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\title{Lecture 4-17: Specht modules and standard tableaux}
\date{April 17, 2024}
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Given a partition $\lambda$ of $n$, last time we defined the Specht module of $S_n$ corresponding to $\lambda$ to be the span of the $v_T=b_T\cdot\{T\}$ as $T$ runs over all tableaux with shape $\lambda$, where $b_T=\sum\limits_{q\in C(T)}\epsilon_qq$, where $\epsilon_q$ denotes the sign of $q$ and $\{T\}$ the tabloid of $T$.  We showed that $b_Tv_T\ne0$ and that $b_T\{T'\}\in\mathbb Qv_T$ for any tabloid $\{T'\}$ if $T'$ has shape $\lambda$.  Hence $b_TM^\lambda=b_TS^\lambda=\mathbb Qv_T\ne0$, while $b_TM^{\lambda'}=0$ if $\lambda'>\lambda$.  
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\begin{alertblock}{Theorem}
The Specht modules $S^\lambda$ are irreducible, as are their complexifications $S^\lambda_{\mathbb C}$ (obtained by replacing the basefield $\mathbb Q$ by $\mathbb C$). Every irreducible complex representation of $S_n$ is isomorphic to $S^\lambda_{\mathbb C}$ for a unique partition $\lambda$ of $n$.
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\begin{proof}
Irreducibility over either $\mathbb C$ or $\mathbb Q$ is equivalent to indecomposability. If we had $S^\lambda=V\oplus W$, then $\mathbb Qv_T=b_TS^{\lambda} = b_T\cdot V\oplus b_T\cdot W$, forcing one of $V$ or $W$, say $V$, to contain $v_T$, whence $V=\mathbb QS_nv_T = S^\lambda$ and $S^\lambda$ is irreducible; similarly so is $S^\lambda_{\mathbb C}$.  Since $<$ is a total order on tableaux, no two $S^\lambda$ are equivalent, by the above formulas for $b_T\cdot M^\lambda$.  Since the number of $S^\lambda$ matches the number of conjugacy classes in $S_n$, we have found all of the irreducible complex representations of $S_n$.
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From this it can be shown that the rational group algebra $\mathbb QS_n$ is isomorphic to a sum of matrix rings $M_{n_i}(\mathbb Q)$, in the same way that $\mathbb CS_n$ is a sum of matrix rings $M_{n_i}(\mathbb C)$.
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 Now we want to work out the degree of $S^\lambda$.  We first show that this degree is at least the number $f^\lambda$ of standard tableaux of shape $\lambda$.
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 For fixed $\lambda$ the elements $v_T$ are linearly independent as $T$ runs through the standard tableaux of shape $\lambda$.
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 \begin{proof}
 We mention that the argument in Chapter 7 of Fulton is inadequate, using only as it does the previously defined total order on tableaux.  Instead one needs a total order $\prec$ on {\sl tabloids} $\{T\},\{T'\}$ of shape $\lambda$, defined by decreeing that $\{T\}\prec \{T'\}$ if the largest number occurring in different rows occurs higher in $\{T\}$ than in $\{T'\}$.  By the remark after Lemma 1 of the last lecture, we have $\{q\cdot T\}\prec\{T\}$ if $T$ is standard and $q\in C(T)$.  Thus for $T$ standard each $v_T$ is a combination of tabloids of which the $\prec$-largest term is $\{T\}$.  The independence of such $v_T$ then follows at once by considering the $\prec$-maximal term occurring with nonzero coefficient in a dependence relation.
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Now we want to show that the $v_T$ for $T$ standard also span $S^\lambda$.  To do this we need a different presentation of $S^\lambda$, realizing it as a quotient rather than a submodule of an $S_n$-module.  We therefore define a \alert{column tabloid} $[T]$ to be an equivalence class of tableaux $\pm T$ with signs attached, where $\pm T$ is identified with $\pm(\epsilon_qT)$ whenever $q\in C(T)$.  There is an obvious action of $S_n$ on column tabloids of shape $\lambda$ and thus on the space $\tilde M^\lambda$ spanned by them; the definitions of $[T]$ and $v_T$ show that the map $\alpha:\tilde M^\lambda\rightarrow S^\lambda$ sending $[T]$ to $v_T$ is a well defined surjective $S_n$-module map.  
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It turns out that there are certain operations $\pi_{j,k}$ on column tabloids which are such that the differences $[T]-\pi_{j,k}[T]$ span the kernel of $\alpha$.  Here the parameter $j$ is a column of $T$ lying to the left of its rightmost column and $k$ is a positive integer at most equal to the length of the $(j+1)$st column of $T$.  Then $\pi_{j,k}[T]$ is the sum of the column tabloids $[S]$ obtained from $[T]$ by interchanging the top $k$ numbers in the $(j+1)$st column with all possible subsets of $k$ numbers in the $j$th column, preserving the vertical order of these elements throughout.  
