A stabiliser of a smooth action is a Lie subgroup

Question

I have got the question at my Lie Groups course.

The Lie subgroup is a submanifold of a Lie Group which is also a subgroup.

Assume the Lie Group $G$ acts on a smooth manifold $M$. The stabilizer of a point $G_x$ is a subgroup, I need to show that it is a Lie subgroup as well, i.e. a submanifold.

The statement is mentioned in this question.

I know that every subgroup of a topological group is closed, so $G_x$ is closed, but looks like we should use that M is a manifold and that the acting is continuous, but I have no clue how to make the coordinate definition of a submanifold work here.

UPD Constant rank argument, Stabilizer for smooth action, Generalised question for closed orbit

Looks like this question was answered two times already but I didn't find it in a brief inspection.

It is NOT true that every subgroup of a topological group is closed. For example, on $T^2$ there are dense subgroups. — Jason DeVito - on hiatus, Sep 30 '19 at 13:52

score 3 · Accepted Answer · answered Sep 30 '19 at 10:17

3

By Cartan's theorem, since $G_x$ is closed, it is automatically a Lie subgroup.

answered Sep 30 '19 at 10:17

José Carlos Santos

440,053

2

While I agree with your answer, the OP seems to be under the mistaken impression that $G_x$ is closed because it is a subgroup of a topological group. Instead, one should use the continuity of the action to show that $G_x$ must be closed. – Jason DeVito - on hiatus Sep 30 '19 at 14:10
Yes, you are right. – José Carlos Santos Sep 30 '19 at 14:11

A stabiliser of a smooth action is a Lie subgroup

1 Answers1