Smooth actions and stabilizer

Question

Let $G$ be a Lie group acting smoothly on a smooth manifold $M$. We consider a point $x$ of $M$ and $g$ an element of its stabilizer $G_x$. The smooth diffeomorphism $\theta_g$ of $M$ defined by $\theta_g(y)=g\cdot y$ induces a linear isomorphism $d_x\theta_g$ of $T_xM$.

I am trying to prove that $d_x\theta_g$ restricts to the identity on $T_x(G\cdot x)$ (where $G\cdot x$ is the orbit through $x$). There is certainly a very simple argument but I fail to see it.

Can someone give me a hint?

Answer 1

For a counterexample, how about

• $M = \mathbb{R}$,
• $G =$ the group of affine isomorphisms $g \cdot y = ay+b$ of $\mathbb{R}$ and so $G \cdot y = \mathbb{R}$ for all $y$,
• $x=0$,
• $g \in G_x$ is given by $g \cdot y = \theta_g(y) = 2y$.

For these values of $x$ and $g$, the linear map $$d_x\theta_g : \mathbb{R}=T_x(G \cdot x) \to T_x(G \cdot x)=\mathbb{R} $$ is given by $v \mapsto 2v$ which is not the identity.