Timeline for For a representation V of a finite group G, when is Hom(W, W⊗V) trivial for all irreps W?
Current License: CC BY-SA 2.5
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Apr 26, 2010 at 4:44 | comment | added | Qiaochu Yuan | I can prove that a nontrivial finite subgroup of SU(2) has nontrivial center without the classification, which means that your answer is what I need; thanks! The proof is as follows: if V is irreducible, then dim V = 2 divides the order of G, so -I is in G. Otherwise, V is a sum of two one-dimensional representations and G is abelian. | |
| Apr 26, 2010 at 4:19 | vote | accept | Qiaochu Yuan | ||
| Apr 25, 2010 at 23:22 | comment | added | Qiaochu Yuan | I'm using this result as a lemma in the proof of the classification, so I'm afraid I can't use that result. | |
| Apr 25, 2010 at 23:12 | comment | added | Dylan Thurston | Sorry, I misread your question. I edited my answer, but I did resort to classification of the finite subgroups of SU(2). | |
| Apr 25, 2010 at 23:12 | history | edited | Dylan Thurston | CC BY-SA 2.5 |
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| Apr 25, 2010 at 22:25 | comment | added | Qiaochu Yuan | Ah. Yes; I had also come to this conclusion by playing around with characters. Can it be shown that a finite subgroup of SU(2) has nontrivial center without already knowing what they are? | |
| Apr 25, 2010 at 22:14 | comment | added | David Jordan | He's answering the contra-positive. If $Hom(V,W^*\otimes W)$ is non-zero, then Z(G) acts trivially on $V$, you agree? | |
| Apr 25, 2010 at 22:13 | comment | added | Dylan Thurston | Perhaps you were confused about which direction I was proving? I tried to clarify. As a special case of what I wrote (or what Ben explained), for any non-trivial rep $V$ of any abelian group, $\mathrm{Hom}(W, W \otimes V)$ is always trivial. | |
| Apr 25, 2010 at 22:12 | history | edited | Dylan Thurston | CC BY-SA 2.5 |
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| Apr 25, 2010 at 22:02 | comment | added | Qiaochu Yuan | The center of the cyclic group of order n certainly doesn't act trivially, but the result is still true for that case...? | |
| Apr 25, 2010 at 21:53 | history | answered | Dylan Thurston | CC BY-SA 2.5 |