Posted by: Steven Sam | April 19, 2010

## Rectangle rule for Kronecker coefficients

I recently learned a neat combinatorial way of producing nonzero Kronecker coefficients. As a reminder, Kronecker coefficients can be defined as follows. Let ${\mathfrak{S}_N}$ be the symmetric group on ${N}$ letters, and let ${\chi^\lambda}$ denote its irreducible representations, where ${\lambda}$ ranges over all partitions of ${N}$. The Kronecker coefficients are defined as

$\displaystyle \chi^\lambda \otimes \chi^\mu = \bigoplus_\nu g_{\lambda, \mu, \nu} \chi^\nu.$
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 ${n \times m}$ rectangle. Fill the parts of ${\nu}$ 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 ${\lambda_i}$ be the sum of the entries in the ${i}$th row, and let ${\mu_j}$ be the sum of the entries in the ${j}$th column. Because of the way we filled the rectangle, both ${\mu = (\mu_1, \dots, \mu_n)}$ and ${\lambda = (\lambda_1, \dots, \lambda_m)}$ are partitions, and they all have the same size: ${|\lambda| = |\mu| = |\nu|}$. The rule is that ${g_{\lambda, \mu, \nu} \ne 0}$!

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 ${\mathbf{S}_\lambda(U)}$ of ${\mathbf{GL}(U)}$ are indexed by partitions ${\lambda}$ with at most ${\dim U}$ parts. The tensor product ${U^{\otimes N}}$ has commuting actions of ${\mathbf{GL}(U)}$ and ${\mathfrak{S}_N}$ (the latter acting on indices), and Schur–Weyl duality says that

$\displaystyle U^{\otimes N} \cong \bigoplus_{\lambda: \ell(\lambda) \le \dim U, |\lambda| = N} \mathbf{S}_\lambda(U) \otimes \chi^\lambda,$

where the isomorphism is as ${\mathbf{GL}(U) \times \mathfrak{S}_N}$-representations. For simplicity of notation, we’ll omit the conditions on ${\lambda}$ in the direct sum (but they’re always there). How do we use this? Well, let’s apply this in the case that ${U = V \otimes W}$ is a tensor product of two vector spaces. Then on the one hand, we get

$\displaystyle (V \otimes W)^{\otimes N} \cong \bigoplus_{\nu} \mathbf{S}_\nu(V \otimes W) \otimes \chi^\nu,$

as ${\mathbf{GL}(V \otimes W) \times \mathfrak{S}_N}$ representations. But we also have ${(V \otimes W)^{\otimes N} \cong V^{\otimes N} \otimes W^{\otimes N}}$, and

$\displaystyle V^{\otimes N} \otimes W^{\otimes N} \cong (\bigoplus_{\lambda} \mathbf{S}_\lambda(V) \otimes \chi^\lambda) \otimes (\bigoplus_{\mu} \mathbf{S}_\mu(W) \otimes \chi^\mu)$
$\displaystyle = \bigoplus_{\lambda, \mu} \mathbf{S}_{\lambda}(V) \otimes \mathbf{S}_{\mu}(W) \otimes \chi^\lambda \otimes \chi^\mu$
$\displaystyle = \bigoplus_{\lambda, \mu, \nu} g_{\lambda, \mu, \nu} \mathbf{S}_\lambda(V) \otimes \mathbf{S}_{\mu}(W) \otimes \chi^\nu$

where the isomorphism is as ${\mathbf{GL}(V) \times \mathbf{GL}(W) \times \mathfrak{S}_N}$ representations. The conclusion is that we have the branching rule from ${\mathbf{GL}(V \otimes W)}$ restricting to ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$:

$\displaystyle \mathbf{S}_\nu(V \otimes W) \cong \bigoplus_{\lambda, \mu} g_{\lambda, \mu, \nu} \mathbf{S}_\lambda(V) \otimes \mathbf{S}_\mu(W).$

The next step is to get a better handle on this branching rule. To show that ${g_{\lambda, \mu, \nu} \ne 0}$, it is enough to exhibit a nonzero ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$-equivariant map

$\displaystyle \mathbf{S}_\lambda(V) \otimes \mathbf{S}_\mu(W) \rightarrow \mathbf{S}_\nu(V \otimes W),$

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 ${\mathbf{Flag}(V)}$ be the complete flag variety, so its points consist of increasing sequences of subspaces ${V_1 \subset V_2 \subset \cdots \subset V_n = V}$ where ${\dim V_i = i}$. There is a trivial vector bundle ${V \times \mathbf{Flag}(V)}$, and it has subbundles ${\mathcal{R}_i = \{(v, V_\bullet) \mid v \in V_i \}}$. Let

$\displaystyle \mathcal{L}(\alpha) = \mathcal{R}_1^{\otimes -\alpha_1} \otimes (\mathcal{R}_2/\mathcal{R}_1)^{\otimes -\alpha_2} \otimes \cdots \otimes (\mathcal{R}_n / \mathcal{R}_{n-1})^{\otimes -\alpha_n}.$

The ${\mathcal{R}_i}$ have a natural ${\mathbf{GL}(V)}$ action, and hence so does ${\mathcal{L}(\alpha)}$, and its space of sections. If ${\alpha}$ is a partition, then the Borel–Weil theorem says that

$\displaystyle \mathrm{H}^0(\mathbf{Flag}(V); \mathcal{L}(\alpha))^* \cong \mathbf{S}_\alpha(V)$

where the isomorphism is as ${\mathbf{GL}(V)}$ representations, and the star denotes dual.

