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\newcommand{\kk}{\mathbf{k}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\A}{\kk[S_n]}
\newcommand{\Gm}{\mathcal{G}}
\newcommand{\Am}{\mathcal{A}}
\newcommand{\Lie}{\mathcal{L}}
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\title{A counterexample to the Burman--Kulishov conjecture on Lie elements}
\author{GPT-5.6 Sol, edited by Darij Grinberg}
\date{\today}

\begin{document}

\maketitle

\begin{abstract}
Burman and Kulishov defined Lie elements in the group algebra
$\kk[S_n]$ by comparing, on every exterior power of the reflection
representation $V$, the usual action of $\kk[S_n]$ with the
infinitesimal action induced by its action on $V$.  They conjectured
that the Lie algebra $\Lie_n$ of all Lie elements is generated by the
Kirchhoff differences $1-(ij)$.  We disprove this conjecture for
$n=4$ by exhibiting an explicit counterexample arising from the
$(2,2)$-block of $\kk[S_4]$.  More generally, we describe $\Lie_n$ in
terms of the Artin--Wedderburn decomposition of $\kk[S_n]$: its hook
blocks are determined by the action on $V$, whereas its non-hook blocks
are arbitrary.  Consequently, the primitive central idempotents of the
non-hook blocks yield linearly independent obstructions to the
conjecture.  We also identify the Lie algebra generated by the
Kirchhoff differences in terms of the derived algebra of the Lie
algebra generated by transpositions.  Along the way, we give an
integral, and hence characteristic-free, proof that the exterior powers
of $V$ are the hook-shaped Specht modules.
\medskip

\textbf{Manifest.} {This is a negative answer to \cite[Conjecture~2.2]{BurKul23}, generated by GPT-5.6 Sol (Plus). It includes a proof of Proposition~\ref{prop.hook-exterior} for its expository value. The output was edited and checked by me.}\footnote{This work is in the public domain. -Darij}
\end{abstract}

\section{The conjecture}

Let $\kk$ be a field of characteristic $0$, let $n\geq 2$, and let
$V$ be the reflection representation of the symmetric group $S_n$.  Thus,
\[
 V=\left\{(x_1,\ldots,x_n)\in\kk^n\ \middle|\
 x_1+\cdots+x_n=0\right\};
\]
the group $S_n$ acts on this by permuting the coordinates.
%Set $\A=\kk[S_n]$.
For every $m\in\mathbb{N}$, Burman and Kulishov (in \cite{BurKul23})
define $\kk$-linear maps
\[
 \Gm_m,\Am_m:\A\longrightarrow\End_{\kk}\left(\Lambda^mV\right)
\]
by extending the following rules linearly from $S_n$ to $\A$:
\begin{align*}
 \Gm_m(g)(v_1\wedge\cdots\wedge v_m)
 &=g(v_1)\wedge\cdots\wedge g(v_m),\\
 \Am_m(g)(v_1\wedge\cdots\wedge v_m)
 &=\sum_{i=1}^m
 \underbrace{v_1\wedge\cdots\wedge g(v_i)\wedge\cdots\wedge v_m}_{\substack{\text{the $i$-th factor is $g(v_i)$;}\\\text{all other factors are the respective $v_j$}}}.
\end{align*}
Here $\Gm_0(g)=\id_{\kk}$ and $\Am_0(g)=0$.  Note that
$\Gm_m(a)$ (for any $a \in \A$) is the action of $a$ on the
representation $\Lambda^mV$ of $S_n$.

An element $a\in\A$ is
called a \emph{Lie element} if
\[
 \Gm_m(a)=\Am_m(a)
 \qquad\text{for every }m\in\mathbb{N}.
 \tag{1}\label{eq.Lie-condition}
\]
It is enough to require \eqref{eq.Lie-condition} for
$0\leq m\leq n-1$, since all $m \geq n$ satisfy $\Lambda^m V=0$.
The Lie elements form a Lie subalgebra of $\A$ (see
\cite[Corollary 1.6]{BurKul23}); we denote this Lie subalgebra
by $\Lie_n$.

For distinct $i,j\in[n]$, define the \emph{Kirchhoff difference}
\[
 \varkappa_{ij}=1-(ij)\in\A,
\]
where $(ij)$ denotes the transposition in $S_n$ that swaps
$i$ with $j$.
We write
\[
 \HLie_n=
 \operatorname{Lie}\left\langle\varkappa_{ij}\ \middle|\
 1\leq i<j\leq n\right\rangle
 \subseteq\A
\]
for the Lie algebra generated by the Kirchhoff differences.
Burman and Kulishov proved that every $\varkappa_{ij}$ is a Lie element
and conjectured that the Lie algebra $\Lie_n$ is generated by the
$\varkappa_{ij}$, or equivalently that $\Lie_n=\HLie_n$; see
\cite[Conjecture~2.2]{BurKul23}.  We shall show
that this conjecture is false, first structurally and then by an explicit
counterexample in $\kk[S_4]$.

