The Unapologetic Mathematician

Mathematics for the interested outsider

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


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 | Algebra, Representation Theory, Representations of Symmetric Groups


  1. […] 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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