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.

notmean "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 subgroupsareembedded Lie is a good one. – Theo Johnson-Freyd Mar 1 '10 at 3:02