Now that we’ve shown the intertwinors that come from semistandard tableaux are independent, we want to show that they span the space . This is a bit fidgety, but should somewhat resemble the way we showed that standard polytabloids span Specht modules.
So, let be any intertwinor, and write out the image
Here we’re implicitly using the fact that .
First of all, I say that if and , then the coefficients of and differ by a factor of . Indeed, we calculate
This tells us that
Comparing coefficients on the left and right gives us our assertion.
As an immediate corollary to this lemma, we conclude that if has a repetition in some column, then . Indeed, we can let be the permutation that swaps the places of these two identical entries. Then , while the previous result tells us that , and so .
Indeed, let’s index the semistandard generalized tableaux as . We will take our reference tableau and show that the vectors are independent. This will show that the are independent, since any linear dependence between the operators would immediately give a linear dependence between the for all .
Anyway, we have
Since we assumed to be semistandard, we know that for all summands . Now the permutations in do not change column equivalence classes, so this still holds: for all summands . And further all the are distinct since no column equivalence class can contain more than one semistandard tableau.
But now we can go back to the lemma we used when showing that the standard polytabloids were independent! The are a collection of vectors in . For each one, we can pick a basis vector which is maximum among all those having a nonzero coefficient in the vector, and these selected maximum basis elements are all distinct. We conclude that our collection of vectors in independent, and then it follows that the intertwinors are independent.
Sorry I forgot to post this yesterday afternoon.
You could probably have predicted this: we’re going to have orders on generalized tabloids analogous to the dominance and column dominance orders for tabloids without repetitions. Each tabloid (or column tabloid) gives a sequence of compositions, and at the th step we throw in all the entries with value .
For example, the generalized column tabloid
gives the sequence of compositions
while the semistandard generalized column tabloid
gives the sequence of compositions
and we find that since for all .
We of course have a dominance lemma: if , occurs in a column to the left of in , and is obtained from by swapping these two entries, then . As an immediate corollary, we find that if is semistandard and is different from , then . That is, is the "largest" (in the dominance order) equivalence class in . The proofs of these facts are almost exactly as they were before.
We want to take our intertwinors and restrict them to the Specht modules. If the generalized tableau has shape and content , we get an intertwinor . This will eventually be useful, since the dimension of this hom-space is the multiplicity of in .
Anyway, if is our standard “reference” tableau, then we can calculate
We can see that it will be useful to know when . It turns out this happens if and only if has two equal elements in some column.
Indeed, if , then
Thus for some with we must have . But then we must have all the elements in each cycle of the same, and these cycles are restricted to the columns. Since is not the identity, we have at least one nontrivial cycle and at least two elements the same.
On the other hand, assume in the same column of . Then . But then the sign lemma tells us that is a factor of , and thus .
This means that we can eliminate some intertwinors from consideration by only working with things like standard tableaux. We say that a generalized tableau is semistandard if its columns strictly increase (as for standard tableaux) and its rows weakly increase. That is, we allow repetitions along the rows, but only so long as we never have any row descents. The tableau
is semistandard, but
Given any generalized Young tableau with shape and content , we can construct an intertwinor . Actually, we’ll actually go from to , but since we’ve seen that this is isomorphic to , it’s good enough. Anyway, first, we have to define the row-equivalence class and column-equivalence class . These are the same as for regular tableaux.
So, let be our reference tableau and let be the associated tabloid. We define
Continuing our example, with
we we define
Now, we extend in the only way possible. The module is cyclic, meaning that it can be generated by a single element and the action of . In fact, any single tabloid will do as a generator, and in particular generates .
So, any other module element in is of the form for some . And so if is to be an intertwinor we must define
Remember here that acts on generalized tableaux by shuffling the entries by place, not by value. Thus in our example we find
Now it shouldn’t be a surprise that since so much of our construction to this point has depended on an aribtrary choice of a reference tableau , the linear combination of generalized tableaux on the right doesn’t quite seem like it comes from the tabloid on the left. But this is okay. Just relax and go with it.
We can obviously create vector spaces out of generalized Young tableaux. Given the collection of tableaux of shape and content , we get the vector space . We want to turn this into an -module.
