Billiards

My main research is in polygonal billiards. Consider a single billiard ball, bouncing around inside a polygon. (We make all the standard ridiculous mathematical assumptions of course—it has radius 0, there is no friction, and collisions are perfectly elastic.) Will the ball end up repeating the same pattern over and over, or will it continue to trace out slightly different paths forever?

It turns out that for almost every starting condition, the pattern never repeats—in fact, with the right tools (namely countability) this is easy to prove. But are there any paths which repeat? This turns out to be much harder. Even the case of triangles has been open since 1775.

Below is a basic simulation for exploring these dynamics in triangles. The simulation comes with a few presets to explore important phenomena, such as periodic trajectories, aperiodic trajectories, and how very similar trajectories will eventually diverge.

The most important tool in the study of polygonal billiards is called unfolding. It stems from the observation that when the billiard ball hits the edge of a polygon, instead of imagining the billiard ball reflecting off that edge, you can imagine the polygon reflecting across that edge while the ball continues in a straight line. This is illustrated below whenever you run the demonstration above for a periodic trajectory.

Taking this further, you can imagine the surface built by gluing together copies of a polygon along their edges. Billiard trajectories are exactly the straight lines (or geodesics) on this surface, and the question of periodic trajectories becomes: are there any closed geodesics on this surface? We can further note that any two polygons with the same orientation are identical for the purposes of trajectories, so we can glue them together, which in some cases gives us a finite surface.

It is not immediately obvious that this is an easier question. But while nobody has come up with any good techniques for characterizing billiard trajectories directly, there are a number of advanced techniques known for characterizing geodesics on surfaces, especially finite surfaces. Of these, the most fruitful is known as Teichmüller theory. However, my research has focused on algebraic geometry, specifically the fundamental group.

Computational Representation Theory

I have done research on the representation theory of the cohomology of the pure configuration space on $\mathbb C^n$. Church, Ellenburg and Farb prove in this preprint that the $S_n$-representation of $H^i(PConf_n(\mathbb C))$ stabilizes as a function of $n$, in a sense they define precisely. I've written a library for computing the stable value of this decomposition, which is available on my GitHub. The project of actually performing these computations is chronicled on my programming page.

Good Problems

These are a few good problems I have collected over the years. Click to show/hide the solutions.

• Consider the sequence $a_n=2^n$. As $n\to\infty$, what fraction of the terms begin with a $1$?

  Note that $2^n$ begins with $1$ iff $\log_{10}2^n\,\mathrm{mod} 1 < \log_{10}2$. Since $\log_{10}2$ is irrational, $\log_{10}2^n=n\log_{10}2$ is equidistributed $\!\mathrm{mod} 1$, so the asymptotic fraction of terms with $\log_{10}2^n \,\mathrm{mod} 1 < \log_{10}2$ is $\log_{10}2$.

• Let $X$ be a metric space. Show that $X$ is compact iff every continuous function $f:X\to \mathbb R$ is bounded.

  If $X$ is compact, then the open cover $\{f^{-1}((-n,n)):n\in\mathbb N\}$ has a finite subcover, hence $f^{-1}((-n,n))=X$ for some $n$, so $f$ is bounded between $-n$ and $n$. If $X$ is not compact, we have some sequence $x_n$ with no convergent subsequence, hence $\{x_n:n\in \mathbb N\}$ is a closed, discrete set. Define $f:\{x_n:n\in \mathbb N\}\to \mathbb R$ by $f(x_n)=n$, which is continuous by discreteness. By the Tietze Extension Theorem, this extends to a (clearly unbounded) function on $X$.

• Take an $8\times 8$ chessboard and remove two opposite corner squares. Can the resulting board be covered by dominoes, which each cover 2 adjacent squares?

  No. Each domino covers one square of each color, but we removed 2 squares of the same color.

The Quadratic Equation and Beyond

This is an experiment in writing math understandably and accessibly, inspired by the displays at the National Museum of Math. This page is customized based on your familiarity with the relevant mathematics. Curiosity is a pre-requisite for understanding the below; a high-school diploma is not.

We all learned the quadratic formula in school, maybe we even proved that it works by plugging it in. But how does it work? How is one to come up with it?

Where to even begin? Suppose we know only about integers and arithmetic. Our "numbers" all look like $a/b$ where $a$ and $b$ are integers. But this alone is not sufficient to talk about the roots of quadratics, even simple ones like $x^2 - 2$. If $(a/b)^2 - 2 = 0$, then $a^2 = 2b^2$, and since any odd number squared is odd then $a$ must be even, i.e. we can decompose it into $a = 2u$. But then $4u^2 = 2b^2$ so $2u = b^2$, hence by the same argument $b$ is even and we can write $u^2 = 2v^2$ where $2v = b$. This is exactly what we started with, so we can keep on dividing $a$ and $b$ by $2$ indefinitely, which is clearly nonsense.

