A new presentation for Specht modules with distinct parts
Friedmann]Tamar Friedmann
Department of Mathematics, Colby College

Hanlon]Phil Hanlon
Department of Mathematics, Dartmouth College

Wachs]Michelle L. Wachs$^*$
Department of Mathematics, University of Miami
$^{*}$Supported in part by NSF grant
DMS 2207337.

We obtain a new presentation for Specht modules whose conjugate shapes have strictly decreasing parts by introducing a linear operator on the space generated by column tabloids. The generators of the presentation are column tabloids and the relations form a proper subset of the Garnir relations of Fulton.  The results in this paper  generalize  earlier results of the authors and Stanley on Specht modules of staircase shape.

§ INTRODUCTION

The Specht modules $S^\lambda$, where $\lambda$ is a partition of $n$, give a complete set of  irreducible
representations of the symmetric group $\sg_n$ over a field  of characteristic $0$, say $\mathbb C $.  They can be constructed as
subspaces of the regular representation $\mathbb C \sg_n$  or  as presentations given in terms of generators and relations,
known as Garnir relations.   This paper deals primarily with the latter type of construction.

Let $\lambda= (\lambda_1 \geq\dots \geq \lambda_l)$ be a partition of $n$.  A Young tableaux of shape $\lambda$ is a filling of the Young diagram of shape $\lambda$ with distinct entries from the set $[n]:=\{1,2,\dots,n\}$.  Let $\mathcal T_\lambda$ be the set of  Young tableaux of shape

To construct the Specht module as a submodule of the regular representation, one can use  Young symmetrizers.  For $t \in \mathcal T_\lambda$, the Young symmetrizer is defined by
$$e_t:=\sum_{\alpha \in R_{t}}  \alpha  \\  \sum_{\beta \in C_t} \sgn(\beta) \beta ,$$

where $C_t$ is the column stabilizer of $t$ and $R_t$ is the row stabilizer.
The Specht module $S^{\lambda}$ is the submodule of the regular representation  $\C \sg_{n}$ spanned by $\{\tau e_t : \tau \in \sg_{n} \} $.

To construct the Specht module as a presentation, one can use column tabloids and Garnir relations. Let $M^\lambda$ be the vector space (over $\C$) generated by
$\mathcal T_\lambda$ subject only to column relations, which are of
the form $t+s$, where $s\in \mathcal T_\lambda$ is obtained from $t\in \mathcal T_\lambda$ by switching two
entries in the same column.
Given $t \in \mathcal T_\lambda$, let $\bar t$ denote the coset of $t$
in $M^\lambda$. These cosets, which are called column tabloids, generate $M^{\lambda}$.
A Young tableau is column strict if the entries of each of its columns increase from top to bottom.  Clearly,
$\{\bar t  :  t \mbox{ is a column strict Young tableau of shape } \lambda\}$ is a basis for $M^\lambda$.

$$ \ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 3 & \scriptstyle 5  \\
\scriptstyle 1 &  \scriptstyle 4  \\
\scriptstyle 2 \\
\end{ytableau}}
\,\, = \,\, - \,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 5  \\
\scriptstyle 3 &  \scriptstyle 4  \\
\scriptstyle 2 \\
\end{ytableau}}
\,\, = \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 5  \\
\scriptstyle 2 &  \scriptstyle 4  \\
\scriptstyle 3 \\
\end{ytableau}}
\,\, = \,\, - \,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 4  \\
\scriptstyle 2 &  \scriptstyle 5  \\
\scriptstyle 3 \\
\end{ytableau}}\,\,,$$

There are various different presentations of the Specht module $S^\lambda$ in the literature that involve  column tabloids and Garnir relations, see eg. [4, 6]. Here we are interested in a presentation of $S^\lambda$ discussed in Fulton <cit.>.  The generators are the column tabloids $\bar t$, where $t \in \mathcal T_\lambda$.
The Garnir relations are of the form
$\bar t-\sum \bar s$, where the sum is over all $s\in \mathcal
T_\lambda$ obtained from $t\in \mathcal T_\lambda$ by exchanging any
$k$ entries of a fixed  column  with the top $k$ entries of the next
column, while maintaining the vertical order of each of the exchanged
sets.  There is a Garnir relation $g^t_{c,k}$ for every $t \in
\mathcal T_\lambda$, every column $c \in [\lambda_1-1]$,  and  every
$k$ from $1$ to the length $l_{c+1}$ of  column $c+1$.

