Let $V$ be the irreducible representation of $GL_n$ with highest $\lambda$, and $|\lambda|=n$. It is well known that the representation of $S_n$ on the $(1,1,\ldots,1)$ weight space is the Specht module $S_{\lambda}$. What is known about the representation of $S_n$ on the rest of $V$?

I am also interested in the case where $|\lambda| < n$.


Each direct sum of an $S_n$-orbit of weight spaces is an induction from the representation of the stabilizer of one of the weight spaces, which is always a Young subgroup (just coming from how many of your entries are the same). That representation of the stabilizer should be a sum of Specht modules depending on how your representation branches to the semi-simple part of the Levi (block diagonal matrices where each block as determinant 1) corresponding to the Young subgroup.

When the weight space is (1,...,1), then everything is in the stabilizer, so you're looking at the branching from the group to itself, and nothing happens. On the other hand, if you take a generic weight, its stabilizer will be trivial, and you'll just get a regular representation tensored with the weight space (the semi-simple part of the Levi is trivial in this case).

UPDATE: This is David Speyer abusing the edit power to warn people that I say a lot of false things in the comments below. I don't want to delete them because that will make Ben's responses look like nonsense, but you shouldn't rely on anything I say here.

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  • $\begingroup$ Thanks, that's helpful! It looks to me like the action of the Levi on the weight space is always trivial. (More precisely, let W \subset V be a weight space, and let L be the largest Levi taking W to itself. It looks to me like W fixes V pointwise.) Is that true, or am I missing something? $\endgroup$ – David E Speyer Nov 11 '09 at 16:57
  • $\begingroup$ I don't follow. You don't want the Levi to fix the weight space, just its torus. $\endgroup$ – Ben Webster Nov 11 '09 at 17:33
  • $\begingroup$ Ah, I see. You ask that the torus fix the weight space pointwise. Explicitly, if the weight is (mu_1, ... mu_n), you look at the subtorus generated by the coweights e_i-e_j, where mu_i = mu_j. The corresponding Levi then does take the weight space to itself, but that wasn't your point. I lied when I said it acted trivially. The right statement is that, for each m in the set {mu_1, ... mu_n}, the corresponding component of the Levi acts by det^m. $\endgroup$ – David E Speyer Nov 11 '09 at 17:41
  • $\begingroup$ I was thinking of the semi-simple part of the Levi (it's a bit unclear to me if that or the whole reductive quotient should be called the Levi. Springer Online Reference says the latter). $\endgroup$ – Ben Webster Nov 11 '09 at 17:56
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    $\begingroup$ Wait a sec, I'm confused. How does your answer reproduce the classical one in the case of the $(1,1,\ldots,1)$ weight space? $\endgroup$ – David E Speyer Nov 11 '09 at 20:16

I keep meaning to post this and forgetting: Richard Stanley sent me the following two references

Enumerative Combinatorics, Volume II Exercise 7.74: If $V_{\lambda}$ is a representation of $GL_n$, and $S_{\mu}$ a representation of $S_n$, then the multiplicity of $S_{\mu}$ in $\mathrm{Restriction}^{GL_n}_{S_n} V_{\lambda}$ is the coefficient of the Schur function $s_{\lambda}$ in the symmetric formal power series $$s_{\mu}(1, x_1, x_2, \ldots, x_1^2, x_1 x_2, x_1 x_3, \ldots, x_1^3, x_1^2 x_2, \ldots).$$ That is to say, we take the Schur function $s_{\mu}$ and plug in every monomial. (Yes, for those who know the term, this is an example of plethysm.)

Gay, David A.
"Characters of the Weyl group of $SU(n)$ on zero weight spaces and centralizers of permutation representations."
Rocky Mountain J. Math. 6 (1976), no. 3, 449—455.

Determines the representation of $S_n$ on the zero weight space of $V_{\lambda}$ when $|\lambda|$ is a multiple of $n$.

I just found another reference, which I haven't digested yet: Scharf, Thibon and Wybourne (1997).

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In the Lie group book by Procesi, at page 272,there is a result of (11...1) weight space I think it is better than Gay's paper.

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  • $\begingroup$ If you could paraphrase what this result says, it might make your answer easier to decipher... $\endgroup$ – Yemon Choi Feb 22 '10 at 7:31
  • $\begingroup$ Just to check, you mean C. Procesi, Lie groups: an approach through invariants and representations, Springer-Verlag 2007, right? Thanks, I'll check it out later today. $\endgroup$ – David E Speyer Feb 22 '10 at 11:56

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