The Unapologetic Mathematician

Mathematics for the interested outsider

Characters of Young Tabloid Modules (first pass)

Let’s try to calculate the characters of the Young tabloid modules M^\lambda we’ve constructed. Since these come from actions of S_n on various sets, we have our usual shortcut to calculate their characters: count fixed points.

So, let’s write \mu^\lambda for the character of the representation M^\lambda corresponding to the partition \lambda\vdash n. For a permutation \pi\in S_n, the character value \mu^\lambda(\pi) is the number of Young tabloids \{t\} such that \pi\{t\}=\{t\}. This might be a little difficult to count on its face, but let’s analyze it a little more closely.

First of all, pick a canonical Young tableau Y. The easiest one just lists the numbers from 1 to n in order from left to right on rows from top to bottom of the tableau, like


but it really doesn’t matter which one we choose. The important thing is that any other tableau t has the form t=\tau Y for some unique \tau\in S_n. Now our fixed-point condition reads \pi\tau\{Y\}=\tau\{Y\}, or \tau^{-1}\pi\tau\{Y\}=\{Y\}. But as \tau runs over S_n, the conjugate \tau^{-1}\pi\tau runs over the conjugacy class K_\pi of \pi. What’s more, it runs evenly over the conjugacy class — exactly n!/\lvert K_\pi\rvert values of \tau give each element in K_\pi. So what we need to count is how many elements \sigma\in K_\pi give a tableau \sigma Y that is row-equivalent to Y. We multiply this by n!/\lvert K_\pi\rvert to get \mu^\lambda(\pi), right?

Well, no, because now we’ve overcounted. We’ve counted the number of tableaux \tau Y with \pi\{\tau Y\}=\{\tau Y\}. But we want the number of tabloids \{\tau Y\} with this property. For example, let’s try to count \mu^\lambda(e): there’s only one element in K_e, and it leaves Y fixed. Our rule above would have us multiply this 1 by n! to get n!=\mu^\lambda(e)=\dim(M^\lambda), but there are not always n! tabloids of shape \lambda!

The story is evidently more complicated than we might have hoped. Instead of letting \tau above range over all of S_n, we could try letting it only range over a transversal for the subgroup of S_n that preserves the rows of Y. But then there’s no obvious reason to assume that the conjugates of \pi should be evenly distributed over K_\pi, which complicates our counting. We’ll have to come back to this later.

December 15, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 1 Comment



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