Let $G$ be a Lie group and $H$ be a subgroup generated by some one parameter unipotent subgroups (in group sense). Is it true that $H$ has a Lie group structure which makes it a Lie subgroup of $G$? Is $H$ closed?
How about for general subgroups?
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$\begingroup$ 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'. $\endgroup$– David Roberts ♦Commented Nov 24, 2010 at 6:18
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$\begingroup$ how do you know that you get to stop at some finite $\mathbb{R}^n$ ? $\endgroup$– Vivek ShendeCommented Nov 24, 2010 at 6:27
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2$\begingroup$ 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...) $\endgroup$– Mariano Suárez-ÁlvarezCommented Nov 24, 2010 at 6:41
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1$\begingroup$ 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? $\endgroup$– BCnrdCommented Nov 24, 2010 at 6:43
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2$\begingroup$ 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. $\endgroup$– BCnrdCommented Nov 24, 2010 at 7:04
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3 Answers
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The many comments as well as the answer by zroslav probably add to the confusion resulting from the original unfocused formulation of the question. First, it's not really a question about Lie groups but about algebraic groups: In a Lie group there is generally no intrinsic Jordan decomposition, hence no intrinsic notion of unipotent subgroup as seen already in dimension 1.
The algebraic group theory was mostly shaped by Borel and Chevalley, both of whom were motivated in part by an interest in Lie groups. But a Lie group is initially a real manifold, whereas algebraic groups start out over algebraically closed fields and then get more complicated relative to smaller fields of definition. For an algebraic group, taken in the Zariski topology, "irreducible" = "connected". Here a finite matrix group can be viewed as an algebraic group but is connected only when trivial. In any case, an affine algebraic group can always be embedded in some general linear group, so over $\mathbb{C}$ you get a Lie subgroup of the complex general linear group (viewed as real). The point is that being Zariski-closed implies being closed in the usual sense.
An early result of Chevalley (treated in several books with the title Linear Algebraic Groups, including section 7.5 of my book) shows the importance of conditions on irreducibility for study of a group generated by closed subsets of an algebraic group. One consequence of Chevalley's theorem is that two closed connected subgroups will generate a closed connected subgroup in a nice way, but the theorem has other applications as well. All of this comes very early in the theory, before Jordan decomposition and the detailed structure of reductive groups, where you can ask more interesting specific questions about what various subgroups generate.
answered Mar 31, 2011 at 12:35
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The answer is "yes" for solvable groups: there is a decomposition of any solvable group on its maximal tori and unipotent radical. For semi-simple groups the answer is "yes": I've read this statement in Humphrey, "Introduction to Lie Algebras and Representation Theory" (H coincides with G). In the general case the answer is also "yes": there is a Levy decomposition of any Lie group on its maximal reductive subgroup and its unipotent radical.
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$\begingroup$ 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. $\endgroup$– BCnrdCommented Nov 25, 2010 at 1:02
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$\begingroup$ Ah, I understand... See my answer below $\endgroup$– zroslavCommented Nov 25, 2010 at 14:53
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$\begingroup$ 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... $\endgroup$– zroslavCommented Nov 25, 2010 at 16:15
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$\begingroup$ Dear Zroslav: Which theorem do you mean in the english edition: Lie groups and algebraic groups. $\endgroup$– ronggangCommented Nov 27, 2010 at 5:13
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$\begingroup$ 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? $\endgroup$– zroslavCommented Nov 27, 2010 at 20:38
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A good reference to Rongguang's questions is Proposition 2.2.6 in Springer's book: Linear Algebraic Groups 2nd Edition.
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1$\begingroup$ Please add details. A reference-only answer is not encouraged on MO. $\endgroup$ Commented Feb 23, 2021 at 17:47