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May 23, 2019 at 10:10 history edited Martin Sleziak CC BY-SA 4.0
minor typo
Apr 12, 2018 at 17:45 comment added Pierre Dehornoy What a coincidence, I am just reading some of your papers carefully these days. Here is an updated version www-fourier.ujf-grenoble.fr/~dehornop/maths/Billiard.pdf
Mar 23, 2018 at 19:02 comment added Lee Mosher @PierreDehornoy: Do you (or does anyone else) have an updated link to that short note? The current link is dead.
Sep 17, 2010 at 22:05 history edited Bill Thurston CC BY-SA 2.5
pointer to Joseph's pictures.
Sep 17, 2010 at 13:02 comment added Joseph O'Rourke @Pierre: Beautiful idea! I love the "church shapes" figures! You may recommend me as a referee. :-) Congratulations!
Sep 17, 2010 at 2:34 comment added Pierre Dehornoy It looks like one can use Ghrist's template in order to construct a non convex polyhedron containing all knots. Here is a short note on this. umpa.ens-lyon.fr/~pdehorno/maths/BilliardKnots.pdf
Sep 17, 2010 at 0:01 history edited Bill Thurston CC BY-SA 2.5
added 2948 characters in body
Sep 16, 2010 at 17:32 comment added Bill Thurston @Pierre Dehornoy: That sounds like a promising idea. Since it's a particular flow, it might actually be possible to construct a convex polyhedron, but I don't remember (if I once knew) the specific construction for the knotholder. One could simulate splits in a branched surface by adding a narrow beam at a corner of a polyhedron. Horizontal splits are similar. Shifter units should be installed that perform unknotted operations to shift beams so they don't intersect. If the dynamics is or can be made PL (seems likely), then a polyhedral billiard realization sounds very plausible.
Sep 16, 2010 at 16:32 comment added Joseph O'Rourke @Bill: Both your sketch on 3(a), and Pierre's idea, sound plausible to me---cool! Sorry for conflating the two different problems; corrected now.
Sep 16, 2010 at 15:55 comment added Pierre Dehornoy I wonder if one could use Ghrist template to construct a (non convex) tube containing all knots: Ghrist constructed a branched surface equipped with a flow in $\mathbb R^3$ such that every link appears as a collection of periodic orbits of this flow. Could we thicken this knot-holder to get a billiard (made of glued square boxes) containing all knots?
Sep 16, 2010 at 15:27 history answered Bill Thurston CC BY-SA 2.5