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Cartan's theorem states that any topologically closed subgroup of a Lie group is an embedded Lie subgroup.

This leads us to ask the following question:

Can we replace "topologically closed" with a different topological property and achieve the same result? For instance, is a semi-locally simply connected subgroup of a Lie group an embedded Lie subgroup? Is a locally connnected and semi-locally simply connected subgroup of a Lie group an embedded Lie subgroup?

Some observations: An arcwise connected subgroup of a Lie group is not always an embedded Lie subgroup. For instance, consider the following example taken from http://en.wikipedia.org/wiki/Lie_subgroup:

"...take G to be a torus of dimension ≥ 2, and let H be a one-parameter subgroup of irrational slope, i.e. one that winds around in G. Then there is a Lie group homomorphism φ : R → G with H as its image. The closure of H will be a sub-torus in G."

This example is an arc-wise connected (but not locally connected) subgroup of a Lie group that is not an embedded Lie subgroup. The issue is that in the definition of an embedded Lie subgroup you require that the subgroup be nice with respect to the subset topology, in order for the Lie subgroup to be an embedded submanifold. See the section on embedded submanifolds in

http://en.wikipedia.org/wiki/Submanifold

So whatever topological constraint we use to replace "closed" it has to be stronger than arcwise-connectedness.

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At the risk of asking a silly question: why would you expect "semi-locally simply connected" to be able to replace "closed"? The only ever time I've come across that property is in the theorem about the existence of universal coverings, but probably I'm missing something... –  José Figueroa-O'Farrill Feb 27 '10 at 19:15
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Thanks for the question. My explanation is that Lie subgroups locally look like a single plane. So one might expect that as long as we rule out "local disconnectivity" and "small loops" we might have a subgroup that locally resembles a plane enough to be a Lie subgroup. –  Khalid Bou-Rabee Feb 27 '10 at 20:39
    
Thanks for the clarification. I believe that this condition is known. (See below for a possible answer.) –  José Figueroa-O'Farrill Feb 28 '10 at 2:22
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I think in general the word "Lie subgroup" should not mean "embedded Lie subgroup". Rather, "embedded Lie subgroup" should continue to carry that extra adjective. The reason for preferring this terminology is because there is a bijection between Lie subalgebras of the Lie algebra of a give Lie group G and (immersed, but not embedded) connected Lie subgroups of G. But as the irrational line in the torus shows, not every Lie subalgebra integrates to an embedded Lie subgroup. But anyway, the question of when subgroups are embedded Lie is a good one. –  Theo Johnson-Freyd Mar 1 '10 at 3:02
    
I believe I have seen the terminology "Virtual Lie Subgroup" for the sitation where we have an immersion like the irrational slope subgroup. See "maths.dept.shef.ac.uk/magic/course_files/43/…;. I personally don't like the word, but it could be a useful mnemonic to say "beware: the group topology for the subgroup is not generally the same as the relative topology inherited from the surrounding group". –  WetSavannaAnimal aka Rod Vance Apr 22 '11 at 8:41
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3 Answers 3

Edit This does not actually answer the question. I do not assume that Lie subgroups are embedded submanifolds, whereas the OP does.


I believe that this question has a classical answer. In Knapp's review of Wulf Rossmann's book Lie groups: an introduction through linear groups, he mentions that this problem was solved by Chevalley in the 1940s and the condition is that of what Chevalley called an analytic subgroup. Despite the name, the notion is topological.

Analytic subgroups are precisely those which are connected in what Rossmann calls the group topology. This is the topology generated by the image under the exponential map of the $\epsilon$-balls in the Lie algebra. In other words, a subset $U$ of a Lie group $G$ is open if and only if for every $a \in U$ there is some $\epsilon>0$ such that the set $$ \left\lbrace a \exp(X) \mid \|X\|<\epsilon \right\rbrace $$ is contained in $U$.

(The asymmetry in the definition is fictitious: either $a\exp(X)$ or $\exp(X) a$ define the same topology.)

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Thank you for your post. I am asking for another form of Cartan's theorem. Let H be a subgroup of a Lie group G. In Cartan's theorem the topological property, closed, refers to H with the subset topology from G. I am not asking for you to invent a new topology (I ruled out irrationally sloped lines on a flat torus, which are analytic subgroups but not Lie subgroups). Recall that stating that H is Lie subgroup of G means more than just that H is a Lie group that is also a subgroup of G. Please see the wikipedia article en.wikipedia.org/wiki/Lie_subgroup for a complete definition. –  Khalid Bou-Rabee Feb 28 '10 at 16:02
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Why would you want to rule out the irrationally sloped lines on the torus? They are perfectly good Lie subgroups. In fact, you don't get a good version of the inverse Lie correspondence unless you take such subgroups into account. The lesson from Chevalley's work, in my opinion, is that demanding that the subgroups be embedded submanifolds is too restrictive. –  José Figueroa-O'Farrill Feb 28 '10 at 16:33
    
Thank you, I should clarify the question. –  Khalid Bou-Rabee Feb 28 '10 at 19:11
    
I added "embedded" in front of Lie subgroup. The reason why I ask this is because I am curious of whether there are other versions of Cartan's theorem. –  Khalid Bou-Rabee Feb 28 '10 at 19:13
    
Fair enough. My answer is then not relevant. Sorry! –  José Figueroa-O'Farrill Mar 1 '10 at 0:44
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Edit This answer is invalid to the question because the OP wanted embedded submanifolds. I'm leaving the answer up because it does handle a related question and contains a reference to a paper.

An arcwise connected subgroup of a Lie group is a Lie subgroup at least in the analytic case; cf. this. I recall that it's in the appendix to volume 1 of Kobayashi and Nomizu's Foundations of Differential Geometry.

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I added a few observations to the question. On your post: I don't believe this answers the question. Please let me know if I am misunderstanding you. –  Khalid Bou-Rabee Feb 28 '10 at 0:23
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Perhaps we are using different definitions of a Lie subgroup. In my case, I allowed immersions and not just embeddings. I'm not sure, but I think you're looking only for embedded ones. Am I correct? –  Akhil Mathew Feb 28 '10 at 1:01
    
You are correct. Sorry for the confusion. –  Khalid Bou-Rabee Feb 28 '10 at 1:17
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Whoops! Then this answer is indeed invalid. I'll leave it up in case anyone else finds it interesting as a random tidbit of differential geometry and edit it. –  Akhil Mathew Feb 28 '10 at 1:32
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For analytic subgroups (that is immersed and connected) to be embedded (i.e. to be a regular submanifold) is equivalent to being topologically closed. This equivalence is true only for Lie groups. In one sense this is Cartan's theorem.I dont remember exactly the proof of the converse, but is something like this: if $H$ is an embedded Lie subgroup its closure K is a closed subgroup, hence embedded too. If the dimension of K is greater than that of H there is a contradiction because in regular submanifolds you can take charts with an isolated slice... I guess this is done in Warner's book.

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