My mathematical philosophy is that mathematical truths exist outside any human action, but the key activity of mathematics is not in discovery but understanding. This has several of important consequences. First, it reaffirms the doctrine that "proofs are more important than theorems, and definitions are more important than proofs". Second, it implies that teaching is just as central to mathematics as research, and arguable moreso.


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 Cn\mathbb C^n. Church, Ellenburg and Farb prove in this preprint that the SnS_n-representation of Hi(PConfn(C))H^i(PConf_n(\mathbb C)) stabilizes as a function of nn, 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.

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/ba/b where aa and bb are integers. But this alone is not sufficient to talk about the roots of quadratics, even simple ones like x22x^2 - 2. If (a/b)22=0(a/b)^2 - 2 = 0, then a2=2b2a^2 = 2b^2, and since any odd number squared is odd then aa must be even, i.e. we can decompose it into a=2ua = 2u. But then 4u2=2b24u^2 = 2b^2 so 2u=b22u = b^2, hence by the same argument bb is even and we can write u2=2v2u^2 = 2v^2 where 2v=b2v = b. This is exactly what we started with, so we can keep on dividing aa and bb by 22 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 root(a,b)\mathrm{root}(a,b) to be a root of x2+ax+bx^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 x2ax^2 - a for positive integers aa to our number system. We use a\sqrt{a} to denote a solution to x2ax^2 - a, but pretty quickly we run into a problem: a-\sqrt{a} is also a solution to x2ax^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 a\sqrt{a} for a>0a \gt 0 to be the unique positive real number[0] satisfying a2=a\sqrt{a}^2=a. Can we use arithmetic plus our new \sqrt{} operator to solve arbitrary quadratics of the form x2+ax+bx^2 + ax + b?

Let uu and vv denote its roots. We can factor x2+ax+bx^2 + ax + b as (xu)(xv)(x - u)(x - v)[1], which gives us the relationships u+v=auv=b\begin{array}{rcr} u + v &= &-a\\ uv &= &b \end{array}

We would like to use these to find uu and vv individually. The equation u+v=au + v = -a is tempting, since if we can find an expression for the very similar-looking uvu - v in terms of aa and bb we can combine them to determine aa and bb. But there's a problem—since our choice of the labels uu and vv 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)f(u,v) in terms of aa and bb unless ff is symmetric—i.e. f(u,v)=f(v,u)f(u, v) = f(v, u). Unfortunately uvu - v is not symmetric, but as luck would have it (uv)2(u - v)^2 is! And expanding this, we see that (uv)2=u22uv+v2=(u+v)24uv=a24b\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 uvu - v are ±a24b\pm\sqrt{a^2 - 4b}. Thus we have u=(u+v)+(uv)2=a±a24b2\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 x3+ax2+bx+cx^3 + ax^2 + bx + c be a cubic polynomial and u,vu, v and ww 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, x31x^3 - 1. It is easy to see that its only real root is 11, but this root doesn't have multiplicity 33 because (x1)3=x33x2+3x1x31(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 jj such that j3=1j^3 = 1 but j1j \ne 1. The three roots of x31x^3 - 1 are then j,j2j, j^2 (since (j2)3=(j3)2=12=1(j^2)^3=(j^3)^2=1^2=1) and 11. As it turns out we can find a number system with such a jj that still obeys the normal rules of arithmetic, namely the complex numbers, and j=1+3i2j=\frac{-1 + \sqrt3 i}{2}. Using jj and the real solutions a1/3a^{1/3} to x3ax^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, x3+ax2+bx+cx^3 + ax^2 + bx + c can be factored as (xu)(xv)(xw)(x - u)(x - v)(x - w)[3]. Multiplying this out gives us x3+ax2+bx+c=x3(u+v+w)x2+(uv+uw+vw)xuvwx^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 u+v+w=auv+uw+vw=buvw=c\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,vu, v and ww that we can combine with u+v+wu + v + w to obtain u,vu, v and ww individually. This trick comes from linear algebra, and is called finding a basis for the span of u,vu, v and ww. This requires three linear combinations of u,vu, v and ww, of which we already have one (u+v+wu + 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,vu, v, but there are six permutations of u,v,wu, 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+j2v+j3wf = ju + j^2v + j^3w. Rotating the three roots (known as a cylic permutation)Cycling gives us jw+j2u+j3v=j(ju+j2v+j3w)=jfjw + j^2u + j^3v = j(ju + j^2v + j^3w) = jf and jv+j2w+j3u=j2(ju+j2v+j3w)=j2fjv + j^2w + j^3u = j^2(ju + j^2v + j^3w) = j^2f. Note that this means f3f^3 does not change when we rotate the roots (or in math terms, is invariant under cyclic permutations), as (jf)3=j3f3=f3(jf)^3=j^3f^3=f^3 and (j2f)3=j6f3=f3(j^2f)^3=j^6f^3=f^ invariant under cyclic permutations. Similarly, for g=j2u+j4v+j6wg = j^2u + j^4v + j^6w, cyclic permutations amount to multiplying by j2j^2 or j4j^4, and so g3g^3 is invariant under cyclic permutations. Note that g=j2u+j4v+j6w=jv+j2u+j3wj2g=j4u+j6v+j8w=ju+j2w+j3vj4g=j6u+j8v+j10w=jw+j2v+j3u\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 ff with two of the roots swapped, so any of the 66 permutations of u,v,wu,v,w will either leave f3f^3 and g3g^3 invariant, or interchange f3f^3 with g3g^3.S3S_3 acts on {f3,g3}\{f^3, g^3\}.

