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Let $G$ be a Lie group and $M$ a manifold such that $G$ acts on $M$. Let $G_a$ denote the stabilizer of an element $a \in M$.

We can define a map $f: G/G_a \to M$ which sends the coset $gG_a$ to $g\cdot a \in M$, and this map is well-defined. I also need to show that it is smooth, but I'm not sure how to do this.

Ordinarily I would choose charts and write the map in coordinates. However, I don't know what a chart on $G/G_a$ looks like. All I know is that the smooth structure on $G/G_a$ is such that the quotient map $p:G \to G/G_a$ is a submersion. I also know that $f$ is related to $p$ by the formula $f\circ p=\tilde{f}$, where $\tilde{f}:G \to M$ is $\tilde{f}(g)=g\cdot a$. I know $p$ and $\tilde{f}$ are smooth, but this doesn't imply $f$ is smooth as well.

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    $\begingroup$ Hint: submersions locally admit smooth sections. $\endgroup$
    – Pedro
    Nov 8, 2016 at 23:56

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In fact, it does. Assume you have a commutative diagram of the form

$$ \begin{array}{ccc} G & \xrightarrow{\tilde{f}} & M \\ \downarrow{p} & \nearrow{f} & \\ N & & \end{array} $$

(sorry for the awful rendering, I have no idea how to fix it) where $p$ is a smooth submersion. Then $\tilde{f}$ is smooth if and only if $f$ is smooth. If $f$ is smooth then $\tilde{f} = f \circ p$ is smooth as a composition of such. For the interesting direction, note that the constant rank theorem implies that $p$ has smooth local sections. That is, for each point $q \in N$ we can find an open neighborhood $q \in U \subseteq N$ and a smooth map $s \colon U \rightarrow G$ such that $p \circ s = \operatorname{id}|_{U}$. Then $f|_{U} = \tilde{f} \circ s$ is smooth as a composition of smooth maps.

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  • $\begingroup$ The constant rank theorem says that in suitable coordinates the map $p$ looks like $(x_1,\dots,x_k) \mapsto (x_1,\dots,x_n)$ where $k$ is the dimension of $G$ and $n$ is the dimension of $n$ ($n \leq k$). So I can construct a local section which maps $(x_1,\dots,x_n)$ to $(x_1,\dots,x_n,0,\dots,0)$, and I know this is smooth because it is smooth in coordinates. Is that the idea? $\endgroup$
    – kccu
    Nov 9, 2016 at 1:05
  • $\begingroup$ @kccu: Precisely. $\endgroup$
    – levap
    Nov 9, 2016 at 1:06

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