Let ${X}$ be a projective variety over the algebraically closed field ${k}$, endowed with a basepoint ${\ast}$. In the previous post, we saw how to define the Picard scheme ${\mathrm{Pic}_X}$ of ${X}$: a map from a ${k}$-scheme ${Y}$ into ${\mathrm{Pic}_X}$ is the same thing as a line bundle on ${Y \times_k X}$ together with a trivialization on ${Y \times \ast}$. Equivalently, ${\mathrm{Pic}_X}$ is the sheafification (in the Zariski topology, even) of the functor

$\displaystyle Y \mapsto \mathrm{Pic}(X \times_k Y)/\mathrm{Pic}(Y),$

so we could have defined the functor without a basepoint.

We’d like to understand the local structure of ${\mathrm{Pic}_X}$ (or, equivalently, of ${\mathrm{Pic}^0_X}$), and, as with moduli schemes in general, deformation theory is a basic tool. For example, we’d like to understand the tangent space to ${\mathrm{Pic}_X}$ at the origin ${0 \in \mathrm{Pic}_X}$. The tangent space (this works for any scheme) can be identified with

$\displaystyle \hom_{0}( \mathrm{Spec} k[\epsilon]/\epsilon^2, \mathrm{Pic}_X).$

In other words, we have to look at maps ${\mathrm{Spec} k[\epsilon]/\epsilon^2 \rightarrow \mathrm{Pic}_X}$ which restrict to zero on the closed point. By the modular interpretation of ${\mathrm{Pic}_X}$, we find that

$\displaystyle T_0 \mathrm{Pic}_X = \mathrm{ker}( \mathrm{Pic}(X \times_k \mathrm{Spec} k[\epsilon]/\epsilon^2) \rightarrow \mathrm{Pic}(X));$

in other words, we have to understand the possible deformations of the trivial line bundle over ${X}$ to the dual numbers. We can do that using the exponential sequence

$\displaystyle 0 \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_{X \times \mathrm{Spec} k[\epsilon]/\epsilon^2}^* \rightarrow \mathcal{O}_X^* \rightarrow 0,$

where the first map sends ${f \mapsto 1 + \epsilon f}$. Applying the long exact sequence in cohomology (notice that the last map has a section), we conclude:

Theorem 3 The tangent space to ${\mathrm{Pic}_X}$ is identified with ${H^1(X; \mathcal{O}_X)}$.

In particular, we should expect the dimension of ${\mathrm{Pic}^0_X}$ to be exactly the Hodge number ${H^1(X; \mathcal{O}_X)}$, as we found earlier over ${\mathbb{C}}$ using transcendental methods. Unfortunately, this is not true in general. The dimension of the tangent space of a scheme takes into account only the infinitesimal structure, but does not see the actual dimension (unless the scheme is smooth).

However, in characteristic zero, a theorem of Cartier (see for instance this post) states that group schemes are smooth, and in this case we can conclude that ${\dim \mathrm{Pic}^0_X = H^1(X; \mathcal{O}_X)}$. In characteristic ${p}$, this is not true in general. Later in this post, I’ll describe a counterexample of Igusa.

3. Smoothness in dimension one

As a test case, let’s check that the Picard scheme of a one-dimensional scheme is always smooth: for a smooth curve, it is the classical Jacobian variety. In fact, smoothness can be checked by the infinitesimal lifting criterion. It suffices to show that whenever ${A}$ is a local artinian ring, and ${A' \twoheadrightarrow A}$ is a square-zero extension by some square-zero ideal ${I \subset A'}$, then we can always solve a lifting problem

Equivalently, we need to show that the map

$\displaystyle \mathrm{Pic}(X \times \mathrm{Spec} A') \rightarrow \mathrm{Pic}(X \times \mathrm{Spec} A)$

is a surjection: we can lift families of line bundles along square-zero extensions.

To do this, we consider the short exact sequence

$\displaystyle 0 \rightarrow I \rightarrow \mathcal{O}_{X \times \mathrm{Spec} A'}^* \rightarrow \mathcal{O}_{X \times \mathrm{Spec} A} \rightarrow 1,$

where the first map is ${f \mapsto 1 + f}$. Applying ${H^1}$ and looking at the long exact sequence, we find that since ${H^2\equiv 0}$ on the one-dimensional scheme ${X}$, the map

$\displaystyle H^1(X; \mathcal{O}_{X \times \mathrm{Spec} A'}^*) \rightarrow H^1(X; \mathcal{O}_{X \times \mathrm{Spec} A})$

is surjective. But this is exactly what we wanted to prove. We conclude:

Theorem 4 For a one-dimensional, reduced, proper scheme ${X}$ over ${k}$, ${\mathrm{Pic}_X}$ is a smooth scheme whose components are quasi-projective.

When ${X}$ itself is smooth, then an argument via the valuative criterion shows that ${\mathrm{Pic}^0_X}$ is projective, hence an abelian variety: the Jacobian variety of ${X}$.

4. The Albanese variety

Let ${(X, \ast)}$ be a pointed, smooth variety. As before, there is a scheme ${\mathrm{Pic}_X}$ such that maps ${Y \rightarrow \mathrm{Pic}_X}$ are in natural bijection with line bundles over ${Y \times X}$ trivialized over ${Y \times \ast}$. We can consider the connected component ${\mathrm{Pic}_X^0 \subset \mathrm{Pic}_X}$ at the identity.

