Posted by: Steven Sam | April 5, 2010

## Stability of Kronecker coefficients

In 1938, Murnaghan proved a remarkable stability property for the tensor product multiplicities of the irreducible representations of the symmetric groups ${\mathfrak{S}_n}$. In order to state it, we need some notation. Given a sequence ${\lambda = (\lambda_1, \lambda_2, \dots)}$, let ${\lambda[N] = (N - |\lambda|, \lambda_1, \lambda_2, \dots)}$. Let ${\chi^\lambda}$ denote the irreducible character of ${\mathfrak{S}_{|\lambda|}}$ indexed in the usual way (which we make precise later). Then there are coefficients ${\overline{g}_{\lambda, \mu, \nu}}$ such that for ${N \gg 0}$, we have

$(*) \quad \chi^{\lambda[N]} \chi^{\mu[N]} = \sum_\nu \overline{g}_{\lambda, \mu, \nu} \chi^{\nu[N]}$.

These coefficients are known as stable (or reduced) Kronecker coefficients. Since the characters of the symmetric group are real-valued (in fact, integer-valued), the coefficients ${\overline{g}}$ are invariant under all permutations of indices. I’ll present a proof due to Thibon of this stability fact, but in order to do so, we’ll first make sense of the above equation for arbitrary ${N}$.

Background for symmetric functions can be found in Chapter 7 of Stanley’s Enumerative Combinatorics Vol. 2 and Chapter 1 of Macdonald’s Symmetric Functions and Hall Polynomials. And here is a link to Thibon’s paper.

First things first, let ${R(\mathfrak{S}_n)}$ be the group of class functions on ${\mathfrak{S}_n}$, let ${\Lambda_n}$ be the space of symmetric functions of degree ${n}$ in infinitely many variables ${x_1, x_2, \dots}$, and set ${\Lambda = \bigoplus_{n \ge 0} \Lambda_n}$ and ${\hat{\Lambda} = \prod_{n \ge 0} \Lambda_n}$. Let ${1_\lambda \in R(\mathfrak{S}_n)}$ be the class function which is 1 on permutations whose cycle lengths are ${\{\lambda_1, \lambda_2, \dots\}}$ and 0 on all other permutations. The homogeneous symmetric functions are defined by

$\displaystyle h_m = \sum_{i_1 \le \cdots \le i_m} x_{i_1} \cdots x_{i_m}.$

Given sequences ${I = (I_1, \dots, I_n)}$ and ${J = (J_1, \dots, J_n)}$, we define the Schur function

$\displaystyle s_{I/J} = \det(h_{I_i - J_j - i + j})_{i,j=1}^n.$

We’ll write ${s_I}$ when ${J = (0,0, \dots)}$. As ${I}$ ranges over all partitions, the ${s_I}$ form a basis for ${\Lambda}$. One other important class of symmetric functions for us will be the power sum symmetric functions

$\displaystyle p_m = \sum_i x_i^m, \quad p_I = p_{I_1} \cdots p_{I_m}.$

We define a linear isomorphism ${F \colon \bigoplus_{n \ge 0} R(\mathfrak{S}_n) \rightarrow \Lambda}$ via ${F(1_\lambda) = p_\lambda}$. We get an inner product on ${\Lambda}$ and ${\hat{\Lambda}}$ coming from the multiplication of characters by

$\displaystyle f * g = F(F^{-1}(f) F^{-1}(g)).$

The irreducible characters of the symmetric group can be defined by ${\chi^\lambda = F^{-1}(s_\lambda)}$, and we can also define ${\chi^I = F^{-1}(s_I)}$ for arbitrary sequences ${I}$. Now we can state Murnaghan’s theorem as follows.

Theorem (Murnaghan). The equation (*) holds for all ${N \in \mathbf{Z}}$.

