# 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.