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This should be true in a more general setting, but for simplicity consider billiards that are connected, compact subsets of the plane with boundary $C^2$ except at finitely many points. A ball (or a ray of light) rolls inside, going in straight lines, and upon collision with the wall, the orbit is reflected.

It is intuitive that a statement like the following is true:

For almost every billiard, there exists an orbit that is dense everywhere inside it.

However, as far as I know this is still open. In fact, the last thing I heard was that it had just been proven for the case in which the billiard was an obtuse triangle with certain restrictions (but I have since forgotten the source, unfortunately).

Question: What is the current status of the problem?

Thank you!

Clarification The question is not about rectangular billiard tables, but in general about the balls rolling in more general shapes. 'Almost always' would then have to be given a meaning within the space of curves. (In fact, the problem is trivial in rectangles because an orbit with irrational slope will do.) Also, this is not about having dense families of orbits, but a single orbit that is dense in the billiard.

I think the way 'almost always' should be defined is by requiring some generic property to hold. Think, for example, of the definition R. Abraham give of bumpy metrics.

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Should that be billiard table - peculiarly shaped billiards would be a different (and maybe interesting) question! –  Mark Bennet Jan 28 '11 at 18:56
Sorry, but what is a 'billiard table'? Is that the case when the boundary of the billiard is a polygon? –  Zatrapilla Jan 28 '11 at 20:06
I think there are languages in which "billiard" is the normal term for what we Americans (and perhaps the British also?) call a billiard table. –  Andreas Blass Jan 28 '11 at 20:48
@andreas Thanks for that explanation. Here in England the billiard table is what the game of billiards is played on, and it makes sense to talk of the shape of the table. So with an elliptical table, if you put the ball at one focus and hit it in any direction, it will have a path through the other focus (so need not be a polygon). Normally we'd assume convex: a connected compact subset which was not convex could have some curious properties even in two dimensions (think Alexander Horned Sphere). –  Mark Bennet Jan 28 '11 at 21:02
Actually, non-simply-connected tables HAVE been considered, by such luminaries as Sinai, I believe. –  Igor Rivin Jan 28 '11 at 22:38
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3 Answers

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Do you mean to ask whether the trajectories in almost all cases (in {shapes X trajectories} are dense in the set of {positions, directions} on the table, or just in the set positions? The first question seems more natural to me; the answer is no: If there are two convex portions of the boundary curve pointing toward each other, they're like convex mirrors, they tend to focus. For open sets of shapes and distances, the return map from the tangent line bundle along a mirror back to itself has eigenvalues of the first derivative a pair of complex conjugate points on the unit circle. Because of the KAM theory, there are typically rings of positive measure consisting of invariant circles for the return map. The orbits of these rings under the billiard flow enclose an open set in phase space.

Another physical example of this effect what happens when you wind something like kite string around a flat object, perhaps a piece of cardboard or a board. The string tends to build up in the middle, and once a bulge gets started, the configuration is stable---the string prefers to wind back and forth across the bulge. (Note that the shortest paths of winding string follow geodesics, which are the same as trajectories as billiards on a table of the same shape).

Even when the KAM situation isn't obvious from the geometry of the table, I think experimental evidence shows that it's commonplace. There are known constructions of Riemannian metrics on $S^2$ with ergodic geodesic flow, but they took a long time before someone found them (sorry, I don't remember the reference). Similarly, I think it's tricky to find examples of simply-connected billiard tables with smooth boundary that are ergodic: you somehow have to systematically eliminate the KAM phenomenon. It's much easier if the table either has angles, or is multiply-connected with two or more obstacles in the middle (so that doubling it produces a surface of negative Euler characteristic).

It's not obvious to me how to use this phenomenon to capture all trajectories that pass through a particular point, but maybe that's not really the most natural question: after all, in a game of billiards, direction and position both matter.

