%%% =================================================
%%% @TeX-file{
%%% title = "Classification of the
%%% Irreducible Modules of the
%%% Monomial Matrix Group G(2,1,n)",
%%% version = "1.0",
%%% author = "Ben Galin",
%%% date = "2007/10/19",
%%% filename = "representationMonomialMatrix.tex",
%%% copyright = "Copyright 2007 Ben Galin.
%%% This work is licensed under the
%%% Creative Commons Attribution 2.5
%%% License. To view a copy of this
%%% license, visit
%%% http://creativecommons.org/
%%% licenses/by/2.5/
%%% or send a letter to
%%% Creative Commons,
%%% 543 Howard Street, 5th Floor,
%%% San Francisco,
%%% California, 94105, USA."
%%% }
%%% =================================================
\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry} % Page layout
\usepackage{amsmath} % Math notation and enviroments
\usepackage{amssymb} % Math symbols
\usepackage{amsthm} % Math theorems and lemmas
\usepackage{setspace} % Line spacing
\usepackage{url} % URL in copyright notes
\newcommand{\C}{\mathbb{C}}
\newcommand{\G}{G(2,1,n)}
\DeclareMathOperator{\Ima}{Im}
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\hspan}{-span}
\DeclareMathOperator{\GL}{GL}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{conjecture}[theorem]{Conjecture}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\author{Ben Galin\thanks{Graded Work. This work is licensed under the Creative Commons Attribution 2.5 License. To view a copy of this license, visit \protect\url{http://creativecommons.org/licenses/by/2.5}. Source code with limited rights can be found at \protect\url{http://www.bens.ws/professional.php}.}}
\title{Classification of the Irreducible Modules of the Monomial Matrix Group $\G$}
\bibliographystyle{amsplain}
\setcounter{secnumdepth}{3}
\begin{document}
\onehalfspacing
\maketitle
Let $\G \cong C_2 \wr S_n$, where $n \geq 2$, be the monomial matrix group of $n \times n$ matrices with exactly one nonzero entry in every column and row, and the nonzero entries are in $C_2$, the cyclic group of order two. In this paper, we classify the irreducible representations of $\G$.
Recall that $\G$ is generated by $\{e_1, s_1, \ldots, s_{n-1}\}$ with the following relations:
\begin{align}\label{eq:relations}
&e_1s_1e_1s_1 = s_1e_1s_1e_1, \qquad s_is_j = s_js_i \text{ for $|i-j| \geq 1$}, \qquad s_is_{i+1}s_i = s_{i+1}s_is_{i+1}, \\
&e_1^2 = s_1^2 = \cdots = s_{n-1}^2 = 1 \nonumber \:.
\end{align}
It is often more convenient, however, to define the elements $e_j = s_{j-1}e_{j-1}s_{j-1}$ for $j = 2, \ldots, n$ and to think of $\G$ as generated by $\{e_1, \ldots, e_n, s_1, \ldots, s_{n-1}\}$ with the relations
\begin{align*}
&e_ie_j = e_je_i, \qquad e_is_j = s_je_{is_j}, \qquad s_is_j = s_js_i \text{ for $|i-j| \geq 1$}, \qquad s_is_{i+1}s_i = s_{i+1}s_is_{i+1}, \\
&e_1^2 = s_1^2 = \cdots = s_{n-1}^2 = 1 \:.
\end{align*}
We shall switch back and forth between the two presentations of $\G$ in the following discussion.
Before we can state the classification theorem, we must extend the definitions of a \emph{tableau of shape $\lambda$} and a \emph{standard tableau of shape $\lambda$} to an ordered pair $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ of partitions with $n$ total boxes. We also provide a natural extension to the definition of the \emph{tableau content $C(T(i))$} of the box containing $i$ in a standard tableau $T$ of shape $\lambda$ to the case when $\lambda$ is an ordered pair $(\lambda^{(1)}, \lambda^{(2)})$ of partitions with $n$ total boxes.
\begin{definition}
A \emph{tableau of shape $\lambda$}, where $\lambda$ is an order pair $(\lambda^{(1)}, \lambda^{(2)})$ of partitions with $n$ total boxes, is a filling of the boxes of $\lambda$ by $1, 2, \ldots, |\lambda^{(1)}| + |\lambda^{(2)}|$.
\end{definition}
\begin{definition}\label{def:standard}
Let $T$ be a tableau of shape $\lambda = (\lambda^{(1)}, \lambda^{(2)})$. Then $T$ is said to be a \emph{standard tableau of shape $\lambda$} if the entries in the fillings of $\lambda^{(1)}$ and $\lambda^{(2)}$ each increase in rows and columns.
\end{definition}
\begin{definition}\label{def:content}
Let $T$ be a standard tableau of shape $\lambda = (\lambda^{(1)}, \lambda^{(2)})$. The \emph{tableau content $C(T(i))$} of the box containing $i$ in $T$ is the content $c_{\lambda^{(1)}}\bigl(\, \framebox{$i$}\, \bigr)$ if the box containing $i$ is in the filling of $\lambda^{(1)}$ and $c_{\lambda^{(2)}}\bigl(\, \framebox{$i$}\, \bigr)$ otherwise.