Thus we have to move beyond arithmetic to incorporate solutions to quadratics, and the only question is how far. We could simplify define $\mathrm{root}(a,b)$ to be a root of $x^2 + ax + b$, but this is unsatisfying. Is there a middle ground—some partial leap we can make, and then leverage arithmetic to get the rest of the way? A natural question is whether it is sufficient to introduce the solutions to $x^2 - a$ for positive integers $a$ to our number system. We use $\sqrt{a}$ to denote a solution to $x^2 - a$, but pretty quickly we run into a problem: $-\sqrt{a}$ is also a solution to $x^2 - a$! We must be more specific, and to do so we reach far beyond arithmetic, beyond algebra even, to the concept of the real numbers. We define $\sqrt{a}$ for $a \gt 0$ to be the unique positive real number^[0] satisfying $\sqrt{a}^2=a$. Can we use arithmetic plus our new $\sqrt{}$ operator to solve arbitrary quadratics of the form $x^2 + ax + b$?

Let $u$ and $v$ denote its roots. We can factor $x^2 + ax + b$ as $(x - u)(x - v)$^[1], which gives us the relationships $\begin{array}{rcr} u + v &= &-a\\ uv &= &b \end{array}$

We would like to use these to find $u$ and $v$ individually. The equation $u + v = -a$ is tempting, since if we can find an expression for the very similar-looking $u - v$ in terms of $a$ and $b$ we can combine them to determine $a$ and $b$. But there's a problem—since our choice of the labels $u$ and $v$ for the two presumably distinct roots of the quadratic are entirely arbitrary, we won't find any simple expression for any function $f(u,v)$ in terms of $a$ and $b$ unless $f$ is symmetric—i.e. $f(u, v) = f(v, u)$. Unfortunately $u - v$ is not symmetric, but as luck would have it $(u - v)^2$ is! And expanding this, we see that $\begin{array}{cl} (u - v)^2 &= u^2 -2uv + v^2\\ &= (u + v)^2 - 4uv\\ &= a^2 - 4b \end{array}$ and so the two possible values of $u - v$ are $\pm\sqrt{a^2 - 4b}$. Thus we have $\begin{array}{cl} u &= \frac{(u + v) + (u - v)}{2}\\ &= \frac{-a \pm\sqrt{a^2 - 4b}}{2} \end{array}$ which is the quadratic formula^[2].

Now what about the cubic formula? If the quadratic formula seemed like a surprising amount of work, buckle up. While Ancient Greeks knew how to solve quadratic equations (although they had not yet even invented the modern notation for equations), it was not until the Italian Renaissance that mathematicians managed to solve cubic equations. The approach I use below would not have been recognizable until the work of Lagrange in the late 18th century, which finally allowed mathematicians to solve the quartic as well.

Let $x^3 + ax^2 + bx + c$ be a cubic polynomial and $u, v$ and $w$ be its roots. Can we find these roots using what we learned from the quadratic? We run into a problem on even the simplest example, $x^3 - 1$. It is easy to see that its only real root is $1$, but this root doesn't have multiplicity $3$ because $(x - 1)^3 = x^3 - 3x^2 + 3x - 1 \ne x^3 - 1$. Given that we have no technique for distinguishing between the three roots of a cubic, it's a pretty serious problem that we seem to be missing two of them.

We have to go even beyond the real numbers now, and invent some larger number system which contains a number $j$ such that $j^3 = 1$ but $j \ne 1$. The three roots of $x^3 - 1$ are then $j, j^2$ (since $(j^2)^3=(j^3)^2=1^2=1$) and $1$. As it turns out we can find a number system with such a $j$ that still obeys the normal rules of arithmetic, namely the complex numbers, and $j=\frac{-1 + \sqrt3 i}{2}$. Using $j$ and the real solutions $a^{1/3}$ to $x^3 - a$ (which unlike the quadratic case are unique, since a negative to the third power does not become positive), can we solve the cubic?

