# Lie group action induces an immersion of $G$ mod the stabilizer

Let $G$ be a Lie group acting smoothly on a manifold $X$. Assume the action $G \times X \rightarrow X$ is smooth. Then for any $x_0 \in X$, the map $G \rightarrow X$, $g \mapsto g.x_0$ has constant rank by homogeneity, so by a variation of the submersion principle, $H = \operatorname{Stab}x_0$ is a submanifold of $G$, hence a Lie subgroup. Then quotient $\pi: G \rightarrow G/H$ is a submersion, and the injection $\bar{\phi_0}: G/H \rightarrow X, gH \mapsto g.x_0$ is a smooth map.

I'm trying to understand why $\bar{\phi_0}$ is an immersion, i.e. injective on tangent spaces. This is the content of a theorem in Lie Groups and Lie Algebras by Serre.

Serre is working in a more general setting than smooth real manifolds. The assumption that $\phi_0$ is a subimmersion is redundant if we are in characteristic zero.

For the proof, Serre is considering the submanifold $gH$ of $G$. Since $\pi: G \rightarrow G/H$ is a submersion, the preimage $gH$ of the coset $\{gH\}$ in $G/H$ is a submanifold, and $T_g(gH)$ is the kernel of $T_g(\pi): T_g(G) \rightarrow T_{gH}(G/H)$.

However, he claims that $T_g(gH)$ is moreover the kernel of $T_g(\phi_0)$. I don't see why this is. It is only clear to me that this tangent space is contained in the kernel of $T_g(\phi_0)$.

• Can you show it for $g= e$? You need to prove that if $X\in \frak g$ goes to zero under $d\phi|_e$, then the whole one parameter subgroup $\{e^{tX}: t\in \mathbb{R}\}$ stabilizes $x_0$. Do this by showing $\mathbb{R}\rightarrow X, t\mapsto e^{tX}x_0$ has derivative zero on all of $\mathbb{R}$ by using the fact that it has derivative zero at $0$. The case of general $g$ should follow from $G$-equivariance of $\phi$. – Tim kinsella Dec 31 '17 at 4:13