\end{definition}
We can now state the main theorem of this paper.
\begin{theorem}\label{thm:classification}
Let $\G$ be the monomial matrix group of $n \times n$ matrices with exactly one nonzero entry in every column and row, and the nonzero entries are in $C_2$, the cyclic group of order two.
\begin{enumerate}
\item The irreducible $\G$-modules $V^\lambda$ are indexed by ordered pairs $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ of partitions with $n$ total boxes.
\item Their dimensions are $\dim(V^\lambda) = n_\lambda$, where $n_\lambda$ is the number of standard tableaux of shape $\lambda$.
\item If $V^\lambda = \C\hspan\{v_T \mid T$ is a standard tableau of shape $\lambda\}$, then
\begin{align*}
v_Te_i &= (-1)^{m+1} v_T\:, \\
v_Ts_i &= C_T(i)v_T + (1+C_T(i))v_{Ts_i} \:,
\end{align*}
where \emph{(i)} the box containing $i$ is in the partition $\lambda^{(m)}$; \emph{(ii)} $v_{Ts_i} = 0$ if $v_{Ts_i}$ is not a standard tableau; and \emph{(iii)} $C_T(i) = \frac{1}{C(T(i+1)) - C(T(i))}$ if the boxes containing $i$ and $i+1$ are in the same partition $\lambda^{(k)}$ and zero otherwise.
\end{enumerate}
\end{theorem}
Before we proceed with the proof of the theorem, we outline the crucial steps of the proof. We will show that (i) the $V^\lambda$ are indeed $\G$-modules; (ii) any two different such modules are non-isomorphic; (iii) these modules are irreducible; and (iv) this classification exhausts the list of irreducible $\G$-modules, up to isomorphism.
To prove (i), we will check that the relations of $\G$ are preserved under the action of the group on $V^\lambda$. The proof of (ii) will rely on generating elements analogous to the Murphy elements. We will show that their actions on a standard tableau completely determine the filling of the tableau. An immediate result of this would be (ii). For (iii), we will show that $V^\lambda$ does not contain nonzero submodules besides itself by deriving some properties of the standard tableaux. Lastly, the proof of (iv) will involve a counting argument.
The following general observation will assist us through out the rest of the proof: The definition of a standard tableau implies that if neither of the boxes immediately to the right and immediately below a box containing $i$ contain $i+1$, then they contain values greater than $i+1$. Moreover, the box containing $i+1$ is either in a different partitions $\lambda^{(k)}$ or it is strictly below and strictly to the right of the box containing $i$. In either case, the boxes above and to the left of the box containing $i+1$ contain values less than $i$. In such a case, the tableau resulting from permuting the boxes containing $i$ and $i+1$ is a standard tableau.
We summarize this result with the following remark:
\begin{remark}\label{rem:standard}
Let $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ be an order pair of partitions with $n$ total boxes, and let $T$ be a standard partition of shape $\lambda$. Then whenever the boxes containing $i$ and $i+1$ in $T$ are not adjacent, $Ts_i$ is also a standard tableau of shape $\lambda$. In particular, if the boxes containing $i$ and $i+1$ are in different partitions $\lambda^{(1)}$ and $\lambda^{(2)}$, then $Ts_i$ is a standard tableau of shape $\lambda$.
\end{remark}
Proving Theorem~\ref{thm:classification} will be significantly easier to follow if we establish a few lemmas to help us later. The first lemma proves item (i) of the outline we presented above: the $V^\lambda$ are $\G$-modules. The proof is rather long and tedious, though not particularly involved. In the interest of a smooth reading flow, the reader may wish the assume the lemma for the time being and return to its proof later.
\begin{lemma}\label{lem:modules}
Let $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ be an order pair of partitions with $n$ total boxes. Then $V^\lambda = \C\hspan\{v_T \mid T$ is a standard tableau of shape $\lambda\}$ is a $\G$-module with respect to the actions defined in Theorem~\ref{thm:classification}(3).
\end{lemma}
\begin{proof}
We need to check that the relations defined in~\eqref{eq:relations} are satisfied by the speficied actions. In the rest of the proof, let $v_T \in V^\lambda$ for a standard tableau $T$.
\subsubsection*{1. The relation $e_1^2 = 1$}
We start by showing that $v_Te_1^2 = v_T$. Suppose first that the box containing 1 is in $\lambda^{(1)}$. Then $v_Te_1^2 = v_Te_1 = v_T$. Otherwise, the box containing 1 is in $\lambda^{(2)}$. Then it follows that $v_Te_1^2 = -v_Te_i = v_T$, as needed.
\subsubsection*{2. The relation $s_i^2 = 1$}
Next, we show that $v_Ts_i^2 = v_T$ for $1 \leq i \leq n-1$. We consider the following cases regarding the relative positions of the boxes containing $i$ and $i+1$ in $T$:
\begin{enumerate}
\item[(a)] the two boxes are adjacent;
\item[(b)] the two boxes are not adjacent.
