# The Unapologetic Mathematician

## The Branching Rule, Part 3

“Part 3”? Didn’t we just finish proving the branching rule? Well, yes, but there’s another part we haven’t mentioned yet. Not only does the branching rule tell us how representations of $S_n$ decompose when they’re restricted to $S_{n-1}$, it also tells us how representations of $S_{n-1}$ decompose when they’re induced to $S_n$.

Now that we have the first statement of the branching rule down, proving the other one is fairly straightforward: it’s a consequence of Frobenius reciprocity. Indeed, the branching rule tells us that

$\displaystyle\dim\left(\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\right)=\bigg\{\begin{array}{cl}1&\mu=\lambda^-\\{0}&\mathrm{otherwise}\end{array}$

That is, there is one copy of $S^\mu$ inside $S^\lambda$ (considered as an $S_{n-1}$-module) if $\mu$ comes from $\lambda$ by removing an inner corner, and there are no copies otherwise.

So let’s try to calculate the multiplicity of $S^\lambda$ in the induced module $S^\mu\!\!\uparrow$:

\displaystyle\begin{aligned}\hom_{S_n}(S^\lambda,S^\mu\!\!\uparrow)^*&\cong\hom_{S_n}(S^\mu\!\!\uparrow,S^\lambda)\\&\cong\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\end{aligned}

Taking dimensions, we find

\displaystyle\begin{aligned}\dim\left(\hom_{S_n}(S^\lambda,S^\mu\!\!\uparrow)\right)&=\dim\left(\hom_{S_{n-1}}(S^\mu,S^\lambda\!\!\downarrow)\right)\\&=\bigg\{\begin{array}{cl}1&\mu=\lambda^-\\{0}&\mathrm{otherwise}\end{array}\\&=\bigg\{\begin{array}{cl}1&\lambda=\mu^+\\{0}&\mathrm{otherwise}\end{array}\end{aligned}

since if $\mu$ comes from $\lambda$ by removing an inner corner, then $\lambda$ comes from $\mu$ by adding an outer corner.

We conclude that

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

which is the other half of the branching rule.