What does the following manifold look like?

In hyperbolic geometry, we know that any complete hyperbolic surface is the quotient $$\mathbb{H}^{2}/\Gamma$$ where $$\Gamma < \text{Isom}(\mathbb{H}^{2})$$ a discrete subgroup.

We know $$\text{Isom}^{+}(\mathbb{H}) \cong PSL(2,\mathbb{R})$$ which contains $$PSL(2,\mathbb{Z})$$ as a discrete subgroup. Also $$\Gamma(m)$$ which is the kernel of the natual map $$PSL(2,\mathbb{Z}) \to SL(2,\mathbb{Z}/m\mathbb{Z})$$ acts freely and properly discontinuously on $$\mathbb{H}^{2}$$ so $$\mathbb{H}^{2}/\Gamma(m)$$ is a hyperbolic surface. I want to see a geometric picture of this manifold. How can I proceed to see what this manifold look like (at least topologically, if not isometrically)?

From a naive point of view it seems reasonable. You can start with a fundamental domain for the fractional linear action of $$PSL(2,\mathbb Z)$$ on the upper half plane; the conventional fundamental domain is given by intersecting $$-\frac{1}{2} \le \text{Real}(z) < \frac{1}{2}$$ with $$\text{abs}{z} \ge 1$$. Then you can carefully choose coset representatives of $$\Gamma(m)$$ in $$PSL(2,\mathbb Z)$$, use them to translate the fundamental domain around, and hope that you chose those representatives carefully enough so that the union of those translated fundamental domains is a polygon which you can then glue up to get $$\mathbb H^2 / \Gamma(m)$$. For the case $$m=2$$ this can probably be done by hand. But the index of $$\Gamma(m)$$ in $$PSL(2,\mathbb Z)$$ grows uncomfortably fast: it is cubic in $$m$$. So this method is not practical in general.
The general heading of your question is the topic of congruence subgroups of $$PSL(2,\mathbb Z)$$. In that reference you'll see, for example, a link to a 2006 paper by Long, Machlachlan and Reid proving that there are only finitely many values of $$m$$ such that the quotient surface $$\mathbb H^2 / \Gamma(m)$$ has genus zero, meaning that it is homeomorphic to a sphere with some finite number of points removed. You'll also see described some amazing connections between the topology of $$\mathbb H^2 / \Gamma(m)$$ (more properly, not that surface itself but a closely related quotient orbifold) and the prime factors of the order of the monster group!
• I was not expecting that surfaces that can be described so easily could be so monstrous. But is there really no way to tell anything about the surface $\mathbb{H}^{2}/\Gamma(m)$? – Prakhar Gupta Dec 1 '18 at 17:04
• Well, that's what the second paragraph of my answer does: it tells something about the surfaces $\mathbb H^2 / \Gamma(m)$. And there's still more to learn about them from the theory of congruence subgroups. But you have to work hard to understand the statements and their proofs. It's true: sometimes mathematical objects that are relatively simple to define are very difficult to say something substantial about. – Lee Mosher Dec 3 '18 at 22:01