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\begin{document}

\title{The Peel exact sequence for hook Specht modules via exterior algebra}
\date{Darij Grinberg, \today}
\maketitle

This is a mostly expository note devoted to one of the first nontrivial
results in the modular representation theory of the symmetric groups: Peel's
hook exact sequence. This sequence has been introduced by Peel in
\cite{Peel71} and revisited by K\"{u}nzer in \cite[Propositions 4.2.3 and
4.2.4]{Kuenze15}. The aim of this note is to reprove its most basic properties
in maximum generality (over arbitrary commutative rings satisfying $n=0$ when
necessary, not just finite fields) and more or less conceptually (using basic
homological algebra rather than ad-hoc computations involving Young tableaux).

It took me a while to figure it out, and I suspect that I ended up
rediscovering known properties of Koszul complexes. Yet I could not easily
locate any of it in the literature, so I have written up my proof in some
reasonable level of detail (not approaching that of my lecture notes
\cite{sga}, however, as this is somewhat more advanced material).

I should note that Peel's article \cite{Peel71} goes significantly beyond
constructing the exact sequence; the present note does not supersede it.

\section{Introduction}

Peel's hook exact sequence is an exact sequence consisting of Specht modules.
The classical way to define them is in terms of Young tableaux. But we will
use an equivalent definition using Vandermonde determinants, since it is
easier and self-contained:

Let $\mathbf{k}$ be any commutative ring, and $n$ a positive integer.

Consider the symmetric group $S_{n}$ of the set $\left[  n\right]  =\left\{
1,2,\ldots,n\right\}  $. It acts from the left on the polynomial ring
$\mathcal{P}_{n}=\mathbf{k}\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ by
$\mathbf{k}$-algebra automorphisms that permute the variables ($\sigma\cdot
x_{i}=x_{\sigma\left(  i\right)  }$ for all $\sigma\in S_{n}$ and $i\in\left[
n\right]  $). For any $k$ elements $a_{1},a_{2},\ldots,a_{k}$ of any
commutative ring, we let $V\left(  a_{1},a_{2},\ldots,a_{k}\right)  $ denote
their Vandermonde determinant
\[
V\left(  a_{1},a_{2},\ldots,a_{k}\right)  :=\det\left(  a_{i}^{j-1}\right)
_{i,j\in\left[  k\right]  }=\prod_{1\leq i<j\leq k}\left(  a_{j}-a_{i}\right)
.
\]
Note that this is clearly an alternating function in the inputs $a_{1}%
,a_{2},\ldots,a_{k}$.

For any $k\in\left[  n\right]  $, we let $\mathcal{S}^{\lambda^{k}}$ denote
the $\mathbf{k}$-linear span of the Vandermonde determinants $V\left(
x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  $ in $\mathcal{P}_{n}$ (where
$\left(  i_{1},i_{2},\ldots,i_{k}\right)  $ ranges over $\left[  n\right]
^{k}$). This is an $S_{n}$-subrepresentation of $\mathcal{P}_{n}$, and is
well-known to be isomorphic to the Specht module of the hook partition
$\lambda^{k}:=\left(  n-k+1,1^{k-1}\right)  $ (where \textquotedblleft%
$1^{k-1}$\textquotedblright\ means $k-1$ many $1$s in sequence). (The explicit
isomorphism can be found, e.g., in \cite[Corollary 5.6 \textbf{(b)}]{sga}.
Note that the Vandermonde determinant $V\left(  x_{i_{1}},x_{i_{2}}%
,\ldots,x_{i_{k}}\right)  $ is an alternating function in the inputs
$i_{1},i_{2},\ldots,i_{k}$; thus, the $\mathbf{k}$-module $\mathcal{S}%
^{\lambda^{k}}$ is spanned by the $V\left(  x_{i_{1}},x_{i_{2}},\ldots
,x_{i_{k}}\right)  $ with $i_{1},i_{2},\ldots,i_{k}$ distinct, and even just
by those with $i_{1}<i_{2}<\cdots<i_{k}$. Note that this is a particular case
of how Specht defined Specht modules in the first place.) Peel (in
\cite{Peel71}) denotes $\mathcal{S}^{\lambda^{k}}$ as $S\left(  k-1,n\right)
$.

Note that by the standard basis theorem for Specht modules, the $\mathbf{k}%
$-module $\mathcal{S}^{\lambda^{k}}$ has a basis consisting of those $V\left(
x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  $ with $1=i_{1}<i_{2}%
<\cdots<i_{k}$ (since these correspond to the standard Young tableaux of shape
$\lambda^{k}$).

Now, assume that $n=0$ in $\mathbf{k}$. Then, Peel (in \cite[\S 3]{Peel71})
and K\"{u}nzer (in \cite[Proposition 4.2.3]{Kuenze15}) show that for each
$k\in\left[  n-1\right]  $, there is an $S_{n}$-representation homomorphism
\begin{align}
f_{k}:\mathcal{S}^{\lambda^{k}}  &  \rightarrow\mathcal{S}^{\lambda^{k+1}%
},\nonumber\\
V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)   &  \mapsto\sum
_{s\in\left[  n\right]  }V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}%
,x_{s}\right)  . \label{eq.darij1.Vtosum}%
\end{align}
Note that the sum can just as well be restricted to the $s\in\left[  n\right]
\setminus\left\{  i_{1},i_{2},\ldots,i_{k}\right\}  $ only, since the addends
for $s\in\left\{  i_{1},i_{2},\ldots,i_{k}\right\}  $ are $0$. Furthermore,
they show that these maps $f_{k}$ form an exact sequence
\[
0\rightarrow\mathcal{S}^{\lambda^{1}}\overset{f_{1}}{\rightarrow}%
\mathcal{S}^{\lambda^{2}}\overset{f_{2}}{\rightarrow}\cdots\overset{f_{n-1}%
}{\rightarrow}\mathcal{S}^{\lambda^{n}}\rightarrow0
\]
that has a $\mathbf{k}$-linear (but not $\mathbf{k}\left[  S_{n}\right]
$-linear) chain contraction.

The purpose of this note is to prove this in a conceptual and readable way.
The proof in Peel's \cite{Peel71} is not fully clear to me, and only considers
the case when $\mathbf{k}$ is a field. K\"{u}nzer in \cite{Kuenze15} only
shows the existence of the maps $f_{k}$, not the exactness of the sequence; it
is also computational and intransparent (though very elementary). Thus, I hope
that this note has some use to others.

\section{The proof}

As I said, the proof I am giving uses just basic homological algebra
(morphisms and chain contractions of complexes) and exterior powers. But we
need to get some notation introduced and auxiliary results proved.

In the following, we do \textbf{not} assume that $n=0$ in $\mathbf{k}$ unless
we explicitly say so.

Let $N$ be the free $\mathbf{k}$-module $\mathbf{k}^{n}$ with its standard
basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $. This is the natural
representation of $S_{n}$, where $S_{n}$ acts on $\mathbf{k}^{n}$ by permuting
the basis ($\sigma\cdot e_{i}=e_{\sigma\left(  i\right)  }$ for each
$\sigma\in S_{n}$ and $i\in\left[  n\right]  $).

Consider the $S_{n}$-invariant element
\begin{equation}
e:=e_{1}+e_{2}+\cdots+e_{n}\in N.
\end{equation}


Consider the exterior algebra $\Lambda N=\bigoplus\limits_{i=0}^{n}\Lambda
^{i}N$. It has a basis $\left(  e_{I}\right)  _{I\subseteq\left[  n\right]  }%
$, where we set
\begin{align*}
e_{\left\{  i_{1}<i_{2}<\cdots<i_{k}\right\}  }  &  :=e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\\
&  \qquad\text{for each }\left\{  i_{1}<i_{2}<\cdots<i_{k}\right\}
\subseteq\left[  n\right]  .
\end{align*}
The exterior algebra $\Lambda N$ as well as each exterior power $\Lambda^{i}N$
is an $S_{n}$-representation, equipped with the diagonal $S_{n}$-action:
\begin{align*}
g\left(  v_{1}\wedge v_{2}\wedge\cdots\wedge v_{i}\right)  =gv_{1}\wedge
gv_{2}\wedge\cdots\wedge gv_{i}%
\end{align*}
for all $g\in S_{n}$ and $v_{1},v_{2},\ldots,v_{i}\in N$.