\section{The structure of the Lie-element algebra}

For $m\in\mathbb{N}$, define a linear map
\[
 D_m:\End_{\kk}(V)\longrightarrow\End_{\kk}\left(\Lambda^mV\right)
\]
by
\[
 \left(D_m(X)\right)(v_1\wedge\cdots\wedge v_m)
 =\sum_{i=1}^m
 v_1\wedge\cdots\wedge X(v_i)\wedge\cdots\wedge v_m.
 \tag{2}\label{eq.Dm-def}
\]
Thus, $D_m$ is the differential at the identity of the $m$-th
exterior-power representation
$\operatorname{GL}(V)\to\operatorname{GL}(\Lambda^mV)$.
Equivalently, $D_m$ is the canonical representation of the
Lie algebra $\glie(V)$ on the $m$-th exterior power of its
natural representation $V$.
In particular, $D_m$ is a Lie algebra morphism.

\begin{lemma}\label{lem.Am-Dm}
For every $a\in\A$ and every $m\in\mathbb{N}$, one has
\[
 \Am_m(a)=D_m\left(\Gm_1(a)\right).
 \tag{3}\label{eq.Am-Dm}
\]
\end{lemma}

\begin{proof}
Write $a=\sum_{g\in S_n}a_g g$.  For $v_1,\ldots,v_m\in V$, the
definition of $\Am_m$ gives
\begin{align*}
 \Am_m(a)(v_1\wedge\cdots\wedge v_m)
 &=\sum_{g\in S_n}a_g\sum_{i=1}^m
 v_1\wedge\cdots\wedge g(v_i)\wedge\cdots\wedge v_m\\
 &=\sum_{i=1}^m
 v_1\wedge\cdots\wedge
 \left(\sum_{g\in S_n}a_g g\right)(v_i)
 \wedge\cdots\wedge v_m\\
 &=\sum_{i=1}^m
 v_1\wedge\cdots\wedge
 \left(\Gm_1(a)\right)(v_i)
 \wedge\cdots\wedge v_m\\
 &=D_m\left(\Gm_1(a)\right)
 (v_1\wedge\cdots\wedge v_m).
\end{align*}
This proves \eqref{eq.Am-Dm}.
\end{proof}

For $0\leq m\leq n-1$, define the partition
\[
 h_m=(n-m,1^m)\qquad \text{of }n.
\]
Thus, $h_0=(n)$ and $h_1=(n-1,1)$ and
$h_{n-1}=(1,1,\ldots,1)$.
The partitions $h_0, h_1, \ldots, h_{n-1}$ are called
\emph{hook partitions}, due to their shape:
\begin{center}
\begin{tikzpicture}[x=0.48cm,y=0.48cm]
 \fill[gray!35] (0,0) rectangle (1,-1);
 \draw (0,0) rectangle (1,-1);
 \draw (1,0) rectangle (2,-1);
 \draw (2,0) rectangle (3,-1);
 \node at (3.6,-0.5) {$\cdots$};
 \draw (4.2,0) rectangle (5.2,-1);
 \draw (0,-1) rectangle (1,-2);
 \draw (0,-2) rectangle (1,-3);
 \node at (0.5,-3.6) {$\vdots$};
 \draw (0,-4.2) rectangle (1,-5.2);
 \draw[decorate,decoration={brace,amplitude=4pt}]
       (0,0.2)--(5.2,0.2)
       node[midway,yshift=9pt] {$n-m$};
 \draw[decorate,decoration={brace,mirror,amplitude=4pt}]
       (-0.2,-1)--(-0.2,-5.2)
       node[midway,xshift=-10pt] {$m$};
 \node[anchor=west] at (7.8,-2.6) {$h_m=(n-m,1^m)$};
\end{tikzpicture}
\end{center}
Every partition of $n$ that is not one of $h_0, h_1, \ldots, h_{n-1}$
will be called a \emph{non-hook partition} (or, for short,
a \emph{non-hook}).

For any partition $\lambda$ of $n$, let $M^\lambda$ denote the
\emph{Young permutation module} of shape $\lambda$ (the free
$\kk$-vector space with basis formed by the tabloids $\{T\}$ of shape
$\lambda$).  We let $S^\lambda$ denote the corresponding Specht module,
that is, the submodule of $M^\lambda$ spanned by the polytabloids
$\mathbf{e}_T$.  The standard polytabloids $\mathbf{e}_T$ form a basis
of $S^\lambda$.  Since $\kk$ has characteristic $0$, the $S_n$-module
$S^\lambda$ is irreducible.

We record a standard fact (in a characteristic-free version,
because we can)\footnote{I (Darij) learned it from
Mark Wildon.}.

\begin{proposition}\label{prop.hook-exterior}
Let $R$ be any commutative ring, and let
\[
 V_R=\left\{(x_1,\ldots,x_n)\in R^n\ \middle|\
 x_1+\cdots+x_n=0\right\}
\]
be the reflection representation of $S_n$ over $R$
(where $S_n$ acts by permuting the coordinates).
For $0\leq m\leq n-1$, let
$S_R^{h_m}$ be the Specht module of shape $h_m$ defined over $R$.  Then,
there is an $R[S_n]$-module isomorphism
\[
 \Lambda_R^m V_R \cong S_R^{h_m}.
 \tag{4}\label{eq.hook-exterior}
\]
In particular, this holds over every field, without any restriction on
its characteristic.
\end{proposition}

\begin{proof}
We shall first construct the isomorphism over $\ZZ$ and then
apply base change.

