The Picard scheme of an abelian variety

Let ${k}$ be an algebraically closed field, and ${X}$ a projective variety over ${k}$ . In the previous two posts, we’ve defined the Picard scheme ${\mathrm{Pic}_X}$ , stated (without proof) the theorem of Grothendieck giving conditions under which it exists, and discussed the infinitesimal structure of ${\mathrm{Pic}_X}$ (or equivalently of the connected component ${\mathrm{Pic}^0_X}$ at the origin).

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

$\displaystyle T \mathrm{Pic}^0_X = H^1(X, \mathcal{O}_X),$

by studying deformations of a line bundle over the dual numbers. In particular, in characteristic zero, a simply connected smooth variety has trivial ${\mathrm{Pic}^0_X}$ . To get interesting ${\mathrm{Pic}_X^0}$ ‘s, we should be looking for non-simply connected varieties: abelian varieties are a natural example.

Let ${X}$ be an abelian variety over ${k}$ . The goal in this post is to describe ${\mathrm{Pic}^0_X}$ , 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, ${\mathrm{Pic}^0}$ 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 ${\mathrm{Pic}^0}$ : they are the translation-invariant ones.

Theorem 7 Given an abelian variety ${X}$ , the line bundles on ${X}$ algebraically equivalent to zero are precisely the translation-invariant ones: that is, the line bundles ${\mathcal{L}}$ such that ${t_x^* \mathcal{L} \simeq \mathcal{L}}$ for ${x \in X}$ where ${t_x: X \rightarrow X}$ is translation by ${x}$ .

Before proving the theorem, let’s try to interpret it. Suppose given a translation-invariant line bundle ${\mathcal{L}}$ . Then if ${m: X \times X \rightarrow X}$ is the multiplication map and ${p_1, p_2: X \times X \rightarrow X}$ are the projections, then the line bundle

$\displaystyle m^* \mathcal{L} \otimes p_1^* \mathcal{L}^{-1} \otimes p_2^* \mathcal{L}^{-1}$

on ${X \times X}$ is trivial. Indeed, since it is trivialized on ${\left\{0\right\} \times X}$ , it is classified by a map ${X \rightarrow \mathrm{Pic}^0_X}$ 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 ${X \times X}$ is trivial, and we can even choose a canonical trivialization once we’ve chosen an isomorphism ${\mathcal{L}(0) \simeq k}$ . (Here ${\mathcal{L}(x)}$ denotes the fiber above ${x}$ .)

In other words, we conclude that the one-dimensional ${k}$ -vector spaces ${\mathcal{L}(x), x \in X}$ are equipped with isomorphisms

$\displaystyle \mathcal{L}(0) \simeq k, \quad \mathcal{L}({x+y}) \stackrel{f_{x,y}}{\simeq} \mathcal{L}(x) \otimes \mathcal{L}(y), \quad x,y \in X.$

Moreover, these isomorphisms satisfy the natural coherences. For example, given ${x \in X}$ , 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 ${X \times X \times X}$ which agree at the origin.

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

$\displaystyle X(k) \rightarrow \mathrm{Pic}(k)$

where ${\mathrm{Pic}(k)}$ is the symmetric monoidal category of one-dimensional ${k}$ -vector spaces, and ${X(k)}$ is regarded as a discrete category. (We could replace ${k}$ by an arbitrary ${k}$ -scheme.)

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

Theorem 8 If ${\mathcal{L}}$ is translation-invariant on ${X}$ and nontrivial, then ${H^*(X, \mathcal{L}) \equiv 0}$ .

Proof: In fact, we know that if ${m: X \times X \rightarrow X}$ is the multiplication map, then

$\displaystyle m^* \mathcal{L} \simeq \mathcal{L} \boxtimes \mathcal{L} \equiv p_1^* \mathcal{L } \otimes p_2^* \mathcal{L} \quad \in \mathrm{Pic}(X \times X),$

so that we get a natural map of graded vector spaces

$\displaystyle H^*(X, \mathcal{L} ) \rightarrow H^*(X \times X, m^* \mathcal{L}) \simeq H^*(X, \mathcal{L}) \otimes H^*(X, \mathcal{L})$

by the Künneth formula. Since ${m}$ has a section, this map is necessarily injective.

