Let be an algebraically closed field, and a projective variety over . In the previous two posts, we’ve defined the Picard scheme , stated (without proof) the theorem of Grothendieck giving conditions under which it exists, and discussed the infinitesimal structure of (or equivalently of the connected component at the origin).

We saw in particular that the tangent space to the Picard scheme could be computed via

by studying deformations of a line bundle over the dual numbers. In particular, in characteristic zero, a simply connected smooth variety has trivial . To get interesting ‘s, we should be looking for non-simply connected varieties: abelian varieties are a natural example.

Let be an abelian variety over . The goal in this post is to describe , which we’ll call the **dual abelian variety** (we’ll see that it is in fact smooth). We’ll in particular identify the line bundles that it parametrizes. Most of this material is from David Mumford’s *Abelian varieties *and Alexander Polischuk’s *Abelian varieties, theta functions, and the Fourier transform. *I also learned some of it from a class that Xinwen Zhu taught last spring; my (fairly incomplete) notes from that class are here.

**7. Translation-invariant line bundles**

In general, by definition, of a variety classifies line bundles that are algebraically equivalent to zero: those which can be connected to the trivial line bundle by a connected algebraic family.

For an abelian variety, however, there is another description of line bundles in : they are the translation-invariant ones.

Theorem 7Given an abelian variety , the line bundles on algebraically equivalent to zero are precisely the translation-invariant ones: that is, the line bundles such that for where is translation by .

Before proving the theorem, let’s try to interpret it. Suppose given a translation-invariant line bundle . Then if is the multiplication map and are the projections, then the line bundle

on is trivial. Indeed, since it is trivialized on , it is classified by a map which is constant at zero, since the restriction to each fiber is trivial. (This argument is actually a general “rigidity lemma” that can be proved directly, but we’re assuming existence of the Picard scheme anyway.) This line bundle on is trivial, and we can even choose a *canonical* trivialization once we’ve chosen an isomorphism . (Here denotes the fiber above .)

In other words, we conclude that the one-dimensional -vector spaces are equipped with isomorphisms

Moreover, these isomorphisms satisfy the natural coherences. For example, given , we have a commutative diagram

The commutativity follows because the two ways of going around the diagram represent two different trivializations of a trivial line bundle on which agree at the origin.

The intuition is that a translation-invariant line bundle is therefore a **categorified character** on the abelian variety . (A general line bundle is more like a quadratic form.) More precisely, we get a symmetric monoidal functor

where is the symmetric monoidal category of one-dimensional -vector spaces, and is regarded as a discrete category. (We could replace by an arbitrary -scheme.)

The following result generalizes the classical orthogonality of characters and will be important in the general theory.

Theorem 8If is translation-invariant on and nontrivial, then .

*Proof:* In fact, we know that if is the multiplication map, then

so that we get a natural map of graded vector spaces

by the Künneth formula. Since has a section, this map is necessarily injective.

But now we have a problem. Unless , there is no way for this to happen. Namely, if has its first nonvanishing entry at , then the above injection implies that its first nonvanishing entry occurs at least at grading . The conclusion is that .

Now however corresponds to under the “shearing” automorphism of , and so we actually have two isomorphisms:

which together imply that must be one-dimensional. Since is translation-invariant, this implies it must be trivial.

**8. Proof of the main result: first step**

Let’s start by showing that a line bundle algebraically equivalent to zero is translation-invariant. In fact, to say that is algebraically equivalent to zero is to say that there is a connected -variety and a line bundle restricting to at and restricting to the trivial bundle at . Without loss of generality, we can assume is trivialized along .

We have to show that is translation-invariant, e.g. that on , with notation as above. To do this, we can form this construction in a family: that is, we can form the line bundle

If we show that this line bundle is trivial, then we will be done. This line bundle has the property that it is trivialized on , on , and on . The theorem of the cube thus implies that it is trivial everywhere.

Alternatively (and this is equivalent to the theorem of the cube), we could think of this line bundle as defining a map

and arguing that it had to be the trivial map. (Without loss of generality, is projective and integral.) In fact, the map is a pointed map, and hence, by a rigidity lemma, a homomorphism of abelian varieties. Since it annihilates and , it is the zero map.

**9. The converse direction**

Conversely, we would like to show that *every* line bundle on which is translation-invariant is also algebraically equivalent to zero. The argument will also give us smoothness of as a by-product. The strategy is to choose an ample line bundle on and consider the map

where is translation by . More precisely, this comes from the family on . This map is a pointed map, hence a homomorphism: this is the theorem of the cube again. Observe that since is connected, it takes values in .

Our main goal is:

Theorem 9The scheme is smooth and therefore an abelian variety. For an ample line bundle , is an isogeny, and every translation-invariant line bundle is in the image.

Let’s start by showing that has finite kernel. In other words, we have to show that if is an abelian subvariety, does not vanish identically on . But then we would have a translation-invariant ample line bundle on , which is absurd: replacing by a sufficiently high tensor power, we always have lots of global sections and in particular a nonvanishing .

Next, we need to show that if is any translation-invariant line bundle on , then there exists such that

Let’s suppose the contrary. In this case, the strategy is to consider the family of line bundles parametrized by given by

which by assumption is never trivial. More precisely, we consider the line bundle

We will consider the cohomology of along the vertical and on the horizontal fibers:

- On the fibers , the bundle restricts to and is by assumption never trivial: in particular the cohomology vanishes identically.
- On the fibers , the bundle restricts to and has nonvanishing cohomology for a finite number of (i.e., those where ).

On the one hand, the first bullet point implies that is identically zero, by the theory of cohomology and base change. By the Leray spectral sequence, we conclude that has no cohomology at all.

On the other hand, the second bullet point (together with cohomology and base change) implies that is supported on a nonempty *finite* set of closed points, and in particular its (derived) global sections are nontrivial. In other words, the Leray spectral sequence degenerates and . This is a contradiction, and we conclude that

is a surjection, and that every translation-invariant line bundle is in .

In other words, we have a surjective map with finite kernel onto , and the consequently has the same dimension as . The last thing to check is smoothness. This will follow if we show that

(and we’ll in fact be able to conclude equality). To see this, let’s appeal to a technique common in algebraic topology: note that the cohomology ring

is a *Hopf algebra* that comes from dualizing the multiplication law on . It is in fact a graded Hopf algebra with in dimension zero: such so-called connected Hopf algebras behave regularly and were studied in a well-known paper of Milnor and Moore. (In characteristic zero, the commutative ones are tensor products of exterior and polynomial algebras.) In particular, the desired inequality on the tangent space will follow from the fact that this Hopf algebra vanishes above and the next result:

Lemma 10Let be a graded, connected commutative Hopf algebra with for . Then .

*Proof:* In fact, if are linearly independent, then I claim that

In fact, suppose that this vanished. A key observation is that the are *primitive* simply for dimensional reasons: the comultiplication carries them to . Therefore, each induces a map from the exterior Hopf algebra . Tensoring them together, we get a map

This map is injective. In fact, if this map were not injective, it would have a kernel, and the element of smallest dimension in the kernel of a map of connected coalgebras is always primitive. So if were not injective, then would have to kill a primitive element. However, the only primitive elements in the tensor product of exterior algebras are the linear combinations of the : this follows by the dual space to the primitives consists of the *indecomposables* in the dual Hopf algebra (which is another exterior algebra). Given that the are linearly independent in , we conclude that the map is injective.

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