# numerically approximating the conformal map between two curvilinear triangles to high precision

Here is a triangular region $$T$$ whose two curved edges are complicated analytic curves that I know only numerically, but can compute to any desired precision:

And here is a simpler region $$H$$ whose one curved side is $$\pi/20=9^\circ$$ of a circular arc:

Both regions have vertex angles of $$\pi/5$$, $$\pi/4$$, and $$\pi/2$$; so there is a unique conformal map $$f\colon T\to H$$ that takes each vertex of $$T$$ to the corresponding vertex of $$H$$. And the map $$f$$ is holomorphic, not only on the interior of $$T$$, but also on its boundary.

I have managed to approximate $$f$$ numerically by a degree-13 polynomial $$p(z)$$ of a complex variable $$z$$ whose coefficients are real numbers. So $$p$$ maps the real axis precisely to itself. And $$p$$ maps the two curved boundaries of $$T$$ to analytic curves that are never farther than 0.00003 from the corresponding boundary of $$H$$. I have thus approximated $$f$$ to roughly four digits.

But I would like to approximate $$f$$ much more precisely --- to at least ten digits, maybe twenty. Can anyone suggest a technique by which I could achieve this? Replacing $$p$$ by a polynomial of higher degree isn't an attractive strategy, since the precision achieved is likely to grow only slowly with increasing degree, while solving for the coefficients of a high-degree polynomial is numerically challenging. Indeed, polynomials may not be a good library of functions to use as approximations to $$f$$, since polynomials are entire, while $$f$$ probably has singularities at finite points outside of $$T$$.

There are Fortran libraries for numerically approximating conformal maps. But they typically use a fixed precision, either single or double; and they deal with conformal maps either to or from the upper half-plane. I am hoping to implement something in Mathematica, so that I can specify a high working precision. And approximating $$f$$ directly, rather than detouring through the upper half-plane, seems like it should be easier, since $$f$$ is holomorphic also on the boundary of $$T$$.

I will append a few remarks about how this problem arose, but I won't give all the details, since the full story is rather complicated.

Suppose that we identify labeled equilateral pentagons $$ABCDE$$ in the Euclidean plane that differ by any combination of translation, magnification, and rotation --- but that we don't identify a pentagon with its reflection. The resulting moduli space $$M$$ is known, topologically, to be a compact, orientable, smooth 2-manifold of genus 4. The region $$T$$ is the triangular subset of $$M$$ in which the pentagons are convex, are traversed counterclockwise, and have $$A$$ and $$B$$ as their narrowest and widest vertices. The region $$T$$ turns out to be $$1/240^\text{th}$$ of $$M$$. (Along the bottom boundary, the widest vertex $$B$$ straightens out; along the upper-left boundary, the vertices $$B$$ and $$E$$ are tied for widest; and along the right-hand curved boundary, the vertices $$A$$ and $$C$$ are tied for narrowest.)

A metric on $$M$$ needs to assign some length to each tangent vector, where a tangent vector at a pentagon $$P$$ is a recipe for altering the shape of $$P$$ as function of time. Extend all pairs of the edges of $$P$$ until they intersect. We can then define the length of that tangent vector as the Euclidean norm of the vector in $$\mathbb{R}^{10}$$ whose coordinates are the rates of change of the ten resulting angles, say in radians per time step. Call that the ten-angles metric. (We use all ten intersections, rather than just the five vertex angles, to achieve greater symmetry.)

Equipping the moduli space $$M$$ with this ten-angles metric gives us a Riemannian manifold with an isometry group of order 240, a manifold that is partitioned into 240 regions, each isometric to $$T$$. The Gaussian curvature of this manifold is everywhere negative, but is not constant.

Pulling the metric of the hyperbolic plane back onto $$M$$ through the conformal map $$f$$ will allow us to view $$M$$, with the conformal structure of the ten-angles metric, as the quotient of the hyperbolic plane by a Fuchsian group, an index-240 subgroup of the (2,4,5)-triangle group. (Indeed, $$M$$ with this conformal structure is conformally equivaent to the Bring sextic, the most symmetric compact Riemann surface of genus 4.)

The opening illustration of $$T$$ is drawn using a coordinate system that is isothermal for the ten-angles metric. I compute my isothermal coordinate system using Gauss's technique for reducing the Beltrami partial differential equation, in the real-analytic case, into an ordinary differential equation over the complex numbers. That ODE has the form $$q'(t)=F(t,q(t))$$ where the function $$F$$ is algebraic, but is so complicated that solving the ODE in closed form seems quite unlikely. Instead, I have Mathematica solve it numerically.

Thus, while I don't know the curved boundaries of $$T$$ in closed form, I could easily compute thousands of points along each of them, each point to any desired precision (within reason). Note also that I don't need to refine my approximation to $$f$$ efficiently; I'll be happy to spend weeks of CPU cycles, is that's what it takes to get $$f$$ to twenty digits.

The sample pentagons in the drawing of $$H$$ are positioned using $$p$$, my degree-13 polynomial approximation to $$f$$. I would like to approximate $$f$$ more accurately since I would like to know, for example, just where the two pentagons $$P_\text{adj}$$ and $$P_\text{nadj}$$ end up along the boundary of $$H$$, those being the equilateral pentagons with two right angles, either adjacent or nonadjacent.