# The Unapologetic Mathematician

## 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

## 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 $U\subseteq{S^\mu}^\perp$.

January 4, 2011