Let $U_{\ZZ}=\ZZ^n$ be the natural
permutation representation of $S_n$, with standard basis
$\varepsilon_1,\ldots,\varepsilon_n$. Thus,
\[
 V_{\ZZ}=\left\{(x_1,\ldots,x_n)\in U_{\ZZ}\ \middle|\
 x_1+\cdots+x_n=0\right\}.
\]
The coordinate-sum map $U_{\ZZ}\to\ZZ$ (which sends each
basis vector $\varepsilon_i$ to $1$) has a $\ZZ$-linear right inverse
(namely, the $\ZZ$-linear map $\ZZ\to U_{\ZZ}$ that sends
$1\mapsto\varepsilon_1$).
Thus, $V_{\ZZ}$ (being the kernel of this map) is a direct summand of
$U_{\ZZ}$ as a $\ZZ$-module (not as an $S_n$-representation!);
in particular, we can regard
$\Lambda_{\ZZ}^mV_{\ZZ}$ as a $\ZZ$-submodule of
$\Lambda_{\ZZ}^mU_{\ZZ}$.

Let $M_{\ZZ}^{h_m}$ be the integral Young permutation module of
shape $h_m$.  If $T$ is a tableau of shape $h_m$ (with entries
$1,2,\ldots,n$), write
\[
 c_r=T(r+1,1)\quad\text{(for $0 \leq r \leq m$)}
\]
for the entries in its first column, read from top to bottom
(in particular, $c_0 = T(1,1)$).  Define a
$\ZZ$-linear map
\[
 \Phi:M_{\ZZ}^{h_m}\longrightarrow
 \Lambda_{\ZZ}^mU_{\ZZ}
\]
on the tabloid basis by
\[
 \Phi\left(\{T\}\right)
 =\varepsilon_{c_1}\wedge\varepsilon_{c_2}
 \wedge\cdots\wedge\varepsilon_{c_m}.
 \tag{4a}\label{eq.Phi-tabloid}
\]
This is well-defined.  Indeed, the rows $2,3,\ldots,m+1$ of $h_m$ are
singletons, so the tabloid $\{T\}$ determines the entries $c_1,\ldots,c_m$.
The possibly ambiguous entry $c_0$ does not occur in
\eqref{eq.Phi-tabloid}.  It is also immediate from the definition that
$\Phi$ is $S_n$-equivariant.

Let $T$ be any tableau of shape $h_m$ (not necessarily standard), and let
\[
 \mathbf{e}_T
 =\sum_{\pi\in C_T}\sgn(\pi)\{\pi T\}
\]
be the polytabloid associated with $T$, where $C_T$ is the column group
of $T$ (that is, the group of permutations in $S_n$ that fix each
column of $T$ as a set).
Note that this column group $C_T$ is the symmetric group
$S_{\{c_0,c_1,\ldots,c_m\}}$, since the only column that (possibly)
contains more than one entry is the first column.
We claim that
\[
 \Phi(\mathbf{e}_T)
 =m!\,
 \left(\varepsilon_{c_1}-\varepsilon_{c_0}\right)
 \wedge\left(\varepsilon_{c_2}-\varepsilon_{c_0}\right)
 \wedge\cdots\wedge
 \left(\varepsilon_{c_m}-\varepsilon_{c_0}\right).
 \tag{4b}\label{eq.Phi-polytabloid}
\]
Indeed, the definition of $\Phi$ gives
\[
 \Phi(\mathbf{e}_T)
 =\sum_{\pi\in C_T}\sgn(\pi)\,
 \varepsilon_{\pi(c_1)}\wedge\cdots\wedge
 \varepsilon_{\pi(c_m)}.
\]
This is the antisymmetrization of
$\varepsilon_{c_1}\wedge\cdots\wedge\varepsilon_{c_m}$ with respect
to the $m+1$ elements $c_0,c_1,\ldots,c_m$.  To check
\eqref{eq.Phi-polytabloid} explicitly, for each $0\leq r\leq m$
compare the coefficient of the pure wedge
\[
 \varepsilon_{c_0}\wedge\cdots\wedge
 \widehat{\varepsilon_{c_r}}\wedge\cdots\wedge
 \varepsilon_{c_m}
\]
(since, after reordering the wedge factors and removing
the wedges that have two equal factors,
both sides of \eqref{eq.Phi-polytabloid} can be expressed as linear
combinations of such pure wedges).
On each side, this coefficient is $m!(-1)^r$: on the left, the $m!$
relevant permutations $\pi$ are those satisfying $\pi(c_0)=c_r$; on the
right, this follows by expanding the wedge product.  This proves
\eqref{eq.Phi-polytabloid}.

