## 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.