Proving Murnaghan’s theorem will come from a study of the map ${f \mapsto \langle f \rangle \colon \Lambda \rightarrow \hat{\Lambda}}$ defined by

$\displaystyle \langle s_I \rangle = \sum_{p \in \mathbf{Z}} s_{Ip},$

where ${Ip = (I_1, \dots, I_n, p)}$, and extending linearly. In particular, I claim that Murnaghan’s theorem will follow from showing that the image of ${\langle\ \rangle}$ is closed under the inner product. First note that if ${J = (I_1, \dots, I_{k-1}, I_{k+1} - 1, I_k + 1, I_{k+1}, \dots, I_n)}$, then ${s_J = -s_I}$, which follows from our definition. Hence ${s_{Ip} = (-1)^n s_{(p-n, I_1+1, \dots, I_n+1)}}$. So given ${\lambda = (\lambda_1, \dots, \lambda_n)}$, set ${I(\lambda) = (\lambda_1 - 1, \dots, \lambda_n - 1)}$. Then ${(-1)^n s_{\lambda[N]} = s_{I(\lambda)p(\lambda)}}$ where ${p(\lambda) = n + N - |\lambda|}$.

So if we suppose that

$\displaystyle \langle s_{I(\lambda)} \rangle * \langle s_{I(\mu)} \rangle = \sum_\nu a^\nu_{\lambda, \mu} \langle s_{I(\nu)} \rangle$

for some coefficients ${a^\nu_{\lambda, \mu}}$, then by matching up degrees (inner product is degree preserving), we get

$\displaystyle s_{\lambda[N]} * s_{\mu[N]} = \sum_\nu a^\nu_{\lambda, \mu} s_{\nu[N]}$

for all ${N \in \mathbf{Z}}$, and the coefficients don’t depend on ${N}$. Applying ${F^{-1}}$ gives us Murnaghan’s theorem. So we’ll focus on proving the previously mentioned fact:

Proposition. The image of ${\langle\ \rangle}$ is closed under the inner product.

To prove this proposition, first we need a better understanding of the map. We’ll introduce a bit more notation to make the analysis cleaner. First, we have the Hall form ${(,)}$ on ${\Lambda}$ which is defined by making the basis ${(s_\lambda)}$ (where ${\lambda}$ ranges over all partitions) orthonormal, i.e., ${(s_\lambda, s_\mu) = \delta_{\lambda, \mu}}$. This extends to a form on ${\hat{\Lambda}}$. Given ${f \in \Lambda}$, we define the skewing operator ${D_f}$ to be the adjoint of multiplication by ${f}$, i.e., ${(D_fg, h) = (g, fh)}$ for all ${g,h}$. This also works for ${f \in \hat{\Lambda}}$. We’ll use without proof the fact that ${D_{s_J} s_I = s_{I / J}}$.

We’ll also use some ${\lambda}$-ring notation: for a symmetric function ${f}$, we’ll use ${f(X-1)}$ to mean ${\sum_{k \ge 0} (-1)^k D_{s_{(1^k)}}f}$, where ${(1^k)}$ just means the partition with ${k}$ parts equal to 1. We’ll assume the facts that ${D_{s_{(1^k)}} s_\lambda = 0}$ if ${\lambda}$ has less than ${k}$ nonzero parts and that the inverse of ${f \mapsto f(X-1)}$ is given by ${g \mapsto g(X+1) = \sum_{k \ge 0} D_{s_k} g}$. Finally, set ${\sigma_1 = \sum_{m \ge 0} h_m \in \hat{\Lambda}}$. Now we will prove:

Lemma. ${\langle f \rangle = \sigma_1 f(X - 1)}$

Proof. The operator ${f \mapsto \sigma_1 f(X-1)}$ is linear, so we only need to prove this for ${f = s_\lambda}$ for ${\lambda}$ a partition. Let ${n}$ be the number of nonzero parts of ${\lambda}$. Then ${s_{\lambda p} = \det(h_{(\lambda p)_i - i + j})_{i,j=1}^{n+1}}$. If we expand the determinant along the last column, this becomes

$\displaystyle s_{\lambda p} = \sum_{i=0}^n (-1)^i s_{p + i} s_{\lambda/(1^i)} = \sum_{i=0}^n (-1)^i s_{p+i} D_{s_{(1^i)}} s_\lambda.$

We can replace ${n}$ with ${\infty}$ in the last sum since the terms with ${i > n}$ are 0 by a previous remark. Hence

$\displaystyle \langle s_\lambda \rangle = \sum_{p \in \mathbf{Z}} s_{\lambda p} = \sum_{p \in \mathbf{Z}, i \ge 0} (-1)^i s_{p+i} D_{s_{(1^i)}} s_\lambda = (\sum_{p \in \mathbf{Z}} s_p) s_\lambda(X-1),$

which is the desired result. QED

We’ll combine this lemma with the following identity to get the proof of the proposition.