The right hand side of \eqref{eq.Phi-polytabloid} belongs to
$m!\Lambda_{\ZZ}^mV_{\ZZ}$.  Since the integral Specht
module $S_{\ZZ}^{h_m}$ is spanned by the polytabloids, we conclude
that $\Phi(S_{\ZZ}^{h_m}) \subseteq m!\Lambda_{\ZZ}^mV_{\ZZ}$.
Thus, there is a $\ZZ$-linear map
\[
 \Psi:S_{\ZZ}^{h_m}\longrightarrow
 \Lambda_{\ZZ}^mV_{\ZZ}
\]
characterized by
\[
 \Phi(s)=m!\Psi(s)
 \qquad\text{for every }s\in S_{\ZZ}^{h_m}.
 \tag{4c}\label{eq.Psi-def}
\]
Indeed, we can simply define $\Psi$ by $\Psi(s) = \Phi(s) / m!$;
this is well-defined because of
$\Phi(S_{\ZZ}^{h_m}) \subseteq m!\Lambda_{\ZZ}^mV_{\ZZ}$ and
since $\Lambda_{\ZZ}^mU_{\ZZ}$ is torsion-free.
This shows furthermore that the map
$\Psi$ is $S_n$-equivariant (since $\Phi$ is).
Explicitly, from \eqref{eq.Phi-polytabloid} and
\eqref{eq.Psi-def}, we conclude that each tableau $T$ satisfies
\[
 \Psi(\mathbf{e}_T)
 =\left(\varepsilon_{c_1}-\varepsilon_{c_0}\right)
 \wedge\cdots\wedge
 \left(\varepsilon_{c_m}-\varepsilon_{c_0}\right).
 \tag{4d}\label{eq.Psi-polytabloid}
\]

The standard tableaux $T$ of shape $h_m$ all have
$c_0 = T(1,1) = 1$
(since $T(1,1)$ is smaller than all other entries of $T$)
and $2 \leq c_1 < c_2 < \cdots < c_m$,
and they
are in bijection with the $m$-element subsets
$\{c_1<\cdots<c_m\}$ of $\{2,3,\ldots,n\}$ (since the entries
in the first column determine all the remaining entries).
The corresponding
polytabloids $\mathbf{e}_T$ form a $\ZZ$-basis of
$S_{\ZZ}^{h_m}$.  On the other hand,
\[
 \varepsilon_2-\varepsilon_1,\ldots,
 \varepsilon_n-\varepsilon_1
\]
is a $\ZZ$-basis of $V_{\ZZ}$, so the pure wedges
\[
\left(\varepsilon_{c_1}-\varepsilon_{c_0}\right)
 \wedge\cdots\wedge
 \left(\varepsilon_{c_m}-\varepsilon_{c_0}\right)
\]
with $2 \leq c_1 < c_2 < \cdots < c_m$
form a $\ZZ$-basis of $\Lambda_{\ZZ}^mV_{\ZZ}$.  Thus,
\eqref{eq.Psi-polytabloid} shows that $\Psi$ maps a basis to a basis and
is therefore an isomorphism.
So we have shown that
$\Psi : S_{\ZZ}^{h_m}\longrightarrow
 \Lambda_{\ZZ}^mV_{\ZZ}$ is an isomorphism of
representations of $S_n$ over $\ZZ$.

% Finally, the sequence
% $0\to V_{\ZZ}\to U_{\ZZ}\to\ZZ\to0$ (where $\ZZ^n \to \ZZ$
% is the above-mentioned coordinate-sum map) is a split exact
% sequence of $\ZZ$-modules
% (since the coordinate-sum map has a $\ZZ$-linear right
% inverse).
% Hence,
Finally, the canonical $\ZZ$-module homomorphism
\[
R\otimes_{\ZZ}V_{\ZZ}\to V_R
\]
is an isomorphism
(since it sends the basis
$ \varepsilon_2-\varepsilon_1,\ldots,
 \varepsilon_n-\varepsilon_1$ of $V_{\ZZ}$ to the
analogous basis of $V_R$).
Since exterior powers commute with base change,
we thus obtain
\[
\Lambda_R^m V_R \cong \Lambda_R^m (R\otimes_{\ZZ} V_{\ZZ})
\cong R\otimes_{\ZZ} (\Lambda_{\ZZ}^m V_{\ZZ}).
\]
This isomorphism is furthermore $S_n$-equivariant.
Since we also have an $S_n$-equivariant isomorphism
$S_R^{h_m} \cong R\otimes_{\ZZ} S_{\ZZ}^{h_m}$ (because
a Specht module has ``the same'' basis of standard
polytabloids over any commutative ring), we thus
obtain an $S_n$-equivariant isomorphism
\[
 \Psi_R:S_{R}^{h_m}\longrightarrow
 \Lambda_{R}^mV_{R}
\]
by tensoring $\Psi$ with $R$. This proves \eqref{eq.hook-exterior}.
\end{proof}

We now return to our characteristic-$0$ field $\kk$.
For every $0\leq m\leq n-1$, fix an $S_n$-module isomorphism
\[
 \theta_m:S^{h_m}\xrightarrow{\ \sim\ }\Lambda^mV
\]
(this exists by Proposition~\ref{prop.hook-exterior}).

