\\\bottomrule
\end{tabular}
\end{center}
The coefficients $v_\nu$ cancel from the difference of the two orders.
Using \eqref{eq:R3} and \eqref{eq:Newton}, the $s_\nu$ coefficient of
$\mathcal R^{(3)}_{a,b}-\mathcal R^{(3)}_{b,a}$ is
$u_\nu(E_bC_a-E_aC_b)+(1+q)\ell_\nu(A-B)C_aC_b$. We divide by
$1-q$ and use the elementary Gaussian identities
$E_bC_a-E_aC_b=C_aC_bq^{a-2}[b-a]_q/[3]_q$ and $A-B=-q^a[b-a]_q$; the
coefficient becomes
\[
 C_aC_bq^{a-2}[b-a]_q\,
 \frac{u_\nu/[3]_q-(1+q)q^2\ell_\nu}{1-q}.
\]
For the three rows of the table, the final factor is respectively
$q^2(1+q)^2/[3]_q$, $q(1+q)^2(1+q^2)/[3]_q$, and $q^2(1+q)^2/[3]_q$.
The lower layers have already canceled, so
Proposition~\ref{prop:triangular} identifies the three resulting
products with the Schur coefficients of $F_{a,b}$ at depth three. They are therefore
$q\Gamma_{a,b}$, $(1+q^2)\Gamma_{a,b}$, and $q\Gamma_{a,b}$, as
Theorem~\ref{thm:main} asserts, because
$C_aC_bq^{a-2}(1+q)^2[b-a]_q/[3]_q=q^{-1}\Gamma_{a,b}$.

\emph{\textbf{Positivity of $\Gamma_{a,b}$.}} It remains to prove that
division by $[3]_q$ in \eqref{eq:Gamma} does not introduce negative
coefficients. If $a\equiv b\equiv2\pmod3$, then $3$ divides $b-a$, and
$[b-a]_q/[3]_q=[(b-a)/3]_{q^3}\in\N[q]$. Otherwise some $n\in\{a,b\}$ satisfies $n\ge3$ and $n\equiv0$ or
$1\pmod3$, and we show
$\qbin n2/[3]_q\in\N[q]$. Write $\qbin n2=[n]_q[n-1]_q/[2]_q$; one of
$n$ and $n-1$ is divisible by three. If that integer is odd, the other one is even, so $[3]_q$
divides the first $q$-integer and $[2]_q$ the second, each with a
geometric sum as quotient. If it is even, write it
as $6r$ and the other as $m=6r\pm1$; the quotient becomes
$[r]_{q^6}(1-q+q^2)[m]_q$, and the last two factors multiply to
$1+\nlsum_{j=2}^{m-1}q^j+q^{m+1}$, which has nonnegative coefficients. All remaining factors of \eqref{eq:Gamma} are in $\N[q]$.
\end{proof}

\subsection{Stability of the depth-three coefficients}

Bergeron's $q$-stability conjecture asks that each coefficient with a
fixed tail increase with $b$. At depth three the common factor in
Theorem~\ref{thm:main} reduces this to the differences of
$\Gamma_{a,b}$. The first difference in $b$ still carries the factor
$1/[3]_q$. The mixed difference in $a$ and $b$ cancels it and factors
positively, and summing mixed differences over $a$ recovers the first
difference. Extend $\Gamma$ by
$\Gamma_{1,b}=\Gamma_{a,a}=0$ and write
$D^{\Gamma}_{a,b}=\Gamma_{a,b+1}-\Gamma_{a,b}$.

