Lecture 1: Monday, August 19, 2013

For the first lecture for the course, I would like to focus on a simple question: Why would anyone care about Dessins d’Enfants anyway? I won’t quite get to answer this question, but I will at least motivate the study of Belyĭ maps.

From Pythagorean Triples to Beal’s Conjecture

Let me begin with a simple thought. Are there integers $a$, $b$, and $c$ such that $a^2 + b^2 = c^2$? There are two ways we can answer this question. First, we can simply say “yes” because we can exhibit some Pythagorean triples right away: $(a,b,c) = (3,4,5)$ and $(a,b,c) = (8, 15, 17)$ are easy examples. We can be a bit more sophisticated, though. There are actually infinitely many such integers because for any integer $z$ we can choose $a = z^2 - 1$, $b = 2 \, z$, and $c = z^2 + 1$.

There is no reason to stop with the Diophantine equation $a^2 + b^2 = c^2$. Indeed, fix positive integers $m$, $n$, and $k$. Are there integers $a$, $b$, and $c$ such that $a^m + b^n = c^k$? If $m = n = k \geq 3$, then this is a question Pierre de Fermat considered in 1637. Sir Andrew Wiles showed in 1994 that there are only trivial solutions:

Fermat’s Last Theorem.
Fix an integer $m = n = k \geq 3$. The only integers $a$, $b$, and $c$ such that $a^n + b^n = c^n$ must satisfy $a \, b \, c = 0$.

There is even a generalization of this! In 1993, businessman Andrew Beal, and independently in 1994, mathematicians Robert Tijdeman and Don Zagier formulated the following.

Beal’s Conjecture / Tijdeman-Zagier Conjecture.
Fix integers $m, \, n, \, k \geq 3$. The only integers $a$, $b$, and $c$ such that $a^m + b^n = c^k$ either satisfy $a \, b \, c = 0$ or $gcd(a, b, c) \geq 2$.

Initially, Beal offered $5,000 to anyone who could prove this conjecture, but this prize was recently increased to$1,000,000. Just last week, actress Danica McKellar and mathematician Jordan Ellenberg appeared on the Today Show to discuss this conjecture!

We have some remaining cases though: what if at least one of $m$, $n$, and $k$ is either 1 or 2? The Conjecture is definitely false. For example, $10^2 + (-7)^3 = (-3)^5$ for $(m,n,k) = (2,3,5)$ while $13^2 + 7^3 = 2^9$ for $(m,n,k) = (2,3,9)$. Since $3^2 + (-2)^3 = 1^k$ for $(m,n,k) = (2,3,k)$, we see that the Conjecture is false for infinitely many exponents with at least one of $m$, $n$, or $k$ is either 1 or 2. In fact, for certain fixed exponents $(m,n,k)$, there are infinitely many nontrivial counterexamples!

• Consider $(m,n,k) = (1,n,k)$. For any integer $z$, denote the relatively prime integers $a = (z+1)^k - z^n$, $b = z$, and $c = z+1$; then $a^1 + b^n = c^k$ and $gcd(a,b,c) = 1$. We can think of $a$, $b$, and $c$ as being “relatively prime” if we view these as elements in $\mathbb C[z]$ rather than $\mathbb Z$.

• Consider $(m,n,k) = (2,n,2)$. Similar to above, denote the polynomials $a = 2^{n-1} \, (z^n - 1)$, $b = 4 \, z$, and $c = 2^{n-1} \, (z^n + 1)$; then $a^2 + b^n = c^2$ and $gcd(a,b,c) = 1$ as elements of $\mathbb C[z]$.

• Consider $(m,n,k) = (2,3,3)$. Similar to above, denote $a = z^6 - 20 \, z^3 - 8$, $b = 4 \, (z^3 - 1)$, and $c = z \, (z^3 + 8)$; then $a^2 + b^3 = c^3$ and $gcd(a,b,c) = 1$.

• Consider $(m,n,k) = (2,3,4)$. Similar to above, denote $a = 648 \, \sqrt{3} \, (z^{12} - 33 \, z^8 - 33 \, z^4 + 1)$, $b = -108 \, (z^8 + 14 \, z^4 + 1)$, and $c = -108 \, z \, (z^4 - 1)$; then $a^2 + b^3 = c^4$ and $gcd(a,b,c) = 1$.

