Canonical quantum cohomology conjecture for hyperplane sections of type $A_n$ adjoint varieties

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\title[A discriminant Jordan chain]{A Discriminant Jordan Chain for Type-$A$ Adjoint Hyperplane Sections} \author{} \date{August 24, 2026} \hypersetup{ pdftitle={A Discriminant Jordan Chain for Type-A Adjoint Hyperplane Sections}, pdfsubject={Explicit non-semisimplicity on the quantum-parameter discriminant}, pdfkeywords={quantum cohomology, adjoint variety, hyperplane section, semisimplicity, Jordan chain} }

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\begin{abstract} Let $Y$ be a general hyperplane section of the adjoint variety of type $A_n$, with quantum parameters $q_1,q_2$. Using only the quantum Chevalley formulas of Benedetti--Perrin, we construct explicit classes $v_n,w_n$ satisfying

$$h_1\star v_n=w_n, \qquad h_1\star w_n=-q_1\bigl(q_1+(-1)^nq_2\bigr)(h_1+h_2).$$

It follows that every fiber on $q_1+(-1)^nq_2=0$ is non-semisimple. This recovers the odd-rank canonical case of Benedetti--Perrin's Conjecture~1.12, adds an even-rank anti-diagonal family, and proves that the known non-semisimple locus of the ambient adjoint variety is contained in that of its hyperplane section. The argument is complete relative to the published Chevalley formulas, but this research note has not yet received external expert review. \end{abstract}

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\section{Setup and statement}

Fix $n\geq 3$. Let $X$ be the adjoint variety of type $A_n$, and let $Y\subset X$ be a general hyperplane section. We use the Schubert basis and the two quantum parameters $q_1,q_2$ from Benedetti--Perrin \cite[Theorem~6.2]{BP}. Put

$$\begin{gathered} A_i=\sigma_{\alpha_i},\qquad B_i=\sigma_{-\alpha_i},\qquad C_i=\sigma_{\alpha_{i,i+2}},\\ L_1=\sigma_{-\alpha_{1,n}},\qquad L_2=\sigma_{-\alpha_{2,n+1}},\qquad \one=\sigma_\Theta,\qquad p=\sigma_{-\Theta}=[\mathrm{pt}]. \end{gathered}$$

Let $s=(-1)^n$ and define \begin{align} v_n&=q_1\sum_{i=1}^{n-1}(-1)^iC_i+s(L_1-L_2),\label{eq:v}\ w_n&=q_1\sum_{i=1}^{n}(-1)^iA_i \notag\ &\quad +q_1\sum_{i=2}^{n-1}(-1)^{i+1}B_i-q_1^2\one-sp. \label{eq:w} \end{align} The point-class coefficient of $w_n$ is $-s\in\{\pm1\}$; hence $w_n$ remains nonzero after every specialization of $q_1,q_2$.

\begin{theorem}[Universal discriminant identity]\label{thm:identity} In $\QH(Y)$ over $\mathbb Q[q_1,q_2]$, for every $n\geq3$, \begin{equation}\label{eq:universal} \boxed{h_1\star v_n=w_n}, \qquad \boxed{h_1\star w_n =-q_1\bigl(q_1+(-1)^nq_2\bigr)(h_1+h_2)}. \end{equation} Consequently, let $k$ be a characteristic-zero field and let $(a,b)\in k^2$ satisfy $a+(-1)^n b=0$. Then the specialized algebra

$$\QH(Y)_{(a,b)} :=\QH(Y)\otimes_{\mathbb Q[q_1,q_2]} k, \qquad q_1\longmapsto a,\quad q_2\longmapsto b,$$

is not semisimple. \end{theorem}

\section{Proof of the universal identity}

We use only the following instances of \cite[Theorem~6.2]{BP}. With the conventions $B_0=B_{n+1}=0$, \begin{align} h_1\star C_i &=A_i+B_i+B_{i-1}+B_{i+1}+q_1\delta_{i,1}\one,\label{eq:C}\ h_1\star L_1 &=q_1(A_n+B_n+B_{n-1})+q_1q_2\one,\label{eq:L1}\ h_1\star L_2 &=p+q_1q_2\one. \label{eq:L2} \end{align}

\subsection*{The first identity} Apply \eqref{eq:C}--\eqref{eq:L2} to \eqref{eq:v}. The two $q_1q_2$ terms cancel in $s(h_1\star L_1-h_1\star L_2)$. The coefficients of the $A_i$ are

$$q_1(-1)^i\quad(1\leq i\leq n-1), \qquad sq_1=q_1(-1)^n\quad(i=n).$$

For the $B_i$, the alternating sum gives zero at $B_1$; $q_1(-1)^{i+1}$ for $2\leq i\leq n-2$; zero at $B_{n-1}$ before the additional contribution $sq_1$ from $sL_1$; and $-sq_1+sq_1=0$ at $B_n$. The correction from $C_1$ is $-q_1^2\one$, while $-sL_2$ contributes $-sp$. These are exactly the coefficients in \eqref{eq:w}. Therefore

$$h_1\star v_n=w_n.$$

The same boundary calculation applies when $n=3$; the interior range is then empty.