\begin{corr}[Depth-three stability]\label{cor:depth-three-stability}
For integers $1\le a\le b$,
\begin{equation}\label{eq:gamma-first-difference}
 D^{\Gamma}_{a,b}=[2]_q[b]_q\nlsum_{i=1}^{a-1}
 q^{2i+b-3}[i]_q[b-i]_q\in\N[q].
\end{equation}
For integers $1\le a<b$,
\begin{equation}\label{eq:gamma-mixed-difference}
 D^{\Gamma}_{a+1,b}-D^{\Gamma}_{a,b}
 =q^{2a+b-3}[2]_q[a]_q[b]_q[b-a]_q\in\N[q].
\end{equation}
For $\nu\vdash3$, write $G_{a,b,\nu}=[s_{(ab-3,\nu)}]F_{a,b}$ for the
fixed-tail coefficients. Consequently, for $2\le a\le b$ with $ab\ge6$,
\[
 G_{a,b+1,\nu}-G_{a,b,\nu}\in\N[q].
\]
For $2\le a<b$, moreover,
\[
 G_{a+1,b+1,\nu}-G_{a+1,b,\nu}
 -G_{a,b+1,\nu}+G_{a,b,\nu}\in\N[q].
\]
\end{corr}
\begin{proof}
Abbreviate $[j]=[j]_q$ and put $d=b-a$. Cancelling the two factors
$[2]$ in \eqref{eq:Gamma} gives
\[
 \Gamma_{a,b}=
 \frac{q^{a-1}[a][a-1][b][b-1][d]}{[3]}.
\]
Since $[b+1]-[b-1]=q^{b-1}[2]$ and
$[d+1]-[d]=q^d$, it follows that
\[
 D^{\Gamma}_{a,b}=\frac{q^{b-1}[a][a-1][b]}{[3]}
 \bigl([b+1]+q^{a-1}[2][d]\bigr).
\]
For $a<b$, we subtract this expression from the one with $a+1$
in place of $a$ and obtain
\[
 D^{\Gamma}_{a+1,b}-D^{\Gamma}_{a,b}
 =\frac{q^{a+b-2}[2][a][b]}{[3]}
 \bigl([b+1]+q[a+1][d-1]-[a-1][d]\bigr).
\]
Using $q[a+1]-[a-1]=q^{a-1}[3]-1$ and
$[d-1]+q^{d-1}[a-1]=[b-2]$, the final bracket is
\[
 [b+1]-[b-2]+q^{a-1}[3][d-1]
 =q^{a-1}[3][d],
\]
which proves \eqref{eq:gamma-mixed-difference}. We sum it with $a$
replaced by $i=1,\ldots,a-1$ and use $D^{\Gamma}_{1,b}=0$, which proves
\eqref{eq:gamma-first-difference}; the sum is empty for $a=1$.
The three fixed-tail coefficients are $q\Gamma_{a,b}$,
$(1+q^2)\Gamma_{a,b}$, and $q\Gamma_{a,b}$, so both stated
comparisons follow from the corresponding differences of $\Gamma$.
\end{proof}

Theorem~\ref{nw:layers} also asserts nonnegative Newton coefficients
based at $b=a$. At depth three they are $q$, $1+q^2$, and $q$ times
those of $\Gamma_{a,b}$ in the basis $\qbin{b-a}j$, which are positive
for every inner size. Put $T_a=[a-1]_q[a]_q[a+1]_q/[3]_q$. One of the three consecutive
integers is divisible by three, so $T_a\in\mathbb N[q]$. The factor
$\Gamma_{a,b}$ in \eqref{eq:Gamma} satisfies
\begin{align}\label{nw:gamma}
 \Gamma_{a,b}={}&q^{a-1}[a]_qT_a\qbin{b-a}1
 +q^{2a-1}[2]_q^2T_a\qbin{b-a}2\nonumber\\
 &+q^{3a+1}[2]_q[a]_q[a-1]_q\qbin{b-a}3.
\end{align}
To prove the identity, put $n=b-a$ and write
\[
 \Gamma_{a,b}=
 \frac{q^{a-1}[a]_q[a-1]_q}{[3]_q}
 [n]_q([a]_q+q^a[n]_q)([a-1]_q+q^{a-1}[n]_q).
\]
Repeatedly apply
$[n]_q\qbin nj=[j]_q\qbin nj+q^j[j+1]_q\qbin n{j+1}$.
The index-two coefficient simplifies by
$[a]_q+q[a-1]_q+q^a(2+q)=[2]_q[a+1]_q$, giving
\eqref{nw:gamma}.

