# Conformal Mappings for hyperbolic polygon

I am searching for a conformal mapping from the upper halfplane onto a hyperbolic polygon, i.e. the sides of the polygon have to be geodesics.

The classical Schwarz Christoffel theorem does the job for euclidean polygons (see e.g. http://en.wikipedia.org/wiki/Schwarz-Christoffel_mapping).

Does anybody know of a similar constructions in hyperbolic geometry?

Does anybody know of similiar constructions for any other domains?

Any idea will be very wellcomed! I am far from being an expert in conformal mappings and do only know some isolated examples!

• I'm no expert on the subject, but presumably you could take normal polygons and and map them into the unit disc in the usual way. Nov 15 '10 at 7:00
• @Adam: a hyperbolic polygon means one whose sides are geodesics for the hyperbolic metric, so they are not usually straight lines. If you take a polygon with straight edges in the upper half plane then the edges are not geodesic for the hyperbolic metric on the upper half plane, and the images in the unit disc are not geodesic for the hyperbolic metric there either. Nov 15 '10 at 8:32
• That is exactly, what I am searching. Thanks Neil for the clarification. Nov 15 '10 at 8:36

See Harmer and Martin's work on Conformal Mappings from the Upper Half Plane to Fundamental Domains on the Hyperbolic Plane.

Some of the ideas developed by Christopher Bishop in the context of computational geometry may also be of interest. See his talks and papers on conformal maps.

• So the work is done already;) Thanks for the refernce! Nov 17 '10 at 8:28
• I have now found a better reference for the question considered above: Nehari - Conformal Mappings - Chapter V - page 198 Nov 24 '10 at 15:42
• Good question, I also happen to need an explicit formula for conformally mapping a hyperbolic polygon to the unit disk or the upper half plane. Is that the 1st edition of Nehari's book ? Is this formula explicit ? Mar 7 '12 at 0:12

There is a theory of conformal map for circular polygons (polygons bounded by arcs of circles). But in this case, instead of an integral in the Schwarz-Christoffel formula, you obtain a linear differential equation. In the case of a circular triangle, the equation is hypergeometric and you have an explicit representation of your mapping. The paper of Harmer and Martin mentioned in the previous answer deals mainly with the case of a triangle. The most comprehensive treatment of triangles is in the second volume of Caratheodory's textbook on complex variables, and in other books on hypergeometric functions. The case of quadrilateral is the simplest case when there is no explicit formula. It was subject of much research. See, for example, arXiv:1110.2696, arXiv:1111.2296, and references in these papers.

In the Schwartz Christoffel differential vector equation, just use higher derivatives instead of first derivatives.

• I have no idea what you mean here. Could you please elaborate on your answer? Aug 9 '11 at 17:45