# The Lie superalgebra of transpositions

Christopher M. Drupieski  Department of Mathematical Sciences, DePaul
University, Chicago, IL 60614, USA<EMAIL_ADDRESS>and  Jonathan R.
Kujawa  Department of Mathematics
University of Oklahoma
Norman, OK 73019, USA<EMAIL_ADDRESS>
###### Abstract.

We consider the group algebra of the symmetric group as a superalgebra, and
describe its Lie subsuperalgebra generated by the transpositions.

###### 1991 Mathematics Subject Classification:

Primary 17B10. Secondary 20B30.

CMD was supported in part by Simons Collaboration Grant for Mathematicians No.
426905.

JRK was supported in part by Simons Collaboration Grant for Mathematicians No.
525043.

##  1\. Introduction

###  1.1. Questions of WunderNatur

This paper answers a series of questions originally posed by the MathOverflow
user WunderNatur in August 2022 [12]: Considering the group algebra of the
symmetric group $\mathbb{C}S_{n}$ as a superalgebra (by considering the even
permutations in $S_{n}$ to be of even superdegree and the odd permutations in
$S_{n}$ to be of odd superdegree), and considering $\mathbb{C}S_{n}$ as a Lie
superalgebra via the super commutator,

$[x,y]=xy-(-1)^{\overline{x}\cdot\overline{y}}yx,$

what is the structure of $\mathbb{C}S_{n}$ as a Lie superalgebra, and what is
the structure of the Lie subsuperalgebra of $\mathbb{C}S_{n}$ generated by the
transpositions? The non-super analogues of these questions were previously
answered by Marin [11]; we describe Marin’s motivation in Section 1.3.

###  1.2. Main results and methods

Fix an integer $n\geq 2$, and let $\mathcal{P}(n)$ denote the set of all
integer partitions of $n$. Given $\lambda\in\mathcal{P}(n)$, write
$\lambda^{\prime}$ for the partition that is conjugate (or transpose) to
$\lambda$, let $S^{\lambda}$ be the simple Specht module for $\mathbb{C}S_{n}$
labeled by $\lambda$, and set $f^{\lambda}=\dim_{\mathbb{C}}(S^{\lambda})$.

Let $\overline{\mathcal{P}}(n)$ be any fixed set of representatives in
$\mathcal{P}(n)$ for the equivalence relation generated by
$\lambda\sim\lambda^{\prime}$. Up to parity change, the simple
$\mathbb{C}S_{n}$-supermodules are labeled by the elements of
$\overline{\mathcal{P}}(n)$, and are described as ungraded
$\mathbb{C}S_{n}$-modules by

$W^{\lambda}=\begin{cases}S^{\lambda}&\text{if $\lambda=\lambda^{\prime}$
(Type M)},\\\ S^{\lambda}\oplus S^{\lambda^{\prime}}&\text{if
$\lambda\neq\lambda^{\prime}$ (Type Q)}.\end{cases}$

The supermodule $W^{\lambda}$ is absolutely irreducible (i.e., irreducible as
an ordinary $\mathbb{C}S_{n}$-module) if $\lambda=\lambda^{\prime}$, and is
self-associate (hence is naturally equipped with an odd involution
$J^{\lambda}:W^{\lambda}\to W^{\lambda}$) if $\lambda\neq\lambda^{\prime}$.
Since $\mathbb{C}S_{n}$ is semisimple as a superalgebra, the graded version of
the Artin–Wedderburn theorem provides a corresponding direct sum decomposition
of $\mathbb{C}S_{n}$ into matrix superalgebras of types M and Q:

$\mathbb{C}S_{n}\cong\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda=\lambda^{\prime}\end{subarray}}\operatorname{End}_{\mathbb{C}}\left(W^{\lambda}\right)\big{]}\oplus\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda\neq\lambda^{\prime}\end{subarray}}Q\left(W^{\lambda}\right)\big{]};$

see cor:CSn-as-superalgebra. This of course also describes $\mathbb{C}S_{n}$
as a Lie superalgebra:

$\mathbb{C}S_{n}\cong\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda=\lambda^{\prime}\end{subarray}}\mathfrak{gl}\left(W^{\lambda}\right)\big{]}\oplus\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda\neq\lambda^{\prime}\end{subarray}}\mathfrak{q}\left(W^{\lambda}\right)\big{]}.$

Write $\mathfrak{g}_{n}$ for the Lie subsuperalgebra of $\mathbb{C}S_{n}$
generated by the transpositions, let $T_{n}$ be the sum in $\mathbb{C}S_{n}$
of all transpositions, and let $\mathfrak{D}(\mathbb{C}S_{n})$ be the derived
subsuperalgebra of the Lie superalgebra $\mathbb{C}S_{n}$. Then
theorem:finaltheorem states that

$\mathfrak{g}_{n}=\mathfrak{D}(\mathbb{C}S_{n})+\mathbb{C}T_{n},$

where

$\mathfrak{D}(\mathbb{C}S_{n})\cong\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda=\lambda^{\prime}\end{subarray}}\mathfrak{sl}\left(W^{\lambda}\right)\big{]}\oplus\big{[}\bigoplus_{\begin{subarray}{c}\lambda\in\overline{\mathcal{P}}(n)\\\
\lambda\neq\lambda^{\prime}\end{subarray}}\mathfrak{sq}\left(W^{\lambda}\right)\big{]}.$

Here $\mathfrak{sl}(W^{\lambda})$ denotes the special linear Lie superalgebra
on the superspace $W^{\lambda}$, and $\mathfrak{sq}(W^{\lambda})$ denotes the
subspace of $\mathfrak{q}(W^{\lambda})$ of elements whose ‘odd trace’ is zero;
see (4.1.1).

The description of $\mathbb{C}S_{n}$ as a Lie superalgebra is straightforward
from the classification of the irreducible $\mathbb{C}S_{n}$-supermodules.
Sections 2 and 3 gather together various results on super representation
theory and then apply them to $\mathbb{C}S_{n}$ to produce this
classification, as well as to show that each $W^{\lambda}$ admits a
restriction to $\mathbb{C}S_{n-1}$ that is multiplicity free if one accounts
for parity shifts; see cor:Wlambda-restriction and rem:multiplicity-free for
details. Many of the results in Sections 2 and 3 can be found in the
literature and are certainly not surprising to experts.

It takes considerably more effort to confirm the claimed description of
$\mathfrak{g}_{n}$. The argument is by a “grand loop” induction on $n$,
wherein the results of Sections 4.2–4.7 are sequentially proven for the value
$n$ under the assumption that the results in these sections have already been
established for the value $n-1$. The arguments require intricate calculations
and considerable case-by-case analysis. An important role is played by the
Gelfand–Zeitlin bases for the $S^{\lambda}$ given by the simultaneous
eigenvectors for the action of the Jucys–Murphy elements.

###  1.3. The results of Marin

The questions answered here were first considered by Marin [10, 11] in the
classical (non-super) setting. As an ungraded algebra, $\mathbb{C}S_{n}$ is
again a direct sum of matrix algebras thanks to the classical Artin–Wedderburn
theorem. Marin showed that the Lie subalgebra of $\mathbb{C}S_{n}$ generated
by the transpositions is reductive with semisimple part isomorphic to a direct
sum of special linear, orthogonal, and symplectic Lie algebras. In particular,
he showed that the transpositions generate a Lie algebra that is roughly half
the dimension of the Lie superalgebra $\mathfrak{g}_{n}$. Thus, the graded and
classical settings are quite different.

One of Marin’s motivations was the representation theory of the braid group,
$B_{n}$. For example, representations of the Type A Iwahori–Hecke algebra,
$H_{d}(q)$, can be inflated to $B_{n}$ via a canonical surjective algebra
homomorphism $\mathbb{C}B_{n}\to H_{d}(q)$. The algebra $H_{d}(q)$ does not
have a natural coproduct and the tensor product of two $H_{d}(q)$-modules,
$V\otimes W$, is not in general again a module over $H_{d}(q)$. However, it is
is a module for the braid group via the coproduct on $\mathbb{C}B_{n}$. Marin
showed that the decomposition of $V\otimes W$ into irreducible
$\mathbb{C}B_{n}$-modules could be determined from the Lie algebra of
transpositions. Marin also calculated the algebraic envelope of the braid
group in the irreducible representations that arise via inflation through the
map $\mathbb{C}B_{n}\to H_{d}(q)$.

The representation theory of the braid group is a rich area of study with
connections to topology, combinatorics, algebraic geometry, and
categorification. The braid group admits evident $\mathbb{Z}$\- and
$\mathbb{Z}_{2}$-gradings (defined by declaring the generators to be of degree
$1$ or $\overline{1}$, respectively), but as far as we are aware the graded
representation theory of the braid group is rather neglected. While this paper
focuses on the questions raised by WunderNatur, it does suggest that the
graded representation theory of the braid group should be notably different
from the classical setting and is worth further study. For example, if one
considers the algebra $\mathcal{A}=\mathbb{C}[q,q^{-1}]$ as a superalgebra
where $q$ is declared to be of odd superdegree (i.e., if we consider
$\mathcal{A}$ as a ‘graded field’), then the Iwahori–Hecke algebra defined
over $\mathcal{A}$, $H_{d}(q)_{\mathcal{A}}$, is a superalgebra when the
generators are taken to be of odd superdegree. There is a surjective
superalgebra homomorphism from $\mathcal{A}B_{n}$ to $H_{d}(q)_{\mathcal{A}}$
and it would be interesting to study the supermodules for the braid group
afforded by this map.

##  2\. Preliminaries

###  2.1. Conventions

Set $\mathbb{Z}_{2}=\mathbb{Z}/2\mathbb{Z}=\\{\overline{0},\overline{1}\\}$.
Following the literature, we use the prefix ‘super’ to indicate that an object
is $\mathbb{Z}_{2}$-graded. We denote the decomposition of a vector superspace
into its $\mathbb{Z}_{2}$-homogeneous components by $V=V_{\overline{0}}\oplus
V_{\overline{1}}$, calling $V_{\overline{0}}$ and $V_{\overline{1}}$ the even
and odd subspaces of $V$, respectively, and writing
$\overline{v}\in\mathbb{Z}_{2}$ to denote the superdegree of a homogeneous
element $v\in V_{\overline{0}}\cup V_{\overline{1}}$. If we state a formula in
which homogeneous degrees of elements are specified, we mean that the formula
is true as written for homogeneous elements, and that it extends by linearity
to non-homogeneous elements. When written without additional adornment, we
consider the field $\mathbb{C}$ to be a superspace concentrated in even
superdegree. All superspaces are assumed to be vector spaces over the field
$\mathbb{C}$, all linear maps are $\mathbb{C}$-linear, and except when
indicated by a modifier (e.g., ‘Lie’), all superalgebras are assumed to be
associative and unital. Given a superspace $V$, let
$\dim(V)=\dim_{\mathbb{C}}(V)$ be the ordinary dimension of $V$ as a
$\mathbb{C}$-vector space.

A linear map between superspaces is _even_ if it preserves homogeneous
degrees, and is _odd_ if it reverses homogeneous degrees. Given superspaces
$V$ and $W$, let
$\operatorname{Hom}(V,W)=\operatorname{Hom}_{\mathbb{C}}(V,W)$ be the
superspace of all $\mathbb{C}$-linear maps $\phi:V\to W$, and let
$\operatorname{End}(V)=\operatorname{Hom}_{\mathbb{C}}(V,V)$. Let
$V^{*}=\operatorname{Hom}(V,\mathbb{C})$ be the usual linear dual of $V$. In
general, isomorphisms between superspaces will be denoted using the symbol
‘$\cong$’ and, except when stated otherwise, should be understood as arising
via an even linear map. We write ‘$\simeq$’ rather than ‘$\cong$’ to emphasize
when an isomorphism arises via an odd linear map. Given a superspace $V$, let
$\Pi(V)=\left\\{v^{\pi}:v\in V\right\\}$ be its parity shift. As a superspace,
$\Pi(V)_{\overline{0}}=V_{\overline{1}}$ and
$\Pi(V)_{\overline{1}}=V_{\overline{0}}$, with
$\overline{v^{\pi}}=\overline{v}+\overline{1}$. Then the map
$(-)^{\pi}:v\mapsto(-1)^{\overline{v}}v^{\pi}$ defines an odd isomorphism
$V\simeq\Pi(V)$.

Given a superalgebra $A$ and (left) $A$-supermodules $M$ and $N$, we say that
a linear map $f:M\to N$ is an $A$-supermodule homomorphism if
$f(a.m)=(-1)^{\overline{a}\cdot\overline{f}}a.f(m)$ for all $a\in A$ and $m\in
M$, and we write $\operatorname{Hom}_{A}(M,N)$ for the set of all
$A$-supermodule homomorphisms from $M$ to $N$. The parity shift $\Pi(M)$ of an
$A$-supermodule is again an $A$-supermodule, with action defined by
$a.m^{\pi}=(a.m)^{\pi}$. Then the function
$(-)^{\pi}:m\mapsto(-1)^{\overline{m}}m^{\pi}$ is an odd $A$-supermodule
isomorphism $M\simeq\Pi(M)$.

Let $\mathbb{N}=\left\\{0,1,2,3,\ldots\right\\}$ be the set of non-negative
integers.

###  2.2. Semisimple superalgebras

Most of the material in this section comes from [2, §2] and [4, §3.1]. For the
authors’ benefit, we write out some of the details that were left to the
reader in [2, 4]. As in [2, 4], we make the standing assumption that each
superalgebra is finite-dimensional.

A superalgebra $A$ is _simple_ if it has no nontrivial superideals.

######  Example 2.2.1 (Type M simple superalgebras).

ex:M(m—n) Given a finite-dimensional superspace $V$, the endomorphism algebra
$\operatorname{End}(V)$ is a simple superalgebra. Fixing a homogeneous basis
for $V$, and making the identification
$V\cong\mathbb{C}^{m|n}:=\mathbb{C}^{m}\oplus\Pi(\mathbb{C}^{n})$ for some
$m,n\in\mathbb{N}$ via this choice of basis, $\operatorname{End}(V)$
identifies with the matrix superalgebra

$M(m|n):=\left\\{\left[\begin{array}[]{c|c}A&B\\\ \hline\cr
C&D\end{array}\right]:A\in M_{m}(\mathbb{C}),B\in M_{m\times
n}(\mathbb{C}),C\in M_{n\times m}(\mathbb{C}),D\in
M_{n}(\mathbb{C})\right\\}.$

As an ungraded associative algebra, $M(m|n)=M_{m+n}(\mathbb{C})$.

######  Example 2.2.2 (Type Q simple superalgebras).

ex:Q(n) Let $V$ be a finite-dimensional vector superspace equipped with an odd
involution $J:V\to V$; i.e., an odd linear map such that $J\circ
J=\operatorname{id}_{V}$. Then the set

(2.2.1)
$Q(V)=Q(V,J)=\left\\{\theta\in\operatorname{End}(V):J\circ\theta=\theta\circ
J\right\\}$

is a simple subsuperalgebra of $\operatorname{End}(V)$. Fix a basis
$\left\\{v_{1},\ldots,v_{n}\right\\}$ for $V_{\overline{0}}$, and set
$v_{i}^{\prime}=J(v_{i})$ for $1\leq i\leq n$, so that
$\left\\{v_{1}^{\prime},\ldots,v_{n}^{\prime}\right\\}$ is a basis for
$V_{\overline{1}}$. Via this choice of homogeneous basis, one has
$V\cong\mathbb{C}^{n|n}$ and $Q(V)$ identifies with the set of supermatrices

(2.2.2) $Q(n):=\left\\{\left[\begin{array}[]{c|c}A&B\\\ \hline\cr
B&A\end{array}\right]:A\in M_{n}(\mathbb{C}),B\in M_{n}(\mathbb{C})\right\\}.$

As an ungraded associative algebra, $Q(n)\cong M_{n}(\mathbb{C})\oplus
M_{n}(\mathbb{C})$ via the map $[\begin{smallmatrix}A&B\\\
B&A\end{smallmatrix}]\mapsto(A+B,A-B)$.

######  Remark 2.2.3.

In the literature, the definition (2.2.1) is frequently stated with the
requirement that the graded commutator
$J\circ\theta-(-1)^{\overline{\theta}}\cdot\theta\circ J$ be equal to $0$,
rather than the requirement that the ordinary commutator
$J\circ\theta-\theta\circ J$ be equal to $0$. We find it more convenient to
use the version stated here. Through appropriate choices of homogeneous bases,
both versions admit the matrix realization (2.2.2). For related discussion,
see [4, §1.1.4].

Given an associative superalgebra $A$, let $|A|$ denote the underlying
associative algebra obtained by forgetting the superspace structure on $A$.
Let

$Z(A)=\left\\{a\in A:ab=(-1)^{\overline{a}\cdot\overline{b}}ba\text{ for all
}b\in A\right\\}$

be the graded center of $A$ (i.e., the center in the sense of superalgebras),
and let

$Z(|A|)=\left\\{a\in A:ab=ba\text{ for all }b\in A\right\\}$

be the ungraded center of $A$ (i.e., the center in the ordinary, non-super
sense). Then both $Z(A)$ and $Z(|A|)$ are subsuperspaces of $A$. That is,
$Z(B)=Z(B)_{\overline{0}}\oplus Z(B)_{\overline{1}}$ for
$B\in\left\\{A,|A|\right\\}$. Also note that
$Z(A)_{\overline{0}}=Z(|A|)_{\overline{0}}$.

######  Example 2.2.4.

Let $m,n\in\mathbb{N}$.

  1. (1)

$Z(M(m|n))=Z(M(m|n))_{\overline{0}}=Z(|M(m|n)|)$, spanned by the identity
matrix $I_{m|n}$.

  2. (2)

$Z(Q(n))_{\overline{0}}$ is spanned by the identity matrix $I_{n|n}$.

  3. (3)

$Z(Q(n))_{\overline{1}}=0$, but $Z(|Q(n)|)_{\overline{1}}$ is nonzero, spanned
by the ‘odd identity matrix’ $[\begin{smallmatrix}0&I_{n}\\\
I_{n}&0\end{smallmatrix}]$.

######  Theorem 2.2.5 ([4, Theorem 3.1]).

Let $A$ be a finite-dimensional simple associative superalgebra.

  1. (1)

If $Z(|A|)_{\overline{1}}=0$, then $A\cong M(m|n)$ for some
$m,n\in\mathbb{N}$.

  2. (2)

If $Z(|A|)_{\overline{1}}\neq 0$, then $A\cong Q(n)$ for some
$n\in\mathbb{N}$.

If $V$ is an irreducible $A$-supermodule, then either $V$ is irreducible as an
$|A|$-module, in which case $V$ is said to be _absolutely irreducible_ (or of
Type M), or else $V$ is reducible as an $|A|$-module, in which case $V$ is
said to be _self-associate_ (or of Type Q).

Given a superspace $V$, let $\pi_{V}:V\to V$ be the parity automorphism,
defined by

$\pi_{V}(v)=(-1)^{\overline{v}}v.$

In particular, $\pi_{A}:A\to A$ is a superalgebra automorphism. A subspace $U$
of a vector superspace $V$ is a sub _super_ space of $V$ if and only if
$\pi_{V}(U)=U$. Given an $|A|$-module $U$, let $\pi_{A}^{*}(U)$ be the
$|A|$-module obtained by pulling back the module structure along $\pi_{A}$. If
$U$ is an $|A|$-submodule of an $A$-supermodule $V$, then $\pi_{V}(U)$ is also
an $|A|$-submodule of $V$, and the map $\pi_{V}(U)\to\pi_{A}^{*}(U)$,
$\pi_{V}(u)\mapsto\pi_{A}^{*}(u)$, is an $|A|$-module isomorphism. In
particular, for each $A$-supermodule $V$, one has
$V=\pi_{V}(V)\cong\pi_{A}^{*}(V)$ as $A$-supermodules.

######  Lemma 2.2.6 ([2, Lemma 2.3]).

lemma:self-associate Let $V$ be a finite-dimensional self-associate
irreducible $A$-supermodule, and let $U$ be an irreducible $|A|$-submodule of
$V$. Then as an $|A|$-module,

$V=U\oplus\pi_{V}(U)\cong U\oplus\pi_{A}^{*}(U),$

with $U\not\cong\pi_{V}(U)$ as $|A|$-modules, and the homogeneous subspaces of
$V$ are

$V_{\overline{0}}=\left\\{u+\pi_{V}(u):u\in U\right\\}\quad\text{and}\quad
V_{\overline{1}}=\left\\{u-\pi_{V}(u):u\in U\right\\}.$

In particular, if $u_{1},\ldots,u_{n}$ is a basis for $U$, then

$\left\\{u_{1}+\pi_{V}(u_{1}),\ldots,u_{n}+\pi_{V}(u_{n})\right\\}\quad\text{and}\quad\left\\{u_{1}-\pi_{V}(u_{1}),\ldots,u_{n}-\pi_{V}(u_{n})\right\\}$

are bases for $V_{\overline{0}}$ and $V_{\overline{1}}$, respectively.

The linear map $J=J_{V}:V\to V$, defined for $u\in U$ by
$J(u\pm\pi_{V}(u))=u\mp\pi_{V}(u)$, is an $|A|$-module homomorphism.
Considered as a function $J:V\to\Pi(V)$,
$u\pm\pi_{V}(u)\mapsto[u\mp\pi_{V}(u)]^{\pi}$, the map $J$ is an even
$A$-supermodule isomorphism $V\cong\Pi(V)$.

