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In Thurston's book The Geometry and Topology of Three-Manifolds it is proven that the underlying space of a two-dimensional orbifold is always a topological surface.

Are there any easy examples of higher dimensional orbifolds whose underlying spaces are not topological manifolds?

Stefan Kohl
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user88649
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    The quotient of $\mathbb R^3$ by the order $2$ group generated by $-id$ is an open cone over $RP^2$. It is contractible, and if it were a 3-manifold, then removing a point would not change its fundamental group, which is trivial. Removing the cone point gives a non-simply-connected space. – Igor Belegradek Mar 07 '16 at 22:46
  • Perhaps it's worth pointing out that, even in dimension 1 (and 2), the underlying topological space of a closed orbifold can be a manifold with boundary. For instance, the closed interval $[0,1]$ with both endpoints labelled $\mathbb{Z}/2$ is a closed 1-dimensional orbifold, since it's double covered by the circle. – HJRW Mar 08 '16 at 11:20

2 Answers2

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It is quite easy to give an example in real dimension $4$.

In fact, it was shown by D. Mumford in the paper

The topology of normal singularities of an algebraic surface and a criterion for simplicity, Inst. Hautes Etudes Sci. Publ. Math. (1961), no. 9, 5 - 22

that a normal algebraic surface over $\mathbb{C}$, whose underlying topological space (in the usual topology) is a topological manifold, must be nonsingular.

Now take the orbifold $X:=\mathbb{C}^2/G$, where $G$ is the group of order $2$ whose generator $\xi$ acts as $\xi \cdot (x, \, y) = (-x, \, -y)$.

The $G$-invariant subalgebra is generated by $A:=x^2, \, B=xy, \, C:=y^2$, so the underlying algebraic surface $X$ is isomorphic to the affine quadric cone $\{B^2-AC=0 \} \subset \mathbb{C}^3$, which has a $A_1$-singularity at its vertex. Such a singularity is normal, because it is a codimension two hypersurface singularity.

Then $X$ cannot be a topological manifold by Mumford's result mentioned above.

Francesco Polizzi
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    Is my intuition correct that the boundary of any small neighborhood of the singular point would have a nontrivial fundamental group, and this should not happen for topological manifolds? Or am I being naive here? – Lev Borisov Mar 07 '16 at 22:28
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    If I remember correctly, Mumford strategy of proof is precisely based on the analysis of the local fundamental group of the singularity. – Francesco Polizzi Mar 07 '16 at 22:29
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    very small neighborhood minus the singular point itself. Every orbifold is locally contractible. – YCor Mar 07 '16 at 22:30
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    @YCor Thank you, this makes more sense than trying to somehow think of the boundary. – Lev Borisov Mar 07 '16 at 22:31
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To give a compact version of Igor's example, consider the antipodal map in the tangent space of a round 3-sphere at a point. This extends to an isometry of the 3-sphere with a pair of fixed points. The quotient by the isometry is a manifold with two conical singularities which is not homeomorphic to a topological 3-manifold.

Mikhail Katz
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