Theorem (Thibon). Let ${(U_I), (V_J)}$ be dual bases of ${\Lambda}$ with respect to the Hall form, and let ${f,g \in \Lambda}$. Then

$\displaystyle (\sigma_1 f) * (\sigma_1 g) = \sigma_1 \sum_{H, K} (D_{V_H} f) (D_{V_K} g) (U_H * U_K)$

where the sum is over all partitions ${H,K}$.

I was going to give a proof for this, but the post is already long and the proof is a long calculation. The reader can find details in Theorem 2.1 of Thibon’s paper.

An immediate application is that

$\displaystyle \langle s_\lambda \rangle * \langle s_\mu \rangle = \langle \sum_{ \eta, \xi} s_{\lambda / \eta}(X) s_{\mu / \xi}(X) (s_\eta * s_\xi)(X+1) \rangle,$

where ${f \mapsto f(X+1)}$ is the inverse to the map ${g \mapsto g(X-1)}$. This tells us a few things. First, expand the right hand side as ${\pm \sum_\nu \overline{g}_{\lambda, \mu, \nu} \langle s_\nu \rangle}$ (the sign comes from the fact that ${(-1)^n s_{\lambda[N]} = s_{I(\lambda) p(\lambda)}}$). By definition, ${s_\nu * s_\eta}$ is only nonzero when ${|\nu| = |\eta|}$. Thus the maximal degree term of ${s_{\lambda / \eta}(X) s_{\mu / \xi}(X) (s_\nu * s_\eta)(X+1)}$ has degree ${|\lambda| - |\eta| + |\mu| - |\xi| + |\eta| = |\lambda| + |\mu| - |\xi|}$.

Hence ${\overline{g}_{\lambda, \mu, \nu}}$ is nonzero only if $|\lambda| + |\mu| \ge |\nu|$ (we also have ${|\lambda| + |\nu| \ge |\mu|}$ and ${|\mu| + |\nu| \ge |\lambda|}$ by the fact that the coefficients ${\overline{g}_{\lambda, \mu, \nu}}$ are invariant under permuting the indices). Furthermore, when ${|\lambda| + |\mu| = |\nu|}$, one has ${|\eta| = |\xi| = 0}$, so the corresponding term is ${s_\lambda(X) s_\mu(X)}$, which means that ${\overline{g}_{\lambda, \mu, \nu}}$ is the Littlewood–Richardson coefficient ${c_{\lambda, \mu}^\nu}$ (these coefficients are defined by ${s_\lambda(X) s_\mu(X) = \sum_\nu c_{\lambda, \mu}^\nu s_\nu(X)}$).

Let’s summarize these results.

Theorem (Murnaghan, Littlewood). The coefficient ${\overline{g}_{\lambda, \mu, \nu}}$ is nonzero only if the triangle inequalities ${|\lambda| + |\mu| \ge |\nu|}$, ${|\lambda| + |\nu| \ge |\mu|}$, and ${|\mu| + |\nu| \ge |\lambda|}$ hold. Furthermore, when ${|\lambda| + |\mu| = |\nu|}$, they agree with the Littlewood–Richardson coefficients: ${\overline{g}_{\lambda, \mu, \nu} = c_{\lambda, \mu}^\nu}$.

Unfortunately, the above proof still feels like black magic. If anyone knows of a more conceptual proof, I’d certainly like to hear it.

-Steven

## Responses

1. I understand the stability, if one thinks instead about the definition of
Kronecker coefficients in terms of branching from GL(ab) to GL(a) x GL(b).
There is a similar stability for any branching from G to H, considering
the multiplicity of V_H(lambda + N mu’) inside V_G(nu + N mu), where mu
maps to mu’ under restriction of dominant weights. As I recall, one is
essentially replacing the G-irrep with something like a Verma module.

I couldn’t figure out why the stable coefficients reduce to L-R, though,
which was very annoying.

2. Then I guess the next question is how one proves the more general stability statement coming from branching. Do you have a reference for this?