# Equilaterally triangulated surfaces with prescribed boundary

There is a problem in Richard Kenyon's list which I would like to post here, because although I have thought about it from time to time, I have not been able to make the slightest progress on it:

Question: "Given a closed polygonal path $p$ in $R^3$ composed of unit segments, is there an immersed polygonal surface whose faces are equilateral triangles of edge length $1$, spanning $p$?"

The condition that the configuration is an "immersed" surface means that the whole configuration may be parametrized by a continuous locally one-to-one mapping from a compact connected 2-dimensional manifold with boundary into $R^3$. In other words at most two triangles may share an edge, and triangles sharing a vertex form an embedded topological disk, but two triangles which do not share an edge or a vertex may intersect.

I cannot answer the question even for the case where $p$ is a quadrilateral. Has there been any partial results for this problem?

• The back of the envelope calculation is that the moduli space of an equilateral surface should be $\leq \frac12$ the dimension of the moduli space of the boundary (under the strong assumption that all of the equations cut down the space of solutions by the expected codimension). This seems difficult to make precise. However, Kapovich and Millson showed that the space of polygons has a symplectic structure projecteuclid.org/euclid.jdg/1214459218 Maybe one can show that the image of the moduli space of an equilateral surface is isotropic in the moduli space of the boundary? – Ian Agol Oct 20 '17 at 4:57
• Can you provide a definition of what an "immersed polygonal surface" is? I guess it means that at most two triangles can share a a side, but what else, especially are the triangles allowed to penetrate each other? (sorry for the simple question). – Manfred Weis Oct 20 '17 at 10:08
• @Manfred Weis: Yes, at most two triangles can share an edge and the star of each vertex must be an embedded topological disk. I just made this explicit. – Mohammad Ghomi Oct 20 '17 at 11:29
• Tangentially relevant: It is NP-hard to determine if a closed 3D polygonal path can be triangulated. Barequet, Gill, Matthew Dickerson, and David Eppstein. "On triangulating three-dimensional polygons." Proc. the 12th Symposium Computational geometry. ACM, 1996, pp.38-47. Journal link. – Joseph O'Rourke Oct 20 '17 at 12:14
• @Joseph: Are there examples of polygonal paths that cannot be triangulated? – Mohammad Ghomi Oct 20 '17 at 12:26

## 2 Answers

We resolve Kenyon's problem in this paper. We discuss in Section 5 a number of further conjectures and open problems.

• Maybe good to point out that you resolve his problem in the negative. – Sam Hopkins May 8 '20 at 22:42

(It is intended as an extended comment and thinking out loud rather than an answer)

That seems that the case when $$p$$ is a quadrilateral is exactly the bottleneck, the general case will hopefully follow from it by an induction argument.

Without loss of generality we can prove the statement for paths of even length (glue a triangle to any edge of an odd-length path and span the resulting even-length path).

Definition: The waist $$w(p)$$ of a closed polygonal path $$p$$ of lenght $$l$$ is the shortest line segment connecting two vertices of $$p$$ that separate $$p$$ into pieces of equal lenght ($$=\frac{l}{2}$$)

Statement: $$|w(p)|\le C\cdot l$$, where $$C$$ is the ratio of the diameter of the regular polygon with $$l$$ edges of length 1 to its perimeter $$l$$.

In other words, a waist of a path of length $$l$$ cannot be longer than a diagonal of the regular $$l$$-gon. In particular it is quite short for large $$l$$, shorter than $$\frac{l}{2}$$.

We can turn the waist into a polygonal path (with unit segments) of length $$\lceil{|w|}\rceil$$ (smallest integer greater than $$|w|$$). Starting from $$l=6$$, when $$C=\frac{1}{3}$$ and a poligonized waist (let's call it $$W$$) has lenght at worst $$2$$ -- everything is more or less fine: span the path consisting of the first half of $$p$$ and $$W$$ (its length at worst $$5$$ -- 5 is not even, of course, but at least less than 6 :). For $$l=8$$ it's surely curable) and in the same manner span the second half of $$p$$ and $$W$$ and take the union of these two spanning surfaces, doing an induction step.

For $$l=4$$ that does't work, however, since $$W$$ can have length $$2$$.

Even for a transcendentally angled flat rhombus it is not clear to me whether one can span it. If no, the answer can be probably given by some sort of extension of fields argument (in any construction I tried to think of the transcendence of the angle survived in some form preventing to complete it). In the positive direction, there is the industry of flexible polyhedra, which potentially could provide an example of a polyhedra with movable quadrilateral living on it that could be a hope for a construction, but I could not readily find in the literature ones with equilateral triangle sides with enough flexibility to do something like we want.