Upper half plane quotient by a discrete group 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?
 A: 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.
A: If $A \in PSL_2(\mathbb{R}$ then $tr^2(A)>4$ means its hyperbolic. There is a unique geodesic in the hyperbolic plane that is invariant under $A$ and $A$ acts a translation along that geodesic. If $tr^2(A)<4$ then $A$ is elliptic.  It has a unique fixed point in the hyperbolic plane and it acts as rotation about that fixed point. If $tr^2(A)=4$ then it could be the identity, or it could be parabolic. If it is parabolic it has a unique fixed point on the circle at infinity in the hyperbolic plane. If you put that fixed point at infinity, then it looks like translation $z\rightarrow z+c$ and with scaling you can make that $z\rightarrow z+1$.
If you take a quotient of the hyperbolic plane by the group generated by $z+1$ you get an open once punctured disk. You should think of this as an open neighborhood of the point
"missing" from the surface.
If you have a surface of finite type, that is it is homeomorphic to the result of removing n points from a closed oriented surface, then the space of all conformal structures on that space  is larger than the space of all structures with parabolic ends which is larger than the space of structures on the closed surface.
Here is an elementary example. The conformal maps from $S^2$ to itself are exactly the group of Mobius transformations. You know  a Mobius transformation is determined by where it sends three points.  Let $C$ be the space of choices of 4 points from $S^2$. Two choices are equivalent if and only if there is a Mobius transformation taking one to the other.  I can choose to send three points to three points, but I have no control of where the $4$th point goes so the space of conformal equivalence classes is one complex, or two real dimensional.
