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Apr 26, 2010 at 3:31 comment added pasquale zito @David: you are absolutely right, silly me!
Apr 26, 2010 at 0:36 comment added David Jordan Thank you for clarifying, but I'm still confused. I see that the Hom condition implies that V is in the identity class. But I don't see why every V in the identity class must be contained in W^* ot W for an irreducible W. It seems that V could be contained in W^* ot W ot W^* ot W (here "ot" is shorthand for \otimes). Is there a statement about strings of length 2 in the irreps? Sorry if I'm being dense.
Apr 26, 2010 at 0:27 comment added pasquale zito @Qiaochu: I replaced "compare" with "appear" (sorry, I mixed words from different languages). @David: it is enough to find one such W to conclude that V is in the identity class, you don't need to check it for all strings.
Apr 25, 2010 at 23:58 history edited pasquale zito CC BY-SA 2.5
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Apr 25, 2010 at 23:28 comment added David Jordan I'm concerned. He only wants to allow $W^*\otimes W$, and not a string of products of such. If he allows strings of size larger than two, then I think it's well known for $G$ finite that $V$ descends to a rep of $G/Z$, since $G/Z$ acts faithfully on $C[G]$, so that every $G/Z$ rep appears in some tensor power of $C[G]$.
Apr 25, 2010 at 23:21 comment added Qiaochu Yuan This sounds great, but I'm not sure what the meaning of "compare" is in your first sentence.
Apr 25, 2010 at 22:57 history answered pasquale zito CC BY-SA 2.5