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For a partition $\lambda$ let $S^{\lambda}$ be the corresponding Schur functor. Is it true that for every $\lambda$ there exists an irreducible representation $V$ of a finite nonabelian group $G$ such that $S^{\lambda}(V)$ is still irreducible?

This is not obvious to me even for the symmetric and exterior powers (although maybe I'm not thinking hard enough), so any partial results would be appreciated.

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I hope you don't mind too much that I made a little edit for clarity. I found the original statement very confusing. –  Ben Webster Jul 18 '10 at 22:03
Ben's edit makes the question more precise, but I'm unclear about the motivation for it. Would there be interesting consequences (for general linear groups or for finite groups) if $f$ can be computed explicitly or if the boundedness is somehow shown to be true or false? The question is intriguing, but also puzzling. Classical Schur-Weyl theory gives good algorithmic knowledge of the constituents of tensor powers of $V$` for general linear groups over $\mathbb{C}$; but neither this nor the finite subgroups are known in closed form as $n$ grows (eventually all finite groups appear). –  Jim Humphreys Jul 18 '10 at 22:08
@Ben: Thanks. @Jim: The motivation is vaguer than the question, which is why I wasn't sure whether to put it in. For any representation V of a finite group G we know that V^{\otimes k} decomposes into parts corresponding to the irreps of GL(V), which by Schur-Weyl duality can be labeled by some irreps of S_k. I want to know if this is the best one can do for a "generic" group G, e.g. whether each of these representations is actually irreducible for some finite group G and some representation V. If I've got the statement of the problem right, this is equivalent to f being unbounded. –  Qiaochu Yuan Jul 18 '10 at 22:31
I have edited the statement of the question to reflect my motivation more accurately. –  Qiaochu Yuan Jul 18 '10 at 22:40
The answer is no, for the sixth symmetric power in characteristic zero. But I don't know if there is an easy proof. See "Symmetric powers and a problem of Kollar and Larsen," by Guralnick and Thiep. –  moonface Jul 18 '10 at 22:57
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up vote 3 down vote accepted

Since the Guralnick and Tiep paper is very long, I thought I would summarize my understanding of it. Note that I only learned about this result from moonface last night, so I am hardly an expert. This answer is community wiki, in case anyone can improve on my summary.

We want to establish the following result: Let $G$ be a finite noncommutative subgroup of $GL(V)$, with $V$ a $\mathbb{C}$ vector space. Let $k \geq 6$. Then $\mathrm{Sym}^k(V)$ is reducible. (If $G$ is commutative and $V$ is one dimensional then, of course, $\mathrm{Sym}^k(V)$ is always one dimensional and hence irreducible.)

Our proof is by induction on $|G|$. Our base case will be when $G$ is a central extension of a simple group.

Choose a nontrivial normal subgroup $H$ of $G$ such that $G/H$ is simple. If we can't do this then $G$ is simple and we are in the base case. Let $V \cong \bigoplus U_i$ be the decomposition of $V$ into $H$-isotypic components. If there is more than one summand, then $\bigoplus \mathrm{Sym}^k(U_i)$ is a nontrivial $G$-subrep of $\mathrm{Sym}^k(V)$. So we may assume that $V \cong W \otimes X$, where $W$ is an $H$-irrep and $H$ acts trivially on $X$. Then one can show (proof of lemma 2.5) then $G$ is contained in $GL(W) \times GL(X)$. (This is nontrivial but not deep; you could give it as a problem in a graduate-level representation theory course.)

Then $\mathrm{Sym}^k(W) \otimes \mathrm{Sym}^k(X)$ is a subrepresentation of $\mathrm{Sym}^k(V)$. This subrep is proper unless $\dim X=1$ or $\dim W=1$. If $\dim X=1$, then $V$ is an irrep of $H$ and $\mathrm{Sym}^k(V)$ is irreducible as an $H$-rep, so we are done by induction.

If $W$ is one dimensional, then $H$ acts on $V$ by scalars, so $H$ is central. So we are in our base case: a central extension of a simple group. This case is done by group cohomology and the classification of simple groups. See section 4 for groups of Lie type, section 6 for alternating groups and section 7 for sporadic groups.

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By averaging, over $\mathbb R$ every representation of a finite group can be given a positive definite inner product so that the action is orthogonal. And my memory is that Schur functurs of representations of $\operatorname{O}(n)$ are not usually irreducible. For example, $\operatorname{Sym}^2(\mathbb R^n)$ has a one-dimensional summand over $\operatorname{O}(n)$, given by the trace w.r.t. the inner product on $\mathbb R^n$. This suggests to me that over any finite group, the essential image of $\operatorname{Sym}^2$ contains very few irreps. But maybe I am in error.

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It sounds like you mean that Sym^2(V) is reducible unless dim(V) = 1? SL(2,3) has some 2 dim reps with Sym^2 irreducible. This appears somewhat common. Checking a standard list of character tables, one sees there are examples for Sym^2 in dimensions 2,3,4,5,6,8,9,10,11,12,13,14,18,20,21,26,28,32,41,42,43,45,60,342,1333; examples for Sym^3 in dimensions 2,3,4,5,6,8,9,10,12,13,14,18,20,32; examples for Sym^4 in dimensions 2,4,6,12; and examples for Sym^5 in dimensions 2,4,6,12. There were no examples for Sym^6 as indicated by Guralnick and Thiep. Maybe this is a U(n) versus O(n) problem? –  Jack Schmidt Jul 19 '10 at 12:33
Yeah, I think all this is saying is that the Frobenius–Schur indicator indicates whether the trivial rep is a summand of the 2nd symmetric power, the 2nd exterior power, or neither. O(n) reps means it is a summand of the 2nd symmetric power. It's pretty common for an irrep not to be realized in O(n). –  Jack Schmidt Jul 19 '10 at 13:21
Ah, so I was in error. It was late last night, after some wine, that I wrote my answer :) –  Theo Johnson-Freyd Jul 19 '10 at 17:03
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