Timeline for Quotient of an algebraic group by a closed algebraic subgroup
Current License: CC BY-SA 3.0
14 events
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| Sep 1, 2016 at 7:11 | answer | added | Akira Masuoka | timeline score: 3 | |
| Feb 7, 2014 at 20:46 | vote | accept | Jesko Hüttenhain | ||
| Feb 7, 2014 at 20:46 | comment | added | Jesko Hüttenhain | Anways, since this is in Mr. Humphreys' book, my question is void (and more importantly, my confusion lifted). I will accept Peter Crooks' answer so this is no longer an open question. Thanks everyone. | |
| Feb 7, 2014 at 20:14 | history | edited | Jesko Hüttenhain | CC BY-SA 3.0 |
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| Feb 7, 2014 at 20:11 | comment | added | Jesko Hüttenhain | @Jim Humphreys and Piotr Achinger: My previous reference was this paper by Chevalley, see Proposition 8 and his comment in the very last paragraph below that. I apparently skipped over the part where he requires the subgroup to be normal. | |
| Feb 7, 2014 at 20:04 | history | edited | Jesko Hüttenhain | CC BY-SA 3.0 |
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| Feb 7, 2014 at 19:14 | comment | added | Jim Humphreys | @Daniel: I basically copied Chevalley/Borel. The idea is elementary (once Chevalley thought of it): realize $H$ as isotropy group of a line in a representation $G \rightarrow \mathrm{GL}(V)$, then pass to the projective space $P(V)$ and embed $G/H$ there. | |
| Feb 7, 2014 at 18:12 | comment | added | Daniel Barter | This is proved in linear algebraic groups by Humphreys. The basic idea is that you can find a representation of $G$ such that the stabilizer of a specific line is exactly $H$ | |
| Feb 7, 2014 at 17:43 | comment | added | Piotr Achinger | And for general $G$ it's true if $H$ is normal ;) | |
| Feb 7, 2014 at 16:45 | comment | added | Jim Humphreys | @Jesko: Note that working over the complex field is unimpotant here. In general, the quotient of a reductive group by a closed subgroup is affine if and only if the subgroup is also reductive. | |
| Feb 7, 2014 at 16:20 | answer | added | Peter Crooks | timeline score: 4 | |
| Feb 7, 2014 at 16:17 | comment | added | Peter Crooks | I think you want to claim that $G/H$ must be quasi-projective. | |
| Feb 7, 2014 at 15:43 | comment | added | David E Speyer | The statement isn't true. Take $G=SL_2$ and $H = \left( \begin{smallmatrix} 1 & \ast \\ 0 & 1 \end{smallmatrix} \right)$. The quotient $G/H$ is isomorphic to $\mathbb{A}^2 \setminus \{ (0,0) \}$, which is not affine. | |
| Feb 7, 2014 at 15:33 | history | asked | Jesko Hüttenhain | CC BY-SA 3.0 |