Let $\rho$ be a group action by a compact group $G$


\begin{equation}
 \rho:G\times M \rightarrow M \\
 \rho:(g,p) \rightarrow \rho\_g(p) 
\end{equation}


Denote the orbit of $p\in M$ by $\mathcal{O}\_p$ and the isotropy group of $p$ by $G\_p$. We have a natural representation $G\_p$ on the vector space $T\_p(M)/T\_p\mathcal{O}\_p$ given by the action of $d\rho\_h(p)$ on $T\_p(M)/T\_p\mathcal{O}\_p$ for $h\in G\_p$. Consider the action of $G\_p$ on the product $G \times T\_p(M)/T\_p\mathcal{O}\_p$


\begin{equation}
 \sigma(h)(g,v)=(h^{-1}g,d\rho\_h(p)(v)) 
\end{equation}


This action is free and therefore the quotient is a manifold. It is a principal bundle over $G/G\_p$ denoted $G \times\_{G\_p} T\_p(M)/T\_p\mathcal{O}\_p$. The slice theorem tells us that this bundle, the action on the fiber being the trivial action is equivariantly diffeomorphic to a neighbourhood of $\mathcal{O}\_p$.


I tried computing some simple examples to get a feeling for this, but of course for the simple examples this vector bundle is trivial. I feel I could understand much better if I had an example where the bundle was not trivial. Is there an "easy" example of this?



Is there an easy example of group action where the slice theorem produces a non-trivial principal bundle?
Consider the usual $G = S^1$ action on $S^2$ given by rotations. This action respects the antipodal map, so descends to a $G$ action on $M = \mathbb{R}P^2$. Let $p\in M$ be any point on the "equator", where the isotropy group is $G\_p = \mathbb{Z}/2\mathbb{Z}\subseteq S^1$.


Then the $G\_p$ action on the slice at $p$ acts as multiplication by $-1$. In particular, $G\times\_{G\_p} T\_p M/T\_p\mathcal{O}\_p$ is a Mobius band.


Let me also mention one other thing: the fact that in many examples the bundle is trivial is no coincidence. If $p\in M$ is regular point (meaning the isotropy group $G\_p$ is conjugate to a subgroup of any other isotropy group), then the action of $G\_p$ on the slice $T\_p M/T\_p \mathcal{O}\_p$ at $p$ is trivial. This clearly implies the bundle $G\times\_{G\_p} T\_p M /T\_p\mathcal{O}\_p$ is trivial. Since the regular points are always open and dense in $M$, in some sense, the bundle is generically trivial.


