The Picard scheme I

Let ${X}$ be an abelian variety over an algebraically closed field ${k}$ . If ${k = \mathbb{C}}$ , then ${X}$ corresponds to a complex torus: that is, ${X}$ can be expressed complex analytically as ${V/\Lambda}$ where ${V}$ is a complex vector space of dimension ${\dim X}$ and ${\Lambda \subset V}$ is a lattice (i.e., a ${\mathbb{Z}}$ -free, discrete submodule of rank ${2g}$ ). In this case, one can form the dual abelian variety

$\displaystyle X^{\vee} = \hom(X, S^1) = \hom_{\mathrm{cont}}(V/\Lambda, \mathbb{R}/\mathbb{Z}) \simeq \hom_{\mathbb{R}}(V, \mathbb{R})/2\pi i \hom(\Lambda, \mathbb{Z}).$

At least, ${X^{\vee}}$ as defined is a complex torus, but it turns out to admit the structure of an abelian variety.

The purpose of the next few posts is to describe an algebraic version of this duality: it turns out that ${X^{\vee}}$ can be constructed as a scheme, purely algebraically. I’d like to start with a couple of posts on Picard schemes. A useful reference here is this article of Kleiman.

1. The Picard scheme analytically

Let ${X }$ be a smooth projective variety over the complex numbers ${\mathbb{C}}$ . The collection of line bundles ${\mathrm{Pic}(X)}$ is a very interesting invariant of ${X}$ . Usually, it splits into two pieces: the “topological” piece and the “analytic” piece. For instance, there is a first Chern class map

$\displaystyle c_1: \mathrm{Pic}(X) \rightarrow H^2(X; \mathbb{Z}) ,$

which picks out the topological type of a line bundle. (Topologically, line bundles on a space are classified by their first Chern class.) The admissible topological types are precisely the classes in ${H^2(X; \mathbb{Z})}$ which project to ${(1,1)}$ -classes in ${H^2(X; \mathbb{C})}$ under the Hodge decomposition.

But there is more to the classification of (algebraic or equivalently holomorphic) line bundles on ${X}$ than the first Chern class: a line bundle may be topologically trivial without being holomorphically trivial. A good way to see this is to use the exponential sequence

$\displaystyle 0 \rightarrow \mathbb{Z} \stackrel{2 \pi i }{\rightarrow}\mathcal{O}_{\mathrm{hol}} \rightarrow \mathcal{O}^*_{\mathrm{hol}} \rightarrow 0,$

where ${\mathcal{O}_{\mathrm{hol}}}$ is the sheaf of analytic functions on the complex points of ${X}$ . The exponential sequence yields after applying cohomology an exact sequence

$\displaystyle H^1(X; \mathbb{Z}) \rightarrow H^1(X; \mathcal{O}) \stackrel{\exp}{\rightarrow} H^1(X; \mathcal{O}^{*}_{\mathrm{hol}}) \stackrel{c_1}{\rightarrow} H^2(X; \mathbb{Z}) .$

Here ${H^1(X; \mathcal{O}_{\mathrm{hol}}^*)}$ is precisely ${\mathrm{Pic}(X)}$ and we get an exact sequence

$\displaystyle 0 \rightarrow H^1(X; \mathcal{O})/H^1(X; \mathbb{Z}) \rightarrow \mathrm{Pic}(X) \rightarrow H^2(X; \mathbb{Z}) \cap H^{1,1}(X; \mathbb{C}) \rightarrow 0.$

This shows that the group ${ H^1(X; \mathcal{O})/H^1(X; \mathbb{Z})}$ (which is a complex torus) precisely picks out the topologically trivial holomorphic line bundles on ${X}$ : they fit into a connected complex manifold called ${\mathrm{Pic}^0(X)}$ (the Picard variety) of dimension ${h^1(X; \mathcal{O}) = \frac{1}{2}b_1}$ .

Example 1 Let ${X = V/\Lambda}$ be a complex abelian variety. Then ${H^1(X; \mathbb{Z})}$ is identified with ${\hom(\pi_1(X); \mathbb{Z}) = \hom_{\mathbb{Z}}(\Lambda, \mathbb{Z})}$ while ${H^1(X; \mathcal{O})}$ is identified (using Hodge theory) with the translation-invariant ${(0,1)}$ -forms on ${X}$ . (These are the harmonic ${(0,1)}$ -forms with respect to a translation-invariant metric.) Such a form is given by an element in the ${(0,1)}$ -cotangent space to ${X}$ , i.e. by an element of

$\displaystyle \hom_{\mathrm{antilinear}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}) = \hom_{\mathbb{R}}(V, \mathbb{R}).$

We find that the Picard variety is exactly ${\hom_{\mathbb{R}}(V, \mathbb{R})/2\pi i \hom(\Lambda, \mathbb{Z})}$ , i.e. the complex torus we encountered before.

In particular, we conclude from this analysis that the Picard group is the extension of a finitely generated abelian group (the so-called Néron-Severi group) by the complex torus ${\mathrm{Pic}^0(X)}$ , i.e. the Picard variety. We also conclude that the tangent space to the Picard variety is precisely ${H^1(X; \mathcal{O}_X)}$ : this is something that holds in the algebraic context as well, but via deformation theory.

Example 2 On a rational variety (or even a unirational variety), the group ${H^1(X; \mathcal{O}_X)}$ vanishes. More generally, this is true for any simply connected variety (over ${\mathbb{C}}$ ) by the Hodge theoretic equality ${h^1(\mathcal{O}_X) = \frac{1}{2}b_1}$ , and consequently the Picard group is discrete for such varieties. (Recall from the previous post that unirational varieties are simply connected by a theorem of Serre; as Jason Starr points out in the comments, this result has been extended significantly, e.g. to rationally connected varieties.)