Example: For

$$t= \ytableausetup
{mathmode, boxsize=1em}
\begin{ytableau}
\scriptstyle 1 & \scriptstyle 5 & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}\,\,,$$

we have

$$ g^t_{1,1} =
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 5 & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 5 & \scriptstyle 1  & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 2  & \scriptstyle 7\\
\scriptstyle 5 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 3 & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 5 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 4 & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 5
\end{ytableau}}

$$ g^t_{1,2} =
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 5 & \scriptstyle 7 \\
\scriptstyle 2 &  \scriptstyle 6  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 5 & \scriptstyle 1 & \scriptstyle 7 \\
\scriptstyle 6 &  \scriptstyle 2  \\
\scriptstyle 3 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 5 & \scriptstyle 1 & \scriptstyle 7\\
\scriptstyle 2 &  \scriptstyle 3  \\
\scriptstyle 6 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 5 & \scriptstyle 1 & \scriptstyle 7\\
\scriptstyle 2 &  \scriptstyle 4  \\
\scriptstyle 3 \\
\scriptstyle 6
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 2  & \scriptstyle 7\\
\scriptstyle 5 &  \scriptstyle 3  \\
\scriptstyle 6 \\
\scriptstyle 4
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 2  & \scriptstyle 7\\
\scriptstyle 5 &  \scriptstyle 4  \\
\scriptstyle 3 \\
\scriptstyle 6
\end{ytableau}}
\,\,- \,\,
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 3  & \scriptstyle 7\\
\scriptstyle 2 &  \scriptstyle 4  \\
\scriptstyle 5 \\
\scriptstyle 6
\end{ytableau}}\, \, .

Let $G^{\lambda}$ be the subspace of $M^\lambda$ generated by the Garnir relations in

\begin{equation} \label{garnireq} \{ g^t_{c,k} : c \in [\lambda_1-1], k \in [l_{c+1}], t \in \mathcal T_\lambda \}  .\end{equation}

The symmetric group $\sg_n$ acts on $\mathcal T_\lambda$ by replacing
each entry of a tableau by its image under the permutation in $\sg_n$.
This induces a representation of $\sg_n$ on $M^\lambda$. Clearly $G^\lambda$ is invariant under the action of
The  presentation of $S^\lambda$ obtained in Section 7.4 of [4] is given by
M^λ/G^λ≅__n S^λ.

On page 102 (after Ex. 15) of [4], a presentation of $S^\lambda$ with a smaller set of relations is given.  In this presentation, the index $k$ in $g^t_{c,k}$ of (<ref>) is restricted to a single value:  $k=\min [l_{c+1}]= 1$. More precisely, the presentation is
M^λ / G^λ,min ≅__n S^λ ,where
$G^{\lambda,\min}$ is the subspace of $G^\lambda$ generated by the subset of Garnir relations,
$$\{ g^t_{c,1} : c \in [\lambda_1-1], t \in \mathcal T_\lambda \}  .$$

In this paper we obtain an analogous presentation of $S^\lambda$, also with a smaller set of relations, when the conjugate $\lambda^*$ has distinct parts. The index $k$ in $g^t_{c,k}$ of (<ref>) is again restricted to a single value, but now it's the maximum value:  $k=\max [l_{c+1}]= l_{c+1}$.
The presentation
is given in our main result:
Let $\lambda$ be a partition whose conjugate has distinct parts. Then as $\sg_n$-modules,

\begin{equation} \label{maineq} M^{\lambda} / G^{\lambda,\max} \cong S^{\lambda} ,\end{equation}
$G^{\lambda,\max}$ is the subspace of $G^\lambda$ generated by
$$\{ g^t_{c,l_{c+1}} : c \in [\lambda_1-1], t \in \mathcal T_\lambda \}  .$$

Moreover, we can further reduce this set of relations by restricting $t$ to the set of column strict tableaux.

Theorem <ref> generalizes  a result in an earlier paper of the authors and Stanley [3] (see also [2]) for staircase partitions,  i.e. partitions whose conjugate  has the form
$(k,k-1,k-2,\dots,k-r)$.  Our proof  generalizes  the proof for staircase partitions,  which relied on a computation of the
eigenvalues and eigenspaces of a certain linear operator on $M^{2^{n-1}1}$.
The work in [2, 3] was presented in the language of the generalized Jacobi relations that define  the LAnKe or
Filippov algebra. An observation in   [2, 3], that the  restricted class of Garnir relations that generate $G^{\lambda,\max}$  (for staircase partitions $\lambda$) corresponds to the generalized Jacobi relations,  led to the  work in the current paper.

