- Out of curiosity, what would be an example for the following situation:
Let $G\curvearrowright X$ be a free action of a locally compact Hausdorff group $G$ on a Hausdorff space $X$.
By the freeness condition, the tranformation groupoid is algebraically principal,
$$X\times G\subseteq X\times X\text{ where }(x,g)\mapsto(x,gx).$$
Moreover one has $X\times G=X\times X$ precisely if the action is transitive.
But the groupoid topology is generally finer than the product topology induced by $X\times X$.
More precisely, convergence in the groupoid topology is given by
$$(x_n,g_n)\to (x,g)\iff x_n\to x\text{ and }g_n\to g,$$
while convergence induced by the product topology is given by
$$(x_n,g_nx_n)\to (x,gx)\iff x_n\to x\text{ and }g_nx_n\to gx,$$
whence the former is generally finer the latter.
What would be an example where the topology is strictly finer?
What I tried so far were examples along the lines $\mathbb{C}\setminus0\curvearrowright\mathbb{C}\setminus 0$,
but from here it looks one requires maybe some sort of "hyperbolic" behavior?
- Parallely out of curiosity, I learned from Ian Putnam the following interesting example:
Inside the unit square, consider the diagonal plus the other two corner points, $$\Delta\cup\{(0,1),(1,0)\}\subseteq [0,1]\times[0,1].$$ I think this example cannot arise as transformation groupoid, but I'm not sure.
So my question, is there maybe a simple argument which forbids such a representation resp. does it admit one?
Consider the action of $G=\mathbb Z$ on $X=S^1$ given by $$ \alpha_n(z)=e^{2\pi i n\theta }x, \quad \forall n\in\mathbb Z, \quad \forall x\in S^1, $$ where $\theta$ is an irrational number. Can you show that the two topologies on the transformation groupoid are very different?