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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?

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  • $\begingroup$ At least in the cases with which I'm familiar, X exists because H/Gamma is a punctured Riemann surface, and after plugging the relevant hole charts one gets a compact Riemann surface, which is a smooth projective complex curve. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 29, 2010 at 21:24
  • $\begingroup$ elliptic points under group action means usually means stabilizer of the point is a nontrivial subgroup (eg, \Gamma = \Gamma(1), then $i$ and $\pho=cube root of unity$ are elliptic points and parabolic fixed point means points stable under the action of parabolic elts(means absolute value of trace is 2, but element is not I or -I , eg [1 1; 0 1]), I usually think of them as cusps $\endgroup$
    – Dipramit Majumdar
    Commented Mar 29, 2010 at 21:27
  • $\begingroup$ I am still not sure whether I understand the words elliptic and parabolic or not. So an elliptic point means a point on $X$ with non-trivial stablizer, and a parabolic point means a point that is not in $X$, but comes from a natural completion of the quotient. I am not familiar with the automorphisms of the complex unit disk. Is there a book that I can look at and get familiar with these basic concepts? $\endgroup$
    – Botong Wang
    Commented Mar 31, 2010 at 21:28

2 Answers 2

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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.

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  • $\begingroup$ Thanks for your answer too. I am not sure how is the Riemannian metric obtained? Do you just choose one which gives finite measure? Is there any relation with the conformal structure? $\endgroup$
    – Botong Wang
    Commented Mar 31, 2010 at 21:19
  • $\begingroup$ The Riemannian metric came from the original hyperbolic metric: you started with the assumption you had $\Gamma$ acting on $H$ so the quotient had finite measure. A Riemannian metric on a 2-manifold gives you a conformal structure, where multiplication by $i$ is rotation by $90$ degrees. $\endgroup$
    – Dylan Thurston
    Commented Apr 1, 2010 at 0:03
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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.

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  • $\begingroup$ Thanks for you answer, the third paragraph is really helpful. What will be the space of all structures with parabolic ends in $S^2$ minus four points? $\endgroup$
    – Botong Wang
    Commented Mar 31, 2010 at 21:17
  • $\begingroup$ Botong, You need to read up on Teichmuller theory. The moduli space of all complex structures on $S^2$ with parabolic ends is an orbifold, it has a canonical branched cover called Teichmuller space. I liked Zieschang's "Finite Groups and Mapping Classes of Surfaces" or Fathi, Laudenbach and Poenaru's "Travaux de' Thurston" as entry point texts. Charlie $\endgroup$
    – Charlie Frohman
    Commented Apr 4, 2010 at 16:21

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