The method of introducing a linear operator on the space of column tabloids was subsequently used in  [1]  to obtain a different presentation of $S^\lambda$ with a reduced number of relations. This presentation works for all
shapes, but rather than using a subset of the Garnir relations,  it uses a set consisting of sums of  the Garnir relations that
generate $G^{\lambda, \min}$.

In the next section we prove  a slightly more general version of  Theorem <ref>  along with a converse by defining a linear operator  for two-column shapes $2^m1^{n-m}$, which generalizes the $m=n-1$ case in [3], and then studying its eigenvalues and eigenspaces.

§ THE PROOF OF THEOREM <REF>: REDUCTION TO THE TWO-COLUMN CASE

Theorem <ref> holds more generally when we allow the conjugate $\lambda^*$ to have multiple parts equal to $1$, while still requiring the parts greater than $1$ to be distinct.   Note that a partition $\lambda=(\lambda_1\ge \lambda_2 \ge  \dots\ge  \lambda_l)$ meets this requirement if  and only if $\lambda_i -\lambda_{i+1} \le 1$ for all $i=2,\dots l-1$ and $\lambda_{l} =1$. For example,  $\lambda = (5,4,2,1,1,1)^*= (6,3,2,2,1)$ meets this requirement, but  $\lambda = (4,3,2,2)^*= (4,4,2,1)$ does not.  The following result implies Theorem <ref>.

Let $\lambda = (\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_\ell)$ be a partition of $n$.  If  $\lambda_i -\lambda_{i+1} \le 1$ for each $i = 2,\dots,l-1$ and $\lambda_l=1$ then  as $\sg_n$-modules,

\begin{equation} \label{maineq} M^{\lambda} / G^{\lambda,\max} \cong S^{\lambda} ,\end{equation}
$G^{\lambda,\max}$ is the subspace of $G^\lambda$ generated by
$$\{ g^t_{c,l_{c+1}} : c \in [\lambda_1-1], t \in \mathcal T_\lambda \}  .$$

Moreover, we can further reduce this set of relations by restricting $t$ to the set of column strict tableaux.  Conversely, if (<ref>) holds then $\lambda_i -\lambda_{i+1} \le 1$ for each $i = 2,\dots,l-1$ and $\lambda_l=1$.

It  follows from Fulton's presentation given in (<ref>) that we need only  prove Theorem <ref> for shapes  with   just two columns.  Indeed, let $\lambda$ be a partition of $n$. By (<ref>) $M^\lambda / G^{\lambda, \max} = S^\lambda$ if and only if $G^{\lambda, \max} = G^\lambda$.
Since $G^{\lambda, \max}$ equals the direct sum over all columns $ c$ of the subspaces spanned by  $\{g^t_{c,l_{c+1}}:  t \in \mathcal T_\lambda\}$, and $G^{\lambda}$ equals the direct sum over all columns $ c$ of the subspaces spanned by  the larger set $\{g^t_{c,k}: k \in [l_{c+1}], t \in \mathcal T_\lambda\}$, we have that $G^{\lambda, \max} = G^\lambda$ if and only if

\begin{equation} \label{twocoleq} \langle g^t_{c,l_{c+1}}:  t \in \mathcal T_\lambda\rangle= \langle g^t_{c,k}: k \in [l_{c+1}], t \in \mathcal T_\lambda\rangle\end{equation}
for all columns $c$.
By applying (<ref>)  to the  conjugate of $(l_c,l_{c+1})$ we have that (<ref>) holds if and only if

$$M^{(l_c,l_{c+1})^*} / G^{(l_c,l_{c+1})^*,\max} = M^{(l_c,l_{c+1})^*} / \langle g^t_{c,l_{c+1}}:  t \in \mathcal T_{(l_c, l_{c+1} )^*}\rangle = S^{(l_c,l_{c+1})^*}.$$

Putting all this together yields $M^\lambda / G^{\lambda, \max} = S^\lambda$ if and only if
$$M^{(l_c,l_{c+1})^*} / G^{(l_c,l_{c+1})^*,\max} = S^{(l_c,l_{c+1})^*}$$
for all columns $c$ of $\lambda$.  Hence our claim that we  need only consider two column shapes holds.