\begin{theorem}\label{thm.structure}
There is a Lie algebra isomorphism
\[
 \Lie_n\cong
 \glie(V)\oplus
 \bigoplus_{\substack{\lambda\vdash n;\\
                      \lambda\text{ is a non-hook}}}
 \glie(S^\lambda).
 \tag{5}\label{eq.Lie-structure}
\]
More precisely, consider the Artin--Wedderburn isomorphism
\[
 \kk[S_n]\xrightarrow{\ \sim\ }
 \bigoplus_{\lambda\vdash n}\End_{\kk}(S^\lambda),
 \qquad
 a\longmapsto\left(\rho_\lambda(a)\right)_{\lambda\vdash n}
 \tag{6}\label{eq.Wedderburn}
\]
(where $\rho_\lambda(a)$ denotes the action of $a$
on $S^\lambda$ as an element of $\End_{\kk}(S^\lambda)$).
For each hook partition $h_m$, write
\[
 \widetilde\rho_{h_m}(a)
 =\theta_m\rho_{h_m}(a)\theta_m^{-1}
 \in\End_{\kk}(\Lambda^mV).
 \tag{6a}\label{eq.rho-transported}
\]
(This is simply the action of $a$ on $\Lambda^m V$.)
Then,
the Lie-element algebra $\mathcal{L}_n$ consists of the
$a \in \A$ satisfying
\begin{align}
 \widetilde\rho_{h_0}(a)&=0,
 \tag{7}\label{eq.trivial-block}\\
 \widetilde\rho_{h_m}(a)&=D_m\left(\widetilde\rho_{h_1}(a)\right)
 \qquad(2\leq m\leq n-1),
 \tag{8}\label{eq.hook-blocks}
\end{align}
while all non-hook components $\rho_\lambda(a)$ are arbitrary.
\end{theorem}

\begin{proof}
The group algebra $\kk[S_n]$ is split semisimple, so
\eqref{eq.Wedderburn} is an isomorphism (of algebras, and
thus also of Lie algebras).  By the definition of
$\widetilde\rho_{h_m}$, the map $\Gm_m$ is precisely the representation
map $a\mapsto\widetilde\rho_{h_m}(a)$.  By \eqref{eq.Am-Dm}, the condition
$\Gm_m(a)=\Am_m(a)$ is therefore
\[
 \widetilde\rho_{h_m}(a)
 =D_m\left(\widetilde\rho_{h_1}(a)\right).
\]
For $m=0$, this says $\widetilde\rho_{h_0}(a)=0$, since $D_0=0$.  For $m=1$,
it is tautological, since $D_1$ is the identity map.  For
$2\leq m\leq n-1$, it gives \eqref{eq.hook-blocks}.

No non-hook representation occurs among the exterior powers of $V$.
Consequently, the conditions defining $\Lie_n$ impose no restriction on
the non-hook components of $a$.  Thus, an element $a$ of $\Lie_n$ is
uniquely specified by the arbitrary component
$X(a)=\widetilde\rho_{h_1}(a)\in\glie(V)$ and arbitrary components
$X_\lambda(a)=\rho_\lambda(a)\in\glie(S^\lambda)$ for the non-hook
partitions $\lambda$.  Its remaining hook components are forced to be
$0,D_2(X(a)),\ldots,D_{n-1}(X(a))$.  Since every $D_m$ is a Lie algebra
morphism, we obtain a Lie algebra isomorphism
\begin{align*}
 \Lie_n &\to
 \glie(V)\oplus
 \bigoplus_{\substack{\lambda\vdash n;\\
                      \lambda\text{ is a non-hook}}}
 \glie(S^\lambda),
 \\
 a &\mapsto \left(X(a), \left(X_\lambda(a)\right)_{\lambda \text{ is a non-hook}}\right).
\end{align*}
This proves \eqref{eq.Lie-structure}.
\end{proof}

\begin{corollary}\label{cor.dimension}
Let $f^\lambda=\dim S^\lambda$.  Then
\begin{align*}
 \dim\Lie_n
 &=(n-1)^2+
 \sum_{\substack{\lambda\vdash n;\\
                  \lambda\text{ is a non-hook}}}(f^\lambda)^2\\
 &=(n-1)^2+n!-\binom{2n-2}{n-1}.
 \tag{9}\label{eq.Lie-dimension}
\end{align*}
\end{corollary}

\begin{proof}
The first equality follows from \eqref{eq.Lie-structure}.  Moreover,
every $0\leq m\leq n-1$ satisfies
\[
 f^{h_m}=\binom{n-1}{m}.
\]
Hence, the sum of the squares of the
dimensions of the hook representations is
\[
 \sum_{m=0}^{n-1}\binom{n-1}{m}^2
 =\binom{2n-2}{n-1}
\]
(by Vandermonde's convolution; equivalently, compare the coefficient of
$t^{n-1}$ in $(1+t)^{n-1}(1+t)^{n-1}=(1+t)^{2n-2}$).
Now use the equality $\sum_{\lambda\vdash n}(f^\lambda)^2=n!$
(which follows from \eqref{eq.Wedderburn} being an isomorphism).
\end{proof}

