The Birch and Swinnerton-Dyer Conjecture:

An Unsolved Problems with Roots in Ancient Times

Picture I took of Birch and Swinnerton-Dyer in 2000 in Utrecht, Holland

Review/Motivation: Nonsingular Plane Curves

A nonsingular plane algebraic curve is the set of solutions to a (nonsingular) polynomial:

$$F(X,Y) = 0$$

A rational point is $(x,y) \in \mathbf{Q}\times\mathbf{Q}$ such that $F(x,y) = 0$.

• Theorem (old): A curve of degree $\leq 2$ has no rational points ($x^2 + y^2 = -1$) or infinitely many rational points ($x^2 + y^2 = 1$), and there is a way to decide which and enumerate all solutions.


• Faltings Theorem (1985): A curve of degree $\geq 4$ has finitely many rational points.


• Birch and Swinnerton-Dyer Conjecture (1960s): A curve of degree 3 has either finitely many rational points ($x^3 + y^3 = 1$) or infinitely many rational points $y^2 + y = x^3 - x$. The BSD Conjecture provides a way to decide which and enumerate all solutions.

Enumerating Pythagorean Triples

Ancient Problem: Find all solutions $x,y\in\mathbf{Q}$ to the equation $x^2 + y^2 = 1$. Equivalently, by clearing denominators, find all Pythagorean triples $(a,b,c)$ such that $a^2+b^2 = c^2$.   This problem goes back thousands of years!

Enumerating Pythagorean Triples

There is a nice construction that completely solves this problem.  Just draw a line of rational slope through (-1,0) and find the unique other point of intersection with the circle.  It will have to be rational, as you can verify with some algebra.  Moreover, this gives every point!  If $(x,y)$ is any rational solution, then the line through $(-1,0)$ and $(x,y)$ has rational slope $\frac{y}{x+1}$, so we would find it via the above process.

Enumerating Pythagorean Triples: Live Demo

Point (x,y) = $%s$'%latex((x,y))) html('Pythagorean (a,b,c) = $%s$

The Congruent Number Problem

Definition: An integer $n$ is a congruent number if $n$ is the area of a right triangle with rational side lengths.

Major Unsolved Problem in Mathematics: Give an algorithm to decide whether or not an integer $n$ is a congruent number.

This is a 1000-year old open problem; it may be the oldest open problem in mathematics.

Congruent Numbers and the BSD Conjecture

Theorem: A proof of the Birch and Swinnerton-Dyer Conjecture would also solve the congruent number problem.

Proof: Suppose $n$ is a positive integer.  Consider the cubic curve $y^2 = x^3 - n^2 x$.  Using algebra, one sees that this cubic curve has infinitely many rational points if and only if there are rationals $a,b,c$ such that $n=ab/2$ and $a^2 + b^2 = c^2$.  The Birch and Swinnerton-Dyer conjecture gives an algorithm to decide whether or not any cubic curve has infinitely many solutions.

The $L$-function

Let $E$ be an elliptic curve. For each prime number $p$, let $N_p=\#E(\mathbf{F}_p)$ be the number of solutions modulo $p$.

Definition: Let $a_p = p+1-N_p$.

Theorem (Hasse): $|a_p| < 2\sqrt{p}$.

Let $\Delta = -16(4a^3+27b^2)$. We have the following very deep theorem:

Theorem (Breuil, Conrad, Diamond, Taylor, Wiles -- 1999): The function $$L^*(E,s) = \prod_{{\rm primes\, } p\nmid \Delta}\left(\frac{1}{1-a_p p^{-s} + p^{1-2s}}\right)$ extends (uniquely) to an entire complex-analytic function on the complex plane $\mathbf{C}$.

NOTE: In practice, one usually works with a slightly more complicated function $L(E,s)$.  When $\Delta$ is "minimal", this function is $$L(E,s) = L^*(E,s) \cdot \prod_{{\rm primes\, } p\mid \Delta}\left(\frac{1}{1-a_p p^{-s}}\right),$$ where $a_p=p+1-\#E_{ns}(\mathbf{F}_p)$ and $\#E_{ns}(\mathbf{F}_p)$ is the subgroup of nonsingular points.

The Birch and Swinnerton-Dyer Conjecture

Heuristic Observation: If $E$ has infinitely many rational points, then the numbers $N_p$ will tend to be "large", since you can reduce all those points modulo $p$.  Since $$L(E,1) "=" \prod_p \frac{p}{N_p} \qquad {\rm (not\, really\, true)},$$ the number $L(E,1)$ will tend to be small.

Theorem (Mordell): There is a finite set $P_1, \ldots, P_r$ of rational points on $E$ so that all rational points (modulo points of finite order) can be generated from these using the group law.

We call the smallest $r$ in Mordell's theorem the rank of $E$.

Conjecture (Birch and Swinnerton-Dyer): $$\text{ord}_{s=1} L(E,s) = \text{rank}(E)$$

This problem, exactly as stated above, is the Clay Math Institute Million Dollar prize problem in algebraic number theory.  Its solution would also resolve the 1000-year old congruent number problem.

Example

The Kolyvagin -- Gross-Zagier Theorem

Theorem: If $\text{ord}_{s=1} L(E,s) \leq 1$ then the Birch and Swinnerton-Dyer conjecture is true for $E$.

The proof involves Heegner points, modular curves, Euler systems and Galois cohomology.

A Wide Open Problem

Open Problem: Show that the Birch and Swinnerton-Dyer conjecture holds for any specific curve with $r=4$ (or higher).  For example, for the curve $$y^2 + xy = x^3 - x^2 - 79x + 289.$$

The following command does in fact verify with certainty that the rank is $4$:

The $L$-series looks like it vanishes to order $4$, but we can't be sure since we get only 0.000000....  It certainly doesn't vanish to order greater than $4$.

Some of My Current Research

• Study the mathematical structures (Heegner points, modular curves, Euler systems, etc.) that appear in the proof of the Kolyvagin-Gross-Zagier theorem in order to understand how to generalize anything to elliptic curves with $\text{ord}_{s=1} L(E,s) \geq 2$. 
  
  
  
• This involves a combination of technical theory and explicit machine computation.