Timeline for Is the subgroup generated by a conjugacy class of semisimple elements Zariski closed?
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| Feb 11, 2019 at 0:21 | history | edited | YCor |
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| Aug 3, 2018 at 12:45 | comment | added | LSpice | Per @YCor's answer, you should probably specify that you're speaking of the $G(k)$- (i.e., rational) conjugacy class of $s$, not its algebraic conjugacy class. | |
| Feb 5, 2018 at 9:09 | vote | accept | Michiel Van Couwenberghe | ||
| Feb 1, 2018 at 11:02 | answer | added | YCor | timeline score: 4 | |
| Feb 1, 2018 at 9:49 | comment | added | Michiel Van Couwenberghe | @YCor Do you mean that in that case it will never be true of that there exist counterexamples? Can you be more precise with your counterexample in $PGL_2$? I find it hard to see whether a subgroup is closed but that is probably because I am not aware of some relevant Theorems... Probably you want $X$ to generate $PSL_2$ or so which is not closed? | |
| Feb 1, 2018 at 9:45 | comment | added | Michiel Van Couwenberghe | @S.Carnahan Is there nothing more to it? Is an algebraic subgroup really the same as a closed subgroup? Or do you use something like the Theorem 1.45 that I mentioned in my question? | |
| Jan 31, 2018 at 17:28 | comment | added | YCor | If $s$ belongs to the subgroup of $G(k)$ generated by unipotents and the latter is not equal to $G(k)$ then the answer is no. For instance, $PGL_2$ over any field in which some element is not a square yields counterexamples. | |
| Jan 31, 2018 at 14:05 | comment | added | S. Carnahan♦ | You can certainly speak of the algebraic subgroup generated by $X$. It is the intersection of all closed subgroups containing $X$. | |
| Jan 31, 2018 at 13:21 | review | First posts | |||
| Jan 31, 2018 at 13:28 | |||||
| Jan 31, 2018 at 13:18 | history | asked | Michiel Van Couwenberghe | CC BY-SA 3.0 |