\section{The explicit counterexample for \texorpdfstring{$n=4$}{n=4}}

In $\kk[S_4]$, define
\begin{align}
 x={}&2\left(1+(12)(34)+(13)(24)+(14)(23)\right)\notag\\
 &-\left((123)+(132)+(124)+(142)
 +(134)+(143)+(234)+(243)\right).
 \tag{10}\label{eq.x-explicit}
\end{align}

\begin{proposition}\label{prop.explicit-counterexample}
The element $x$ is a Lie element, but it does not belong to the Lie
algebra generated by the six Kirchhoff differences $\varkappa_{ij}$.
Consequently, \cite[Conjecture~2.2]{BurKul23} is false already for $n=4$.
\end{proposition}

\begin{proof}
Let $e_{(2,2)}$ be the primitive central idempotent of $\kk[S_4]$
corresponding to $S^{(2,2)}$.  The character values of $S^{(2,2)}$ on
the conjugacy classes of cycle types
\[
 1^4,\quad 2\,1^2,\quad 2^2,\quad 3\,1,\quad 4
\]
are, respectively,
\[
 2,\quad 0,\quad 2,\quad -1,\quad 0.
\]
If we let $\chi^\lambda$ denote the character of the
Specht module $S^\lambda$, then
the central-idempotent formula therefore gives
\begin{align*}
 e_{(2,2)}
 &=\frac{\dim S^{(2,2)}}{4!}
 \sum_{\sigma\in S_4}\chi^{(2,2)}(\sigma^{-1})\sigma\\
 &=\frac{x}{12}.
 \tag{11}\label{eq.x-idempotent}
\end{align*}

For the three-dimensional reflection representation $V$ of $S_4$,
Proposition~\ref{prop.hook-exterior} yields
\[
 \Lambda^0V\cong S^{(4)},\qquad
 \Lambda^1V\cong S^{(3,1)},\qquad
 \Lambda^2V\cong S^{(2,1,1)},\qquad
 \Lambda^3V\cong S^{(1,1,1,1)};
\]
none of these $S_4$-representations contains an $S^{(2,2)}$.
Thus, $e_{(2,2)}$ acts as $0$ on every $\Lambda^mV$.  It follows that
\[
 \Gm_m(x)=0
 \qquad\text{for every }m.
 \tag{12}\label{eq.Gmx-zero}
\]
In particular, $\Gm_1(x)=0$, and Lemma~\ref{lem.Am-Dm} yields
\[
 \Am_m(x)=D_m\left(\Gm_1(x)\right)=D_m(0) = 0
 \qquad\text{for every }m.
 \tag{13}\label{eq.Amx-zero}
\]
Equations \eqref{eq.Gmx-zero} and \eqref{eq.Amx-zero} show that
$x\in\Lie_4$.

It remains to prove that $x$ is not generated by the Kirchhoff
differences.  Suppose the contrary.  Every element of the Lie algebra
generated by the $\varkappa_{ij}$ is a linear combination of the
$\varkappa_{ij}$ themselves and of nested commutators of bracket-length
at least $2$.  Hence, we could write
\[
 x=\sum_{1\leq i<j\leq4}c_{ij}\varkappa_{ij}+u,
 \tag{14}\label{eq.x-assumed}
\]
where $u$ is a linear combination of nested commutators of bracket-length
at least $2$.

Apply the sign representation $\sgn:\kk[S_4]\to\kk$.  Since its target
is commutative, it annihilates every commutator.  Furthermore,
\[
 \sgn(\varkappa_{ij})=1-(-1)=2,
 \qquad
 \sgn(x)=0.
\]
The last equality also follows from \eqref{eq.x-idempotent}, since
$e_{(2,2)}$ annihilates the sign representation.  Applying $\sgn$ to
\eqref{eq.x-assumed}, we obtain
\[
 \sum_{1\leq i<j\leq4}c_{ij}=0.
 \tag{15}\label{eq.coefficient-sum-zero}
\]

Now let
\[
 \tau_{(2,2)}:\kk[S_4]\longrightarrow\kk,
 \qquad
 a\longmapsto\chi^{(2,2)}(a) = \Tr_{S^{(2,2)}}(a).
\]
We have $\tau_{(2,2)}(u) = 0$, since the trace of every commutator is $0$.
Since $\chi^{(2,2)}((ij))=0$, we have
\[
 \tau_{(2,2)}(\varkappa_{ij})
 =\dim S^{(2,2)}-\chi^{(2,2)}((ij))
 =2.
 \tag{16}\label{eq.trace-kappa}
\]
Consequently, \eqref{eq.x-assumed}, \eqref{eq.coefficient-sum-zero},
and \eqref{eq.trace-kappa} would imply
\[
 \tau_{(2,2)}(x)=0.
 \tag{17}\label{eq.false-trace}
\]
On the other hand, \eqref{eq.x-idempotent} shows that $x$ acts as
$12\id$ on the two-dimensional module $S^{(2,2)}$.  Hence,
\[
 \tau_{(2,2)}(x)=12\cdot2=24,
\]
contradicting \eqref{eq.false-trace}.  This proves the proposition.
\end{proof}

\section{The general obstruction}

The preceding counterexample is the smallest instance of a general
obstruction.

\begin{proposition}\label{prop.general-obstruction}
Let $\lambda\vdash n$ be a non-hook partition, and let $e_\lambda$ be
the corresponding primitive central idempotent of $\kk[S_n]$.  Then
\[
 e_\lambda\in\Lie_n,
\]
but $e_\lambda$ does not belong to the Lie algebra generated by the
$\varkappa_{ij}$.

