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:

\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.

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

2 Comments »

  1. [...] we’ve described the Specht modules, and we’ve shown that they give us a complete set of irreducible representations for the symmetric groups. But we [...]

    Pingback by Standard Tableaux « The Unapologetic Mathematician | January 5, 2011 | Reply

  2. [...] tableaux of shape and content . This number we call the “Kostka number” . We’ve seen that there is a [...]

    Pingback by Kostka Numbers « The Unapologetic Mathematician | February 17, 2011 | Reply


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