# The Unapologetic Mathematician

## 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