# If two group actions lead to the same orbifold, are they conjugate?

In The Geometry and Topology of Three-Manifolds, Thurston says: "In these examples, it was not hard to construct the quotient space from the group action. In order to go in the opposite direction, we need to know not only the quotient space, but also the singular locus and appropriate data concerning the local behavior of the group action above the singular locus."

Thus, I gather that the purpose of an orbifold is to "remember" the group action that gave rise to it. Here is my question: Suppose $$M$$ is a simply connected smooth manifold, and $$\Gamma, \Lambda$$ are properly discontinuous subgroups of $$\mathrm{Diff}(M)$$. If the orbifolds $$M/\Gamma$$ and $$M/\Lambda$$ are diffeomorphic (as orbifolds), are $$\Gamma$$ and $$\Lambda$$ conjugate subgroups of $$\mathrm{Diff}(M)$$? I would greatly appreciate confirmation that this is true, and references that talk about it in user-friendly detail.

Notes:

(1) Let $$\pi_1\colon M \to M/\Gamma$$ and $$\pi_2\colon M \to M/\Lambda$$ be the projections. To answer the above question, it suffices to show that an orbifold diffeomorphism $$f\colon M/\Gamma \to M/\Lambda$$ must lift to a diffeomorphism $$\tilde{f}\colon M \to M$$ such that $$\pi_2\circ\tilde{f} = f\circ\pi_1$$. For then we will have $$\Lambda = \tilde{f}\Gamma\tilde{f}^{-1}$$.

(2) If the words "diffeomorphism" and "covering" mean what I think they do in the orbifold context, then point (1) must hold. Indeed, since $$\pi_1$$ is a universal cover of $$M/\Gamma$$, and $$f^{-1}\circ\pi_2$$ (being the composition of a covering with a diffeomorphism) is another covering, there must be a covering-map $$F\colon M \to M$$ such that $$\pi_1 = f^{-1}\circ\pi_2\circ F$$. By the same logic, $$\pi_2\colon M \to M/\Lambda$$ is a universal cover, so there must be a covering map $$G\colon M \to M$$ such that $$\pi_2 = f\circ\pi_1\circ G$$.

Now note that $$\pi_1\circ G\circ F = f^{-1}\circ (f\circ\pi_1\circ G)\circ F = f^{-1}\circ \pi_2\circ F = \pi_1$$, so $$G\circ F$$ is a Deck transformation. If we chose $$F$$ and $$G$$ to do the right thing to base points, we can assume $$G\circ F = id$$, and let $$F = \tilde{f}, G = \tilde{f}^{-1}$$.

(3) I hope this question isn't too basic. I asked it on the Math Stackexchange and didn't get any responses.

Thank you!

• The argument you give in $(2)$ seems correct to me, and shouldn't be too hard to formalise: morphisms M->N/G can be lifted locally everywhere, essentially by definition, hence if $M$ is simply connected all the lift along different path will agree hence giving a unique and well defined lift. Though I don't know the litterature about Orbifold well enough to say whether this can be found somewhere or not, moreover there have been historically several non-equivalent definitions of morphisms of Orbifolds, I'm not sure this works for all these definitions (but it definitely works for some of them). – Simon Henry Sep 17 '18 at 13:58
• I have no doubt your question has a positive answer, which should follow more or less from the definitions, but locating the definitions is another matter... You could try reading chapter 18 of Noohi's Foundations of Topological Stacks I - in particular Proposition 18.18 seems to answer your question, if you're okay with thinking about orbifolds as stacks. (I unfortunately don't have a good grasp on the classical orbifold language) – Will Chen Sep 17 '18 at 17:20