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\title{A Counterexample to a Conjecture on Fused Specht Polynomials}
\author{GPT-5.6 Sol, prompted by Darij Grinberg}
\date{\today}

\begin{document}

\maketitle

\begin{abstract}
Lafay, Peltola and Roussillon conjectured that their realization of simple
modules of the fused Hecke algebra by fused Specht polynomials extends from
Young diagrams with two columns to Young diagrams of arbitrary shape.  We
give a counterexample for
\[
 n=8,\qquad \lambda=(3,2,2,1),\qquad
 \varsigma=(2,2,2,2).
\]
More precisely, we exhibit an explicit linear dependence among three fused
Specht polynomials indexed by row-strict Young tableaux.  The same example
also yields an infinite family of counterexamples.
\end{abstract}

We use the notation of \cite{LPR}.  In particular, $\p_\varsigma$ denotes the
$\varsigma$-antisymmetrizer idempotent, $P_N$ denotes the Specht polynomial
associated with a numbering $N$, and $\mathcal F_T$ denotes the fused Specht
polynomial associated with a filling $T$.

\section{The counterexample}

Let
\[
 \lambda=(3,2,2,1)
 \quad\text{and}\quad
 \varsigma=(2,2,2,2).
\]
Consider the following three elements of $\RSYT^\lambda_\varsigma$:
\begin{equation}
\label{eq:tableaux}
 A=
 \begin{array}{ccc}
  1&2&3\\
  1&4&{}\\
  2&4&{}\\
  3&{}&{}
 \end{array},
 \qquad
 B=
 \begin{array}{ccc}
  1&2&4\\
  1&3&{}\\
  2&3&{}\\
  4&{}&{}
 \end{array},
 \qquad
 C=
 \begin{array}{ccc}
  1&3&4\\
  1&3&{}\\
  2&4&{}\\
  2&{}&{}.
 \end{array}
\end{equation}
Their numberings from \cite[Definition 2.9]{LPR} are
\begin{equation}
\label{eq:numberings}
 \widetilde A=
 \begin{array}{ccc}
  1&4&6\\
  2&7&{}\\
  3&8&{}\\
  5&{}&{}
 \end{array},
 \qquad
 \widetilde B=
 \begin{array}{ccc}
  1&4&8\\
  2&5&{}\\
  3&6&{}\\
  7&{}&{}
 \end{array},
 \qquad
 \widetilde C=
 \begin{array}{ccc}
  1&5&8\\
  2&6&{}\\
  3&7&{}\\
  4&{}&{}.
 \end{array}
\end{equation}

To avoid using the same variables before and after fusion, we temporarily
write $z_1,\ldots,z_8$ for the variables of the Specht polynomials and
$x_1,\ldots,x_4$ for the variables of the fused Specht polynomials.  For a
subset $J\subseteq[8]$, set
\[
 \Delta_J:=\prod_{\substack{i,j\in J\\i<j}}(z_j-z_i).
\]
It follows immediately from the columns of the numberings in
\eqref{eq:numberings} that
\begin{align}
 P_{\widetilde A}&=\Delta_{\{1,2,3,5\}}\Delta_{\{4,7,8\}},
 \label{eq:PA}\\
 P_{\widetilde B}&=\Delta_{\{1,2,3,7\}}\Delta_{\{4,5,6\}},
 \label{eq:PB}\\
 P_{\widetilde C}&=\Delta_{\{1,2,3,4\}}\Delta_{\{5,6,7\}}.
 \label{eq:PC}
\end{align}

The following elementary observation makes the fusion calculation short.

\begin{lemma}
\label{lem:differential}
For $Q\in\CC[z_1,\ldots,z_8]$, let
\[
 D:=\prod_{k=1}^4
 \left(\frac{\partial}{\partial z_{2k}}
       -\frac{\partial}{\partial z_{2k-1}}\right).
\]
Then
\begin{equation}
\label{eq:fusion-differential}
 \psi(\p_\varsigma.Q)
 =\frac1{16}
 \left[DQ\right]_{z_{2k-1}=z_{2k}=x_k\text{ for all }k\in[4]}.
\end{equation}
\end{lemma}

\begin{proof}
For two variables $z_a,z_b$, Taylor expansion at $z_a=z_b=x$ gives
\begin{equation}
\label{eq:pair-differential}
 \left[
 \frac{Q-(a\ b).Q}{z_b-z_a}
 \right]_{z_a=z_b=x}
 =
 \left[
 \left(\frac{\partial}{\partial z_b}
       -\frac{\partial}{\partial z_a}\right)Q
 \right]_{z_a=z_b=x}.
\end{equation}
Since
\[
 \p_\varsigma
 =\frac1{16}
 \prod_{k=1}^4\bigl(1-(2k-1\ \ 2k)\bigr),
\]
applying \eqref{eq:pair-differential} successively to the four pairs proves
\eqref{eq:fusion-differential}.
\end{proof}

We can now state the counterexample.

