Using topology to characterize embedded Lie subgroups of Lie groups. - MathOverflow

Using topology to characterize embedded Lie subgroups of Lie groups.

2010-02-27T18:27:36Z

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.

Answer by Akhil Mathew for Using topology to characterize embedded Lie subgroups of Lie groups.

2010-02-27T22:29:04Z

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.

Answer by José Figueroa-O'Farrill for Using topology to characterize embedded Lie subgroups of Lie groups.

2010-02-28T02:22:18Z

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.)

Answer by Enrique Macias for Using topology to characterize embedded Lie subgroups of Lie groups.

2011-04-11T22:37:23Z

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.