Let $\varepsilon:\Lambda N\to\Lambda N$ be the $\mathbf{k}$-linear map sending
each $w$ to $e\wedge w$.

Let $\varepsilon_{1}:\Lambda N\to\Lambda N$ be the $\mathbf{k}$-linear map
sending each $w$ to $e_{1}\wedge w$.

The exterior algebra $\Lambda N$ is known to be a supercommutative
superalgebra with $\mathbb{N}$-grading given by placing $\Lambda^{i}N$ in
degree $i$. A \emph{superderivation} of $\Lambda N$ shall mean a $\mathbf{k}%
$-linear map $d:\Lambda N\rightarrow\Lambda N$ that satisfies the
\emph{super-Leibniz rule}
\begin{align*}
d\left(  ab\right)   &  =d\left(  a\right)  \cdot b+\left(  -1\right)
^{i}a\cdot d\left(  b\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }i\in\mathbb{N}\text{ and }a\in
\Lambda^{i}N\text{ and }b\in\Lambda N.
\end{align*}


Let $\partial:\Lambda N\rightarrow\Lambda N$ be the $\mathbf{k}$-linear map
defined by
\begin{align*}
\partial\left(  e_{\left\{  i_{1}<i_{2}<\cdots<i_{k}\right\}  }\right)   &
=\sum_{p=1}^{k}\left(  -1\right)  ^{p-1}e_{\left\{  i_{1}<i_{2}<\cdots
<\widehat{i_{p}}<\cdots<i_{k}\right\}  }\\
&  \qquad\text{for each }\left\{  i_{1}<i_{2}<\cdots<i_{k}\right\}
\subseteq\left[  n\right]  .
\end{align*}
Here (and in the following), the \textquotedblleft magician's
hat\textquotedblright\ $\ \widehat{}$\ \ is understood to vanish whatever
stands under it; thus, \textquotedblleft$i_{1}<i_{2}<\cdots<\widehat{i_{p}%
}<\cdots<i_{k}$\textquotedblright\ means \textquotedblleft$i_{1}<i_{2}%
<\cdots<i_{p-1}<i_{p+1}<i_{p+2}<\cdots<i_{k}$\textquotedblright.

Let $\partial_{1}:\Lambda N\rightarrow\Lambda N$ be the $\mathbf{k}$-linear
map defined by
\[
\partial_{1}\left(  e_{I}\right)  =%
\begin{cases}
e_{I\setminus\left\{  1\right\}  }, & \text{if }1\in I\\
0, & \text{if }1\notin I
\end{cases}
\qquad\text{for each }I\subseteq\left[  n\right]  .
\]


We note that both maps $\partial$ and $\partial_{1}$ can be described in more
canonical ways. Namely, if we let $\left(  e_{1}^{\ast},e_{2}^{\ast}%
,\ldots,e_{n}^{\ast}\right)  $ denote the dual basis to the basis $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ of $N$ (that is, each $e_{i}^{\ast}\in
N^{\ast}$ is the $\mathbf{k}$-linear map $N\rightarrow\mathbf{k}$ that sends
each vector to its $i$-th coordinate), and if we set $e^{\ast}:=e_{1}^{\ast
}+e_{2}^{\ast}+\cdots+e_{n}^{\ast}\in N^{\ast}$, then all $k\in\mathbb{N}$ and
$v_{1},v_{2},\ldots,v_{k}\in N$ satisfy
\begin{align}
&  \partial\left(  v_{1}\wedge v_{2}\wedge\cdots\wedge v_{k}\right)
\nonumber\\
&  =\sum_{p=1}^{k}\left(  -1\right)  ^{p-1}e^{\ast}\left(  v_{p}\right)  \cdot
v_{1}\wedge v_{2}\wedge\cdots\wedge\widehat{v_{p}}\wedge\cdots\wedge v_{k}
\label{eq.darij1.del-direct}%
\end{align}
and
\begin{align}
&  \partial_{1}\left(  v_{1}\wedge v_{2}\wedge\cdots\wedge v_{k}\right)
\nonumber\\
&  =\sum_{p=1}^{k}\left(  -1\right)  ^{p-1}e_{1}^{\ast}\left(  v_{p}\right)
\cdot v_{1}\wedge v_{2}\wedge\cdots\wedge\widehat{v_{p}}\wedge\cdots\wedge
v_{k}. \label{eq.darij1.del1-direct}%
\end{align}
In other words, in terms of interior products (see, e.g., \cite[\S 2.3.1]%
{Winitz23}), we have $\partial=\iota_{e^{\ast}}$ and $\partial_{1}%
=\iota_{e_{1}^{\ast}}$.

Both maps $\partial$ and $\partial_{1}$ are superderivations. The maps
$\varepsilon$ and $\varepsilon_{1}$ shift the degree by $1$ upwards, while the
maps $\partial$ and $\partial_{1}$ shift it by $1$ downwards. Hence, their
images and kernels are $\mathbb{N}$-graded $\mathbf{k}$-submodules of $\Lambda
N$, and their cokernels inherit the $\mathbb{N}$-grading from $\Lambda N$.
When we shall speak of $\left(  \operatorname*{Coker}\partial\right)  _{i}$,
we will mean the $i$-th graded component of this grading on
$\operatorname*{Coker}\partial$. Explicitly,
\[
\left(  \operatorname*{Coker}\partial\right)  _{i}=\left(  \Lambda
^{i}N\right)  \diagup\partial\left(  \Lambda^{i+1}N\right)  \qquad\text{for
each }i\in\mathbb{N}.
\]


Note that the maps $\varepsilon$ and $\partial$ are morphisms of $S_{n}%
$-representations, whereas $\varepsilon_{1}$ and $\partial_{1}$ are merely
$\mathbf{k}$-module morphisms. Thus, $\operatorname*{Coker}\partial$ and its
graded components $\left(  \operatorname*{Coker}\partial\right)  _{i}$ are
$S_{n}$-representations.

The following is well-known:

\begin{proposition}
\label{prop.1} Each of these four maps $\varepsilon,\partial,\varepsilon
_{1},\partial_{1}$ makes the graded algebra $\Lambda N$ into a long exact
sequence: i.e., we have
\begin{align*}
\operatorname*{Ker}\varepsilon &  =\operatorname{Im}\varepsilon
,\ \ \ \ \ \ \ \ \ \ \operatorname*{Ker}\partial=\operatorname{Im}\partial,\\
\operatorname*{Ker}\varepsilon_{1}  &  =\operatorname{Im}\varepsilon
_{1},\ \ \ \ \ \ \ \ \ \ \operatorname*{Ker}\partial_{1}=\operatorname{Im}%
\partial_{1}.
\end{align*}
Moreover, these four exact sequences are each other's chain contractions:
i.e., we have
\begin{align*}
\partial\varepsilon_{1}+\varepsilon_{1}\partial &  =\operatorname*{id}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\partial_{1}\varepsilon+\varepsilon\partial_{1}  &  =\operatorname*{id}.
\end{align*}

\end{proposition}

\begin{proof}
Clearly, $\varepsilon^{2}=0$ (since $\underbrace{e\wedge e}_{=0}\wedge\,w=0$
for each $w\in\Lambda N$) and $\varepsilon_{1}^{2}=0$ (likewise). Very easily,
$\partial_{1}^{2}=0$. By a standard computation from homological algebra, we
have $\partial^{2}=0$ as well. (This can also be proved without getting one's
hands dirty, by defining the symmetric bilinear form $\left\langle \cdot
,\cdot\right\rangle $ on $\Lambda N$ by $\left\langle e_{I},e_{J}\right\rangle
=%
\begin{cases}
1, & \text{if }I=J;\\
0, & \text{if }I\neq J
\end{cases}
$ for all $I,J\subseteq\left[  n\right]  $. Indeed, this bilinear form
$\left\langle \cdot,\cdot\right\rangle $ is nondegenerate and is easily seen
to satisfy
\[
\left\langle \varepsilon\left(  e_{I}\right)  ,\ e_{J}\right\rangle
=\left\langle e_{I},\ \partial\left(  e_{J}\right)  \right\rangle
\ \ \ \ \ \ \ \ \ \ \text{for all }I,J\subseteq\left[  n\right]  ;
\]
thus, the map $\partial$ is the adjoint of $\varepsilon$ with respect to this
bilinear form. Hence, $\partial^{2}=0$ follows from $\varepsilon^{2}=0$.)