\begin{proposition}
\label{prop:counterexample}
The three fused Specht polynomials associated with the tableaux in
\eqref{eq:tableaux} satisfy
\begin{equation}
\label{eq:dependence}
 2\mathcal F_A-2\mathcal F_B+\mathcal F_C=0.
\end{equation}
Consequently, the map $\psi$ in \cite[(2.16)]{LPR} is not injective on
$\p_\varsigma(P^\lambda)$.  In particular, Conjecture~2.17 of \cite{LPR} is
false.
\end{proposition}

\begin{proof}
Applying Lemma~\ref{lem:differential} to the factorizations
\eqref{eq:PA}--\eqref{eq:PC} gives
\begin{align}
 \mathcal F_A
 &=-\frac12(x_1-x_2)(x_1-x_3)(x_2-x_4)
 \nonumber\\
 &\qquad\cdot
 \left(x_1^2+x_1x_2-3x_1x_3+x_2^2-3x_2x_4+3x_3x_4\right),
 \label{eq:FA}\\
 \mathcal F_B
 &=-\frac12(x_1-x_2)(x_1-x_4)(x_2-x_3)
 \nonumber\\
 &\qquad\cdot
 \left(x_1^2+x_1x_2-3x_1x_4+x_2^2-3x_2x_3+3x_3x_4\right),
 \label{eq:FB}\\
 \mathcal F_C
 &=(x_1-x_2)^4(x_3-x_4).
 \label{eq:FC}
\end{align}
A direct subtraction of \eqref{eq:FA} from \eqref{eq:FB} yields
\begin{equation}
\label{eq:difference}
 \mathcal F_B-\mathcal F_A
 =\frac12(x_1-x_2)^4(x_3-x_4)
 =\frac12\mathcal F_C.
\end{equation}
This proves \eqref{eq:dependence}.

It remains to check that this dependence comes from a nonzero element of
$\p_\varsigma(P^\lambda)$.  By \cite[Proposition 2.11]{LPR} and the
isomorphism $\phi:V^\lambda\to P^\lambda$ from
\cite[Lemma 2.3]{LPR}, the collection
\[
 \left\{\p_\varsigma.P_{\widetilde T}
 \ \middle|\ T\in\RSYT^\lambda_\varsigma\right\}
\]
is a basis of $\p_\varsigma(P^\lambda)$.  In particular,
\[
 2\p_\varsigma.P_{\widetilde A}
 -2\p_\varsigma.P_{\widetilde B}
 +\p_\varsigma.P_{\widetilde C}\ne0,
\]
whereas its image under $\psi$ is zero by \eqref{eq:dependence}.  Thus
$\psi$ is not injective.
\end{proof}

\section{Further remarks}

\begin{remark}[An infinite family]
\label{rem:infinite-family}
For each $r\geq0$, set
\[
 \lambda^{(r)}=(r+3,2,2,1)
 \quad\text{and}\quad
 \varsigma^{(r)}=(2,2,2,2,\underbrace{1,\ldots,1}_{r\text{ times}}).
\]
Append the entries $5,6,\ldots,4+r$, in this order, to the first row of each
of $A,B,C$.  This gives three elements
\[
 A^{(r)},B^{(r)},C^{(r)}
 \in\RSYT^{\lambda^{(r)}}_{\varsigma^{(r)}}.
\]
The added columns have one box each, so they contribute no factors to the
associated Specht polynomials.  The added blocks of the antisymmetrizer also
have size one.  Therefore
\[
 \mathcal F_{A^{(r)}}=\mathcal F_A,
 \qquad
 \mathcal F_{B^{(r)}}=\mathcal F_B,
 \qquad
 \mathcal F_{C^{(r)}}=\mathcal F_C,
\]
where the right hand sides are regarded as polynomials independent of the
additional variables.  Thus
\[
 2\mathcal F_{A^{(r)}}-2\mathcal F_{B^{(r)}}
 +\mathcal F_{C^{(r)}}=0
\]
for every $r\geq0$, giving counterexamples in every size $n\geq8$.
\end{remark}

\begin{remark}[Small sizes]
There are seven tableaux in $\RSYT^{(3,2,2,1)}_{(2,2,2,2)}$.  Exact row
reduction gives
\[
 \dim\p_{(2,2,2,2)}(P^{(3,2,2,1)})=7
 \quad\text{and}\quad
 \dim\operatorname{span}_{\CC}
 \left\{\mathcal F_T\mid
 T\in\RSYT^{(3,2,2,1)}_{(2,2,2,2)}\right\}=6.
\]
Thus the kernel in Proposition~\ref{prop:counterexample} is one-dimensional.
An exhaustive computation for $n\leq8$, up to permuting the parts of
$\varsigma$, found no counterexample for $n\leq7$ and found this as the only
counterexample for $n=8$.  Thus the example above is computationally minimal.
\end{remark}

\begin{remark}[The collision in Remark 2.33 of \cite{LPR}]
The two tableaux in \cite[Remark 2.33]{LPR}, which have shape
$(3,2,2,1)$ and content $(3,1,2,2)$, do not themselves give a dependence.
Indeed, their fused Specht polynomials are, up to nonzero scalar factors,
\[
 (x_3-x_1)^2(x_4-x_2)^2
 \quad\text{and}\quad
 (x_4-x_1)^2(x_3-x_2)^2,
\]
respectively, and are therefore linearly independent.  The dependence in
Proposition~\ref{prop:counterexample} occurs for the more symmetric content
$(2,2,2,2)$.
\end{remark}

\begin{thebibliography}{1}

\bibitem{LPR}
Augustin Lafay, Eveliina Peltola and Julien Roussillon,
\emph{Fused Specht Polynomials and $c=1$ Degenerate Conformal Blocks},
arXiv:2410.09798v3, 2025.
\url{https://arxiv.org/abs/2410.09798v3}.

\end{thebibliography}

\end{document}