Timeline for is the subgroup generated by one-parameter unipotent subgroups a Lie subgroup?
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2021 at 18:14 | comment | added | YCor | How is defined "unipotent" in a purely Lie context? | |
| Feb 23, 2021 at 17:02 | answer | added | Zongzhu Lin | timeline score: 0 | |
| Mar 31, 2011 at 12:35 | answer | added | Jim Humphreys | timeline score: 3 | |
| Nov 24, 2010 at 22:20 | comment | added | David Roberts♦ | @BCnrd Ah, of course. That what you get for rushing in like a fool. | |
| Nov 24, 2010 at 20:26 | answer | added | zroslav | timeline score: 0 | |
| Nov 24, 2010 at 13:25 | history | edited | ronggang | CC BY-SA 2.5 |
added 10 characters in body
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| Nov 24, 2010 at 9:26 | comment | added | André Henriques | The word "unipotent" appears in the title of the question, but not in the body... (?) | |
| Nov 24, 2010 at 7:47 | comment | added | ronggang | I am reading Ratner's results on the rigidity theorems and do not know whether the generated there is in group sense or the closure of it. | |
| Nov 24, 2010 at 7:04 | comment | added | BCnrd | Dear David: even with a finite number, they will hardly ever commute with each other. So there is no homomorphism from $\mathbf{R}^n$ built from the given collection of 1-parameter subgroups. One has to use nontrivial facts from the structure theory of Lie groups (e.g., relation of root datum to birational group law on open cell in the reductive case) to analyze this kind of question. | |
| Nov 24, 2010 at 6:50 | comment | added | David Roberts♦ | Actually I meant to say, assuming that you only are dealing with a finite number of 1-parameter subgroups, but forgot. If you are having to deal with an infinite number of such, then I have nothing constructive to say. | |
| Nov 24, 2010 at 6:43 | comment | added | BCnrd | If you work with Lie groups associated to split connected reductive algebraic groups over $\mathbf{R}$ and you assume that the 1-parameter subgroups are normalized by a common split maximal torus then there are nice affirmative answers expressed in terms of "positivity" and "closedness" of the corresponding sets of roots. So one is led to request a clarification: are you willing to impose some structure on the situation (e.g., reductivity and being normalized by the action of a suitable maximal torus), or do you insist on working in total generality? Briefly, what is your motivation? | |
| Nov 24, 2010 at 6:41 | comment | added | Mariano Suárez-Álvarez | Since not all one-parameter subgroups are closed, the answer to your last question is no (consider the subgroup generated by exactly one one parameter subgroup which is not closed...) | |
| Nov 24, 2010 at 6:27 | comment | added | Vivek Shende | how do you know that you get to stop at some finite $\mathbb{R}^n$ ? | |
| Nov 24, 2010 at 6:18 | comment | added | David Roberts♦ | I would think of H as really being a (smooth) homomorphism $\phi:\mathbb{R}^n \to G$ (much as a 1-parameter subgroup is really a homomorphism $\mathbb{R} \to G$ en.wikipedia.org/wiki/One-parameter_group). The image of $\phi$ should be a Lie subgroup. I don't know what you mean by 'general subgroups'. | |
| Nov 24, 2010 at 6:04 | history | asked | ronggang | CC BY-SA 2.5 |