# slice theorem for proper actions

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?

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 $U$.
• Since $G$ acts properly on $M$, $H$ is automatically compact. Lets denote the shrinked "vector-subspace" by $D_0$. So if $H$ acts as the identity on $D_0$, we can take a basis $e_1, \dots, e_m$ of $D$ and the corresponding dual basis of $D^*$ denoted by $\tau_1, \dots, \tau_m$. Now we can restrict these linear functions to $D_0$. Since $H$ acts on $D_0$ as the identity, these functions are $H$-invariant. So the last thing is to show, that the map $\pi \colon G \times_H D_0 \to D_0, \ \pi([g,v])=v$ is well-defined and a submersion. Jun 12, 2016 at 17:15
• So $\pi$ maps $[g,v] \mapsto H.v$ and since $H$ acts as the identity, we have $\pi([g,v])=H.v=v$. Is this right? If it is, then $\pi$ is clearly a submersion and we constructed these functions I was searching for. Jun 12, 2016 at 17:19
• If the action of $G$ on $D_0$ is trivial then $G\times_HD_0=G/H\times D_0$ and the projection $\pi:G\times_HD_0=G/H\times D_0\rightarrow D_0$ is a submersion. Jun 12, 2016 at 17:23
• One more question: You said in your answer, that "in the compact case" such an $D$ with $H$ acting trivially on it, exists. Do you mean, that $G$ is compact or $H$? And what could be the problem, if $G$ is not compact? Jun 12, 2016 at 21:40
• You can assume only that $H$ is compact. Jun 12, 2016 at 22:06