On contractible orbifolds

We prove that a contractible orbifold is a manifold.


Introduction
Following [Dav10], we call an orbifold X contractible if all of its orbifold homotopy groups π orb i (X), i ≥ 1 vanish. We refer the reader to [Dav10] and the literature therein for basics about orbifolds. Davis has asked in [Dav10], whether any contractible orbifold X must be developable. In this note we answer this question affirmatively. THEOREM 1.1. Let X be a smooth contractible orbifold. Then it is a manifold.
Proof. Since X is contractible, it is it is orientable. Let n be the dimension of X. Define on X a Riemannian metric and let the smooth manifold M be the bundle of oriented orthonormal frames on X (cf. [Hae84]). Then G = SO(n) acts effectively and almost freely on M with X = M/G.
Let E denote a contractible CW complex on which G act freely, with quotient E/G = BG, the classifying space of G. ThenX = (M ×E)/G is the classifying space of X (cf. [Hae84]). By definition, the orbifold homotopy groups of X are the usual homotopy groups ofX. Thus, by our assumption, the topological spaceX is contractible.
The projection M × E →X is a homotopy fibration. Thus the contractibility ofX implies that the embedding of any orbit of G into M × E is a homotopy equivalence between G and M × E. Since E is contractible, the projection M × E → M is a homotopy equivalence as well. Therefore, for any p ∈ M, the composition o p : G → G · p → M given by orbit map o p (g) := g · p is a homotopy equivalence.
Assume now that X is not a manifold. Then G does not act freely on M. Thus, for some p ∈ M, the stabilizer G p of p is a finite non-trivial group. Then the orbit map o p : G → M factors through the quotient map π p : G → G/G p . Since the orbit map is a homotopy equivalence, there must exist some map i : G/G p → G such that i • π : G → G is a homotopy equivalence. However, the manifolds G and G/G p are orientable and the map G → G/G p is a covering of degree |G p |. Thus, for m = dim(G) = n(n − 1)/2, the image of π * p in H m (G, Z) = Z is a subgroup of H m (G, Z) of index |G p |. In particular, (i • π) * cannot be surjective. Contradiction.
We add the following not surprising improvement: PROPOSITION 1.2. Let X be a smooth n-dimensional orbifold. If the orbifold homotopy groups π orb i (X) are trivial, for all 0 ≤ i ≤ n, then X is a contractible manifold.
Proof. Let M, G andX = M × E/G be as in the proof of Theorem 1.1. Consider the induced vector bundleŶ =X ×R n /G over the classifying spaceX. By assumption,X is n-connected, hence the vector bundlê Y →X is a trivial vector bundle. But M × E is by construction ofŶ the space of oriented frames inŶ . Thus M × E is homotopy equivalent toX × G as a principal G-bundle.
We deduce again, that the orbit map o p : G → M used in the proof of Theorem 1.1 admits an inverse i : M → G, such that i • o p is homotopy equivalent to the identity. Again, as above, this provides a contradiction if the stabilizer G p is not trivial. Therefore, X is an n-dimensional manifold. Since X is n-connected, it is contractible. I wonder whether Proposition 1.2 is optimal in high dimensions. In fact, I do not know a single example of a 4-connected bad orbifold. I would like to finish the note by formulating two problems.