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:
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.