As in the quadratic case, $x^3 + ax^2 + bx + c$ can be factored as $(x - u)(x - v)(x - w)$^[3]. Multiplying this out gives us $x^3 + ax^2 + bx + c = x^3 - (u + v + w)x^2 + (uv + uw + vw)x - uvw$ thus we have three equations relating the roots to the coefficients $\begin{array}{rcr} u + v + w &= &-a\\ uv + uw + vw &= &b\\ uvw &= &-c \end{array}$ We want to pull the same trick as before, where we find other combinations of $u, v$ and $w$ that we can combine with $u + v + w$ to obtain $u, v$ and $w$ individually. This trick comes from linear algebra, and is called finding a basis for the span of $u, v$ and $w$. This requires three linear combinations of $u, v$ and $w$, of which we already have one ($u + v + w$). We want the other two to be as close to symmetric as possible, but this is much harder than in the quadratic case, since there are only two possible orderings (or permutations) of two roots $u, v$, but there are six permutations of $u, v, w$. I used a Sage notebook to help me with the following calculations, and I encourage you to follow along with it and check my work.

Experience and instinct guide me to try $f = ju + j^2v + j^3w$. Rotating the three roots (known as a cylic permutation)Cycling gives us $jw + j^2u + j^3v = j(ju + j^2v + j^3w) = jf$ and $jv + j^2w + j^3u = j^2(ju + j^2v + j^3w) = j^2f$. Note that this means $f^3$ does not change when we rotate the roots (or in math terms, is invariant under cyclic permutations), as $(jf)^3=j^3f^3=f^3$ and $(j^2f)^3=j^6f^3=f^3$.is invariant under cyclic permutations. Similarly, for $g = j^2u + j^4v + j^6w$, cyclic permutations amount to multiplying by $j^2$ or $j^4$, and so $g^3$ is invariant under cyclic permutations. Note that $\begin{array}{rclcl} g &= &j^2u + j^4v + j^6w &= &jv + j^2u + j^3w\\ j^2g &= &j^4u + j^6v + j^8w &= &ju + j^2w + j^3v\\ j^4g &= &j^6u + j^8v + j^{10}w &= &jw + j^2v + j^3u \end{array}$ and the right-hand sides are just $f$ with two of the roots swapped, so any of the $6$ permutations of $u,v,w$ will either leave $f^3$ and $g^3$ invariant, or interchange $f^3$ with $g^3$.$S_3$ acts on $\{f^3, g^3\}$.

This should remind you of the quadratic case: we need to find expressions for $f^3$ and $g^3$ in terms of symmetric polynomials of them. This amounts to finding the roots of the polynomial $x^2 - (f^3 + g^3)x + f^3g^3$ in terms of its coefficients. For convenience, we use $h = f^3 + g^3$ and $t = fg$ to rewrite this as $x^2 - hx + t^3$.

Expanding $h$ and collecting terms gives us: $h = 2(u^3 + v^3 + w^3) + (3j + 3j^2)(u^2v + u^2w + uv^2 + v^2w + uw^2 + vw^2) + 12uvw$

We can simplify this by recalling from our factoring of $x^3 + ax^2 + bx + c$ that the sum of the roots is always the negative of the coefficient of $x^2$, so in the case of $x^3 - 1$ the sum $1 + j + j^2$ of its roots is $0$ hence $j + j^2 = -1$. This gives us $h = 2(u^3 + v^3 + w^3) - 3(u^2v + u^2w + uv^2 + v^2w + uw^2 + vw^2) + 12uvw$

Note that each monomial on the right is homogeneous and has degree $3$. We say this polynomial is homogeneous because all of its monomials have the same degree. Similarly $a,b$ and $c$ are homogeneous with degree $1,2$ and $3$ respectively. The product of two homogeneous polynomials is itself homogeneous, and its degree is the sum of their degrees. The products of $a,b,c$ which are homogeneous polynomials of $u,v,w$ of degree $3$ are $a^3, ab$ and $c$. Thus if we are looking for a polynomial of $a,b,c$ that equals $h$, we should look at linear combinations of $a^3, ab$ and $c$: $\begin{array}{cl} a^3 &= -(u + v + w)^3\\ &= -(u^3 + v^3 + w^3) - 3(u^2v + u^2w + uv^2 + uw^2 + v^2w + vw^2) - 6uvw\\ ab &= (u + v + w)(uv + uw + vw)\\ &= u^2v + u^2w + uv^2 + uw^2 + v^2w + vw^2 + 3uvw\\ c &= -uvw \end{array}$ We start by using $a^3$ to cancel out the $u^3 + v^3 + w^3$ term in $f^3+g^3$. This gives us $h + 2a^3 = -9(u^2v + u^2w + uv^2 + uw^2 + v^2w + vw^2)$ and using $ab$ to cancel out this term we get $\begin{array}{cl} h + 2a^3 + 9ab &= 27uvw\\ &= -27c \end{array}$ hence $h = -2a^3 - 9ab - 27c$