• Consider $(m,n,k) = (2,3,5)$. Similar to above, denote $a = 12^6 \, \bigl[ (z^{30} + 1) - 522 \, (z^{25} - z^5) - 10005 \, (z^{20} + z^{10}) \bigr]$, $b = -12^4 \, \bigl[ (z^{20} + 1) + 228 \, (z^{15} - z^5) + 494 \, z^{10} \bigr]$, and $c = -12^3 \, z \, (z^{10} - 11 \, z^5 - 1)$; then $a^2 + b^3 = c^5$ and $gcd(a,b,c) = 1$.

(It’s no coincidence that these $m$, $n$, and $k$ are precisely those positive integers such that $1/m + 1/n + 1/k > 1$.) The question you should be asking right now is: Where are these polynomials coming from? Felix Klein worked them out over 100 years ago! What we’re going to see later is $\beta(z) = b(z)^n / c(z)^k$ is actually a Belyĭ map.

Transcendental Galois Extensions

Let’s return to the integral domain $\mathbb C[z]$. Its quotient field $\mathbb C(z)$ consists of ratios $\beta(z) = p(z) / q(z)$ of polynomials $p(z), \, q(z) \in \mathbb C[z]$. Then $\mathbb C(\beta)$ is a subfield of $\mathbb C(z)$, so that we can think of $\mathbb C(z) / \mathbb C(\beta)$ as a finite, separable extension. We are motivated by a few questions:

• The collection of Möbius transformations $\text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr) \simeq PSL_2(\mathbb C)$ acts on the field $\mathbb C(z)$ via $z \mapsto (a \, z + b)/(c \, z + d)$. What is the collection $G \subseteq PSL_2(\mathbb C)$ of those Möbius transformations which fix $\beta(z)$?

• We can define the automorphisms $\text{Aut} \bigl( \mathbb C(z) / \mathbb C(\beta) \bigr)$. How does this group differ from $G$?

• What are the possibilities for $G$? Can $G$ be nonabelian? Must $G$ be solvable?

Throughout the course of the semester, we’ll consider the first two questions, but Felix Klein answered the third question.

Theorem.
The only finite subgroups $G$ of $PSL_2(\mathbb C)$ are the cyclic group $Z_n$, the dihedral group $D_n$, the tetrahedral group $A_4$, the octahedral group $S_4$, and the icosahedral group $A_5$.

Of course, there is no reason to just focus on the automorphisms of the Riemann sphere $X = \mathbb P^1(\mathbb C)$ or its global sections $\mathcal O_X = \mathbb C[z]$. We will attempt to generalize all of these to any Riemann surface $X$. To gain intuition, we’ll start by only considering those finite separable extensions $\mathbb C(z) / \mathbb C(\beta)$ which are unramified outside of $\{ 0, 1, \infty \}$. The primitive element $\beta(z)$ is what we’ll define as a Belyĭ map. Once we write down some examples, we can attempt to compute the groups $G \subseteq PSL_2(\mathbb C)$ associated with them. For example, $\beta(z) = \dfrac {\bigl[ (z^{20} + 1) + 228 \, (z^{15} - z^5) + 494 \, z^{10} \bigr]^3}{1728 \, z^5 \, \bigl[z^{10} - 11 \, z^5 - 1 \bigr]^5}$ has $G \simeq A_5$.

Modular Functions and Riemann Surfaces

How did all of this come about anyway? To answer this, we’ll need to go back to original ideas of Felix Klein, Bernhard Riemann, and Karl Weierstrass. There were interested in complex functions with many symmetries. To be precise, they focused on modular functions: meromorphic functions $f: \mathbb H^2 \cup \mathbb P^1(\mathbb Q) \to \mathbb P^1(\mathbb C)$ from the extended upper-half plane to the Riemann sphere which satisfy $f(\tau + 1) = f(-1/\tau) = f(\tau)$, that is, $f \left( \dfrac {a \, \tau + b}{c \, \tau + d} \right) = f(\tau)$ for all Möbius transformations coming from $SL_2(\mathbb Z)$.

Theorem.
There exists a nontrivial modular function for $SL_2(\mathbb Z)$, namely $J(\tau) = \dfrac {\displaystyle \left[ 1 + 240 \sum_{n=1}^{\infty} \sigma_3(n) \, q_1^n \right]^3}{\displaystyle 1728 \, q_1 \prod_{n=1}^{\infty} \bigl( 1 - q_1^n \bigr)^{24}}$ satisfies $J \left( \dfrac {a \, \tau + b}{c \, \tau + d} \right) = J(\tau)$ for all Möbius transformations coming from $SL_2(\mathbb Z)$, where $q_h = e^{2 \pi i \tau/h}$ and $\sigma_h(n) = \sum_{d | n} d^h$ is the divisor function. In fact, denoting the compact, connected Riemann surface $X(1) = SL_2(\mathbb Z) \backslash \bigl( \mathbb H^2 \cup \mathbb P^1(\mathbb Q) \bigr)$ as the quotient via Möbius transformations, $J: X(1) \to \mathbb P^1(\mathbb C)$ is an isomorphism.

