I'm trying to understand the slice-theorem for proper Lie-group actions.
Having a smooth manifold $M$ and a Liegroup $G$ acting on $M$ in a proper way, we have the slice theorem, saying that at each point $x \in M$ we find a slice and a tube. 


So choosing $x \in M$, denote by $H = G\_x$ the isotropy-group of $G$ at $x$. Using the tube-theorem we find a $G$-invariant neighborhood $U \subset M$ of the orbit $G \cdot x$, a vectorspace $V \cong T\_xM / T\_x(G \cdot x)$ where $H$ acts linearly on $V$, a $H$-invariant vector-subspace $D \subset V$ and a $G$-equivariant diffeomorphism 
$$\phi \colon G\times\_H D \to U$$
such that $\phi([e,0])=x, \ \phi([G,0])=G \cdot x,\ \phi([e,D])=S$ where $S$ is a slice of $G \cdot x$.


From now on I'm considering the case, that $M$ is connected and we choose $x \in M$ such that the $G$-orbit is a principal orbit.


Now I want to construct $k = \dim M -\dim G \cdot x$ functions $f\_1, \dots, f\_k$ on $U$, such that they are $G$-invariant, with $d\_xf\_1 \wedge \dots \wedge d\_xf\_k \neq 0$ and such that $G \cdot x = \{y \in U \ |\ f\_1(y) = \dots = f\_k(y)=0\}$.


Using the slice theorem I have to find functions $\tilde{f}\_1, \dots, \tilde{f}\_k \in C^\infty(D)$ which are $H$-invariant, and satisfy $\tilde{f}\_j(0)=0$ then I could take the extension 
\begin{align}
\tilde{F}\_j \colon G \times\_H V \to \mathbb{R}, \quad F\_j([g,v])= \tilde{f}\_j(v)
\end{align}


and the functions $$F\_j = \tilde{F}\_j \circ \phi^{-1} \colon U \to \mathbb{R}$$ are $G$-invariant and satisfy $G \cdot x = \{y\in U \ | \ F\_1(y)=\dots = F\_k(y)=0\}$.


But I'm not sure how to construct these functions $\tilde{f}\_1,\dots, \tilde{f}\_k$ on $D\subset V$.


Edit: What I forgot to post: Since $x$ is a regular point, i.e. the dimension of $G \cdot x$ is maximal, we find a neighborhood $W \subset U \subset M$ of $x$, such that $\dim G \cdot y = \dim G \cdot x$ for all $y \in W$. Since now $T\_yM = T\_yS + T\_y (G\cdot y)$ and $T\_yS \cap T\_y (G \cdot y)=T\_y (H \cdot y)$ we have 
\begin{align}
 \dim M &= \dim G \cdot y + \dim S - \dim H \cdot y
\\&= \dim G \cdot x +\dim S - \dim H \cdot y 
\\&= \dim M -\dim H \cdot y
\end{align}
we have $\dim H \cdot y =0$ for all $y \in S \cap W$. From that I concluded, that $H$ acts as the identity on $D$. But is this true?



slice theorem for proper actions
Yes, it is true that there exists $D$ such that $H$ acts as the identity on $D$ if the orbit is principal (at least in the compact case)


If the orbit of $x$ is principal, then the union $W\_H$ of elements of $M$ whose stabilizer is conjugated to $H$ is an open and dense subset of $M$. We can just analyze the situation in a Koszul neigborhood:


let $[g,v]$ be an element of $G\times\_HD$, for $g'\in G$, we have $g'.[g,v]=g'g,g'v]$. Suppose that $g'$ fixes $[g,v]$, then there exists $h\in H$ such that $v=h.v$ and $gg'=gh^{-1}$. This is equivalent to saying that $g'\in gH\_vg^{-1}$. Suppose that the orbit of $[g,0]$ is principal, then you have an open subset $U\subset G\times\_HD$ such that the stabilizer $G\_{[g,v]}$ of every element $[g,v]\in U$ is conjugated to $H$, since $G\_{[g,v]}$ is conjugated to the subgroup $H\_v$ of $H$, you deduce that $H\_v=H$. This is equivalent to saying that $H.v=v$ for every $[g,v]\in U$ Now you can shrink $D$ such that the Koszul neighborhood is $U$. 