Since $\partial$ is a superderivation, each $w\in\Lambda N$ satisfies
$\partial\left(  e_{1}\wedge w\right)  =\underbrace{\partial\left(
e_{1}\right)  }_{=1}w-e_{1}\wedge\partial\left(  w\right)  =w-e_{1}%
\wedge\partial\left(  w\right)  $, that is, $\partial\left(  \varepsilon
_{1}\left(  w\right)  \right)  =w-\varepsilon_{1}\left(  \partial\left(
w\right)  \right)  $, that is, $\partial\left(  \varepsilon_{1}\left(
w\right)  \right)  +\varepsilon_{1}\left(  \partial\left(  w\right)  \right)
=w$. Thus, $\partial\varepsilon_{1}+\varepsilon_{1}\partial=\operatorname*{id}%
$ holds. Similarly, $\partial_{1}\varepsilon+\varepsilon\partial
_{1}=\operatorname*{id}$ holds (since $\partial_{1}\left(  e\right)  =1$).

The \textquotedblleft Ker=Im\textquotedblright\ identities now follow easily:
For instance, let us prove $\operatorname*{Ker}\varepsilon=\operatorname{Im}%
\varepsilon$. The $\operatorname{Im}\varepsilon\subseteq\operatorname*{Ker}%
\varepsilon$ inclusion follows from $\varepsilon^{2}=0$. To prove the converse
inclusion, fix $w\in\operatorname*{Ker}\varepsilon$; then, $\varepsilon\left(
w\right)  =0$; but $\partial_{1}\varepsilon+\varepsilon\partial_{1}%
=\operatorname*{id}$ yields $\partial_{1}\left(  \varepsilon\left(  w\right)
\right)  +\varepsilon\left(  \partial_{1}\left(  w\right)  \right)  =w$,
whence $w=\partial_{1}\underbrace{\left(  \varepsilon\left(  w\right)
\right)  }_{=0}+\,\varepsilon\left(  \partial_{1}\left(  w\right)  \right)
=\varepsilon\left(  \partial_{1}\left(  w\right)  \right)  \in
\operatorname{Im}\varepsilon$. So $\operatorname*{Ker}\varepsilon
\subseteq\operatorname{Im}\varepsilon$ is proved, and with it
$\operatorname*{Ker}\varepsilon=\operatorname{Im}\varepsilon$. Likewise, the
other three \textquotedblleft Ker=Im\textquotedblright\ identities can be
shown. Altogether, the proof of Proposition \ref{prop.1} is complete.
\end{proof}

\begin{proposition}
\label{prop.2}We have
\begin{equation}
\partial_{1}\partial=-\partial\partial_{1}.\label{eq.prop.peel.p1b.dd=-dd}%
\end{equation}
Hence,
\begin{equation}
\partial_{1}\left(  \operatorname{Im}\partial\right)  \subseteq
\operatorname{Im}\partial.\label{eq.prop.peel.p1b.dImd}%
\end{equation}
Thus, the map $\partial_{1}:\Lambda N\rightarrow\Lambda N$ descends to a
$\mathbf{k}$-linear map $\partial_{1}^{\prime}:\operatorname*{Coker}%
\partial\rightarrow\operatorname*{Coker}\partial$ such that the diagram
\begin{equation}
\xymatrix@C=4pc{
\Lambda^k N \ar[r]^{\partial_1} \arsurj[d] & \Lambda^{k-1} N \arsurj[d] \\
\tup{\Coker \partial}_k \ar[r]_-{\partial'_1} & \tup{\Coker \partial}_{k-1}
}\label{eq.prop.peel.p1b.diagram}%
\end{equation}
(where the vertical arrows are graded parts of the canonical projection
$\Lambda N\rightarrow\operatorname*{Coker}\partial$) is commutative for each
$k>0$.
\end{proposition}

\begin{proof}
The equality $\partial_{1}\partial=-\partial\partial_{1}$ is easy to check
directly from the definitions of $\partial$ and $\partial_{1}$. Alternatively,
we can prove it abstractly using the fact that the superderivations of a
superalgebra form a Lie superalgebra, so that the supercommutator of two
superderivations is again a superderivation. Indeed, this shows that
$\partial_{1}\partial+\partial\partial_{1}$ (being the supercommutator of
$\partial_{1}$ and $\partial$) is a superderivation. Since it sends the
generators $\Lambda^{1}N$ of $\Lambda N$ to $0$, it thus is $0$ everywhere (by
the super-Leibniz rule). Either way, \eqref{eq.prop.peel.p1b.dd=-dd} is proved.

Of course, \eqref{eq.prop.peel.p1b.dImd} follows immediately from \eqref{eq.prop.peel.p1b.dd=-dd}.

The existence of $\partial_{1}^{\prime}$ for which the diagram
\eqref{eq.prop.peel.p1b.diagram} commutes is an immediate consequence of \eqref{eq.prop.peel.p1b.dImd}.
\end{proof}

So far, this all was true for any commutative ring $\mathbf{k}$.

Now, we note that $\partial\left(  e\right)  =n$. Hence, if $n=0$ in
$\mathbf{k}$, then we gain extra properties:

\begin{proposition}
\label{prop.3}Assume that $n=0$ in $\mathbf{k}$. We have
\begin{equation}
\varepsilon\partial=-\partial\varepsilon. \label{eq.prop.peel.p2.ed}%
\end{equation}
Thus, up to the usual $\left(  -1\right)  ^{\deg}$ sign twist, $\varepsilon$
is a degree-shifting endomorphism of the complex $\left(  \Lambda
N,\ \partial\right)  $, and vice versa.

We furthermore have
\begin{equation}
\varepsilon\left(  \operatorname{Im}\partial\right)  \subseteq
\operatorname{Im}\partial.\label{eq.prop.peel.p2.eImd}%
\end{equation}
Consequently, the map $\varepsilon:\Lambda N\rightarrow\Lambda N$ descends to
a $\mathbf{k}$-linear endomorphism $\varepsilon^{\prime}:\operatorname*{Coker}%
\partial\rightarrow\operatorname*{Coker}\partial$ of $\operatorname*{Coker}%
\partial=\left(  \Lambda N\right)  /\operatorname{Im}\partial$ such that the
diagram
\begin{equation}
\xymatrix@C=4pc{
\Lambda^k N \ar[r]^{\varepsilon} \arsurj[d] & \Lambda^{k+1} N \arsurj[d] \\
\tup{\Coker \partial}_k \ar[r]_-{\varepsilon'} & \tup{\Coker \partial}_{k+1}
}\label{eq.prop.peel.p2.diagram}%
\end{equation}
(where the vertical arrows are graded parts of the canonical projection
$\Lambda N\rightarrow\operatorname*{Coker}\partial$) is commutative for each
$k\in\mathbb{N}$.
\end{proposition}

\begin{proof}
For each $w\in\Lambda N$, we have
\begin{align*}
\partial\left(  \varepsilon\left(  w\right)  \right)   &  =\partial\left(
e\wedge w\right)  \qquad\left(  \text{by the definition of }\varepsilon\right)
\\
&  =\underbrace{\partial\left(  e\right)  }_{=n=0}w-e\wedge\partial\left(
w\right)  \qquad\left(  \text{since }\partial\text{ is a superderivation}%
\right) \\
&  =-e\wedge\partial\left(  w\right)  =-\varepsilon\left(  \partial\left(
w\right)  \right)  \qquad\left(  \text{by the definition of }\varepsilon
\right)  .
\end{align*}
Hence, $\partial\varepsilon=-\varepsilon\partial$, so that $\varepsilon
\partial=-\partial\varepsilon$. This proves \eqref{eq.prop.peel.p2.ed}. Thus,
\eqref{eq.prop.peel.p2.eImd} immediately follows, and from it flows the
existence of $\varepsilon^{\prime}$ that makes the diagram
\eqref{eq.prop.peel.p2.diagram} commutative. $\blacksquare$
\end{proof}

\begin{theorem}
\label{thm.4}Assume that $n=0$ in $\mathbf{k}$. Then, the sequence
\[
\cdots\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k-1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\left(  \operatorname*{Coker}\partial\right)
_{k}\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k+1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\cdots
\]
(where $\varepsilon^{\prime}$ was defined in Proposition \ref{prop.3}) is
exact and has a $\mathbf{k}$-linear chain contraction.
\end{theorem}

\begin{proof}
Recall the $\partial_{1}^{\prime}$ map from Proposition \ref{prop.2}. Clearly,
$\left(  \varepsilon^{\prime}\right)  ^{2}=0$ because $\varepsilon^{2}=0$.
Moreover, projecting the equality $\partial_{1}\varepsilon+\varepsilon
\partial_{1}=\operatorname*{id}$ (from Proposition \ref{prop.1}) onto
$\operatorname*{Coker}\partial$, we obtain $\partial_{1}^{\prime}%
\varepsilon^{\prime}+\varepsilon^{\prime}\partial_{1}^{\prime}%
=\operatorname*{id}$, so that $\partial_{1}^{\prime}$ is a chain contraction
for the sequence
\[
\cdots\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k-1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\left(  \operatorname*{Coker}\partial\right)
_{k}\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k+1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\cdots.
\]
Thus, this sequence is exact, and Theorem \ref{thm.4} is proved.
\end{proof}

