Let $M$ be a smooth manifold on which a Lie group $G$ acts properly and freely. Suppose there exists a set $D_M \subset M$ which is a smooth submanifold and which contains one and only one representative from each orbit of $M$. I am trying to see whether from a given chart on $M$ which is adapted to the group action, we can construct a chart on $D_M$.
Let $n = \dim M – \dim G$ and $k= \dim G$. Around each point in $M$ we can find an adapted chart of the form $(\tilde U, \tilde \phi = (x^1, \ldots, x^k, y^1 ,\ldots y^n))$ such that for any orbit $qG$ we either have $qG \cap \tilde U = \emptyset$ or:
$$ qG \cap \tilde U = \{ y^i = c^i_{qG} \} $$
for $c^i_{qG}$ some constants depending on the orbit $qG$. This is shown in Theorem 21.10 in Lee's "Introduction to smooth manifolds".
Of course, the constants $c^i_{qG}$ are different for different orbits, because the orbits are either the same or disjoint.
Now let $p \in D_M$ and consider a chart $\tilde U$ as above around $p$. Let $\textrm{pr}_2: \mathbb R^k \times \mathbb R^n \rightarrow \mathbb R^n$ be the projection.
My question is: is it true that $\phi:=\textrm{pr}_2 \circ \tilde \phi: \tilde U \cap D_M \rightarrow \mathbb R^n$ is a chart in $D_M$ around $p$, eventually restricting $\tilde U$ to a smaller open subset? If so, how could we show it?
Since $D_M$ only contains points from different orbits, we have that $\phi: \tilde U \cap D_M \rightarrow \mathbb R^n$ is injective. It is also smooth, since $\textrm{pr}_2$ is smooth and the restriction of the smooth map $\tilde \phi$ to the submanifold $D_M \cap \tilde U$ is smooth.
But beyond this, I don't know how to show that $\phi$ is a diffeomorphism on its image. For example, I can't even show that $\phi$ is an open map.
Best Answer
I think this is a not difficult but very technical proof. Here is an outline of what I think you can do:
First, your map $\tilde{\phi}$ is a diffeomorphism from $\tilde{U}$ to $\tilde{\phi}(\tilde{U})$, obviously.
Second, the submanifolds are preserved under diffeomorphism. So since $D_M$ is a submanifold, $N:=\tilde{\phi}(D_M)$ is a submanifold of $\tilde{\phi}(\tilde{U})\subset R^{n+k}$. From now on, we will work, then, in $\tilde{\phi}(\tilde{U})$.
Take a parametrization $\sigma: R^n \to N$ in a neighbourhood of $p$. If the map $f=pr_2 \circ \sigma$ is a diffeomorphism (even in a srunken domain) then you are done, because then $\sigma \circ f^{-1}$ gives you the desired chart (a kind of "inverse" of the projection).
To show that $f=pr_2 \circ \sigma$ is a diffeomorphism, you have to assume an additional transversality hypothesis, I think. Let's see. If $\sigma(a)=\phi(p)\in\phi(D_M)$ then to show is a diffeomorphism you need to prove that $$ d(f)_a=d(pr_2 \circ \sigma)_a=d(pr_2)_{\phi(p)} \cdot d(\sigma)_a $$ is non-singular. But $d(pr_2)_{\phi(p)}=(0_{k\times n} | I_n)$, and $$ d(\sigma)_a=\begin{pmatrix} \dfrac{\partial (x_1,\ldots,x_k)}{\partial (t_1,\ldots,t_n)}\\ \dfrac{\partial (y_1,\ldots,y_n)}{\partial (t_1,\ldots,t_n)}\\ \end{pmatrix} $$ where $t_1,\ldots,t_n$ are the parameters for $\sigma$, so you have to require that $$ det(\dfrac{\partial (y_1,\ldots,y_n)}{\partial (t_1,\ldots,t_n)})\neq 0 $$ So assuming this last condition you are done. Anyway, my own picture tell me that it should be possible to prove your statement even without this requirement, but I don't get the way.