Illuminating piecewise-flat manifolds with geodesics - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T16:03:28Zhttp://mathoverflow.net/feeds/question/32797http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/32797/illuminating-piecewise-flat-manifolds-with-geodesicsIlluminating piecewise-flat manifolds with geodesicsJoseph O'Rourke2010-07-21T13:08:06Z2010-08-04T23:34:45Z
<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#32853Answer by Dmitri for Illuminating piecewise-flat manifolds with geodesicsDmitri2010-07-21T21:43:50Z2010-07-21T22:52:32Z<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#33552Answer by Alex for Illuminating piecewise-flat manifolds with geodesicsAlex2010-07-27T18:14:03Z2010-07-27T18:14:03Z<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>