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

$\displaystyle \mathrm{H}^0(\mathbf{Flag}(V) \times \mathbf{Flag}(W); p_1^*\mathcal{L}(\alpha) \otimes p_2^*\mathcal{L}(\beta))^* \cong \mathbf{S}_\alpha(V) \otimes \mathbf{S}_\beta(W),$

where the isomorphism is as ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$ representations. So the plan is this: let’s find a ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$-equivariant embedding ${i \colon \mathbf{Flag}(V) \times \mathbf{Flag}(W) \rightarrow \mathbf{Flag}(V \otimes W)}$. Then if we form the line bundle ${\mathcal{L}(\nu)}$ on ${\mathbf{Flag}(V \otimes W)}$, we can restrict it to ${\mathbf{Flag}(V) \times \mathbf{Flag}(W)}$ and we get a ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$-equivariant restriction map

$\displaystyle \mathrm{H}^0(\mathbf{Flag}(V \otimes W); \mathcal{L}(\nu)) \rightarrow \mathrm{H}^0(\mathbf{Flag}(V) \times \mathbf{Flag}(W); i^*\mathcal{L}(\nu)).$

Well, the domain is ${\mathbf{S}_\nu(V \otimes W)^*}$ if ${\nu}$ is a partition, so we would just need to identify what ${i^*\mathcal{L}(\nu)}$ is. Of course, this will depend on the embedding ${i}$.

Here’s one way to get an embedding. Let ${n = \dim V}$ and ${m = \dim W}$, and let ${T}$ be a standard Young tableau of the ${n \times m}$ rectangle. This means that we fill in the numbers ${1, 2, \dots, mn}$ into the rectangle such that they are decreasing along rows and columns. For a number ${i=1,2,\dots, mn}$, let ${r(i)}$ and ${c(i)}$ be the row and column numbers of the square that contains it. Given such a ${T}$, we define an embedding ${i_T}$, where given two flags ${V_\bullet}$ and ${W_\bullet}$, we get a flag ${F_\bullet}$ of ${V \otimes W}$ defined by

$\displaystyle F_i = \sum_{j \le i} V_{r(j)} \otimes W_{c(j)}.$

By the fact that ${T}$ is a standard Young tableau, ${\dim F_i = i}$ for all ${i}$. Then ${i_T}$ is ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$-equivariant. To understand what ${i_T^*\mathcal{L}(\nu)}$ is, it’s enough to see what happens to ${\mathcal{R}_j / \mathcal{R}_{j-1}}$. From our definition above, we see that the fiber of ${\mathcal{R}_j / \mathcal{R}_{j-1}}$ over ${i_T(V_\bullet, W_\bullet)}$ is ${V_{r(j)} / V_{r(j) - 1} \otimes W_{c(j)} / W_{c(j) - 1}}$. Putting this all together, this means that ${i_T^*\mathcal{L}(\nu) \cong p_1^* \mathcal{L}(\lambda) \otimes p_2^* \mathcal{L}(\mu)}$ where ${p_i}$ are the two projections from ${\mathbf{Flag}(V) \times \mathbf{Flag}(W)}$ and ${\lambda_i = \sum_{k: r(k) = i} \nu_k}$ and ${\mu_j = \sum_{k : c(k) = j} \nu_k}$.

So we get the following ${\mathbf{GL}(V) \times \mathbf{GL}(W)}$-equivariant restriction map:

$\displaystyle \mathrm{H}^0(\mathbf{Flag}(V \otimes W); \mathcal{L}(\nu)) \rightarrow \mathrm{H}^0(\mathbf{Flag}(V) \times \mathbf{Flag}(W); p_1^* \mathcal{L}(\lambda) \otimes p_2^* \mathcal{L}(\mu)).$

Dualizing, we get the map

$\displaystyle \mathbf{S}_\lambda(V) \otimes \mathbf{S}_\mu(W) \rightarrow \mathbf{S}_\nu(V \otimes W).$

The last step is to see that this map is nonzero. This is equivalent to saying that there is some section of ${\mathcal{L}(\nu)}$ on ${\mathbf{Flag}(V \otimes W)}$ which is nonzero on some point of the image of ${\mathbf{Flag}(V) \times \mathbf{Flag}(W)}$. If this weren’t true, this would mean that the line bundle ${\mathcal{L}(\nu)}$ has base points, but this never happens when ${\nu}$ is a partition. Hence the map is nonzero!

-Steven

## Responses

1. 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 $\chi^{n \lambda} \ otimes \chi^{n \nu}$.