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Schur functors

I’m going to describe the basic ideas of the Schur functors, \mathbb{S}^\lambda(V), where \lambda is a partition and V is a vector space. These will turn out to be the complete set of irreducible polynomial representations of GL_n (for all n). The main facts to strive for are:

  • Every irreducible representation of GL_n is a unique Schur functor. Conversely, every Schur functor is irreducible.
  • The character of \mathbb{S}^\lambda(V) is the Schur polynomial s_\lambda.
  • The dimension of \mathbb{S}^\lambda(V) is the number of SSYTs of shape \lambda and entries from 1, \ldots, n (where n = \dim(V).) 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 Rep(\text{GL}) \to \lambda_n between the representation ring and the ring of symmetric polynomials.


GL-representations, symmetric polynomials, and geometry

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 \Lambda are involved in the representation theory of the symmetric group S_n, and the general linear group GL_n, 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 GL_n, so I’ll be focusing on those.


Everything you wanted to know about symmetric polynomials, part V

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 \Lambda and its interaction with the \omega-involution. As a side benefit, we’ll get the “dual” Jacobi-Trudi and Pieri rules, with e and h swapped.


Everything you wanted to know about symmetric polynomials, part IV

Alternating and Symmetric Polynomials

Consider the following recipe for building symmetric polynomials, using alternating polynomials. Consider the Vandermonde determinant

\displaystyle{  \Delta(x_1, \ldots, x_n) = \left|\begin{matrix}  1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\  1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\  & \vdots & & \vdots & \\  1 & x_n & x_n^2 & \cdots & x_n^{n-1}  \end{matrix}\right| = \prod_{i < j} (x_j - x_i).  }

To see the last equality, note that the determinant is zero if x_i = x_j for any i \ne j, so it is divisible by (x_j-x_i), and all these factors are distinct. By degree-counting (the polynomial is evidently homogeneous of degree n \choose 2), 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 x_i = i).

If, instead of the row (1 \ x_i\ x_i^2\ \cdots\ x_i^{n-1}), we used some other sequence of polynomials, like (f_1(x_i)\ f_2(x_i)\ \cdots \ f_{n-1}(x_i)), the result would still be alternating in the x_i‘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 (x_i\ x_i^2\ \cdots\ x_i^n), then the result is the original Vandermonde determinant times x_1 \cdots x_n.)

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.