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Let $\Gamma$ be the cyclic group generated by the matrix $$\begin{pmatrix} \cos(2\pi/3) & \sin(2\pi/3) & 0 \\ -\sin(2\pi/3) & \cos(2\pi/3) & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ Show that $S^2/\Gamma$ is not a differentiable manifold. Any hint? I can prove that $C/\Gamma$ is a manifold (where C is $S^2$ without the north and south pole), but I don't know how to show that $S^2/\Gamma$ is not a manifold. Maybe it is not locally euclidean? How to prove it? Thanks $\ \ \ \ \ $

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  • $\begingroup$ Well you've shown that $C/\Gamma$ is a manifold, so you've shown that each point has an neighborhood isomorphic to a neighborhood of any other point. I would imagine that looking at a neighborhood of one of the pole points and lifting it to $S^2$ could answer your question. $\endgroup$ Commented Dec 19, 2015 at 15:52

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Without further context, it's not even clear that the statement is true. What is true that the usual quotient manifold theorem (which says that a smooth, free, and proper action of a Lie group on a smooth manifold yields a smooth manifold as quotient space) doesn't apply here, because the action is not free.

But that doesn't imply that the quotient is not a smooth manifold with some smooth structure. First of all, $S^2/\Gamma$ with the quotient topology is in fact homeomorphic to $S^2$, so it is a topological manifold, and of course $S^2$ has a smooth structure making it into a smooth manifold.

Moreover, it's possible to choose that smooth structure in such a way that the quotient map $q\colon S^2\to S^2/\Gamma$ is a smooth map. Here's how. View $\mathbb R^3$ as $\mathbb C^2\times\mathbb R$, with coordinates $(w,z) = (x+iy,z)$, and define a map $F\colon S^2 \to S^2$ by $$F(w,z) = \frac{(w^3,z)}{\sqrt{ (w\bar w)^3 + z^2}}. $$ This map is the restriction of a smooth map from the complement of the origin in $\mathbb C^2\times\mathbb R$ to itself, so it's smooth. It's a quotient map by the closed map lemma, and it makes the same identifications as $q\colon S^2\to S^2/\Gamma$, so it descends to a homeomorphism $\widetilde F\colon S^2/\Gamma\to S^2$. If we give the quotient space the unique smooth structure that makes $\widetilde F$ a diffeomorphism, then $q$ is a smooth map because it's equal to the composition $\widetilde F^{-1}\circ F$.

Another context in which to view the question is that of orbifolds. The quotient $S^2/\Gamma$ is an orbifold, more or less by definition, so we can give it the orbifold smooth structure induced by $q$. (This just boils down to saying that a function $f\colon S^2/\Gamma\to\mathbb R$ is smooth if and only if $f\circ q$ is smooth.) With this structure, the orbifold has two singular points and thus is not a smooth manifold.

The orbifold context might very well be what the statement is referring to; but then it should have said "Show that $S^2/\Gamma$ is not a differentiable manifold when given the orbifold smooth structure induced by $\Gamma$."

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In fact it is a manifold, $S^2$. Look at $S^2$ as the extended complex plane. $$S^2 = \mathbb{C} \cup \{ \infty\} ( = P^1(\mathbb{C}) )$$ Consider the surjective map $$p \colon \mathbb{C} \cup \{ \infty\} \to \mathbb{C} \cup \{ \infty\}, z \mapsto z^3$$ Then $p(z_1) =p(z_2) \iff z_1 \equiv_{\Gamma} z_2$. Therefore we have $$S^2/\Gamma \simeq \mathbb{C} \cup \{ \infty\} \simeq S^2$$

(works for every $n$ instead of $3$).

Obs: I think @Jack Lee: pointed this out in his answer.

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