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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?

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    $\begingroup$ There is no conformal map between the plane and open disk. Rather, it would be between a half-plane and the open unit disk. This is the Cayley transform. In general, you can map any disk to any other disk, and half-planes are just disks! Indeed, you can map any three points to any other three points using the cross-ratio formula as a Mobius transformation of one of its variables, which is useful because any circle is determined by three points. $\endgroup$
    – anon
    Commented Jun 11, 2021 at 4:10
  • $\begingroup$ Thanks. My understanding of the hyperbolic tesselations of Escher was a little lacking. The Poincare disk gives a model in which the circles in question are images of the geodesics in the hyperbolic plane, not the geodesics in the Euclidean plane. With this metric then, I believe it is conformal. One thing I do not understand in your answer is how the upper half plane is a "disk" and and the full plane is not. Both are topologically disks. So, what do you mean by disk when you say that you can map any disk to any other disk? $\endgroup$
    – rihartley
    Commented Jun 11, 2021 at 6:38
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    $\begingroup$ I guess if you want a technical definition (making it tautological) I mean the image of the unit disk under mobius transformations, which is exactly all familiar disks as well as all half-planes. A less restrictive condition would be regions conformally equivalent to the open unit disk, but then that would include a massively infinite collection of non-disk shapes as well. Even then, though, the unit disk is not conformally equivalent to the plane! In certain kinds of geometry, lines and circles together are called "generalized circles"; this is just the filled-in version of that. $\endgroup$
    – anon
    Commented Jun 11, 2021 at 8:10
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    $\begingroup$ Under stereographic projection, in which the complex plane is the Riemann sphere minus a "point at infinity," generalized circles just become circles. The lines in the plane correspond to those circles in the Riemann sphere through the point at infinity that we stereographically project from. The Riemann sphere is the proper place to interpret the action Mobius transformations, not the plane really. With this viewpoint, we treat lines and circles as being one and the same. $\endgroup$
    – anon
    Commented Jun 11, 2021 at 8:10
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    $\begingroup$ I think, your question is based on a misunderstanding: There is no such thing as "the plane." If you mean "the hyperbolic plane" then it is unnatural to identify it with the Cartesian plane, the unit disk is one of the standard models of the hyperbolic plane and there is no need to map anything to the disk. Given the highly symmetric nature of Escher's drawings, it is very unlikely that he worked with some other model of the hyperbolic plane. $\endgroup$ Commented Jun 11, 2021 at 19:26

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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 smaller 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.

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    $\begingroup$ Thanks. The paper looks very interesting (and has some great pictures). $\endgroup$
    – rihartley
    Commented Jun 14, 2021 at 9:36

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