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?


1 Answer 1


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 $U$.

  • $\begingroup$ 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. $\endgroup$
    – Feanoris
    Jun 12, 2016 at 17:15
  • $\begingroup$ 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. $\endgroup$
    – Feanoris
    Jun 12, 2016 at 17:19
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 12, 2016 at 17:23
  • $\begingroup$ 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? $\endgroup$
    – Feanoris
    Jun 12, 2016 at 21:40
  • $\begingroup$ You can assume only that $H$ is compact. $\endgroup$ Jun 12, 2016 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.