# The Unapologetic Mathematician

## Dual Billiards

I’m not sure when I’ll get to post tomorrow, so I’m giving a little extra tonight: something I realized at about 3 in the morning last night.

The last time I talked about billiards I was linking to Rich Schwartz’ paper on “outer billiards”. I noted that it seemed to me there should be some sort of “duality” between outer and inner billiards, turning problems in one into problems in the other. I think I’ve figured it out. I haven’t checked through all the details, but it looks good enough to satisfy my curiosity. If I were going to write a paper and use this fact, of course, I’d rake it over the coals.

So here it is: inner and outer billiards are related by projective duality. Those of you who know what this is probably are already thinking either “ah, I see” or “of course it is. you didn’t see that before now?” For the rest of you, I’ll skim a bit about the projective plane and formal geometry. I’m sure I’ll eventually come back and write more about them, but for now I can give enough of the gist.

First of all, projective geometry tweaks the familiar axioms from Euclid’s Elements. Euclid says that given a line $l$ and a point $p$ off the line there is exactly one line through $p$ parallel to $l$. In projective geometry, though, any two lines intersect, and moreover they intersect exactly once. That seems nutty at first, but we can make it work by adding a “line at infinity”, with one point for each direction parallel lines could run. If lines seem parallel, they’ll run into each other at that point.

The other ingredient is a formal approach to geometry. Remember when I defined the natural numbers, I said that we don’t care what it is that satisfies these properties, just that anything satisfying these properties will do whatever we say the natural numbers will. Well the same goes for geometry. We have an intuitive idea of “point”, “line”, and “plane”, but that doesn’t really matter. David Hilbert famously said that all of Euclidean geometry should still be true if we replaced “point”, “line”, and “plane” with “table”, “chair”, “beer mug”, wherever they occur. Here: “Any two tables intersect in a unique chair”.

So all the axioms of projective geometry do is set up a system of referents and relations like the Peano axioms do. Any things that fill those referential slots and relations between those things implementing the axioms will do. The points and lines of the regular Euclidean plane, plus those points and the line “at infinity”, satisfy the right axioms, and so everything projective geometry says will hold true for them.

Here’s the trick: The lines and points of the projective plane also satisfy those axioms. Did you miss that? The axioms for “points” and “lines” of projective geometry are satisfied by the lines and points of the projective plane. We can switch lines and points and everything still works out! For example, we’ve talked about the axiom that any two “lines” share a unique “point”. There’s also an axiom that any two “points” share a unique “line” through them. Switching lines and points swaps these two axioms. Any result for projective geometry is really two results: one for the points and lines and one for the lines and points.

Okay, here’s how this all ties back to billiards: don’t think of a ball moving along the table and bouncing off the edge. Think of the line the ball is traveling on and the line of the edge it’s moving towards. They share a unique point, where the lines intersect. Then there’s another line intersecting the edge line in the same point at the same angle, but “on the other side”. That’s the line the ball follows after the bounce, and so on. In outer billiards, we have a point and the edge point it’s heading towards. There’s a unique line between them, and another point on the same line the same distance away, but on the other side of the edge point. We interchange points and lines, lengths and angles, and transform inner billiards into outer billiards and vice versa.

Of course, the calculations strike me as being pretty horrendous in all but the simplest situations. I don’t know that it would be useful to use this duality in practice, but maybe it can come in handy. Actually, for all I know the experts are already well aware of it. Still, it’s nice to have figured it out.

March 9, 2007 Posted by | Billiards | 2 Comments

## Outer billiards has an unbounded orbit

This just popped up on the arXiv, so I thought I should mention it: Richard Schwartz has a paper up showing that there is a shape for an “outer billiards” table and a starting point whose path gets as far away from the shape as you want. Even better, it’s one of the shapes from the Penrose tiling. Curiouser and curiouser. The first section or so of the paper are very readable, and gives a much better explanation (with pictures!) of outer billiards than I could manage. The proof itself is heavily aided by computer calculations, but seems to be tightly reasoned apart from needing help to handle a lot of cases.

I’m not quite sure how billiards and outer billiards are related. My intuition is that there’s some sort of duality going on, which would exchange lengths of segments in outer billiards with angles in billiards. On the other hand, if there were such a straightforward translation, couldn’t the enormous machinery of billiards have been brought to bear on this problem before now? Do any billiard theorists in the audience know anything about outer billiards?

February 23, 2007 Posted by | Billiards | 4 Comments

## Billiards 2

After looking for images of billiards I could use with no luck I just sat down and sketched an example.

The billiard table itself is the lower left square. I’ve drawn a path moving towards the upper right and bouncing around a few times. The other three squares are the reflections of the original table that I spoke of. We can imagine the path continuing into them in a straight line, and wrapping from one side of the big square to the other, and from the top to the bottom. The sides marked “A” are identified, as are the sides marked “B”. When we wrap up the sides as marked we get a donut-shaped surface called a “torus”.

Now we didn’t have to start with a square table. If we start with a rectangle we’ll get pretty much the same picture, except the torus we get will either be longer and thinner or shorter and fatter. There are a lot of different tori out there, but they all come from taking some parallelogram and identifying the opposite sides like we did here. So, what sorts of parallelograms are there?

February 12, 2007 Posted by | Billiards | 2 Comments

## Billiards course

This semester I’m sitting in on Jayadev Athreya’s course on billiards. “Billiards?” I hear you cry, “The game like pool but with all the red balls?” No, that’s snooker. Besides, the course is on (not surprisingly) a mathematical model inspired by balls bouncing around on a table.

When playing mathematical billiards, we place a ball down on a polygonal table and send it off in some direction. When it hits the edge of the table it rebounds, making the same angle as it leaves as it did when it came in. We simplify a bit by assuming that the ball isn’t spinning, has no friction with the table, and so travels at a constant speed in a straight line between bounces.

So let’s start with a square table. How do we determine how the path behaves as time goes on? Imagine the ball approaching a side of the table. Instead of the ball reflecting off the edge, let’s reflect the whole table through the edge and let the ball continue on its straight-line path. Whenever the ball’s about to hit an edge, we have a reflection of the table ready for it.

But we really don’t need all that many reflections. In fact, four will do it: the first table, a horizontal reflection, a vertical one, and one reflected in both directions. We can put these four squares together into one bigger square. Instead of reflecting the ball at an edge, we can just let it wrap around to the other side of the table travelling in the same direction, just like a game of Asteroids. That means we’re really just rolling a ball across the surface of a torus.

Now the whole picture is pretty clear: by Weyl’s criterion, if the tangent of the angle between the path and an edge of the table is rational the path eventually closes back up and runs over itself again. If not, the path covers every point on the table equally. By this I mean that if the whole table has area $T$, then given a section of the table of area $A$, the ball spends $\frac{A}{T}$ of its time in that section.

There’s a lot of interesting material here, and a lot of it can be broken down to bite-sized chunks, tying into all sorts of other areas of mathematics. I’m starting a category (in the WordPress sense) so if you’re interested, there will be plenty more.

January 30, 2007 Posted by | Billiards | 7 Comments