\subsection{The two-parameter Gaussian expansion}\label{app:newton-double}
To handle every inner size at once, we treat $q^a$ like $q^b$. The
point to check is that the second expansion terminates, that is, the
coefficients of the first expansion have degree in $q^a$ bounded
independently of $a$.
For $\nu\vdash r$, the fixed-depth formula writes
$[s_{(ab-r,\nu)}]F_{a,b}$ as a polynomial $G_{r,\nu}(x,z;q)$ in
$x=q^a$ and $z=q^b$ over $\mathbb Q(q)$. Applied to each variable, the occupancy count of
Section~\ref{nw:section} bounds its bidegree
by $(r,r)$, and $G(x,z)=-G(z,x)$. Its total degree is
therefore at most $2r-1$. If $E$ shifts $n$ to $n+1$, then
$\nlprod_{i=0}^{j-1}(E-q^i)$ extracts the $j$th Gaussian coefficient at
$n=0$. Applied to $x^uz^v$ after $z=xq^n$, it gives
$x^{u+v}\nlprod_{i=0}^{j-1}(q^v-q^i)$.
The product vanishes for $v<j$, so the $j$th coefficient is divisible by
$x^j$, and its quotient has $x$-degree at most $2r-1-j$.
For an integer $a_0\ge2$ with $a_0^2\ge2r$, a second Newton expansion
gives
\begin{equation}\label{nw:double}
 [s_{(ab-r,\nu)}]F_{a,b}
 =\nlsum_{j=1}^r\nlsum_{k=0}^{2r-1-j}
 q^{j(a-a_0)}B_{r,\nu,j,k}(q)
 \qbin{a-a_0}k\qbin{b-a}j,
 \qquad b\ge a\ge a_0.
\end{equation}
The index $j=0$ is absent because $G(x,x)=0$. The signs of the
$B_{r,\nu,j,k}$ come from the certificates of Section~\ref{app:layers}.

\subsection{Exact certificates for depths four to ten}\label{app:layers}
Each certificate is a finite set of identities over $\Q(q)$ between the
fixed-depth formula and a Gaussian Newton expansion whose coefficients
are visibly nonnegative. Finitely many identities cover all $a$ and
$b$, because the degree bounds of Section~\ref{nw:section} and
Appendix~\ref{app:newton-double} determine each polynomial from finitely
many values, provided every identity holds exactly in $q$. All
computations work in the power-sum basis, where multiplication
concatenates cycle partitions and an Adams operation sends $p_j$ to
$p_{tj}$ and every grading variable to its $t$th power. They start from
$h_d=\nlsum_{\rho\vdash d}p_\rho/z_\rho$ and, by \eqref{eq:III-kappa},
$H_d=\nlsum_{\rho\vdash d}\kappa_d(\rho;q)p_\rho/z_\rho$, and return to
Schur functions by $[s_\nu]f=\nlsum_\rho[p_\rho]f\,\chi^\nu(\rho)$, with
the characters computed both by the rim-hook rule and by the Frobenius
formula.

For an occupancy $\lambda=(1^{n_1}2^{n_2}\cdots)\vdash k$ of length
$m$, the restriction term is
\begin{equation}\label{nw:certificate-occupancy}
 \qbin bm
 \begin{bmatrix}m\\n_1,n_2,\ldots\end{bmatrix}_q
 \nlprod_{j\ge1}H_{n_j}\!\left[\qbin ajH_j\right].
\end{equation}
These terms sum to $\mathcal R^{(k)}_{a,b}$, and
Proposition~\ref{prop:triangular} recovers the tail character as
\begin{equation}\label{nw:certificate-recovery}
 \mathcal G_{a,r}(b;Y)
 =\frac1{1-q}\nlsum_{i=0}^r(-1)^ie_i(Y)
 \bigl(\mathcal R^{(r-i)}_{a,b}-\mathcal R^{(r-i)}_{b,a}\bigr).
\end{equation}
For fixed $a,r,B$, we evaluate this expression at $b=B,\ldots,B+r$,
solve \eqref{nw:inversion}, and verify exact divisibility by $1-q$,
integrality, and the sign of every coefficient and monomial exponent.
The $r+1$ formal identities determine the full polynomial, since the
result has degree at most $r$ in $q^b$.