Now we come to the Peel modules $\mathcal{S}^{\lambda^{k}}$. As we recall, for
each $k\in\left[  n\right]  $, the $S_{n}$-representation $\mathcal{S}%
^{\lambda^{k}}$ is defined as the $\mathbf{k}$-linear span of the Vandermonde
determinants $V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  $ in
$\mathcal{P}_{n}$. We furthermore set
\begin{align*}
\mathcal{S}^{\lambda^{k}}:=0\ \ \ \ \ \ \ \ \ \ \text{for all }k>n\text{ and
also for }k=0.
\end{align*}
Note that this agrees with the original definition of $\mathcal{S}%
^{\lambda^{k}}$ as the span of the $V\left(  x_{i_{1}},x_{i_{2}}
,\ldots,x_{i_{k}}\right)  $ for $k>n$ (since $V\left(  x_{i_{1}},x_{i_{2}
},\ldots,x_{i_{k}}\right)  $ is alternating as a function in $i_{1}
,i_{2},\ldots,i_{k}$), but not for $k=0$. Yet it is the right way to define
$\mathcal{S}^{\lambda^{0}}$, as we will see.

For each $k>0$, we define the $\mathbf{k}$-linear map
\begin{align*}
\omega_{k}:\Lambda^{k}N  &  \to\mathcal{S}^{\lambda^{k}},\\
e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}  &  \mapsto V\left(
x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  .
\end{align*}
This map $\omega_{k}$ is well-defined, since $V\left(  x_{i_{1}},x_{i_{2}
},\ldots,x_{i_{k}}\right)  $ is an alternating function in its inputs
$i_{1},i_{2},\ldots,i_{k}$. Moreover, $\omega_{k}$ is a morphism of $S_{n}
$-representations (since $\sigma\cdot e_{i}=e_{\sigma\left(  i\right)  }$ and
$\sigma\cdot x_{i}=x_{\sigma\left(  i\right)  }$ for all $\sigma\in S_{n}$ and
$i\in\left[  n\right]  $).

We furthermore define $\omega_{k}:\Lambda^{k}N\to\mathcal{S}^{\lambda^{k}}$ to
be the zero map $0$ for $k>n$ and for $k=0$.

We now claim the following:

\begin{lemma}
\label{lem.5}Let $k\in\mathbb{N}$. Then, $\omega_{k}\partial=0$ on
$\Lambda^{k+1}N$.
\end{lemma}

\begin{proof}
We WLOG assume that $k\in\left[  n\right]  $, since otherwise the claim is
made obvious by the fact that $\omega_{k}=0$.

The $\mathbf{k}$-module $\Lambda^{k+1}N$ is spanned by the vectors $e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k+1}}$ for all $\left(  i_{1}%
,i_{2},\ldots,i_{k+1}\right)  \in\left[  n\right]  ^{k+1}$. Hence, by
linearity, it suffices to show that
\[
\omega_{k}\left(  \partial\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k+1}}\right)  \right)  =0\ \ \ \ \ \ \ \ \ \ \text{for all }\left(
i_{1},i_{2},\ldots,i_{k+1}\right)  \in\left[  n\right]  ^{k+1}.
\]


Let us thus do this. Fix $\left(  i_{1},i_{2},\ldots,i_{k+1}\right)
\in\left[  n\right]  ^{k+1}$. Then, the equality \eqref{eq.darij1.del-direct}
shows that
\begin{align*}
&  \partial\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k+1}%
}\right) \\
&  =\sum_{p=1}^{k+1}\left(  -1\right)  ^{p-1}\underbrace{e^{\ast}\left(
e_{i_{p}}\right)  }_{=1}\cdot\,e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots
\wedge\widehat{e_{i_{p}}}\wedge\cdots\wedge e_{i_{k+1}}\\
&  =\sum_{p=1}^{k+1}\left(  -1\right)  ^{p-1}e_{i_{1}}\wedge e_{i_{2}}%
\wedge\cdots\wedge\widehat{e_{i_{p}}}\wedge\cdots\wedge e_{i_{k+1}}.
\end{align*}
Thus,
\begin{align*}
&  \omega_{k}\left(  \partial\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge
\cdots\wedge e_{i_{k+1}}\right)  \right) \\
&  =\omega_{k}\left(  \sum_{p=1}^{k+1}\left(  -1\right)  ^{p-1}e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge\widehat{e_{i_{p}}}\wedge\cdots\wedge e_{i_{k+1}%
}\right) \\
&  =\sum_{p=1}^{k+1}\underbrace{\left(  -1\right)  ^{p-1}}_{=\left(
-1\right)  ^{p+1}}\underbrace{\omega_{k}\left(  e_{i_{1}}\wedge e_{i_{2}%
}\wedge\cdots\wedge\widehat{e_{i_{p}}}\wedge\cdots\wedge e_{i_{k+1}}\right)
}_{\substack{=V\left(  x_{i_{1}},x_{i_{2}},\ldots,\widehat{x_{i_{p}}}%
,\ldots,x_{i_{k+1}}\right)  \\=\det\left(
\begin{array}
[c]{ccccccc}%
x_{i_{1}}^{0} & x_{i_{2}}^{0} & \cdots & \widehat{x_{i_{p}}^{0}} & \cdots &
x_{i_{k}}^{0} & x_{i_{k+1}}^{0}\\
x_{i_{1}}^{1} & x_{i_{2}}^{1} & \cdots & \widehat{x_{i_{p}}^{1}} & \cdots &
x_{i_{k}}^{1} & x_{i_{k+1}}^{1}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots\\
x_{i_{1}}^{k-1} & x_{i_{2}}^{k-1} & \cdots & \widehat{x_{i_{p}}^{k-1}} &
\cdots & x_{i_{k}}^{k-1} & x_{i_{k+1}}^{k-1}%
\end{array}
\right)  \\\text{(since }V\left(  a_{1},a_{2},\ldots,a_{k}\right)
=\det\left(  a_{j}^{i-1}\right)  _{i,j\in\left[  k\right]  }\text{ for all
}a_{1},a_{2},\ldots,a_{k}\text{)}}}\\
&  =\sum_{p=1}^{k+1}\left(  -1\right)  ^{p+1}\det\left(
\begin{array}
[c]{ccccccc}%
x_{i_{1}}^{0} & x_{i_{2}}^{0} & \cdots & \widehat{x_{i_{p}}^{0}} & \cdots &
x_{i_{k}}^{0} & x_{i_{k+1}}^{0}\\
x_{i_{1}}^{1} & x_{i_{2}}^{1} & \cdots & \widehat{x_{i_{p}}^{1}} & \cdots &
x_{i_{k}}^{1} & x_{i_{k+1}}^{1}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots\\
x_{i_{1}}^{k-1} & x_{i_{2}}^{k-1} & \cdots & \widehat{x_{i_{p}}^{k-1}} &
\cdots & x_{i_{k}}^{k-1} & x_{i_{k+1}}^{k-1}%
\end{array}
\right)  .
\end{align*}
On the other hand, if we let $A$ denote the $\left(  k+1\right)  \times\left(
k+1\right)  $-matrix
\[
\left(
\begin{array}
[c]{ccccc}%
1 & 1 & \cdots & 1 & 1\\
x_{i_{1}}^{0} & x_{i_{2}}^{0} & \cdots & x_{i_{k}}^{0} & x_{i_{k+1}}^{0}\\
x_{i_{1}}^{1} & x_{i_{2}}^{1} & \cdots & x_{i_{k}}^{1} & x_{i_{k+1}}^{1}\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
x_{i_{1}}^{k-1} & x_{i_{2}}^{k-1} & \cdots & x_{i_{k}}^{k-1} & x_{i_{k+1}%
}^{k-1}%
\end{array}
\right)  ,
\]
then
\[
\det A=\sum_{p=1}^{k+1}\left(  -1\right)  ^{p+1}\det\left(
\begin{array}
[c]{ccccccc}%
x_{i_{1}}^{0} & x_{i_{2}}^{0} & \cdots & \widehat{x_{i_{p}}^{0}} & \cdots &
x_{i_{k}}^{0} & x_{i_{k+1}}^{0}\\
x_{i_{1}}^{1} & x_{i_{2}}^{1} & \cdots & \widehat{x_{i_{p}}^{1}} & \cdots &
x_{i_{k}}^{1} & x_{i_{k+1}}^{1}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots\\
x_{i_{1}}^{k-1} & x_{i_{2}}^{k-1} & \cdots & \widehat{x_{i_{p}}^{k-1}} &
\cdots & x_{i_{k}}^{k-1} & x_{i_{k+1}}^{k-1}%
\end{array}
\right)
\]
(by Laplace expansion along the first row of $A$). Comparing these two
equalities, we find
\[
\omega_{k}\left(  \partial\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k+1}}\right)  \right)  =\det A.
\]


