# What if a group action on a smooth manifold is proper, smooth BUT not free?

For simplicity, let $$G$$ be a finite group with discrete topology and $$M$$ be a smooth manifold.

Suppose that $$G$$ acts on $$M$$ smoothly. Then, it is kind of trivial to see that the action of $$G$$ is proper.

However, I further assume that $$G$$ does NOT act freely on $$M$$. That is, there exists some $$g (\neq e_G) \in G$$ and a point $$p \in M$$ such that $$g \cdot p =p$$.

Then, can I still define the quotient space $$M/G$$ as some kind of a manifold?

If I remember right, $$M/G$$ is NOT Hausdorff. Nevertheless, is $$M/G$$ still locally Euclidean and given a smooth structure in lieu with the projection map $$\pi : M \to M/G$$?

The link https://en.wikipedia.org/wiki/Non-Hausdorff_manifold somewhat gives hope to me in this aspect, but I cannot judge on my own..

• If a finite group $G$ acts on a Hausdorff topological space $M$, then the quotient $M/G$ is always Hausdorff. Here is a sketch of the proof: Suppose $O_1,O_2\in M/G$. Take $x\in O_1$. For each $y\in O_2$, there is a neighbourhood that separate $x$ from $y$. Take the intersection of all of them ($O_2$ is finite so the intersection is still an open set) and call it $U_x$. Do the same for a fixed point $y\in O_2$ which results in a neighbourhood $U_y$. Then $U_x, U_y$ separate $O_1$ and $O_2$ in $M/G$. Commented May 4, 2023 at 12:15
• Oh I see...then is it possible to give $M/G$ some kind of a smooth manifold structure? I do not assume any linearity or faithfulness in the action, so I cannot employ the notion of "orbifold"... So, I just want to stick to the formalism of manifold for my problem. Commented May 4, 2023 at 12:30
The simplest example of a smooth finite group action such that the quotient space is not a topological manifold is $$M={\mathbb R}^3, G\cong {\mathbb Z}_2,$$ and the generator $$g$$ of $$G$$ acts on $$M$$ by $$g({\mathbf x})= -{\mathbf x}.$$ The quotient space $$M/G$$ is homeomorphic to the (open) cone over $${\mathbb R} P^2$$, hence, is not a topological manifold.
• If you take $M = \mathbb{R}$, then $M/(x \sim -x) \cong [0, 1)$ is only a manifold with boundary. Commented May 5, 2023 at 8:30