The Unapologetic Mathematician

Mathematics for the interested outsider

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:


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 | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments

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 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 1 Comment



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