###### Proof.

Most of the details of the proof are given in [2], though one point that is
not explicitly explained is the fact that $U\not\cong\pi_{V}(U)$. Here is a
justification for this statement. Let $\pi=\pi_{V}$.

Suppose for the sake of argument that there exists an $|A|$-module isomorphism
$\psi:U\to\pi(U)$. Let $\phi=\pi\circ\psi:U\to U$. Then for all $u\in U$ one
has $\psi(u)=\pi(\phi(u))$, and $\phi$ is a linear bijection such that for all
$a\in A$ one has $\phi(a\cdot u)=(-1)^{\overline{a}}a\cdot\phi(u)$.
Consequently, $\phi^{2}:U\to U$ is an $|A|$-module isomorphism, so by Schur’s
Lemma it is a nonzero scalar multiple of the identity. Rescaling $\phi$ if
necessary, we may assume that $\phi^{2}=\operatorname{id}_{U}$.

Now since $V=U\oplus\pi(U)$ and $V=V_{\overline{0}}\oplus V_{\overline{1}}$,
it follows that also $V=U^{+}\oplus U^{-}$, where

$\displaystyle U^{+}$ $\displaystyle=\left\\{\phi(u)+\pi(u):u\in
U\right\\}=\left\\{u-\pi(\phi(u)):u\in U\right\\},\quad\text{and}$
$\displaystyle U^{-}$ $\displaystyle=\left\\{\phi(u)-\pi(u):u\in
U\right\\}=\left\\{u-\pi(\phi(u)):u\in U\right\\}.$

For $u\in U$, the decomposition of $\phi(u)+\pi(u)$ into its even and odd
components is

$\displaystyle\phi(u)+\pi(u)$
$\displaystyle=\Big{(}\tfrac{1}{2}[\phi(u)+\pi(\phi(u))]+\tfrac{1}{2}[\phi(u)-\pi(\phi(u))]\Big{)}+\Big{(}\tfrac{1}{2}[u+\pi(u)]-\tfrac{1}{2}[u-\pi(u)]\Big{)}$
$\displaystyle=\tfrac{1}{2}\Big{(}[\phi(u)+\pi(\phi(u))]+[u+\pi(u)]\Big{)}+\tfrac{1}{2}\Big{(}[\phi(u)-\pi(\phi(u))]-[u-\pi(u)]\Big{)}$
$\displaystyle=\tfrac{1}{2}\Big{(}[\phi(u)+\pi(u)]+[u+\pi(\phi(u))]\Big{)}+\tfrac{1}{2}\Big{(}[\phi(u)+\pi(u)]-[u+\pi(\phi(u))]\Big{)}.$

After the second and third equals signs, the expressions within the big
parentheses are homogeneous of even and odd superdegree, respectively. This
shows that $U^{+}$ is a subsuperspace of $V$. Finally, for $a\in A$ one has

$a\cdot[\phi(u)+\pi(u)]=(-1)^{\overline{a}}\cdot[\phi(a\cdot u)+\pi(a\cdot
u)],$

so $U^{+}$ is a proper $A$-subsupermodule of $V$. In an entirely similar
fashion, one can also show that $U^{-}$ is a proper $A$-subsupermodule of $V$.
Then the irreducible $A$-supermodule $V$ is a direct sum of two proper
subsupermodules, a contradiction. ∎

######  Remark 2.2.7.

The decomposition of a self-associate irreducible $A$-supermodule into a
direct sum of non-isomorphic $|A|$-modules is canonical, by the uniqueness of
isotypical components.

######  Lemma 2.2.8 (Super Schur Lemma).

Let $V$ be a finite-dimensional irreducible $A$-supermodule. Then

$\operatorname{End}_{|A|}(V)=\begin{cases}\operatorname{span}\left\\{\operatorname{id}_{V}\right\\}&\text{if
$V$ is absolutely irreducible,}\\\
\operatorname{span}\left\\{\operatorname{id}_{V},J_{V}\right\\}&\text{if $V$
is self-associate,}\end{cases}$

where $J_{V}$ is defined as in lemma:self-associate. In particular, if $V$ is
self-associate, then $J_{V}$ is the unique $|A|$-module homomorphism (up to
scalar multiples) that is homogeneous of odd superdegree.

###### Proof.

If $V$ is absolutely irreducible, the lemma is true by the classical Schur’s
Lemma. If $V$ is self-associate, the classical Schur’s Lemma gives
$\operatorname{End}_{|A|}(V)=\operatorname{span}\left\\{\operatorname{id}_{U},\operatorname{id}_{\pi(U)}\right\\}$,
with notation as in lemma:self-associate. Since
$\operatorname{id}_{V}=\operatorname{id}_{U}+\operatorname{id}_{\pi(U)}$ and
$J_{V}=\operatorname{id}_{U}-\operatorname{id}_{\pi(U)}$, the result follows.
∎

######  Remark 2.2.9.

rem:J_V-understood Henceforward, if $V$ is a finite-dimensional self-associate
irreducible $A$-supermodule, we will write $Q(V)$ to denote $Q(V,J_{V})$.

An $A$-supermodule $V$ is _semisimple_ if every subsupermodule of $V$ is a
direct summand, or equivalently, if $V$ is a (direct) sum of irreducible
$A$-supermodules.

######  Theorem 2.2.10 (Super Artin–Wedderburn Theorem [4, Theorem 3.3]).

theorem:super-wedderburn The following statements are equivalent for a finite-
dimensional associative superalgebra $A$:

  1. (1)

Every $A$-supermodule is semisimple.

  2. (2)

The left regular $A$-module is a direct sum of minimal left superideals.

  3. (3)

The superalgebra $A$ is a direct sum of simple superalgebras. Specifically, if
$\left\\{V_{1},\ldots,V_{n}\right\\}$ is a complete, irredundant set of
irreducible $A$-supermodules (up to homogeneous isomorphism), such that
$V_{1},\ldots,V_{n}$ are absolutely irreducible and $V_{m+1},\ldots,V_{n}$ are
self-associate, then the natural maps $A\to\operatorname{End}(V_{i})$, arising
from the $A$-supermodule structures on the $V_{i}$, induce a superalgebra
isomorphism

$A\cong\left(\bigoplus_{i=1}^{m}\operatorname{End}(V_{i})\right)\oplus\left(\bigoplus_{i=m+1}^{n}Q(V_{i})\right).$

A superalgebra that satisfies these conditions is called _semisimple_.

######  Lemma 2.2.11.

lemma:ss-super-ordinary Let $A$ be a finite-dimensional superalgebra. Then $A$
is semisimple (as a superalgebra) if and only if $|A|$ is semisimple (as an
ordinary algebra).

###### Proof.

If $A$ is a direct sum of simple superalgebras, then $|A|$ is a direct sum of
simple algebras, and hence is semisimple, by Examples LABEL:ex:M(m|n) and
LABEL:ex:Q(n). Conversely, suppose $|A|$ is semisimple. Let
$I_{1},\ldots,I_{2m},I_{2m+1},\ldots,I_{n}$ be a complete set of pairwise non-
isomorphic irreducible $|A|$-modules, ordered so that $\pi_{A}(I_{2j})\cong
I_{2j-1}$ for $1\leq j\leq m$, and $\pi_{A}(I_{i})\cong I_{i}$ for $2m<i\leq
n$. For $1\leq i\leq n$, let $A^{I_{i}}$ be the sum of all minimal left ideals
in $|A|$ that are isomorphic to $I_{i}$ as left $|A|$-modules. Then
$|A|=\bigoplus_{i=1}^{n}A^{I_{i}}$, and one has
$\pi_{A}(A^{I_{2j}})=A^{I_{2j-1}}$ for $1\leq j\leq m$, and
$\pi_{A}(A^{I_{i}})=A^{I_{i}}$ for $2m<i\leq n$. This implies for $1\leq j\leq
m$ and $2m<i\leq n$ that $A^{I_{2j-1}}\oplus A^{I_{2j}}$ and $A^{I_{i}}$ are
each subsupermodules of the left regular representation of $A$. Given $1\leq
j\leq m$, fix a decomposition $A^{I_{2j}}=U_{1}\oplus\cdots\oplus U_{t}$ of
$A^{I_{2j}}$ into a direct sum of copies of $I_{2j}$. Then $A^{I_{2j-1}}\oplus
A^{I_{2j}}=\bigoplus_{i=1}^{t}[\pi_{A}(U_{i})\oplus U_{i}]$ is a direct sum
decomposition of $A^{I_{2j-1}}\oplus A^{I_{2j}}$ into (self-associate)
irreducible $A$-supermodules.

Now fix an integer $2m<i\leq n$, and set $I=I_{i}$. We will show that $A^{I}$
is a sum—hence a direct sum—of (absolutely) irreducible $A$-supermodules.
First, $A^{I}$ is a sum of minimal left ideals $U$ such that $U\cong
I\cong\pi_{A}(I)$ as $|A|$-modules, and for each of these ideals $U$ one has
$U+\pi_{A}(U)\subseteq A^{I}$ because $\pi_{A}(A^{I})=A^{I}$. Then
$U+\pi_{A}(U)$ is an $A$-subsupermodule of $A^{I}$. Since $U$ is irreducible
as an $|A|$-module, one has either $U=\pi_{A}(U)$, in which case $U$ is an
irreducible $A$-supermodule, or the sum $U+\pi_{A}(U)$ is a direct sum. In the
latter case, one can argue exactly as in the proof of lemma:self-associate
(but now, without reaching a contradiction) to show that $U+\pi_{A}(U)$ is a
direct sum of two $A$-subsupermodules $U^{+}$ and $U^{-}$, each isomorphic as
$A$-supermodules to $U$. ∎

Given a superalgebra $A$, one can check that
$\operatorname{Ann}_{A}(\pi_{A}^{*}(M))=\pi_{A}(\operatorname{Ann}_{A}(M))$
for each $|A|$-module $M$. This implies that the Jacobson radical of $|A|$ is
closed under the parity map $\pi_{A}$, and hence is a superideal in $A$. Then
the next lemma follows from lemma:ss-super-ordinary.

######  Lemma 2.2.12 ([2, Lemma 2.6]).

Let $A$ be a finite-dimensional superalgebra, and let
$J=\operatorname{rad}(|A|)$ be the Jacobson radical of $|A|$. Then $J$ is the
unique smallest superideal of $A$ such that $A/J$ is a semisimple
superalgebra.

Finally, since each irreducible $A$-supermodule $M$ is a sum of irreducible
$|A|$-modules, one gets
$\operatorname{rad}(|A|)\subseteq\operatorname{Ann}_{A}(M)$, which implies
that the superalgebras $A$ and $A/\operatorname{rad}(|A|)$ have the same
irreducibles. Then the next lemma follows by passing to the quotient
$A/\operatorname{rad}(|A|)$, considering the left regular representations of
$A$ and $|A|$, and applying the Super Artin–Wedderburn Theorem.

######  Lemma 2.2.13 ([2, Corollary 2.8]).

lemma:complete-irreducibles Let $A$ be a finite-dimensional superalgebra, and
let $\left\\{V_{1},\ldots,V_{n}\right\\}$ be a complete, irredundant set of
irreducible $A$-supermodules (up to homogeneous isomorphism) such that
$V_{1},\ldots,V_{m}$ are absolutely irreducible and $V_{m+1},\ldots,V_{n}$ are
self-associate. For $m+1\leq i\leq n$, write $V_{i}=V_{i}^{+}\oplus V_{i}^{-}$
as a direct sum of irreducible $|A|$-modules. Then

(2.2.3) $\left\\{V_{1},\ldots,V_{m},V_{m+1}^{\pm},\ldots,V_{n}^{\pm}\right\\}$

is a complete set of pairwise non-isomorphic irreducible $|A|$-modules.

###  2.3. Finite supergroups

In this section, let $G$ be a finite group, and suppose $G$ contains a normal
subgroup $H$ of index $2$. Let $\operatorname{sgn}:G\to G/H\cong\\{\pm 1\\}$
be the quotient homomorphism, considered also as a representation of $G$.
Define a $\mathbb{Z}_{2}$-grading on $G$ by
$G_{\overline{0}}=H=\ker(\operatorname{sgn})$ and
$G_{\overline{1}}=G\backslash H$. This grading is multiplicative and it makes
$G$ into a _supergroup_. The $\mathbb{Z}_{2}$-grading on $G$ extends by
linearity to a $\mathbb{Z}_{2}$-grading on the group algebra $\mathbb{C}G$,
making $\mathbb{C}G$ into a superalgebra that we call the _group superalgebra_
of $G$. Since $\mathbb{C}G$ is semisimple as an ordinary algebra by Maschke’s
Theorem, then $\mathbb{C}G$ is semisimple as a superalgebra by lemma:ss-super-
ordinary.

Given a $\mathbb{C}H$-module $W$ and an element $t\in G_{\overline{1}}$, let
${}^{t}W=\\{{}^{t}w:w\in W\\}$ be the _conjugate_ representation in which the
action of an element $h\in H$ is defined by $h.{}^{t}w={}^{t}[(tht^{-1}).w]$.
Up to isomorphism, the conjugate representation does not depend on the
particular choice of element in $G_{\overline{1}}$. We say that two
$\mathbb{C}H$-modules $W$ and $W^{\prime}$ are conjugate if
$W^{\prime}\cong{}^{t}W$ for some $t\in G_{\overline{1}}$. If $V$ is a
$\mathbb{C}G$-module, we write $\operatorname{Res}_{H}^{G}(V)$ for the
$\mathbb{C}H$-module obtained by restriction, and if $U$ is a
$\mathbb{C}H$-module, we denote the induced $\mathbb{C}G$-module
$\mathbb{C}G\otimes_{\mathbb{C}H}U$ by $\operatorname{Ind}_{H}^{G}(U)$.

######  Proposition 2.3.1 ([5, Proposition 5.1]).

prop:index-2-restriction Let $V$ be an irreducible $\mathbb{C}G$-module. Then
exactly one of the following holds:

  1. (1)

$V\not\cong V\otimes\operatorname{sgn}$ as $\mathbb{C}G$-modules,
$\operatorname{Res}_{H}^{G}(V)$ is irreducible and isomorphic to its
conjugate, and $\operatorname{Ind}_{H}^{G}(\operatorname{Res}_{H}^{G}(V))\cong
V\oplus(V\otimes\operatorname{sgn})$.

  2. (2)

$V\cong V\otimes\operatorname{sgn}$ as $\mathbb{C}G$-modules,
$\operatorname{Res}_{H}^{G}(V)=U^{\prime}\oplus U^{\prime\prime}$ for
$\mathbb{C}H$-submodules $U^{\prime}$ and $U^{\prime\prime}$ that are
irreducible and conjugate but not isomorphic, and
$\operatorname{Ind}_{H}^{G}(U^{\prime})\cong\operatorname{Ind}_{H}^{G}(U^{\prime\prime})\cong
V$.

Each irreducible $\mathbb{C}H$-module arises uniquely in this way, noting that
in case (1) the irreducible $\mathbb{C}G$-modules modules $V$ and
$V\otimes\operatorname{sgn}$ each determine the same $\mathbb{C}H$-module.

######  Remark 2.3.2.

Given a $\mathbb{C}G$-(super)module $V$, it is immediate from the definitions
that $V\otimes\operatorname{sgn}\cong\pi_{\mathbb{C}G}^{*}(V)$ as
$\mathbb{C}G$-(super)modules. We emphasize however that the sign
representation is _not_ a $\mathbb{C}G$-_super_ module, nor is the one-
dimensional trivial $\mathbb{C}G$-module. However, their direct sum is
naturally a self-associate simple $G$-supermodule.

###  2.4. Example: The group superalgebra of the dihedral group

In this section, fix a positive integer $n\geq 3$ and let $D_{n}$ be the
corresponding dihedral group of order $2n$. Write

$D_{n}=\langle
r,s:r^{n}=s^{2}=(sr)^{2}=1\rangle=\\{1,r,r^{2},\ldots,r^{n-1},s,sr,\ldots,sr^{n-1}\\}$

and let $R_{n}=\\{1,r,r^{2},\ldots,r^{n-1}\\}$ be the subgroup of rotations in
$D_{n}$. Then $R_{n}$ is a normal subgroup of index $2$ in $D_{n}$, so
$\mathbb{C}D_{n}$ is a superalgebra with
$(\mathbb{C}D_{n})_{\overline{0}}=\mathbb{C}R_{n}$.

The irreducible complex representations of the group $D_{n}$ are given as
follows:

  * •

Let $\zeta=e^{2\pi i/n}\in\mathbb{C}$. Given an integer $k$, define
$\rho_{k}:D_{n}\to GL_{2}(\mathbb{C})$ by

$\displaystyle\rho_{k}(r)$ $\displaystyle=\begin{pmatrix}\zeta^{k}&0\\\
0&\zeta^{-k}\end{pmatrix},$ $\displaystyle\rho_{k}(s)$
$\displaystyle=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}.$

These representations are irreducible and pairwise non-isomorphic provided
that $1\leq k<\frac{n}{2}$.

  * •

The trivial representation $\rho_{0}:D_{n}\to GL_{1}(\mathbb{C})$, defined by
$\rho_{0}(r)=\begin{pmatrix}1\end{pmatrix}$ and
$\rho_{0}(s)=\begin{pmatrix}1\end{pmatrix}$.

  * •

The sign representation $\operatorname{sgn}:D_{n}\to GL_{1}(\mathbb{C})$,
defined by $\operatorname{sgn}(r)=\begin{pmatrix}1\end{pmatrix}$ and
$\operatorname{sgn}(s)=\begin{pmatrix}-1\end{pmatrix}$.

  * •

If $n$ is even, then there are two additional $1$-dimensional representations
of $D_{n}$:

    * –

$\rho_{0}^{-}:D_{n}\to GL_{1}(\mathbb{C})$, defined by
$\rho_{0}^{-}(r)=\begin{pmatrix}-1\end{pmatrix}$ and
$\rho_{0}^{-}(s)=\begin{pmatrix}1\end{pmatrix}$.

    * –

$\operatorname{sgn}^{-}:D_{n}\to GL_{1}(\mathbb{C})$, defined by
$\operatorname{sgn}^{-}(r)=\begin{pmatrix}-1\end{pmatrix}$ and
$\operatorname{sgn}^{-}(s)=\begin{pmatrix}-1\end{pmatrix}$.

Now define subspaces of $\mathbb{C}D_{n}$ as follows:

  * •

Given an integer $k$, let $\lambda=e^{2\pi ik/n}\in\mathbb{C}$, and let
$V_{k}$ be the subspace of $\mathbb{C}D_{n}$ spanned by

$\sum_{i=0}^{n-1}\lambda^{-i}\cdot
r^{i}\quad\text{and}\quad\sum_{j=0}^{n-1}\lambda^{-j}\cdot sr^{j}.$

  * •

Let $V_{0}$ be the subspace of $\mathbb{C}D_{n}$ spanned by

$\left(\sum_{i=0}^{n-1}r^{i}\right)+\left(\sum_{i=0}^{n-1}sr^{i}\right)\quad\text{and}\quad\left(\sum_{i=0}^{n-1}r^{i}\right)-\left(\sum_{i=0}^{n-1}sr^{i}\right).$

Then it is straightforward to check the following statements:

  * •

For all integers $k$, $V_{k}$ is a subsuperspace of $\mathbb{C}D_{n}$, and

$\mathbb{C}D_{n}=\begin{cases}V_{0}\oplus V_{1}\oplus\cdots\oplus
V_{n/2}&\text{if $n$ is even,}\\\ V_{0}\oplus V_{1}\oplus\cdots\oplus
V_{\lfloor n/2\rfloor}&\text{if $n$ is odd.}\end{cases}$

  * •

For each integer $1\leq k<\frac{n}{2}$, $V_{k}$ is an absolutely irreducible
$\mathbb{C}D_{n}$-supermodule affording the representation $\rho_{k}$ of
$D_{n}$.

  * •

$V_{0}$ is a self-associate irreducible $\mathbb{C}D_{n}$-supermodule, whose
restriction to $|\mathbb{C}D_{n}|$ is $\rho_{0}\oplus\operatorname{sgn}$.

  * •

If $n$ is even, then $V_{n/2}$ is a self-associate irreducible
$\mathbb{C}D_{n}$-supermodule, whose restriction to $|\mathbb{C}D_{n}|$ is the
direct sum of $\rho_{0}^{-}$ (spanned by
$\sum_{i=0}^{n-1}(-1)^{i}r^{i}+\sum_{i=0}^{n-1}(-1)^{i}sr^{i}$) and
$\operatorname{sgn}^{-}$ (spanned by
$\sum_{i=0}^{n-1}(-1)^{i}r^{i}-\sum_{i=0}^{n-1}(-1)^{i}sr^{i}$).