\end{enumerate}
In case (a), we have $C_T(i) = \pm 1$ and $Ts_i$ is not standard. Thus, $v_Ts_i^2 = \pm v_Ts_i = v_T$. In case (b), we recall from Remark~\ref{rem:standard} that $Ts_i$ is standard. We also clearly have $C_{Ts_i}(i) = -C_T(i)$. Hence,
\begin{align*}
v_Ts_i^2 &= C_T(i)v_Ts_i + \bigl[1 + C_T(i)\bigr]v_{Ts_i}s_i \\
&= C_T(i)^2v_T + C_T(i)\bigl[1 + C_T(i)\bigr]v_{Ts_i} - C_T(i)\bigl[1 + C_T(i)\bigr]v_{Ts_i} + \bigl[1 - C_T(i)\bigr]\bigl[1 + C_T(i)\bigr]v_T = v_T \:.
\end{align*}
\subsubsection*{3. The relation $s_is_j = s_js_i$}
Next, we show that $v_Ts_is_j = v_Ts_js_i$ for $1 \leq i,j \leq n-1$ and $|i-j| > 1$. We observe that $C_T(i) = C_{Ts_j}(i)$ and $C_T(j) = C_{Ts_i}(j)$. Thus,
\begin{align*}
v_Ts_is_j &= C_T(i)v_Ts_j + \bigl[1 + C_T(i)\bigr]v_{Ts_i}s_j \\
&= C_T(i)C_T(j)v_T + C_T(i)\bigl[1 + C_T(j)\bigr]v_{Ts_j} + \bigl[1 + C_T(i)\bigr]C_T(j)v_{Ts_i} + \bigl[1 + C_T(i)\bigr]\bigl[1 + C_T(j)\bigr]v_{Ts_is_j} \\
&= C_T(j)C_T(i)v_T + C_T(j)\bigl[1 + C_T(i)\bigr]v_{Ts_i} + \bigl[1 + C_T(j)\bigr]C_T(i)v_{Ts_j} + \bigl[1 + C_T(j)\bigr]\bigl[1 + C_T(i)\bigr]v_{Ts_js_i} \\
&= C_T(j)v_Ts_i + \bigl[1 + C_T(j)]v_{Ts_j}s_i = v_Ts_js_i \:,
\end{align*}
where we recall that for any $\sigma \in S_n$, we define $v_{T\sigma} = 0$ if $v_{T\sigma}$ is not a standard tableau.
\subsubsection*{4. The relation $s_1 e_1 s_1 e_1 = e_1 s_1 e_1 s_1$}
The next relation we establish is $v_Ts_1e_1s_1e_1 = v_Te_1s_1e_1s_1$. The box containing 1 must be at the top left corner of one of the partitions $\lambda^{(k)}$. The box containing 2 can only be underneath the box containing 1, to the right of the box containing 1, or at the top left corner of the other partition $\lambda^{(l)}$. In each of these case we see that $v_Ts_1 = \pm v_T$. Thus, for simplicity we write $v_Ts_1 = (-1)^av_T$ for some integer $a$. Also, depending on whether the box containing 1 is in the partition $\lambda^{(1)}$ or $\lambda^{(2)}$, we have $v_Te_1 = \pm v_T$. Again for simplicity, we write $v_Te_i = (-1)^b v_T$ for some integer $b$. Then
\begin{align*}
v_Ts_1e_1s_1e_1 &= (-1)^a v_T e_1s_1e_1 = (-1)^a(-1)^b v_T s_1e_1 = (-1)^b v_Te_1 = v_T \\
&= (-1)^a v_T s_1 = (-1)^b(-1)^a v_T e_1s_1 = (-1)^b v_T s_1e_1s_1 = v_T e_1s_1e_1s_1 \:,
\end{align*}
as needed.
\subsubsection*{5. The relation $s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}$}
The last relation from the list in~\eqref{eq:relations} we wish to establish is $v_Ts_is_{i+1}s_i = v_Ts_{i+1}s_is_{i+1}$, for $1 \leq i \leq n-2$. Here we have the following cases regarding the relative positions of boxes containing $i$ and $i+1$ and $i+2$:
\begin{enumerate}
\item[(a)] the three boxes are in one column or in one row;
\item[(b)] only two of the boxes are adjacent;
\item[(c)] none of the boxes is adjacent to another.
\end{enumerate}
In case (a), note that $v_Ts_i$ and $v_Ts_{i+1}$ are not standard. Hence, $C_T(i) = C_T(i+1) = \pm 1$, depending on whether the boxes are lined up in a column or in a row. Thus, for simplicity, we write $C_T(i) = (-1)^a$ for some integer $a$. Then,
\begin{align*}
v_T s_i s_{i+1} s_i &= (-1)^a v_Ts_{i+1}s_i = (-1)^{2a}v_Ts_i = (-1)^{3a} = (-1)^{2a}v_Ts_{i+1} = (-1)^a v_Ts_is_{i+1} \\
&= v_T s_{i+1} s_i s_{i+1} \:.