For the remainder of this section we consider only two-column shapes $2^m1^{n-m}$.
For $n \ge m$, let $V_{n,m} := M^{2^m1^{n-m}}$. For each $n$-element subset $S$ of $[n+m]$,  let $t_S$ be  the column strict Young tableau of shape $2^m1^{n-m}$ whose first column consists of the elements of $S$ and   second column consists of the elements of $[n+m] \setminus S $, and let $v_S := \bar t_S$.

\begin{equation} \label{basiseq} \left \{ v_S : S \in \binom{[n+m]}{n}\right \}\end{equation}

is a basis for $V_{n,m}$.  Thus $V_{n,m}$ has dimension $\binom{n+m}{n}$.

For each $S \in \binom{[n+m]}{n}$,  let $g_S$ denote the Garnir relation $g^{t_S}_{1,m}$, that is

$$g_S :=  v_S \, -\sum \bar t ,$$
where the sum is over all tableaux
$t$ obtained from $t_S$ by  choosing an $m$-subset $S^\prime$ of $S$ and for each $i=1,2,\dots,m$, exchanging the $i$th smallest  element of $S^\prime$  (which is in column 1 of $t_S$) with the $i$th element of column 2 of $t_S$.
For example, if $n=3$ and $m=2$  then

\begin{align*}g_{\{2,4,5\}} &=
\ytableausetup
{mathmode, boxsize=1em}
\overline{\begin{ytableau}
\scriptstyle 2 & \scriptstyle 1  \\
\scriptstyle 4 &  \scriptstyle 3  \\
\scriptstyle 5
\end{ytableau}}
\,\,- \,\,
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 2  \\
\scriptstyle 3 &  \scriptstyle 4  \\
\scriptstyle 5
\end{ytableau}}
\,\, - \,\,
\overline{\begin{ytableau}
\scriptstyle 1 & \scriptstyle 2  \\
\scriptstyle 4 &  \scriptstyle 5  \\
\scriptstyle 3
\end{ytableau}}
\,\, - \,\,
\overline{\begin{ytableau}
\scriptstyle 2 & \scriptstyle 4  \\
\scriptstyle 1 &  \scriptstyle 5  \\
\scriptstyle 3
\end{ytableau}} \,\,.
\\
&= v_{\{2,4,5\} }-v_{\{1,3,5\}} +v_{\{1,3,4\}} +v_{\{1,2,3\}}.
\end{align*}

For all $v \in V_{n,m}$ and $S \in \binom{[n+m]}n$, let $\langle v,v_S \rangle$ denote the coefficient of $v_S$ in the expansion of $v$ in the basis given in (<ref>).  Also let $p:= n-m$. Then for all $S,T \in \binom{[n+m]}{n}$,

$$\langle g_S,v_T \rangle = \left\{
\begin{array}{l l}

1& \quad  \text{if } S=T  \\
\\

(-1)^{d_1+ \dots + d_{p} +\binom{p +1}2+1}&  \quad  \text{if } S\cap T = \{d_1,\dots, d_{p}\}\\
\\
0 & \quad \text{if } S\neq T \text{ but } |S\cap T| >p . \
\end{array} \right.$$

It is easy to see that  the only possible cases are: $S=T$, $|S\cap T| = p$  and $|S\cap T|  > p$.
It is straightforward that the  result for the first and third cases holds.  For the second case, let $S = \{a_1,a_2,\dots,a_n \}$
and $[n+m] \setminus S = \{b_1,b_2,\dots,b_m \}$, where the $a$'s and $b$'s are listed in increasing order.
Then the terms of $g_S$  other than $v_S$ are of the form $-\bar t$, where $t$ is a tableau whose   first column listed from top
to bottom is
$$b_1, \dots, b_{i_1-1}, a_{i_1}, b_{i_1}, \dots, b_{i_2-2}, a_{i_2}, b_{i_2-1},\dots,
b_{i_p-p}, a_{i_p}, b_{i_p-p+1}, \dots ,b_m $$
with  $1\le i_1<i_2<\dots<i_p\le n$, and whose second column is $[n+m] \setminus T$ listed in increasing order,
where $T=\{  a_{i_1}, \dots, a_{i_p}, b_1, \dots, b_m\}$.
Clearly $\bar t$ is equal to $(-1)^{i_1+\dots+i_p-\binom {p+1}2}$ times the tabloid   whose first column from top to bottom is

$$ a_{i_1}, \dots, a_{i_p}, b_1, \dots, b_m$$
and whose second column is $[n+m] \setminus T$ listed in increasing order.  This tabloid is equal to
$(-1)^{j_1+j_2+\dots+j_p} v_T$, where  each  $j_k$ is the number of $b$'s  that
are less than $a_{i_k}$.   Since
$i_k + j_k= a_{i_k}$ for each $k$,  we have

$$\langle g_S, v_T\rangle = (-1)^{a_{i_1} + a_{i_2} + \dots +a_{i_p}-\binom{p+1}2 +1}.$$