We also know that $t^3=f^3g^3$ is invariant under all permutations of the roots, so we expect^[4] be able to express it in terms of the coefficients of our original cubic. We actually get a bit lucky—you can check that $t=fg$ is itself invariant under all permutations of $u, v$ and $w$. Expanding $t$ and collecting terms gives $t = (u^2 + v^2 + w^2) - (uv + uw + vw)$ and cancelling terms we get $\begin{array}{cl} t - a^2 &= -3(uv + uw + vw)\\ &= -3b \end{array}$ hence $t = a^2 - 3b$

Working backwards, we apply the quadratic formula to $x^2 - hx + t^3$ to get $\begin{array}{cl} f &= \left(\frac{h + \sqrt{h^2 - 4t^3}}{2}\right)^{1/3}\\ g &= \left(\frac{h - \sqrt{h^2 - 4t^3}}{2}\right)^{1/3} \end{array}$ Note that we've arbitrarily chosen $f$ to correspond to the positive square root and $g$ to the negative square root. Finally, we get the system of equations $\begin{array}{rcr} u + v + w &= &-a\\ ju + j^2v + w &= &f\\ j^2u + jv + w &= &g\\ \end{array}$ to which the solution is represented as $\begin{pmatrix} u\\ v\\ w \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1\\ j & j^2 & 1\\ j^2 & j & 1 \end{pmatrix}^{-1} \begin{pmatrix} -a\\ f\\ g \end{pmatrix}$

This method extends to the quartic polynomial $x + ax^3 + bx^2 + cx + d$ as well. We want to build off of our solution to the cubic by finding a basis for the roots, where powers of each expression are invariant under a subset of permutations, and where the remaining permutations in turn permute these expressions. the action of some normal subgroup of $S_4$ which has quotient $S_3$. Luckily the Klein 4-group $K_4$ consisting of the permutations $\{(), (12)(34), (13)(24), (14)(23)\}$ is just such a group. A natural such basis is $\begin{matrix} -a &= (u + v) + (w + z)\\ p &= (u + v) - (w + z)\\ q &= (u - v) + (w - z)\\ r &= (u - v) - (w - z) \end{matrix}$ Note that $p^2, q^2$ and $r^2$ are all invariant under switching any pair of roots at the same time as any others, e.g. $u,v,w,z \to v,u,z,w$ $K_4$. Thus we can reduce the quartic case to finding the roots of $x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + q^2r^2)x - p^2q^2r^2$ Repeating the same process we used for the cubic will allow us to solve for $p^2,q^2,r^2$ in terms of the coefficients $a, b, c, d$, and taking square roots and solving the system of linear equations will produce the roots to our original quartic.

However, this method breaks down when we get to the quintic, as was first proven at the turn of the 19th century and fully understood with the development of Galois theory. There is no normal subgroup of $S_5$ from which we can generate such a basis. In fact the only nontrivial normal subgroup of $S_5$ is the alternating group $A_5$, which in turn has no nontrivial normal subgroups, so there is no way for us to recurse the process we used for lower degree polynomials to produce such a basis for $A_5$. It can be shown using Galois theory that our process will always work for some choice of subgroups if the roots can be constructed using arithmetic operations and taking solutions to $x^n - a$, thus in general the roots of quintic polynomials cannot be constructed this way.

0. ^

   Since $f(x) = x^2 - a$ is a continuous function, and for $a \gt 0$ we have $f(0) \lt 0$ and $f(a + 1) = (a + 1)^2 - a = a^2 + a + 1 \gt 0$, by the Intermediate Value Theorem it has some real root between $0$ and $a + 1$.

1. ^

   To prove this, first note that $u + v = -a$ is equivalent to $-v = u + a$, so we are trying to show that $(x - u)(x + (u + a))=x^2+ax+b$. To see that this holds, note that $(x - u)(x + (u + a)) = x^2 + ax - (u^2 + au)$ and $-(u^2 + au) = b - (u^2 + au + b) = b$ as $u^2 + au + b = 0$ by definition of $u$ as a root.

2. ^

   The quadratic formula as usually taught is for quadratics in the form $ax^2 + bx + c$. Dividing this by $a$ and re-labelling the coefficients gives our form.

3. ^

   In general, any polynomial with leading coefficient $1$ can be factored into $(x - r_1)\cdots(x - r_n)$ where the $r_i$ are its roots; this is known as the Factor Theorem.

4. ^

   In fact, any symmetric polynomial in the roots of a polynomial can be written as a polynomial in its coefficients. This surprising fact is known as the Fundamental Theorem of Symmetric Polynomials.