But now we have a problem. Unless ${H^0(X, \mathcal{L}) \neq 0}$ , there is no way for this to happen. Namely, if ${H^*(X, \mathcal{L})}$ has its first nonvanishing entry at ${n > 0}$ , then the above injection implies that its first nonvanishing entry occurs at least at grading ${2n}$ . The conclusion is that ${H^0(X, \mathcal{L}) \neq 0}$ .

Now however ${m^* \mathcal{L} \simeq \mathcal{L} \boxtimes \mathcal{L}}$ corresponds to ${\mathcal{L} \boxtimes \mathcal{O}}$ under the “shearing” automorphism ${(x,y) \mapsto (x+y, y)}$ of ${X \times X}$ , and so we actually have two isomorphisms:

$\displaystyle H^0(X \times X, \mathcal{L} \boxtimes \mathcal{L}) \simeq H^0(X, \mathcal{L})^{\otimes 2}, \quad H^0(X \times X, \mathcal{L} \boxtimes \mathcal{L}) \simeq H^0(X, \mathcal{L}),$

which together imply that ${H^0(X, \mathcal{L})}$ must be one-dimensional. Since ${\mathcal{L}}$ is translation-invariant, this implies it must be trivial. $\Box$

8. Proof of the main result: first step

Let’s start by showing that a line bundle ${\mathcal{L} \in \mathrm{Pic}(X)}$ algebraically equivalent to zero is translation-invariant. In fact, to say that ${\mathcal{L}}$ is algebraically equivalent to zero is to say that there is a connected ${k}$ -variety ${T}$ and a line bundle ${\mathcal{M} \in \mathrm{Pic}(X \times_k T)}$ restricting to ${\mathcal{L}}$ at ${t_0 \in T}$ and restricting to the trivial bundle at ${t_1 \in T}$ . Without loss of generality, we can assume ${\mathcal{M}}$ is trivialized along ${\left\{0\right\} \times T}$ .

We have to show that ${\mathcal{L}}$ is translation-invariant, e.g. that ${m^* \mathcal{L} \simeq \mathcal{L} \boxtimes \mathcal{L}}$ on ${X \times X}$ , with notation as above. To do this, we can form this construction in a family: that is, we can form the line bundle

$\displaystyle \mathcal{N} = m^* \mathcal{M} \otimes p_{13} \mathcal{M}^{-1} \otimes p_{23}^* \mathcal{M}^{-1} \in \mathrm{Pic}(X \times_k X \times_k T).$

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 ${\left\{0\right\} \times_k X \times_k T}$ , on ${X \times_k \left\{0\right\} \times_k T}$ , and on ${X \times_k X \times_k \left\{t_1\right\}}$ . 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 ${\mathcal{N}}$ as defining a map

$\displaystyle X \times X \rightarrow (\mathrm{Pic}^0_{T})_{\mathrm{red}}$

and arguing that it had to be the trivial map. (Without loss of generality, ${T}$ is projective and integral.) In fact, the map ${X \times X \rightarrow (\mathrm{Pic}^0_T)_{\mathrm{red}}}$ is a pointed map, and hence, by a rigidity lemma, a homomorphism of abelian varieties. Since it annihilates ${X \times \left\{0\right\}}$ and ${\left\{0\right\} \times X}$ , it is the zero map.

9. The converse direction

Conversely, we would like to show that every line bundle on ${X}$ which is translation-invariant is also algebraically equivalent to zero. The argument will also give us smoothness of ${\mathrm{Pic}^0_X}$ as a by-product. The strategy is to choose an ample line bundle ${\mathcal{L}}$ on ${X}$ and consider the map

$\displaystyle \phi_{\mathcal{L}}: X \rightarrow \mathrm{Pic}^0_X, \quad x \mapsto t_x^* \mathcal{L} \otimes \mathcal{L}^{-1},$

where ${t_x}$ is translation by ${x}$ . More precisely, this comes from the family ${m^* \mathcal{L} \otimes (\mathcal{L}^{-1} \boxtimes \mathcal{L}^{-1})}$ on ${X \times_k X}$ . This map is a pointed map, hence a homomorphism: this is the theorem of the cube again. Observe that since ${X}$ is connected, it takes values in ${\mathrm{Pic}^0_X}$ .

Our main goal is:

Theorem 9 The scheme ${\mathrm{Pic}^0_X}$ is smooth and therefore an abelian variety. For an ample line bundle ${\mathcal{L}}$ , ${\phi_{\mathcal{L}}}$ is an isogeny, and every translation-invariant line bundle is in the image.