Let $G_{n,m}$ be the subspace of $V_{n,m}$ generated by $\{g_S : S \in \binom{[n+m]} n\}$.  It is not difficult to see that $G_{n,m}$ is invariant under the action of $\sg_{n+m}$ on $V_{n,m}$. The two column case of  Theorem <ref> can now be stated as follows.
Let $m \le n$. Then  as $\sg_{n+m}$-modules,

$$V_{n,m}/G_{n,m} \cong S^{2^m1^{n-m}}$$
if and only if $m <n$ or $m=n=1$.

We will  introduce a linear operator that will enable us to prove this result (and thereby prove  Theorems <ref> and <ref>).  First we note that
due to the column relations, as  an $\sg_{n+m}$–module, $V_{n,m}$ is isomorphic to the representation of $\sg_{n+m}$ induced from the sign representation of the Young subgroup $\sg_n \times \sg_m$:

\[ V_{n,m} \cong_{\sg_{n+m}}  (\sgn_n \times \sgn_{m} ) \uparrow _{\sg_n \times \sg_{m}}^{\sg_{n+m}}.  \]

Hence by  Pieri's rule,

\begin{equation} \label{freeeq} V_{n,m} \cong_{\sg_{n+m}} \bigoplus _{i=0}^{m} S^{2^i1^{n+m-2i}}.\end{equation}

Now consider the linear operator $\varphi: V_{n,m} \to V_{n,m}$ defined on  basis elements by

$$\varphi(v_S) = g_S.$$
It is not difficult to see that $\varphi$ is an $\sg_{n+m}$-module homomorphism whose image is $G_{n,m}$.

Let $m \le n$.
The operator $\varphi$ acts as multiplication by a scalar on each irreducible submodule of $V_{n,m}$.
Moreover, as  $\sg_{n+m}$–modules,
\begin{equation} \label{kereq} \nullsp \varphi \cong V_{n,m} / G_{n,m}. \end{equation}

This follows from the fact that $V_{n,m}$ is multiplicity-free, as indicated in  (<ref>), and Schur's Lemma.

It follows from (<ref>) that to prove Theorem <ref>, we need only show that the null space of $\varphi$ is isomorphic to $S^{2^m1^{n-m}}$ if and only if $m <n$  or $m=n=1$.  This is handled by the following theorem and its corollary.

Let $m \le n$.
The operator $\varphi:V_{n,m} \to V_{n,m}$ acts as multiplication by the scalar
w _i := 1-n-i m-i(-1)^m-i,on the irreducible submodule  isomorphic to $S^{2^i1^{(n+m)-2i}}$ for $i=0,1,\dots,m$.

* For $m<n$, the operator $\varphi$ has $m+1$ distinct eigenvalues $w_0, w_1,\dots,w_m$.
Moreover, if $E_i$ is the eigenspace corresponding to $w _i$ then as $\sg_{n+m}$-modules,

$$E_i\cong S^{2^i1^{(n+m)-2i}}$$

for each $i=0,1, \ldots , m$. Consequently,

\begin{equation} \label{nulleq} \nullsp \varphi \,\, \cong_{\sg_{n+m}}\,\, S^{2^m1^{n-m}}.
\end{equation}

* For $m=n$ the operator $\varphi$ has the eigenvalues $0$ and $2$, with

$$ w_i=
\begin{cases}
0 & \text{if } n-i \text{ is even}, \\
2& \text{if } n-i \text{ is odd},
\end{cases}

for each $i=0, \ldots , n$. Consequently,
φ   ≅__n+m     ⊕^n_
c i=0
n-i  even  S^2^i1^2n-2i,
and the eigenspace with eigenvalue 2 is given by

\[  \bigoplus^n_
{\begin{subarray}{c} i=0 \\ n-i \text{ odd} \end{subarray}} S^{2^i1^{2n-2i}}.
\]

By Lemma <ref>, $\varphi$ acts as a scalar on each irreducible submodule.
To compute the scalar, we start by letting $t$ be the standard Young tableau of shape $2^i1^{n+m-2i}$ given by

$$t\, = \,
\ytableausetup
{mathmode, boxsize=2em}
\begin{ytableau}
\scriptstyle1 &\scriptstyle n+1  \\
\scriptstyle 2 & \scriptstyle n+2  \\
\vdots & \vdots \\
\scriptstyle i &\scriptstyle n+i\\
\scriptstyle i + 1  \\
\vdots \\
\scriptstyle n \\
\scriptstyle n+i+1\\
\vdots \\
\scriptstyle n+m
\end{ytableau}\hskip .5cm.