But the matrix $A$ has two equal rows: In fact, its second row \newline%
$\left(
\begin{array}
[c]{ccccc}%
x_{i_{1}}^{0} & x_{i_{2}}^{0} & \cdots & x_{i_{k}}^{0} & x_{i_{k+1}}^{0}%
\end{array}
\right)  $ agrees with its first row $\left(
\begin{array}
[c]{ccccc}%
1 & 1 & \cdots & 1 & 1
\end{array}
\right)  $, since $x_{j}^{0}=1$ for all $j$. Hence, $\det A=0$ (since a matrix
with two equal rows must always have determinant $0$). Thus,
\[
\omega_{k}\left(  \partial\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k+1}}\right)  \right)  =\det A=0.
\]


As explained above, this proves Lemma \ref{lem.5}.
\end{proof}

Next comes the most technical lemma in this note:

\begin{lemma}
\label{lem.6}Let $k>0$. A monomial in $\mathcal{P}_{n}$ is called \emph{nice}
if it can be written as $x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots x_{i_{k-1}}^{k-1}$
with $1<i_{1}<i_{2}<\cdots<i_{k-1}\leq n$. Let $\eta:\mathcal{P}%
_{n}\rightarrow\Lambda^{k-1}N$ be the $\mathbf{k}$-linear map that%
\begin{align*}
&  \text{sends each nice monomial }x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots
x_{i_{k-1}}^{k-1}\text{ to }e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k-1}}\in\Lambda^{k-1}N\text{,}\\
&  \text{and sends each monomial that is not nice to }0.
\end{align*}
(This is well-defined, since the monomials form a basis of the $\mathbf{k}%
$-module $\mathcal{P}_{n}$, and since a nice monomial can be written in the
form $x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots x_{i_{k-1}}^{k-1}$ uniquely.)

Then,
\[
\varepsilon_{1}\eta\omega_{k}=\operatorname*{id}\text{ on }\varepsilon
_{1}\left(  \Lambda^{k-1}N\right)  .
\]

\end{lemma}

\begin{proof}
The $\mathbf{k}$-module $\Lambda^{k-1}N$ is spanned by the vectors $e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}$ for all $\left(  i_{1}%
,i_{2},\ldots,i_{k-1}\right)  \in\left[  n\right]  ^{k-1}$ satisfying
$i_{1}<i_{2}<\cdots<i_{k-1}$. Hence, its image $\varepsilon_{1}\left(
\Lambda^{k-1}N\right)  $ is spanned by the images $\varepsilon_{1}\left(
e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}\right)  $ of these
vectors. Since $\varepsilon_{1}\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge
\cdots\wedge e_{i_{k-1}}\right)  =e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}%
\wedge\cdots\wedge e_{i_{k-1}}$ (by the definition of $\varepsilon_{1}$), we
can restate this as follows: The image $\varepsilon_{1}\left(  \Lambda
^{k-1}N\right)  $ is spanned by the vectors $e_{1}\wedge e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}$ for all $\left(  i_{1},i_{2}%
,\ldots,i_{k-1}\right)  \in\left[  n\right]  ^{k-1}$ satisfying $i_{1}%
<i_{2}<\cdots<i_{k-1}$.

However, many of these vectors vanish: Indeed, $e_{1}\wedge e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}=0$ whenever $i_{1}=1$, since an
exterior product with two equal factors is $0$. Obviously, these vanishing
vectors are unneccessary for spanning $\varepsilon_{1}\left(  \Lambda
^{k-1}N\right)  $. Hence, we can remove them from our list, and conclude that
the image $\varepsilon_{1}\left(  \Lambda^{k-1}N\right)  $ is spanned by the
vectors $e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}$
for all $\left(  i_{1},i_{2},\ldots,i_{k-1}\right)  \in\left[  n\right]
^{k-1}$ satisfying $1<i_{1}<i_{2}<\cdots<i_{k-1}$.

Thus, by linearity, it suffices (towards our goal of proving that
$\varepsilon_{1}\eta\omega_{k}=\operatorname*{id}$ on $\varepsilon_{1}\left(
\Lambda^{k-1}N\right)  $) to show that
\[
\left(  \varepsilon_{1}\eta\omega_{k}\right)  \left(  e_{1}\wedge e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}\right)  =e_{1}\wedge e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}%
\]
for all $\left(  i_{1},i_{2},\ldots,i_{k-1}\right)  \in\left[  n\right]
^{k-1}$ satisfying $1<i_{1}<i_{2}<\cdots<i_{k-1}$. Let us do this.

Let $\left(  i_{1},i_{2},\ldots,i_{k-1}\right)  \in\left[  n\right]  ^{k-1}$
be such that $1<i_{1}<i_{2}<\cdots<i_{k-1}$. Then, the definition of
$\omega_{k}$ yields
\[
\omega_{k}\left(  e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k-1}}\right)  =V\left(  x_{1},x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k-1}%
}\right)  .
\]
Setting $i_{0}:=1$, we can rewrite this as
\begin{align}
\omega_{k}\left(  e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k-1}}\right)   &  =V\left(  x_{i_{0}},x_{i_{1}},x_{i_{2}},\ldots
,x_{i_{k-1}}\right) \nonumber\\
&  =\det\left(  x_{i_{u-1}}^{v-1}\right)  _{u,v\in\left[  k\right]
}\nonumber\\
&  =\sum_{\sigma\in S_{k}}\left(  -1\right)  ^{\sigma}\prod_{p=1}%
^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1} \label{pf.lem.peel.l2b.3}%
\end{align}
by the Leibniz formula for the determinant.

The products $\prod_{p=1}^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1}$ that
appear on the right hand side are monomials, but I claim that only one of them
is nice: viz., the one obtained for $\sigma=\operatorname*{id}$. Indeed, the
latter product is
\[
\prod_{p=1}^{k}x_{i_{p-1}}^{\operatorname*{id}\left(  p\right)  -1}%
=\prod_{p=1}^{k}x_{i_{p-1}}^{p-1}=\underbrace{x_{i_{0}}^{0}}_{=1}x_{i_{1}}%
^{1}x_{i_{2}}^{2}\cdots x_{i_{k-1}}^{k-1}=x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots
x_{i_{k-1}}^{k-1},
\]
which is clearly nice. Now let me show that none of the remaining products
$\prod_{p=1}^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1}$ (with $\sigma
\neq\operatorname*{id}$) is nice. Indeed, a nice monomial cannot contain the
variable $x_{1}$ at all, and must contain the exponents $1,2,\ldots,k-1$ on
variables with increasing subscripts (i.e., the larger the exponent, the
larger the subscript). But a product $\prod_{p=1}^{k}x_{i_{p-1}}%
^{\sigma\left(  p\right)  -1}$ with $\sigma\neq\operatorname*{id}$ cannot
satisfy these two properties, since it either contains $x_{i_{0}}=x_{1}$ (when
$\sigma\left(  1\right)  \neq1$), or contains the other exponents
$1,2,\ldots,k-1$ in the \textquotedblleft wrong order\textquotedblright\ (when
$\sigma\left(  1\right)  =1$, so that $\sigma$ has an inversion among its
values $\sigma\left(  2\right)  ,\sigma\left(  3\right)  ,\ldots,\sigma\left(
k\right)  $). Thus, a monomial of the form $\prod_{p=1}^{k}x_{i_{p-1}}%
^{\sigma\left(  p\right)  -1}$ with $\sigma\neq\operatorname*{id}$ cannot be
nice. Hence, such a monomial will always satisfy
\begin{equation}
\eta\left(  \prod_{p=1}^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1}\right)  =0
\label{pf.lem.peel.l2b.eta0}%
\end{equation}
(since $\eta$ kills all non-nice monomials, by definition).

