# Abelian Varieties: an outline

An Abelian variety is a complete group variety over an algebraically closed field; in particular it comes with a multiplication and inverse map $m,i$.  Naturally we get translation maps $t_x(y) = m(x,y)$ which is probably better denoted $y \mapsto x + y$ and a multiplication by $n$ map $[n](x) = m(x,[n-1]x)$ with $[0](x) = e$, the identity.

In what follows varieties ( = integral finite type and separated) are understood to be varieties over some algebraically closed field.

In what follows these are things I discussed in an earlier post on Abelian Varieties.  These things are most relevant things regarding the Poincare line bundle and these things are fairly easy properties that can be proved along the way.

I think one of the most important aspects of Abelian varieties is the

Poicare Line Bundle ; in order to describe I need the notion of the

Dual Abelian variety and there is a natural construction of the dual via a certain group called

Pic Zero $Pic^0$, over the complex numbers these are topologically trivial line bundles.  The definition of $Pic^0$ is easy once I have the Theorem of the square which is itself an easy consequence of the

Theorem of the Cube.  The proof of this theorem depends on an intuitive but nontrivial result that says on any variety of dimension at least 1, any two points can be connected by an irreducible curve.  I’ll say a little more about this in coming posts.

This theorem gives some easy properties:

a) $[n]^*L = L^{n^2}[L \otimes [-1]^*L]^{\frac{n^2 - n}{2}}$

b) $[n] \colon A \to A$ is surjective.

With $Pic^0$ defined the following are readily proved

1) $L \in Pic^0$ iff $\Lambda(L):= m^*L\otimes p_1^*L^{-1}\otimes p_2^*L^{-1}$

For $L \in Pic^0$

2) $(f+g)^*L \cong f^*L\otimes g^*L$

3) $[n]^*L \cong L^n$

4) $L\otimes [-1]^*L \in Pic^0$ for arbitrary $L$.

5) $L^n \cong O$ implies $L \in Pic^0$.

To make the dual Abelian Variety $\hat A$ of $A$ its necessary to show that a natural map $A \to Pic^0A$ is surjective.   This uses

i) a computation of the cohomology of $Pic^0$ line bundles

ii) Leray Spectral Sequence

iii) Cohomology with base change

I wont discuss the proofs of ii), iii) I’ll only apply them.  Nevertheless their utility here is a good demonstration of the power of concepts like derived functors and flatness.

With all of this in place I can define the Poincare bundle on $A \times \hat A$ and the functor that $\hat A$ represents.

The next couple of posts will be about adding some of the details to this outline.  But the above theory would implicitly be incomplete without the niffty fact that

The group law on an Abelian variety is always commutative.  This is proved using the

Rigidity lemma: If $g \colon X \times Y \to Z$ is map between varieties with $X$  complete and if $g(X \times y) = z \in Z$ for some $y \in Y$ then $g$ factors as $X \times Y \xrightarrow{p_2} Y \xrightarrow{f} Z$.

pf: Eventually I’ll use the result that a map from a complete variety to an affine variety $X \to U = \bar U \subset \mathbb{A}^n$ has image a point.

pf: $X \to U$ can be thought of as $x \mapsto (f_1(x), ..., f_n(x))$, i.e. has $n$ functions $X \to \mathbb{A}^1$.  Now general theory say $f_i(X)$ is a closed connected complete subvariety of $\mathbb{A}^1$; the choice are the whole space of a point, and only a point is complete. qed

Now let $U$ be an open affine nbd of $z$.  The set $C = X\times Y - g^{-1}(U)$ is closed. As $X$ is complete, the projection $p\colon X\times Y \to Y$ is closed, so $p(C)$ is closed.  As $X \times y \mapsto z$, I have $y \not \in p(C)$.  Therefor $U' = Y - p(C),\ p^{-1}U'$ are open an nonempty.

Pick and arbitrary $x_0 \in X$. Define $f \colon Y \to Z$ by $f(y) = g(x_0,y)$.  Note that for $q \in U’$, I have $g(X \times q) \subset U$ hence is a point; i.e. $g(x,q) = g(x_o,q) = f(q)$.  It follows that $g$ and $f \circ p$ agree on a dense open set hence are equal.

Note $Y$ needed to be a variety so any open set is dense.  To conclude $g = f \circ p$ I need $X\times Y$ reduced and $Z$ separated.  See ex II.4.2 for more.

Let $m$ denote the multiplication map for $A$ and $i$ the inverse. The Rigidity lemma implies that if $f \colon A \to A$ fixes the identity element then it is a homomorphism

pf: apply the lemma to $A \times A \to A$ to the map $(x,y) \mapsto m(f(m(x,y)), i(m(f(x),f(y))$.  In more pleasant notation this is $x,y \mapsto f(x+y) - f(x) - f(y)$. It collapses $e \times A$ and $A \times e$ hence is constant.

Finally $i \colon A \to A$ fixes the identity hence is homomorphism, i.e. $b^{-1}a^{-1} = (ab)^{-1} = a^{-1}b^{-1}$, so the group law is commutative.