I was reading Mehta and Seshadri's paper "Moduli of vector bundles on curves with parabolic structures".
In the second paragraph, they wrote:
"Suppose that $H$ mod $\Gamma$ has finite measure ($H$ is the complex upper half plane, and $\Gamma$ is a discrete group). Let $X$ be the smooth projective curve containing $H$mod$\Gamma$ as an open subset and $S$ the finite subset of $X$ corresponding to parabolic and elliptic fixed points under $\Gamma$."
I am not sure about what parabolic and elliptic mean here. And why does such an $X$ exist? If I take a Riemann surface and remove several small balls from it, when is it biholomorphic to another Riemann surface removing several points?
Best Answer
I like to think about this geometrically. $H/\Gamma$ is a topological metric space. At most points of $H$ (that are not fixed by any element of $\Gamma$, the quotient looks just like the hyperbolic plane $H$ itself. The singularities come from elliptic elements of $\Gamma$, i.e., (locally) rotations, where you get a cone metric (locally like the metric on a piece of paper rolled into a cone). Remove those points and consider the conformal structure coming from the resulting Riemannian metric. The ends of this surface look like removable singularities (locally like $\mathbf{C}$ minus a point in their conformal structure), and so can be filled in uniquely. $X$ is the result of filling in the points, and $S$ is the set of filled in points.
The points in $S$ come both from the removed cone points and from some points at infinity, the elliptic points as Charlie explains above.
You have to be careful to distinguish between removing a point (leaving a removable singularity) and removing a small ball (which is different conformally). Removing a small ball always gives a surface with infinite hyperbolic area when uniformized, and is never equivalent to a compact surface minus points.