# Action on weight spaces

Let $\lambda$ and $\mu$ be partitions with at most $n$ parts. Write $\mu = 0^{n_0}1^{n_1}2^{n_2}...$ with $\sum_i n_i = n$, to mean $\mu$ has $n_0$ parts of size $0$, $n_1$ parts of size $1$, and so on. Let $V(\lambda)$ be the irreducible finite dimensional polynomial representation of $GL_n(\mathbb{C})$ of highest weight $\lambda$, and let $V(\lambda)_\mu$ be its $\mu$-weight space.

$V(\lambda)_\mu$ naturally carries an action of $S_{n_0} \times S_{n_1} \times S_{n_2} \times \dots$ (a subgroup of $S_n$, conjugate to the obvious inclusion). I am wondering what is known about these spaces as $S_{n_0} \times S_{n_1} \times S_{n_2} \times \dots$ representations?

Here are a few initial things I've noticed:

• The dimension of these spaces are Kostka numbers.

• For polynomial representations the $S_{n_0}$ factor acts trivially, so we can just think of them as $S_{n_1} \times S_{n_2} \times S_{n_3} \times \dots$ representations.

• If we fix partitions $\lambda$ and $\mu$ and let $n$ vary the $S_{n_1} \times S_{n_2} \times S_{n_3} \times \dots$ representations we get don't change (provided $n$ is large enough for this all to make sense).

Ideally I'd like some description of these representations in terms of some combinatorics on semistandard Young tableaux, but I'd be interested in hearing whatever is known.

• It may be useful to look at an older question and the references given: mathoverflow.net/questions/185797/… – Jim Humphreys Jan 25 '17 at 0:11
• In your third bullet point, how can you vary $n$ for fixed partition $\mu$? According to your setup, $n$ is determined by $\mu$. – Victor Protsak Jan 25 '17 at 6:37
• @JimHumphreys Thanks, I did look at that a bit before posting this. My takeaway was that for the zero weight space (with a slightly different labeling of weights than I was using above) we know that any irreducible symmetric group representation can occur, but in general we don't know how it decomposes. For more general weights there doesn't seem to be anything in the references there. In particular, for weights that are "close" to the highest weight I would expect the combinatorics to be more tractable and I was wondering if anyone had looked at that. – Nate Jan 25 '17 at 16:24
• @VictorProtsak In the convention I am using polynomial $GL_n$ weights are indexed by $n$-tuples of non-negative integers. I can think of a partition $\mu = (\mu_1, \mu_2, ... \mu_\ell)$ as an $n$-tuple for any $n \ge \ell$ by appending a bunch of zeroes to the end of it. – Nate Jan 25 '17 at 16:29
• Nate, you wrote $\sum_i n_i =n$ and $n_i$ are uniquely determined by $\mu$. – Victor Protsak Jan 25 '17 at 17:08