For the two-parameter expansion, we recover the $j$th Newton
coefficient at baseline $b=a$, divide it by $q^{j(a-a_0)}$, and expand
again in $a-a_0$ by \eqref{nw:inversion}. Its degree is at most
$2r-1-j$ in $q^a$, so $2r-j$ distinct formal nodes determine it. The
computation uses all these nodes and further nodes as a check, always
with $q$ as an indeterminate, so it establishes identities over
$\mathbb Q(q)$. The certificates uniform in $a$ have these sizes.
\begin{center}
\begin{tabular}{@{}rrrrr@{}}
\toprule
Depth $r$ & Base $a_0$ & Tails & Newton polynomials & Coefficients\\
\midrule
5 & 4 & 7 & 245 & 6,830\\
6 & 4 & 11 & 561 & 22,658\\
7 & 5 & 15 & 1,050 & 62,649\\
8 & 6 & 22 & 2,024 & 166,851\\
\bottomrule
\end{tabular}
\end{center}
Every polynomial $B_{r,\nu,j,k}$ is nonzero, and its coefficients are
positive integers from its lowest to its highest degree. For $a=B=7$,
the additional depths nine and ten have respectively 270 and 420
nonconstant Newton polynomials, with 23,810 and 42,850 positive
coefficients; all 30 and 42 constant terms vanish. These fixed-parameter identities hold for every $b\ge7$ by
the degree-$r$ bound. Rational evaluations with the second character construction confirm
these identities inside and beyond the interpolation ranges, including
$b=41$.

At depth four the certificate is symbolic in both parameters. We
replace $\qbin aj$ and $\qbin bj$ in \eqref{nw:certificate-occupancy}
by their product formulas in $x=q^a$ and $z=q^b$ and expand twice in
Gaussian binomials, as in \eqref{nw:double}. The 110 polynomials
$B_{4,\nu,j,k}$ give the expression
\[
 \nlsum_{j=1}^4q^{-4j}
 \left(\nlsum_{k=0}^{7-j}B_{4,\nu,j,k}(q)
 \frac{\nlprod_{i=0}^{k-1}(1-xq^{-4-i})}{(q;q)_k}\right)
 \frac{\nlprod_{i=0}^{j-1}(x-zq^{-i})}{(q;q)_j}.
\]
This expression equals \eqref{nw:certificate-recovery} in
$\mathbb Q(q)[x,z]$ for each of the five tails, and all 1,908
coefficients and all monomial shifts are nonnegative.

\begin{proof}[Proof of Theorem~\ref{nw:layers}]
For $(r,a_0)=(4,4),(5,4),(6,4),(7,5),(8,6)$, the exact
identities above give nonnegative
coefficients in \eqref{nw:double}. These identities hold in
$\mathbb Q(q)[x,z]$, so they apply for every $b\ge a\ge a_0$.
For $a=7$ and $r=9,10$, the certificates for \eqref{nw:expansion}
based at $B=7$ have nonnegative coefficients and zero constant term.
The degree bound proves that they hold for every $b\ge7$.
Depths at most three follow from Theorem~\ref{thm:main} and
\eqref{nw:gamma}. Formula~\eqref{nw:monotonicity} proves coefficientwise
monotonicity in $b$.
\end{proof}

\section{Construction of the current maps}\label{gka:free-currents}
The current maps of Section~\ref{cur:section} must respect the
coinvariant relations of both modules, and the polynomial extension of
the canonical map does not (Proposition~\ref{gka:quadratic-descent}).
The construction avoids the relations by differentiating only on the
side of $\mathcal P$ and multiplying only on the side of $\mathcal Q$.
Derivatives preserve the harmonic complement of every ideal, and
multiplication preserves every ideal, so a single orthogonal
projection at the end suffices, which gives the block formula
\eqref{gka:harmonic-block}. The differentiations contribute operators
$A_k^U(d)$ on full polynomial modules, which we call free factors.
They are positive for $d\ge N$, hence invertible, so the seed and the
final projection determine each block.

