# The Unapologetic Mathematician

## The Branching Rule, Part 4

What? More!?

Well, we got the idea to look for the branching rule by trying to categorify a certain combinatorial relation. So let’s take the flip side of the branching rule and decategorify it to see what it says!

Strictly speaking, decategorification means passing from a category to its set of isomorphism classes. That is, in the case of our categories of $S_n$-modules we should go from the Specht module $S^\lambda$ to its character $\chi^\lambda$. And that is an interesting question, but since the original relation turned into the dimensions of the modules in the branching rule, let’s do the same thing in reverse.

So the flip side of the branching rule tells us how a Specht module decomposes after being induced up to the next larger symmetric group. That is:

$\displaystyle S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}\cong\bigoplus\limits_{\lambda^+}S^{\lambda^+}$

To “decategorify”, we take dimensions

\displaystyle\begin{aligned}\dim\left(S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}\right)&=\dim\left(\bigoplus\limits_{\lambda^+}S^{\lambda^+}\right)\\&=\sum\limits_{\lambda^+}\dim\left(S^{\lambda^+}\right)\\&=\sum\limits_{\lambda^+}f^{\lambda^+}\end{aligned}

As we see, the right hand side is obvious, but he left takes a little more work. We consult the definition of induction to find

$\displaystyle S^\lambda\!\!\uparrow_{S_n}^{S_{n+1}}=\mathbb{C}\left[S_{n+1}\right]\otimes_{S_n}S^\lambda$

Calculating its dimension is a little easier if we recall what induction looks like for matrix representations: the direct sum of a bunch of copies of $S^\lambda$, one for each element in a transversal of the subgroup. That is, there are

$\displaystyle\lvert S_{n+1}/S_n\rvert=\lvert S_{n+1}\rvert/\lvert S_n\rvert=\frac{(n+1)!}{n!}=n+1$

copies of $S^\lambda$ in the induced representation. We find our new relation:

$\displaystyle(n+1)f^\lambda=\sum\limits_{\lambda^+}f^{\lambda^+}$

That is, the sum of the numbers of standard tableaux of all the shapes we get by adding an outer corner to $\lambda$ is $n+1$ times the number of standard tableaux of shape $\lambda$.

This is actually a sort of surprising result, and there should be some sort of combinatorial proof of it. I’ll admit, though, that I don’t know of one offhand. If anyone can point me to a good one, I’d be glad to post it.

February 1, 2011 -

## 1 Comment »

1. Hi John,

I *think* I have a combinatorial proof- I’d be happy to write it up and send you a copy if you’re interested.

Cheers,

Andy

Comment by Andrew Poulton | February 4, 2011 | Reply