Example 3 Since we are over the complex numbers, we have $h^1(X; \mathcal{O}_X) = h^0(X; \Omega_X)$ and the latter is a birational invariant. Therefore, the dimension of the Picard variety is a birational invariant.

2. The Picard scheme

Let’s now move to an algebraic setting. If ${X}$ is a projective variety over the algebraically closed field ${k}$ , we would like to put on the group ${\mathrm{Pic}(X)}$ the structure of an algebraic variety. Or rather, it should be a scheme, which is the disjoint union of countably many algebraic varieties.

In order to turn ${\mathrm{Pic}(X)}$ into a scheme, we’ll need to say what it means for a scheme ${Y}$ to map into ${\mathrm{Pic}(X)}$ . Roughly, this should be the same as a “family” of line bundles on ${X}$ parametrized by ${Y}$ . Here is a reasonable definition of that:

Definition 1 A family of line bundles on ${X}$ parametrized by ${Y}$ is simply a line bundle on ${X \times Y}$ .

We could thus try to hope that maps from ${Y \rightarrow \mathrm{Pic}(X)}$ would be isomorphism classes of such line bundles. Unfortunately, this does not work: the functor thus defined fails to satisfy flat descent, because line bundles admit automorphisms. For example, if ${X = \ast}$ , then the functor thus defined is

$\displaystyle Y \mapsto \left\{\mathrm{isomorphism\ classes \ of \ line \ bundles \ on \ Y}\right\}$

which is clearly not a sheaf in the flat topology (or even Zariski topology!). The reason it fails to be a sheaf is precisely that in order to do flat descent, you need to keep track of automorphisms rather than quotient by them.

But one could instead define a functor

$\displaystyle \mathrm{Pic}^{\mathrm{gpd}}\mathrm{Sch}_{/k} \rightarrow \mathbf{Gpds}, \quad Y \mapsto \{ \text{line bundles on } X \times Y \text{ and isomorphisms }\},$

sending any ${Y}$ to the groupoid of line bundles on ${X \times Y}$ (rather than the set of isomorphism classes). This functor is a sheaf in the flat topology: this is equivalent to the theory of flat descent for quasi-coherent sheaves. We can use this to get a Picard stack of line bundles.

However, in this case, there’s no real need to form the Picard stack, because there aren’t any interesting automorphisms to keep track of. The only automorphisms come from multiplication by invertible functions on ${Y}$ . In fact, given a line bundle ${\mathcal{L}}$ on ${X \times Y}$ , we have

$\displaystyle \hom_{X \times Y}(\mathcal{L}, \mathcal{L}) = H^0( X \times Y, \mathrm{End}(\mathcal{L})) = H^0(X \times Y, \mathcal{O}_{X \times Y}) = H^0(Y, \mathcal{O}_Y),$

because ${X}$ is reduced and projective. In particular, the automorphisms of ${\mathcal{L}}$ are exactly ${H^0(Y, \mathcal{O}_Y)^{\ast}}$ .

This suggests that if we rigidify the moduli problem, we can eliminate the need for stacks without losing any interesting information. In other words, we can define the improved Picard functor via

$\displaystyle \mathrm{Pic}_X: Y \mapsto \left\{\text{line bundles on } X \times Y \text{trivialized on } \ast \times Y\right\}$

where ${\ast \in X}$ is a fixed basepoint.

The main deep result is the following:

Theorem 2 (Grothendieck) ${\mathrm{Pic}_X}$ is representable by the countable union of quasi-projective schemes.

Moreover, the Picard stack described previously is simply ${\mathrm{Pic}_X \times B \mathbb{G}_m}$ (mistaken assertion corrected and recorrected here): it’s what you get by adding a ${\mathbb{G}_m}$ ‘s worth of automorphisms to each object in ${\mathrm{Pic}_X}$ (and stackifying). Observe that ${\mathrm{Pic}_X}$ is a group scheme by tensor product of line bundles.

We can define ${\mathrm{Pic}^0_X}$ to be the connected component at the identity of ${\mathrm{Pic}_X}$ : this represents line bundles which are algebraically equivalent to zero (i.e., which can be connected to the zero bundle by a connected family). The quotient ${\mathrm{Pic}_X/\mathrm{Pic}^0_X}$ is the Néron-Severi group, which over ${\mathbb{C}}$ picks out the topological type of a line bundle. In characteristic ${p}$ , it is a deep theorem that the group is still finitely generated.

hilbertthm90 Says:

March 19, 2013 at 12:46 pm

One might say the Picard stack is a $\mathbb{G}_m$ -gerbe over the Picard scheme!

Akhil Mathew Says:

March 19, 2013 at 2:09 pm
Actually, I think my assertion was mistaken: there’s no natural map from the stack to the scheme. It’s a $\mathbb{G}_m$ -quotient rather than a product.

1. hilbertthm90 Says:
  
  March 19, 2013 at 4:08 pm
  Hmm…I think what you had is OK if interpreted properly. The sheafification of the Picard stack is the Picard scheme, so as you said this is just a rigidification by killing off the $\mathbb{G}_m$ -automorphisms. Unless I have something backwards ….
  
  1. Akhil Mathew Says:
    
    March 20, 2013 at 7:00 am
    I was thinking about this a bit more, and as you say I now think the original assertion was correct. That is, the Picard stack *does* map to the Picard scheme, it’s just that you have to twist the line bundle on $X \times_k Y$ by the pull-back of the line bundle on $Y$ given by restricting to $\ast \times Y$ so that you can make the trivialization. In other words, I believe it is the product stack as claimed.

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