Fix $N\ge2$ and
$W_0=\{w\in\mathbb C^N:\nlsum_iw_i=0\}$. For a finite-dimensional
unitary $S_N$-module $U$, set
\[
 \mathscr P_k(U)=\Sym^k(W_0^*)\otimes U,\qquad
 \mathscr P(U)=\bigoplus_{k\ge0}\mathscr P_k(U).
\]
The group acts simultaneously on coordinates and on $U$. Use the
Euclidean Fischer inner product on the polynomial factor. Let $C_i$
multiply by $w_i$ and let $D_i=C_i^*$ differentiate in the direction
$e_i-N^{-1}\nlsum_je_j$. If $\rho(g)$ denotes the simultaneous action, then
\begin{equation}\label{gka:free-euler}
 \nlsum_iC_iD_i=kI\quad\hbox{on }\mathscr P_k(U),\qquad
 \rho(g)C_i\rho(g)^{-1}=C_{g(i)},\quad \rho(g)D_i\rho(g)^{-1}=D_{g(i)}.
\end{equation}
Put $Y_i=\nlsum_{j\ne i}\rho((ij))$. Each transposition is self-adjoint
and unitary, so $-(N-1)I\preceq Y_i\preceq(N-1)I$.
When $Y_i$ follows $D_i$, it acts on the remaining polynomial degree
and on $U$.

Let $V$ be another finite-dimensional unitary $S_N$-module. For an
equivariant map $T\colon\mathscr P(U)\to\mathscr P(V)$, define
\begin{equation*}
 \mathscr C_d(T)=\nlsum_i C_iT(dI+Y_i)D_i.
\end{equation*}
The operator $\mathscr C_d$ differentiates the input once and
multiplies the output once, and it models the contraction of a current
operator with tensors of dimension $d$. Restoring the differentiated
tensor occurrence gives $d$ terms, while restoring a different
occurrence exchanges its label with the first, which accounts for
$Y_i$.
The displayed order of the factors follows from the identity
$\rho((ij))C_j=C_i\rho((ij))$ and equivariance of $T$. The formula is polynomial in $d$;
its tensor interpretation uses an integer $d\ge N$.

Iterating $\mathscr C_d$ on the identity produces the free factors.
Their positivity is not automatic, because the operators $Y_i$ need
not be positive; they are only bounded below by $-(N-1)I$. Taking $d\ge N$ makes every factor $dI+Y_i$
positive, and equivariance makes these factors commute.

\begin{proposition}[Positive free factors]\label{gka:free-factor}
For $k\ge0$ define
\begin{equation}\label{gka:free-recursion}
 A_0^U(d)=I_U,\qquad
 A_k^U(d)=\frac1k\nlsum_i C_iA_{k-1}^U(d)(dI+Y_i)D_i
 \quad(k\ge1).
\end{equation}
The operator $A_k^U(d)$ is $S_N$-equivariant and monic of degree $k$
in $d$. For real $d\ge N$,
\begin{equation}\label{gka:free-positive}
 A_k^U(d)\succeq(d-N+1)^kI.
\end{equation}
If $\mu\colon U\to V$ is equivariant, its polynomial-linear extension
$\widetilde\mu=I\otimes\mu$ satisfies
$A_k^V(d)\widetilde\mu=\widetilde\mu A_k^U(d)$.
\end{proposition}
\begin{proof}
Covariance in \eqref{gka:free-euler} proves equivariance inductively.
It follows that $A_{k-1}^U(d)$ commutes with every $Y_i$.
For $d\ge N$, the induction hypothesis therefore makes
$A_{k-1}^U(d)(dI+Y_i)$ a product of commuting self-adjoint operators with
\[
 A_{k-1}^U(d)(dI+Y_i)\succeq(d-N+1)^kI.
\]
Sandwich this inequality between $D_i$ and $C_i=D_i^*$, sum over
$i$, and use \eqref{gka:free-euler} to obtain
\eqref{gka:free-positive}. The same Euler identity makes the leading
term of \eqref{gka:free-recursion} equal to $d^kI$.
Finally, $\widetilde\mu$ intertwines the group actions and commutes
with coordinate multiplication and differentiation, so induction in
\eqref{gka:free-recursion} proves the last identity.
\end{proof}

