Billiard tables – and what is mathematical research

Mathematical research is something that most people don’t really understand. They can imagine someone in a lab mixing chemicals or doing experiments with some scientific machinery, but mathematical research?

The goal of this post is to share with you a small part of this process – how we start with a simple problem, what happens when we complicate this problem, and how eventually we get what mathematicians call a research problem. For those interested, I made a video about this subject which you can see here and a Hebrew version here.

The square billiard table

This problem’s origin is in the well known game of billiard. While in a general billiard game there are many balls, we will simplify it and have a square billiard table and a single ball. We then hit the ball and assume that it moves in a constant speed (so it doesn’t lose any momentum), and when it hit the edges of the table, it bounces back they we expect it to.

In the example above there is a very special type of movement – after bouncing 10 times from the edges, the ball return to the same place with the same direction, resulting in a periodic movement. Our first question is how can we find directions which result in periodic movements?

There are some obvious directions as above which produce simple periodic movement with small period (=number of bounces), but there are many more complicated movements, so how can we find them?

Adding mirrors

The trick is to use mirror images. In the animation below you can see the original table together with a mirror image to its right. What’s interesting in this new point of view is that :

• Each mirror image contains all the information of our original billiard table. For example, the mirror image of the ball is in the center of the mirrored table exactly when the original ball is in the center of its table.
• Second, we also have the illusion that the balls can pass through the mirrors, thus moving longer on straight line before bouncing of the edges.

In order to use these two properties, we added an extra “transmirror” ball which can move though the mirrors. We can now use the location of this transmirror ball to know the exact location of our original ball, and note that this new transmirror ball moves on longer straight lines.

Consider now our starting location (in the center of the table) before the balls begins to move, where the transmirror ball is exactly where our original red ball. We can now direction the transmirror ball (and the original red ball) in the direction of the blue mirror image, and then let the balls start to move. This choice of direction means that eventually the transmirror ball will get to the starting location of the blue ball (the center of the mirrored table), which means that the blue ball will return to the center. But this happens exactly when the red ball return to its starting location at the center of the original table.

This is already the first step towards finding a periodic movement, but it is not enough. While the red ball returns to its starting location, it returns with a different direction (it started by moving to the right and finished by moving to the left). In order to get a periodic movement it need to return to the same place and moving in the same direction. Its true that if we just let it continue to move, it will return again to the center moving to the right, and we will return to this observation later. In order to overcome this direction problem for now, we add more mirror images.

We now have a special phenomenon where all the mirror images with the same color above move exactly the same way. Hence if we do the same trick as before, but moving towards one of the red mirror images, our original ball will return to the same place with the same direction. Indeed, when the transmirror ball reaches to the red mirror image, it will have the same location and direction as that mirror image, and by our choice of the red mirror image, it also have the same direction as the original ball. Finally, our transmirror ball moved on a long straight line, so it didn’t change directions, so the red ball starts and end moving at the same direction also!

To give a brief summation:

• Going towards the mirror images of the ball implies that the original ball will return to its starting location.
• All of the red mirror images always move in the same direction.
• The transmirror ball moves on a straight line until he gets to the red mirror image – or in other words it doesn’t change direction.

Tiling the plane

Now that we know how to use these mirrors, we can add more and more of them to produce more complicated periodic movements. We can actually add infinitely many mirror images until we tile the whole plane with squares.

Now our transmirror ball doesn’t bounce at all and can move on the same straight line all the time. If we want our original ball to have a periodic movement, we need to direct this line towards one of the red points (the red mirror images) in the picture above. This is a very special set which we can identify with the integer vectors. We can use this to show several interesting results.

For example, suppose that we move on the line towards the vector $(3,2)$. When the transmirror ball get to this point, it means that we finished the periodic movement. If we want to ask how many times the original ball bounces on the edges, then this is the same as asking how many times the transmirror ball passes through the horizonal and vertical lines. This can be easily seen to be $2\times (3+2) = 10$.

We can define the period to be the number of times the ball hit the edges until it returns to its starting location with the same direction. So for the vector $(m,n)$ the period will be $2\times(|m|+|n|)$. However, if it moves like this, then it will return to the same place with the same direction infinitely many times, so we will usually want the first time that it happens. In our geometric point of view above, this means that we want the first red point on the line.

For example, the point $(6,4)$ is on the same line as $(3,2)$ but it is not the first point on that line. This means that by the time the transmirror ball reaches $(6,4)$ the original ball finishes two periods.

More generally, these “smallest periods” correspond to points $(m,n)$ where $gcd(m,n)=1$ – we cannot divide both $m$ and $n$ with the same number $\geq 2$.