Let $C_t$ be the column stabilizer of $t$ and let $R_t$ be the row stabilizer.
Recall that the Young symmetrizer associated with $t$ is defined by
$$e_t:=\sum_{\alpha \in R_{t}}  \alpha  \\  \sum_{\beta \in C_t} \sgn(\beta) \beta $$

and that the Specht module $S^{2^i1^{n+m-2i} }$ is the submodule of the regular representation  $\C \sg_{n+m}$ spanned by $\{\tau e_t : \tau \in \sg_{n+m} \} $.

Now set $T:=[n]$, $r_t:= \sum_{\alpha \in R_t} \alpha$ and factor
$$\sum_{\beta \in C_t} \sgn(\beta) \beta = f_t d_t,$$
where $d_t$ is the signed sum of permutations in $C_t$ that stabilize $\{1, 2, \ldots , n\}$, $\{ n+1, \ldots , n+i\}$, $\{ n+i+1, \ldots , n+m\}$ and $f_t$ is the signed sum of permutations in $C_t$ that maintain the vertical order of these sets.
So $e_t v_T =  r_t f_t d_t v_T$. Because of the antisymmetry of the columns, we have

\begin{align} \non d_t v_T &=n!\, ((n+i)-(n+1)+1)!\, ((n+m)-(n+i+1)+1)!\, \,\, v_T
\\ \non &=n!\,i!(m-i)! \,\,\, v_T \, .
\end{align}

Hence $r_t f_t v_T$  is a  scalar multiple of $e_t v_T$ .  Since   the coefficient of $v_T$ in the expansion of $r_t f_t v_T$ is  $1 $,  we have $e_t v_T \ne 0$.

Let $\psi:\C \sg_{n+m} \to V_{n,m}$ be the $\sg_{n+m}$-module
homomorphism defined  by $\psi(\sigma) = \sigma v_T$, where $\sigma
\in \sg_{n+m}$ and $T:=[n]$.   Now consider the  restriction of $\psi$
to the Specht module  $S^{2^i1^{m+n-2i} }$.    By the irreducibility
of  the Specht module and  the fact that $e_t v_T \ne 0$, this
restriction  is  an  isomorphism from  $S^{2^i1^{n+m-2i} }$ to the
subspace of $V_{n,m}$ spanned by $\{\tau e_t v_T : \tau \in \sg_{n+m}
\} $.  This subspace is therefore the unique subspace of $V_{n,m}$
isomorphic to $S^{2^i1^{(n+m)-2i}}$.  From here on, we will abuse
notation by letting $S^{2^i1^{(n+m)-2i}}$ denote the subspace of
$V_{n,m}$ spanned by $\{\tau e_t v_T : \tau \in \sg_{n+m} \} $.

Since $r_t f_t v_T$ is a scalar multiple of $e_tv_T$, it is in $S^ {2^i1^{(n+m)-2i}}$.  It follows that

$$\varphi(r_t f_t v_T ) = c r_t f_t v_T ,$$
for some scalar $c$, which we want to show equals $w_i$.
Using the fact that the coefficient of $v_T$ in $r_t f_t v_T$ is 1, we conclude that
$c$  is the coefficient of $v_T$ in $\varphi(r_t f_t v_T)$.  Hence to complete the proof we need only show that
⟨φ(r_t f_t v_T), v_T ⟩= w_i := 1-n-i m-i(-1)^m-i.Consider the expansion,

$$r_t f_t v_T = \sum_{S \in \binom{[n+m]}{n}} \langle r_t f_t v_T , v_S\rangle v_S,$$

which by linearity yields,

$$\varphi(r_t f_t v_T )= \sum_{S \in \binom{[n+m]}{n}} \langle r_t f_t v_T , v_S\rangle \varphi(v_S).$$

Hence the coefficient of $v_T$ is given by

$$\langle \varphi(r_t f_t v_T), v_T \rangle = \sum_{S \in \binom{[n+m]}{n}} \langle r_t f_t v_T , v_S\rangle \langle\varphi(v_S),v_T\rangle.$$

