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I was reading this post and wondered about a similar thing.

Let $\mathbb{Z}_2$ act on $S^2$ by letting the non-trivial element take $$(x,y,z)\mapsto(-x,-y,z).$$ This action is not free, so one would expect something to go "wrong" with the quotient space $S^2/\mathbb{Z}_2$.

On the surface, it seems like $S^2/\mathbb{Z}_2$ is a nice space; it is homeomorphic to $S^2$, which is a topological manifold. However, the north and south ends of $S^2/\mathbb{Z}_2$ seem somewhat crushed and "conical", instead of smooth. But I'm not sure how to match this intuition with something rigorous.

Question 1: Does $S^2/\mathbb{Z}_2$ inherit a smooth structure from $S^2$? If not, why not?

Question 2: Is there some precise sense in which the north and south ends of $S^2/\mathbb{Z}_2$ are conical? (This would seem to assume that $S^2/\mathbb{Z}_2$ has a natural Riemannian metric; why is this the case?)

geometricK
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    Take this with a huge grain of salt: I don't think the quotient inherits a smooth structure from $\Bbb S^2$, since the fixed points of the action may correspond in the quotient to "boundary points" or to a point which is not in the domain of a coordinate system. I do think it should turn out to be an orbifold, and although I'm really not familiar with those things, I understand that one can define tangent spaces and Riemannian metrics for orbifolds. Since $\Bbb Z_2$ acts by isometries, there should be a "Riemannian metric" (whatever that means now) on the quotient. – Ivo Terek Mar 08 '22 at 05:03
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    As Ivo Terek said, this is called an orbifold. It is similar to the sphere but the two poles are replaces with singular points with neighbourhoods of the form $\Bbb R^2 / \Bbb Z_2$. – Didier Mar 08 '22 at 08:31
  • The linked question is now answered. tl; dr: The poles of the Riemannian quotient are conical, but there is a smooth holomorphic quotient (the sphere again). – Andrew D. Hwang Mar 08 '22 at 13:04

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