Consequences of the Submodule Theorem

We have a number of immediate consequences of the submodule theorem. First, and most important, the Specht modules form a complete list of irreducible modules for the symmetric group $S_n$ . We know that they’re irreducible, and that there’s one of them for each partition $\lambda\vdash n$ , which is the number of modules we’re looking for. But we need to show that the Specht modules corresponding to distinct partitions are themselves distinct. For this, we’ll use a lemma.

If $\theta\in\mathrm{hom}_{S_n}(S^\lambda,M^\mu)$ is a nonzero intertwinor, then $\lambda\trianglerighteq\mu$ . Further, if $\lambda=\mu$ , then $\theta$ must be multiplication by a scalar. Indeed, since $\theta\neq0$ there must be some polytabloid $e_t$ with $\theta(e_t)\neq0$ . We decompose $M^\lambda=S^\lambda\oplus{S^\lambda}^\perp$ , and extent $\theta$ to all of $M^\lambda$ by sending every vector in ${S^\lambda}^\perp$ to $0$ . That is:

$\displaystyle\begin{aligned}0&\neq\theta(e_t)\\&=\theta(\kappa_t\{t\})\\&=\kappa_t\theta(\{t\})\\&=\kappa_t\left(\sum\limits_ic_i\{s_i\}\right)\\&=\sum_ic_i\kappa_t(\{s_i\})\end{aligned}$

where the $s_i$ are $\mu$ -tableaux. Now, the $\kappa_t(\{s_i\})$ can’t all be zero, so we must have at least one $\mu$ -tableau $s^\mu$ so that $\kappa_t\{s\}\neq0$ . But then our corollary of the sign lemma tells us that $\lambda\trianglerighteq\mu$ , as we asserted!

Further, if $\lambda=\mu$ , then our other corollary shows us that $\theta(e_t)=ce_t$ for some scalar $c$ . We can thus calculate

$\displaystyle\begin{aligned}\theta(e_{\pi t})&=\theta(\pi e_t)\\&=\pi\theta(e_t)\\&=\pi ce_t\\&=ce_{\pi t}\end{aligned}$

and so $\theta$ multiplies every vector by $c$ .

As a consequence, the $S^\mu$ must be distinct for distinct permutations, since if $S^\lambda=S^\mu\subseteq M^\mu$ then there is a nonzero homomorphism $S^\lambda\to M^\mu$ , and thus $\lambda\trianglerighteq\mu$ . But the same argument shows that $\mu\trianglerighteq\lambda$ , and thus $\lambda=\mu$ .

More particularly, we have a decomposition

$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}m_{\lambda\mu}S^\lambda$

where the diagonal multiplicities are $m_{\lambda\lambda}=1$ . The rest of these multiplicities will eventually have a nice interpretation.

January 4, 2011 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

2 Comments »

[…] we’ve described the Specht modules, and we’ve shown that they give us a complete set of irreducible representations for the symmetric groups. But we […]

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[…] tableaux of shape and content . This number we call the “Kostka number” . We’ve seen that there is a […]

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