As a consequence of these observations and theorem:super-wedderburn, we deduce
the existence of a superalgebra isomorphism

$\mathbb{C}D_{n}\cong\begin{cases}M(1|1)^{\oplus(\frac{n}{2}-1)}\oplus
Q(1)^{\oplus 2}&\text{if $n$ is even,}\\\ M(1|1)^{\oplus\lfloor
n/2\rfloor}\oplus Q(1)&\text{if $n$ is odd.}\end{cases}$

##  3\. The symmetric group as a supergroup

In this section, fix an integer $n\geq 2$ and let $S_{n}$ be the symmetric
group on $n$ letters. The sign representation
$\operatorname{sgn}:S_{n}\to\left\\{\pm 1\right\\}$,
$\sigma\mapsto(-1)^{\sigma}$, makes $S_{n}$ into a supergroup such that
$(S_{n})_{\overline{0}}=A_{n}$, the alternating group on $n$ letters, and
$(S_{n})_{\overline{1}}=S_{n}\backslash A_{n}$ is the set of odd permutations.
Then the group algebra $\mathbb{C}S_{n}$ becomes a superalgebra with
$(\mathbb{C}S_{n})_{\overline{0}}=\mathbb{C}A_{n}$, the group algebra of
$A_{n}$.

###  3.1. The irreducible supermodules of the symmetric group

Write $\lambda\vdash n$ to denote that $\lambda$ is a partition of $n$, and
let $\mathcal{P}(n)=\left\\{\lambda:\lambda\vdash n\right\\}$ be the set of
all partitions of $n$. Given $\lambda\in\mathcal{P}(n)$, write
$\lambda^{\prime}$ for the partition that is conjugate (or transpose) to
$\lambda$, and let $\sim$ be the equivalence relation on $\mathcal{P}(n)$ with
equivalence classes
$\left\\{\left\\{\lambda,\lambda^{\prime}\right\\}:\lambda\in\mathcal{P}(n)\right\\}$.

######  Definition 3.1.1.

Let $\overline{\mathcal{P}}(n)$ be any fixed set of representatives for the
distinct equivalence classes in $\mathcal{P}(n)$ under the relation $\sim$.
Then $\overline{\mathcal{P}}(n)$ is a disjoint union of sets $E_{n}$ and
$F_{n}$, where

$E_{n}=\\{\lambda\in\overline{\mathcal{P}}(n):\lambda\neq\lambda^{\prime}\\}\quad\text{and}\quad
F_{n}=\\{\lambda\in\overline{\mathcal{P}}(n):\lambda=\lambda^{\prime}\\}.$

For $\lambda\vdash n$, let $S^{\lambda}$ be the corresponding Specht module.
Then the set $\\{S^{\lambda}:\lambda\vdash n\\}$ is a complete set of pairwise
non-isomorphic irreducible $\mathbb{C}S_{n}$-modules. It is well-known that
$S^{\lambda}\otimes\operatorname{sgn}\cong S^{\lambda^{\prime}}$; see [7,
Theorems 4.12 and 6.7]. If $\lambda\neq\lambda^{\prime}$, then
prop:index-2-restriction implies that $S^{\lambda}$ and $S^{\lambda^{\prime}}$
are irreducible (and isomorphic) as $\mathbb{C}A_{n}$-modules, while for
$\lambda=\lambda^{\prime}$ one gets that
$\operatorname{Res}_{A_{n}}^{S_{n}}(S^{\lambda})=S^{\lambda^{+}}\oplus
S^{\lambda^{-}}$ for two irreducible, conjugate, non-isomorphic
$\mathbb{C}A_{n}$-modules $S^{\lambda^{+}}$ and $S^{\lambda^{-}}$. In
particular, if $\tau\in S_{n}$ is any odd permutation, then multiplication by
$\tau$ defines a linear isomorphism $S^{\mu^{+}}\to S^{\mu^{-}}$.

Since $S^{\lambda^{\prime}}\otimes\operatorname{sgn}\cong S^{\lambda}$,
Schur’s Lemma implies that
$\operatorname{Hom}_{\mathbb{C}S_{n}}(S^{\lambda},S^{\lambda^{\prime}}\otimes\operatorname{sgn})\cong\mathbb{C}$.
For each $\lambda\vdash n$, choose a nonzero element $\phi^{\lambda}$ of this
space, and interpret it as a linear isomorphism $\phi^{\lambda}:S^{\lambda}\to
S^{\lambda^{\prime}}$ such that

(3.1.1) $\phi^{\lambda}(\sigma\cdot
v)=(-1)^{\sigma}\sigma\cdot\phi^{\lambda}(v)\quad\text{for all}\quad v\in
S^{\lambda}\text{ and }\sigma\in S_{n}.$

Then
$\phi^{\lambda^{\prime}}\circ\phi^{\lambda}\in\operatorname{Hom}_{\mathbb{C}S_{n}}(S^{\lambda},S^{\lambda})=\mathbb{C}\cdot\operatorname{id}_{S^{\lambda}}$.
Rescaling our choice of $\phi^{\lambda}$ if necessary, we may assume that
$\phi^{\lambda^{\prime}}\circ\phi^{\lambda}=\operatorname{id}_{S^{\lambda}}$.
This implies that
$\phi^{\lambda}\circ\phi^{\lambda^{\prime}}=\operatorname{id}_{S^{\lambda^{\prime}}}$,
as well. Now for $\lambda=\lambda^{\prime}$, we deduce that $\phi^{\lambda}$
is the unique self-inverse linear map satisfying (3.1.1), while for
$\lambda\neq\lambda^{\prime}$ we deduce that up to mutual rescalings of the
form
$(\phi^{\lambda},\phi^{\lambda^{\prime}})\mapsto(c\cdot\phi^{\lambda},\frac{1}{c}\cdot\phi^{\lambda^{\prime}})$,
$\phi^{\lambda}$ and $\phi^{\lambda^{\prime}}$ are the unique mutually-inverse
linear maps each satisfying (3.1.1).

Now for each symmetric partition $\lambda$, one has
$(\phi^{\lambda})^{2}=\operatorname{id}_{S^{\lambda}}$, and hence
$S^{\lambda}$ decomposes into $+1$ and $-1$ eigenspaces for $\phi^{\lambda}$.
These eigenspaces are $A_{n}$-stable (because $\phi^{\lambda}$ is a
$\mathbb{C}A_{n}$-homomorphism), and hence are $\mathbb{C}A_{n}$-submodules of
$S^{\lambda}$. Moreover, neither eigenspace is equal to all of $S^{\lambda}$,
since otherwise (3.1.1) would imply for all $v\in S^{\lambda}$ that
$\sigma.v=0$ for all odd permutations (which is false). Combining these
observations with those made two paragraphs ago, and using the uniqueness of
isotypical components, one deduces that the $\pm 1$ eigenspaces of
$\phi^{\lambda}$ are the irreducible $\mathbb{C}A_{n}$-constituents of
$\operatorname{Res}_{A_{n}}^{S_{n}}(S^{\lambda})$. One can take
$S^{\lambda^{+}}$ and $S^{\lambda^{-}}$ to be the $+1$ and $-1$ eigenspaces of
$\phi^{\lambda}$, respectively.

######  Lemma 3.1.2.

lemma:super-CSn-as-ordinary Let $n>1$, and let $W$ be an irreducible
$\mathbb{C}S_{n}$-supermodule.

  1. (1)

If $W$ is absolutely irreducible, then $W\cong S^{\lambda}$ as a
$|\mathbb{C}S_{n}|$-module, for some symmetric partition $\lambda\vdash n$.
Under this identification, the homogeneous subspaces of $W$ are
$S^{\lambda^{+}}$ and $S^{\lambda^{-}}$.

  2. (2)

If $W$ is self-associate, then $W\cong S^{\lambda}\oplus S^{\lambda^{\prime}}$
as a $|\mathbb{C}S_{n}|$-module, for some non-symmetric partition
$\lambda\vdash n$. Under this identification, the homogeneous subspaces of $W$
are

$W_{\overline{0}}=\\{u+\phi^{\lambda}(u):u\in
S^{\lambda}\\}\quad\text{and}\quad
W_{\overline{1}}=\\{u-\phi^{\lambda}(u):u\in S^{\lambda}\\}.$

###### Proof.

First suppose $W$ is an absolutely irreducible $\mathbb{C}S_{n}$-supermodule.
Then as a $|\mathbb{C}S_{n}|$-module, $W\cong S^{\lambda}$ for some
$\lambda\vdash n$. Since

$W=\pi_{W}(W)\cong\pi_{\mathbb{C}S_{n}}^{*}(W)\cong
W\otimes\operatorname{sgn}\cong S^{\lambda}\otimes\operatorname{sgn}\cong
S^{\lambda^{\prime}}$

as $|\mathbb{C}S_{n}|$-modules, this implies that $\lambda=\lambda^{\prime}$.
Next, since the odd permutations in $S_{n}$ do not annihilate $S^{\lambda}$,
$W$ cannot be simply a purely even or a purely odd superspace. Then
$W_{\overline{0}}$ and $W_{\overline{1}}$ are nonzero
$\mathbb{C}A_{n}$-submodules of $W$. Since $S^{\lambda}=S^{\lambda^{+}}\oplus
S^{\lambda^{-}}$ as a $\mathbb{C}A_{n}$-module, the uniqueness of isotypic
components implies that
$\\{S^{\lambda^{+}},S^{\lambda^{-}}\\}=\\{W_{\overline{0}},W_{\overline{1}}\\}$.

Now suppose $W$ is self-associate as a $\mathbb{C}S_{n}$-supermodule. Then by
lemma:self-associate, there exists $\lambda\vdash n$ such that, as a
$|\mathbb{C}S_{n}|$-module,

$W=S^{\lambda}\oplus\pi_{W}(S^{\lambda})\cong
S^{\lambda}\oplus\pi_{\mathbb{C}S_{n}}^{*}(S^{\lambda})\cong S^{\lambda}\oplus
S^{\lambda^{\prime}},$

and $S^{\lambda}\not\cong S^{\lambda^{\prime}}$ as
$|\mathbb{C}S_{n}|$-modules. Then $\lambda\neq\lambda^{\prime}$. Making the
identification $\pi_{W}(S^{\lambda})=S^{\lambda^{\prime}}$, the parity map
$\pi=\pi_{W}:W\to W$ restricts to mutually-inverse linear maps
$\pi^{\lambda}:S^{\lambda}\to S^{\lambda^{\prime}}$ and
$\pi^{\lambda^{\prime}}:S^{\lambda^{\prime}}\to S^{\lambda}$ satisfying
(3.1.1). Then by uniqueness (up to mutual rescaling) of $\phi^{\lambda}$ and
$\phi^{\lambda^{\prime}}$, we may assume that $\pi^{\lambda}=\phi^{\lambda}$
and $\pi^{\lambda^{\prime}}=\phi^{\lambda^{\prime}}$. Now the identification
of $W_{\overline{0}}$ and $W_{\overline{1}}$ follows from lemma:self-
associate. ∎

######  Proposition 3.1.3.

prop:Wlambda-decomposition Let $n>1$.

  1. (1)

For each $\lambda\in E_{n}$, there exists a self-associate irreducible
$\mathbb{C}S_{n}$-supermodule $W^{\lambda}$ such that $W^{\lambda}\cong
S^{\lambda}\oplus S^{\lambda^{\prime}}$ as a $|\mathbb{C}S_{n}|$-module, with

$W^{\lambda}_{\overline{0}}=\\{u+\phi^{\lambda}(u):u\in
S^{\lambda}\\}\quad\text{and}\quad
W^{\lambda}_{\overline{1}}=\\{u-\phi^{\lambda}(u):u\in S^{\lambda}\\}.$

The $|\mathbb{C}S_{n}|$-module decomposition $W^{\lambda}\cong
S^{\lambda}\oplus S^{\lambda^{\prime}}$ is unique. We denote by
$J^{\lambda}:W^{\lambda}\to W^{\lambda}$ the odd involution defined for $u\in
S^{\lambda}$ by $J^{\lambda}(u\pm\phi^{\lambda}(u))=u\mp\phi^{\lambda}(u)$.

  2. (2)

For each $\lambda\in F_{n}$, there exists an absolutely irreducible
$\mathbb{C}S_{n}$-supermodule such that $W^{\lambda}\cong S^{\lambda}$ as a
$|\mathbb{C}S_{n}|$-module, with $W^{\lambda}_{\overline{0}}=S^{\lambda^{+}}$
and $W^{\lambda}_{\overline{1}}=S^{\lambda^{-}}$.

The set $\\{W^{\lambda}:\lambda\in\overline{\mathcal{P}}(n)\\}$ is a complete
set of pairwise non-isomorphic irreducible $\mathbb{C}S_{n}$-supermodules.

###### Proof.

Using lemma:complete-irreducibles, lemma:super-CSn-as-ordinary, and the
classification of the irreducible $|\mathbb{C}S_{n}|$-modules, one deduces for
each $\lambda\in\overline{\mathcal{P}}(n)$ that there exists an irreducible
$\mathbb{C}S_{n}$-supermodule $W^{\lambda}$ with the given restriction to
$|\mathbb{C}S_{n}|$. In particular, if $\lambda\in E_{n}$, and if $W$ and
$W^{\prime}$ are self-associate irreducible $\mathbb{C}S_{n}$-supermodules
that are both isomorphic as $|\mathbb{C}S_{n}|$-modules to $S^{\lambda}\oplus
S^{\lambda^{\prime}}$, then $W\cong W^{\prime}$, so the notation $W^{\lambda}$
does not depend on the choice of representative for the equivalence class
$\left\\{\lambda,\lambda^{\prime}\right\\}$.

For $\lambda\in E_{n}$, the decomposition $W^{\lambda}\cong S^{\lambda}\oplus
S^{\lambda^{\prime}}$ is unique by the uniqueness of isotypic components and
the fact that $S^{\lambda}\not\cong S^{\lambda^{\prime}}$ as
$|\mathbb{C}S_{n}|$-modules. For $\lambda\in F_{n}$, one can replace
$W^{\lambda}$ with its parity shift if necessary (to which $W^{\lambda}$ is
odd-isomorphic) to ensure that $W^{\lambda}_{\overline{0}}=S^{\lambda^{+}}$
and $W^{\lambda}_{\overline{1}}=S^{\lambda^{-}}$. ∎

######  Remark 3.1.4.

remark:self-dual It is well-known that the irreducible complex representations
of $S_{n}$ are self-dual. From this and prop:Wlambda-decomposition, it follows
for all $\lambda\vdash n$ that $(W^{\lambda})^{*}$ is isomorphic (via an even
supermodule homomorphism) to either $W^{\lambda}$ or $\Pi(W^{\lambda})$. For
$\lambda\in E_{n}$, one always has $(W^{\lambda})^{*}\cong W^{\lambda}$, while
for $\lambda\in F_{n}$, one has $(W^{\lambda})^{*}\cong W^{\lambda}$ if and
only if the $\mathbb{C}A_{n}$-modules $S^{\lambda^{+}}$ and $S^{\lambda^{-}}$
are each self-dual.

The next result is an immediate consequence of [8, Theorem 2.4.10].

######  Lemma 3.1.5.

lemma:lower-bound-dim Let $n\geq 2$, and let
$\lambda\in\overline{\mathcal{P}}(n)$.

  1. (1)

Suppose $\lambda\in E_{n}$. If $\lambda=(n)$ or $\lambda=(1^{n})$, then
$\dim(W^{\lambda})=2$. Otherwise, $\dim(W^{\lambda})\geq 2n-2$.

  2. (2)

Suppose $\lambda\in F_{n}$. If $n=3$ and $\lambda=(2,1)$, or if $n=4$ and
$\lambda=(2,2)$, then $\dim(W^{\lambda})=2$. If $n=5$ and $\lambda=(3,1,1)$,
then $\dim(W^{\lambda})=6=n+1$. Otherwise, $\dim(W^{\lambda})\geq n+3$.

Given a partition $\lambda\vdash n$ and a permutation $\sigma\in S_{n}$, let
$S^{\lambda}(\sigma)\in\operatorname{End}(S^{\lambda})$ and
$W^{\lambda}(\sigma)\in\operatorname{End}(W^{\lambda})$ denote the
corresponding linear maps $u\mapsto\sigma.u$. For $\sigma\in A_{n}$, let
$S^{\lambda^{+}}(\sigma)\in\operatorname{End}(S^{\lambda^{+}})$ and
$S^{\lambda^{-}}(\sigma)\in\operatorname{End}(S^{\lambda^{-}})$ be defined
similarly. By abuse of notation, we will also write $S^{\lambda}(\sigma)$,
$W^{\lambda}(\sigma)$, etc., for the corresponding matrices when bases for the
underlying modules are fixed, and we extend the notation $S^{\lambda}(\sigma)$
to arbitrary elements $\sigma\in\mathbb{C}S_{n}$ by linearity.

The next result is an immediate consequence of theorem:super-wedderburn and
prop:Wlambda-decomposition.

######  Corollary 3.1.6.

cor:CSn-as-superalgebra Let $n\geq 2$. The map
$\mathbb{C}S_{n}\to\bigoplus_{\lambda\in\overline{\mathcal{P}}(n)}\operatorname{End}(W^{\lambda})$,
$\sigma\mapsto\bigoplus_{\lambda\in\overline{\mathcal{P}}(n)}W^{\lambda}(\sigma)$,
induces a superalgebra isomorphism

$\displaystyle\mathbb{C}S_{n}$
$\displaystyle\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}Q\left(W^{\lambda}\right)\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}\operatorname{End}\left(W^{\lambda}\right)\Bigg{]}\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}Q\left(f^{\lambda}\right)\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}M\left(\tfrac{1}{2}f^{\lambda},\tfrac{1}{2}f^{\lambda}\right)\Bigg{]},$

where $f^{\lambda}=\dim(S^{\lambda})$.

Let $\lambda\in E_{n}$. For $u\in S^{\lambda}$, the expression
$u\pm\phi^{\lambda}(u)\in W^{\lambda}$ is linear in $u$, and one has

(3.1.2)
$\sigma.\big{(}u\pm\phi^{\lambda}(u)\big{)}=(\sigma.u)\pm(-1)^{\sigma}\phi^{\lambda}(\sigma.u)$

for all $\sigma\in S_{n}$. Then making the identification
$Q(W^{\lambda})=Q(f^{\lambda})$ via a choice of homogeneous basis as in
lemma:self-associate, the identity (3.1.2) implies that

(3.1.3)
$W^{\lambda}(\sigma)=\begin{cases}\left[\begin{array}[]{c|c}S^{\lambda}(\sigma)&0\\\
\hline\cr 0&S^{\lambda}(\sigma)\end{array}\right]&\text{if $\sigma$ is an even
permutation,}\\\ \left[\begin{array}[]{c|c}0&S^{\lambda}(\sigma)\\\ \hline\cr
S^{\lambda}(\sigma)&0\end{array}\right]&\text{if $\sigma$ is an odd
permutation.}\end{cases}$

On the other hand, let $\lambda\in F_{n}$. Choose a basis
$\left\\{u_{1},\ldots,u_{m}\right\\}$ for
$W^{\lambda}_{\overline{0}}=S^{\lambda^{+}}$, and let $\tau\in S_{n}$ be a
fixed odd permutation. Then $\left\\{\tau.u_{1},\ldots,\tau.u_{m}\right\\}$ is
a basis for $W^{\lambda}_{\overline{1}}=S^{\lambda^{-}}$. Now identifying
$\operatorname{End}(W^{\lambda})$ with
$M(\frac{1}{2}f^{\lambda},\frac{1}{2}f^{\lambda})$ via this choice of
homogeneous basis, one gets

(3.1.4)
$W^{\lambda}(\sigma)=\begin{cases}\left[\begin{array}[]{c|c}S^{\lambda^{+}}(\sigma)&0\\\
\hline\cr 0&S^{\lambda^{+}}(\tau^{-1}\sigma\tau)\end{array}\right]&\text{if
$\sigma$ is an even permutation,}\\\
\left[\begin{array}[]{c|c}0&S^{\lambda^{+}}(\sigma\tau)\\\ \hline\cr
S^{\lambda^{+}}(\tau^{-1}\sigma)&0\end{array}\right]&\text{if $\sigma$ is an
odd permutation.}\end{cases}$

###  3.2. Weight space decompositions of Specht modules

Our primary references for this section are [9, §2] and [3, §3].

The Jucys–Murphy elements $L_{1},\ldots,L_{n}\in\mathbb{C}S_{n}$ are defined
by $L_{j}=\sum_{1\leq i<j}(i,j)$. In particular, $L_{1}=0$. The elements
$L_{1},\ldots,L_{n}$ generate a commutative, semisimple subalgebra of
$\mathbb{C}S_{n}$. Since this subalgebra is semisimple, each finite-
dimensional $\mathbb{C}S_{n}$-module $V$ decomposes into a direct sum of
simultaneous eigenspaces for $L_{1},\ldots,L_{n}$. Given
$\alpha=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{C}^{n}$, the $\alpha$-weight
space of $V$ is defined by

$V_{\alpha}=\left\\{v\in V:L_{i}\cdot v=\alpha_{i}v\text{ for all }1\leq i\leq
n\right\\}.$

Given $\alpha=(\alpha_{1},\ldots,\alpha_{n})$, we may write
$\alpha(L_{i})=\alpha_{i}$. The nonzero elements of $V_{\alpha}$ are called
_weight vectors_. If $V_{\alpha}\neq 0$, then we say that $\alpha$ is a weight
of $V$. Let

$\mathcal{W}(\lambda)=\\{\alpha\in\mathbb{C}^{n}:\text{$\alpha$ is a weight of
$S^{\lambda}$}\\},$

and let $\mathcal{W}(n)=\bigcup_{\lambda\vdash n}\mathcal{W}(\lambda)$.