To extend a map from degree zero, let
$\sigma_r\colon U\to\mathscr P_r(V)$ be equivariant for each $r\ge0$,
with only finitely many nonzero. Define
\[
 \widehat\sigma_r(f\otimes u)=f\,\sigma_r(u),\qquad
 T_d^{\mathrm{free}}=
 \exp(\mathscr C_d/N)\bigl((\textstyle\nlsum_r\sigma_r)\Pi_0\bigr),
\]
where $\Pi_0$ projects onto polynomial degree zero. The exponential
is locally finite because each contraction differentiates the domain
once. The extension $\widehat\sigma_r$ commutes with multiplication;
it need not commute with differentiation. The asymmetry is harmless,
because in every block all differentiations act before the seed.

\begin{proposition}[Block formula]\label{gka:free-normal}
For every $k,r\ge0$ the block from degree $k$ to degree $k+r$ is
\begin{equation}\label{gka:free-block}
 (T_d^{\mathrm{free}})^{k+r,k}
       =N^{-k}\widehat\sigma_r A_k^U(d).
\end{equation}
\end{proposition}
\begin{proof}
After $k$ contractions, $\Pi_0$ kills $\mathscr P_j(U)$ for every
$j\ne k$. Induction gives
\[
 \mathscr C_d^k(\sigma_r\Pi_0)|_{\mathscr P_k(U)}
       =k!\,\widehat\sigma_r A_k^U(d).
\]
Indeed, at the inductive step its left side is
\[
 (k-1)!\nlsum_i C_i\widehat\sigma_r
          A_{k-1}^U(d)(dI+Y_i)D_i.
\]
Move $C_i$ through $\widehat\sigma_r$ and apply
\eqref{gka:free-recursion}. No derivative is moved through the seed.
The factor $1/(N^kk!)$ in the exponential proves
\eqref{gka:free-block}.
\end{proof}

\subsection{The two modules as harmonic spaces}
To apply the construction to Bergeron's modules, we realize them as
harmonic polynomials. Set $N=ab$ and let the wreath products $W_{a,b}$ and $W_{b,a}$ stabilize
the rows and columns of a $b\times a$ rectangle. Put
$U=\mathbb C[S_N/W_{a,b}]$ and $V=\mathbb C[S_N/W_{b,a}]$.
For a row partition $\pi$, write
$s_r=a^{-1}\nlsum_cw_{rc}$ and $x_{rc}=w_{rc}-s_r$.
In $\mathbb C[W_0]$, let $I_\pi$ be generated by
\[
 e_j(x_{r1},\ldots,x_{ra})\quad(2\le j\le a,\ 1\le r\le b),
 \qquad e_\ell(s_1,\ldots,s_b)\quad(2\le\ell\le b).
\]
For a column partition $\kappa$, define $J_\kappa$ in the same way
using $\eta_c=b^{-1}\nlsum_rw_{rc}$ and $y_{rc}=w_{rc}-\eta_c$.
These quotients are the nested coinvariant coefficient algebras of
\eqref{eq:III-modules}. Their Fischer orthogonal
complements give the induced harmonic modules
\[
 \mathcal P=\bigoplus_\pi I_\pi^\perp\otimes e_\pi
       \subset\mathscr P(U),\qquad
 \mathcal Q=\bigoplus_\kappa J_\kappa^\perp\otimes e_\kappa
       \subset\mathscr P(V),
\]
with characters $P_{a,b}$ and $Q_{a,b}$, respectively.
Here $e_\pi,e_\kappa$ are the coset basis vectors, and all complements
use the global Euclidean Fischer metric.

