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\title{Lecture 4-15: Representations of the symmetric group}
\date{April 15, 2024}
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We turn now to a particular group whose representations turn out to be particularly nice (and were historically among the first ones studied).  This is the symmetric group $S_n$.  We will follow the development in Chapter 7 of Fulton's wonderful 1997 book entitled \lq\lq Young tableaux"; throughout we will work over the rational field $\mathbb Q$ rather than the complex one.   First recall that the conjugacy class of a permutation $\sigma$ is determined by the lengths of the cycles in its cycle decomposition (including the 1-cycles).  These form a partition $\lambda=(\lambda_1,\ldots,\lambda_m)$ of $n$, so that $\sum\lambda_i=n$; by convention we arrange the parts $\lambda_i$ of $\lambda$ so that $\lambda_1\ge\lambda_2\ge\ldots$.  We also write $|\lambda|=n$.
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Given a partition $\lambda$ of $n$, we define a \alert{(Young) diagram of shape $\lambda$} to be a an arrangement of boxes in rows, lined up on the left, so that the $i$th row of the arrangement has $\lambda_i$ boxes.  Filling in the boxes with the numbers 1 through $n$, using each number exactly once, we get a \alert{(Young) tableau} of this shape, which is called \alert{standard} if the numbers in the boxes increase across rows and down columns.
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Thus for example
\[
\begin{matrix} 1&2&6\\3&5& \\4\end{matrix}
\]
\noindent is a standard tableau of shape $(3,2,1)$.  There is an obvious action of $S_n$ on tableaux of shape $\lambda$, obtained by permuting the numbers in the boxes.  Given such a tableau $T$, denote by $R(T)$ the subgroup of $S_n$ consisting of permutations permuting the elements of each row among themselves.  Then $R(T)$ is a direct product of symmetric groups, one for each part of $\lambda$.  
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Similarly denote by $C(T)$ the subgroup of permutations preserving the columns of $T$.  In the above example $R(T)$ and $C(T)$ are both isomorphic to $S_3\times S_2\times S_1$.  Note that $R(T)\cap C(T)=1$, since a permutation in the intersection cannot move any number from its row or column in $T$.  Given two partitions $\lambda=(\lambda_1,\ldots,\lambda_m)$ and $\lambda'=(\lambda_1',\ldots,\lambda_r')$ of the same integer $n$, we say that $\lambda$ {\sl dominates} $\lambda'$ if for all $i$ we have $\sum\limits_{j=1}^i \lambda_j\ge\sum\limits_{j=1}^i \lambda_j'$, defining $\lambda_j=0$ if $j>m$ and $\lambda_k'=0$ if $k>r$.  This is a partial order on partitions of $n$.  The following lemma provides the basic tool we need.
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\begin{alertblock}{Lemma 1}
Let $T,T'$ be tableaux of shapes $\lambda,\lambda'$ with $|\lambda|=|\lambda'|=n$.  Assume that $\lambda$ does not strictly dominate $\lambda'$.  Then exactly one of the following holds.
\begin{itemize}
\item There are two distinct integers in the same row of $T'$ and the same column of $T$. 
\item $\lambda'=\lambda$ and there are $p'\in R(T'),q\in C(T)$ with $p'\cdot T'=q\cdot T$.
\end{itemize}
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\begin{proof}
If the first assertion fails, then the numbers in the first row of $T'$ all occur in different columns of $T$, so there is $q_1\in C(T)$ such that these numbers occur in the first row of $q_1\cdot T$.  The numbers in the second row of $T'$ then occur in different columns of $T$, so also of $q_1\cdot T$, so there is $q_2\in C(q_1\cdot T) = C(T)$ not moving the numbers equal to those in the first row of $T'$, such that these numbers all occur in the first two rows of $q_2q_1\cdot T$.  Continuing in this way we get $q_1,\ldots,q_k\in C(T)$ such that the numbers in the first $k$ rows of $T'$ all occur in the first $k$ rows of $q_k\cdots q_1\cdot T$.  Since $T$ and $q_k\ldots q_1\cdot T$ have shape $\lambda$, the sum of the first $k$ parts of $\lambda'$ can be at most the corresponding sum for $\lambda$ and $\lambda$ dominates $\lambda'$.
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\begin{proof}
Since we have assumed the $\lambda$ does not strictly dominate $\lambda'$, we must have $\lambda=\lambda'$; taking $k$ to be the number of rows of $\lambda$ and $q=q_k\cdots q_1$, we see that $q\cdot T$ and $T'$ have the same numbers in each row, so there is $p'\in R(T')$ with $p'\cdot T'=q\cdot T$, as desired; conversely, if such $p',q$ exist, then the first assertion must fail.
