Timeline for Extension of unipotent algebraic groups
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2013 at 17:13 | vote | accept | Xingting | ||
| Feb 15, 2013 at 2:54 | comment | added | user30379 | What are the implicit smoothness hypotheses on $G$ and $N$? (One doesn't need any such hypotheses for the affirmative answer to the question, but it affects the extent to which one can give an entirely elementary proof or not.) Also, the OP should clarify what they wish to take as the initial definition of "unipotence" (and under what smoothness hypotheses). As many readers here know well, there are multiple different-looking definitions, all ultimately equivalent, but the choice of initial definition affects which basic facts are easy to prove and which require more development. | |
| Feb 15, 2013 at 1:24 | comment | added | anon | @Jim In order to define the unipotent radical, you need to know enough about unipotent groups to answer the question. And connectedness is not a problem in any characteristic, at least, not if you are talking about algebraic group schemes. | |
| Feb 15, 2013 at 1:06 | comment | added | Jim Humphreys |
@Xingting: It's best to specify whether or not $G$ is connected though it's not an issue in characteristic 0. Anyway, the answer is definitely yes. If you already have the full Borel-Chevalley structure theory of (connected) linear algebraic groups in hand, it's an easy consequence: modulo the unipotent radical of $G$, you get a reductive group (which can't be unipotent if nontrivial). Naturally you might prefer short-cuts.
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| Feb 15, 2013 at 0:36 | answer | added | anon | timeline score: 3 | |
| Feb 15, 2013 at 0:24 | comment | added | anon | Yes, see for example Milne's notes AGS, XV, 2.5. | |
| Feb 15, 2013 at 0:05 | history | asked | Xingting | CC BY-SA 3.0 |