Fix a partition $\lambda\vdash n$, and let $\mathbb{T}(\lambda)$ be the set of
all standard $\lambda$-tableaux. The nonzero weight spaces of the irreducible
$\mathbb{C}S_{n}$-module $S^{\lambda}$ are each one-dimensional, spanned by
vectors $v_{T}$ for $T\in\mathbb{T}(\lambda)$. Given $T\in\mathbb{T}(\lambda)$
and an integer $1\leq i\leq n$, let $T_{i}$ be the box in $T$ that is occupied
by $i$, and let $\operatorname{cont}(T_{i})$ be the content (or residue) of
the box $T_{i}$. Then $v_{T}$ is of weight

$\alpha(T):=(\operatorname{cont}(T_{1}),\ldots,\operatorname{cont}(T_{n})).$

In particular, $\mathcal{W}(\lambda)\subseteq\mathbb{Z}^{n}$. For example, if
$n=7$, $\lambda=(4,2,1)$, and

$T=\ytableau 1&245\\\ 37\\\ 6\;,$

then $\alpha(T)=(0,1,-1,2,3,-2,0)$. This description implies that the union
$\mathcal{W}(n)=\bigcup_{\lambda\vdash n}\mathcal{W}(\lambda)$ is disjoint,
and for $\alpha\in\mathcal{W}(n)$ one has $\alpha=-\alpha$ only if $n=1$,
which we have excluded by assumption.

We may denote a weight vector in $S^{\lambda}$ by $v_{T}$, for a standard
$\lambda$-tableau $T$, or by $v_{\alpha}$, where $\alpha=\alpha(T)$ is the
corresponding weight. Conversely, if $\alpha\in\mathcal{W}(\lambda)$ is
specified, let $T(\alpha)$ be the corresponding standard $\lambda$-tableau.
Then $v_{\alpha(T)}=v_{T}$ for all $T\in\mathbb{T}(\lambda)$, and
$v_{T(\alpha)}=v_{\alpha}$ for all $\alpha\in\mathcal{W}(\lambda)$. Given a
(standard) $\lambda$-tableau $T$, let $T^{\prime}$ be its transpose, which is
then a (standard) $\lambda^{\prime}$-tableau. Then for all
$T\in\mathbb{T}(\lambda)$, one has $\alpha(T^{\prime})=-\alpha(T)$, and for
all $\alpha\in\mathcal{W}(\lambda)$, one has $T(-\alpha)=T(\alpha)^{\prime}$.

######  Proposition 3.2.1 ([9, Corollary 2.2.3]).

prop:si-on-weights Let $\alpha\in\mathcal{W}(\lambda)$. Given $1\leq i<n$, let
$s_{i}$ be the transposition $(i,i+1)\in S_{n}$, and let
$\beta=s_{i}.\alpha=(\alpha_{1},\ldots,\alpha_{i-1},\alpha_{i+1},\alpha_{i},\alpha_{i+2},\ldots,\alpha_{n})$.
Then:

  1. (1)

$\alpha_{i}\neq\alpha_{i+1}$.

  2. (2)

If $\alpha_{i+1}=\alpha_{i}\pm 1$, then $s_{i}\cdot v_{\alpha}=\pm v_{\alpha}$
and $\beta\notin\mathcal{W}(\lambda)$.

  3. (3)

Suppose $\alpha_{i+1}\neq\alpha_{i}\pm 1$, and let
$c_{i}=(\alpha_{i+1}-\alpha_{i})^{-1}$. Then $\beta\in\mathcal{W}(\lambda)$
and $w_{\beta}:=(s_{i}-c_{i})\cdot v_{\alpha}$ is a nonzero scalar multiple of
$v_{\beta}$; the elements $L_{i}$, $L_{i+1}$, and $s_{i}$ leave
$S^{\lambda}_{\alpha}\oplus S^{\lambda}_{\beta}$ invariant; and they act in
the basis $\left\\{v_{\alpha},w_{\beta}\right\\}$ of
$S^{\lambda}_{\alpha}\oplus S^{\lambda}_{\beta}$ via the matrices

$L_{i}=\begin{bmatrix}\alpha_{i}&0\\\ 0&\alpha_{i+1}\end{bmatrix},\qquad
L_{i+1}=\begin{bmatrix}\alpha_{i+1}&0\\\ 0&\alpha_{i}\end{bmatrix},\qquad
s_{i}=\begin{bmatrix}c_{i}&1-c_{i}^{2}\\\ 1&-c_{i}\end{bmatrix}.$

###  3.3. Weight space decompositions of irreducible supermodules

In this section we describe the actions of the odd operators
$L_{1},\ldots,L_{n}$ and the transpositions $s_{1},\ldots,s_{n-1}$ on the
irreducible $\mathbb{C}S_{n}$-supermodules in terms of the weight vectors
described in Section 3.2.

Given $\lambda\vdash n$ and $\alpha\in\mathcal{W}(\lambda)$, it follows from
the intertwining condition (3.1.1) that the function
$\phi^{\lambda}:S^{\lambda}\to S^{\lambda^{\prime}}$ specified in Section 3.1
defines a linear isomorphism
$\phi^{\lambda}:S^{\lambda}_{\alpha}\xrightarrow{\cong}S^{\lambda^{\prime}}_{-\alpha}$.
We will assume that the spanning vectors $v_{\alpha}\in S^{\lambda}_{\alpha}$
and $v_{-\alpha}\in S^{\lambda^{\prime}}_{-\alpha}$ are chosen so that

(3.3.1) $v_{-\alpha}=\phi^{\lambda}(v_{\alpha}).$

This can be done for all $\lambda\vdash n$ and $\alpha\in\mathcal{W}(\lambda)$
because the union $\mathcal{W}(n)=\bigcup_{\lambda\vdash
n}\mathcal{W}(\lambda)$ is disjoint, because $\alpha\neq-\alpha$ for all
$\alpha\in\mathcal{W}(n)$ by the assumption that $n>1$, and because
$\phi^{\lambda^{\prime}}\circ\phi^{\lambda}=\operatorname{id}_{S^{\lambda}}$
and
$\phi^{\lambda}\circ\phi^{\lambda^{\prime}}=\operatorname{id}_{S^{\lambda^{\prime}}}$.
In terms of standard tableaux, one has $v_{T^{\prime}}=\phi^{\lambda}(v_{T})$
for all $T\in\mathbb{T}(\lambda)$.

The preceding discussion implies that the elements of
$\mathcal{W}(\lambda)\cup\mathcal{W}(\lambda^{\prime})$ occur in $\pm$ pairs.
Let

$\overline{\mathcal{W}}(\lambda)=[\mathcal{W}(\lambda)\cup\mathcal{W}(\lambda^{\prime})]/{\pm}$

be the set of all such pairs. For
$\lambda\in\overline{\mathcal{P}}(n)=E_{n}\cup F_{n}$, we will write
$\pm\alpha$ to denote an element of $\overline{\mathcal{W}}(\lambda)$. Note
that this notation implies a fixed choice for the ‘positive’ element $\alpha$
of the pair $\pm\alpha$. If $\lambda\in E_{n}$, we will assume that
$\alpha\in\mathcal{W}(\lambda)$; if $\lambda\in F_{n}$, we will assume that
$\alpha_{n}\geq 0$. This uniquely determines the choice of the positive
element $\alpha$, except when $\lambda\in F_{n}$ and $\alpha_{n}=0$. Now given
$\lambda\in\overline{\mathcal{P}}(n)$, we will describe bases for
$W^{\lambda}_{\overline{0}}$ and $W^{\lambda}_{\overline{1}}$ that are indexed
by $\overline{\mathcal{W}}(\lambda)$.

First let $\lambda\in E_{n}$, so that $W^{\lambda}\cong S^{\lambda}\oplus
S^{\lambda^{\prime}}$ as a $|\mathbb{C}S_{n}|$-module. Given a pair
$\pm\alpha\in\overline{\mathcal{W}}(\lambda)$, set

$\displaystyle v_{\alpha}^{+}$
$\displaystyle=\tfrac{1}{2}(v_{\alpha}+v_{-\alpha})=\tfrac{1}{2}\big{(}v_{\alpha}+\phi^{\lambda}(v_{\alpha})\big{)},$
$\displaystyle v_{\alpha}^{-}$
$\displaystyle=\tfrac{1}{2}(v_{\alpha}-v_{-\alpha})=\tfrac{1}{2}\big{(}v_{\alpha}-\phi^{\lambda}(v_{\alpha})\big{)}.$

Then by lemma:self-associate, the sets
$\\{v_{\alpha}^{+}:\pm\alpha\in\overline{\mathcal{W}}(\lambda)\\}$ and
$\\{v_{\alpha}^{-}:\pm\alpha\in\overline{\mathcal{W}}(\lambda)\\}$ are bases
for $W^{\lambda}_{\overline{0}}$ and $W^{\lambda}_{\overline{1}}$,
respectively. One has $v_{\alpha}=v_{\alpha}^{+}+v_{\alpha}^{-}$ and
$v_{-\alpha}=v_{\alpha}^{+}-v_{\alpha}^{-}$.

Next let $\lambda\in F_{n}$, so that $W^{\lambda}\cong S^{\lambda}$ as a
$|\mathbb{C}S_{n}|$-module. Then $W^{\lambda}_{\overline{0}}=S^{\lambda^{+}}$
and $W^{\lambda}_{\overline{1}}=S^{\lambda^{-}}$ are the $+1$ and $-1$
eigenspaces, respectively, for the function $\phi^{\lambda}:S^{\lambda}\to
S^{\lambda}$. For each pair $\pm\alpha\in\overline{\mathcal{W}}(\lambda)$,
write $v_{\alpha}=v_{\alpha}^{+}+v_{\alpha}^{-}$, with $v_{\alpha}^{+}\in
S^{\lambda^{+}}$ and $v_{\alpha}^{-}\in S^{\lambda^{-}}$. Then
$v_{-\alpha}=\phi^{\lambda}(v_{\alpha})=v_{\alpha}^{+}-v_{\alpha}^{-}$. Now

$\displaystyle v_{\alpha}^{+}$
$\displaystyle=\tfrac{1}{2}(v_{\alpha}+v_{-\alpha})=\tfrac{1}{2}\big{(}v_{\alpha}+\phi^{\lambda}(v_{\alpha})\big{)},$
$\displaystyle v_{\alpha}^{-}$
$\displaystyle=\tfrac{1}{2}(v_{\alpha}-v_{-\alpha})=\tfrac{1}{2}\big{(}v_{\alpha}-\phi^{\lambda}(v_{\alpha})\big{)},$

and the sets
$\\{v_{\alpha}^{+}:\pm\alpha\in\overline{\mathcal{W}}(\lambda)\\}$ and
$\\{v_{\alpha}^{-}:\pm\alpha\in\overline{\mathcal{W}}(\lambda)\\}$ are bases
for $W^{\lambda}_{\overline{0}}$ and $W^{\lambda}_{\overline{1}}$,
respectively.

With notation as above, one gets $L_{i}\cdot
v_{\alpha}^{+}=\alpha_{i}v_{\alpha}^{-}$ and $L_{i}\cdot
v_{\alpha}^{-}=\alpha_{i}v_{\alpha}^{+}$ for each $1\leq i\leq n$. For
$\pm\alpha\in\overline{\mathcal{W}}(\lambda)$, set

$W^{\lambda}_{\pm\alpha}=\operatorname{span}\left\\{v_{\alpha}^{+},v_{\alpha}^{-}\right\\}=\operatorname{span}\left\\{v_{\alpha},v_{-\alpha}\right\\}.$

Then
$W^{\lambda}=\bigoplus_{\pm\alpha\in\overline{\mathcal{W}}(\lambda)}W^{\lambda}_{\pm\alpha}$.
We may refer to $W^{\lambda}_{\pm\alpha}$ as the $\pm\alpha$-weight space of
$W^{\lambda}$.

The next result follows directly from prop:si-on-weights.

######  Proposition 3.3.1.

prop:si-on-weights-super Let
$\alpha=(\alpha_{1},\ldots,\alpha_{n})\in\mathcal{W}(\lambda)$. Let $1\leq
i<n$, and set $\beta=s_{i}.\alpha$.

  1. (1)

If $\alpha_{i+1}=\alpha_{i}\pm 1$, then the transposition $s_{i}$ leaves the
superspace $W^{\lambda}_{\pm\alpha}$ invariant, and it acts in the homogeneous
basis $\left\\{v_{\alpha}^{+},v_{\alpha}^{-}\right\\}$ of
$W^{\lambda}_{\pm\alpha}$ via the matrix

$\pm\begin{bmatrix}0&1\\\ 1&0\end{bmatrix},$

where the $\pm$ sign is the same as in prop:si-on-weights(2).

  2. (2)

Suppose $\alpha_{i+1}\neq\alpha_{i}\pm 1$. Let
$c_{i}=(\alpha_{i+1}-\alpha_{i})^{-1}$, and set

$\displaystyle w_{\beta}$ $\displaystyle=(s_{i}-c_{i})\cdot v_{\alpha},$
$\displaystyle w_{-\beta}$ $\displaystyle=(s_{i}+c_{i})\cdot
v_{-\alpha}=-\phi^{\lambda}(w_{\beta}),$ $\displaystyle w_{\beta}^{+}$
$\displaystyle=\tfrac{1}{2}(w_{\beta}+\phi^{\lambda}(w_{\beta}))=\tfrac{1}{2}(w_{\beta}-w_{-\beta}),$
$\displaystyle w_{\beta}^{-}$
$\displaystyle=\tfrac{1}{2}(w_{\beta}-\phi^{\lambda}(w_{\beta}))=\tfrac{1}{2}(w_{\beta}+w_{-\beta}).$

Then $\\{w_{\beta}^{+},w_{\beta}^{-}\\}$ is a homogeneous basis for
$W^{\lambda}_{\pm\beta}$, the elements $L_{i}$, $L_{i+1}$, and $s_{i}$ leave
the space $W^{\lambda}_{\pm\alpha}\oplus W^{\lambda}_{\pm\beta}$ invariant,
and they act in the homogeneous basis
$\\{v_{\alpha}^{+},w_{\beta}^{+},v_{\alpha}^{-},w_{\beta}^{-}\\}$ of this
space via the following supermatrices:

$\displaystyle L_{i}$
$\displaystyle=\left[\begin{array}[]{cc|cc}0&0&\alpha_{i}&0\\\
0&0&0&\alpha_{i+1}\\\ \hline\cr\alpha_{i}&0&0&0\\\
0&\alpha_{i+1}&0&0\end{array}\right],$ $\displaystyle L_{i+1}$
$\displaystyle=\left[\begin{array}[]{cc|cc}0&0&\alpha_{i+1}&0\\\
0&0&0&\alpha_{i}\\\ \hline\cr\alpha_{i+1}&0&0&0\\\
0&\alpha_{i}&0&0\end{array}\right],$

and

$s_{i}=\left[\begin{array}[]{cc|cc}0&0&c_{i}&1-c_{i}^{2}\\\ 0&0&1&-c_{i}\\\
\hline\cr c_{i}&1-c_{i}^{2}&0&0\\\ 1&-c_{i}&0&0\end{array}\right].$

###  3.4. Restriction of irreducible supermodules

Given partitions $\lambda\vdash n$ and $\mu\vdash(n-1)$, write
$\mu\prec\lambda$ if the Young diagram of $\mu$ is obtained by removing a box
from the Young diagram of $\lambda$. In this case, let
$\operatorname{cont}(\lambda/\mu)$ denote the content of the box that is
removed from $\lambda$ to obtain $\mu$. Let $\operatorname{cont}(\lambda)$
denote the sum of the contents of all the boxes in the Young diagram for
$\lambda$.

Identify $S_{n-1}$ with the subgroup of $S_{n}$ consisting of all permutations
that leave the integer $n$ fixed. Then $\sigma\cdot
L_{n}\cdot\sigma^{-1}=L_{n}$ for each $\sigma\in S_{n-1}$, and hence $L_{n}$
commutes (in the ordinary, non-super sense) with each element of $S_{n-1}$.
This implies for each partition $\lambda\vdash n$ that
$\operatorname{Res}_{S_{n-1}}^{S_{n}}(S^{\lambda})$ decomposes into
eigenspaces for the action of $L_{n}$. In fact, one has

(3.4.1)
$\operatorname{Res}_{S_{n-1}}^{S_{n}}(S^{\lambda})=\bigoplus_{\mu\prec\lambda}\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu)\end{subarray}}S^{\lambda}_{\alpha}\Big{]},$

and the summand indexed by $\mu$ is isomorphic as a $\mathbb{C}S_{n-1}$-module
to $S^{\mu}$.

Making the $\mathbb{C}S_{n-1}$-module identifications
$S^{\lambda}=\bigoplus_{\mu\prec\lambda}S^{\mu}$ and
$S^{\lambda^{\prime}}=\bigoplus_{\mu^{\prime}\prec\lambda^{\prime}}S^{\mu^{\prime}}$,
and using the fact that $\operatorname{Hom}_{S_{n-1}}(S^{\mu},S^{\nu})=0$
unless $\mu=\nu$, one can show that the functions $\phi^{\lambda}$ and
$\phi^{\lambda^{\prime}}$ must restrict for each $\mu\prec\lambda$ to linear
isomorphisms $S^{\mu}\to S^{\mu^{\prime}}$ and $S^{\mu^{\prime}}\to S^{\mu}$
that satisfy the intertwining relation (3.1.1) for all $\sigma\in S_{n-1}$,
and whose composites are the respective identity functions. Then by uniqueness
(up to mutual rescaling), we may assume that
$\phi^{\lambda}|_{S^{\mu}}=\phi^{\mu}$ and
$\phi^{\lambda^{\prime}}|_{S^{\mu^{\prime}}}=\phi^{\mu^{\prime}}$.

Now let $\lambda\in\overline{\mathcal{P}}(n)=E_{n}\cup F_{n}$. As a
superspace, one has

(3.4.2)
$W^{\lambda}=\bigoplus_{k\in\mathbb{Z}}W_{k}^{\lambda},\quad\text{where}\quad
W_{k}^{\lambda}=\bigoplus_{\begin{subarray}{c}\pm\alpha\in\overline{\mathcal{W}}(\lambda)\\\
\alpha_{n}=k\end{subarray}}W^{\lambda}_{\pm\alpha}.$

By our conventions for the choice of the ‘positive’ weight $\alpha$ from each
pair $\pm\alpha\in\overline{\mathcal{W}}(\lambda)$, if $\lambda\in F_{n}$ and
$W_{k}^{\lambda}\neq 0$, then $k\geq 0$. In general, if $W_{k}^{\lambda}\neq
0$, then there exists a unique partition $\mu\vdash(n-1)$ such that
$\mu\prec\lambda$ and $\operatorname{cont}(\lambda/\mu)=k$. Specifically,
$\mu$ is the partition obtained by removing a box of content $k$ from the
outer edge of the Young diagram of $\lambda$. Indeed, a box of content $k$ can
be removed from the outer edge of the Young diagram of $\lambda$ to produce a
new partition $\mu$ if and only if there exists a weight
$\alpha\in\mathcal{W}(\lambda)$ with $\alpha_{n}=k$, and for any given $k$
there is at most one removable box of content $k$ in the Young diagram of
$\lambda$. For any $\lambda\vdash n$, the boxes in the Young diagram of
$\lambda$ have contents bounded between $-(n-1)$ and $n-1$, so in (3.4.2) one
has $W_{k}^{\lambda}\neq 0$ only if $0\leq k<n$.

Since $L_{n}$ commutes with $S_{n-1}$, it follows that $W_{k}^{\lambda}$ is a
$\mathbb{C}S_{n-1}$-subsupermodule of $W^{\lambda}$.

######  Proposition 3.4.1.

prop:Wklambda-restriction Let $\lambda\in\overline{\mathcal{P}}(n)$, let
$k\in\mathbb{Z}$ such that $W_{k}^{\lambda}\neq 0$, and let $\mu\vdash(n-1)$
be the unique partition such that $\mu\prec\lambda$ and
$\operatorname{cont}(\lambda/\mu)=k$. Then as a
$\mathbb{C}S_{n-1}$-supermodule,

$W_{k}^{\lambda}\cong\begin{cases}W^{\mu}\oplus\Pi(W^{\mu})&\text{if
$\lambda\in E_{n}$ and $\mu=\mu^{\prime}$,}\\\
W^{\mu}&\text{otherwise.}\end{cases}$

###### Proof.

First suppose $\lambda\in E_{n}$. Then

$W_{k}^{\lambda}=\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=k\end{subarray}}\left[S^{\lambda}_{\alpha}\oplus
S^{\lambda^{\prime}}_{-\alpha}\right]=\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu)\end{subarray}}S^{\lambda}_{\alpha}\Big{]}\oplus\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda^{\prime})\\\
\alpha_{n}=\operatorname{cont}(\lambda^{\prime}/\mu^{\prime})\end{subarray}}S^{\lambda^{\prime}}_{\alpha}\Big{]}.$

By (3.4.1), this is isomorphic as a $|\mathbb{C}S_{n-1}|$-module to
$S^{\mu}\oplus S^{\mu^{\prime}}$. For $\mu\neq\mu^{\prime}$, this implies that
$W_{k}^{\lambda}\cong W^{\mu}$ as a $\mathbb{C}S_{n-1}$-supermodule, so
suppose that $\mu=\mu^{\prime}$. Making the $|\mathbb{C}S_{n-1}|$-module
identification

$S^{\mu}=\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu)\end{subarray}}S^{\lambda}_{\alpha},$

one sees that $W_{k}^{\lambda}$ decomposes into the direct sum of two
$\mathbb{C}S_{n-1}$-supermodules,

(3.4.3) $\begin{split}W^{\mu}&\cong\\{u+\phi^{\lambda}(u):u\in
S^{\mu^{+}}\\}\oplus\\{w-\phi^{\lambda}(w):w\in
S^{\mu^{-}}\\},\quad\text{and}\\\
\Pi(W^{\mu})&\cong\\{u-\phi^{\lambda}(u):u\in
S^{\mu^{+}}\\}\oplus\\{w+\phi^{\lambda}(w):w\in S^{\mu^{-}}\\}.\end{split}$

In particular, $W^{\mu}$ and $\Pi(W^{\mu})$ are interchanged by the odd
involution $J^{\lambda}:W^{\lambda}\to W^{\lambda}$.