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Thank you for your answer. I meant to ask whether for almost every shape there exists a trajectory that is dense. In other words, whether for almost every $S\in\{\textrm{shapes}\}$ there is a $(p,v)\in T(S^\circ)$, with $\|v\|=1$ such that the orbit that goes through $(p,v)$ is dense in $S$. –  Zatrapilla Jan 29 '11 at 7:32
Especially in the case of convex billiard tables, this subject has a long and distinguished history, often phrased as "twist maps of the annulus", started early last century I think with Birkhoff, and I think Poincaré as well. The set of non-inward-facing directions around the boundary circle is topologically a closed annulus; the return map (direction -> direction after first reflection) is an area-preserving map, the identity at the boundary that is isotopic to a Dehn twist. These also have bands (Cantor-sets-worth) of invariant circles near the boundary. Look for pictures in the literature. –  Bill Thurston Jan 29 '11 at 12:15
I can see why what you are saying is true.I am familiar with the theory of twist maps all the way to Mather.I think this does not answer my question, because what you get is a result about what happens in the boundary only.My interest is about density in the open set $S^\circ=\textrm{interior of the shape}\,S$ of the projected orbits (i.e., if the orbit is a set of points in the tangent bundle $TS^\circ$, then you project them down to $S^\circ$ and see if that is dense).I believe you can get such density without having any density in the annulus $\subset T\partial S$ you describe.Thanks again! –  Zatrapilla Jan 29 '11 at 15:34
(Thurston was right.) –  Zatrapilla Jul 16 '13 at 2:37
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Assuming that the OP is about billiard tables, not strangely shaped billiard balls:

The result for (some) obtuse triangles is the opposite of what the OP wants: Rich Schwartz has proved that (some) obtuse-angled triangles have periodic orbits, see:

MR2549685 (2010g:37060) Schwartz, Richard Evan(1-BRN) Obtuse triangular billiards. II. One hundred degrees worth of periodic trajectories. (English summary) Experiment. Math. 18 (2009), no. 2, 137–171.

That in polygonal billiards with angles rational multiples of $\pi$ the generic orbit is dense follows from the classic result

MR0855297 (88f:58122) Kerckhoff, Steven(1-STF); Masur, Howard(1-ILCC); Smillie, John(1-CUNY7) Ergodicity of billiard flows and quadratic differentials. Ann. of Math. (2) 124 (1986), no. 2, 293–311. 58F17 (30F30 58F11)

See the following for more along the lines of specifically dense orbits:

MR1458298 (98k:58179) Boshernitzan, M.(1-RICE); Galperin, G.(D-BLF-BB); Krüger, T.(D-BLF-BB); Troubetzkoy, S.(D-BLF-BB) Periodic billiard orbits are dense in rational polygons. (English summary) Trans. Amer. Math. Soc. 350 (1998), no. 9, 3523–3535. 58F20

Serge Tabachnikov's book "Geometry and Billiards" is highly recommended.

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Do I understand correctly that you are not in any sense talking about "almost every billiard" as asked for in the OP? (I don't dispute the interest of the results you mention, just whether they answer the actual question asked.) –  gowers Jan 28 '11 at 22:40
@gowers: you are correct, although the ergodicity results do say that almost every orbit in a fixed billiard of the type considered (eg, rational polygon) is dense. My impression was that the OP question was intentionally vague ("almost every" = "anything that one can think of"), and that does appear to be true if one replaces the rhs with "anything one can prove anything about". –  Igor Rivin Jan 28 '11 at 22:50
Ah yes -- I do sometimes use "almost every" not in anything like a measure-theoretic sense but rather in a psychological sense. It's not quite obvious what the OP intends but I interpreted it more as asking about "generic" billiards. –  gowers Jan 29 '11 at 16:22
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It is possible for a billiard path to be dense and yet not be uniformly distributed. This is true for the triangle with angles $0.4\pi,0.3\pi,0.3\pi$. (See, for example, theorem 1.3 of http://homepages.math.uic.edu/~demarco/billiards.pdf.)

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