Timeline for Taking roots in simple linear algebraic groups
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Oct 26, 2016 at 18:20 | comment | added | LSpice | I guess that, over another algebraically closed field of characteristic 0 (say, $\overline{\mathbb Q}$), one can replace your use of the usual exponential map with something like Bourbaki, Groupes de Lie, Théorème III.4.3.4, which guarantees the existence of an exponential map for any 'groupuscule'. | |
| Feb 10, 2010 at 18:52 | comment | added | Kevin McGerty | Oops yes thanks, I meant to spell that out in my answer -- it's a simpler example than the one in the paper mentioned in Kovalev's comment (which uses the $SL_2\times SL_2$ subgroup in $Sp_4$). | |
| Feb 10, 2010 at 17:41 | comment | added | Pavel Etingof | Now I understand. So, the simplest example is the non-semisimple element of $SL(2)$ with eigenvalue -1. Thanks! | |
| Feb 10, 2010 at 15:23 | comment | added | Kevin McGerty | My post yesterday was pretty incoherent, but hopefully the edit is better! | |
| Feb 10, 2010 at 15:20 | history | edited | Kevin McGerty | CC BY-SA 2.5 |
Removed the fuzzy train of thought which led me to the condition about centralizers, and added why this (should!) give counterexamples.
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| Feb 10, 2010 at 1:59 | comment | added | Pavel Etingof | I also did not understand this argument. The group $Z$ could be smaller than $Z_G(x)$, so it could contain a central torus, even if $Z_G(x)$ does not. E.g. take $x$ to be central in $G$ (say, $-1$ in $SL(2)$). | |
| Feb 10, 2010 at 0:41 | comment | added | blt | I may be being stupid, but this appears to contradict P. Etingof's answer above. Is there something I'm missing here? | |
| Feb 9, 2010 at 21:43 | vote | accept | blt | ||
| Feb 9, 2010 at 21:43 | |||||
| Feb 9, 2010 at 21:42 | vote | accept | blt | ||
| Feb 9, 2010 at 21:43 | |||||
| Feb 9, 2010 at 21:41 | history | answered | Kevin McGerty | CC BY-SA 2.5 |