On connectivity of Lie subgroups

differential-geometrylie-groupssmooth-manifolds

Here we assume that all manifolds are second-countable.

Let $ G $ be a Lie group and $ H $ be its Lie subgroup. That is, $ H $ is a subgroup of $ G $ equipped with a differential structure with which $ H $ is a Lie group and the inclusion map $ \iota\colon H \to G $ is an immersion. Let $ H' $ denote $ H $ as a topological subspace of $ G $. Note that the topology of $ H $ is generally different from that of $ H' $.

I am thinking about the relationship between the following four conditions:

  1. $ H $ is connected;
  2. $ H $ is path-connected;
  3. $ H' $ is connected;
  4. $ H' $ is path-connected.

Here is what I understood:

  • 1 and 2 are equivalent since $ H $ is a manifold.
  • 1 (or, equivalently, 2) implies 3 and 4 since a continuous image of a connected (resp. path-connected) is connected (resp. path-connected).
  • 3 does not imply 1. To see this, let $ G = \mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2 $ and
    $$ H = \{(x, \sqrt{2} x) + \mathbb{Z}^2 \mid x \in \mathbb{R}\} \cup \{(x + 1/2, \sqrt{2} x) + \mathbb{Z}^2 \mid x \in \mathbb{R}\}. $$
    Then $ H $ is a Lie subgroup of $ G $ which is isomorphic to $ \mathbb{R} \times \mathbb{Z}/2\mathbb{Z} $, but $ H' \subseteq G $ is connected.
  • 4 implies 3 (generality from General Topology), but the converse is not true as the above example shows.

Now my question is:

Does 4 imply 1? That is, if the subspace $ H' \subseteq G $ is path-connected, is the Lie group $ H $ (path-)connected?

Remark. Without second-countability, there are trivial counterexamples such as $ (\mathbb{R}, \text{discrete}) \to (\mathbb{R}, \text{usual}) $.

Best Answer

Yes, 4 does imply 1.

The key observation is that every Lie subgroup $H\subseteq G$ is an integral manifold of an involutive distribution (see Theorem 19.25 in my Introduction to Smooth Manifolds, 2nd ed.), which implies that a continuous map into $G$ whose image lies in $H$ is actually continuous as a map into $H$. (This is proved in exactly the same way as Theorem 19.17 in ISM. That proof assumed the map was smooth, but the same proof goes through with "smooth" replaced by "continuous" throughout.)

To prove 4 $\implies$ 1, suppose $H\subseteq G$ is a Lie subgroup and $H'$ is path-connected. Given any two points $p,q\in H$, there is a continuous path $\gamma\colon [0,1]\to H'$ with $\gamma(0)=p$ and $\gamma(1)=q$. Then $\gamma$ is continuous as a map into $G$ by composition with the inclusion map $H'\hookrightarrow G$, and therefore it's continuous as a map into $H$ by the observation in the previous paragraph.

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