# Riemann surface construction

1. May 29, 2012

### angy

I can't understand how to recover a Riemann surface from the branch data. In particular, given a group acting on the Riemann sphere with some points removed, how can I construct a Riemann surface?

2. May 30, 2012

### mathwonk

It is easier to describe how to go backwards, from a Riemann surface to a punctured sphere plus branching behavior.
The whole construction is based on the fact that an unbranched cover of a simply connected region is trivial, i.e. just a finite number of disjoint copies of that region.

Now let a Riemann surface X be mapped as a branched cover of the sphere S. That means at every point p of X, the map looks locally topologically like z→z^k for some k ≥ 1. The branch order is k-1, so z→z^2 has simple branch order.

There are a finite number of points of X where k > 1, called ramification points. Their images are called branch points. Now choose a non branch point O in S, and join O to each of the branch points b1,…,br, by a sequence of disjoint arcs. Then if we assume the points b1,…bn are arranged around a circle with O at the center, the arcs are like r spokes emanating from O to the various branch points.

Now cut the sphere S along these arcs, leaving a sphere minus O and the r spokes joined only at O. The complement of these arcs in the sphere is homeomorphic to a disc (assuming r ≥ 1), hence simply connected. Thus the part of the Riemann surface over that simply connected complement looks like a finite disjoint union of say n copies of that slit sphere. Thus the Riemann surface can be retrieved from a disjoint collection of n slit spheres, essentially polygons, by gluing the slits back together. How to do that is given by the branching behavior at the n branch points b1,…,bn.

Now lets describe the branching behavior at the points b1,…,bn. Now for each of the r slits on the sphere, there were n copies of that slit lying over it on the Riemann surface. Paint each of the n slits on the Riemann surface lying over the same slit on S a different color. Then when we cut the Riemann surface along those n slits to separate it into n disjoint discs, the paint sticks to each side of the slit so we can see where it needs to be glued back.

The whole secret is that when the Riemann surface disconnects into n copies of a slit sphere, i.e. after cutting the Riemann surface along all the slits, there are n copies of the slit sphere with edges colored, but there is no reason for the two apparently opposing edges on a given copy to be the same color!

I.e. if there was simple branching at a point upstairs at the end of a slit, then the two edges of that slit are glued to the two edges of a different copy of the slit sphere. I.e. the two edges of one copy of the disc ending at that point are colored say red and green, while the two edges of the other disc are colored green and red. so when you glue red to red and green to green you connect up the two discs.

I.e. locally to recover the Riemann surface, we take two copies of a disc, cut both discs from 3 o’clock to the center of the two discs, then we glue each edge of one disc to the opposing edge of the other disc! Thus topologically, after gluing, we get something homeomorphic to a single disc.

I.e. just as a single disc can map to a single disc two to one, with one branch point t the center, by the map z→z^2, we can recover the domain disc of this map by cross gluing two slit copies of a disc in this way.
OK?

So after cutting apart our Riemann surface, we have n copies of a slit sphere, each looking like a polygon with jag tooth edges, and we have colors on all the edges. The colors tell us how to match up the slits to recover the original Riemann surface.

Now we can encode this into a group action, by the “group of loops”, if we number the discs. I.e. each loop around a branch point on S, gives us a local permutation of the n sheets.

So suppose we choose a loop going once counterclockwise around a branch point and ask how it acts on sheet/disc # 1. Think of the branch point as the origin. We look upstairs over this, at the disc numbered 1, and see what color the edge is that lies on the edge of the part lying over the part of the plane below the positive x axis.

Then we look at the color of the facing edge lying over the quadrant above the x axis. If it is the same color, then our loop sends 1 to 1. If the facing edge is a different color, we look for the disc that has its facing edge of that color. Say that is disc #4, then the loop sends 1 to 4.

Thus each loop around each branch point defines a permutation of the integers {1,…,n}. Since the fundamental group of the punctured sphere is generated by these loops, we get an action by that fundamental group on the set {1,…,n}, i.e. a homomorphism from the fundamental group to S(n).

Actually we should note that the fundamental group satisfies one relation: the product of all r loops is the identity. This means the product of the actions by all the loops must be the identity permutation.

So defining a Riemann surface by branching data, or by a group action, means giving a homomorphism from the fundamental group of the punctured sphere, to S(n). If the Riemann surface should be connected, we need the image of the homomorphism to be a transitive subgroup.

Ok? Now you know why no one answered right away.

Last edited: May 30, 2012
3. Jun 4, 2012

### angy

Thank you very very very much!!!!
If I can, I would like to ask another question... In my notes I wrote the following: let X be a topological space and let p:Y->X be a covering. Let p* be the induced homomorphism between the fundamental groups. Then Y can be constructed as the universal cover of X quotiented by the image of p*. Is it true? In this case, is there a way to apply this result to our construction?

4. Jun 4, 2012

### mathwonk

you can find thiS in any topology book. pleaSE FORGIVE ME as I AM TIRED AND NOT ABLE to write this up well. try hatcher.