Timeline for Conjugacy classes in the automorphism group of a simple Lie algebra
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2022 at 12:29 | vote | accept | Youness EL KHARRAF | ||
| Apr 24, 2022 at 11:27 | answer | added | YCor | timeline score: 4 | |
| Apr 23, 2022 at 23:26 | comment | added | Youness EL KHARRAF | Dear @YCor, if you repost your comment as an answer for the infinity of the image of the trace, and so for the existence of infinity conjugacy classes in $\operatorname{Aut}(\mathfrak{s})$. I will accept it as an anwer. | |
| Apr 23, 2022 at 22:51 | answer | added | Youness EL KHARRAF | timeline score: 2 | |
| Apr 22, 2022 at 23:09 | comment | added | YCor | Over a perfect field, $\mathrm{Aut}(\mathfrak{s})^0$ is Zariski dense in the underlying algebraic group (Rosenlicht), so if the trace had a finite image, it would be constant and also constant over the algebraic closure. But this would contradict the assertion for split $\mathfrak{s}$. | |
| Apr 22, 2022 at 22:02 | comment | added | Youness EL KHARRAF | Thanks for the counterexample. Instead, in characateristic zero or for an infinite field, $\operatorname{im}\operatorname{tr}(\operatorname{Aut}(\mathfrak{s}))$ is it infinite? | |
| Apr 22, 2022 at 5:55 | comment | added | YCor | For the real field and a compact Lie group, the trace is bounded, hence not surjective. | |
| Apr 22, 2022 at 5:54 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals
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| S Apr 22, 2022 at 3:05 | review | First questions | |||
| Apr 22, 2022 at 7:24 | |||||
| S Apr 22, 2022 at 3:05 | history | asked | Youness EL KHARRAF | CC BY-SA 4.0 |