## The 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 inner corner is . If the tableau (or the tabloid has its in row , then (or ) is the result of removing that .

We’re looking for a chain of subspaces

such that as -modules. I say that we can define to be the subspace of spanned by the standard polytabloids where the shows up in row or above in .

For each , define the map by removing an in row . That is, if latex M^\lambda$ has its in row , set ; otherwise set . These are all homomorphisms of -modules, since the action of always leaves the in the same row, and so it commutes with removing an from row .

Similarly, I say that if is in row of , and we get if it’s in row with . Indeed if shows up above row , then since it’s the bottommost entry in its column that column can have no entries at all in row . Thus as we use to shuffle the columns, all of the tabloids that show up in will be sent to zero by . Similar considerations show that if is in row , then of all the tabloids that show up in , only those leaving in that row are not sent to zero by . The permutations in leaving fixed are, of course, exactly those in , and our assertion holds.

Now, since each standard polytabloid comes from some polytabloid , we see they’re all in the image of . Further, these all have their s in row , so they’re all in . That is, . On the other hand, if has its above row , then , and so .

So now we’ve got a longer chain of subspaces:

But we also know that

So the steps from to give us all the as we add up dimensions. Comparing to the formula we’re categorifying, we see that this accounts for all of . And so there are no dimensions left for the steps from to , and these containments must actually be equalities!

And thus

as asserted. The branching rule then follows.

[…] 3″? Didn’t we just finish proving the branching rule? Well, yes, but there’s another part we haven’t mentioned yet. Not […]

Pingback by The Branching Rule, Part 3 « The Unapologetic Mathematician | January 31, 2011 |

Hi, I really like your proof where everything is explained so well. But I dont get why the V^(i) as you defined them here are S^(n-1) modules. I tried showing it and the actual point missing is that the map from S_{n-1} x V^(i) gives results in V^(i). I think i.e. linear combinations of the polytabloids you used for V^(i) do still have n in one of the required rows after a permutation of S_{n-1} was applied. But how can this be true for an arbitrary permutation?

Comment by pani puri | December 4, 2013 |

Sorry, it’s been a long time since I looked at this (almost three years now), so I might be a little rusty. I think the problem might be that I’m asserting that the quotient space is an $S_{n-1}$-module, not that itself is. Does that help?

Comment by John Armstrong | December 4, 2013 |

Hi, thanks a lot for your answer! But I think there I have the same problem since now I need that n stays in the same row…

Comment by pani puri | December 9, 2013 |

What I wrote was wrong. Now I need that all n stay in their line or are in a higher one by considering only polytabloids where the n in t is in one special line.

Comment by pani puri | December 10, 2013 |