Let a lightray bounce around inside a cube whose faces are (internal) mirrors. If its slopes are rational, it will eventually form a cycle. For example, starting with a point $p_0$ in the interior of the $-Y$ face of an axis-aligned cube, and initially heading in a direction $v_0=(1,1,1)$, the ray will rejoin $p_0$ after 5 reflections, forming a hexagon. The figure below shows a more complicated 16-cycle.
![alt text][1]
Assume that $p_0$ and $v_0$ are chosen so that (a) the ray never directly hits an edge or corner of the cube, and (b) the ray path never self-intersects inside the cube.
Can every knot type be realized by a lightray reflecting inside in a cube?
The figure above is an unknot.
I believe (but am not certain) the 31-cycle below is knotted:
alt text http://cs.smith.edu/%7Eorourke/MathOverflow/Bounces30.jpg
Any such knotted path is a stick representation of the knot, but perhaps the many unsolved problems in stick representations are not relevant to this situation. My question is related to the probability of random knots forming under various models, but usually those models are aimed at polymers or DNA. I have not seen this lightray model explored, but would be interested to know of related models.
The choice of $(p_0,v_0)$ allows considerable freedom to "design" a knot, but it seems difficult to control the structure of the path to achieve a particular result. I've explored tiling space by reflected cubes so that the lightray may be viewed as a straight segment between two images of $p_0$, but this viewpoint is not yielding me insights.
If anyone has ideas, however partial, I would appreciate hearing them. Thanks!
Edit1 (15Sep10). I have not been able to yet access the Jones-Przytycki paper that Pierre cites, but knowing the keywords he kindly provided, I did find related work by Christoph Lamm ("There are infinitely many Lissajous knots" Manuscripta Mathematica 93(1): 29-37 (1997)) that provides useful information:
Theorem: Billiard knots in a cube are isotopic to Lissajous knots.
As Pierre said, many knots are unachievable in these models. In particular, algebraic knots cannot be achieved. The technical result is this. Theorem: The Alexander polynomial of a billiard knot is a square mod 2.
In 1997, there were several intriguing open problems, including these two. (a) Is every knot a billiard knot in some convex polyhedron? (b) Can the unknot be achieved in every convex polyhedron that supports periodic paths?
Edit2 (15Sep10). Here is a little more information on open problems ca. 2000, found in a list by Jozef H. Przytycki in the book that resulted from Knots in Hellas '98:
Is there a manifold that supports every knot? (By "supports every knot" he means there is a billiard path isotopic to every knot type.)
Is there a finite polyhedron that supports every knot? There apparently is an "infinite polyhedron" that supports every knot.
More specifically, is there a convex polyhedron that supports every knot?
(This is 3(a) above.)[See Bill Thurston's correction below!]Even more specifically, is every knot supported by one of the Platonic solids?
I have not been successful in finding information on this topic later than 2000. If anyone knows later status information, I would appreciate a pointer. Thanks for the interest!
Edit3 (5Jul11). The question has been answered (affirmatively) in a paper by Pierre-Vincent Koseleff and Daniel Pecker recently (28Jun11) posted to the arXiv: "Every knot is a billiard knot":
"We show that every knot can be realized as a billiard trajectory in a convex prism. ... Using a theorem of Manturov [M], we first prove that every knot has a diagram which is a star polygon. [...Manturov’s theorem tells us that every knot (or link) is realized as the closure of a quasitoric braid...] Then, perturbing this polygon, we obtain an irregular diagram of the same knot. We deduce that it is possible to suppose that 1 and the arc lengths of the crossing points are linearly independent over $\mathbb{Q}$. Then, it is possible to use the classical Kronecker density theorem to prove our result."