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Jul 20, 2010 at 0:44 history edited Qiaochu Yuan CC BY-SA 2.5
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Jul 19, 2010 at 18:45 vote accept Qiaochu Yuan
Jul 19, 2010 at 17:17 comment added Jim Humphreys The restated question still isn't quite clear to me, being worded in terms of an irreducible representation V of a finite (rather than general linear) group. Anyway, the arXiv preprint by Guralnick and Tiep is located at math.GR/0502080, to appear in the electronic journal Representation Theory.
Jul 19, 2010 at 14:32 answer added David E Speyer timeline score: 3
Jul 19, 2010 at 4:30 answer added Theo Johnson-Freyd timeline score: 1
Jul 18, 2010 at 22:57 comment added moonface 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.
Jul 18, 2010 at 22:40 comment added Qiaochu Yuan I have edited the statement of the question to reflect my motivation more accurately.
Jul 18, 2010 at 22:40 history edited Qiaochu Yuan CC BY-SA 2.5
Improved the statement of the question.
Jul 18, 2010 at 22:31 comment added Qiaochu Yuan @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.
Jul 18, 2010 at 22:08 comment added Jim Humphreys 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).
Jul 18, 2010 at 22:03 comment added Ben Webster I hope you don't mind too much that I made a little edit for clarity. I found the original statement very confusing.
Jul 18, 2010 at 21:26 history edited Ben Webster CC BY-SA 2.5
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Jul 18, 2010 at 20:49 history asked Qiaochu Yuan CC BY-SA 2.5