The Unapologetic Mathematician

Mathematics for the interested outsider

The Branching Rule, Part 1

We want to “categorify” the relation we came up with last time:

\displaystyle f^\lambda=\sum\limits_{\lambda^-}f^{\lambda^-}

That is, we want to replace these numbers with objects of a category, replace the sum with a direct sum, and replace the equation with a natural isomorphism.

It should be clear that an obvious choice for the objects is to replace f^\lambda with the Specht module S^\lambda, since we’ve seen that f^\lambda=\dim(S^\lambda). But what category are they in? On the left side, S^\lambda is an S_n-module, but on the right side all the S^{\lambda^-} are S_{n-1}-modules. Our solution is to restrict S^\lambda, suggesting the isomorphism

\displaystyle S^\lambda\downarrow^{S_n}_{S_{n-1}}\cong\bigoplus\limits_{\lambda^-}S^{\lambda^-}

This tells us what happens to any of the Specht modules as we restrict it to a smaller symmetric group. As a side note, it doesn’t really matter which S_{n-1}\subseteq S_n we use, since they’re all conjugate to each other inside S_n. So we’ll just use the one that permutes all the numbers but n.

Anyway, say the inner corners of \lambda occur in the rows r_1<\dots<r_k, and of course they must occur at the ends of these rows. For each one, we’ll write \lambda^i for the partition that comes from removing that inner corner. Similarly, if t is a standard tableau with n in the ith row, we write t^i for the (standard) tableau with n removed. And the same goes for the standard tabloids \{t\} and \{t^i\}.

Our method will be to find a tower of subspaces

\displaystyle0=V^{(0)}\subseteq V^{(1)}\subseteq\dots\subseteq V^{(k)}=S^\lambda

so that at each step we have V^{(i)}/V^{(i-1)}\cong S^{\lambda^i} as S_{n-1}-modules. Then we can see that

\displaystyle S^\lambda\downarrow^{S_n}_{S_{n-1}}=V^{(k)}\cong V^{(k-1)}\oplus(V^{(k)}/V^{(k-1)}\cong V^{(k-1)}\oplus S^{\lambda^k}

And similarly we find V^{(k-1)}\cong V^{(k-2)}\oplus S^{\lambda^{k-1}}, and step by step we go until we find the proposed isomorphism. The construction itself will be presented next time.


January 27, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups


  1. […] Branching Rule, Part 2 We pick up our proof of the branching rule. We have a partition with inner corners in rows . The partitions we get by removing each of the […]

    Pingback by The Branching Rule, Part 2 « The Unapologetic Mathematician | January 28, 2011 | Reply

  2. […] 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 […]

    Pingback by The Branching Rule, Part 4 « The Unapologetic Mathematician | February 1, 2011 | Reply

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