Illuminating piecewise-flat manifolds with geodesics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:03:28Z http://mathoverflow.net/feeds/question/32797 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32797/illuminating-piecewise-flat-manifolds-with-geodesics Illuminating piecewise-flat manifolds with geodesics Joseph O'Rourke 2010-07-21T13:08:06Z 2010-08-04T23:34:45Z <p>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. </p> <p>I am using the term <em>piecewise-flat manifold</em> in the sense that David Glickenstein uses it, e.g., in "<a href="http://math.arizona.edu/~asp/2010/piecewiseflatmflds.pdf" rel="nofollow">Introduction to piecewise flat manifolds</a>:" a gluing of Euclidean triangles edge-to-edge. This is also called a <em>polyhedral manifold</em>. 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 <em>vertices</em> (or <em>cone points</em>) at which the surrounding angle differs from $2 \pi$.</p> <p>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" [<em>Amer. Math. Monthly</em>, 102:867-879 (1995)]. Mathworld has a <a href="http://mathworld.wolfram.com/IlluminationProblem.html" rel="nofollow">nice description</a>, including this figure: <br /> <img src="http://mathworld.wolfram.com/images/eps-gif/TokarskyRoom_850.gif" alt="alt text"> <br /> 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.</p> <p>One can ask many questions here, but these three interest me:</p> <ol> <li>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?</li> <li>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? </li> <li>Are there examples of these same phenomena in piecewise-flat 3-manifolds (gluings of Euclidean tetrahedra)?</li> </ol> <p><b>Edit</b>. 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!</p> <p><b>Addendum</b>: Thanks for the interest and help! I have much to learn on the topic of translation surfaces!</p> http://mathoverflow.net/questions/32797/illuminating-piecewise-flat-manifolds-with-geodesics/32853#32853 Answer by Dmitri for Illuminating piecewise-flat manifolds with geodesics Dmitri 2010-07-21T21:43:50Z 2010-07-21T22:52:32Z <p>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.</p> <p>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 $(\pm \frac{1}{4},\pm\frac{1}{4})$.</p> <p>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.</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/32797/illuminating-piecewise-flat-manifolds-with-geodesics/33552#33552 Answer by Alex for Illuminating piecewise-flat manifolds with geodesics Alex 2010-07-27T18:14:03Z 2010-07-27T18:14:03Z <p>This illumination problem has been studied for special kinds of polygonal surfaces, called (pre-) lattice translation surfaces. See</p> <p><a href="http://front.math.ucdavis.edu/0602.5394" rel="nofollow">http://front.math.ucdavis.edu/0602.5394</a>.</p> <p>For these surfaces, the paper proves that the set of non-illuminated points is countable. </p>