Timeline for If a representation has enough reductive stabilizers, is it a direct sum of characters?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2011 at 20:12 | answer | added | Anton Geraschenko | timeline score: 4 | |
| Mar 26, 2011 at 2:33 | vote | accept | Anton Geraschenko | ||
| Mar 25, 2011 at 22:08 | comment | added | Anton Geraschenko | I realized that my Remark 1 is cryptic/wrong. In my situation, $C$ is the closure of a $G$-orbit which is isomorphic to $G$. Just knowing that $C$ is the closure of a $G$-orbit doesn't allow you to reduce to this case in general. | |
| Mar 25, 2011 at 19:22 | comment | added | Anton Geraschenko | @Peter: sorry about that. I really care about the char 0 case. I "removed it" to emphasize that the proof in the case $C=V$ does not use it. If I had thought more carefully when moving the algebraically closed condition, I guess I would have said "geometric point," but again, I really care about the algebraically closed case. | |
| Mar 25, 2011 at 18:13 | comment | added | Peter McNamara | I wrote my answer before alg. closed. char 0 condition was dropped. If k is not alg.closed, what do you mean by a point of V? If you only mean k-point, then G=units of division algebra acts on division algebra with reductive stabalisers. | |
| Mar 25, 2011 at 18:06 | answer | added | Peter McNamara | timeline score: 6 | |
| Mar 25, 2011 at 17:54 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
added 152 characters in body; edited title
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| Mar 25, 2011 at 17:39 | comment | added | Anton Geraschenko | @Jim: You're right that if the answer is "yes" then $G$ is a torus if the representation is faithful. I suppose I should have mentioned this, but I don't see how to actually use this fact to modify the phrasing of the problem. Thanks for the references. I only knew the result in the case when $G$ is linearly reductive, which is why I left that word there. I suppose I'm simultaneously broadcasting my general preference not to assume characteristic 0 and my lack of understanding of (non-linearly) reductive groups. I'll edit the question. | |
| Mar 25, 2011 at 11:07 | comment | added | Jim Humphreys |
Side remark: For a closed subgroup $H$ of a reductive group $G$ in any characteristic, it's true that $G/H$ is affine iff $H$ is reductive. The history is complicated, but in char 0 the hard direction goes back to Matsushima and in char $p$ Mumford's Conjecture (proved by Haboush) is needed. See R.W. Richardson, Bull. London Math. Soc. 9 (1977).
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| Mar 25, 2011 at 10:46 | comment | added | Jim Humphreys | I'm confused by the formulation of the question. If the answer is yes (as expected), then the group must be just a torus if the representation is faithful (?) The wording is also too elaborate, since "linearly reductive" = "reductive" in characteristic 0. | |
| Mar 25, 2011 at 7:00 | comment | added | Angelo | Your proof for the case $C = V$ ia quite nice. | |
| Mar 25, 2011 at 4:37 | history | asked | Anton Geraschenko | CC BY-SA 2.5 |