Timeline for When does an irreducible representation remain irreducible after restriction to a semi-simple subgroup?
Current License: CC BY-SA 3.0
14 events
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| S Dec 11, 2016 at 2:16 | history | edited | David Handelman | CC BY-SA 3.0 |
add tag correct title; fixed/deleted articles.
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| S Dec 11, 2016 at 2:16 | history | suggested | Henry.L | CC BY-SA 3.0 |
add tag correct title
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| Dec 11, 2016 at 1:34 | review | Suggested edits | |||
| S Dec 11, 2016 at 2:16 | |||||
| Dec 11, 2016 at 1:00 | answer | added | anonymous | timeline score: 1 | |
| Sep 15, 2010 at 16:15 | vote | accept | genshin | ||
| Sep 15, 2010 at 14:46 | comment | added | Victor Protsak | I suggest that you re-post your comment from 13:29 as a new question. It is very clear and does not contain several paragraphs of heavy notation interspersed with wrong claims to wade through (and MO is not the best place to check proofs). | |
| Sep 15, 2010 at 14:41 | comment | added | Victor Protsak | The claim is false. To amplify on the comments, consider $G=SL_n, H=SO_n$ for $n\geq 3$ (for any form of the orthogonal group), then $Z(H,G)$ is 0-dim and hence anisotropic over any field. A simple $G$-module with highest weight $\lambda$ remains simple upon restriction to $H$ is and only if $\lambda$ is a fundamental weight (so that the module is an exterior power of the defining module of $SL_n$). Certainly the restriction of $S^2$ to $H$ contains an invariant vector. A similar example is $G=SL_{2n}, H=Sp_{2n}:$ only symmetric powers of the defining module or its dual remain simple. | |
| Sep 15, 2010 at 10:21 | answer | added | Sheikraisinrollbank | timeline score: 4 | |
| Sep 15, 2010 at 10:13 | comment | added | Sheikraisinrollbank | I don't understand your comment (in the last paragraph) about $SO_3 \subseteq SO_4$; certainly the standard representation of $SO_4$ on $\mathbb{R}^4$ is not irreducible when restricted to $SO_3$. The next question about the branching rule doesn't seem to make much sense either: of course, if every irreducible for $G$ remains irreducible when restricted to $H$ then the branching rule is not very interesting. Can you clarify? | |
| Sep 15, 2010 at 8:57 | history | edited | genshin | CC BY-SA 2.5 |
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| Sep 15, 2010 at 8:32 | comment | added | Bruce Westbury | I don't see how every irreducible representation of $G$ could restrict to an irreducible representation of $H$. Could you give an example? | |
| Sep 15, 2010 at 8:26 | history | edited | Bruce Westbury | CC BY-SA 2.5 |
Spelling corrected
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| Sep 15, 2010 at 8:17 | history | edited | genshin | CC BY-SA 2.5 |
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| Sep 15, 2010 at 8:12 | history | asked | genshin | CC BY-SA 2.5 |