Now suppose $\lambda\in F_{n}$. If $k>0$, then the partition $\mu$ is non-
symmetric, $\operatorname{cont}(\lambda/\mu^{\prime})=-k$, and

$W_{k}^{\lambda}=\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu)\end{subarray}}S^{\lambda}_{\alpha}\Big{]}\oplus\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu^{\prime})\end{subarray}}S^{\lambda}_{\alpha}\Big{]}\cong
S^{\mu}\oplus S^{\mu^{\prime}}$

as $|\mathbb{C}S_{n-1}|$-modules. This implies that $W_{k}^{\lambda}\cong
W^{\mu}$ as a $\mathbb{C}S_{n-1}$-supermodule. On the other hand, if $k=0$,
then $\mu$ is symmetric, and

$W_{k}^{\lambda}=\Big{[}\bigoplus_{\begin{subarray}{c}\alpha\in\mathcal{W}(\lambda)\\\
\alpha_{n}=\operatorname{cont}(\lambda/\mu)\end{subarray}}S^{\lambda}_{\alpha}\Big{]}\cong
S^{\mu}$

as a $|\mathbb{C}S_{n-1}|$-module. Since $\phi^{\lambda}:S^{\lambda}\to
S^{\lambda}$ restricts to $\phi^{\mu}:S^{\mu}\to S^{\mu}$ via this
identification, one deduces that the $+1$-eigenspace of $\phi^{\mu}$ is
contained in the $+1$-eigenspace of $\phi^{\lambda}$. Then $S^{\mu^{+}}$ is
concentrated in even superdegree, so $W_{k}^{\lambda}\cong W^{\mu}$ as a
$\mathbb{C}S_{n-1}$-supermodule. ∎

Let $W^{\lambda}\\!\\!\downarrow_{\mathbb{C}S_{n-1}}$ denote the restriction
of $W^{\lambda}$ to the subalgebra $\mathbb{C}S_{n-1}$ of $\mathbb{C}S_{n}$.

######  Corollary 3.4.2.

cor:Wlambda-restriction Let $\lambda\in\overline{\mathcal{P}}(n)$. Then

$W^{\lambda}\\!\\!\downarrow_{\mathbb{C}S_{n-1}}\cong\begin{cases}\displaystyle\Bigg{[}\bigoplus_{\begin{subarray}{c}\mu\prec\lambda\\\
\mu\neq\mu^{\prime}\end{subarray}}W^{\mu}\Bigg{]}\oplus\Bigg{[}\bigoplus_{\begin{subarray}{c}\mu\prec\lambda\\\
\mu=\mu^{\prime}\end{subarray}}W^{\mu}\oplus\Pi(W^{\mu})\Bigg{]}&\text{if
$\lambda\in E_{n}$,}\\\
\displaystyle\bigoplus_{\begin{subarray}{c}\mu\prec\lambda\\\
\operatorname{cont}(\lambda/\mu)\geq 0\end{subarray}}W^{\mu}&\text{if
$\lambda\in F_{n}$.}\end{cases}$

######  Remark 3.4.3.

rem:multiplicity-free The corollary implies that, if one allows only even
supermodule homomorphisms (so that a supermodule and its parity shift are not
necessarily isomorphic), then the restriction
$W^{\lambda}\\!\\!\downarrow_{\mathbb{C}S_{n-1}}$ is multiplicity free, just
as in the classical (non-super) situation for Specht modules. If, on the other
hand, one allows odd isomorphisms as well (so that a supermodule and its
parity shift are _odd_ isomorphic), then the restriction
$W^{\lambda}\\!\\!\downarrow_{\mathbb{C}S_{n-1}}$ is multiplicity free if
$\lambda\in F_{n}$, but may have a (unique) repeated composition factor if
$\lambda\in E_{n}$.

##  4\. The Lie superalgebra generated by transpositions

###  4.1. The setup

Recall the associative superalgebras defined in ex:M(m—n) and ex:Q(n). Given a
vector superspace $V\cong\mathbb{C}^{m|n}$, write $\mathfrak{gl}(V)$ and
$\mathfrak{gl}(m|n)$ for the sets $\operatorname{End}(V)$ and $M(m|n)$,
respectively, considered as Lie superalgebras via the super commutator

$[x,y]=xy-(-1)^{\overline{x}\cdot\overline{y}}yx.$

If $V$ is a vector superspace equipped with an odd involution $J:V\to V$ (so
in particular, the even and odd subspaces of $V$ must be of the same
dimension), write $\mathfrak{q}(V)$ and $\mathfrak{q}(n)$ for the sets $Q(V)$
and $Q(n)$, respectively, considered as Lie superalgebras via the super
commutator. For an arbitrary Lie superalgebra $\mathfrak{g}$, we denote its
derived subalgebra $[\mathfrak{g},\mathfrak{g}]$ by
$\mathfrak{D}(\mathfrak{g})$. Then

$\mathfrak{D}(\mathfrak{gl}(m|n))=\mathfrak{sl}(m|n):=\left\\{\left[\begin{array}[]{c|c}A&B\\\
\hline\cr
C&D\end{array}\right]\in\mathfrak{gl}(m|n):\operatorname{tr}(A)-\operatorname{tr}(D)=0\right\\}.$

Let $V$ be a vector superspace equipped with an odd involution $J:V\to V$, let
$\theta\in\mathfrak{q}(V)$, and let
$\theta=\theta_{\overline{0}}+\theta_{\overline{1}}$ be the decomposition of
$\theta$ into its even and odd components. Then
$J\circ\theta_{\overline{1}}=\theta_{\overline{1}}\circ J$ restricts to an
even linear map
$(J\circ\theta_{\overline{1}})|_{V_{\overline{0}}}:V_{\overline{0}}\to
V_{\overline{0}}$. Identifying $V_{\overline{0}}$ and $V_{\overline{1}}$ via
$J$, this is equal to the even linear map $(\theta_{\overline{1}}\circ
J)|_{V_{\overline{1}}}:V_{\overline{1}}\to V_{\overline{1}}$. Now define the
_odd trace_ of $\theta$, denoted $\operatorname{otr}(\theta)$, by

(4.1.1)
$\operatorname{otr}(\theta)=\operatorname{tr}\big{(}(J\circ\theta_{\overline{1}})|_{V_{\overline{0}}}\big{)}=\operatorname{tr}\big{(}(\theta_{\overline{1}}\circ
J)|_{V_{\overline{1}}}\big{)},$

and define the subsuperspace $\mathfrak{sq}(V)\subseteq\mathfrak{q}(V)$ by

$\mathfrak{sq}(V)=\left\\{\theta\in\mathfrak{q}(V):\operatorname{otr}(\theta)=0\right\\}.$

Then one can show that $\mathfrak{D}(\mathfrak{q}(V))=\mathfrak{sq}(V)$.
Fixing a basis for $V$ as in ex:Q(n), one has

(4.1.2)
$\mathfrak{D}(\mathfrak{q}(n))=\mathfrak{sq}(n):=\left\\{\left[\begin{array}[]{c|c}A&B\\\
\hline\cr
B&A\end{array}\right]\in\mathfrak{q}(n):\operatorname{tr}(B)=0\right\\}.$

######  Remark 4.1.1.

rem:generated-by-odd One can check that the Lie superalgebra
$\mathfrak{sl}(m|m)$ is generated by its odd subspace
$\mathfrak{sl}(m|m)_{\overline{1}}$ if $m\geq 2$, and that $\mathfrak{sq}(m)$
is generated by its odd subspace $\mathfrak{sq}(m)_{\overline{1}}$ if $m\geq
3$.

The super commutator endows the superalgebra $\mathbb{C}S_{n}$ with the
structure of a Lie superalgebra. cor:CSn-as-superalgebra then gives the Lie
superalgebra isomorphism

(4.1.3) $\mathbb{C}S_{n}\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}\mathfrak{q}(W^{\lambda})\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}\mathfrak{gl}(W^{\lambda})\Bigg{]}\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}\mathfrak{q}(f^{\lambda})\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}\mathfrak{gl}(\tfrac{1}{2}f^{\lambda},\tfrac{1}{2}f^{\lambda})\Bigg{]},$

where $f^{\lambda}=\dim(S^{\lambda})$. Taking derived subalgebras, one has

(4.1.4) $\mathfrak{D}(\mathbb{C}S_{n})\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}\mathfrak{sq}(W^{\lambda})\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}\mathfrak{sl}(W^{\lambda})\Bigg{]}\cong\Bigg{[}\bigoplus_{\lambda\in
E_{n}}\mathfrak{sq}(f^{\lambda})\Bigg{]}\oplus\Bigg{[}\bigoplus_{\lambda\in
F_{n}}\mathfrak{sl}(\tfrac{1}{2}f^{\lambda},\tfrac{1}{2}f^{\lambda})\Bigg{]}.$

From this one sees that

$\displaystyle\dim(\mathfrak{D}(\mathbb{C}S_{n})_{\overline{0}})$
$\displaystyle=\dim((\mathbb{C}S_{n})_{\overline{0}})-|F_{n}|=\tfrac{n!}{2}-|F_{n}|,\quad\text{and}$
$\displaystyle\dim(\mathfrak{D}(\mathbb{C}S_{n})_{\overline{1}})$
$\displaystyle=\dim((\mathbb{C}S_{n})_{\overline{1}})-|E_{n}|=\tfrac{n!}{2}-|E_{n}|.$

In total, $\dim(\mathfrak{D}(\mathbb{C}S_{n}))=n!-|E_{n}\cup F_{n}|$.

Let $T_{n}=\sum_{j=1}^{n}L_{j}=\sum_{j=1}^{n}\sum_{i=1}^{j-1}(i,j)$, the sum
in $\mathbb{C}S_{n}$ of all transpositions $\tau\in S_{n}$. Let $\lambda\vdash
n$. Since all transpositions are conjugate in $S_{n}$, the trace of the map
$S^{\lambda}(\tau):S^{\lambda}\to S^{\lambda}$ is independent of $\tau$. This
implies for each transposition $\tau$ that
$\operatorname{tr}(S^{\lambda}(T_{n}))=\binom{n}{2}\cdot\operatorname{tr}(S^{\lambda}(\tau))$,
and hence $\tau-\frac{2}{n(n-1)}\cdot T_{n}$ is a traceless operator on
$S^{\lambda}$. Then (3.1.3) implies for $\lambda\in E_{n}$ that

$\textstyle W^{\lambda}\left(\tau-\frac{2}{n(n-1)}\cdot
T_{n}\right)\in\mathfrak{sq}(W^{\lambda})=\mathfrak{D}(\mathfrak{q}(W^{\lambda})),$

while for $\lambda\in F_{n}$ one gets

$\textstyle W^{\lambda}\left(\tau-\frac{2}{n(n-1)}\cdot
T_{n}\right)\in\mathfrak{sl}(W^{\lambda})=\mathfrak{D}(\mathfrak{gl}(W^{\lambda})),$

because for any finite-dimensional superspace $V$ one has
$\mathfrak{gl}(V)_{\overline{1}}=\mathfrak{sl}(V)_{\overline{1}}=\mathfrak{D}(\mathfrak{gl}(V))_{\overline{1}}$.
Combining these observations, one gets

(4.1.5) $\textstyle\left\\{\tau-\frac{2}{n(n-1)}\cdot T_{n}:\tau\text{ is a
transposition in $S_{n}$}\right\\}\subseteq\mathfrak{D}(\mathbb{C}S_{n}).$

On the other hand, for any partition $\lambda\vdash n$, $T_{n}$ acts on
$S^{\lambda}$ as scalar multiplication by $\operatorname{cont}(\lambda)$, the
sum of the contents of the boxes in the Young diagram of $\lambda$. For $n>1$,
there are non-symmetric partitions for which this scalar is nonzero, so this
implies by (3.1.3) that $T_{n}\notin\mathfrak{D}(\mathbb{C}S_{n})$, and hence
$\tau\notin\mathfrak{D}(\mathbb{C}S_{n})$, as well, for each transposition
$\tau$. Since $T_{n}$ acts on $S^{\lambda}$ as scalar multiplication by
$\operatorname{cont}(\lambda)$, it follows that

(4.1.6) $W^{\lambda}(T_{n})=\begin{cases}\operatorname{cont}(\lambda)\cdot
J^{\lambda}&\text{if $\lambda\in E_{n}$,}\\\ 0&\text{if $\lambda\in
F_{n}$,}\end{cases}$

where $J^{\lambda}:W^{\lambda}\to W^{\lambda}$ is the odd involution defined
in prop:Wlambda-decomposition(1).

######  Remark 4.1.2.

If $\operatorname{cont}(\lambda)\neq 0$, then $\lambda\neq\lambda^{\prime}$.
The converse of this statement is false. For example, if
$\lambda=(5,5,5,3,1,1)$, then $\lambda\neq\lambda^{\prime}$ but
$\operatorname{cont}(\lambda)=0$.

######  Definition 4.1.3.

Let $\mathfrak{g}_{n}\subseteq\mathbb{C}S_{n}$ be the Lie superalgebra
generated by all transpositions in $S_{n}$.

Evidently, $T_{n}\in\mathfrak{g}_{n}$. Then (4.1.5) implies that

(4.1.7)
$\mathfrak{g}_{n}\subseteq\mathfrak{D}(\mathbb{C}S_{n})+\mathbb{C}\cdot
T_{n}.$

Our goal by the end of the paper is to show that (4.1.7) is an equality for
all $n\geq 2$.

######  Lemma 4.1.4.

lemma:gn-equality If $n\in\left\\{2,3,4,5\right\\}$, then
$\mathfrak{g}_{n}=\mathfrak{D}(\mathbb{C}S_{n})+\mathbb{C}\cdot T_{n}$.

###### Proof.

Since $T_{n}\notin\mathfrak{D}(\mathbb{C}S_{n})$, the sum
$\mathfrak{D}(\mathbb{C}S_{n})+\mathbb{C}\cdot T_{n}$ is direct, and hence

$\dim(\mathfrak{D}(\mathbb{C}S_{n})+\mathbb{C}\cdot
T_{n})=\dim(\mathfrak{D}(\mathbb{C}S_{n}))+1=n!-|E_{n}\cup F_{n}|+1.$

For $n\leq 5$, we have verified that $\dim(\mathfrak{g}_{n})\geq n!-|E_{n}\cup
F_{n}|+1$, and hence (4.1.7) is an equality, via calculations in GAP [6]. ∎

######  Remark 4.1.5.

It is straightforward, if somewhat tedious, to check by hand for $n\leq 4$
that $\dim(\mathfrak{g}_{n})\geq n!-|E_{n}\cup F_{n}|+1$. Later in Section
4.4, we will find it convenient to assume that (4.1.7) is an equality for
$n=5$, as well, to help avoid certain annoying special cases. In fact, we have
verified that lemma:gn-equality is also true for $n=6$ and $n=7$ using GAP,
but there is nothing to be gained in our induction argument by taking these
cases for granted.

######  Lemma 4.1.6.

lemma:lie-algebra-centers Let $n\geq 2$. Set $\mathfrak{g}=\mathfrak{g}_{n}$.

  1. (1)

$Z(\mathbb{C}S_{n})=Z(\mathbb{C}S_{n})_{\overline{0}}$.

  2. (2)

$Z(\mathfrak{g})\subseteq Z(\mathbb{C}S_{n})$. In particular,
$Z(\mathfrak{g})\subseteq\mathfrak{g}_{\overline{0}}$.

  3. (3)

If $n\geq 5$, then
$Z(\mathfrak{g}_{\overline{0}})=\mathfrak{g}_{\overline{0}}\cap
Z(\mathbb{C}A_{n})$, and the projection map $p:\mathbb{C}A_{n}\to
Z(\mathbb{C}A_{n})$,

$p(z)=\frac{2}{n!}\sum_{\sigma\in A_{n}}\sigma z\sigma^{-1},$

restricts to a projection map $p:\mathfrak{g}_{\overline{0}}\to
Z(\mathfrak{g}_{\overline{0}})$.

###### Proof.

Under the superalgebra isomorphism (4.1.3), one can deduce that the only
homogeneous elements of $\mathbb{C}S_{n}$ that commute (in the super sense)
with all other elements of $\mathbb{C}S_{n}$ correspond to linear combinations
of the identity elements from the various matrix factors. In particular,
$Z(\mathbb{C}S_{n})$ is a purely even superspace.

Next, for each $z\in\mathbb{C}S_{n}$, the map
$\operatorname{ad}_{z}:x\mapsto[z,x]=zx-(-1)^{\overline{z}\cdot\overline{x}}xz$
is a superalgebra derivation on $\mathbb{C}S_{n}$. If $z\in Z(\mathfrak{g})$,
then $\operatorname{ad}_{z}(x)=0$ for each transposition $x$, since those
elements generate $\mathfrak{g}$ as a Lie superalgebra. But the transpositions
also generate $\mathbb{C}S_{n}$ as an associative superalgebra, so this
implies that $\operatorname{ad}_{z}:\mathbb{C}S_{n}\to\mathbb{C}S_{n}$ is the
zero map, and hence $z\in Z(\mathbb{C}S_{n})$.

Now suppose $n\geq 5$. In this case, it is well-known that $A_{n}$ is
generated as a group by the set

(4.1.8) $\left\\{(i,j)(k,\ell):\text{$i,j,k,\ell$ distinct}\right\\}$

of all products of two disjoint transpositions. These are all elements of
$\mathfrak{g}_{\overline{0}}$ because

(4.1.9) $[(i,j),(k,\ell)]=(i,j)(k,\ell)+(k,\ell)(i,j)=2(i,j)(k,\ell)$

whenever $i,j,k,\ell$ are distinct. Then reasoning as in the previous
paragraph, it follows for $z\in\mathfrak{g}_{\overline{0}}$ that $z\in
Z(\mathfrak{g}_{\overline{0}})$ if and only if $z\in Z(\mathbb{C}A_{n})$.
Finally, since the set of transpositions in $S_{n}$ is closed under
conjugation by arbitrary elements of $S_{n}$, it follows that $\mathfrak{g}$
is closed under conjugation. Conjugation is an even linear map, so
$\mathfrak{g}_{\overline{0}}$ is also closed under conjugation. Then the
projection map must sends elements of $\mathfrak{g}_{\overline{0}}$ to
elements of $\mathfrak{g}_{\overline{0}}\cap
Z(\mathbb{C}A_{n})=Z(\mathfrak{g}_{\overline{0}})$. ∎

###  4.2. Image of $\mathfrak{g}_{n}$ in $\operatorname{End}(W^{\lambda})$

Given $\lambda\in\overline{\mathcal{P}}(n)$, let
$W^{\lambda}(\mathfrak{g}_{n})$ denote the image of $\mathfrak{g}_{n}$ under
the supermodule structure map
$\mathbb{C}S_{n}\to\operatorname{End}(W^{\lambda})$. Our goal in Sections 4.3
and 4.4 is to establish theorem:Wlambda(gn), stated below.

As described in the introduction, we prove the results in Sections 4.2–4.7 by
induction on $n$. First, for the base case of induction, observe that
theorem:finaltheorem is true for $n\in\left\\{2,3,4,5\right\\}$, by lemma:gn-
equality. This implies that theorem:Wlambda(gn) is true for
$n\in\left\\{2,3,4,5\right\\}$ by (4.1.4) and (4.1.6). Hence cor:g0-reductive
is true for $n$ in this range as well. In the case $n=5$, one can then work
sequentially through Sections 4.6 and 4.7 to deduce that all subsequent
results in the paper leading up to theorem:finaltheorem are also true for
$n=5$. Now for the general inductive step of this argument we make the
following assumptions:

  * •

$n\geq 6$, and

  * •

all results in Sections 4.2–4.7 are true as stated for the value $n-1$.

The inductive step is then completed by working sequentially through Sections
4.2–4.7, starting with theorem:Wlambda(gn), to establish that each result is
true as stated for the value $n$.

######  Theorem 4.2.1.

theorem:Wlambda(gn) Let $n\geq 2$, and let
$\lambda\in\overline{\mathcal{P}}(n)$. Then

(4.2.1)
$W^{\lambda}(\mathfrak{g}_{n})=\begin{cases}\mathfrak{sq}(W^{\lambda})+\mathbb{C}\cdot(\operatorname{cont}(\lambda)\cdot
J^{\lambda})&\text{if $\lambda\in E_{n}$,}\\\
\mathfrak{sl}(W^{\lambda})&\text{if $\lambda\in F_{n}$.}\end{cases}$

The “$\subseteq$” direction of (4.2.1) follows from (4.1.7), (4.1.4), and
(4.1.6).

