Is a Polish subgroup of a Lie group Lie?

Question

Let $G$ be a Lie group. Suppose that $H\le G$ is a subgroup such that endowed with the subspace topology it is a Polish space.

Does this imply that $H$ is a Lie group?

I know that in case $H$ is closed the answer is yes by Cartan's theorem, but I do not know the answer in this relaxed case.

Answer 1

The answer is positive, this can be proved by showing that $H$ must be closed and using what you already know. In order to show that $H$ is closed we need two things:

1. Lie groups are Polish.
2. Polish subgroups of Polish groups are closed

Luckily 1. is obvious, so let's prove 2.

Let $G$ be a Polish group and $H\leq G$ a subgroup which is Polish in the induced topology. Since a subspace of a Polish space is Polish iff it is $G_\delta$ we can assume that $H$ is $G_\delta$ in G and we only need to upgrade it to closed. By replacing $G$ with $\overline{H}$ (which is a closed, hence Polish, subgroup of $G$) we can assume that $H$ is a comeager $G_\delta$ subgroup. If now $H\neq\overline{H}$, there must be a coset $hH\neq H$ in $\overline{H}$, but this is a contradiction as $hH$ and $H$ are now disjoint comegear sets in $\overline{H}$.