In 1938, Murnaghan proved a remarkable stability property for the tensor product multiplicities of the irreducible representations of the symmetric groups . In order to state it, we need some notation. Given a sequence , let . Let denote the irreducible character of indexed in the usual way (which we make precise later). Then there are coefficients such that for , we have

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 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 .

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 be the group of class functions on , let be the space of symmetric functions of degree in infinitely many variables , and set and . Let be the class function which is 1 on permutations whose cycle lengths are and 0 on all other permutations. The **homogeneous symmetric functions** are defined by

Given sequences and , we define the **Schur function**

We’ll write when . As ranges over all partitions, the form a basis for . One other important class of symmetric functions for us will be the **power sum symmetric functions**

We define a linear isomorphism via . We get an **inner product** on and coming from the multiplication of characters by

The irreducible characters of the symmetric group can be defined by , and we can also define for arbitrary sequences . Now we can state Murnaghan’s theorem as follows.

**Theorem (Murnaghan).** The equation (*) holds for all .

Proving Murnaghan’s theorem will come from a study of the map defined by

where , and extending linearly. In particular, I claim that Murnaghan’s theorem will follow from showing that the image of is closed under the inner product. First note that if , then , which follows from our definition. Hence . So given , set . Then where .

So if we suppose that

for some coefficients , then by matching up degrees (inner product is degree preserving), we get

for all , and the coefficients don’t depend on . Applying gives us Murnaghan’s theorem. So we’ll focus on proving the previously mentioned fact:

**Proposition.** The image of 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 which is defined by making the basis (where ranges over all partitions) orthonormal, i.e., . This extends to a form on . Given , we define the **skewing operator** to be the adjoint of multiplication by , i.e., for all . This also works for . We’ll use without proof the fact that .

We’ll also use some -ring notation: for a symmetric function , we’ll use to mean , where just means the partition with parts equal to 1. We’ll assume the facts that if has less than nonzero parts and that the inverse of is given by . Finally, set . Now we will prove:

**Lemma.**

*Proof.* The operator is linear, so we only need to prove this for for a partition. Let be the number of nonzero parts of . Then . If we expand the determinant along the last column, this becomes

We can replace with in the last sum since the terms with are 0 by a previous remark. Hence

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 be dual bases of with respect to the Hall form, and let . Then

where the sum is over all partitions .

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

where is the inverse to the map . This tells us a few things. First, expand the right hand side as (the sign comes from the fact that ). By definition, is only nonzero when . Thus the maximal degree term of has degree .

Hence is nonzero only if (we also have and by the fact that the coefficients are invariant under permuting the indices). Furthermore, when , one has , so the corresponding term is , which means that is the Littlewood–Richardson coefficient (these coefficients are defined by ).

Let’s summarize these results.

**Theorem (Murnaghan, Littlewood).** The coefficient is nonzero only if the triangle inequalities , , and hold. Furthermore, when , they agree with the Littlewood–Richardson coefficients: .

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

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.

By:

Allen Knutsonon April 5, 2010at 10:38 PM

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?

By:

Steven Samon April 11, 2010at 2:23 PM