If $\lambda\in E_{n}$, then (4.1.6) also implies that
$\operatorname{cont}(\lambda)\cdot J^{\lambda}\in
W^{\lambda}(\mathfrak{g}_{n})$. Next, if $\tau\in S_{n}$ is any transposition,
then

$\operatorname{id}_{W^{\lambda}}=W^{\lambda}(1_{\mathbb{C}S_{n}})=W^{\lambda}\left(\tfrac{1}{2}[\tau,\tau]\right)\in
W^{\lambda}(\mathfrak{g}_{n}).$

For $\lambda\in\left\\{(n),(1^{n})\right\\}$, one has
$\mathfrak{sq}(W^{\lambda})=\mathbb{C}\cdot\operatorname{id}_{W^{\lambda}}$,
so the theorem is true in this case. For all other partitions
$\lambda\in\overline{\mathcal{P}}(n)$, lemma:lower-bound-dim implies (by the
assumption $n\geq 6$, and the fact that all $W^{\lambda}$ are even-
dimensional) that $\dim(W^{\lambda})\geq 10$. Then to finish proving the
“$\supseteq$” direction of (LABEL:theorem:Wlambda(gn)), it will suffice by
rem:generated-by-odd to show that

(4.2.2)
$W^{\lambda}(\mathfrak{g}_{n})\supseteq\begin{cases}\mathfrak{sq}(W^{\lambda})_{\overline{1}}&\text{if
$\lambda\in E_{n}$,}\\\ \mathfrak{sl}(W^{\lambda})_{\overline{1}}&\text{if
$\lambda\in F_{n}$.}\end{cases}$

In the notation of Section 3.4, one has
$W^{\lambda}=\bigoplus_{k\in\mathbb{Z}}W_{k}^{\lambda}$, where
$W_{k}^{\lambda}\neq 0$ if and only if there exists a (unique) partition
$\mu_{k}\vdash(n-1)$ such that $\mu_{k}\prec\lambda$ and
$\operatorname{cont}(\lambda/\mu_{k})=k$. To simplify notation, for the rest
of this section we will fix a partition $\lambda\in\overline{\mathcal{P}}(n)$,
and we will write

$W^{k}=W_{k}^{\lambda}.$

By prop:Wklambda-restriction, if $W^{k}\neq 0$, then $W^{k}$ identifies as a
$\mathbb{C}S_{n-1}$-supermodule with either the irreducible supermodule
$W^{\mu_{k}}$, or the direct sum of $W^{\mu_{k}}$ and its parity shift
$\Pi(W^{\mu_{k}})$. In any case, if $k\neq\ell$, then $W^{k}$ and $W^{\ell}$
have no irreducible $\mathbb{C}S_{n-1}$-constituents in common. This implies
by the analogue of (4.1.4) for $\mathbb{C}S_{n-1}$ and the induction
hypothesis for $\mathfrak{g}_{n-1}^{\prime}:=\mathfrak{D}(\mathfrak{g}_{n-1})$
that

(4.2.3)
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})=\bigoplus_{k\in\mathbb{Z}}W^{k}(\mathfrak{g}_{n-1}^{\prime}),$

where $W^{k}(\mathfrak{g}_{n-1}^{\prime})$ denotes the image of
$\mathfrak{g}_{n-1}^{\prime}$ in $\operatorname{End}(W^{k})$.

Conceptually, our strategy for the proof of theorem:Wlambda(gn) runs roughly
as follows. First, we show that $W^{\lambda}(\mathfrak{g}_{n})$ contains a
large semisimple Lie subalgebra $\mathfrak{h}$—specifically, a direct sum of
special linear Lie algebras—over which

(4.2.4) $\displaystyle\operatorname{End}(W^{\lambda})_{\overline{1}}$
$\displaystyle=\bigoplus_{k,\ell\in\mathbb{Z}}\operatorname{Hom}(W^{k},W^{\ell})_{\overline{1}}$
(4.2.5)
$\displaystyle=\bigoplus_{k,\ell\in\mathbb{Z}}\left[\operatorname{Hom}(W^{k}_{\overline{0}},W^{\ell}_{\overline{1}})\oplus\operatorname{Hom}(W^{k}_{\overline{1}},W^{\ell}_{\overline{0}})\right]$

is a semisimple $\mathfrak{h}$-module. Next, the transposition
$s_{n-1}=(n-1,n)$ defines an element
$W^{\lambda}(s_{n-1})\in\operatorname{End}(W^{\lambda})_{\overline{1}}$ that
has nonzero components in various irreducible $\mathfrak{h}$-module summands
of $\operatorname{End}(W^{\lambda})_{\overline{1}}$. Using the semisimplicity
of $\mathfrak{h}$, we deduce that certain $\mathfrak{h}$-module summands of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$ must be contained in the Lie
superalgebra generated by $W^{\lambda}(\mathfrak{g}_{n-1})$ and
$W^{\lambda}(s_{n-1})$, and hence must be contained in
$W^{\lambda}(\mathfrak{g}_{n})$. These summands in turn generate a large
enough Lie superalgebra for us to deduce the inclusion (4.2.2).

###  4.3. Proof of Theorem LABEL:theorem:Wlambda(gn): the case $\lambda\in
F_{n}$

#### 4.3.1.

First suppose $\lambda\in F_{n}$. In this case one has $W^{k}\neq 0$ only if
$k\geq 0$, and the decomposition (4.2.3) takes the form

(4.3.1)
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})=\bigoplus_{k=0}^{n-1}W^{k}(\mathfrak{g}_{n-1}^{\prime})=\mathfrak{sl}(W^{0})\oplus\mathfrak{sq}(W^{1})\oplus\mathfrak{sq}(W^{2})\oplus\cdots\oplus\mathfrak{sq}(W^{n-1}).$

By convention, $\mathfrak{sl}(W^{k})=0$ whenever $W^{k}=0$. For example, since
$n\geq 6$ by hypothesis, and since $\lambda\in F_{n}$ is a symmetric
partition, the Young diagram of $\lambda$ cannot have a removable box of
content $n-1$, so the summand $\mathfrak{sq}(W^{n-1})$ must be zero.
Furthermore, lemma:lower-bound-dim implies that if $W^{k}\neq 0$, then

(4.3.2) $\dim(W^{k})\geq\min\left\\{2(n-1)-2,(n-1)+1\right\\}\geq 6.$

#### 4.3.2.

If $\mathfrak{sl}(W^{0})$ is the only nonzero summand in (4.3.1), then
$W^{\lambda}=W^{0}$, and hence

$\mathfrak{sl}(W^{\lambda})=\mathfrak{sl}(W^{0})=W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})\subseteq
W^{\lambda}(\mathfrak{g}_{n}),$

establishing (4.2.2). So assume that $W^{k}\neq 0$ for at least one value
$k>0$. For such $k$, $W^{k}\cong W^{\mu_{k}}$ is a self-associate irreducible
$\mathbb{C}S_{n-1}$-supermodule, and the even and odd subspaces
$W^{k}_{\overline{0}}$ and $W^{k}_{\overline{1}}$ of $W^{k}$ can be identified
via the odd involution $J^{\mu_{k}}:W^{\mu_{k}}\to W^{\mu_{k}}$. Making the
identification $W^{k}_{\overline{0}}\simeq W^{k}_{\overline{1}}$ via
$J^{\mu_{k}}$, and writing $W_{k}$ for this new common space (considered just
as an ordinary vector space, without any superspace structure), the diagonal
maps

(4.3.3)
$\mathfrak{gl}(W_{k})\to\mathfrak{gl}(W^{k}_{\overline{0}})\oplus\mathfrak{gl}(W^{k}_{\overline{1}})\quad\text{and}\quad\mathfrak{sl}(W_{k})\to\operatorname{Hom}(W^{k}_{\overline{0}},W^{k}_{\overline{1}})\oplus\operatorname{Hom}(W^{k}_{\overline{1}},W^{k}_{\overline{0}})$

induce vector space isomorphisms
$\mathfrak{gl}(W_{k})\cong\mathfrak{sq}(W^{k})_{\overline{0}}$ and
$\mathfrak{sl}(W_{k})\cong\mathfrak{sq}(W^{k})_{\overline{1}}$ that are
compatible with the adjoint action. At the risk of confusing the reader, we
will immediately change the meaning of our notation and will write
$\mathfrak{sl}(W_{k})$ to mean the evident Lie subalgebra of
$\mathfrak{gl}(W_{k})\cong\mathfrak{sq}(W^{k})_{\overline{0}}$. With this
notation, we see that

$\mathfrak{f}:=[\mathfrak{sl}(W^{0}_{\overline{0}})\oplus\mathfrak{sl}(W^{0}_{\overline{1}})]\oplus\mathfrak{sl}(W_{1})\oplus\mathfrak{sl}(W_{2})\oplus\cdots\oplus\mathfrak{sl}(W_{n-1})$

naturally identifies with a semisimple Lie subalgebra of
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})_{\overline{0}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$.111Since $\dim(W^{k})\geq 6$ whenever
$W^{k}\neq 0$, the nonzero summands in $\mathfrak{f}$ are each of the form
$\mathfrak{sl}(m)$ for some $m\geq 3$. Further, (4.2.4) and (4.2.5) give
$\mathfrak{f}$-module decompositions of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$ under the adjoint action.

#### 4.3.3.

We will write elements of $\operatorname{End}(W^{k})$ in the supermatrix block
form

(4.3.4) $\left[\begin{array}[]{c|c}A&B\\\ \hline\cr C&D\end{array}\right]$

where $A\in\operatorname{Hom}(W^{k}_{\overline{0}},W^{k}_{\overline{0}})$,
$B\in\operatorname{Hom}(W^{k}_{\overline{1}},W^{k}_{\overline{0}})$,
$C\in\operatorname{Hom}(W^{k}_{\overline{0}},W^{k}_{\overline{1}})$, and
$D\in\operatorname{Hom}(W^{k}_{\overline{1}},W^{k}_{\overline{1}})$. In this
notation, the inclusion $\mathfrak{sq}(W^{k})\subseteq
W^{\lambda}(\mathfrak{g}_{n})$ for $k\geq 1$ translates into the statement
that $W^{\lambda}(\mathfrak{g}_{n})$ contains all supermatrices such that
$A=D$, $B=C$, and $\operatorname{tr}(B)=0$, while the summand
$\mathfrak{sl}(W_{k})$ of the algebra $\mathfrak{f}$ identifies with those
supermatrices such that $A=D$, $B=C=0$, and $\operatorname{tr}(A)=0$. For each
$k\geq 1$, one sees that, as an $\mathfrak{f}$-module,
$\operatorname{End}(W^{k})_{\overline{1}}$ is the direct sum of:

  1. (4.3.5)

a two-dimensional trivial $\mathfrak{f}$-submodule, spanned by the odd
supermatrices222A supermatrix of the form (4.3.4) is even if $B=C=0$, and is
odd if $A=D=0$. of the form (4.3.4) such that $B$ and $C$ are arbitrary scalar
matrices; and

  2. (4.3.6)

two nontrivial isomorphic irreducible $\mathfrak{f}$-modules
$\mathfrak{sl}(W_{k})^{+}$ and $\mathfrak{sl}(W_{k})^{-}$, spanned by the odd
supermatrices of the form (4.3.4) such that $\operatorname{tr}(B)=0$ and
$C=0$, and such that $B=0$ and $\operatorname{tr}(C)=0$, respectively. Via the
projection $\mathfrak{f}\twoheadrightarrow\mathfrak{sl}(W_{k})$, these
irreducibles each identify with the adjoint representation of
$\mathfrak{sl}(W_{k})$. Moreover, these two nontrivial irreducibles do not
occur in any other summands in (4.2.4).

#### 4.3.4.

Our first main goal is to show that $W^{\lambda}(\mathfrak{g}_{n})$ contains
the semisimple Lie algebra

(4.3.7)
$\mathfrak{h}:=[\mathfrak{sl}(W^{0}_{\overline{0}})\oplus\mathfrak{sl}(W^{0}_{\overline{1}})]\oplus[\mathfrak{sl}(W^{1}_{\overline{0}})\oplus\mathfrak{sl}(W^{1}_{\overline{1}})]\oplus\cdots\oplus[\mathfrak{sl}(W^{n-1}_{\overline{0}})\oplus\mathfrak{sl}(W^{n-1}_{\overline{1}})].$

Given $k\geq 1$ such that $W^{k}\neq 0$, we will show that
$[\mathfrak{sl}(W^{k}_{\overline{0}})\oplus\mathfrak{sl}(W^{k}_{\overline{1}})]\subseteq
W^{\lambda}(\mathfrak{g}_{n})$ by considering the component of the map
$W^{\lambda}(s_{n-1})$ that lies in the summand
$\operatorname{End}(W^{k})_{\overline{1}}$ of (4.2.4).

By definition, $W^{k}$ is spanned by the vectors $v_{\alpha}^{+}$ and
$v_{\alpha}^{-}$ for $\alpha\in\overline{\mathcal{W}}(\lambda)$ of the form
$\alpha=(\cdots,j,k)$. If $|k-j|=1$, then $W^{\lambda}(s_{n-1})$ acts on the
vectors $v_{\alpha}^{+}$ and $v_{\alpha}^{-}$ via the matrix given in prop:si-
on-weights-super(1). If $|k-j|\geq 2$ and $j\neq-k$, then
$W^{\lambda}(s_{n-1})$ maps the vectors $v_{\alpha}^{+}$ and $v_{\alpha}^{-}$
into the subspace $W^{|j|}=W^{\lambda}_{|j|}$, and $W^{|j|}\neq W^{k}$. Thus,
if $|k-j|\geq 2$ and $j\neq-k$, then the action of $W^{\lambda}(s_{n-1})$ on
$v_{\alpha}^{+}$ and $v_{\alpha}^{-}$ does not arise from a map in
$\operatorname{End}(W^{k})_{\overline{1}}$.

Next consider a weight in $\overline{\mathcal{W}}(\lambda)$ of the form
$\alpha=(\alpha^{\prime\prime},-k,k)=(\alpha_{1},\ldots,\alpha_{n-2},-k,k)$.
Weights of this form do exist: If the Young diagram of $\lambda$ has a
removable box $B_{k}$ of content $k\geq 1$, then by symmetry of $\lambda$ it
also has a removable box $B_{-k}$ of content $-k$, and the two boxes can be
removed in either order to produce a partition $\lambda^{\prime\prime}$ of
$n-2$. Let $T^{\prime\prime}$ be any standard
$\lambda^{\prime\prime}$-tableau, and let
$\alpha^{\prime\prime}=\alpha(T^{\prime\prime})\in\mathcal{W}(\lambda^{\prime\prime})$
be the weight of $T^{\prime\prime}$. We can extend $T^{\prime\prime}$ to a
standard $\lambda$-tableau in two ways: by putting $n-1$ in box $B_{-k}$ and
$n$ in box $B_{k}$, to get a standard $\lambda$-tableau $T_{k}$ of weight
$(\alpha^{\prime\prime},-k,k)$, or by putting $n-1$ in box $B_{k}$ and $n$ in
box $B_{-k}$ to get a standard $\lambda$-tableau of weight $T_{-k}$ of weight
$(\alpha^{\prime\prime},k,-k)$. Every weight in
$\overline{\mathcal{W}}(\lambda)$ of the form $(\alpha^{\prime\prime},-k,k)$
arises in this way.

Now if $\alpha=(\alpha^{\prime\prime},-k,k)$ is a weight in
$\overline{\mathcal{W}}(\lambda)$, then
$\gamma:=(-\alpha^{\prime\prime},-k,k)$ is also a weight in
$\overline{\mathcal{W}}(\lambda)$, and $\alpha,-\alpha,\gamma,-\gamma$ are
four distinct weights in $\mathcal{W}(\lambda)$. Let $\beta=-\gamma$, let
$c=(k-(-k))^{-1}=1/(2k)$, and let $w_{\beta}=(s_{n-1}-c)\cdot v_{\alpha}$ and
$w_{-\beta}=(s_{n-1}+c)\cdot v_{-\alpha}$ be defined as in prop:si-on-weights-
super(2). Then $w_{\beta}=a\cdot v_{-\gamma}$ for some $0\neq a\in\mathbb{C}$,
hence $w_{-\beta}=-\phi^{\lambda}(w_{\beta})=-a\cdot v_{\gamma}$,

$\displaystyle w_{\beta}^{+}$
$\displaystyle=\tfrac{1}{2}(w_{\beta}-w_{-\beta})=a\cdot\tfrac{1}{2}(v_{-\gamma}+v_{\gamma})=a\cdot
v_{\gamma}^{+},\quad\text{and}$ $\displaystyle w_{\beta}^{-}$
$\displaystyle=\tfrac{1}{2}(w_{\beta}+w_{-\beta})=a\cdot\tfrac{1}{2}(v_{-\gamma}-v_{\gamma})=-a\cdot
v_{\gamma}^{-}.$

By prop:si-on-weights-super(2), $W^{\lambda}(s_{n-1})$ leaves invariant the
span of $v_{\alpha}^{+},v_{\gamma}^{+},v_{\alpha}^{-},v_{\gamma}^{-}$, and
acts in this homogeneous basis via the supermatrix

$\left[\begin{array}[]{cc|cc}0&0&c&-(1-c^{2})/a\\\ 0&0&a&c\\\ \hline\cr
c&(1-c^{2})/a&0&0\\\ -a&c&0&0\end{array}\right].$

Combined with the observations two paragraphs ago, this implies that the
component of $W^{\lambda}(s_{n-1})$ in
$\operatorname{End}(W^{k})_{\overline{1}}$ can be written as an odd
supermatrix of the form (4.3.4) such that $B\neq C$, but each pair of
corresponding diagonal entries in $B$ and $C$ are equal.

Since $k\geq 1$, the partition $\mu_{k}\prec\lambda$ is not symmetric. Then by
(4.1.6), $W^{k}(T_{n-1})=\operatorname{cont}(\mu_{k})J^{\mu_{k}}$, and
$\operatorname{cont}(\mu_{k})=\operatorname{cont}(\lambda)-k=-k\neq 0$. Now it
follows that for some scalar $r$, the component in
$\operatorname{End}(W^{k})_{\overline{1}}$ of the operator
$W^{k}(s_{n-1}-r\cdot T_{n-1})=W^{k}(s_{n-1})-r\cdot W^{k}(T_{n-1})$ has the
form

$\left[\begin{array}[]{c|c}0&B^{\prime}\\\ \hline\cr
C^{\prime}&0\end{array}\right]$

where $B^{\prime}$ and $C^{\prime}$ are nonzero traceless matrices. Recall
that $\mathfrak{sq}(W^{k})_{\overline{1}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$ consists of all supermatrices
$[\begin{smallmatrix}0&X\\\ X&0\end{smallmatrix}]$ such that
$\operatorname{tr}(X)=0$. Then $\phi:=W^{k}(s_{n-1}-r\cdot
T_{n-1})-[\begin{smallmatrix}0&C^{\prime}\\\ C^{\prime}&0\end{smallmatrix}]$
is an odd element of $W^{\lambda}(\mathfrak{g}_{n})$ whose component in
$\operatorname{End}(W^{k})_{\overline{1}}$ is a nonzero element of the
$\mathfrak{f}$-module $\mathfrak{sl}(W_{k})^{+}$ described in (4.3.6). Since
$\mathfrak{sl}(W_{k})^{+}$ does not occur in any other summand of (4.2.4), the
semisimplicity of the Lie algebra $\mathfrak{f}$ implies that the entire
module $\mathfrak{sl}(W_{k})^{+}$ must be contained in the
$\mathfrak{f}$-submodule of $W^{\lambda}(\mathfrak{g}_{n})$ generated by
$\phi$, i.e., $\mathfrak{sl}(W_{k})^{+}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$. By similar reasoning, one gets
$\mathfrak{sl}(W_{k})^{-}\subseteq W^{\lambda}(\mathfrak{g}_{n})$. Now using
the observation just after (4.3.2) that $\dim(W^{k})\geq 6$, and hence
$\dim(W^{k}_{\overline{0}})=\dim(W^{k}_{\overline{1}})\geq 3$, one can show
that the Lie superalgebra generated by the subspaces
$\mathfrak{sl}(W_{k})^{+}$ and $\mathfrak{sl}(W_{k})^{-}$ of
$\operatorname{End}(W^{k})$ must contain nonzero elements
$x\in\mathfrak{sl}(W^{k}_{\overline{0}})$ and
$y\in\mathfrak{sl}(W^{k}_{\overline{1}})$. Then the Lie algebra generated by
$x$, $y$, and $\mathfrak{gl}(W_{k})=\mathfrak{sq}(W^{k})_{\overline{0}}$ must
contain
$\mathfrak{sl}(W^{k}_{\overline{0}})\oplus\mathfrak{sl}(W^{k}_{\overline{1}})$.
Thus,
$[\mathfrak{sl}(W^{k}_{\overline{0}})\oplus\mathfrak{sl}(W^{k}_{\overline{1}})]\subseteq
W^{\lambda}(\mathfrak{g}_{n})$.

#### 4.3.5.

We have shown that the semisimple Lie algebra $\mathfrak{h}$ of (4.3.7) is
contained in $W^{\lambda}(\mathfrak{g}_{n})$. Next observe that (4.2.5) is a
multiplicity-free decomposition of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$ into irreducible
$\mathfrak{h}$-modules. In particular, each irreducible summand is equal to
its own isotypical component. Together with the semisimplicity of
$\mathfrak{h}$, this implies that if
$\psi\in\operatorname{End}(W^{\lambda})_{\overline{1}}$, and if $\psi$ has a
nonzero component in some irreducible $\mathfrak{h}$-module summand of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$, then the entire summand in
question must be contained in the $\mathfrak{h}$-submodule of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$ generated by $\psi$. By
induction, we already know that
$\operatorname{End}(W^{0})_{\overline{1}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$. Then to show that
$\operatorname{End}(W^{\lambda})_{\overline{1}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$, it suffices to show for all
$k,\ell\in\mathbb{Z}$ for which $W^{k}\neq 0$ and $W^{\ell}\neq 0$, and for
which at least one of $k$ or $\ell$ is nonzero, that $W^{\lambda}(s_{n-1})$
has nonzero components in both
$\operatorname{Hom}(W^{k}_{\overline{0}},W^{\ell}_{\overline{1}})$ and
$\operatorname{Hom}(W^{k}_{\overline{1}},W^{\ell}_{\overline{0}})$. We have
already established this is true when $k=\ell\geq 1$, so we may assume that
$k\neq\ell$.

If $W^{k}\neq 0$ and $W^{\ell}\neq 0$, then the Young diagram of $\lambda$ has
removable boxes $B_{k}$ and $B_{\ell}$ of contents $k$ and $\ell$,
respectively, and these boxes can be removed in either order. Moreover, since
$B_{k}$ and $B_{\ell}$ are both removable, it must be the case that
$|k-\ell|\geq 2$. Now reasoning as we did earlier, one can deduce that
$\overline{\mathcal{W}}(\lambda)$ contains a pair of weights of the forms
$\alpha=(\alpha^{\prime\prime},\ell,k)$ and
$\beta=(\alpha^{\prime\prime},k,\ell)$. Finally, applying prop:si-on-weights-
super(2), one sees that $W^{\lambda}(s_{n-1})$ has nonzero components in each
of $\operatorname{Hom}(W^{k}_{\overline{0}},W^{\ell}_{\overline{1}})$,
$\operatorname{Hom}(W^{k}_{\overline{1}},W^{\ell}_{\overline{0}})$,
$\operatorname{Hom}(W^{\ell}_{\overline{0}},W^{k}_{\overline{1}})$, and
$\operatorname{Hom}(W^{\ell}_{\overline{1}},W^{k}_{\overline{0}})$.

###  4.4. Proof of Theorem LABEL:theorem:Wlambda(gn): the case $\lambda\in
E_{n}$

#### 4.4.1.

Now suppose $\lambda\in E_{n}$. In this case one has $W^{k}\neq 0$ only if
$|k|<n$. If there is only one summand in the decomposition
$W^{\lambda}=\bigoplus_{k\in\mathbb{Z}}W^{k}$, i.e., if $\lambda$ has only one
removable box, say $B_{\ell}$ of content $\ell$, then the Young diagram of
$\lambda$ must be a (non-symmetric) rectangle with $B_{\ell}$ in its outer
corner, and hence the partition $\mu_{\ell}$ obtained by removing $B_{\ell}$
must also be non-symmetric. Then by induction, $W^{\lambda}\cong
W^{\mu_{\ell}}$ as a $\mathbb{C}S_{n-1}$-supermodule, and

$\mathfrak{sq}(W^{\lambda})=\mathfrak{sq}(W^{\mu_{\ell}})=W^{\mu_{\ell}}(\mathfrak{g}_{n-1}^{\prime})=W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})\subseteq
W^{\lambda}(\mathfrak{g}_{n}),$

establishing (4.2.2). So assume that $W^{k}\neq 0$ for more than one value of
$k$.

#### 4.4.2.

Since $\lambda$ is not symmetric, there is at most one value of $k$ such that
the partition $\mu_{k}\prec\lambda$ is symmetric; call this value $s$ (if it
exists). If $s$ exists, then $s\neq 0$, because $\lambda$ is not symmetric.

By the induction hypothesis,

(4.4.1)
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})=W^{s}(\mathfrak{g}_{n-1}^{\prime})\oplus\bigoplus_{k\neq
s}\mathfrak{sq}(W^{k}),$

where, by definition, $W^{s}=0$ and the first summand is omitted if a
symmetric partition $\mu_{s}\prec\lambda$ does not exist. The supermodule
$W^{\lambda}$ is equipped with the odd involution
$J=J^{\lambda}:W^{\lambda}\to W^{\lambda}$, which restricts for each
$k\in\mathbb{Z}$ to an odd involution $J^{k}:W^{k}\to W^{k}$. As in Section
4.3, we make the identification $W^{k}_{\overline{0}}\simeq
W^{k}_{\overline{1}}$ via $J^{k}$ and write $W_{k}$ for the common identified
space. Then for $k\neq s$ one has
$\mathfrak{gl}(W_{k})\cong\mathfrak{sq}(W^{k})_{\overline{0}}$ as in (4.3.3).
For $k=s$, we see from (3.4.3) that $J^{s}$ defines an odd isomorphism
$W^{\mu_{s}}\simeq\Pi(W^{\mu_{s}})$. Then conjugation by $J^{s}$, $\phi\mapsto
J^{s}\circ\phi\circ J^{s}$, defines an even isomorphism
$\operatorname{End}(W^{\mu_{s}})\cong\operatorname{End}(\Pi(W^{\mu_{s}}))$,
and $W^{s}(\mathfrak{g}_{n-1}^{\prime})$ is the image in
$\operatorname{End}(W^{s})$ of the diagonal map

(4.4.2)
$\mathfrak{sl}(W^{\mu_{s}})\to\operatorname{End}(W^{\mu_{s}})\oplus\operatorname{End}(\Pi(W^{\mu_{s}})).$

Make the identifications $W^{\mu_{s}}\simeq\Pi(W^{\mu_{s}})$,
$W^{\mu_{s}}_{\overline{0}}\simeq\Pi(W^{\mu_{s}}_{\overline{0}})$, and
$W^{\mu_{s}}_{\overline{1}}\simeq\Pi(W^{\mu_{s}}_{\overline{1}})$ via $J^{s}$,
and write $W_{\mu_{s}}$, $W_{\mu_{s},\overline{0}}$, and
$W_{\mu_{s},\overline{1}}$ for the common identified spaces, respectively
(considered just as ordinary vector spaces, without any superspace
structures).

######  Remark 4.4.1.

remark:sl(1)=0 It could happen that

  1. (1)

$W_{\mu_{s},\overline{0}}\cong W_{\mu_{s},\overline{1}}\neq 0$, but
$\mathfrak{sl}(W_{\mu_{s},\overline{0}})\cong\mathfrak{sl}(W_{\mu_{s},\overline{1}})=0$;
or that

  2. (2)

$W_{k}\neq 0$ for some $k\neq s$, but $\mathfrak{sl}(W_{k})=0$.

These situations occur if and only if $\dim(W^{\mu_{s}})=2$ or
$\dim(W^{\mu_{k}})=2$, respectively. lemma:lower-bound-dim implies for $n\geq
6$ that situation (1) cannot occur, and implies that up to the equivalence
$\lambda\sim\lambda^{\prime}$, situation (2) occurs only if $\lambda=(n-1,1)$.
In this case, $W^{(n-1,1)}=W^{(n-1,1)}_{n-2}\oplus W^{(n-1,1)}_{-1}$, and one
has $\mathbb{C}S_{n-1}$-supermodule isomorphisms

$\displaystyle W^{(n-1,1)}_{n-2}$ $\displaystyle\cong W^{(n-2,1)},$
$\displaystyle\dim(W^{(n-2,1)})$ $\displaystyle=2(n-2),$ $\displaystyle
W^{(n-1,1)}_{-1}$ $\displaystyle\cong W^{(n-1)},$
$\displaystyle\dim(W^{(n-1)})$ $\displaystyle=2.$

In any event, $\operatorname{Hom}(W_{k},W_{\ell})$ remains an irreducible
$\mathfrak{sl}(W_{k})\oplus\mathfrak{sl}(W_{\ell})$-module even if one of
$W_{k}$ or $W_{\ell}$ is one-dimensional (hence even if one of
$\mathfrak{sl}(W_{k})$ or $\mathfrak{sl}(W_{\ell})$ is zero).

#### 4.4.3.

Now from (4.4.1), we see that the semisimple Lie algebra

$\mathfrak{h}:=\left[\mathfrak{sl}(W_{\mu_{s},\overline{0}})\oplus\mathfrak{sl}(W_{\mu_{s},\overline{1}})\right]\oplus\bigoplus_{k\neq
s}\mathfrak{sl}(W_{k})$

identifies with a subalgebra of
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})_{\overline{0}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$. Further, (4.2.4) and (4.2.5) give
$\mathfrak{h}$-module decompositions of
$\operatorname{End}(W^{\lambda})_{\overline{1}}$ under the adjoint action. The
set $W^{\lambda}(\mathfrak{g}_{n})$ is contained in

$\mathfrak{sq}(W^{\lambda})=\operatorname{End}(W^{\lambda})^{J}:=\\{\theta\in\operatorname{End}(W^{\lambda}):J\circ\theta\circ
J=\theta\\},$

and the decomposition (4.2.4) gives rise to the corresponding decomposition of
$J$-invariants

(4.4.3)
$\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}=\bigoplus_{k,\ell\in\mathbb{Z}}\operatorname{Hom}(W^{k},W^{\ell})^{J}_{\overline{1}}.$

For $s\notin\left\\{k,\ell\right\\}$, one sees that the diagonal map

$\operatorname{Hom}(W_{k},W_{\ell})\to\operatorname{Hom}(W^{k}_{\overline{0}},W^{\ell}_{\overline{1}})\oplus\operatorname{Hom}(W^{k}_{\overline{1}},W^{\ell}_{\overline{0}})$

induces an $\mathfrak{h}$-module isomorphism
$\operatorname{Hom}(W_{k},W_{\ell})\cong\operatorname{Hom}(W^{k},W^{\ell})^{J}_{\overline{1}}$.
Similarly, for $k\neq s$ one sees that the diagonal maps

$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{k})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{0}},W^{k}_{\overline{1}})\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{0}}),W^{k}_{\overline{0}}),\quad\text{and}$
$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{k})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{1}},W^{k}_{\overline{0}})\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{1}}),W^{k}_{\overline{1}})$

induce an $\mathfrak{h}$-module isomorphism

$\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{k})\oplus\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{k})\cong\operatorname{Hom}(W^{s},W^{k})^{J}_{\overline{1}}.$

An analogous description holds for
$\operatorname{Hom}(W^{k},W^{s})^{J}_{\overline{1}}$. Finally, as an
$\mathfrak{h}$-module,

$\begin{split}\operatorname{Hom}(W^{s},W^{s})^{J}_{\overline{1}}\cong&\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{1}})\oplus\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})\\\
&\oplus\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{0}})\oplus\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{1}}),\end{split}$

where the summands on the right side of the isomorphism are identified with
the images of the corresponding diagonal maps

(4.4.4)
$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{1}})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{0}},W^{\mu_{s}}_{\overline{1}})\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{0}}),\Pi(W^{\mu_{s}}_{\overline{1}})),$
(4.4.5)
$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{1}},W^{\mu_{s}}_{\overline{0}})\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{1}}),\Pi(W^{\mu_{s}}_{\overline{0}})),$
(4.4.6)
$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{0}})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{0}},\Pi(W^{\mu_{s}}_{\overline{0}}))\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{0}}),W^{\mu_{s}}_{\overline{0}}),$
(4.4.7)
$\displaystyle\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{1}})$
$\displaystyle\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{1}},\Pi(W^{\mu_{s}}_{\overline{1}}))\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{1}}),W^{\mu_{s}}_{\overline{1}}).$

####  4.4.4. Irreducible constituents

Altogether, $\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}$ admits the
$\mathfrak{h}$-module decomposition

(4.4.8)
$\begin{split}&\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}\cong\Big{[}\bigoplus_{s\notin\left\\{k,\ell\right\\}}\operatorname{Hom}(W_{k},W_{\ell})\Big{]}\\\
&\oplus\Big{[}\bigoplus_{k\neq
s}\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{k})\oplus\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{k})\oplus\operatorname{Hom}(W_{k},W_{\mu_{s},\overline{0}})\oplus\operatorname{Hom}(W_{k},W_{\mu_{s},\overline{1}})\Big{]}\\\
&\oplus\Big{[}\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{1}})\oplus\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})\oplus\operatorname{End}(W_{\mu_{s},\overline{0}})\oplus\operatorname{End}(W_{\mu_{s},\overline{1}})\Big{]}.\end{split}$

For $k\neq s$, the term $\operatorname{End}(W_{k})$ in (4.4.8) is either
simply a one-dimensional trivial $\mathfrak{h}$-module, if $\dim(W_{k})=1$, or
else is the direct sum of a one-dimensional trivial $\mathfrak{h}$-module and
a copy of the adjoint module for $\mathfrak{sl}(W_{k})$, the latter of which
is contained in
$W^{\lambda}(\mathfrak{g}_{n-1}^{\prime})_{\overline{1}}\subseteq
W^{\lambda}(\mathfrak{g}_{n})$ by (4.4.1). By (4.4.2), the summands
$\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{\mu_{s},\overline{1}})$ and
$\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})$ are
also contained in $W^{\lambda}(\mathfrak{g}_{n})$. The summands
$\operatorname{End}(W_{\mu_{s},\overline{0}})$ and
$\operatorname{End}(W_{\mu_{s},\overline{1}})$ in (4.4.8) are each direct sums
of a one-dimensional trivial $\mathfrak{h}$-module and a copy of the adjoint
representation for $\mathfrak{sl}(W_{\mu_{s},\overline{0}})$ and
$\mathfrak{sl}(W_{\mu_{s},\overline{1}})$, respectively. The remaining nonzero
summands in (4.4.8) are each nontrivial irreducible $\mathfrak{h}$-modules.
Overall, the non-trivial irreducible $\mathfrak{h}$-modules that occur in
$\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}$ each do so with
multiplicity one.

#### 4.4.5.

If $k$, $\ell$, and $s$ are distinct, and if $W^{k}$ and $W^{\ell}$ are both
nonzero, then one can argue as in the last two paragraphs of Section 4.3 to
show first that $W^{\lambda}(s_{n-1})$ has nonzero components in the
irreducible $\mathfrak{h}$-module summands
$\operatorname{Hom}(W^{k},W^{\ell})^{J}_{\overline{1}}$ and
$\operatorname{Hom}(W^{\ell},W^{k})^{J}_{\overline{1}}$ of
$\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}$, and then to deduce that
these summands must both be contained in $W^{\lambda}(\mathfrak{g}_{n})$.

####  4.4.6. The case $W^{s}=0$

If $W^{s}=0$, then the previous paragraph together with our observations in
Section 4.4.4 imply that each non-trivial $\mathfrak{h}$-module constituent of
$\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}$ is contained in
$W^{\lambda}(\mathfrak{g}_{n})$. Identifying $\mathfrak{sq}(W^{\lambda})$ with
supermatrices as in (4.1.2), it implies that

(4.4.9) $\left\\{\left[\begin{array}[]{c|c}0&B\\\ \hline\cr
B&0\end{array}\right]:\text{the diagonal entries of $B$ are all
zero}\right\\}$

is contained in $W^{\lambda}(\mathfrak{g}_{n})$. More precisely, the direct
sum decomposition (4.4.8) induces a block decomposition of the matrix $B$ such
that the diagonal blocks correspond to
$\oplus_{k\in\mathbb{Z}}\operatorname{End}(W_{k})$, and we deduce that
$W^{\lambda}(\mathfrak{g}_{n})$ contains the (larger) set of all matrices of
the form $[\begin{smallmatrix}0&B\\\ B&0\end{smallmatrix}]$ such that these
diagonal blocks each individually have trace zero. By lemma:lower-bound-dim
and the assumption that $W^{k}\neq 0$ for more than one value of $k$, we have
$\dim(W^{\lambda})\geq 2n-2\geq 10$. Then the inclusion
$\mathfrak{sq}(W^{\lambda})\subseteq W^{\lambda}(\mathfrak{g}_{n})$ in the
case where $W^{s}=0$ is obtained from the following lemma:

######  Lemma 4.4.2.

lemma:sq(m)-generated If $m\geq 5$, then $\mathfrak{sq}(m)$ is generated as a
Lie superalgebra by the identity matrix $I_{m|m}$ and the set (4.4.9).

###### Proof.

Let $\mathfrak{g}\subseteq\mathfrak{sq}(m)$ be the Lie superalgebra generated
by the identity matrix $I_{m|m}$ and the set (4.4.9). First show that all
simple root vectors in
$\mathfrak{sl}(m)\subseteq\mathfrak{gl}(m)\cong\mathfrak{sq}(m)_{\overline{0}}$
are elements of $\mathfrak{g}$ by taking Lie brackets between root vectors in
the set (4.4.9). The simple root vectors generate $\mathfrak{sl}(m)$ as a Lie
algebra, and together with the identity matrix they generate all of
$\mathfrak{gl}(m)\cong\mathfrak{sq}(m)_{\overline{0}}$. Then
$\mathfrak{sq}(m)_{\overline{0}}\subseteq\mathfrak{g}$. Now since
$\mathfrak{sq}(m)_{\overline{1}}$ is irreducible under the adjoint action of
$\mathfrak{sq}(m)_{\overline{0}}$, one deduces that
$\mathfrak{sq}(m)_{\overline{1}}\subseteq\mathfrak{g}$ as well, and hence
$\mathfrak{g}=\mathfrak{sq}(m)$. ∎

####  4.4.7. The case $W^{s}\neq 0$

Now suppose that $k\neq s$ and that $W^{k}$ and $W^{s}$ are both nonzero. Let
$B_{k}$ and $B_{s}$ be the removable boxes of contents $k$ and $s$ in the
Young diagram of $\lambda$. Recall that $s\neq 0$, because $\lambda$ is not
symmetric. Since $B_{k}$ and $B_{s}$ are both removable, we can remove $B_{s}$
and then $B_{k}$, showing that $\mu_{s}$ has removable box of content $k$.
Then by symmetry, $\mu_{s}$ must have a removable box $B_{-k}$ of content
$-k$. (It may happen that $k=0$, in which case $B_{-k}=B_{k}$.) This implies
that $-k\neq s$, because a new box of content $s$ would be removable from
$\mu_{s}$ only if there had originally been boxes both immediately above and
immediately to the left of $B_{s}$ in the Young diagram of $\lambda$, and if
both of those boxes had already been removed. In other words, it requires at
least two intermediate steps to remove two boxes of the same content from
$\lambda$. Thus the Young diagram of $\lambda$ must have removable boxes of
contents $s$, $k$, and $-k$ (the latter two being the same box, if $k=0$),
implying that $|s-k|\geq 2$ and $|s+k|\geq 2$, and implying that $W^{s}$,
$W^{k}$, and $W^{-k}$ are nonzero.

