The Unapologetic Mathematician

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