We can now use this method to count how many directions will result in a periodic movement with small period $\leq M$ – this is exactly the number of integer vectors $(m,n)$ with $2\times(|m|+|n|)\leq M$ (or if we want the smallest period, then $gcd(m,n)=1$ as well). Geometrically, this is just counting the number of integer vector in the diamond below:

The hexagon table

The process that we described in the previous section, already leads to many interesting results. The next question is if we can generalize it to similar more complicated questions. There are many ways to generalize the question, but we will restrict our attention to when we simply just change the shape of the table to other regular polygons. In particular, we can consider the hexagon table.

If we try to do the same trick we had with the square table, we immediately run into our first problem. While we can have mirror images through the edges of the table , they don’t exactly match each other. So we can have two mirror images of the original table, but they will not be mirror images of one another through their joint edge.

Previously, with the square table, because we didn’t have this mismatch problem, if we tile the plane with squares, then any point in the plane correspond to a unique point in the original table (under the mirror reflections). While we can use hexagons to tile the plane, we can no longer use this result.

However, we can go around this problem. In order to do this, first note that no matter how many reflection we take through the edges, there are only 6 types of tables that we can see (in the square case we had 4 tables)

This is not so hard to check by hand, but for those who know some group theory, you should try to prove that the group generated by these reflections is exactly $S_3$ – the symmetry group on 3 element, and of course, the size of this group is 6.

We can still define out “transmirror ball”. Whenever this ball reaches to one of the edges, it teleports to the corresponding edge in the corresponding mirror image. This is basically what we did with the square table, only now we cannot tile the plane using these images.

While we cannot (yet) use the tiling trick, we can use the fact that our transmirror ball moves in the same direction all the time (though on different tables). We now have another trick – we put all the table one over the other and join them to a single hexagon table. We have lost some of the information – this is because a point on the new table can be one of 6 points on our mirror images. However, we gained the fact that (1) we now have a single hexagon table, and (2) the ball teleports between the edges instead of bouncnig, so in a way it moves on a straight line all the time.

A table as above, where the ball teleports between the edges, but keeps the same direction, is called a translation surface. The reason is that the “teleportation” is simply translation of the ball (the same way that “bouncing” was a reflection). We can now do the same tiling trick, because instead of reflections, we have translations. As can you see in the image below – this time the colors do match.

After this tiling, we again have a special set of points which correspond to periodic directions in the translation (teleporation) table. The problem, is that exactly like the square, returning to the starting location in this table, doesn’t mean that the ball returns with the same direction. If we go back to the square example, but this time look on all the points coming from mirror images, then they correspond to vectors of the form $(\frac{m}{2},\frac{n}{2})$ where both $m,n$ are integers. If both of these integers are even, then we are on a red point. In other words, if $(x,y)$ is a point as above, then $2(x,y)$ is a red point. So we might not come back after the first iteration, but two is always enough. A similar phenomenon happens here – if $(x,y)$ is a periodic point for the translation surface, then it might not be periodic in the original billiard table, but $6(x,y)$ is going to be. You should try to check some simple example and make sure that it is correct there, and if you know some group theory, then this should sound familiar to you and you can try and give a full proof.

The octagon table

In the hexagon table, our original idea from the square table didn’t work, but we managed to find a new way to view our problem, using translation surfaces, and even salvage some of the techniques and results. However, when we go to the octagon table, we lose our tiling trick completely – we cannot tile the plane using octagons.

While this is a big problem, our workaround for the hexagon is still very useful. The trick to move to translation surface still works (and it actually works for any 2n-regular polygon table, and with slight adjustments for 2n+1-regular polygons as well).

Understanding the periodic direction for this new translation surface is not as simple as in the square and hexagon tables, mainly because we cannot use the tiling trick. Here we need new ways to view the problem in order to understand it better. When we had the problem with the hexagon table, we returned back to the square, tried to understand it better, and see what we can conclude for the hexagon. This is a very useful approach in general – when you get stuck in a certain case of a problem, go to a previous case that you know very well and try to solve it again with your new case in mind.

For example, we used the 4 to 1 and 6 to 1 covering to be able to go from periodic direction in the translation surface, and the periodic directions in the original billiard table.

We can do something similar with the octagon. The points which correspond to “smallest periods” were $(m,n)$ where $gcd(m,n)=1$. The standard way to check that two numbers are coprime is by using the Euclidean division algorithm which is probably one of the most elementary results in mathematics. There is something very similar in the octagon (and other tables as well), which eventually gives us a way to compute all the periodic direction. This is connected to what is called the Veech group of the translation surface, though I will not get into it here.

While we can find (computationally) all the periodic directions, and the size of each period, there are still many properties that we would like to know about this set. For example, both of the sets for the square and the hexagon were “arithmetic”. They came from a very well understood set with algebraic structure (e.g. the integer vectors $\mathbb{Z}^2$). Is the same true for the octagon as well? If not, is there a better definition for “arithmetic” for this type of sets which encompass the square, hexagon and octagon?

There are many questions still open, and just wait for the next idea which will clarify and give new perspective on this problem. But this is a problem for a future post.

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