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?

  • $\begingroup$ Are you looking for a necessary and sufficient condition, or just a nice sufficient condition? $\endgroup$ – user43326 Feb 2 '14 at 9:57
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    $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

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