More generally, the cosets in the vector-space quotient
$\Lie_n/\HLie_n$ of the elements
\[
 \left(e_\lambda\right)_{\substack{\lambda\vdash n;\\
                                   \lambda\text{ is a non-hook}}}
\]
are linearly independent.
\end{proposition}

\begin{proof}
Proposition~\ref{prop.hook-exterior} shows that every exterior
power $\Lambda^mV$ of $V$ is isomorphic to the Specht module
of a hook shape.
Thus, since $\lambda$ is a non-hook, $e_\lambda$ annihilates every exterior
power $\Lambda^mV$.
That is, $\Gm_m(e_\lambda)=0$ for every $m$.  In
particular, $\Gm_1(e_\lambda)=0$, so
Lemma~\ref{lem.Am-Dm} gives
\[
 \Am_m(e_\lambda)=D_m\left(\Gm_1(e_\lambda)\right)=0 = \Gm_m(e_\lambda).
\]
Hence, $e_\lambda\in\Lie_n$.

It remains to prove the linear-independence claim (since
$e_\lambda \notin \HLie_n$ will then immediately follow).  Suppose that
\begin{align}
 y=\sum_{\substack{\lambda\vdash n;\\
                   \lambda\text{ is a non-hook}}}
 b_\lambda e_\lambda
 \tag{18a}\label{eq.y-1}
\end{align}
belongs to the Lie algebra generated by the $\varkappa_{ij}$.  We can
then write
\[
 y=\sum_{1\leq i<j\leq n}c_{ij}\varkappa_{ij}+u,
 \tag{18b}\label{eq.y-assumed}
\]
where $u$ is a linear combination of nested commutators of bracket-length
at least $2$.  Every non-hook central idempotent annihilates the sign
representation, so $\sgn(y)=0$.  Applying the sign representation to
\eqref{eq.y-assumed} thus yields
\[
 \sum_{1\leq i<j\leq n}c_{ij}=0,
 \tag{19}\label{eq.general-coefficient-sum}
\]
since $\sgn(u) = 0$ (because $\sgn$ sends any commutator to $0$).

Fix a non-hook partition $\mu\vdash n$.  Taking the trace in $S^\mu$
in \eqref{eq.y-1}, we find
\[
\Tr_{S^\mu}(y)
=\sum_{\substack{\lambda\vdash n;\\
                   \lambda\text{ is a non-hook}}}
 b_\lambda \Tr_{S^\mu}(e_\lambda)
= b_\mu f^\mu,
\]
where $f^\mu = \dim S^\mu$
(since $e_\mu$ acts as the identity on $S^\mu$, whereas
every $e_\lambda$ with $\lambda \neq \mu$ acts as zero
on $S^\mu$).
Thus,
\begin{align*}
 b_\mu f^\mu
 &=\Tr_{S^\mu}(y)\\
 &=\sum_{i<j} c_{ij}
 \Tr_{S^\mu}(\varkappa_{ij}) + \Tr_{S^\mu}(u)
 \qquad \left(\text{by \eqref{eq.y-assumed}}\right)
 \\
 &=\sum_{i<j} c_{ij}
 \Tr_{S^\mu}(\varkappa_{ij})
 \qquad \left(\text{since $\Tr_{S^\mu}$ of any commutator is $0$}\right)
 \\
 &=\sum_{i<j}c_{ij}
 \left(f^\mu-\chi^\mu((12))\right)
\end{align*}
(since each $i<j$ satisfies
$\Tr_{S^\mu}(\varkappa_{ij})
= \Tr_{S^\mu}(1 - (ij))
= \Tr_{S^\mu}(1) - \Tr_{S^\mu}((ij))
= \chi^\mu(1) - \chi^\mu((ij))
= f^\mu - \chi^\mu((12))$
due to the conjugacy-invariance of $\chi^\mu$).
This rewrites as
\begin{align*}
 b_\mu f^\mu
 &=\left(f^\mu-\chi^\mu((12))\right)
 \sum_{i<j}c_{ij}\\
 &=0
 \qquad \left(\text{by \eqref{eq.general-coefficient-sum}}\right).
\end{align*}
Since $\kk$ has characteristic $0$, the
scalar $f^\mu$ is nonzero in $\kk$, and therefore $b_\mu=0$.  Since
$\mu$ was arbitrary, all the $b_\lambda$ vanish.
This proves the linear-independence claim and thus completes the proof.
\end{proof}

\begin{corollary}
Let $p(n)$ be the number of all partitions of $n$.
There are at least $p(n)-n$ linearly independent obstructions to the
Burman--Kulishov conjecture.
\end{corollary}

\begin{proof}
There are precisely $n$ hook partitions of $n$, namely
$h_0,h_1,\ldots,h_{n-1}$.  Thus, there are $p(n)-n$ non-hook
partitions, and Proposition~\ref{prop.general-obstruction} applies.
\end{proof}

\section{What the Kirchhoff differences do generate}

The Lie algebra generated by the Kirchhoff differences can be expressed
in terms of the well-studied Lie algebra generated by the transpositions.
Let
\[
 \GLie_n=\operatorname{Lie}\left\langle(ij)\ \middle|\
 1\leq i<j\leq n\right\rangle\subseteq\kk[S_n],
\]
let $\GLie_n'=[\GLie_n,\GLie_n]$, and set
\[
 N=\binom{n}{2},
 \qquad
 T=\sum_{1\leq i<j\leq n}(ij).
\]
Recall that $\HLie_n$ denotes the Lie algebra generated by the
$\varkappa_{ij}$.

\begin{proposition}\label{prop.generated-algebra}
One has
\[
 \HLie_n=\kk(N\cdot1-T)\oplus\GLie_n'.
 \tag{20}\label{eq.generated-algebra}
\]
\end{proposition}

\begin{proof}
The sign representation annihilates $\GLie_n'$ but sends
$N\cdot1-T$ to $2N\neq0$.  Hence, the sum in
\eqref{eq.generated-algebra} is direct.
It remains to prove that this sum equals $\HLie_n$.