Looking back at Lemma <ref>, we see that the   $S=T$ term is $1$, which yields,

$$\langle \varphi(r_t f_t v_T), v_T \rangle = 1+  \sum_{S \in \binom{[n+m]}{n} \setminus \{T\}} \langle r_t f_t v_T , v_S\rangle \langle\varphi(v_S),v_T\rangle.$$

To get a contribution from an $S \ne T$ term, by Lemma <ref>, we must have $|S \cap T| = p$.  Let $S \cap T= \{d_1,d_2,\dots,d_{p}\}$  in which case $ \langle\varphi(v_S),v_T\rangle = (-1)^{d_1+ \dots + d_{p} +\binom {p+1}2+1}$.  Hence
⟨φ(r_t f_t v_T), v_T ⟩= 1+ ∑_D ∈[n] p (-1)^∑_d ∈D d  +p+12+1 ⟨r_t f_t v_T , v_S(D)⟩,where
$$S(D) = D \cup \{n+1,n+2,\dots,n+m\}.$$

To compute $ \langle r_t f_t v_T , v_{S(D)}\rangle$, we must consider how we get $v_{S(D)}$ from the action of permutations appearing  in $r_tf_t$ on $v_T$.
Recall that $f_t$ is a sum of column permutations $\sigma$ of $t$ (with sign) that maintain the vertical order of $ \{1, 2, \ldots , n\} $, $ \{n+1, \ldots , n+i\} $, and $ \{n+i+1, \ldots , n+m\}$.
In order to get $S(D)$, we have that $\sigma$ fixes $ 1, 2, \ldots , i $ and $n+1, \ldots , n+i$ and then the row permutation $\alpha$ is

$$\alpha = (1, n+1)(2, n+2)\cdots (i, n+i) (i+1)\cdots (n+m)$$

and $\sigma$ interchanges $\{n+i+1, \ldots , n+m\}$ with $\{ i+1, \ldots ,n\} \setminus D$, leaving $D$ in rows $i+1,\dots,n$.

Since $\sigma$ maintains the vertical order of $1,2,\ldots , n$, it must be that

Thus the summation in (<ref>) is left only with the $D=\{i+1,i+2,\dots,i+p\}$ term.
Suppose that each $i+k$ goes to row $j_k$ with $i+k\leq j_k\leq n$ and $j_1<j_2<\dots<j_p$. See Figure <ref> below.

$$ t\, = \,
\ytableausetup
{mathmode, boxsize=2.5em}
\begin{ytableau}
\scriptstyle1 &\scriptstyle n+1  \\
\scriptstyle 2 & \scriptstyle n+2  \\
\vdots & \vdots \\
\scriptstyle i &\scriptstyle n+i\\
\scriptstyle i + 1  \\
\vdots \\
\scriptstyle j_1-1 \\
\scriptstyle {\bf \color{red} j_1}\\
\scriptstyle j_1+1\\
\vdots \\
\scriptstyle j_2-1 \\
\scriptstyle {\bf \color{red} j_2}\\
\scriptstyle j_2+1\\
\vdots \\
\scriptstyle j_p-1 \\
\scriptstyle {\bf \color{red} j_p}\\
\scriptstyle j_p+1\\
\vdots \\
\scriptstyle n \\
\scriptstyle n+i+1\\
\vdots \\
\scriptstyle n+m
\end{ytableau}
\hskip 1cm \longrightarrow \hskip 1cm
\sigma t  \, = \,
\ytableausetup
{mathmode, boxsize=2.5em}
\begin{ytableau}
\scriptstyle1 &\scriptstyle n+1  \\
\scriptstyle 2 & \scriptstyle n+2  \\
\vdots & \vdots \\
\scriptstyle i &\scriptstyle n+i\\
\scriptstyle n+i+1\\
\vdots \\
\scriptstyle n+j_1-1 \\
\scriptstyle {\bf \color{red} i+1} \\
\scriptstyle n+j_1 \\
\vdots\\
\scriptstyle n+j_2-2 \\
\scriptstyle {\bf \color{red} i+2} \\
\scriptstyle n+j_2-1 \\
\vdots\\
\scriptstyle n+j_p-p \\
\scriptstyle {\bf \color{red}i+p }\\
\scriptstyle n+j_p-p+1 \\
\vdots\\
\scriptstyle n+m\\
\scriptstyle i + p+1  \\
\vdots \\
\scriptstyle n
\end{ytableau}
\hskip 1cm \longrightarrow \hskip 1cm
\alpha \sigma t  \, = \,
\ytableausetup
{mathmode, boxsize=2.5em}
\begin{ytableau}
\scriptstyle n+1 &\scriptstyle 1  \\
\scriptstyle n+2 & \scriptstyle 2  \\
\vdots & \vdots \\
\scriptstyle n+i &\scriptstyle i\\
\scriptstyle n+i+1\\
\vdots \\
\scriptstyle n+j_1-1 \\
\scriptstyle {\bf \color{red} i+1} \\
\scriptstyle n+j_1 \\
\vdots\\
\scriptstyle n+j_2-2 \\
\scriptstyle {\bf \color{red} i+2} \\
\scriptstyle n+j_2-1 \\
\vdots\\
\scriptstyle n+j_p-p \\
\scriptstyle {\bf \color{red}i+p }\\
\scriptstyle n+j_p-p+1 \\
\vdots\\
\scriptstyle n+m\\
\scriptstyle i + p+1  \\
\vdots \\
\scriptstyle n
\end{ytableau}

