Examples of Specht Modules

Let’s look at a few examples of Specht modules.

First, let $\lambda=(n)$ . The only polytabloid is

$\displaystyle e_{\begin{array}{cccc}1&2&\cdots&n\end{array}}=\begin{array}{cccc}\cline{1-4}1&2&\cdots&n\\\cline{1-4}\end{array}$

on which $S_n$ acts trivially. And so $S^\lambda$ is a one-dimensional space with the trivial group action. This is the only possibility anyway, since $S^\lambda\subseteq M^\lambda$ , and we’ve seen that $M^\lambda$ is itself a one-dimensional vector space with the trivial action of $S_n$ .

Next, consider $\lambda=(1^n)$ — with $n$ parts each of size $1$ . This time we again have one polytabloid. We fix the Young tableau

$\displaystyle t=\begin{array}{c}1\\2\\\vdots\\n\end{array}$

Since every entry is in the same column, the column-stabilizer $C_t$ is all of $S_n$ . And so we calculate the polytabloid

$\displaystyle e_t=\sum\limits_{\sigma\in S_n}\mathrm{sgn}(\sigma)\sigma\{t\}$

We use our relations to calculate

$\displaystyle\begin{aligned}\pi e_t&=\sum\limits_{\sigma\in S_n}\mathrm{sgn}(\sigma)\pi\sigma\{t\}\\&=\sum\limits_{\tau\in S_n}\mathrm{sgn}(\pi^{-1}\tau)\tau\{t\}\\&=\sum\limits_{\tau\in S_n}\mathrm{sgn}(\pi^{-1})\mathrm{sgn}(\tau)\tau\{t\}\\&=\mathrm{sgn}(\pi^{-1})\sum\limits_{\tau\in S_n}\mathrm{sgn}(\tau)\tau\{t\}\\&=\mathrm{sgn}(\pi)e_t\end{aligned}$

We conclude that $S^\lambda\subseteq M^\lambda$ is a one-dimensional space with the signum representation of $S_n$ . Unlike our previous example, there is a huge difference between $S^\lambda$ and $M^\lambda$ ; we’ve seen that $M^\lambda$ is actually the left regular representation, which has dimension $n!$ .

Finally, if $\lambda(n-1,1)$ , then we can take a tableau and write a tabloid

$\displaystyle\left\{\begin{array}{ccc}i&\dots&j\\k&&\end{array}\right\}=\mathbf{k}$

where the notation we’re using on the right is well-defined since each tabloid is uniquely identified by the single entry in the second row. Now, the polytabloid in this case is $e_t=\mathbf{k}-\mathbf{i}$ , since the only column rearrangement is to swap $i$ and $k$ . It’s straightforward to see that these polytabloids span the subspace of $M^\lambda$ where the coefficients add up to zero:

$\displaystyle c_1\mathbf{1}+\dots+c_n\mathbf{n}\in S^\lambda\quad\Leftrightarrow\quad\sum\limits_{i=1}^nc_i=0$

As a basis, we can pick $\{\mathbf{2}-\mathbf{1},\mathbf{3}-\mathbf{1},\dots,\mathbf{n}-\mathbf{1}\}$ . We recognize this pattern from when we calculated the invariant subspaces of the defining representation of $S_3$ . And indeed, $M^\lambda$ is the defining representation of $S_n$ , which contains $S^\lambda$ as the analogous submodule to what we called $V^\perp$ before.

December 28, 2010 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

2 Comments »

[…] is the alternating sum of the elements of . The proof basically runs the same as when we showed that where has shape […]

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