7
$\begingroup$

Assume that $G$ is a Lie group. Is it understood which subgroups of $G$ are Lie groups?

Ideally, I would like to make no extra assumption about $G$. In particular, $G$ can be infinite dimensional.

I am aware that in finite dimensions, Cartan's theorem ensures that any closed subgroup is a Lie group.

In Neeb's notes about infinite dimensional Lie groups, it is mentioned that already for Banach-Lie groups, Cartan's theorem is not true anymore. However locally compact subgroups are Lie subgroups, see p.59.

Is there anything known beyond this?

$\endgroup$
  • $\begingroup$ Are you looking for a necessary and sufficient condition, or just a nice sufficient condition? $\endgroup$ – user43326 Feb 2 '14 at 9:57
  • 1
    $\begingroup$ Anything going beyond the information above would be interesting. Of course, a necessary and sufficient condition would be great, but it's maybe too much to hope for. $\endgroup$ – Samuel Monnier Feb 2 '14 at 10:03
  • $\begingroup$ Well, if you consider things like $\mathbb R$ embedded in the torus, it sounds extremely unlikely that you get a necessary and sufficient condition. Do you suppose that the topology is the induced topology? $\endgroup$ – user43326 Feb 2 '14 at 10:08
  • $\begingroup$ Yes, I assume that the topology is the induced topology. $\endgroup$ – Samuel Monnier Feb 2 '14 at 10:29
2
$\begingroup$

Consider the situation in finite dimension, and assume that Lie groups are second countable.

A second countable, locally Euclidean group can have at most one differentiable structure making it into a Lie group. This follows from the fact that a continuous homomorphism between Lie groups is automatically smooth. Now, the condition that a Lie subgroup $H$ of a Lie group $G$ has the induced topology is very restrictive. It implies that $H$ is closed in $G$, so you are in the situation described by Cartan's theorem (essentially one shows that the inclusion map is proper, so it has a closed image), see the book by F. Warner, 3.29.

A Lie group $G$ has no small subgroups (namely, there is a neighborhood of the identity containing no non-trivial subgroups). If $H$ is a subgroup, it follows that it does not contain small subgroups either. Now if $H$ is locally compact with the induced topology, then it follows fom Gleason-Montgomery-Samelson solution to Hilbert's 5th problem that $H$ is a Lie group (and in particular that $H$ is closed in $G$).

If you relax the condition on the topology of $H$, then $H$ is a Lie subgroup of $G$ if and only if it is a (second countable) submanifold of $G$. The idea of the proof is also related to Frobenius; namely, if $H$ is a submanifold and an abstract group, show that it is the leaf of the involutive left-invariant distribution determined by its tangent space at the identity, see Warner, 3.20.

$\endgroup$
0
$\begingroup$

If you define a manifold to be a paracompact Hausdorff space locally homeomorphic to Euclidean space, and a Lie group to be a manifold and a group with continuous group operation, then every subgroup of a Lie group is a Lie subgroup; see Sharpe, Differential Geometry: Cartan's Generalization of Klein's Erlangen Programme, p xii. Furthermore, if $G$ is a Lie group equipped with a smooth structure for which the multiplication is a smooth map, then every subgroup of $G$ is a smooth submanifold.

The point is that we don't ask here for second countability.

$\endgroup$
  • $\begingroup$ I must not be following something. The subgroup is required by the OP to carry the subspace topology. How are things like an irrational line on a torus Lie groups under the subspace topology? $\endgroup$ – Todd Trimble Feb 2 '14 at 18:57
  • $\begingroup$ They are only immersed submanifolds, not the subspace topology. Sorry, I didn't read the relevant comment under the question. $\endgroup$ – Ben McKay Feb 2 '14 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.