Timeline for Jordan decomposition of elements in non-connected component of algebraic group
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2015 at 4:51 | vote | accept | Anupam Singh | ||
| Nov 15, 2015 at 15:00 | answer | added | Jim Humphreys | timeline score: 2 | |
| Oct 15, 2015 at 4:40 | answer | added | Anupam Singh | timeline score: 1 | |
| Oct 1, 2015 at 3:12 | comment | added | Anupam Singh | That's a good idea. In finite group Jordan decomposition would correspond to looking at the power of the element, the p powers and co-prime to p factor. That's reasonable. | |
| Sep 30, 2015 at 11:21 | comment | added | anon | The homomorphism of algebraic groups $G\to G/G^{\circ}$ is compatible with Jordan decompositions, and so this is just a question of understanding the Jordan decompositions in the finite group (scheme) $G/G^{\circ}$. | |
| Sep 30, 2015 at 9:08 | comment | added | Anupam Singh | right! usually finite order elements are semisimple (in non-modular set up) so one expects that. | |
| Sep 30, 2015 at 8:41 | comment | added | Nick Gill | My first thought: if you have a coset $gG_0$, then you could only hope to have a semisimple representative if the field characteristic doesn't divide the order of $gG_0$ (as an element of $G/G_0$)... I wouldn't like to speculate whether this necessary condition is also sufficient... | |
| Sep 30, 2015 at 8:25 | history | edited | Anupam Singh | CC BY-SA 3.0 |
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| Sep 30, 2015 at 8:22 | comment | added | Anupam Singh | thanks! actually my problem is "can I choose F consisting of semisimple elements only?". Which often looks like the case. | |
| Sep 30, 2015 at 5:42 | comment | added | Venkataramana | there is no reason for $g$ to be semi-simple; you can take $G=G^0\times F$ where $F$ is a finite group, and take $g=(u,f)$ for some unipotent element $u\in G^0$. | |
| Sep 30, 2015 at 4:51 | history | asked | Anupam Singh | CC BY-SA 3.0 |