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Let $G$ be a Lie group acting smoothly on the smooth manifold $M$, all finite-dimensional, given by some map $\rho:G\times M\to M$, where we write $g\cdot m := \rho(g)(m)$. If we set, for $m\in M$, $$\mathrm{Stab}_G(m):=\{g\in G | g\cdot m=m\}$$ then must $\mathrm{Stab}_G(m)$ be a Lie subgroup of $G$ (i.e. must it be a smooth sub manifold)? What about $\mathrm{Stab}_G(M):=\bigcap_{m\in M}\mathrm{Stab}_G(m)$?

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    $\begingroup$ Well we know that Stab$_G(m)$ is a closed subgroup of $G$ (can you prove this?) and that a closed subgroup of a Lie group is a Lie subgroup (en.wikipedia.org/wiki/Closed_subgroup_theorem) $\endgroup$ – user259242 Nov 6 '16 at 5:43
  • $\begingroup$ I'd prefer not to use such heavy machinery. Is there a more elementary proof? $\endgroup$ – Monstrous Moonshine Nov 6 '16 at 5:45
  • $\begingroup$ This question was asked (by me!) here. Though not that extra bit about the "global stabilizer group". $\endgroup$ – Mike F Nov 6 '16 at 6:08
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Maybe try the implicit function theorem.

Here is a rough idea, without precise details. Define the map $f_m : G \to M$ as $f_m : \, g \mapsto \, g\cdot m$. In other words simply write $f_m(g) = g\cdot m$. Then the stabilizer of point $m$ is $$\text{Stab}_G(m) = \{g \in G \, : \, f_m(g) = m\}$$ so it is the solution of a bunch of equations. The fact that the stabilizer has the group property allows you to show that the rank of $Df_m$ is the same everywhere on the stabilizer.

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