# The Geometric Viewpoint

geometric and topological excursions by and for undergraduates

• Monthly Archives April 2014
• ## Periodic Billiard Paths

“When are we actually going to use this in real life?” A student complains to his or her math teacher. It is the question that we’ve all heard, at one point or another, in our high school years. When we are young, it can be difficult to see how mathematics is related to the ‘real world’. As we grow older, those of us who continue to study mathematics realize that one of the beautiful things about mathematics (and geometry in particular) is that it is present all over the place. This can be seen, for example, in the fact that the golden ratio and the Fibonacci numbers show up, well, almost everywhere. It should be no surprise, then, that almost anything can be studied using mathematics. Here we will see how to take a simple game of skill, billiards, and use geometry to study the game mathematically.

Let’s get acquainted with the ‘rules’ of mathematical billiards, which are somewhat different from the game with which many of us are familiar. First off, we play with only one infinitesimally small billiard ball. For this reason, our goal is not to hit other billiard balls into pockets; rather, it’s to see how our ball travels once we’ve set it in motion. We also play on a frictionless table and require all collisions to be elastic (this means that energy is conserved during each collision; i.e. the ball doesn’t ever slow down). So, once our ball starts moving, it doesn’t stop. The only exception is that we say the ball has stopped if it lands precisely in a corner. Our final rule change is that we can make our billiard table whatever shape we want; later on we will discuss what shapes behave ‘nicely’ as billiard tables.

There is also an important law of physics that our ball must obey; in fact, this law is also present in real billiards. We require that when the billiard ball strikes an edge of our table that

angle of incidence = angle of reflection

That is, the angle the ball’s path (prior to collision) and the edge of the table make is equal to the angle the ball’s path (after the collision) and the edge of the table make.

One of the interesting open questions in the study of mathematical billiards is, given a particular shape as our billiard table, whether it is possible to find a billiard path that is periodic; that is, one which repeats itself over and over again. A billiard path is simply the path that a billiard ball takes once it is set in motion. Here are some simple examples of periodic billiard paths. Notice that the table does not necessarily have to have straight edges.

In the case of the circle, notice that the path is periodic because it bounces perpendicularly off the edges; this is a common way to find periodic billiard paths, as we will see. We will, however, consider only polygonal billiard tables from here on. It is unknown whether every polygonal table has a periodic billiard path, and the solution to this question is an active area of research. R. Schwartz has recently proved that every triangular table with largest angle no greater than 100 degrees has a periodic billiard path. So, what other polygons are known to have periodic billiard paths? We will look at a particular class of polygons called rational polygons and prove the following theorem:

Theorem: Every rational polygon has a periodic billiard path.

A rational polygon is one whose angles are all rational multiples of $\pi$. It’s easy to get lost in the subtleties of a full proof of this theorem, so we will not discuss every detail. If you are interested in seeing the details and a more rigorous discussion, most of what follows can be found in chapter 17 of Richard Schwartz’ book, Mostly Surfaces.

The first detail we need to discuss is what we mean by a translation surface. We will demonstrate this concept with a simple example. Consider an isosceles triangle $T$ with angles $\frac{\pi}{4}, \frac{\pi}{4},$ and $\frac{\pi}{2}$. Let’s glue the congruent edges together. We call the ‘glued up’ space $\bar{T}$. $\bar{T}$ is an example of what we call a cone surface. Now, pick a point $x \in T$ that is a vertex of $T$. Then we call $\bar{x}$ (the point of $\bar{T}$ that $x$ became after gluing) a cone point. In our example, we have two cone points since two vertices of our triangle were glued together. For each cone point, we define the cone angle to be

cone angle of $\bar{x}$ = $\sum\limits_{x \in \bar{x}}$ angle at $x$.

The summation is over all points in $T$ which are glued together to form $\bar{x}$. So, in our example, the cone angle of each cone point is $\frac{\pi}{2}$ because we glued the two vertices that have angle $\frac{\pi}{4}$ together, and the vertex with angle $\frac{\pi}{2}$ is not glued to anything but itself. Now, we can finally give a definition of a translation surface: A translation surface is a cone surface such that each cone angle is an integer multiple of $2\pi$. We will not be able to prove this here, but what we need to know in order to proceed is that for any rational polygon $P$, we can take finitely many copies of $P$ and come up with gluings of pairs of edges so we obtain a translation surface. In general, we cannot do this for polygons which are not rational; as we will see shortly, this is the reason why we can only prove that rational polygons have a periodic billiard path.

