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Let $\Gamma$ be a finite subgroup of $SO(3)$ acting on $\Bbb R^3$. What sort of space do we get by taking the quotient $\Bbb R^3/\Gamma$? Is that a manifold?

The group $\Gamma$ is compact since it is finite, so $\Bbb R^3/\Gamma$ is a manifold whenever the action is free. But what if the action is not free? Do we have examples where the action is not free?

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    $\begingroup$ Certainly it won't be free at the origin, and you will have an orbifold singularity there. Maybe as a warm up you should try to visualize $\mathbb{R}^2$ modulo the action of a cyclic group acting by rotation. $\endgroup$ – Nate Mar 27 '16 at 20:06
  • $\begingroup$ For examples "where the action is not free" see here. $\endgroup$ – Dietrich Burde Mar 27 '16 at 20:10
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The quotient is always a 3-dimensional manifold, actually it is always homeomorphic to $R^3$. One way to see this is to note that the quotient is a cone over $S^2/\Gamma$. Now, you have to check that $S^2/\Gamma$ is homeomorphic to $S^2$, whenever $\Gamma$ is a finite subgroup of $SO(3)$. To see this, first, observe that it is a compact 2-dimensional manifold $S$ (I will leave it to you as an exercise). Second, it is a general theorem due to Armstrong that if $G$ is a group acting properly discontinuously on a simply connected simplicial complex $X$, then $\pi_1(X/G)$ is isomorphic to the quotient group of $G$ by the normal subgroup generated by all elements having nonempty fixed point sets in $X$. In your case, every element of $\Gamma$ has a fixed point in $S^2$, hence, $\pi_1(S)= 1$. Therefore, $S\cong S^2$ by the classification of surfaces.

Edit. Just for completeness: If you allow for finite subgroups of $O(3)$, then the quotients are not necessarily manifolds. First of all, they could be manifolds with boundary, in the case when the subgroup $\Gamma$ contains reflections. If $\Gamma=\{I, -I\}$, the quotient is the cone over $RP^2$, and, hence, is not even a manifold with boundary.

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  • $\begingroup$ The theorem of Armstrong probably has some hypothesis you omitted, as otherwise it applies to $G=1$ with weird consequences :-) $\endgroup$ – Mariano Suárez-Álvarez Mar 30 '16 at 22:14
  • $\begingroup$ @Mariano: Yes, I forgot to say that X is simply connected. $\endgroup$ – Moishe Kohan Mar 30 '16 at 23:03

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