Timeline for Are there "unsociable" irreps? (Definition inside)

7 events

when toggle format	what		by	license	comment
Jun 15, 2020 at 7:27	history	edited	CommunityBot		Commonmark migration
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 14, 2012 at 19:16	comment	added	Qiaochu Yuan		Instead of assuming $G$ semisimple you can assume (after assuming WLOG that $R$ is unitary) that the image of $G$ in $\text{Aut}(R)$ is contained in $\text{SU}(R)$. Then $R^{\ast} \cong \Lambda^{d-1}(R)$ where $d = \dim R$ and your argument from the other thread works.
Aug 14, 2012 at 18:45	comment	added	David E Speyer		I agree that looking at $R^{otimes m} \otimes (R^{\vee})^{\otimes n}$ is more useful than looking just at $R^{\otimes m}$, but its not what the OP asked for.
Aug 14, 2012 at 18:44	comment	added	David E Speyer		Putting in the dual rep will allow you to discard the semi-simple hypothesis (and replace it by either reductiveness or compactness), see my answer to mathoverflow.net/questions/58633
Aug 14, 2012 at 18:23	comment	added	Jim Humphreys		@David: Note my reference to Chevalley, where the dual representation is also used. Your approach will need this extra ingredient at some point. Concerning your Comments 4 and 5, I'm unaware of further work in those directions (especially the last).
Aug 14, 2012 at 17:15	history	answered	David E Speyer	CC BY-SA 3.0