\end{align*}
In case (b), consider first the case when the boxes contained $i$ and $i+1$ are adjacent, and again write $C_T(i) = (-1)^a$ for some integer $a$. Then $U = \C\hspan\{v_T, v_Ts_{i+1}, v_Ts_{i+1}s_i\} \subseteq V^\lambda$ contains all the different scenarios satisfying the condition in case (b). Furthermore, $U$ is invariant under $H = \langle s_i, s_{i+1} \rangle$, so it suffices to check the actions of $H$ on $U$. Direct computation shows that we have the representation $\rho: H \to \GL_3(\C)$ given by
\begin{align*}
s_i &\mapsto
\begin{pmatrix}
(-1)^a & 0 & 0 \\
0 & C_{Ts_{i+1}}(i) & 1 + C_{Ts_{i+1}}(i) \\
0 & 1 - C_{Ts_{i+1}}(i) & - C_{Ts_{i+1}}(i)
\end{pmatrix} &
s_{i+1} &\mapsto
\begin{pmatrix}
C_T(i) & 1 + C_T(i) & 0 \\
1 - C_T(i) & - C_T(i) & 0 \\
0 & 0 & (-1)^a
\end{pmatrix} \:.
\end{align*}
It follows that $v_T s_i s_{i+1} s_i = v_T s_{i+1} s_i s_{i+1}$.
We prove case (c) similarly. Let $U = \C\hspan\{v_T, v_{Ts_i}, v_{Ts_{i+1}}, v_{Ts_is_{i+1}}, v_{Ts_is_{i+1}}, v_{Ts_is_{i+1}s_i}\}$ and $H$ as above. Then $U$ is invariant under $H$. The representation we seek is $\rho: H \to \GL_6(\C)$ given by
\begin{align*}
s_i &\mapsto
\begin{pmatrix}
C_T(i) & 1 + C_T(i) & 0 & 0 & 0 & 0 \\
1 - C_T(i) & - C_T(i) & 0 & 0 & 0 & 0 \\
0 & 0 & C_{Ts_{i+1}}(i) & 0 & 1 + C_{Ts_{i+1}}(i) & 0 \\
0 & 0 & 0 & C_{Ts_is_{i+1}}(i) & 0 & 1 + C_{Ts_is_{i+1}}(i) \\
0 & 0 & 1 - C_{Ts_{i+1}}(i) & 0 & - C_{Ts_{i+1}}(i) & 0 \\
0 & 0 & 0 & 1 - C_{Ts_is_{i+1}}(i) & 0 & - C_{Ts_is_{i+1}}(i)
\end{pmatrix} \\
s_{i+1} &\mapsto
\begin{pmatrix}
C_{Ts_is_{i+1}}(i) & 0 & 1 + C_{Ts_is_{i+1}}(i) & 0 & 0 & 0 \\
0 & C_{Ts_{i+1}}(i) & 0 & 1 + C_{Ts_{i+1}}(i) & 0 & 0 \\
1 - C_{Ts_{i+1}}(i) & 0 & - C_{Ts_{i+1}}(i) & 0 & 0 & 0 \\
0 & 1 - C_{Ts_{i+1}}(i) & 0 & - C_{Ts_{i+1}}(i) & 0 & 0 \\
0 & 0 & 0 & 0 & C_T(i) & 1 + C_T(i) \\
0 & 0 & 0 & 0 & 1 - C_T(i) & - C_T(i)
\end{pmatrix} \:.
\end{align*}
And again it follows that $v_T s_i s_{i+1} s_i = v_T s_{i+1} s_i s_{i+1}$ to complete the proof of the lemma.
\end{proof}
The next lemma we prove will assist us in proving item (ii) described in the outline above. We first define a generalization to the Murphy elements, which we will denote by $\mu_1, \ldots, \mu_n$. These elements are members of the group algebra $\C\G$. We will show that their actions on a standard tableau $T$ of shape $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ completely determines the filling of $T$.
\begin{definition}\label{def:murpy}
For $k: 1, \ldots, n$, define the \emph{generalized Murphy element $\mu_k \in \C\G$} to be $\mu_k = ke_k + \frac{1}{2}\sum_{j=1}^{k-1} (1 + e_je_k)(j,k)$, where $(j,k) \in S_n$ is a transposition, and the sum is taken to be zero if $k=1$.
\end{definition}
\begin{lemma}\label{lem:murphy}
Let $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ be an order pair of partitions with $n$ total boxes, and consider the $\G$-module $V^\lambda$ as defined above. Then for any $v_T \in V^\lambda$, where $T$ is a standard tableau of shape $\lambda$, and for any generalized Murphy element $\mu_k$, we have
\begin{align}\label{eq:murphy}
v_T\mu_k &= D_T(k)v_T \:,
\end{align}
where we define $D_T(k)$ as follows:
\begin{align*}
D_T(k) =
\begin{cases}
(C(T(k)) + k) & \text{if the box containing $k$ is in $\lambda^{(1)}$} \\
(C(T(k)) - k) & \text{otherwise}
\end{cases} \:.
\end{align*}
Furthermore, the sequence $D_T(1), \ldots, D_T(n)$ uniquely determines the tableau $T$.