We want to show that $W^{\lambda}(s_{n-1})$ has nonzero components in each of
the terms in the second line of (4.4.8). We will do this simultaneously for
the indices $k$ and $-k$. Since $\lambda$ has removable boxes of contents $s$,
$k$, and $-k$, we can argue as before to deduce that $\mathcal{W}(\lambda)$
contains four (distinct) weights of the form

$\displaystyle\alpha$ $\displaystyle=(\alpha^{\prime\prime},k,s),$
$\displaystyle\alpha^{\prime}$ $\displaystyle=(-\alpha^{\prime\prime},-k,s),$
$\displaystyle\beta$ $\displaystyle=(\alpha^{\prime\prime},s,k),$
$\displaystyle\beta^{\prime}$ $\displaystyle=(-\alpha^{\prime\prime},s,-k).$

Then $\gamma:=(\alpha^{\prime\prime},k)$ and
$-\gamma=(-\alpha^{\prime\prime},-k)$ are elements of $\mathcal{W}(\mu_{s})$.
We may assume that $k\geq 0$ and that
$\gamma\in\overline{\mathcal{W}}(\mu_{s})$, i.e., $\gamma$ is the ‘positive’
element of the pair $\pm\gamma$. As in (3.4.1), $S^{\mu_{s}}$ is a
$\mathbb{C}S_{n-1}$-module summand of $S^{\lambda}$, and under this
identification the weight vectors $v_{\alpha},v_{\alpha^{\prime}}\in
S^{\lambda}$ restrict to a pair of weight vectors $u_{\gamma},u_{-\gamma}\in
S^{\mu_{s}}$. (We use the letter $u$ rather than $v$ to indicate when we are
considering a vector’s restriction to the $\mathbb{C}S_{n-1}$-module
$S^{\mu_{s}}$.) Rescaling the vector $v_{\alpha^{\prime}}=u_{\gamma}$ if
necessary, we may assume that $u_{-\gamma}=\phi^{\mu}(u_{\gamma})$ as in
(3.3.1). Then in the decomposition $S^{\mu_{s}}=S^{\mu_{s}^{+}}\oplus
S^{\mu_{s}^{-}}$, one has $u_{\gamma}=u_{\gamma}^{+}+u_{\gamma}^{-}$ and
$u_{-\gamma}=u_{\gamma}^{+}-u_{\gamma}^{-}$, where

$\displaystyle u_{\gamma}^{+}$
$\displaystyle=\tfrac{1}{2}\left(u_{\gamma}+u_{-\gamma}\right)=\tfrac{1}{2}\left(v_{\alpha}+v_{\alpha^{\prime}}\right)\quad\text{and}\quad
u_{\gamma}^{-}=\tfrac{1}{2}\left(u_{\gamma}-u_{-\gamma}\right)=\tfrac{1}{2}\left(v_{\alpha}-v_{\alpha^{\prime}}\right).$

Then under the $\mathbb{C}S_{n-1}$-supermodule decomposition $W^{s}\cong
W^{\mu}\oplus\Pi(W^{\mu})$, one has

$\displaystyle W^{\mu}_{\overline{0}}$ $\displaystyle\ni
u_{\gamma}^{+}+\phi^{\lambda}(u_{\gamma}^{+})=\tfrac{1}{2}\left(v_{\alpha}+v_{\alpha^{\prime}}\right)+\tfrac{1}{2}\left(v_{-\alpha}+v_{-\alpha^{\prime}}\right)=v_{\alpha}^{+}+v_{\alpha^{\prime}}^{+},$
$\displaystyle W^{\mu}_{\overline{1}}$ $\displaystyle\ni
u_{\gamma}^{-}-\phi^{\lambda}(u_{\gamma}^{-})=\tfrac{1}{2}\left(v_{\alpha}-v_{\alpha^{\prime}}\right)-\tfrac{1}{2}\left(v_{-\alpha}-v_{-\alpha^{\prime}}\right)=v_{\alpha}^{-}-v_{\alpha^{\prime}}^{-},$
$\displaystyle\Pi(W^{\mu}_{\overline{0}})$ $\displaystyle\ni
u_{\gamma}^{+}-\phi^{\lambda}(u_{\gamma}^{+})=\tfrac{1}{2}\left(v_{\alpha}+v_{\alpha^{\prime}}\right)-\tfrac{1}{2}\left(v_{-\alpha}+v_{-\alpha^{\prime}}\right)=v_{\alpha}^{-}+v_{\alpha^{\prime}}^{-},$
$\displaystyle\Pi(W^{\mu}_{\overline{1}})$ $\displaystyle\ni
u_{\gamma}^{-}+\phi^{\lambda}(u_{\gamma}^{-})=\tfrac{1}{2}\left(v_{\alpha}-v_{\alpha^{\prime}}\right)+\tfrac{1}{2}\left(v_{-\alpha}-v_{-\alpha^{\prime}}\right)=v_{\alpha}^{+}-v_{\alpha^{\prime}}^{+}.$

Set $c=(s-k)^{-1}$ and $d=(s+k)^{-1}$. Let $w_{\beta}=(s_{n-1}-c)\cdot
v_{\alpha}$, let $w_{\beta^{\prime}}=(s_{n-1}-d)\cdot v_{\alpha^{\prime}}$,
and let the auxiliary vectors $w_{\beta}^{+},w_{\beta}^{-}\in W^{k}$ and
$w_{\beta^{\prime}}^{+},w_{\beta^{\prime}}^{-}\in W^{-k}$ be defined as in
prop:si-on-weights-super(2). Then $W^{\lambda}(s_{n-1})$ leaves invariant the
span of the homogeneous vectors

$w_{\beta}^{+}$, $w_{\beta^{\prime}}^{+}$,
$u_{\gamma}^{+}+\phi^{\lambda}(u_{\gamma}^{+})$,
$u_{\gamma}^{-}+\phi^{\lambda}(u_{\gamma}^{-})$, $w_{\beta}^{-}$,
$w_{\beta^{\prime}}^{-}$, $u_{\gamma}^{+}-\phi^{\lambda}(u_{\gamma}^{+})$,
$u_{\gamma}^{-}-\phi^{\lambda}(u_{\gamma}^{-})$,

and acts in this homogeneous basis via the matrix

(4.4.10)
$\left[\begin{array}[]{cccc|cccc}0&0&0&0&-c&0&\frac{1}{2}&\frac{1}{2}\\\
0&0&0&0&0&-d&\frac{1}{2}&-\frac{1}{2}\\\
0&0&0&0&1-c^{2}&1-d^{2}&\frac{c+d}{2}&\frac{c-d}{2}\\\
0&0&0&0&1-c^{2}&d^{2}-1&\frac{c-d}{2}&\frac{c+d}{2}\\\
\hline\cr-c&0&\frac{1}{2}&\frac{1}{2}&0&0&0&0\\\
0&-d&\frac{1}{2}&-\frac{1}{2}&0&0&0&0\\\
1-c^{2}&1-d^{2}&\frac{c+d}{2}&\frac{c-d}{2}&0&0&0&0\\\
1-c^{2}&d^{2}-1&\frac{c-d}{2}&\frac{c+d}{2}&0&0&0&0\end{array}\right]$

This shows (for both $k$ and $-k$) that $W^{\lambda}(s_{n-1})$ has nonzero
components in each of the terms in the second line of (4.4.8). Then by the
semisimplicity of $\mathfrak{h}$, and by the fact that all nontrivial
irreducible $\mathfrak{h}$-module summands in
$\operatorname{End}(W^{\lambda})^{J}_{\overline{1}}$ occur with multiplicity
one, we conclude that each of the summands in the second line of (4.4.8) must
be contained in $W^{\lambda}(\mathfrak{g}_{n})$.

####  4.4.8. The case $W^{s}\neq 0$, concluded

Now the only nontrivial $\mathfrak{h}$-module constituents of (4.4.8) that we
have not yet shown are contained in $W^{\lambda}(\mathfrak{g}_{n})$ are the
copies of the adjoint representations of
$\mathfrak{sl}(W_{\mu_{s},\overline{0}})$ and
$\mathfrak{sl}(W_{\mu_{s},\overline{1}})$ in
$\operatorname{End}(W_{\mu_{s},\overline{0}})$ and
$\operatorname{End}(W_{\mu_{s},\overline{1}})$, respectively. Once we show
that these irreducible constituents are contained in
$W^{\lambda}(\mathfrak{g}_{n})$, we can then argue as in Section 4.4.6, using
lemma:sq(m)-generated, to conclude that $\mathfrak{sq}(W^{\lambda})\subseteq
W^{\lambda}(\mathfrak{g}_{n})$.

We know that $W^{\lambda}(\mathfrak{g}_{n})$ contains the terms
$\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{k})$ and
$\operatorname{Hom}(W_{k},W_{\mu_{s},\overline{1}})$ from (4.4.8). Like all of
the terms in (4.4.8), these two terms are concentrated in odd superdegree. We
also know from (4.4.2) that $W^{s}(\mathfrak{g}_{n-1}^{\prime})\subset
W^{\lambda}(\mathfrak{g}_{n})$ contains a copy of
$\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})$, also
concentrated in odd superdegree, equal to the image of the diagonal map

$\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}})\to\operatorname{Hom}(W^{\mu_{s}}_{\overline{1}},W^{\mu_{s}}_{\overline{0}})\oplus\operatorname{Hom}(\Pi(W^{\mu_{s}}_{\overline{1}}),\Pi(W^{\mu_{s}}_{\overline{0}})).$

By remark:sl(1)=0 and lemma:lower-bound-dim, we know that $W_{k}$,
$W_{\mu_{s},\overline{0}}$, and $W_{\mu_{s},\overline{1}}$ are each at least
3-dimensional. Now one can choose appropriate ‘matrix units’

$x\in\operatorname{Hom}(W_{\mu_{s},\overline{1}},W_{\mu_{s},\overline{0}}),\qquad
y\in\operatorname{Hom}(W_{k},W_{\mu_{s},\overline{1}}),\qquad
z\in\operatorname{Hom}(W_{\mu_{s},\overline{0}},W_{k})$

such that the Lie bracket $[x,[y,z]]$ is a nonzero element—of
$W^{\lambda}(\mathfrak{g}_{n})$—in the subspace
$\mathfrak{sl}(W_{\mu_{s},\overline{0}})\subset\operatorname{End}(W_{\mu_{s},\overline{0}})\subset\operatorname{End}(W^{s})^{J}_{\overline{1}}$.
Then by the irreducibility of the adjoint representation
$\mathfrak{sl}(W_{\mu_{s},\overline{0}})$, the $\mathfrak{h}$-submodule of
$W^{\lambda}(\mathfrak{g}_{n})$ generated by $[x,[y,z]]$ must be equal to all
of $\mathfrak{sl}(W_{\mu_{s},\overline{0}})$. Similarly, one can show that
$W^{\lambda}(\mathfrak{g}_{n})$ contains the subspace
$\mathfrak{sl}(W_{\mu_{s},\overline{1}})\subset\operatorname{End}(W_{\mu_{s},\overline{1}})\subset\operatorname{End}(W^{s})^{J}_{\overline{1}}$.

###  4.5. First consequence of Theorem LABEL:theorem:Wlambda(gn)

######  Corollary 4.5.1.

cor:g0-reductive Let $n\geq 2$, and set $\mathfrak{g}=\mathfrak{g}_{n}$.

  1. (1)

For each $\lambda\in E_{n}\cup F_{n}$, the supermodule $W^{\lambda}$ is
semisimple as a $\mathfrak{g}_{\overline{0}}$-module.

  2. (2)

$\mathfrak{g}_{\overline{0}}$ is a reductive Lie algebra.

In particular,
$\mathfrak{g}_{\overline{0}}=Z(\mathfrak{g}_{\overline{0}})\oplus\mathfrak{D}(\mathfrak{g}_{\overline{0}})$,
where $Z(\mathfrak{g}_{\overline{0}})$ is the center of
$\mathfrak{g}_{\overline{0}}$, and $\mathfrak{D}(\mathfrak{g}_{\overline{0}})$
is semisimple.

###### Proof.

By theorem:Wlambda(gn),

$W^{\lambda}(\mathfrak{g}_{\overline{0}})=W^{\lambda}(\mathfrak{g})_{\overline{0}}=\begin{cases}\mathfrak{sq}(W^{\lambda})_{\overline{0}}&\text{if
$\lambda\in E_{n}$,}\\\ \mathfrak{sl}(W^{\lambda})_{\overline{0}}&\text{if
$\lambda\in F_{n}$}.\end{cases}$

In either case, this implies that $W^{\lambda}_{\overline{0}}$ and
$W^{\lambda}_{\overline{1}}$ are irreducible
$\mathfrak{g}_{\overline{0}}$-modules, and hence $W^{\lambda}$ is semisimple.
Now $\bigoplus_{\lambda\in E_{n}\cup F_{n}}W^{\lambda}$ is a faithful, finite-
dimensional, semisimple $\mathfrak{g}_{\overline{0}}$-module. Then
$\mathfrak{g}_{\overline{0}}$ is reductive,
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$ is semisimple, and
$\mathfrak{g}_{\overline{0}}=Z(\mathfrak{g}_{\overline{0}})\oplus\mathfrak{D}(\mathfrak{g}_{\overline{0}})$,
by Proposition 5 of [1, Chapter I, §6, no. 4]. ∎

######  Remark 4.5.2.

$\mathfrak{D}(\mathfrak{g}_{\overline{0}})\neq\mathfrak{D}(\mathfrak{g})_{\overline{0}}$.

###  4.6. Detecting isomorphisms of supermodules

######  Lemma 4.6.1.

lemma:submodule-equivalent Let $n\geq 5$, set $\mathfrak{g}=\mathfrak{g}_{n}$,
and let $W$ be a finite-dimensional $\mathbb{C}A_{n}$-module. Then the
following are equivalent for a subspace $V\subseteq W$:

  1. (1)

$V$ is a $\mathbb{C}A_{n}$-submodule.

  2. (2)

$V$ is a submodule for the action of the Lie subalgebra
$\mathfrak{g}_{\overline{0}}\subseteq\mathbb{C}A_{n}$.

  3. (3)

$V$ is a submodule for the action of the Lie subalgebra
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})\subseteq\mathbb{C}A_{n}$.

###### Proof.

The fact that (1) and (2) are equivalent is immediate from the observation in
the proof of lemma:lie-algebra-centers that for $n\geq 5$, the Lie algebra
$\mathfrak{g}_{\overline{0}}$ contains a set of associative algebra generators
for $\mathbb{C}A_{n}$. The fact that (2) implies (3) is evident. We will show
that (3) implies (2). First, since $\mathfrak{D}(\mathfrak{g}_{\overline{0}})$
is semisimple by cor:g0-reductive, the
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$-submodule $V$ can be written as a
direct sum of irreducible
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$-submodules. Without loss of
generality, we may thus assume that $V$ itself is irreducible. Then by Schur’s
Lemma, each element $z\in Z(\mathfrak{g}_{\overline{0}})$ acts on $V$ as a
scalar multiple of the identity. Since
$\mathfrak{g}_{\overline{0}}=Z(\mathfrak{g}_{\overline{0}})\oplus\mathfrak{D}(\mathfrak{g}_{\overline{0}})$,
this implies that $V$ is closed under the action of
$\mathfrak{g}_{\overline{0}}$. ∎

######  Proposition 4.6.2.

prop:irred-equivalent Let $n\geq 5$, set $\mathfrak{g}=\mathfrak{g}_{n}$, and
let $V_{1}$ and $V_{2}$ be two irreducible $\mathbb{C}A_{n}$-modules. Then the
following are equivalent:

  1. (1)

$V_{1}$ and $V_{2}$ are isomorphic as $\mathbb{C}A_{n}$-modules.

  2. (2)

$V_{1}$ and $V_{2}$ are isomorphic as modules over the Lie subalgebra
$\mathfrak{g}_{\overline{0}}\subseteq\mathbb{C}A_{n}$.

  3. (3)

$V_{1}$ and $V_{2}$ are isomorphic as modules over the Lie subalgebra
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})\subseteq\mathbb{C}A_{n}$.

###### Proof.

Our argument is an adaptation of the proof of [11, Proposition 2]. The fact
that (1) implies (2), and that (2) implies (3), is evident. We will show that
(3) implies (1). For $n\geq 5$, the trivial module is the unique one-
dimensional $\mathbb{C}A_{n}$-module (cf. [8, Theorem 2.5.15]), so we may
assume that $V_{1}$ and $V_{2}$ are each of dimension at least $2$.

Suppose $\phi:V_{1}\to V_{2}$ is an isomorphism of
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$-modules, and let
$\rho_{1}:\mathbb{C}A_{n}\to\operatorname{End}(V_{1})$ and
$\rho_{2}:\mathbb{C}A_{n}\to\operatorname{End}(V_{2})$ be the structure maps
for $V_{1}$ and $V_{2}$, respectively. Then for all
$x\in\mathfrak{D}(\mathfrak{g}_{\overline{0}})$, one has
$\phi\circ\rho_{1}(x)=\rho_{2}(x)\circ\phi$, or equivalently,
$\rho_{2}(x)=\phi\circ\rho_{1}(x)\circ\phi^{-1}$. Let $s=(i,j)(k,\ell)$ be a
generator of $A_{n}$ from the set (4.1.8), and set
$T=p(s)=\frac{2}{n!}\sum_{\sigma\in A_{n}}\sigma s\sigma^{-1}$. The elements
of the set (4.1.8) form a single conjugacy class in $A_{n}$ (because the cycle
type does not consist of distinct odd integers), so $T$ is independent of the
particular choice of $s$. Since $T$ is central in $\mathbb{C}A_{n}$, Schur’s
Lemma implies that $\rho_{1}(T)=c_{1}\operatorname{id}_{V_{1}}$ and
$\rho_{2}(T)=c_{2}\operatorname{id}_{V_{2}}$ for some scalars
$c_{1},c_{2}\in\mathbb{C}$, which also do not depend on the choice of $s$. We
have $p(s-T)=p(s)-p(T)=T-T=0$, so
$s-T\in\mathfrak{D}(\mathfrak{g}_{\overline{0}})$ by cor:g0-reductive and
lemma:lie-algebra-centers. Then

$\displaystyle\rho_{2}(s)-c_{2}\operatorname{id}_{V_{2}}$
$\displaystyle=\rho_{2}(s-T)$
$\displaystyle=\phi\circ\rho_{1}(s-T)\circ\phi^{-1}$
$\displaystyle=\phi\circ\rho_{1}(s)\circ\phi^{-1}-\phi\circ(c_{1}\operatorname{id}_{V_{1}})\circ\phi^{-1}$
$\displaystyle=\phi\circ\rho_{1}(s)\circ\phi^{-1}-c_{1}\operatorname{id}_{V_{2}},$

or equivalently,

(4.6.1)
$\rho_{2}(s)=\phi\circ\rho_{1}(s)\circ\phi^{-1}+\omega\cdot\operatorname{id}_{V_{2}},$

where $\omega=c_{2}-c_{1}$. Squaring both sides of (4.6.1), and using the fact
that $s^{2}=1$, we get

$\operatorname{id}_{V_{2}}={\operatorname{id}_{V_{2}}}+2\omega\cdot\phi\circ\rho_{1}(s)\circ\phi^{-1}+\omega^{2}\cdot\operatorname{id}_{V_{2}},$

or equivalently,
$\omega^{2}\cdot\operatorname{id}_{V_{2}}=2\omega\cdot\phi\circ\rho_{1}(s)\circ\phi^{-1}$.
The scalar $\omega$ does not depend on the choice of $s$, so if $\omega\neq
0$, we would deduce first for all $s$ in the set (4.1.8), and then for all
$s\in A_{n}$ by multiplicativity, that $\rho_{1}(s)$ is equal to a nonzero
scalar multiple of $\operatorname{id}_{V_{1}}$. Since $\dim(V_{1})\geq 2$ by
assumption, this would contradict the irreducibility of $V_{1}$. Then
$\omega=0$, and (4.6.1) implies first for all $s$ in the set (4.1.8), and then
for all $s\in A_{n}$ by multiplicativity, that
$\phi\circ\phi_{1}(s)=\rho_{2}(s)\circ\phi^{-1}$; that is, $\phi$ is an
isomorphism of $\mathbb{C}A_{n}$-modules. ∎

######  Corollary 4.6.3.

module-iso-equivalent Let $n\geq 5$, set $\mathfrak{g}=\mathfrak{g}_{n}$, and
let $V_{1}$ and $V_{2}$ be two finite-dimensional
$\mathbb{C}S_{n}$-supermodules. Then the following statements (in which
‘isomorphic’ is taken to mean ‘isomorphic via a homogeneous homomorphism’) are
equivalent:

  1. (1)

$V_{1}$ and $V_{2}$ are isomorphic as $\mathbb{C}S_{n}$-supermodules.

  2. (2)

$V_{1}$ and $V_{2}$ are isomorphic as $\mathbb{C}A_{n}$-supermodules.

  3. (3)

$V_{1}$ and $V_{2}$ are isomorphic as supermodules over the Lie subalgebra
$\mathfrak{g}_{\overline{0}}\subseteq\mathbb{C}A_{n}$.

  4. (4)

$V_{1}$ and $V_{2}$ are isomorphic as supermodules over the Lie subalgebra
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})\subseteq\mathbb{C}A_{n}$.

###### Proof.

By semisimplicity of the superalgebra $\mathbb{C}S_{n}$, each finite-
dimensional $\mathbb{C}S_{n}$-supermodule is a direct sum of irreducible
supermodules, and two finite-dimensional supermodules are isomorphic if and
only if their collections of irreducible factors are the same (up to overall
parity change). The classification of the irreducible
$\mathbb{C}S_{n}$-supermodules in Section 3.1 shows that each irreducible
supermodule $W^{\lambda}$ is determined by its restriction to
$\mathbb{C}A_{n}$ (and for the absolutely irreducible modules, by the
homogeneous degrees in which the irreducible $|\mathbb{C}A_{n}|$-constituents
are concentrated). Thus (1) and (2) are equivalent, while (2) evidently
implies (3), and (3) evidently implies (4). To show that (4) implies (2), one
can first reduce to the even (resp. odd) subspaces of $V_{1}$ and $V_{2}$,
which are submodules for each of $\mathbb{C}A_{n}$ and
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$. Then apply the semisimplicity of
$\mathfrak{D}(\mathfrak{g}_{\overline{0}})$, lemma:submodule-equivalent, and
prop:irred-equivalent. ∎

###  4.7. Structure of $\mathfrak{g}_{n}$

In this section let $n\geq 2$ and let $\mathfrak{g}=\mathfrak{g}_{n}$. By
cor:g0-reductive, the Lie algebra $\mathfrak{g}_{\overline{0}}$ is reductive
and its derived subalgebra
$\mathfrak{D}:=\mathfrak{D}(\mathfrak{g}_{\overline{0}})$ is semisimple. For
$\lambda\in E_{n}$, make the identification $W^{\lambda}_{\overline{0}}\simeq
W^{\lambda}_{\overline{1}}$ via the odd involution $J^{\lambda}:W^{\lambda}\to
W^{\lambda}$ and write $W_{\lambda}$ for the common identified space, as in
Section 4.3.2. Then by theorem:Wlambda(gn), one has

(4.7.1)
$W^{\lambda}(\mathfrak{D})=\mathfrak{D}(W^{\lambda}(\mathfrak{g})_{\overline{0}})=\begin{cases}\mathfrak{sl}(W_{\lambda})&\text{if