Timeline for is the subgroup generated by one-parameter unipotent subgroups a Lie subgroup?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2021 at 18:15 | comment | added | YCor | What do you mean by "solvable group" in your answer? You probably mean solvable linear algebraic group. | |
| Mar 19, 2011 at 21:18 | comment | added | zroslav | @ronggagng: I don't understand your comment | |
| Nov 29, 2010 at 2:12 | comment | added | ronggang | Another correction: The above is true in case of two parabolic $k$-subgroups containing a common minimal parabolic $k$-subgroup of ...........; | |
| Nov 29, 2010 at 2:11 | comment | added | ronggang | A correction: $(M_1M_2)(k)=M_1(k)M_2(k)$ | |
| Nov 29, 2010 at 2:10 | comment | added | ronggang | Dear Zroslav: This theorem only holds for the rational points over an algebraically closed field. It seems not true that $(M_1M_2)(k)=(M_1M_2)(k)$ which is true in some special cases, e.g. two parabolic $k$-subgroups of a connected reductive $k$-group. | |
| Nov 27, 2010 at 20:38 | comment | added | zroslav | I have no english edition, only russian. The theorem stement is: If subgroup H of algebraic group G is generated by the set of irreducible algebraic subsets $M_a$ (dense in their closures and containing 1) then it is an irreducible algebraic subgroup. Idea of the proof: the consequence $A_n$ of sets of $g_1\ldots g_n$ where $g_i\in M_{a_i}$ or $\in (M_{a_i})^{-1}$ is "growing" (in the sense of including: $A_n\subset A_{n+1}$) and stabilizing on the step N. Every subgroup that dense in its closure is closed. Is that clear? | |
| Nov 27, 2010 at 5:13 | comment | added | ronggang | Dear Zroslav: Which theorem do you mean in the english edition: Lie groups and algebraic groups. | |
| Nov 25, 2010 at 16:15 | comment | added | zroslav | The answer is "yes" for any algebraic group over $\mathbb C$. Your unipotent subgroups generate the closed subgroup $H_{\mathbb C}$ of $G_{\mathbb C}$ (this is the corollary of Theorem 3.1.4 in Vinberg, Onishchik, "Seminar on Lie groups and algebraic groups"). I'd tried to prove the common statement for real Lie groups by complexification, but for an hour I had no proof... | |
| Nov 25, 2010 at 14:53 | comment | added | zroslav | Ah, I understand... See my answer below | |
| Nov 25, 2010 at 1:02 | comment | added | BCnrd | The question asks about using "some" unipotent subgroups, without any specification of their properties (e.g., "algebraicity", normalized by some kind of maximal torus, generation in the algebraic or group-theoretic senses over $\mathbf{R}$, etc.), so it doesn't seem likely (based on the motivation from Ratner's work) that the OP is asking if $G$ is generated by unipotent subgroups (as is the case for split connected reductive groups, for example). The above answer may therefore be addressing a different question than the one intended. | |
| Nov 24, 2010 at 20:26 | history | answered | zroslav | CC BY-SA 2.5 |