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

Advertisements

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?


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Categories

%d bloggers like this: