parametrization of non-convex polygons, flat disks and flat spheres

[videlec]

The EquiangularPolygons class allows to play with polygons where angles are fixed. However, it does not handle gracefully the non-convex case where intersections might occur. The parameter space for the lengths is an a priori complicated space since each intersection of segments gives a quadratic equation.

Perhaps, non-convex polygons is the wrong question and one should look at equiangular flat disks. In many situation, such generalized polygons would be used as a building block for a translation surface or conical surface (such as the double in the billiard construction).

In these three settings (polygons with fixed boundary angles, flat disks with fixed boundary angles, flat spheres with fixed conical angles) it would be nice to implement algorithms for

• producing an example
• producing many examples (possibly involving randomness)
• parametrizing the space (e.g. provide a reasonable cell decomposition in which one can "walk")

A possibly interesting question : let us start with a "piecewise linear curve" given as a sequence of n lengths and interior angles such that the curve closes up and the total interior angle is (n-2) \pi. Does such curve necessarily correspond to the boundary of at least one flat disk? Always to finitely many flat disks? Can we parametrize all flat disks with this boundary by some combinatorial decoration?

The article https://arxiv.org/pdf/2008.10192.pdf might be helpful.
See also #61

[videlec]

@wphooper Pat, do you know the answer to the "possibly interesting question" ? I think it is worth investigating how much the boundary gives you information on a flat disk.

[videlec]

There is a numerical approach to this. Consider a tuple of integers (a_1, a_2, ..., a_k) and let N = a_1 + a_2 + ... + a_k. We consider the associated angles (k-2) pi / N (a_1, a_2, ..., a_k). Then a flat disk with these angles is nothing more than a (multivalued) differential

(z - p_1)^(a_1 - N/k) (z - p_2)^(a_2 - N/k) ... (z - p_k)^(a_k - N/k) dz

with p1, ..., pk reals (and one can fix p_1=0, p_2=1, p_3=oo to have a bijection with flat disks with fixed angles). To obtain the corresponding lengths one has to integrate the differential between consecutive points p_i on the circle. The absolute value of the integral is well defined whatever choice of branch is made.

One can even compute a combinatorial triangulation (close to the Delaunay cell decomposition) by integrating the linear flow.

If there are two flat disks with the same boundary it means that there are two tuples of p_i with the same ordering on the circle for which all integrals coincide (up to a global multiplicative scalar which corresponds to the area) !

[saraedum]

This is related to #170.

I'm not sure what you had in mind here @videlec. Should flat disks (and spheres?) be fundamental building blocks of surfaces like polygons are currently? Or do you just want to create a flat disk with a convenience function that produces a gadget that is built from polygons glued to the flat disk and that can then be used to build more complex surfaces.

[videlec]

I don't think that flat disks should be any special : they are translation surfaces with boundary (that happen to be homeomorphic to a disk). I just wanted some convenience functions. In particular, given boundary segments, it should be possible to build all flat disks whose boundary is made of the segments.

[saraedum]

I don't understand this comment.

There is a numerical approach to this. Consider a tuple of integers (a_1, a_2, ..., a_k) and let N = a_1 + a_2 + ... + a_k. We consider the associated angles (k-2) pi / N (a_1, a_2, ..., a_k). Then a flat disk with these angles is nothing more than a (multivalued) differential

(z - p_1)^(a_1 - N/k) (z - p_2)^(a_2 - N/k) ... (z - p_k)^(a_k - N/k) dz

with p1, ..., pk reals

So, if I take a₁, a₂, a₃ = 1 (an equilateral triangle), I just get dz. From what you wrote it would seem that I can choose the p_k however I want but then I won't get three equal side lengths for my equilateral triangle.

(and one can fix p_1=0, p_2=1, p_3=oo to have a bijection with flat disks with fixed angles).

I don't understand this. Didn't you say above that I get a flat disk with these angles already without fixing the points like that?

Sorry, for asking these questions two years after the fact but I was looking into whether it's worthwhile to implement this now and I would like to understand the idea here.