One can easily compute the sign of the column permutation $\sigma$ by counting inversions.  The number of inversions of $\sigma$ restricted to the first column is

\begin{eqnarray*} &  &(n-(i+1)+1-p)(n+m-(n+i+1)+1) \\ &+& (j_1-(i+1)) + (j_2 -(i+2)) + \dots +(j_p-(i+p)) \\
&=& (m-i)(m-i) + j_1 + \dots + j_p -pi -\binom{p+1}{2} \, ,\end{eqnarray*}

and the number of inversions of $\sigma$ restricted to the second column is $0$. It follows that

$$\sgn(\sigma) = (-1)^{m-i +pi -\binom{p+1}{2}+ j_1 + \dots + j_p }\, .$$

In terms of our basis, $ \alpha \sigma v_T$ has first column equal to

$$n+1,n+2,\dots, n+j_1-1, {\bf \color{red} i+1}, n+j_1,\dots, n+j_2-2, {\bf \color{red} i+2}, n+j_2-1,\dots, $$

$$n+j_p-p, {\bf \color{red} i+p},n+j_p-p+1,\dots, n+m$$
and second column equal to
$$1,2,\dots,i,i+p+1,i+p+2,\dots, n.$$

To put this basis in canonical form, we need to move $i+1, i+2,\dots, i+p$ to the top of the column, which
yields $ \alpha \sigma v_T = (-1)^{j_1+j_2+\dots +j_p - \binom{p+1}2} v_{S(i+1,i+2,\dots,i+p)}$.
Hence $\sgn(\sigma) \alpha \sigma v_T = (-1)^{m-i+pi} v_{S(i+1,i+2,\dots,i+p)}$.
Since there are $\binom{n-i} p$ subsets of positions $\{j_1,\dots,j_p\}$ where $i+1, 1+2,\dots,i+p$ might land,

$$\langle r_t f_t v_T, v_{S(i+1,i+2,\dots,i+p)} \rangle = \binom{n-i} p (-1)^{m-i+pi} .$$

Since  all other terms in the summation in (<ref>) vanish and
$$\sum_{d \in D}d = \sum_{k=1}^p (i+k) = pi + \binom {p+1}2,$$
plugging this into (<ref>), we obtain
(<ref>), which completes the proof of Theorem <ref>.

We can now use  (<ref>), (<ref>), and the $n=1$ case of (<ref>) to conclude that the “if" direction of Theorem <ref> holds.  The “only if" direction also follows from  (<ref>), (<ref>), and   (<ref>).

Final Remarks. Part (b) of Corollary <ref>  has another interesting consequence.  Indeed, the decomposition  of   $\nullsp\varphi$ given in (<ref>) yields a  decomposition of $V_{n,n}/G_{n,n}$ into irreducibles.  Since $V_{n,n}/G_{n,n}$ is clearly isomoprhic to the composition product of the trivial representation of $\sg_2$ and the sign representation  of $\sg_n$, we recover a well-known result on the decomposition of this composition product into irreducibles; see eg. <cit.>.

One may ask what happens if the  $k$ in the Garnir relations  (<ref>) is restricted to a single value other than the minimum 1 and the maximum $l_{c+1}$.   For which partitions $\lambda$ do such Garnir relations generate the entire set of  Garnir relations? Is there a
generalization of Theorem <ref> that answers this question?  We leave these questions open.[Motivated by an earlier version of our paper,
Maliakas and Metzaki address these questions in a recent preprint.]

We are grateful to Mihalis Maliakas for pointing out an error in the statement of Theorem <ref> in an earlier version of this paper.

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