# Is there an easy example of group action where the slice theorem produces a non-trivial principal bundle?

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

$$\rho:G\times M \rightarrow M \\ \rho:(g,p) \rightarrow \rho_g(p)$$

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 $G \times T_p(M)/T_p\mathcal{O}_p$

$$\sigma(h)(g,v)=(h^{-1}g,d\rho_h(p)(v))$$

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?

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.