## Jacobians

I’d like to begin by talking about the Jacobian of a curve.

Let $C$ be a smooth algebraic curve. That is, it is (most of the time) the zero set of a two-variable polynomial such as

$y^2 = x^6 - 1$

Topologically, such objects are uniquely classified by their genus. This is a non-negative integer, and is simply the count of the number of handles that you have to add to a sphere to obtain your curve. Topologically, curves aren’t particularly interesting as individuals.

Algebraically, however, curves have much more structure. To begin with (and in part, to motivate what will come after), let’s begin by looking at the case of genus 1 curves, or Elliptic curves.

The simplest case of a genus 1 curves is given by looking at the topological quotient $E = \mathbb{C}/\langle 1, i\rangle$. That is, we consider the additive subgroup generated by 1 and by $i$, and we quotient out by this. We end up with a compact real two-dimensional surface which is of genus 1. Moreover, such a surface obtains a complex structure which it inherits from the quotient map $\mathbb{C} \to E$. It isn’t immediately obvious, but this can in fact be written as the zero set of the equation

$y^2 = 4x^3 - x$

More generall, the universal cover of a genus 1 curve is the complex line*, $\mathbb{C}$. The group of deck transformations is some lattice $\Lambda = \langle z_1, z_2\rangle \subset \mathbb{C}$ with $z_1, z_2$ not linearly dependent over the real numbers. On any such curve we obtain a complex structure from the quotient map, and so the natural question is raised: When are two such complex structures equivalent?

In order to answer this question, let $C_1, C_2$ be a pair of elliptic curves with associated lattices $\Lambda_1, \Lambda_2$, and suppose that $f : C_1 \to C_2$ is a map between them. This map naturally lifts to a map $F$ between their universal covers which necessarily satisfies

$F(x + \lambda_1) - F(x) = \lambda_2$

for some $\lambda_1 \in \Lambda_1$ and $\lambda_2 \in \Lambda_2$. If we differentiate this equation with respect to $x$, then we see that the derivative of the function $F(x + \lambda) - F(x)$ is zero for all $\lambda$, and so this function must be constant. It follows then that $F(x) = Cx + \lambda$ for some $C \in \mathbb{C}$, and some $\lambda \in \Lambda_2$.

That is, the only maps between elliptic curves are linear. It will follow that we are able to simplify this picture immensely.

*I say complex line, because it is a one-dimensional complex manifold.

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## About charlesflorian

Mathematician. Climber. Living in Copenhagen.
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