First, given any tabloid of shape , we can product a (generalized) tableau by defining to be the number of the row in that contains the entry . As an example, consider the tabloid
This gives us the function , , and . If and we use the usual reference tableau , this gives us the generalized tabloid
The shape of is obviously , and it’s easy to see that the content is exactly . Indeed, there are entries in with the value , just as there are entries in the first row of .
It should also be clear that this correspondence is a bijection. That is, given any generalized tableau of shape and content we can get a tabloid of shape by turning into a function and then putting on row of if .
That means that the basis of generalized tableaux of the vector space is in bijection with the basis of -tabloids of the vector space . And this space carries an action of — the linear extension of the action on tabloids. We want to pull this action across the bijection we just set up to get an action on .
On the one hand, this is as easy as saying it: if corresponds to , we define to be the generalized tableau corresponding to and we’re done. To be a bit more explicit, we define by considering it as a function and setting
So, for example, if
then we can calculate
Even more explicitly, if
then we calculate
We should be clear about a major distinction here: the permutation acts on the entries in — replacing by — but it acts on the places in — moving to the position of .
If we write the correspondence as , then for to be an intertwinor we need . This forces
and so this explicit action is forced on us.
The really interesting thing is that when we use this action on the generalized tableaux in , we always get a module , no matter what shape we start with.
And now we have another generalization of Young tableaux. These are the same, except now we allow repetitions of the entries.
Explicitly, a generalized Young tableau — we write them with capital letters — of shape is an array obtained by replacing the points of the Ferrers diagram of with positive integers. Any skipped or repeated numbers are fine. We say that the “content” of is the composition where is the number of entries in .
As an example, we have the generalized Young tableau
of shape and content .
Notice that if , then as well, since both count up the total number of places in the tableau. Given a partition and a composition , both decomposing the same number , we define to be the collection of generalized Young tableaux of shape and content . All the tableaux we’ve considered up until now have content .
Now, pick some fixed (ungeneralized) tableau . We can use the same one we usually do, numbering the rows from to across each row and from top to bottom, but it doesn’t really matter which we use. For our examples we’ll pick
Using this “reference” tableau, we can rewrite any generalized tableau as a function; define to be the entry of in the same place as is in . That is, any generalized tableau looks like
and in our particular example above we have , , and . Conversely, any such function assigning a positive integer to each number from to can be interpreted as a generalized Young tableau. Of course the particular correspondence depends on exactly which reference tableau we use, but there will always be some such correspondence between functions and generalized tableaux.
Well, 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 branching rule and decategorify it to see what it says!
Strictly speaking, decategorification means passing from a category to its set of isomorphism classes. That is, in the case of our categories of -modules we should go from the Specht module to its character . And that is an interesting question, but since the original relation turned into the dimensions of the modules in the branching rule, let’s do the same thing in reverse.
So the flip side of the branching rule tells us how a Specht module decomposes after being induced up to the next larger symmetric group. That is:
To “decategorify”, we take dimensions
As we see, the right hand side is obvious, but he left takes a little more work. We consult the definition of induction to find
Calculating its dimension is a little easier if we recall what induction looks like for matrix representations: the direct sum of a bunch of copies of , one for each element in a transversal of the subgroup. That is, there are
copies of in the induced representation. We find our new relation:
That is, the sum of the numbers of standard tableaux of all the shapes we get by adding an outer corner to is times the number of standard tableaux of shape .
This is actually a sort of surprising result, and there should be some sort of combinatorial proof of it. I’ll admit, though, that I don’t know of one offhand. If anyone can point me to a good one, I’d be glad to post it.
“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 decompose when they’re restricted to , it also tells us how representations of decompose when they’re induced to .
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
That is, there is one copy of inside (considered as an -module) if comes from by removing an inner corner, and there are no copies otherwise.
So let’s try to calculate the multiplicity of in the induced module :
Taking dimensions, we find
since if comes from by removing an inner corner, then comes from by adding an outer corner.
We conclude that
which is the other half of the branching rule.
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!
as asserted. The branching rule then follows.