Here’s an easy example of how we can construct a translation surface from a rational polygon. Let’s begin with one of the simplest examples of a rational polygon: a square. Call our square $S$. Let $e_i$, $i = 1,2,3,4$ be the edges of $S$ as shown to the left.

For each $e_i$, let $R_i$ be the reflection over that edge. If $e_i$ and $e_j$ are parallel, then $R_i = R_j$ (this isn’t exactly true, but hold on!). So, in our example, $R_1 = R_3$ and $R_2 = R_4$. Let $G$ be the group consisting of all possible compositions of $R_1$ and $R_2$. We can see that $G = \left\{ {I, R_1, R_2, R_1 \circ R_2} \right\}$ since any other composition gives us something already in $G$. Note that the order of $G$ is 4; in particular, $G$ is a finite group! This is very important: going through this process with any rational polygon will yield a finite group. If we generate a group this way for a polygon which is not rational, then the group will not necessarily be finite, and if that happens we cannot continue. We make ‘copies’ of $S$ by applying each of our group elements to $S$. We also shift our copies of $S$ so that they are all disjoint from each other. The fact that we shift is what allows us to say two reflections are equal if the corresponding edges are parallel. We now have four copies of our original square.

We begin with the obvious gluings: $e_1$ to $R_1(e_1)$, $e_2$ to $R_2(e_2)$, $R_2(e_1)$ to $R_1 \circ R_2(e_1)$, and $R_1(e_2)$ to $R_1 \circ R_2(e_2)$.

Now we have a bigger square whose edges we need to glue together. To do this, we will glue opposite edges together. For example, we glue $e_3$ to $R_1(e_3)$, $R_2(e_3)$ to $R_1 \circ R_2(e_3)$, etc. (For general polygons, there are equations for determining which gluings to make, but we won’t go into that). It is easy to verify that all cone angles are integer multiples of $2\pi$. The resulting surface is a torus, which is a doughnut or ring shape, and this is our translation surface. It’s important to note that we still think of our translation surface as being flat. Although our translation surface does look like the surface below, we prefer to visualize it as a flat polygon with with the gluings that we’ve defined.

We need two lemmas before we can tackle our theorem. Before we introduce our first lemma, however, we need a little background information. Define a path $\gamma \in \bar{P}$ (where $\bar{P}$ is a translation surface) to be straight if when we ‘unglue’ our translation surface, the path becomes a straight line. For example, the equator is a straight path on the Earth; although the equator is curved, it looks like a straight line on a map.

We need a way to go between our polygon $P$ and the translation surface $\bar{P}$ that we build from $P$. To accomplish this, we define a function $\pi : \bar{P} \rightarrow P$. $\pi$ takes as an input some point $\bar{p}$ and outputs the point on $P$ corresponding to $\bar{p}$ after we construct our translation surface.

Lemma 1: Let $\hat{\gamma}$ be a straight path on $\bar{P}$ which does not go through any cone points of $\bar{P}$. Then $\gamma = \pi(\hat\gamma)$ is a billiard path on $P$.

We won’t go through a full proof here. However, it is not so difficult to convince yourself of this result: take a piece of paper and fold it in half, creating a crease. Unfold it and draw a straight line which crosses the crease. This straight line is analogous to a straight path which crosses an edge from one copy of $P$ to another. When you refold the paper (folding the paper is analogous to hitting $\bar{P}$ with our $\pi$ function), you’ll see that the straight line now ‘bounces’ off the crease in accordance with the rule angle of incidence = angle of reflection. In the figure below, the dotted line is the image of the straight path after refolding the paper.