\subsection*{The second identity} Theorem~6.2(3) of \cite{BP} gives

$$h_1\star A_i=h_1\star B_i.$$

Thus the terms indexed by $2\leq i\leq n-1$ cancel in $h_1\star w_n$. The remaining formulas are \begin{align*} h_1\star B_1&=q_1h_2,\ h_1\star B_n&=\sigma_{-\alpha_{n-1,n+1}},\ h_1\star p &=q_1\sigma_{-\alpha_{n-1,n+1}}+q_1q_2(h_1+h_2). \end{align*} Writing $N=\sigma_{-\alpha_{n-1,n+1}}$, we obtain \begin{align*} h_1\star w_n &=-q_1(q_1h_2)+sq_1N-q_1^2h_1 -s\bigl(q_1N+q_1q_2(h_1+h_2)\bigr)\ &=-q_1(q_1+sq_2)(h_1+h_2), \end{align*} which is the second identity in \eqref{eq:universal}.

\begin{lemma}\label{lem:linear} Let $A$ be a finite-dimensional commutative semisimple algebra over a characteristic-zero field. After extending scalars to an algebraic closure, every multiplication operator $M_x\colon A\to A$ is diagonalizable. \end{lemma}

\begin{proof} After scalar extension, $A$ is a finite product of copies of the algebraic closure. Multiplication is coordinatewise and hence diagonal in the idempotent basis. \end{proof}

Now specialize to $a+(-1)^n b=0$. Equation~\eqref{eq:universal} becomes

$$M_{h_1}(v_n)=w_n\neq0, \qquad M_{h_1}(w_n)=0.$$

This is a nontrivial length-two Jordan chain. Indeed, if $v_n$ were a multiple of $w_n$, the two displayed relations would contradict $w_n\neq0$. Lemma~\ref{lem:linear} therefore rules out semisimplicity and completes the proof of Theorem~\ref{thm:identity}.

\section{Consequences}

Define the parameter hyperplane

$$D_n=\{(q_1,q_2):q_1+(-1)^nq_2=0\}.$$

\begin{corollary}[Discriminant containment]\label{cor:disc} The hyperplane $D_n$ is contained in the non-semisimple locus of $\QH(Y)$. For the ambient type-$A_n$ adjoint variety, the same equation defines its non-semisimple locus \cite[Proposition~1.8]{SP}. Hence, fiberwise,

$$\Disc\bigl(\QH(X)\bigr) \subseteq \Disc\bigl(\QH(Y)\bigr).$$

\end{corollary}

\begin{corollary}[Canonical parity] On the canonical line $q_1=q_2=q$, the equation of $D_n$ holds identically when $n$ is odd. Thus $\QH(Y)_{\mathrm{can}}$ is not semisimple for every odd $n\geq3$. Together with the even-rank result \cite[Proposition6.4]{BP}, this proves Conjecture1.12 of \cite{BP} for all $n\geq2$. \end{corollary}

\begin{corollary}[Even-rank anti-diagonal] For every even $n\geq4$, the fibers on $q_2=-q_1$ are not semisimple. \end{corollary}

The parity in the canonical conjecture is therefore explained by how the diagonal $q_1=q_2$ meets the uniform discriminant hyperplane $D_n$: it is contained in $D_n$ for odd $n$, whereas for even $n$ it meets $D_n$ only at the origin.

\section{Scope, verification, and next question}

Theorem~\ref{thm:identity} proves the inclusion $D_n\subseteq \Disc(\QH(Y))$. It does \emph{not} prove the converse

$$q_1+(-1)^nq_2\neq0 \quad\Longrightarrow\quad \QH(Y)_{(q_1,q_2)}\ \text{is semisimple}.$$

The even-rank canonical proof in \cite{BP} uses additional information about the common zero eigenspace and its intersection form. Extending that part to an arbitrary parameter ratio requires products not determined by the divisor Chevalley matrices alone.

As reproducibility checks, the two-parameter divisor matrices verify \eqref{eq:universal} symbolically for $3\leq n\leq9$, and an independent sparse implementation verifies it over $\mathbb Q[q_1,q_2]$ for every $3\leq n\leq501$. These finite checks are not the proof; the proof is the uniform coefficient cancellation above. No division by $q_1$, $q_2$, or $q_1+(-1)^nq_2$ occurs, so the argument includes the coordinate boundaries and the origin.

The strongest natural next question is whether the reverse discriminant inclusion holds. Until this note receives independent expert review, the result should be cited as a candidate theorem rather than as an accepted published result.

\begin{thebibliography}{9}

\bibitem{BP} V.~Benedetti and N.~Perrin, \emph{Cohomology of hyperplane sections of (co)adjoint varieties}, arXiv:2207.02089 (2022), \url{https://arxiv.org/abs/2207.02089}.

\bibitem{SP} M.~Smirnov and N.~Perrin, \emph{On the big quantum cohomology of coadjoint varieties}, Selecta Mathematica, \href{https://doi.org/10.1007/s00029-025-01017-w} {doi:10.1007/s00029-025-01017-w}.

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