The Unapologetic Mathematician

Mathematics for the interested outsider

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.

January 31, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 1 Comment

   

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