Now, for any $\bar{x} \in \bar{P}$, and for any direction on $\bar{P}$, define $f(\bar{x})$ to be the point of $\bar{P}$ obtained by starting at $\bar{x}$ and moving one unit in the chosen direction. $f$ is defined everywhere except when the path between $\bar{x}$ and $f(\bar{x})$ hits a cone point. If we think of our path as traveling at unit speed, we can say that $\pi(f^n(\bar{x}))$ is the location of our billiard ball on $P$ after $n$ seconds. We will now prove our second lemma:

Lemma 2: Pick a point $p \in \bar{P}$. Then there is a point $q \in \bar{P}$ that is very close to $p$ and $f^n(q)$ is very close to $p$ (for some $n$).

Proof: Let $D_0$ be a small disk around $p$, and for each $n \in \mathbb{N}$ let $D_n$ be a disk centered around $f^n(p)$ of the same size as $D_0$. Since the area of $\bar{P}$ is finite, the disks $D_0, D_1, D_2, \ldots$ are not disjoint. Hence we can find two disks, $D_i$ and $D_j$ that intersect at a point. Call this point $x_n$ and assume $i < j$. Then we can see that $D_{i-1}$ and $D_{j-1}$ intersect at the point $f^{-1}(x_n) = x_{n-1}$. We can continue this process until we see that $D_0$ and $D_{j-i}$ intersect at the point $x_0$, which is obtained by composing $f^{-1}$ with itself $(j-i)$ times and evaluating it at $x_n$. But then $x_0$ is very close to $p$ and also $f^{j-i}(x_0)$ is very close to $p$ (since $x_o$ is contained in the disks $D_0$ and $D_{j-i}$). This completes the proof.

We now have all the tools we need to prove the theorem. We let $P$ be any rational polygon, and we build a translation surface $\bar{P}$ from $P$. Pick a point $p \in \bar{P}$ and a direction that is perpendicular to and heading towards a nearby edge of $P$. $\hat{\gamma}$ will be a straight path which starts at $p$ and in the chosen direction. By our first lemma, there is a billiard path $\gamma$ in $P$ which corresponds to $\hat{\gamma}$. By our choice of direction, $\gamma$ is traveling perpendicularly to an edge of $P$ at time 0. By our second lemma, there is a $q$ very close to $p$ such that $f^n(q)$ is also very close to $p$ (for some $n$). Let $\beta$ be the path starting at $q$ and in our chosen direction. We make our choice of “very close” to mean that $q$ and $p$ are on the same copy of $P$ used in the construction of $\bar{P}$. Recall that $\beta$ is a straight path, and so $\beta$ is traveling in the same direction at times 0 (when it is at $\pi(q)$) and $n$ (when it is at $\pi(f^n(q))$). But at time 0, $\beta$ is traveling perpendicularly to an edge of $P$, since it is traveling in the same direction as $\gamma$. Hence, $\beta$ is traveling perpendicularly to an edge of $P$ at times 0 and $n$. Therefore, $\beta$ hits the edge perpendicularly twice, and so it periodic, since it just retraces its path each time it hits the edge perpendicularly. Such a path might look like this:

There are two natural questions to ask now. The first is “What about polygons that are not rational? Do these polygons not have periodic billiard paths?” The answer here is ‘no’, there are some non-rational polygons which do have periodic billiard paths. To convince yourself of this, notice that any parallelogram has a periodic path. In fact, there are no known examples of non-rational polygons which do not have a periodic billiard path. The second, and more difficult, question to ask is “Does this result hold in three dimensions as well? Is it true that rational polyhedra have periodic billiard paths?” The intuitive answer might be ‘yes’ because none of our rules change when we move up in dimension. It has been proven that the answer is in fact ‘yes’ for tetrahedra, but the result is unknown for more complicated polyhedra, like this one.

Sources

Mostly Surfaces by Richard Evan Schwartz

Obtuse Triangular Billiards II: 100 Degrees Worth of Periodic Trajectories by Richard Evan Schwartz

Images were either hand drawn or found using a Google Image search

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• ## Completeness on an Incomplete Sphere.

I used to think I had a fairly good understanding of distance. If asked to find the distance between two points, I mentally connect them with a line segment and then find its length using the Pythagorean Theorem. Though I knew how to create fancy mathematical spaces, when it came to the real world I had my real sense of distance: straight lines connecting points.

Figure 1: When measuring the distance between cities, it doesn’t make sense to find the length along the straight line through the center of the Earth. Instead, we look at the length of the curve along the Earth’s surface.

