# Almost free Lie group action

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?

• I don't understand the question. Is the point that you are not insisting that the action of $G_x$ on a neighbourhood of $x$ be locally linear? If the action is locally linear, then surely $M/G$ is an orbifold by definition and the orbifold universal covering of $M/G$ is the ordinary universal covering of $M$.
– IJL
Commented Jan 14, 2022 at 10:04
• I think your definition of almost free is not standard. (I don't think there should be a condition on the amount of $x$'s with a non trivial stabilizer. Commented Jan 14, 2022 at 12:54

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 $$G_x \backslash T_xM / T_x (G\cdot x)$$.
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).