\end{lemma}
\begin{proof}
Note that for $i: 1, \ldots, k-2$ we have $s_{k-1}(1 + e_ie_{k-1})(i, k-1)s_{k-1} = (1 + e_ie_k)(i, k)$ and $s_{k-1}(k-1)e_{k-1}s_{k-1} = (k-1)e_k$. Thus,
\begin{align*}
\mu_k = s_{k-1}\mu_{k-1}s_{k-1} + {\textstyle \frac{1}{2}}(1 + e_{k-1}e_k)s_{k-1} + e_k \:.
\end{align*}
We prove the first part of the lemma by induction. By definition, $\mu_1 = e_1$ so $v_T\mu_1 = v_T$ if the box containing 1 is in $\lambda^{(1)}$ and $v_T\mu_1 = -v_T$ otherwise. Since we must have $C(T(1)) = 0$, the base case is established. In the following, suppose that the lemma is proven for $\mu_{k-1}$.
Suppose that the boxes containing $k$ and $k-1$ are in different partitions $\lambda^{(1)}$ and $\lambda^{(2)}$, not necessarily respectively. Then $v_T\left[\frac{1}{2}(1 + e_{k-1}e_k)\right] = \frac{1}{2}(v_T - v_T) = 0$, and also, by Remark~\ref{rem:standard}, we know that $v_{Ts_{k-1}}$ is standard. If the box containing $k$ is in the partition $\lambda^{(1)}$, then
\begin{align*}
v_T\mu_k &= v_T\left[s_{k-1}\mu_{k-1}s_{k-1} + {\textstyle \frac{1}{2}}(1 + e_{k-1}e_k)s_{k-1} + e_k\right] = v_T\left(s_{k-1}\mu_{k-1}s_{k-1} + e_k\right) \\
&= v_{Ts_{k-1}}\mu_{k-1}s_{k-1} + v_T = (C(T(k)) + k-1)v_{Ts_{k-1}}s_{k-1} + v_T = (C(T(k)) + k)v_T \:.
\end{align*}
Otherwise, if $k$ is in the partition $\lambda^{(2)}$, then
\begin{align*}
v_T\mu_k &= v_T\left[s_{k-1}\mu_{k-1}s_{k-1} + {\textstyle \frac{1}{2}}(1 + e_{k-1}e_k)s_{k-1} + e_k\right] = v_T\left(s_{k-1}\mu_{k-1}s_{k-1} + e_k\right) \\
&= v_{Ts_{k-1}}\mu_{k-1}s_{k-1} - v_T = (C(T(k)) - k+1)v_{Ts_{k-1}}s_{k-1} - v_T = (C(T(k)) - k)v_T \:,
\end{align*}
as needed.
Now suppose that boxes containing $k$ and $k-1$ are in the same partition $\lambda^{(l)}$. Then we have $v_T\left[\frac{1}{2}(1 + e_{k-1}e_k)\right] = \frac{1}{2}(v_T + v_T) = v_T$. Suppose first that $l = 1$. Then,
\begin{align*}
v_T\mu_k &= v_T\left(s_{k-1}\mu_{k-1}s_{k-1} + s_{k-1} + e_k\right) = v_T\left[\left(s_{k-1}\mu_{k-1} + 1\right)s_{k-1} + e_k\right] \\
&= \textstyle \left(\frac{1}{C(T(k)) - C(T(k-1))}v_T\mu_{k-1} + \left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}}\mu_{k-1} + v_T\right)s_{k-1} + v_T \\
&= \textstyle \left(\frac{C(T(k-1)) + k - 1}{C(T(k)) - C(T(k-1))}v_T + \bigl(C(T(k)) + k - 1\bigr)\left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}} + v_T \right)s_{k-1} + v_T \\
&= \textstyle \left(\frac{C(T(k)) + k - 1}{C(T(k)) - C(T(k-1))}v_T + \bigl(C(T(k)) + k - 1\bigr)\left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}} \right)s_{k-1} + v_T \\
&= (C(T(k)) + k - 1)v_Ts_{k-1}s_{k-1} + v_T = (C(T(k)) + k)v_T \:.
\end{align*}
If, on the other hand, $l = 2$, then
\begin{align*}
v_T\mu_k &= v_T\left(s_{k-1}\mu_{k-1}s_{k-1} + s_{k-1} + e_k\right) = v_T\left[\left(s_{k-1}\mu_{k-1} + 1\right)s_{k-1} + e_k\right] \\
&= \textstyle \left(\frac{1}{C(T(k)) - C(T(k-1))}v_T\mu_{k-1} + \left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}}\mu_{k-1} + v_T\right)s_{k-1} - v_T \\
&= \textstyle \left(\frac{C(T(k-1)) - k + 1}{C(T(k)) - C(T(k-1))}v_T + \bigl(C(T(k)) - k + 1\bigr)\left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}} + v_T \right)s_{k-1} - v_T \\
&= \textstyle \left(\frac{C(T(k)) - k + 1}{C(T(k)) - C(T(k-1))}v_T + \bigl(C(T(k)) - k + 1\bigr)\left(1 + \frac{1}{C(T(k)) - C(T(k-1))} \right) v_{Ts_{k-1}} \right)s_{k-1} - v_T \\
&= (C(T(k)) - k + 1)v_Ts_{k-1}s_{k-1} - v_T = (C(T(k)) - k)v_T \:,
\end{align*}
to complete the proof by induction of the first part of the lemma.