Now, the map $\eta$ is $\mathbf{k}$-linear. Hence,
\begin{align*}
\eta\left(  \sum_{\sigma\in S_{k}}\left(  -1\right)  ^{\sigma}\prod_{p=1}%
^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1}\right)   &  =\sum_{\sigma\in
S_{k}}\left(  -1\right)  ^{\sigma}\underbrace{\eta\left(  \prod_{p=1}%
^{k}x_{i_{p-1}}^{\sigma\left(  p\right)  -1}\right)  }_{\substack{=0\text{
whenever }\sigma\neq\operatorname*{id}\\\text{(by
\eqref{pf.lem.peel.l2b.eta0})}}}\\
&  =\underbrace{\left(  -1\right)  ^{\operatorname*{id}}}_{=1}\eta\left(
\underbrace{\prod_{p=1}^{k}x_{i_{p-1}}^{\operatorname*{id}\left(  p\right)
-1}}_{=x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots x_{i_{k-1}}^{k-1}}\right) \\
&  =\eta\left(  x_{i_{1}}^{1}x_{i_{2}}^{2}\cdots x_{i_{k-1}}^{k-1}\right)
=e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}%
\end{align*}
(by the definition of $\eta$). In view of \eqref{pf.lem.peel.l2b.3}, this
rewrites as
\[
\eta\left(  \omega_{k}\left(  e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}%
\wedge\cdots\wedge e_{i_{k-1}}\right)  \right)  =e_{i_{1}}\wedge e_{i_{2}%
}\wedge\cdots\wedge e_{i_{k-1}}.
\]
Hence,
\begin{align*}
\varepsilon_{1}\left(  \eta\left(  \omega_{k}\left(  e_{1}\wedge e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}\right)  \right)  \right)   &
=\varepsilon_{1}\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge
e_{i_{k-1}}\right) \\
&  =e_{1}\wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}%
\end{align*}
(by the definition of $\varepsilon_{1}$). In other words,
\[
\left(  \varepsilon_{1}\eta\omega_{k}\right)  \left(  e_{1}\wedge e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}\right)  =e_{1}\wedge e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k-1}}.
\]
But this is precisely what remained to prove. So the proof of Lemma
\ref{lem.6} is complete.
\end{proof}

\begin{theorem}
\label{thm.7}Let $k\in\mathbb{N}$. Then, the map $\omega_{k}:\Lambda
^{k}N\rightarrow\mathcal{S}^{\lambda^{k}}$ is a surjective morphism of $S_{n}%
$-representations. Its kernel is
\begin{equation}
\operatorname*{Ker}\left(  \omega_{k}\right)  =\operatorname*{Ker}\left(
\partial\mid_{\Lambda^{k}N}\right)  =\partial\left(  \Lambda^{k+1}N\right)  .
\label{eq.thm.peel.t2.KKI}%
\end{equation}
Thus, $\omega_{k}$ induces an isomorphism
\begin{equation}
\mathcal{S}^{\lambda^{k}}\cong\left(  \operatorname*{Coker}\partial\right)
_{k}\ \ \ \ \ \ \ \ \ \ \text{of }S_{n}\text{-representations.}
\label{eq.thm.peel.t2.iso}%
\end{equation}

\end{theorem}

\begin{proof}
We already know that $\omega_{k}$ is a morphism of $S_{n}$-representations.

From Proposition \ref{prop.1}, we know that $\operatorname*{Ker}%
\partial=\operatorname{Im}\partial$. Taking the $k$-th graded component of
this equality, we obtain $\operatorname*{Ker}\left(  \partial\mid_{\Lambda
^{k}N}\right)  =\partial\left(  \Lambda^{k+1}N\right)  $.

Next, we shall prove that $\omega_{k}$ is surjective and that
$\operatorname*{Ker}\left(  \omega_{k}\right)  =\partial\left(  \Lambda
^{k+1}N\right)  $.

We WLOG assume that $k>0$, since the $k=0$ case is easily done by hand
(remember that $\omega_{0}=0$ and $\mathcal{S}^{\lambda^{0}}=0$ by definition,
and observe that $\partial\left(  \Lambda^{1}N\right)  $ contains
$\partial\left(  e_{1}\right)  =1$ and thus is the whole $\Lambda^{0}N$).
Hence, $\mathcal{S}^{\lambda^{k}}$ is defined as the span of all $V\left(
x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  $. Thus, the surjectivity of
$\omega_{k}$ is obvious from its definition.

Proposition \ref{prop.1} yields $\partial\varepsilon_{1}+\varepsilon
_{1}\partial=\operatorname*{id}$. Hence, $\Lambda^{k}N=\varepsilon_{1}\left(
\Lambda^{k-1}N\right)  +\partial\left(  \Lambda^{k+1}N\right)  $.

Lemma \ref{lem.5} says that $\omega_{k}\partial=0$ on $\Lambda^{k+1}N$. In
other words, $\partial\left(  \Lambda^{k+1}N\right)  \subseteq
\operatorname*{Ker}\left(  \omega_{k}\right)  $.

Lemma \ref{lem.6} shows that $\varepsilon_{1}\eta\omega_{k}=\operatorname*{id}%
$ on $\varepsilon_{1}\left(  \Lambda^{k-1}N\right)  $ (where $\eta$ is as
defined in that lemma). Hence, the map $\omega_{k}$ is injective when
restricted to $\varepsilon_{1}\left(  \Lambda^{k-1}N\right)  $ (since it has a
left inverse $\varepsilon_{1}\eta$). In other words, $\operatorname*{Ker}%
\left(  \omega_{k}\right)  \cap\varepsilon_{1}\left(  \Lambda^{k-1}N\right)
=0$.

But a general fact (easy exercise) about modules says the following: If
$A,B,C$ are three $\mathbf{k}$-submodules of a $\mathbf{k}$-module $M$
satisfying $M=A+B$ and $B\subseteq C$ and $C\cap A=0$, then $B=C$. Applying
this to $M=\Lambda^{k}N$ and $A=\varepsilon_{1}\left(  \Lambda^{k-1}N\right)
$ and $B=\partial\left(  \Lambda^{k+1}N\right)  $ and $C=\operatorname*{Ker}%
\left(  \omega_{k}\right)  $, we obtain $\partial\left(  \Lambda
^{k+1}N\right)  =\operatorname*{Ker}\left(  \omega_{k}\right)  $ (since
$\Lambda^{k}N=\varepsilon_{1}\left(  \Lambda^{k-1}N\right)  +\partial\left(
\Lambda^{k+1}N\right)  $ and $\partial\left(  \Lambda^{k+1}N\right)
\subseteq\operatorname*{Ker}\left(  \omega_{k}\right)  $ and
$\operatorname*{Ker}\left(  \omega_{k}\right)  \cap\varepsilon_{1}\left(
\Lambda^{k-1}N\right)  =0$). Hence, $\operatorname*{Ker}\left(  \omega
_{k}\right)  =\partial\left(  \Lambda^{k+1}N\right)  $ is proved. Combining
this with $\operatorname*{Ker}\left(  \partial\mid_{\Lambda^{k}N}\right)
=\partial\left(  \Lambda^{k+1}N\right)  $, we conclude that
\eqref{eq.thm.peel.t2.KKI} holds.