By the universal property, there is a line bundle ${\mathcal{L}}$ over ${X \times \mathrm{Pic}_X^0}$ with a trivialization on ${\ast \times \mathrm{Pic}_X^0 }$. Now, ${\mathrm{Pic}_X^0}$ need not itself be a variety, but it is a quasi-projective ${k}$-group scheme.

Proposition 5 If ${X}$ is smooth, then ${\mathrm{Pic}_X^0}$ is projective.

Proof: One way to prove this is to use the valuative criterion to argue that if ${R}$ is a discrete valuation ring (over ${k}$) with quotient field ${K}$, then ${\mathrm{Pic}(X \times \mathrm{Spec} R) \simeq \mathrm{Pic}(X \times \mathrm{Spec} K)}$ by an argument using divisors; see this MO discussion. $\Box$

In particular, we can form the reduction ${(\mathrm{Pic}_X^0)_{\mathrm{red}}}$, and that will be an abelian variety. We thus get a line bundle ${\mathcal{L}_{\mathrm{red}}}$ on ${X \times (\mathrm{Pic}_X^0)_{\mathrm{red}}}$ which is trivialized on ${\ast \times (\mathrm{Pic}_X^0)_{\mathrm{red}}}$. Moreover, by definition, we can trivialize the line bundle on ${X \times e}$, where ${e}$ is image of the identity in ${(\mathrm{Pic}_X^0)_{\mathrm{red}}}$.

It follows that our situation is completely symmetric. That is, rather than thinking of ${\mathcal{L}_{\mathrm{red}}}$ as being a family of line bundles on ${X}$ parametrized by ${(\mathrm{Pic}_X^0)_{\mathrm{red}}}$, we can think of it as a family of line bundles on ${(\mathrm{Pic}_X^0)_{\mathrm{red}}}$ parametrized by ${X}$. The conclusion is that we get a map

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

which sends ${\ast}$ to zero. This map is natural in ${X}$.

Although we have not proved it yet, for an abelian variety, this map is an isomorphism: the Picard scheme is in that case always smooth, and the map above is the biduality isomorphism.

Definition 6 We define ${(\mathrm{Pic}^0_{(\mathrm{Pic}_X^0)_{\mathrm{red}}})_{\mathrm{red}} }$ to be the Albanese variety ${\mathrm{Alb}_X}$ of ${X}$ (in fact, the second ${\mathrm{red}}$ is redundant).

In particular, we have a natural map

$\displaystyle X \rightarrow \mathrm{Alb}_X,$

of pointed varieties (where an abelian variety is pointed at zero), which is an isomorphism if ${X}$ itself is an abelian variety. It follows that the Albanese variety is the universal abelian variety generated by ${(X, \ast)}$: that is, it is the left adjoint of the forgetful functor from abelian varieties to pointed varieties.

5. Igusa surfaces

Finally, let’s describe an example, due to Igusa, where the Picard scheme is nonreduced (and so has dimension smaller than expected). I learned this example from these notes of Jesse Kass.

Let ${E}$ be an ordinary elliptic curve over an algebraically field ${k}$ of characteristic two. In this case, there is a unique nonzero 2-torsion point ${\theta \in E(k)}$, and we can define the involution

$\displaystyle \iota: (x,y) \mapsto (x + \theta, -y)$

on ${E \times E}$. The involution acts without fixed points, so we can form the quotient ${(E \times E)/\iota}$ and get a smooth projective surface ${S}$.

The basic observation is that it is very difficult for ${S}$ to map to an abelian variety (let’s point ${S}$ with the image of ${(0,0)}$). In fact, a map

$\displaystyle S \rightarrow X,$

for ${X}$ an abelian variety, is the same as a pointed map ${f: E \times E \rightarrow X}$ such that

$\displaystyle f( x+ \theta, -y) = f(x,y).$

Since ${f}$ is a pointed map of abelian varieties, it is a homomorphism. We conclude that ${f(0, -y) = f(0, y)}$, so that ${f}$ entirely kills the second factor and the image of ${E \times E}$ (which is the same as the image of ${S}$) has dimension one.

In fact, it’s even better: we find that ${f(\theta) =0}$, so ${f}$ factors through the quotient ${E/\theta}$ (which is a ${\mathbb{Z}/2}$-quotient of ${E}$, hence an elliptic curve). Since there is a pointed map

$\displaystyle S \rightarrow E/\theta,$

we conclude that ${E/\theta}$ must be the universal abelian variety that ${S}$ maps into: that is,

$\displaystyle \mathrm{Alb}_S = E/\theta.$

However, ${\mathrm{Alb}_S}$ has the same dimension as ${(\mathrm{Pic}^0_S)_{\mathrm{red}}}$: any abelian variety is isogeneous to its dual. It follows that

$\displaystyle \dim \mathrm{Pic}_S^0 = 1.$

Conversely, the expected dimension from deformation theory is at least two. We have a spectral sequence

$\displaystyle H^i(\mathbb{Z}/2; H^j( E \times E; \mathcal{O})) \implies H^{i+j}(S; \mathcal{O})$

which implies that

$\displaystyle \dim H^1(S; \mathcal{O}) \geq \dim H^1(E \times E; \mathcal{O})^{\mathbb{Z}/2} = 2,$

since ${\iota}$ acts trivially in cohomology.