Let $P_{\mathcal Q}$ be the orthogonal projection and suppose
$\sigma_r(U)\subset\mathcal Q_r$. Define $T_d$ on $\mathcal P$
by the same contraction exponential, using compressed codomain
multiplication $P_{\mathcal Q}C_i|_{\mathcal Q}$.
Its blocks satisfy
\begin{equation}\label{gka:harmonic-block}
 T_d^{k+r,k}=N^{-k}P_{\mathcal Q,k+r}\widehat\sigma_r
                 A_k^U(d)|_{\mathcal P_k}.
\end{equation}
To prove \eqref{gka:harmonic-block}, note that derivatives preserve
the orthogonal complement of every ideal $I$, because
$\langle D_ih,f\rangle=\langle h,C_if\rangle=0$ whenever $h\perp I$
and $f\in I$.
The simultaneous group action also preserves $\mathcal P$ and
$\mathcal Q$. The operators $(dI+Y_i)D_i$ therefore preserve harmonicity. Since
multiplication preserves each ideal $J_\kappa$, the identity
$P_{\mathcal Q}C_i=P_{\mathcal Q}C_iP_{\mathcal Q}$ removes the
intermediate projections onto $\mathcal Q$. Applying
Proposition~\ref{gka:free-normal} leaves precisely the final projection
in \eqref{gka:harmonic-block}.

\subsection{Why the final projection is needed}
The final projection in \eqref{gka:harmonic-block} cannot be omitted
by extending the canonical map polynomial-linearly through the two
coefficient quotients. Call a row partition and a column partition transverse if every row
block meets every column block in exactly one position. Let
$\mu\colon U\to V$ be the canonical incidence map, whose nonzero
entries correspond to transverse pairs, and put
$\widetilde\mu=I\otimes\mu$.

\begin{proposition}\label{gka:quadratic-descent}
For $a,b\ge2$, the extension $\widetilde\mu$ does not map
$\bigoplus_\pi I_{\pi,2}\otimes e_\pi$ into
$\bigoplus_\kappa J_{\kappa,2}\otimes e_\kappa$.
\end{proposition}
\begin{proof}
Fix transverse partitions $\pi,\kappa$, a row $R$ of $\pi$, and two
blocks $K_1,K_2$ of $\kappa$. Choose
\[
 i\in R\cap K_1,\quad j\in R\cap K_2,\quad
 i'\in K_1\setminus R,\quad j'\in K_2\setminus R.
\]
These four labels are distinct. Consider the quadratic relation
\[
 f=\nlsum_{\ell\in R}w_\ell^2
       -\frac1a\left(\nlsum_{\ell\in R}w_\ell\right)^2\in I_{\pi,2}.
\]
The relation $f$ satisfies $\Lambda(f)=-2/a$, where
\[
 \Lambda(h)=[w_iw_j]h-[w_iw_{j'}]h
                 -[w_{i'}w_j]h+[w_{i'}w_{j'}]h.
\]
Every inner quadratic generator of $J_\kappa$ has zero cross-column
coefficients, and its outer quadratic generator has constant
coefficients on each pair of distinct column blocks. Hence $\Lambda$
annihilates both types of quadratic generator of $J_\kappa$. The functional $\Lambda$ also annihilates a multiple
$(\nlsum_\ell w_\ell)(\nlsum_\ell c_\ell w_\ell)$ of global centering,
whose coefficient of $w_\ell w_{\ell'}$ for $\ell\ne\ell'$ is $c_\ell+c_{\ell'}$. Thus $\Lambda$ is well
defined on centered quadratics and $f\notin J_{\kappa,2}$.
The $e_\kappa$ component of $\widetilde\mu(f\otimes e_\pi)$ is a
nonzero scalar multiple of $f$, proving the assertion.
\end{proof}

\end{document}