Let’s start by showing that ${\phi_{\mathcal{L}} }$ has finite kernel. In other words, we have to show that if ${B \subset A}$ is an abelian subvariety, ${\phi_{\mathcal{L}}}$ does not vanish identically on ${B}$ . But then we would have a translation-invariant ample line bundle on ${B}$ , which is absurd: replacing ${\mathcal{L}|_{B}}$ by a sufficiently high tensor power, we always have lots of global sections and in particular a nonvanishing ${H^*}$ .

Next, we need to show that if ${\mathcal{M}}$ is any translation-invariant line bundle on ${X}$ , then there exists ${x \in X}$ such that

$\displaystyle \mathcal{M} \simeq t_x^* \mathcal{L} \otimes \mathcal{L}^{-1}.$

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

$\displaystyle x \mapsto t_x^* \mathcal{L} \otimes \mathcal{L}^{-1} \otimes \mathcal{M}^{-1},$

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

$\displaystyle \mathcal{Q} = m^* \mathcal{L} \otimes (\mathcal{L}^{-1} \boxtimes \mathcal{L}^{-1}) \otimes p_2^* \mathcal{M}^{-1} \in \mathrm{Pic}(X \times X).$

We will consider the cohomology of ${\mathcal{Q}}$ along the vertical and on the horizontal fibers:

On the fibers ${\left\{x\right\} \times X}$ , the bundle ${\mathcal{Q}}$ restricts to ${t_x^* \mathcal{L} \otimes \mathcal{L}^{-1} \otimes \mathcal{M}^{-1}}$ and is by assumption never trivial: in particular the cohomology vanishes identically.
On the fibers ${X \times \left\{x\right\}}$ , the bundle restricts to ${t_x^* \mathcal{L} \otimes \mathcal{L}^{-1}}$ and has nonvanishing cohomology for a finite number of ${x \in X}$ (i.e., those where ${t_x^* \mathcal{L} \simeq \mathcal{L}}$ ).

On the one hand, the first bullet point implies that ${R p_{1*} \mathcal{Q}}$ is identically zero, by the theory of cohomology and base change. By the Leray spectral sequence, we conclude that ${\mathcal{Q}}$ has no cohomology at all.

On the other hand, the second bullet point (together with cohomology and base change) implies that ${R p_{2*} \mathcal{Q}}$ 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 ${H^*(\mathcal{Q}) \neq 0}$ . This is a contradiction, and we conclude that

$\displaystyle \phi_L : X \rightarrow \mathrm{Pic}^0_X$

is a surjection, and that every translation-invariant line bundle is in ${\mathrm{Pic}^0_X}$ .

In other words, we have a surjective map ${\phi_{\mathcal{L}}}$ with finite kernel onto ${\mathrm{Pic}^0_X}$ , and the consequently ${\mathrm{Pic}^0_X}$ has the same dimension as ${X}$ . The last thing to check is smoothness. This will follow if we show that

$\displaystyle \dim T_0 \mathrm{Pic}^0_X = \dim H^1(X, \mathcal{O}_X) \leq \dim X$

(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

$\displaystyle H^*(X, \mathcal{O}_X)$

is a Hopf algebra that comes from dualizing the multiplication law on ${X}$ . It is in fact a graded Hopf algebra with ${k}$ 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 $\dim X$ and the next result:

Lemma 10 Let ${A_\bullet}$ be a graded, connected commutative Hopf algebra with ${A_i = 0}$ for ${i > g}$ . Then ${\dim A_1 \leq g}$ .

Proof: In fact, if ${x_1, \dots, x_{g+1} \in A_1}$ are linearly independent, then I claim that

$\displaystyle x_1 \dots x_{g+1} \neq 0.$

In fact, suppose that this vanished. A key observation is that the ${x_i}$ are primitive simply for dimensional reasons: the comultiplication carries them to ${1 \otimes x_i + x_i \otimes 1}$ . Therefore, each ${x_i}$ induces a map from the exterior Hopf algebra ${\bigwedge (x_i) \rightarrow A_\bullet}$ . Tensoring them together, we get a map

$\displaystyle \psi: \bigotimes_{i=1}^{g+1} \bigwedge (x_i) \rightarrow A_\bullet.$

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 ${\psi}$ were not injective, then ${\psi}$ 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 ${x_i}$ : 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 ${x_i}$ are linearly independent in ${A_\bullet}$ , we conclude that the map is injective. $\Box$

Climbing Mount Bourbaki