Despite this intuition, I jettison this definition of distance when it doesn’t make sense. When I pulled out my ruler on Google Maps to find the distance between my hometown of Jay, Maine and Sydney, Australia, I found that there were over ten thousand miles between them. Quite impressive since the diameter of the Earth (the greatest distance between two points) is only about eight thousand miles. Clearly then, this “distance” according to Google is not just the linear displacement. In fact, as we see in Figure 1, this distance follows a path along the exterior of the sphere. Though different that our original idea of distance, this length is more useful if we were to plan a trip.

Suddenly, these fancy mathematical spaces with their unintuitive notions of distance didn’t seem quite so pointless. Depending on the purpose of the measurements, there is very good reason to draw something besides a straight line, or to even cause our sense of length to stretch as depending on our position. When working on my thesis, I encountered an interesting theorem by Paul Montel, which states that a set of analytic functions is normal when their range misses at least three points. In this proof, I was required to work with an entirely new sense of distance on this thrice punctured sphere.

In different contexts, we may define many different ways to measure distance (metrics) so that our measurements will be useful. One property we seek when defining a new metric is the idea of completeness. If we take a sequence of points, $( x_1, x_2,$…), completeness says that if the sum of the distances between each $x_n$ and the next one is finite, then the points must be getting arbitrarily close to some limit in our space. When traveling across the surface of the Earth, if we walk some finite distance and then stop we are always guaranteed to still be on the surface of the Earth. Completeness serves as an important property in real analysis. In fact, it is because we want completeness that we construct the real numbers from the rational numbers.

Figure 2: The sphere is projected onto the plane by drawing a line from N through a point P, sending P to where the line crosses the plane (P’).

If we cut out some small, closed region from our sphere, or even a single point, then our new metric space (under the same definition of distance) ceases to be complete. Consider if a country, on Earth, closed itself to foreigners. I would never be able to travel to this country, despite the fact that the distance between me and its border would be finite. If we want to consider the Earth, without this country, to be a complete system, we must redefine distance so that the distance from anywhere on our space to the border of this country is infinite. To illustrate how we might define such a distance, let us look at a sphere which has a single point $N$ removed from it. We may transform any point of the once-punctured sphere onto the Euclidean ($\mathbb R^2$) plane via the continuous bijection known as the Stereographic Projection (Figure 2). We then say that the distance between two points on the sphere, p and q, is the Euclidean distance between p’ and q’. This metric, it turns out, is complete on the once- punctured sphere. Through a similar process, we may induce a metric on the a sphere with two distinct punctures using the Eucledian metric (we won’t go into detail about that process here).

Figure 3: Starting on the disk, we see that we can connect sides to make a pair of pants.

Figure 4: The pair of pants can be deformed into a sphere with three holes.

So then, if we have a simple example of a metric on the sphere, and a way to induce a metric on the spheres with one or two punctures using Euclidean space, is it possible to find a complete  metric on the thrice-punctured sphere? The answer is yes… but not by relating our back to Euclidean space.

Consider the shape, S, formed by four arcs in a unit disk and its edge connecting them (Figure 3). Getting rid of the unnecessary portion of the disk, we may fold S over the x axis and connect the top two arcs to the bottom two. We then take any two points along these arcs sharing an x-coordinate and define them to be equivalent (”gluing” them together so to speak).  This gluing forms what we call a pair of pants. Note that this shape has three holes in it, suggesting a similarity to the thrice punctured sphere. In fact, we can stretch it so that it exactly resembles such a sphere (Figure 4). By making the arcs intersect the edge of the sphere closer to one another, we may shrink these punctures down to single points.

Recall how we defined a metric on the punctured sphere by sending points to the Euclidean plane. We now define distance on the thrice-punctured using the mapping from the sphere back into the disk. Note that, for any path on the sphere, we may unwrap the sphere back into $S$, giving us some collection of paths in the disk. We then define the length of a path by however long its segments are on the unwrapped space. The distance between two points, then, is the largest value which is shorter than the lengths of all these paths.

Figure 5: Showing how a path on the thrice-punctured sphere can be “unwrapped” back onto our original shape S. The length of the path then, from A to B, is the length of each of the segments of its inverse image in S.