Consider any $1 \leq k \leq n$. The minimal value $C(T(k))$ can attain is $-k+1$, which is possible only if all the numbers $1, \ldots, k$ are placed in boxes of the same column in either $\lambda^{(1)}$ or $\lambda^{(2)}$. Similarly, the maximal value $C(T(k))$ can attain is $k-1$, which is possible only if all the numbers $1, \ldots, k$ are placed in boxes of the same row in either $\lambda^{(1)}$ or $\lambda^{(2)}$. Thus, $D_T(k)$ is bounded between $1$ and $2k-1$ if $k$ is placed in a box in $\lambda^{(1)}$ and is bounded between $-2k+1$ and $-1$ otherwise. That is, knowing the value of $D_T(k)$ determines in which partition $k$ is placed. However, since the tableau $T$ is standard, in any given partition $\lambda^{(l)}$ we know that no two different possible placing of the box containing $k$ can have the same content $C(T(k))$. The second part of the lemma follows.
\end{proof}
While the previous lemma showed that the basis elements $v_T$ are eigenvectors of the generalized Murphy elements $\mu_k$, the next lemma shows that, up to scaling, they are the only such eigenvectors. The two lemmas together will be crucial in proving that two modules $V^\lambda$ and $V^\gamma$ are isomorphic only if $\lambda = \gamma$.
\begin{lemma}\label{lem:eigenvector}
Let $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ be an order pair of partitions with $n$ total boxes, and consider the $\G$-module $V^\lambda$ as defined above. If $v \in V^\lambda$ is an eigenvector for all generalized Murphy elements $\mu_1, \ldots, \mu_n$ then $v = cv_T$ for some standard tableau $T$ of shape $\lambda$ and complex number $c$.
\end{lemma}
\begin{proof}
Let $v = \sum_T a_Tv_T$, where we sum over the standard tableaux of shape $\lambda$ and the coefficients are from the complex field. By hypothesis, we have $v\mu_k = \alpha_kv$ for a complex number $\alpha_k$. On the other hand, from equation~\eqref{eq:murphy} we know that $v\mu_k = \sum_T a_T D(T(k)) v_T$. Thus, all but one $a_T$ must vanish, and $c = \alpha_k = D(T(k))$.
\end{proof}
The last lemma we provide shows that standard tableaux form a sequence of tableaux differing by a single transposition. We will later construct a submodule of $V^\lambda$ and show that it contains one of the basis elements $v_T$. The following lemma will assist us in showing that, in fact, all the basis elements are contained in the submodule, proving that the submodule is the entire module $V^\lambda$. We precede the lemma with an extension to the definition of a \emph{column reading tableau} to the case of an ordered pair of partitions.
\begin{definition}
Let $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ be a partition of $n$ total boxes. Then the \emph{column reading tableau of shape $\lambda$} is a tableau $C$ of shape $\lambda$ satisfying the following conditions: (a) its $\lambda^{(1)}$-filling is the same as the column reading tableau of shape $\lambda^{(1)}$; and (b) its $\lambda^{(2)}$-filling is the same as the column reading tableau of shape $\lambda^{(2)}$, but with each entry $k$ replaced by $k + |\lambda^{(1)}|$.
\end{definition}
\begin{lemma}\label{lem:tableaux}
If $P$ and $Q$ are standard tableaux of shape $\lambda = (\lambda^{(1)}, \lambda^{(2)})$, then there exists a sequence of standard tableaux $P, P_{s_{i_1}}, P_{s_{i_1},s_{i_2}}, \ldots, P_{s_{i_1},s_{i_2}, \ldots, s_{i_r}} = Q$.
\end{lemma}
\begin{proof}
Consider the tableau $P$. Let $i_1$ be the maximal number such that either (a) the box containing $i$ is strictly above and strictly to the right of the box containing $i_1 + 1$; or (b) the box containing $i$ is in $\lambda^{(2)}$, while the box containing $i_1 + 1$ is in $\lambda^{(1)}$. If no such $i_1$ exists, then $P = C$, the column reading tableau of $\lambda$.
Otherwise, note the $P_{s_{i_1}}$ is a standard tableau as well. Again find a maximal number with respect to $P_{s_{i_1}}$ satisfying the conditions above and denote by $i_2$. Continue in this manner to generate $i_j$. The column reading tableau is characterized by the conditions above, and by Lemma~\ref{lem:murphy} it is the only tableau satisfying these conditions. Furthermore, by choosing the maximal element to satisfy the conditions above, we guarantee that at no point will we arrive at a standard tableau we have seen already. Thus, the algorithm must terminate for some $P_{s_{i_1},s_{i_2}, \ldots, s_{i_s}} = C$.
Apply the same algorithm to $Q$ to find $Q_{s_{j_1},s_{j_2}, \ldots, s_{j_t}} = C$. Then we have found a sequence $P, P_{s_{i_1}}, \ldots, P_{s_{i_1},s_{i_2}, \ldots, s_{i_r}} = C = Q_{s_{j_1},s_{j_2}, \ldots, s_{j_t}}, \ldots, Q_{s_{j_1}}, Q$ of standard tableaux, as needed.
