The Submodule Theorem

Sorry for the delay.

Let $U\subseteq M^\mu$ be a submodule of one of the Young tabloid modules. Then I say that either $U$ contains the Specht module $S^\mu\subseteq M^\mu$ , or it is contained in the orthogonal complement ${S^\mu}^\perp\subseteq M^\mu$ . In particular, each Specht module $S^\mu$ is irreducible, since any nontrivial submodule $U\subseteq S^\mu$ cannot be contained in the orthogonal complement, and so it must also contain — and thus be equal to — $S^\mu$ .

To see this, let $u\in U\subseteq M^\mu$ be any vector, and let $t^\mu$ be a $\mu$ -tableau. Our last result showed us that $\kappa_tu=ce_t$ for some $c\in\mathbb{C}$ . First, we’ll assume that there is some pair of $u$ and $t$ so that $c\neq0$ . In this case, $e_t=\frac{1}{c}\kappa_tu\in U$ . Since $e_t$ generates all of $S^\mu$ — the Specht modules are cyclic — we must have $S^\mu\subseteq U$ .

On the other hand, what if there is no such pair of $u$ and $t$ ? That is, $\kappa_tu=0$ for all vectors $u\in U$ and $\mu$ -tableaux $t$ . We can calculate

$\displaystyle\begin{aligned}\langle u,e_t\rangle&=\langle u,\kappa_t\{t\}\rangle\\&=\langle\kappa_tu,\{t\}\rangle\\&=\langle0,\{t\}\rangle\\&=0\end{aligned}$

So each vector $u\in U$ is orthogonal to each polytabloid $e_t$ . Since the polytabloids span $S^\mu$ , we must have $U\subseteq{S^\mu}^\perp$ .

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

1 Comment »

[...] 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 [...]

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