What mapping does Escher use? In Escher's hyperbolic tesselations, he takes (effectively) a tesselation of the plane and maps it to a tesselation of the unit disk, by a mapping that takes straight lines to circles meeting the disk boundary at right angles.  What precisely is this mapping (with a formula)?  I am aware that this is effectively a mapping from the hyperbolic plane to the Poincare disk, taking a mapping from a tesselation of the hyperbolic plane to the disk.  However, what is it as a map from $R^2$ to $D$, with formula.  I believe that it is a conformal map.
What is the correct formula in higher dimensions, also?
 A: In my Ph.D. thesis I described a process for turning a Euclidean ornament into a hyperbolic one. The key point here is to increase the order of some of the centers of rotational symmetry. That decreases the interior angle of a single fundamental domain at that corner, changing from Euclidean interior angles sum equation to hyperbolic angle sum inequality. I'll argue that for aesthetic reasons the interior of the fundamental domain should be mapped conformally, preserving angles.
I'd not so much think of this as a map of the whole Euclidean plane to the hyperbolic plane, because due to that change to centers of rotation you can no longer clearly associate individual copies of the ornament in the two planes with one another. Instead I'd speak about a mapping of the orbifold, which is the thing you  get by identifying points of the plane that contain the same part of the ornament. The transformation I'm describing changes the metric and combinatoric properties of the orbifold while preserving its topology.
You're asking for a formula. The best I have to offer in terms of exact formula would be based on conformally mapping triangles to the unit disc, which works for schneller groups that have a triangle as their fundamental domain. According to the Riemann mapping theorem, such a map has to exist, and there is a Schwarz-Christoffel mapping that can be expressed in closed form with the use of the hypergeometric function $_2F_1$. If you do that mapping forward for one and in reverse for the other triangle, you can get a conformal map from an Euclidean to a hyperbolic triangle.
In practice I found that approach to be numerically very unstable. So instead of a closed form formula I went for an approximation using discrete conformal maps. This maps a triangle mesh without detour via the unit disc. It also works for ornaments whose fundamental domain is not a single triangle.
If I remember to, I'll add some illustrations later on. Till then I'll recommend you have a look at a paper of mine which comes with a bunch of illustrations, including a few with actual Escher material.