This just means that we need to clarify our notion of distance on $S$, inside of the open unit disk. Now, we could use the Euclidean metric again, but this would still fail to give us a complete space. If I take the sequence $0, \frac 12, \frac 23, \frac 34,\frac 45$…, it should not be hard to see that the sum of the distances between consecutive points is just one. Indeed, the first $n$ many pairs of distances will just be the difference between zero and the $n+1$th term. As the sequence goes to one, the sum of these lengths goes to one. However, the fact that the sequence converges to one also causes a problem, as one corresponds with one of the punctures on our sphere. So then, we cannot have a complete sense of distance on the open disk using the Euclidean sense of distance.

Figure 6: Taking three line segments A,B, and C of equal Euclidean length, it turns out that the hyperbolic length of them is increased the closer they are to the boundary. The length of B is greater than that of A, while we say that the length of C is infinite, as it approaches the boundary.

This is where hyperbolic space comes in. On the unit disk model, we think of hyperbolic space as becoming squished near the boundary. Though the Euclidean distance between two points in the disk may only be some value d, if we slide them towards the edge of the disk (without changing their Euclidean displacement) then the hyperbolic distance between them approaches infinity. Though a severe over-simplification, think about taking a 6″ pen and holding it horizontally in front of your face. Though we might say that the pen has a constant Euclidean length, regardless of where you place it, if you slowly move the pen towards one of your eyes (while holding the other shut) the width of the pen appears to grow arbitrarily large, until it is wider than your field of vision.

It is this property that makes our space complete. Looking back at our previous example, of the sequence converging to one. In Euclidean space, the distances between consecutive terms added up to one. However, we now have that the magnitude of these distances gets amplified as they get closer to one. In fact, since the line segment from zero to one has infinite length in hyperbolic space, we infer that the sum of these distances (which are just the pieces of this line segment) must also grow arbitrarily large as we add them. We see then that this fixes the problem we had when we tried to apply a Euclidean manifold.

By repairing this issue, it turns out that our new space is, in fact, complete. The existence of a complete metric induced by the hyperbolic disk is implied by applying the Poincare’ Polygon Theorem to $S$. Some may wonder then, why am I making such a big deal about completeness? It turns out that many forms of advanced mathematics may only work to their full potential on complete spaces. My entire preoccupation with the thrice punctured sphere began when doing research for my honors thesis in Complex Analysis. Despite being a completely different field, I needed this definition of distance. The Schwarz–Ahlfors–Pick Theorem directly relates this metric to the idea of differentiability, allowing us to create differentiable functions on the open disk to this punctured sphere which don’t increase hyperbolic distance.

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• ## Curvy Television

Television screen size became a status symbol in the American household soon after TV’s adoption
in the 1940’s, and remains so today. There are many equivalent ways to characterize screen size; today’s standard is by the diagonal length of the screen. Everyone understands that this length is related in a familiar way to a screen’s area. With today common 16:9 aspect ratio (horizontal length:vertical length), the screen area ${A(l)=\alpha l^2}$ where ${l}$ is the diagonal length and ${\alpha=337/144}$. In other words, the area varies with the square of the diagonal length.

This polynomial relationship, however, is a feature of euclidean space. TV screens placed on hyperbolic and spherical surfaces lead to surprising differences in area relative to diagonal length. Curvature will play an important role in illustrating the discrepancy between euclidean, spherical, and hyperbolic geometries. To see this, let us imagine covering our television screen with a piece of paper. Using Ilmari Karonen’s imagery from Stack Exchange we shall lay a large sheet of paper over our euclidean television, and cut it down to size.

Now imagine that the surface of our TV is lying on a sphere. We will try to lay our cut-to-size sheet over the screen, a curved surface. To anyone who has used paper-mache, it’s immediately apparent that the paper will need to wrinkle and fold up around itself to fit to the surface. There is simply too much paper and too little sphere to go around. What is happening? The intuitive answer is that there is simply not enough space on our spherical surface. This simple fact hints at a sobering idea for cartographers; there exists no flat map that is locally isometric to the Earth at any point. No matter how carefully you restrict your vision, a flat map will be always be wrong. Gauss formally proves this in his Theorema Egregium. It turns out to be a consequence of the constant positive curvature of spherical space, as opposed to zero curvature in euclidean space.

