The isometry group of the 3-dimensional hyperbolic space $\mathbb{H}^{3}$ is $PSL(2,\mathbf{C})$. What are the homotopy groups of the quotient space $\mathbb{H}^{3}/PSL(2,\mathbf{Z})$ ?
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As a topological space, this is homotopy equivalent to $\mathbb{D}^3,$ so the homotopy groups are whatever they are for $\mathbb{D}^3.$ As an orbifold, the fundamental group is $\mathbb{PSL}(2, \mathbb{Z}),$ while the higher homotopy groups vanish, since the universal cover is $\mathbb{H}^3.$
NOTE Thanks to HJRW for the correction.
answered Nov 3, 2015 at 19:28
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$\begingroup$ @HJRW Actually, unless I am confused (possible, due to lack of sleep), the quotient of $\mathbb{H}^2$ by $SL(2, Z)$ is a sphere (topologically), and I think the quotient of $\mathbb{H}^3$ is topologically just that cross a line. No? $\endgroup$ Commented Nov 3, 2015 at 19:36
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2$\begingroup$ Perhaps the confusion is that the topological quotient is not $\mathbb{S}^2$, but rather $\mathbb{R}^2$ (it's the $j$-line...) $\endgroup$ Commented Nov 3, 2015 at 23:40
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1$\begingroup$ @IgorRivin, the quotient of $\mathbb{H}^2$ by $PSL(2,\mathbb{Z})$ is the modular orbifold, which is a punctured 2-sphere with order-2 and -3 cone points. (This follows from the usual picture of the fundamental domain for the action on $\mathbb{H}^2$.) $\endgroup$– HJRWCommented Nov 4, 2015 at 6:34
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1$\begingroup$ @IgorRivin, perhaps it's helpful to remember, which I know you do well, that $PSL(2,\mathbb{Z})$ is virtually free. This couldn't be the case if its action on the hyperbolic plane were cocompact. $\endgroup$– HJRWCommented Nov 4, 2015 at 6:41