# When can a folded polygon be isometrically (locally) embedded into R^3 ?

I am interested in 3-D representations of various things that naturally live in a non-simply-connected compact surface. There is the usual way of producing a compact surface of any orientable or non-orientable genus, by associating a boundary word to the edges of a polygon to indicate how they are to be glued. But there may or may not be an isometric embedding of this polygon quotient into R^3, not even a local one (isometric immersion). As a simple example : "a a^-1 b b^-1 c c^-1" generates a simply connected surface, but if you put this word around a regular hexagon and fold accordingly, the result is a squashed tetrahedron - not even an immersion. You can make it work by altering the angles of the hexagon slightly.

Is there a known method to decide, given the angles at each vertex of the unfolded polygon, and a boundary word, whether an isometric embedding, or immersion, exists? (More ambitiously I would like it to decide whether it is ugly, but that sounds like it might be an AI-complete question.)

The faces are to be pasted as Joseph O'Rourke indicated. Thanks for the clarification.

So yes, there would be a finite number of points with angle other than 2 pi (they could actually be greater - multiple vertices of the original polygon can get identified.) Everywhere else it's flat. Self-intersection is unavoidable for non-orientable cases, but for example the regular hexagon case, the "embedding" is not injective even locally.

And never mind "ugly". A decision procedure for that would probably be a sentient algorithm (AI-complete question).

• What is the exact meaning of "put this word (aa^{-1}bb{-1}cc^{-1}) around a regular hexagon and fold accordingly"? – X.M. Du Jul 5 '10 at 6:20
• @Daniel: "whether it is ugly"??? – Joseph O'Rourke Jul 5 '10 at 12:49
• In what sense to you mean isometric? At the corners, after you fold, since the total angle is less than 2 &pi;, at best you get an orbitfold. My limited understanding of the word isometric usually asks for the tangent space to be well defined. – Willie Wong Jul 5 '10 at 13:08
• Slightly related to my comment above: how does your question compare with, say, the Nash-Kuiper theorem? Any closed Riemann surface can be $C^1$ isometrically embedded into $\mathbb{R}^3$, so as far as I know, any problem can only come from the conical singularities. (Presumeably that's why you asked about folded polygons.) – Willie Wong Jul 5 '10 at 13:18
• @Willie, while Riemannian geometry is a world in itself, there do exist other metric spaces! :) – Mariano Suárez-Álvarez Jul 5 '10 at 18:47

I offer a clarification, one substantive point, and some references, but not a full answer.

The clarification is in response to X.M.Du's question. I assume what Daniel means is this. Start with a geometric hexagon (or any polygon) embedded in the plane, with edge lengths and angles. Label the edges with symbols. Glue edges with the same symbol together, with the convention that $a^{-1}$ glues oppositely oriented to $a$. Daniel is asking if, knowing the angles and gluing instructions, is there a way to decide if the resulting object is embedded/immersed.

The one point I'd like to make is that, if the angles and gluing instructions are such that each point of the resulting manifold is surrounded by at most $2 \pi$, then Alexandrov's theorem answers the question: The result is realized by a unique convex polyhedron. Alexandrov's amazing theorem is discussed in this MO question, in Lectures on Discrete and Polyhedral Geometry, in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, and in Alexandrov's 1950 book, recently translated into English: Convex Polyhedra, Springer, 2005.

I can't resist this tangent. One very neat application of this theorem is the following. Take any convex shape (smooth or polygonal). Choose two boundary points that partition the perimeter into equal halves. Glue the two halves together. The result is a unique convex body. For smooth convex shapes, this is called a D-form. This is discussed in Geometric Folding Algorithms above.

Without the nonnegative curvature condition needed for Alexandrov's theorem, I am not aware of any general method beyond ad hoc techniques applied to small examples. The work of Jügen Bokowski is the closest I know to what you seek. He found the first realization of Dyck's regular map: "On Heuristic Methods for Finding Realizations of Surfaces," 2008. I would be interested to learn more from other responses to Daniel's question.

• Right, but the regular hexagon case satisfies the hypothesis - three vertices have angle 2 pi/3 and three vertices are amalgamated to one point with angle 2 pi. The edges and interior of course have angle 2 pi. But the polyhedron is degenerate. – Daniel Mehkeri Jul 5 '10 at 18:30
• @Daniel: First, Alexandrov's theorem does not permit twisting in the gluing. It only applies to surfaces homeomorphic to a sphere. Second, this theorem explicitly includes flat, doubly covered polygons as degenerate convex polyhedra. Thus folding a regular hexagon over a line through two antipodal vertices produces a flat doubly covered quadrilateral. – Joseph O'Rourke Jul 5 '10 at 18:39
• @Joseph: By "simply connected" I did mean homeomorphic to the sphere. "a a^-1 b b^-1 c c^-1" is simply connected the way I intended it. When the angles are all equal the result is a double covered triangle. So, is there a way to decide if the Alexandrov construction produces a double covered polygon or not? That would at least decide the special case. – Daniel Mehkeri Jul 5 '10 at 20:45
• @Daniel: Yes, I know how to check whether or not a gluing produces a flat doubly covered polygon, with an $O(n^3)$ algorithm, for an $n$ vertex polygon. I never published it, and this comment is too small to contain it... :-) If I find the time, I might describe it in a separate answer. It will likely disappoint you in that, as you can guess from the complexity (which can surely be reduced), it is rather brute-force. – Joseph O'Rourke Jul 5 '10 at 21:00
• @Daniel: I wrote down the algorithm in a note: arxiv.org/abs/1007.2016 . – Joseph O'Rourke Jul 14 '10 at 12:53