It's known that if a compact Lie group $G$ acts freely on a compact manifold $M$, then the orbit space $M/G$ is a manifold. If we only assume that $G$ acts almost freely (i.e. $G\_x$ is finite for any $x\in M$ and there are only finitely many $x$ such that $G\_x$ is not trivial)，then can we deduce that $M/G$ is a orbifold or even a good orbifold (i.e. the universal covering is a manifold)?


And is there any reference for such kind of actions?



Almost free Lie group action
I think the answer to the first question is yes and the answer to the second one is no:


Yes, the quotient is an orbifold. The action of the finite group $G\_x$ in a neighbourhood of $x$ can be linearized (at least if the action is by diffeomorphisms, I don't know about $C^0$ regularity), and the quotient $M/G$ is locally modelled on $G\_x \backslash T\_xM / T\_x (G\cdot x)$.


No this orbifold is not good in general. For instance, you can glue a solid torus with a trivial circle fibration to a solid torus with a Seifert fibration with one singular fiber in the center and get a closed Seifert 3-manifold with one singular fiber. The fibration is given by the orbits of an action of S^1 and the quotient orbifold is a sphere with a single orbifold point, the simplest example of an orbifold not covered by a manifold.


More generally, I think every orbifold $M$ if dimension $n$ is the quotient of a manifold $P$ by an almost free action of the orthogonal group $O\_n$ ($P$ is the principal $O\_n$-bundle associated with the orbifold tangent bundle equipped with an orbifold Riemannian metric).