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We now define two total orders, one on partitions and the other on tableaux.  Given two distinct partitions $\lambda=(\lambda_1,\ldots,\lambda_m)$ and $\lambda'=(\lambda_1',\ldots,\lambda_r')$ we say that $\lambda>\lambda'$ (in the \alert{lexicographic order}) if we have $\lambda_i>\lambda_i'$, where $i$ is the smallest index for which $\lambda_i\ne\lambda_i'$.  Given tableaux $T,T'$ of respective shapes $\lambda,\lambda'$ we write $T>T'$ if either $\lambda>\lambda'$ in the lexicographic order, or $\lambda=\lambda'$ and the largest number occurring in a different position in $T$ and $T'$ occurs either in a column further to the left in $T$ or in the same column but lower down.  Then \alert{for $T$ standard, if $p\in R(T),q\in C(T)$, then $p\cdot T\ge T,q\cdot T\le T$}; indeed, the largest number in $T$ moved by $p$ is must be moved to the left, while the largest number moved by $q$ must be moved up.
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It follows that \alert{if $T,T'$ are standard tableaux with $T'>T$ then there is a pair of numbers in the same row of $T'$ and the same column of $T$}.  For then we must be in the second case of Lemma 1, so that $p'\cdot T'=q\cdot T$ for some $p',q$; but this forces $q\cdot T\le T,p'\cdot T'\ge T'$ by the above observation, a contradiction.
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We now define a \alert{tabloid} $\{T\}$ to be an equivalence class of tableaux, two tableaux being equivalent if they have the same shape and the same numbers in each row.  Thus the tableaux represented by 
\[
\begin{matrix} 1&4&7\\ 3&6& \\2&5& \end{matrix}
\] 
\noindent and
\[
\begin{matrix} 4&7&1\\6&3& \\2&5& \end{matrix}
\]
\noindent are the same.  Clearly $\{T\} = \{T'\}$ if and only if $T'=p\cdot T$ for some $p\in R(T)$.  
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$S_n$ acts on tabloids by the recipe $\sigma\cdot\{T\}=\{\sigma\cdot T\}$; thus the space $M^\lambda$ spanned by all tabloids of shape $\lambda$ is an $S_n$-module.  For a tableau $T$, define $v_T=\sum\limits_{\sigma\in C(T)}\epsilon_\sigma\sigma\{T\} = b_T\{T\}$, where $b_T=\sum\limits_{\sigma\in C(T)}\epsilon_\sigma\sigma\in\mathbb QS_n$, the rational group algebra of $S_n$, where $\epsilon_\sigma$ is the sign of $\sigma$ (1 if $\sigma$ is an even permutation, $-1$ otherwise).  Clearly $v_T\ne0$, since $R(T)\cap C(T)=1$, whence $b_Tv_T = b_T^2\{T\} = \#C(T)v_T\ne0$, where $\#C(T)$ denotes the cardinality of $C(T)$.  We have $\sigma\cdot v_T=v_{\sigma\cdot T}$ for $\sigma\in S_n$ and all tableaux $T$.  Now finally we define the \alert{Specht module} $S^\lambda$ to be the $\mathbb QS_n$-module spanned by the $v_T$ as $T$ runs through tableaux of shape $\lambda$.
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Irreducibility of the $S^\lambda$ will follow from the following lemma.
\begin{alertblock}{Lemma 2}
Let $T,T'$ be tableaux of respective shapes $\lambda,\lambda'$ and assume that $\lambda$ does not dominate $\lambda'$.  If there is a pair of integers in the same row of $T'$ and column of $T$, then $b_T\cdot\{T'\}=0$.  Otherwise we have $b_T\cdot\{T'\} = \pm v_T$.
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\begin{proof}
If there is such a pair of integers, let $t$ be the transposition that swaps them.  Then $b_Tt = -b_T$, since $t\in C(T)$, but $t\cdot \{T'\} = \{T'\}$, since $t\in R(T')$.  It follows that $b_T\cdot\{T'\} = -b_T\cdot \{T'\} = 0$.  If there is no such pair, choose $p'$ and $q$ as in the second case of Lemma 1.  Then $b_T\cdot\{T'\} = b_T\cdot\{p'\cdot T'\} = b_T\cdot\{q\cdot T'\} = b_T\cdot q\cdot \{T\} = \epsilon_qb_T\cdot\{T\} = \epsilon_q\cdot v_T$.
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By the remark right after Lemma 1, we deduce that \alert{if $T,T'$ are standard tableaux with $T'>T$ then $b_T\cdot\{T'\}=0$}.
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