By the homomorphism theorem, we have $\operatorname{Im}\left(  \omega
_{k}\right)  \cong\Lambda^{k}N\diagup\operatorname*{Ker}\left(  \omega
_{k}\right)  $ as $S_{n}$-representations (since $\omega_{k}$ is a morphism of
$S_{n}$-representations). Since $\omega_{k}$ is surjective, we have
$\operatorname{Im}\left(  \omega_{k}\right)  =\mathcal{S}^{\lambda^{k}}$ and
thus
\[
\mathcal{S}^{\lambda^{k}}=\operatorname{Im}\left(  \omega_{k}\right)
\cong\Lambda^{k}N\diagup\underbrace{\operatorname*{Ker}\left(  \omega
_{k}\right)  }_{=\partial\left(  \Lambda^{k+1}N\right)  }=\Lambda^{k}%
N\diagup\partial\left(  \Lambda^{k+1}N\right)  =\left(  \operatorname*{Coker}%
\partial\right)  _{k}%
\]
as $S_{n}$-representations. Thus we have proved \eqref{eq.thm.peel.t2.iso},
and Theorem \ref{thm.7} is proven.
\end{proof}

We can now show that Peel's maps $f_{k}$ are well-defined and form an exact
sequence, and in fact form a commutative diagrams with the maps $\left(
-1\right)  ^{k}\varepsilon$ on $\Lambda N$ and the morphisms $\omega_{k}$:

\begin{theorem}
\label{thm.8}Assume that $n=0$ in $\mathbf{k}$. Then, for each $k>0$, the map
\begin{align*}
f_{k}:\mathcal{S}^{\lambda^{k}} &  \rightarrow\mathcal{S}^{\lambda^{k+1}},\\
V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)   &  \mapsto\sum
_{s\in\left[  n\right]  }V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}%
,x_{s}\right)
\end{align*}
is a well-defined morphism of $S_{n}$-representations. Let furthermore
$f_{0}:\mathcal{S}^{\lambda^{0}}\rightarrow\mathcal{S}^{\lambda^{1}}$ be the
zero map $0$. Then, the diagram
\begin{equation}
\xymatrix@C=4pc{
\Lambda^k N \ar[r]^{\tup{-1}^k \varepsilon} \ar[d]_{\omega_k} & \Lambda^{k+1} N \ar[d]^{\omega_{k+1}} \\
\mathcal{S}^{\lambda^k} \ar[r]_{f_k} & \mathcal{S}^{\lambda^{k+1}}
}\label{eq.thm.peel.t3.diag}%
\end{equation}
commutes for each $k\in\mathbb{N}$. Thus, the sequence
\begin{equation}
\cdots\overset{f_{k-2}}{\longrightarrow}\mathcal{S}^{\lambda^{k-1}%
}\overset{f_{k-1}}{\longrightarrow}\mathcal{S}^{\lambda^{k}}\overset{f_{k}%
}{\longrightarrow}\mathcal{S}^{\lambda^{k+1}}\overset{f_{k+1}}{\longrightarrow
}\cdots\label{eq.thm.peel.t3.fseq}%
\end{equation}
is exact and has a $\mathbf{k}$-linear (but not $S_{n}$-equivariant) chain contraction.
\end{theorem}

\begin{proof}
First, we note that the $S_{n}$-action is not relevant to any claims of the
theorem, except for the easy claim that $f_{k}$ is a morphism of $S_{n}%
$-representations. Thus, we can forget about this action now (although this
does not simplify much, just taking some minor cargo off our backs).

Theorem \ref{thm.7} shows that for each $k\in\mathbb{N}$, the map $\omega
_{k}:\Lambda^{k}N\rightarrow\mathcal{S}^{\lambda^{k}}$ is surjective and has
kernel $\operatorname*{Ker}\left(  \omega_{k}\right)  =\partial\left(
\Lambda^{k+1}N\right)  $. Thus, this map gives rise to a canonical
isomorphism
\begin{align*}
\omega_{k}^{\prime}:\Lambda^{k}N\diagup\partial\left(  \Lambda^{k+1}N\right)
&  \overset{\cong}{\rightarrow}\mathcal{S}^{\lambda^{k}},\\
\overline{v} &  \mapsto\omega_{k}\left(  v\right)
\end{align*}
(where $\overline{v}$ denotes the projection of a vector $v$ onto the
quotient). Since we have $\Lambda^{k}N\diagup\partial\left(  \Lambda
^{k+1}N\right)  =\left(  \operatorname*{Coker}\partial\right)  _{k}$, we can
rewrite this as
\begin{align*}
\omega_{k}^{\prime}:\left(  \operatorname*{Coker}\partial\right)  _{k} &
\overset{\cong}{\rightarrow}\mathcal{S}^{\lambda^{k}},\\
\overline{v} &  \mapsto\omega_{k}\left(  v\right)  .
\end{align*}
Using these isomorphisms $\omega_{k}^{\prime}$, we can turn the morphisms
$\varepsilon^{\prime}:\left(  \operatorname*{Coker}\partial\right)
_{k}\rightarrow\left(  \operatorname*{Coker}\partial\right)  _{k+1}$ from
Proposition \ref{prop.3} into morphisms $\varepsilon_{k}^{\prime\prime
}:\mathcal{S}^{\lambda^{k}}\rightarrow\mathcal{S}^{\lambda^{k+1}}$ so that the
diagram
\begin{equation}
\xymatrix@C=4pc{
\tup{\Coker \partial}_k \ar[r]^{\varepsilon'} \ar[d]_{\omega'_k}^{\cong} & \tup{\Coker \partial}_{k+1} \ar[d]^{\omega'_{k+1}}_{\cong} \\
\mathcal{S}^{\lambda^k} \ar[r]_{\varepsilon''_k} & \mathcal{S}^{\lambda^{k+1}}
}\label{pf.thm.peel.t3.diag1}%
\end{equation}
commutes (explicitly, we set $\varepsilon_{k}^{\prime\prime}:=\omega
_{k+1}^{\prime}\varepsilon^{\prime}\left(  \omega_{k}^{\prime}\right)  ^{-1}$
on $\mathcal{S}^{\lambda^{k}}$). Consider these morphisms $\varepsilon
_{k}^{\prime\prime}$. Let us furthermore define a new morphism
\[
g_{k}:=\left(  -1\right)  ^{k}\varepsilon_{k}^{\prime\prime}:\mathcal{S}%
^{\lambda^{k}}\rightarrow\mathcal{S}^{\lambda^{k+1}}\qquad\text{for each }%
k\in\mathbb{N}.
\]
Thus, $g_{k}$ differs from $\varepsilon_{k}^{\prime\prime}$ only in the sign
factor $\left(  -1\right)  ^{k}$. Hence, the commutative diagram
\eqref{pf.thm.peel.t3.diag1} yields a commutative diagram
\begin{equation}
\xymatrix@C=4pc{
\tup{\Coker \partial}_k \ar[r]^{\tup{-1}^k \varepsilon'} \ar[d]_{\omega'_k}^{\cong} & \tup{\Coker \partial}_{k+1} \ar[d]^{\omega'_{k+1}}_{\cong} \\
\mathcal{S}^{\lambda^k} \ar[r]_{g_k} & \mathcal{S}^{\lambda^{k+1}}
}\label{pf.thm.peel.t3.diag2}%
\end{equation}
(obtained from it by scaling both horizontal arrows by $\left(  -1\right)
^{k}$).

