I recently learned a neat combinatorial way of producing nonzero Kronecker coefficients. As a reminder, Kronecker coefficients can be defined as follows. Let be the symmetric group on letters, and let denote its irreducible representations, where ranges over all partitions of . The Kronecker coefficients are defined as

It is an open problem to find a combinatorial rule for these numbers, but I will explain a quick combinatorial rule which is a sufficient condition for these coefficients to be nonzero.

Take an rectangle. Fill the parts of into this rectangle so that the entries are weakly decreasing along rows going left to right and weakly decreasing along columns going top to bottom. Let be the sum of the entries in the th row, and let be the sum of the entries in the th column. Because of the way we filled the rectangle, both and are partitions, and they all have the same size: . The rule is that !

The proof that I learned is really neat, so I’ll share it here. I don’t know who it is attributed to, but it is hinted at in a lecture that Laurent Manivel gave. The two tools that we’ll use are Schur–Weyl duality and the Borel–Weil theorem. Schur–Weyl duality will let us reformulate Kronecker coefficients in terms of a branching rule for the general linear group, and Borel–Weil lets us use geometry to understand this branching.

First, let’s see what Schur–Weyl duality says. The irreducible polynomial representations of are indexed by partitions with at most parts. The tensor product has commuting actions of and (the latter acting on indices), and Schur–Weyl duality says that

where the isomorphism is as -representations. For simplicity of notation, we’ll omit the conditions on in the direct sum (but they’re always there). How do we use this? Well, let’s apply this in the case that is a tensor product of two vector spaces. Then on the one hand, we get

as representations. But we also have , and

where the isomorphism is as representations. The conclusion is that we have the branching rule from restricting to :

The next step is to get a better handle on this branching rule. To show that , it is enough to exhibit a nonzero -equivariant map

or one going in the other direction. Where can we find such maps? One such place is embeddings of flag varieties into bigger flag varieties. In order to explain this, let me quickly mention the Borel–Weil theorem.

Let be the complete flag variety, so its points consist of increasing sequences of subspaces where . There is a trivial vector bundle , and it has subbundles . Let

The have a natural action, and hence so does , and its space of sections. If is a partition, then the Borel–Weil theorem says that

where the isomorphism is as representations, and the star denotes dual.

We can of course extend this to a product of flag varieties, i.e.,

where the isomorphism is as representations. So the plan is this: let’s find a -equivariant embedding . Then if we form the line bundle on , we can restrict it to and we get a -equivariant restriction map

Well, the domain is if is a partition, so we would just need to identify what is. Of course, this will depend on the embedding .

Here’s one way to get an embedding. Let and , and let be a standard Young tableau of the rectangle. This means that we fill in the numbers into the rectangle such that they are decreasing along rows and columns. For a number , let and be the row and column numbers of the square that contains it. Given such a , we define an embedding , where given two flags and , we get a flag of defined by

By the fact that is a standard Young tableau, for all . Then is -equivariant. To understand what is, it’s enough to see what happens to . From our definition above, we see that the fiber of over is . Putting this all together, this means that where are the two projections from and and .

So we get the following -equivariant restriction map:

Dualizing, we get the map

The last step is to see that this map is nonzero. This is equivalent to saying that there is some section of on which is nonzero on some point of the image of . If this weren’t true, this would mean that the line bundle has base points, but this never happens when is a partition. Hence the map is nonzero!

-Steven

I saw your contribution last week. Your proof is very concise and very clear. With Boris Pasquier and Nicolas Ressayre we made the same remark

in our last paper

http://arxiv.org/abs/1004.2119 . We agree with you that this result is probably already known.

In our paper we study the problem of decomposing the restriction of a

representation from a reductif group to a reductif subgroup. We include components obtained as above in a larger family which generalizes the components obtained by the PRV conjecture proved by S. Kumar and O. Mathieu (The PRV conjecture concern the problem of decomposing tensor product of

irreducible representations of a reductif group).

Returning to the above problem, in our paper we show that if one relaxes

the growth condition on the rows and columns, so the component appears

but asymptotically: there are positive integer $n$ such that $\chi^{n \nu}$ have

non-zero multiplicity in $\chi^{n \lambda} \ otimes \chi^{n \nu}$.

By:

Pierre-Louis Montagardon November 8, 2010at 12:18 PM