quotient manifold theorem

Question

I found a proof for the quotient manifold theorem here : https://www.mathi.uni-heidelberg.de/~lee/StephanSS16.pdf

In the proof of the theorem, we find this

To prove the claim, note first that the G-orbits are properly embedded submanifolds of M dif- feomorphic to G by our characterization of orbits.

I get that they are properly embedded submanifolds of $M$ (proposition 2.6 gives us this) but why diffeomorphic to $G$?

I thought of the following map that would give the diffeomorphism

$$ \psi: O_p\rightarrow G: g\cdot p\rightarrow g$$

But I don't see why this would be a smooth function?

Answer 1

I will use the following:

$M$ and $N$ are diffeomorphic if there exists a map $\psi:M\rightarrow N, C^\infty$ and bijective such that $\psi_{*_q}$ is a linear isomorphism

Here $$\psi:G\rightarrow O_p:g\rightarrow g\cdot p \\ \psi_{*_g}:T_gG\rightarrow T_{g\cdot p}O_p$$ We know $\psi_{*_q}$ is an isomorphism

1. Let's show surjectivity:

$\gamma:I\rightarrow O_p$ with $\gamma(0) = g\cdot p$. Then $\dot{\gamma}(0)\in T_{g\cdot p}O_p$. Define $\Gamma:I\rightarrow G: s\rightarrow g^{-1}\cdot \gamma(s)$ (then $\Gamma(0) = p$ so $\dot{\Gamma}(0)\in T_gG$. Let's show that $\dot{\Gamma}(0)$ is sent to $\dot{\gamma}(0)$

$$\phi_{*g}(\dot{\Gamma}(0))(f) = (\dot{\phi\circ\Gamma})(0)(f) =\frac{d}{ds}f(\phi\circ \Gamma(s))\vert_{s=0} = \frac{d}{ds}f(g\cdot g^{-1}\cdot\gamma(s))\vert_{s=0} \\= \frac{d}{ds}f(\gamma(s))\vert_{s=0} = \dot{\gamma}(0)(f)$$

2. With similar calculations we get infectivity if we can prove that $$ (\dot{g\cdot\gamma})(0) = (\dot{g\cdot\Gamma})(0)$$ implies that $$\dot{\gamma}(0) = \dot{\Gamma}(0)$$ But I don't see how to show this.