This function is known as Klein’s $J$-Invariant. The condition “$f(\tau + 1) = f(-1/\tau) = f(\tau)$” is a bit strict, so we can loosen it a bit. For each positive integer $N$, consider the congruence subgroup $\Gamma(N) = \left \{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{Mat}_{2 \times 2}(\mathbb Z) \ \biggl| \ a \, d - b \, c = 1 \ \text{and} \ c \equiv 0 \ \pmod{N} \right \}$. Then $\Gamma(1) = SL_2(\mathbb Z)$ is the special linear group group, and $\Gamma(N) \subseteq \Gamma(1)$ is a normal subgroup. We will denote the modular curve $X(N) = \Gamma(N) \backslash \bigl( \mathbb H^2 \cup \mathbb P^1(\mathbb Q) \bigr)$ as the quotient via Möbius transformations; it is a compact, connected Riemann surface.

Theorem.
There exists a nontrivial modular function for $\Gamma(2)$, namely $\lambda(\tau) = \displaystyle 16 \, q_2 \, \prod_{n=1}^{\infty} \dfrac {\bigl( 1 + q_2^{2n} \bigr)^{16}}{\bigl( 1 + q_2^{n} \bigr)^{8}} = 16 \, \biggl[ q_2 - 8 \, q_2^2 + 44 \, q_2^3 + \cdots \biggr]$ satisfies $\lambda \left( \dfrac {a \, \tau + b}{c \, \tau + d} \right) = \lambda(\tau)$ for all Möbius transformations in $\Gamma(2)$. In fact, $\lambda: X(2) \to \mathbb P^1(\mathbb C)$ is an isomorphism such that $J(\tau) = \dfrac {4}{27} \, \dfrac {\bigl[ \lambda(\tau)^2 - \lambda(\tau) + 1 \bigr]^3}{\lambda(\tau)^2 \, \bigl[ \lambda(\tau) - 1 \bigr]^2}$.

This is known as the Modular $\lambda$ Function. This result states that $\mathbb C(J)$ is a subfield of $\mathbb C(\lambda)$, so that we can think of $\mathbb C(\lambda) / \mathbb C(J)$ as a finite, separable extension. In fact, $G = \text{Gal} \bigl( \mathbb C(\lambda) / \mathbb C(J) \bigr) \simeq \Gamma(1) / \Gamma(2) \simeq D_3$ is the dihedral group of order 6. This means we can generate Galois extensions $\mathbb C(z) / \mathbb C(\beta)$ by choosing $J = \beta(z)$ to be some rational function of an isomorphism $z: X(N) \to \mathbb P^1(\mathbb C)$! We will return to this idea throughout the course.

Belyĭ’s Theorem

We have seen several examples of compact, connected Riemann surfaces, namely $X(N)$ for small values of $N$. Here is the main result for the semester:

Theorem.
Let $X$ be a compact, connected Riemann surface.

• $X$ is a non-singular, irreducible, projective variety of dimension $d = 1$. In particular, $X$ is an algebraic variety, that is, can be defined by a polynomial equation $\sum_{i,j} a_{ij} \, z^i \, w^j = 0$ with $a_{ij} \in \mathbb C$.

• If $X$ can be defined by a polynomial equation $\sum_{i,j} a_{ij} \, z^i \, w^j = 0$ where the coefficients $a_{ij} \in \overline {\mathbb Q}$, then there exists a rational function $\beta: X \to \mathbb P^1(\mathbb C)$ which has at most three critical values.

• Conversely, if there exists rational function $\beta: X \to \mathbb P^1(\mathbb C)$ which has at most three critical values, then $X$ can be defined by a polynomial equation $\sum_{i,j} a_{ij} \, z^i \, w^j = 0$ where $a_{ij} \in \overline {\mathbb Q}$.

The first statement is standard from Complex Analysis. We will sketch its proof using the Riemann-Roch Theorem. The third statement was shown by André Weil in 1956 in his paper “The Field of Definition of a Variety”. The second statement was shown by the Russian mathematician Gennadiĭ Vladimirovich Belyĭ in 1979 in his paper “Galois Extensions of a Maximal Cyclotomic Field”. We will spend time discussing his proof.

The main issue here is in actually computing examples of Belyĭ maps. We have seen that historically there are several examples, but how do we compute more? This is one of the main questions for the semester!