I’m going to describe the basic ideas of the Schur functors, , where is a partition and is a vector space. These will turn out to be the complete set of irreducible polynomial representations of (for all ). The main facts to strive for are:
- Every irreducible representation of is a unique Schur functor. Conversely, every Schur functor is irreducible.
- The character of is the Schur polynomial .
- The dimension of is the number of SSYTs of shape and entries from (where .) This fact will be explicit: there will be a “tableau basis” for the representation.
As a corollary, we get an improved understanding of the Littlewood-Richardson numbers and the isomorphism between the representation ring and the ring of symmetric polynomials.
One of the many applications of symmetric polynomials is to representation theory, and in this post I want to begin sketching out how.
Symmetric polynomials and the ring are involved in the representation theory of the symmetric group , and the general linear group , in related ways. The precise relationship between the representation theory of these two groups is spelled out in the Schur-Weyl Duality theorem, as well as in explicit constructions of representations of both groups.
I’m mainly interested in the Schur functors, which are representations of , so I’ll be focusing on those.
This will be the last post on symmetric polynomials, at least for now. (They’ll continue to come up when I get to representation theory and my true love, algebraic geometry, but only as part of other theories.)
I want to discuss the Hall inner product on the symmetric function ring and its interaction with the -involution. As a side benefit, we’ll get the “dual” Jacobi-Trudi and Pieri rules, with and swapped.
Alternating and Symmetric Polynomials
Consider the following recipe for building symmetric polynomials, using alternating polynomials. Consider the Vandermonde determinant
To see the last equality, note that the determinant is zero if for any , so it is divisible by , and all these factors are distinct. By degree-counting (the polynomial is evidently homogeneous of degree ), this is the whole thing, up to a scalar. Finally, to get the scalar, we do some computation (e.g. plugging in convenient values like ).
If, instead of the row , we used some other sequence of polynomials, like , the result would still be alternating in the ‘s, so it would still be divisible by the product above. However, the degree-counting argument might no longer be relevant. (For example, if we use , then the result is the original Vandermonde determinant times .)
Still, we can divide out the Vandermonde determinant, and (surprise!) the result will be a symmetric polynomial, since the sign-change is “divided out” as well.