\end{proof}
We are now ready to prove the classification theorem.
\begin{proof}[Proof (of the classification theorem, Theorem~\ref{thm:classification})]
As mentioned above, Lemma~\ref{lem:modules} proves item (i) of the outline: the $V^\lambda$ are $\G$-modules.
Consider any two partitions $\lambda = (\lambda^{(1)}, \lambda^{(2)})$ and $\gamma = (\gamma^{(1)}, \gamma^{(2)})$ with $n$ total boxes. Then we need to show that $V^\lambda$ and $V^\gamma$ are isomorphic as $\G$-modules if and only if $\gamma = \lambda$; this is item (ii) of the outline. The right-to-left implication is trivial. Conversely, suppose $\theta: V^\lambda \to V^\gamma$ is an isomorphism of $\G$-modules. Consider a standard tableau $T$ of shape $\lambda$. Since $\theta$ is a homomorphism, we have $C(T(k))\bigl(v_T\bigr)\theta = \bigl(C(T(k))v_T\bigr)\theta = (v_T\mu_k)\theta = \bigl((v_T)\theta\bigr)\mu_k$. Hence, by Lemma~\ref{lem:eigenvector} we know that $(v_T)\theta = cv_Q$ for a standard tableau $Q$ and some complex number $c$. By Lemma~\ref{lem:murphy} we know that the sequence $D(T(1)), \ldots, D(T(n))$ and $D(Q(1)), \ldots, D(Q(n))$ uniquely determine $T$ and $Q$, which by the isomorphism $\theta$ we conclude to be the same. This completes the proof of (ii).
We have established that the $V^\lambda$ defined in the statement of the theorem are non-isomorphic $\G$-modules. Next we prove item (iii): that each such $V^\lambda$ is irreducible. Fix some nonzero $v_0 = \sum_T a_T v_T \in V^\lambda$, and let $M = \{v_0 x \mid x \in \C\G\}$, a submodule of $V^\lambda$. We show that $M = V$, which in turn implies that the only nonzero submodule of $V^\lambda$ is itself, or equivalently, that $V^\lambda$ is irreducible.
For each standard tableau $T$, consider the element
\begin{align*}
\pi_T =\!\!\! \prod_{\substack{-2n \leq j \leq 2n \\ j \neq D_T(1)}} \frac{\mu_1 - j}{D_T(1) - j} \prod_{\substack{-2n \leq j \leq 2n \\ j \neq D_T(2)}} \frac{\mu_2 - j}{D_T(2) - j} \ \cdots\!\!\! \prod_{\substack{-2n \leq j \leq 2n \\ j \neq D_T(n)}} \frac{\mu_n - j}{D_T(n) - j} \:.
\end{align*}
If we let $\pi_T$ act on a basis element $v_Q$ for $Q \neq T$, then the product vanishes due to $j$ assuming a value $D(Q(k))$ for a $k$ on which $Q$ and $T$ differ. On the other hand, if we let $\pi_T$ act on $v_T$, then the ratios in the product are all 1, so $\pi_T$ fixes $v_T$. That is, we have proved
\begin{align}\label{eq:piFix}
v_Q\pi_T =
\begin{cases}
0 & \text{if $Q \neq T$} \\
v_Q & \text{otherwise}
\end{cases} \:.
\end{align}
Consider again $v_0 = \sum_T a_T v_T \in V^\lambda$. From equation~\eqref{eq:piFix}, for each nonzero $a_T$ we can write $v_T = v_0 \left(\frac{\pi_T}{a_T}\right) \in M$. By construction, there exists such a nonzero $a_Q$, as otherwise $v_0 = 0$. In particular, we have shown that at least one basis element $v_Q$ is a member of $M$.
Suppose $Qs_i$ is a standard tableau for some $s_i$. If the boxes containing $i$ and $i+1$ are in different partitions $\lambda^{(1)}$ and $\lambda^{(2)}$, then $v_Qs_i = v_{Qs_i} \in M$. Otherwise, the boxes containing $i$ and $i+1$ are in the same partitions $\lambda^{(l)}$. Then $v_Qs_i = \frac{1}{C(Q(i+1)) - C(Q(i))} v_Q + \left(1 + \frac{1}{C(Q(i+1)) - C(Q(i))}\right) v_{Qs_i}$. Note that it is impossible that $C(Q(i+1)) - C(Q(i)) = -1$, as this would imply that the two boxes are adjacent, contradicting the hypothesis that $Qs_i$ is a standard tableau. It follows that $v_{Qs_i} \in M$. By Lemma~\ref{lem:tableaux} and induction on the number of transpositions, we have that $v_T \in M$ for all standard tableaux $T$. But then $M = V^\lambda$, so $V^\lambda$ is irreducible to prove (iii).
To prove the last item (iv), we use an indirect counting argument. Let $G$ be any finite group. In \cite{james}, Theorem~15.3, it is proven that the number of irreducible $G$-modules is equal to the number of conjugacy classes of $G$. In \cite{galin}, it is shown that the conjugacy classes of $\G$ are indexed by ordered pairs of partitions $(\lambda^{(1)}, \lambda^{(2)})$ of $n$ total boxes. Thus, there are no other irreducible $\G$-modules than the ones defined in the theorem. This concludes the proof of the theorem.