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For example, 
\[
\pi_{1,2}\begin{pmatrix} 1&2\\4&3\\5&6\end{pmatrix} = \begin{pmatrix}2&1\\3&4\\5&6\end{pmatrix}+\begin{pmatrix} 2&1\\4&5\\3&6\end{pmatrix}+\begin{pmatrix}1&4\\2&5\\3&6\end{pmatrix}
\]
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For the proof that the differences $[T]-\pi_{j,k}[T]$ span the kernel of $\alpha$, see Chapter 7 of Fulton's book; it is not difficult but a bit too much of a digression to include here.  We now introduce one more total ordering, this time on column tabloids (the last one, I promise!)  We decree that $[T]\succ[T']$ if in the rightmost column where $[T],[T']$ differ, the lowest box which has different entries after both columns are rearranged in increasing order is larger in $[T]$ than in $[T']$.  Then we have
\begin{alertblock}{Lemma 3:  straightening law}
If $S$ is a nonstandard tableau of shape $\lambda$ then by repeatedly using the relations $[T]=\pi_{j,k}[T]$ (for various $j,k$) we can write $[S]$ as a combination of $[T_i]$ where the $T_i$ are standard.  
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\begin{proof}
First, we may rearrange the columns of $S$ are in increasing order, possibly changing $[S]$ by a sign.  If $S$ is still not standard, then suppose that the $k$th number in the $j$th column is larger than the $k$th number in the $(j+1)$st column.  Applying $\pi_{j,k}$ to $S$, we find that all column tabloids  appearing are larger than $[S]$ in the ordering, so iteration of this process equates $[S]$ to a combination of the desired form.
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\noindent For example, taking $j=k=1$, we find that 
\[
\begin{pmatrix} 2&1\\3& \end{pmatrix}=\begin{pmatrix} 1&2\\3& \end{pmatrix}+\begin{pmatrix} 2&3\\1& \end{pmatrix}=\begin{pmatrix} 1&2\\3& \end{pmatrix}-\begin{pmatrix} 1&3\\2& \end{pmatrix}
\]
\noindent Hence the $v_T$ for $T$ standard provide a basis of the Specht module $S^\lambda$ and its dimension equals the number $f^\lambda$ of standard tableaux of shape $\lambda$.   Also we see that the Specht module $S^\lambda$ can in fact be defined over the integers, so that every representation $\pi$ of $S_n$ is equivalent to one whose range lies in $GL_n(\mathbb Z)$.  In particular, we recover the result you proved in HW that the character table of $S_n$ consists entirely of integers.
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There is a similar dual construction of a module $\tilde S^\lambda$ isomorphic to $S^\lambda$, obtained by taking the span of the elements $\tilde v_T=a_T\cdot[T]\in\tilde M^\lambda$, where $a_T=\sum\limits_{\sigma\in R(T)}\sigma$.  Thus we have composite maps $S^\lambda\hookrightarrow M^\lambda\rightarrow\tilde S^\lambda$ and $\tilde S^\lambda\hookrightarrow\tilde M^\lambda\rightarrow S^\lambda$.  The composite of these composites sending $S^\lambda$ to itself is multiplication by a scalar, by Schur's Lemma.  It maps $v_T$ to $b_Ta_T\cdot v_T=n_Tv_T$, where $n_T$ equals the cardinality of the set of quadruples $(p_1,q_1,p_2,q_2)$ such that $p_i\in R(T),q_i\in C(T),p_1q_1p_2q_2=1$ and $\epsilon_{q_1}=\epsilon_{q_2}$, minus the cardinality of the set of quadruples $(p_1,q_1,p_2,q_2)$ satisfying the first two conditions but with $\epsilon_{q_1}=-\epsilon_{q_2}$.  (In Fulton's book, the subtracted term is erroneously omitted.)  This is independent of the choice of $T$ (with shape $\lambda$) since replacing $T$ by a different tableau replaces the subgroups $R(T),C(T)$ by conjugates of themselves.  Taking $T$ minimal in the order on tableaux of shape $\lambda$, we see that multiplication by $c_T$ on $\mathbb QS_n$ sends all $c_U$ to 0 for all standard $U\ne T$ (even those not of shape $\lambda$), but acts with trace $n!$ on $\mathbb QS_n$, so finally $n_\lambda=\frac {n!} {\dim S^\lambda}\ne0$. 
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On the other hand, the representations $S^\lambda$ and $\tilde S^\lambda$ are not in general equivalent over the integers, and the Specht modules $S^\lambda$, while still defined over any field $k$, need not be irreducible (or inequivalent) in general.
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