Tangent plane intersecting a surface to create a curve.

Curvature, also know as Gaussian curvature ${K}$ at a point ${\alpha}$ on a surface, is defined as the product ${\kappa_1 \cdot \kappa_2 }$, where ${\kappa_1}$ and ${\kappa_2 }$ are the principal curvatures of ${\alpha}$. Finding the principal curvatures at ${\alpha}$ can be visualized by the following method: First, find the normal vector to the surface at ${\alpha}$. Next, look at the intersection of a plane containing the normal vector with the surface. This intersection is a curve on the surface. As you rotate the plane around the normal vector, the intersection curve will change. The principal curvatures are the maximum and minimum curvature of the intersection curve at ${\alpha}$ as you rotate the plane.

On a plane, the intersection of a normal plane to a point on the surface will be a line. Since the curvature of any line is zero, ${k_1 \cdot k_2 = 0}$ at all points, resulting in constant zero curvature as we have already mentioned. Similarly, on a sphere, any plane containing a normal vector will intersect the sphere along a great circle (geodesic) of the sphere. Since any two great circles have the same (non-zero) curvature, we see that ${\kappa_1 \cdot \kappa_2 = \beta > 0}$ at all points, i.e. a sphere has constant positive Gaussian curvature.

Hyperbolic Spaces locally look like a saddle point.

Hyperbolic space challenges this definition of curvature. How can we use our method on a surface which cannot be embedded in euclidean 3-space? One way to visualize hyperbolic space is as a collection of points each of which locally looks like a saddle point. If we take a plane containing a normal vector to a saddle point, and intersect it with the saddle, we see that the curve on the saddle radically changes as we rotate the plane. The maximal and minimal curvatures of the intersection curves will be positive and negative respectively. Therefore, ${K= \kappa_1 \cdot \kappa_2 < 0}$. Since each point of hyperbolic space locally looks like an identical saddle, we see that hyperbolic space has constant negative curvature.

Largest rectangle possible in hyperbolic space

What happens when we embed our television in a hyperbolic surface? The first thing to realize is that ${2\pi}$ is the maximum area of a rectangular screen. This is similar to how a spherical rectangle is bounded by the size of the sphere. However, we do not want to restrict the size of our television. Bigger TV’s are always better! Instead, we will investigate the area of circular televisions, or hyperbolic disks, which have no upper bound. If we calculate the circumference of a hyperbolic circle in terms of its radius ${r}$, we find that ${Circ(r) = 2\pi\sinh(r)}$. Now we can simply integrate this value from ${0}$ to ${R}$ to find the area ${A(R)}$ of a hyperbolic disk, ${\int_0^R 2\pi\sinh(r) dr = 4\pi\sinh^2(\frac{1}{2}R)}$. This can be rewritten as ${A(R) = \pi(e^{R}+e^{-R})-2\pi}$, where ${R}$ is the radius of the disk.

Notice that in hyperbolic geometry, circular area grows exponentially with the radius of a disk. Our TV’s area, therefore, will grow much faster as we increase its radius than it would in euclidian space. In other words, there is more space in a hyperbolic disk than in a euclidean disk of the same radius. This is the opposite of our result from spherical geometry, and is a good example of the connection between Gaussian Curvature and the “amount of space” in a given surface. The book Crocheting Adventures with Hyperbolic Planes is a good reference for more information about curvature and the hyperbolic plane.

The connection between curvature and space has been used by physicists to investigate our universe. A major question in the field of astronomy is whether or not our universe, on a large scale, has positive, zero, or negative curvature. One method to test this theory is to calculate the density of galaxies at different distances from our own galaxy. A constant density would be reminiscent of a flat (zero-curvature) universe. However, if we observed a growing density of galaxies as we looked further away, this would hint at negatively curved space, because we are observing “more space” with our instruments, and hence we would see more galaxies. Similarly, a shrinking density of galaxies would suggest a positively curved universe. Using this method, physicists have been able to rule out the possibility of a large curvature in our universe, however the difficulty of accurately measuring densities has restricted the ability to make a stronger claim.

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