# Illuminating piecewise-flat manifolds with geodesics

If one issues a geodesic in every direction from a point $p$ on a piecewise-flat 2-manifold, will it necessarily illuminate the entire surface? I know the answer is 'No,' but I would like to explore the question further.

I am using the term piecewise-flat manifold in the sense that David Glickenstein uses it, e.g., in "Introduction to piecewise flat manifolds:" a gluing of Euclidean triangles edge-to-edge. This is also called a polyhedral manifold. For the purposes of this question, whether it is embedded in $\mathbb{R}^3$ is not relevant. More generally, these manifolds are formed by edge-to-edge gluings of planar polygons (each of which could be triangulated). The manifold is flat everywhere but at a finite number of vertices (or cone points) at which the surrounding angle differs from $2 \pi$.

Because geodesics do not pass through vertices (or, more accurately, I stipulate they cannot), it is conceivable that there is some $p$ from which geodesics shot in every direction fail to reach every point of the manifold. This was established in a rather different context in the paper by George Tokarsky, "Polygonal Rooms Not Illuminable from Every Point" [Amer. Math. Monthly, 102:867-879 (1995)]. Mathworld has a nice description, including this figure:

If you glue two copies of either of these polygons back-to-back, it forms a polyhedral 2-manifold with the property that geodesics (light rays) from one red point cannot reach the other red point.

One can ask many questions here, but these three interest me:

1. Tokarsky's example is a doubly covered polygon. If one generalizes instead to arbitrary polyhedral manifolds, are there other, perhaps more straightforward examples where from some $p$ not all the manifold is covered its geodesics?
2. I conjectured long ago that the measure of the "dark points" is zero. Is there an example (of a polyhedral manifold) where more than isolated points are unilluminated? Could a segment be unilluminated? A region of positive area?
3. Are there examples of these same phenomena in piecewise-flat 3-manifolds (gluings of Euclidean tetrahedra)?

Edit. In response to Henrik's example below, I should have said that ideally two further conditions should be satisfied: (a) $p$ is not at a vertex (so it is surrounded by $2\pi$ of surface); and (b) the manifold should be closed, without boundary. This is not to say that $p$ at a vertex and a manifold with boundary are not of interest!

Addendum: Thanks for the interest and help! I have much to learn on the topic of translation surfaces!

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"Because geodesics do not pass through vertices" means that you just don't consider geodesics, that would pass through vertices (because its unclear how you should extend them)? if so, you can just glue 7 regular triangles side by side (to get a filled 7-gon with a funny metric). Take one vertex point. Every geodesic from this point to a point in the opposing triangle must pass through the middle point. I guess one can make this example more elaborate by taking the regular tesselation of hyperbolic plane with 7 (or more) triangles meeting at each vertex etc. –  HenrikRüping Jul 21 '10 at 17:29
@Henrik: Ah, I see I should specify two more aspects: the source of the geodesics should not be a vertex (like those red points in the figure), and the manifold should be closed. Sorry to alter the conditions on the fly! I will edit it ... –  Joseph O'Rourke Jul 21 '10 at 17:52
@Henrick: Good point about geodesics through vertices! They never go through positive-curvature vertices, but they could go through negative-curvature vertices. So, yes, I guess I was simply disallowing that. Apologies for these flaws! –  Joseph O'Rourke Jul 21 '10 at 18:01

I would like to propose a simple example of a flat surface of genus $3$ with dark points. This also gives an example in dimension $3$. It is based on the following simple observation.

Observation. Consider the torus $T^2=\mathbb R^2/\mathbb Z^2$. Suppose that there is a geodesic segment that joins points $(0,0)$ with $(\frac{1}{2},\frac{1}{2})$. Then it passes through one of four points $(\pm \frac{1}{4},\pm\frac{1}{4})$.

Now consider the double ramified cover $S$ of $T^2$ branching at points $(\pm \frac{1}{4},\pm\frac{1}{4})$. Then on $S$ there is no geodesic segment that goes from any of two preimages of the point $(0,0)$ to any of preimages of the point $(\frac{1}{2},\frac{1}{2})$. Indeed if there were such a segment it would project on $T^2$ to a segment that joins $(0,0)$ with $(\frac{1}{2},\frac{1}{2})$. Hence on $S$ it should pass through a branch point, which is forbidden.

Added. One can desribe give an alternative description of this example. Namely, we can take $8$ copies of squares of size $\frac{1}{2}\times \frac{1}{2}$ and glue a surface of genus $3$ from them in such a way, that at each vertex $8$ squars meet.

If you want a 3-dimensional example just multiply this example by $S^1$. But it should be of course possible to construct examples that are not products, using similar idea.

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@Dmitri: Cool, I want to understand this, but the term "double ramified cover" is new to me and I am not finding a definition... –  Joseph O'Rourke Jul 21 '10 at 22:35
@Joseph, this is terminology from the theory of Riemman surfaces, en.wikipedia.org/wiki/Ramified_covering_map . Double ramified cover is a ramified cover of degree two, i.e. generic point has two preimages. An example of such a cover is $z\to z^2$, $z\in \mathbb C^1$. The example that I proposed can be understood without this treminology. You can glue $S$ from 8 squares of the size $1/2 \times 1/2$, at each vertex on $S$ $8$ squares should meet. –  Dmitri Jul 21 '10 at 22:49
@Dmitri: Thanks for the explanation. I will ponder this! –  Joseph O'Rourke Jul 21 '10 at 22:54
Dmitri: Beautiful construction! Joseph: At the branch points the total angle is $2\times 2\pi=4\pi,$ so geodesic is forbidden to pass through them according to your rules. –  Victor Protsak Jul 22 '10 at 0:55
I am accepting this as the answer even though I have to admit I do not entirely understand how the 8 squares are identified along their 32 edges. But this is my fault, not Dmitri's. I need to study translation surfaces! –  Joseph O'Rourke Aug 4 '10 at 23:29

This illumination problem has been studied for special kinds of polygonal surfaces, called (pre-) lattice translation surfaces. See

For these surfaces, the paper proves that the set of non-illuminated points is countable.

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@Alex: Thanks so much for this reference, which I never would have encountered on my own! This conjecture of theirs is I believe equivalent to mine in (2) of my question: "Conjecture 1: In any rational polygon P, any point p illumines all but finitely many points q ∈ P." –  Joseph O'Rourke Aug 4 '10 at 23:24