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

\displaystyle\begin{array}{ccc}1&2&3\\4&5&6\\7&8&\\9&&\end{array}

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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