We claim that the new maps $g_{k}$ are precisely the maps $f_{k}$ defined in
the theorem (and, in particular, the latter maps $f_{k}$ are well-defined).
Indeed, this is obvious for $k=0$, so let us take $k>0$. Let $\left(
i_{1},i_{2},\ldots,i_{k}\right)  \in\left[  n\right]  ^{k}$. Then,
\begin{align*}
&  V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right) \\
&  =\omega_{k}\left(  e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}%
}\right)  \qquad\left(  \text{by the definition of }\omega_{k}\right) \\
&  =\omega_{k}^{\prime}\left(  \overline{e_{i_{1}}\wedge e_{i_{2}}\wedge
\cdots\wedge e_{i_{k}}}\right)  \qquad\left(  \text{by the definition of
}\omega_{k}^{\prime}\right)  ;
\end{align*}
thus,
\begin{align*}
&  g_{k}\left(  V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  \right)
\\
&  =g_{k}\left(  \omega_{k}^{\prime}\left(  \overline{e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}}\right)  \right) \\
&  =\omega_{k+1}^{\prime}\left(  \left(  -1\right)  ^{k}\varepsilon^{\prime
}\left(  \overline{e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}%
}\right)  \right)
\end{align*}
(since the diagram \eqref{pf.thm.peel.t3.diag2} commutes). Since
\begin{align*}
&  \left(  -1\right)  ^{k}\varepsilon^{\prime}\left(  \overline{e_{i_{1}%
}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}}\right) \\
&  =\left(  -1\right)  ^{k}\overline{\varepsilon\left(  e_{i_{1}}\wedge
e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\right)  }\qquad\left(  \text{by the
definition of }\varepsilon^{\prime}\right) \\
&  =\left(  -1\right)  ^{k}\overline{e\wedge e_{i_{1}}\wedge e_{i_{2}}%
\wedge\cdots\wedge e_{i_{k}}}\qquad\left(  \text{by the definition of
}\varepsilon\right) \\
&  =\left(  -1\right)  ^{k}\overline{\left(  \sum_{s\in\left[  n\right]
}e_{s}\right)  \wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}%
}\qquad\left(  \text{since }e=\sum_{s\in\left[  n\right]  }e_{s}\right) \\
&  =\overline{\sum_{s\in\left[  n\right]  }\underbrace{\left(  -1\right)
^{k}e_{s}\wedge e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}%
}_{=e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\wedge e_{s}}}\\
&  =\overline{\sum_{s\in\left[  n\right]  }e_{i_{1}}\wedge e_{i_{2}}%
\wedge\cdots\wedge e_{i_{k}}\wedge e_{s}},
\end{align*}
this rewrites as
\begin{align*}
&  g_{k}\left(  V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  \right)
\\
&  =\omega_{k+1}^{\prime}\left(  \overline{\sum_{s\in\left[  n\right]
}e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{k}}\wedge e_{s}}\right) \\
&  =\omega_{k+1}\left(  \sum_{s\in\left[  n\right]  }e_{i_{1}}\wedge e_{i_{2}%
}\wedge\cdots\wedge e_{i_{k}}\wedge e_{s}\right)  \qquad\left(  \text{by the
definition of }\omega_{k+1}^{\prime}\right) \\
&  =\sum_{s\in\left[  n\right]  }V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}%
},x_{s}\right)  \qquad\left(  \text{by the definition of }\omega_{k+1}\right)
.
\end{align*}
So we have shown that $g_{k}:\mathcal{S}^{\lambda^{k}}\rightarrow
\mathcal{S}^{\lambda^{k+1}}$ is a $\mathbf{k}$-linear map that sends each
$V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}}\right)  $ to $\sum_{s\in\left[
n\right]  }V\left(  x_{i_{1}},x_{i_{2}},\ldots,x_{i_{k}},x_{s}\right)  $. But
this is exactly what the map $f_{k}$ is supposed to do. Thus, it follows that
the map $f_{k}$ is well-defined (it is unique since the $V\left(  x_{i_{1}%
},x_{i_{2}},\ldots,x_{i_{k}}\right)  $ span $\mathcal{S}^{\lambda^{k}}$), and
that $g_{k}=f_{k}$.

We have thus proved that for each $k\in\mathbb{N}$, the map $f_{k}$ is
well-defined and satisfies $g_{k}=f_{k}$. As we said, it is easy to show that
$f_{k}$ is a morphism of $S_{n}$-representations.

Now, the commutative diagram \eqref{pf.thm.peel.t3.diag2} rewrites as
\begin{equation}
\xymatrix@C=4pc{
\tup{\Coker \partial}_k \ar[r]^{\tup{-1}^k \varepsilon'} \ar[d]_{\omega'_k}^{\cong} & \tup{\Coker \partial}_{k+1} \ar[d]^{\omega'_{k+1}}_{\cong} \\
\mathcal{S}^{\lambda^k} \ar[r]_{f_k} & \mathcal{S}^{\lambda^{k+1}}
}\label{pf.thm.peel.t3.diag3}%
\end{equation}
(since $g_{k}=f_{k}$). Now, we consider the diagram
\begin{equation}
\xymatrix@C=4pc{
\Lambda^k N \ar[r]^{\tup{-1}^k \varepsilon} \arsurj[d] & \Lambda^{k+1} N \arsurj[d] \\
\tup{\Coker \partial}_k \ar[r]^{\tup{-1}^k \varepsilon'} \ar[d]_{\omega'_k}^{\cong} & \tup{\Coker \partial}_{k+1} \ar[d]^{\omega'_{k+1}}_{\cong} \\
\mathcal{S}^{\lambda^k} \ar[r]_{f_k} & \mathcal{S}^{\lambda^{k+1}}
}\label{pf.thm.peel.t3.diag4}%
\end{equation}
(where the two topmost vertical arrows are canonical projections $\Lambda
N\rightarrow\operatorname*{Coker}\partial$). This diagram is commutative,
because the top square commutes (indeed, it is just the diagram
\eqref{eq.prop.peel.p2.diagram} with both horizontal arrows scaled by $\left(
-1\right)  ^{k}$) and the bottom square commutes (this square is just the
diagram \eqref{pf.thm.peel.t3.diag3}). Composing the vertical arrows in this
diagram (and recalling that $\omega_{k}^{\prime}\left(  \overline{v}\right)
=\omega_{k}\left(  v\right)  $ for each $v\in\Lambda^{k}N$), we obtain the
commutative diagram
\[
\xymatrix@C=4pc{
\Lambda^k N \ar[r]^{\tup{-1}^k \varepsilon} \ar[d]_{\omega_k} & \Lambda^{k+1} N \ar[d]^{\omega_{k+1}} \\
\mathcal{S}^{\lambda^k} \ar[r]_{f_k} & \mathcal{S}^{\lambda^{k+1}}
}\ \ .
\]
Thus we have shown that the diagram \eqref{eq.thm.peel.t3.diag} commutes.

Recall from Theorem \ref{thm.4} that the sequence
\[
\cdots\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k-1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\left(  \operatorname*{Coker}\partial\right)
_{k}\overset{\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k+1}\overset{\varepsilon^{\prime
}}{\longrightarrow}\cdots
\]
is exact and has a $\mathbf{k}$-linear chain contraction. Hence, the same is
true of the sequence
\[
\cdots\overset{\left(  -1\right)  ^{k-2}\varepsilon^{\prime}}{\longrightarrow
}\left(  \operatorname*{Coker}\partial\right)  _{k-1}\overset{\left(
-1\right)  ^{k-1}\varepsilon^{\prime}}{\longrightarrow}\left(
\operatorname*{Coker}\partial\right)  _{k}\overset{\left(  -1\right)
^{k}\varepsilon^{\prime}}{\longrightarrow}\left(  \operatorname*{Coker}%
\partial\right)  _{k+1}\overset{\left(  -1\right)  ^{k+1}\varepsilon^{\prime
}}{\longrightarrow}\cdots
\]
(because the exactness of a sequence is preserved when we scale every other
arrow by $-1$, and the same holds for the existence of a chain contraction;
indeed, this is just the shift functor $A\mapsto A\left[  1\right]  $ on chain
complexes). Thus, the same is true of the sequence
\[
\cdots\overset{f_{k-2}}{\longrightarrow}\mathcal{S}^{\lambda^{k-1}%
}\overset{f_{k-1}}{\longrightarrow}\mathcal{S}^{\lambda^{k}}\overset{f_{k}%
}{\longrightarrow}\mathcal{S}^{\lambda^{k+1}}\overset{f_{k+1}}{\longrightarrow
}\cdots,
\]
because the commutative diagram \eqref{pf.thm.peel.t3.diag3} reveals that
these two sequences are isomorphic (with isomorphism given by $\omega
_{k}^{\prime}:\left(  \operatorname*{Coker}\partial\right)  _{k}%
\overset{\cong}{\rightarrow}\mathcal{S}^{\lambda^{k}}$). This finishes the
proof of Theorem \ref{thm.8}.
\end{proof}

\bigskip

\begin{thebibliography}{99999999}                                                                                         %


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\bibitem[Kuenze15]{Kuenze15}%
\href{https://pnp.mathematik.uni-stuttgart.de/lexmath/kuenzer/thesis.pdf}{Matthias
K\"{u}nzer, \textit{Ties for the integral group ring of the symmetric group},
thesis, version 2015.}

\bibitem[Peel71]{Peel71}\href{https://doi.org/10.1017/S0017089500001245}{M. H.
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\bibitem[Winitz23]{Winitz23}Sergei Winitzki, \textit{Linear Algebra via
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\end{thebibliography}


\end{document}