Timeline for Nilpotent elements of Lie algebra and unipotent groups
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 21, 2019 at 18:15 | comment | added | Jdoe | @LSpice Ok I'll keep that in mind :) Thanks for you help ! | |
| Jul 21, 2019 at 18:05 | comment | added | LSpice | @Jdoe, the import of Lemma 8.3 which I cite is indeed that you gain nothing in a technical sense from this additional knowledge; whether it helps in your applications I don't know, but it might be convenient to have! | |
| Jul 21, 2019 at 18:04 | comment | added | LSpice | @YCor, you are right. I did not mean to claim that you said that, only that the problem asked for it; but I was wrong. I automatically read "unipotent subgroup" as "unipotent radical of a parabolic subgroup." | |
| Jul 21, 2019 at 15:13 | comment | added | YCor | @LSpice I have no idea why you claim that I said that $H$ is unipotent radical of a parabolic subgroup. There's no parabolic subgroup in the question either. | |
| Jul 21, 2019 at 13:20 | comment | added | Jdoe | Thank you very much YCor ! Thank you also for the comment on the non-intrinsicness of nilpotency it corrected a big misconception of mine :) @LSpice I think I am happy with just having $X$ be in the Lie algebra of a unipotent subgroup. Would I gain something by knowing that it is in the Lie algebra of the unipotent radical of a Parabolic subgroup ? | |
| Jul 21, 2019 at 13:15 | vote | accept | Jdoe | ||
| Jul 21, 2019 at 12:24 | comment | added | LSpice | Why is $H$ the unipotent radical of a (rational) parabolic subgroup of $G$? Or maybe you mean to enlarge it to such a radical, in which case I guess that you would still need something like Lemma 8.3 of Borel and Tits - Groupes réductifs. | |
| Jul 21, 2019 at 12:18 | history | edited | YCor | CC BY-SA 4.0 |
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| Jul 21, 2019 at 12:11 | history | answered | YCor | CC BY-SA 4.0 |