This should remind you of the quadratic case: we need to find expressions for f3f^3 and g3g^3 in terms of symmetric polynomials of them. This amounts to finding the roots of the polynomial x2(f3+g3)x+f3g3x^2 - (f^3 + g^3)x + f^3g^3 in terms of its coefficients. For convenience, we use h=f3+g3h = f^3 + g^3 and t=fgt = fg to rewrite this as x2hx+t3x^2 - hx + t^3.

Expanding hh and collecting terms gives us: h=2(u3+v3+w3)+(3j+3j2)(u2v+u2w+uv2+v2w+uw2+vw2)+12uvw h = 2(u^3 + v^3 + w^3) + (3j + 3j^2)(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 33. We say this polynomial is homogeneous because all of its monomials have the same degree. Similarly a,ba,b and cc are homogeneous with degree 1,21,2 and 33 respectively. The product of two homogeneous polynomials is itself homogeneous, and its degree is the sum of their degrees. Thus the homogeneous products of a,b,ca,b,c of degree 33 are a3,aba^3, ab and cc. If some polynomial in a,b,ca,b,c equals gg, then its terms of degree 33 must equal hh, but these can only come from a3,aba^3, ab and cc. So we are looking for a linear combination of the degree 33 products of the coefficients: a3=(u+v+w)3=(u3+v3+w3)3(u2v+u2w+uv2+uw2+v2w+vw2)6uvwab=(u+v+w)(uv+uw+vw)=u2v+u2w+uv2+uw2+v2w+vw2+3uvwc=uvw\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 a3a^3 to cancel out the u3+v3+w3u^3 + v^3 + w^3 term in f3+g3f^3+g^3. This gives us h(2)a3=(6+3j2+3j)(u2v+u2w+uv2+uw2+v2w+vw2) h - (-2)a^3 = (-6 + 3j^2 + 3j)(u^2v + u^2w + uv^2 + uw^2 + v^2w + vw^2) and using abab to cancel out this term we get h(2)a3(6+3j+3j2)ab=(189j9j2)uvw=(18+9j+9j2)c\begin{array}{cl} h - (-2)a^3 - (-6+3j+3j^2)ab &= (18-9j-9j^2)uvw\\ &= (-18 + 9j + 9j^2)c \end{array} hence h=2a3+(63j3j2)ab+(18+9j+9j2)c h = -2a^3 + (6 - 3j - 3j^2)ab + (-18 + 9j + 9j^2)c

We also know that t3=f3g3t^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=fgt=fg is itself invariant under all permutations of u,vu, v and ww. Expanding tt and collecting terms gives t=(u2+v2+w2)+(j+j2)(uv+uw+vw) t = (u^2 + v^2 + w^2) + (j + j^2)(uv + uw + vw) and cancelling terms we get ta2=(2+j+j2)(uv+uw+vw)=(2+j+j2)b\begin{array}{cl} t - a^2 &= (-2 + j + j^2)(uv + uw + vw)\\ &= (-2 + j + j^2)b \end{array} hence t=a2+(2+j+j2)b t = a^2 + (-2 + j + j^2)b

Working backwards, we apply the quadratic formula to x2hx+t3x^2 - hx + t^3 to get f=(h+h24t32)1/3g=(hh24t32)1/3\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 ff to correspond to the positive square root and gg to the negative square root. Finally, we get the system of equations u+v+w=aju+j2v+w=fj2u+jv+w=g\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 (uvw)=(111jj21j2j1)1(afg) \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+ax3+bx2+cx+dx + 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 S4S_4 which has quotient S3S_3. Luckily the Klein 4-group K4K_4 consisting of the permutations {(),(12)(34),(13)(24),(14)(23)}\{(), (12)(34), (13)(24), (14)(23)\} is just such a group. A natural such basis is a=(u+v)+(w+z)p=(u+v)(w+z)q=(uv)+(wz)r=(uv)(wz)\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 p2,q2p^2, q^2 and r2r^2 are all invariant under switching any pair of roots at the same time as any others, e.g. u,v,w,zv,u,z,wu,v,w,z \to v,u,z,w K4K_4. Thus we can reduce the quartic case to finding the roots of x3(p2+q2+r2)x2+(p2q2+p2r2+q2r2)xp2q2r2 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 p2,q2,r2p^2,q^2,r^2 in terms of the coefficients a,b,c,da, 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 S5S_5 from which we can generate such a basis. In fact the only nontrivial normal subgroup of S5S_5 is the alternating group A5A_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 A5A_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 xnax^n - a, thus in general the roots of quintic polynomials cannot be constructed this way.

  1. ^

    Since f(x)=x2af(x) = x^2 - a is a continuous function, and for a>0a \gt 0 we have f(0)<0f(0) \lt 0 and f(a+1)=(a+1)2a=a2+a+1>0f(a + 1) = (a + 1)^2 - a = a^2 + a + 1 \gt 0, by the Intermediate Value Theorem it has some real root between 00 and a+1a + 1.

  2. ^

    To prove this, first note that u+v=au + v = -a is equivalent to v=u+a-v = u + a, so we are trying to show that (xu)(x+(u+a))=x2+ax+b(x - u)(x + (u + a))=x^2+ax+b. To see that this holds, note that (xu)(x+(u+a))=x2+ax(u2+au)(x - u)(x + (u + a)) = x^2 + ax - (u^2 + au) and (u2+au)=b(u2+au+b)=b-(u^2 + au) = b - (u^2 + au + b) = b as u2+au+b=0u^2 + au + b = 0 by definition of uu as a root.

  3. ^

    The quadratic formula as usually taught is for quadratics in the form ax2+bx+cax^2 + bx + c. Dividing this by aa and re-labelling the coefficients gives our form.

  4. ^

    In general, any polynomial with leading coefficient 11 can be factored into (xr1)(xrn)(x - r_1)\cdots(x - r_n) where the rir_i are its roots; this is known as the Factor Theorem.

  5. ^

    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.