\end{proof}
Using Theorem~\ref{thm:classification}, it is possible to derive the character table of different groups $\G$ for an $n > 2$. We analyze the group of $G(2,1,3)$. We choose the following representatives for the conjugacy classes of $G(2,1,3)$:
\begin{align*}
g_1 &= 1 & g_2 &= e_1e_2e_3 & g_3 &= e_1e_2 & g_4 &= e_1 & g_5 &= s_1 \\
g_6 &= e_1s_1 & g_7 &= e_3s_1 & g_8 &= e_1e_3s_1 & g_9 &= s_1s_2 & g_{10} &= e_1s_1s_2 \:.
\end{align*}
The corresponding sizes of the centralizers $|C_G(g_r)|$ are as follows:
\begin{align*}
|C_G(g_1)| &= 48 & |C_G(g_2)| &= 48 & |C_G(g_3)| &= 16 & |C_G(g_4)| &= 16 & |C_G(g_5)| &= 8 \\
|C_G(g_6)| &= 8 & |C_G(g_7)| &= 8 & |C_G(g_8)| &= 8 & |C_G(g_9)| &= 6 & |C_G(g_{10})| &= 6 \:.
\end{align*}
The character table of $G(2,1,3)$ is the following:
\begin{tabular}{c|cccccccccc}
& $g_1$ & $g_2$ & $g_3$ & $g_4$ & $g_5$ & $g_6$ & $g_7$ & $g_8$ & $g_9$ & $g_{10}$ \\ \hline
$|C_G(g_r)|$ & 48 & 48 & 16 & 16 & 8 & 8 & 8 & 8 & 6 & 6 \\ \hline
$\chi_1$ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
$\chi_2$ & 1 & 1 & 1 & 1 & $-1$ & $-1$ & $-1$ & $-1$ & 1 & 1 \\
$\chi_3$ & 1 & $-1$ & 1 & $-1$ & 1 & $-1$ & $-1$ & 1 & 1 & $-1$ \\
$\chi_4$ & 1 & $-1$ & 1 & $-1$ & $-1$ & 1 & 1 & $-1$ & 1 & $-1$ \\
$\chi_5$ & 2 & 2 & 2 & 2 & 0 & 0 & 0 & 0 & $-1$ & $-1$ \\
$\chi_6$ & 2 & $-2$ & 2 & $-2$ & 0 & 0 & 0 & 0 & $-1$ & 1 \\
$\chi_7$ & 3 & $-3$ & $-1$ & 1 & 1 & 1 & $-1$ & $-1$ & 0 & 0 \\
$\chi_8$ & 3 & $-3$ & $-1$ & 1 & $-1$ & $-1$ & 1 & 1 & 0 & 0 \\
$\chi_9$ & 3 & 3 & $-1$ & $-1$ & 1 & $-1$ & 1 & $-1$ & 0 & 0 \\
$\chi_{10}$ & 3 & 3 & $-1$ & $-1$ & $-1$ & 1 & $-1$ & 1 & 0 & 0
\end{tabular} \\
We finish this discussion with a conjecture of the classification theorem of $G(p,1,n) \cong C_p \wr S_n$, where $n,p \geq 2$. The term a \emph{standard tableau of shape $\lambda$} and the notation $C(T(i))$ appearing in this conjecture are the natural generalizations of the terms defined in Definitions~\ref{def:standard} and~\ref{def:content}.
\begin{conjecture}
Let $G(p,1,n)$ be the monomial matrix group of $n \times n$ matrices with exactly one nonzero entry in every column and row, and the nonzero entries are in $C_p$, the cyclic group of order $p$.
\begin{enumerate}
\item The irreducible $\G$-modules $V^\lambda$ are indexed by $p$-tuples $\lambda = (\lambda^{(1)},\ldots, \lambda^{(p)})$ of partitions with $n$ total boxes.
\item Their dimensions are $\dim(V^\lambda) = n_\lambda$, where $n_\lambda$ is the number of standard tableaux of shape $\lambda$.
\item If $V^\lambda = \C\hspan\{v_T \mid T$ is a standard tableau of shape $\lambda\}$, then
\begin{align*}
v_Te_i &= e^\frac{2\pi i r}{p} v_T\:, \\
v_Ts_i &= C_T(i)v_T + (1+C_T(i))v_{Ts_i} \:,
\end{align*}
where \emph{(i)} the box containing $i$ is in the partition $\lambda^{(r)}$ for some $1 \leq r \leq p$; \emph{(ii)} $v_{Ts_i} = 0$ if $v_{Ts_i}$ is not a standard tableau; and \emph{(iii)} $C_T(i) = \frac{1}{C(T(i+1)) - C(T(i))}$ if the boxes containing $i$ and $i+1$ are in the same partition $\lambda^{(k)}$ and zero otherwise.
\end{enumerate}
\end{conjecture}
\bibliography{bib-representationMonomialMatrix}
\end{document}