First,
\[
 N\cdot1-T=\sum_{1\leq i<j\leq n}\varkappa_{ij}
 \in\HLie_n.
 \tag{21}\label{eq.central-kappa-sum}
\]
For a transposition $s$, we furthermore have
\[
 (1-s)-\frac{N\cdot1-T}{N}
 =-\left(s-\frac{T}{N}\right).
 \tag{22}\label{eq.centered-transposition}
\]
Let
\[
 p:\kk[S_n]\longrightarrow Z(\kk[S_n]),
 \qquad
 p(a)=\frac{1}{n!}\sum_{\sigma\in S_n}\sigma a\sigma^{-1}
\]
be the averaging projection onto the center.  Since the transpositions
form a single conjugacy class, $p(s)=T/N$ for every transposition $s$.
Moreover, the Lie algebra $\GLie_n$ is reductive with center $\kk T$,
and $\GLie_n'=\ker\left(p|_{\GLie_n}\right)$; see
\cite[Proposition~1]{Marin07}.  Hence, each centered transposition
$s-T/N$ belongs to $\GLie_n'$.  Conversely, replacing each
transposition $s$ with $s-T/N$ does not change any bracket, since $T$
is central.  Thus, the centered transpositions $s-T/N$ generate $\GLie_n'$.
It now follows from \eqref{eq.centered-transposition} that
$\GLie_n'\subseteq\HLie_n$.
Combined with \eqref{eq.central-kappa-sum}, this yields
$\kk(N\cdot1-T)\oplus\GLie_n' \subseteq \HLie_n$.

It remains to prove the reverse inclusion.
Recall that $\HLie_n$ is spanned by the Kirchhoff differences
$1-s$ (for $s$ a transposition) and their nested commutators.
But every nested commutator involving the elements $1-s$ is the same,
up to the expected signs, as the corresponding nested commutator involving the
transpositions $s$, and hence belongs to $\GLie_n'$.  Meanwhile,
the differences $1-s$ themselves are linear combinations of $N\cdot1-T$
and the centered transpositions (which belong to $\GLie_n'$),
so they also belong to $\kk(N\cdot1-T)\oplus\GLie_n'$.
This proves \eqref{eq.generated-algebra}.
\end{proof}

Marin completely determined $\GLie_n'$ in \cite[Theorem~A]{Marin07}.
In particular, the hook representations contribute a single factor
$\slie(V)$, acting on the other hooks through the exterior-power maps
$D_m$.  A pair of distinct conjugate non-hook partitions contributes a
single special-linear factor, diagonally embedded in the two Wedderburn
blocks, while a self-conjugate non-hook partition contributes an
orthogonal or symplectic factor.  This is generally much smaller than
the independent full general-linear factors in
\eqref{eq.Lie-structure}.

For $n=4$, Theorem~\ref{thm.structure} and
Proposition~\ref{prop.generated-algebra}, together with Marin's
decomposition, give
\begin{align*}
 \Lie_4&\cong\glie_3(\kk)\oplus\glie_2(\kk),
 &\dim\Lie_4&=9+4=13,\\
 \HLie_4&\cong\kk\oplus\slie_3(\kk)\oplus\slie_2(\kk),
 &\dim\HLie_4&=1+8+3=12.
\end{align*}
Thus, the explicit element $x$ of \eqref{eq.x-explicit} represents the
unique missing direction modulo $\HLie_4$.  For $n=2,3$, there are no
non-hook partitions, and the conjectured equality does hold.

% \begin{remark}
% In the current version of \texttt{sga-work.tex}, the definition of a Lie
% element says that $\Gm_m(a)=\Am_m(a)$ while leaving $m$ free.  It should
% say that this equality holds \emph{for every} $m\in\mathbb{N}$, or,
% equivalently, for every $m\in\{0,1,\ldots,n-1\}$, as in
% \cite[Definition~1.1]{BurKul23}.
% \end{remark}

\begin{thebibliography}{9}

\bibitem[BurKul23]{BurKul23}
Yurii Burman and Valeriy Kulishov,
\textit{Lie elements and the matrix-tree theorem},
Moscow Math. J. \textbf{23} (2023), no.~1, 47--58.

\bibitem[Marin07]{Marin07}
Ivan Marin,
\textit{L'alg\`ebre de Lie des transpositions},
J. Algebra \textbf{310} (2007), 742--774;
\href{https://arxiv.org/abs/math/0502119}{arXiv:math/0502119}.

\end{thebibliography}

\end{document}
