Deformations of abelian varieties III: smoothness

(This is the third in a series of posts on deformations of abelian varieties; the first two posts are here and here.)

Let ${X/k}$ be an abelian variety. In the previous post, we saw that the functor

$\displaystyle \mathrm{Art}_k \rightarrow \mathbf{Sets}$

sending a local artinian algebra ${R}$ with residue field $k$ to the set of isomorphism classes of deformations of ${X}$ over ${R}$ was product-preserving, by showing that there were no infinitesimal automorphisms. The next step is to show that it is smooth: that is, given a deformation over ${R}$ and a surjection ${R' \twoheadrightarrow R}$ , it can be lifted to a deformation over ${R'}$ . Together with Schlessinger’s theorem, this will yield a description of the local moduli space.

1. Facts from deformation theory

Our first step is to study deformations of ${X}$ as a plain variety. There is a well-developed theory of deformations for a smooth variety over a field, a few facts of which we recall below.

Let ${Y/k}$ be a smooth variety.

Isomorphism classes of deformations over ${k[\epsilon]/\epsilon^2}$ are parametrized by ${H^1(T_Y)}$ .
More generally, given a deformation ${\overline{Y}/R}$ of ${Y}$ over ${R \in \mathrm{Art}_k}$ and given an extension
$\displaystyle 0 \rightarrow I \rightarrow R' \twoheadrightarrow R \rightarrow 0$

in ${\mathrm{Art}_k}$ , where ${\mathfrak{m}_{R'} I = 0}$ , then isomorphism classes of extensions of this deformation to ${R'}$ are either empty or parametrized by the vector space ${H^1(Y, T_Y \otimes_k I)}$ .
Given an extension ${0 \rightarrow I \rightarrow R' \twoheadrightarrow R \rightarrow 0}$ as above, there is an obstruction ${o \in H^2(Y, T_Y \otimes_k I)}$ which is natural under isomorphisms in ${I}$ and in ${(\overline{Y}, R)}$ , such that ${o}$ vanishes precisely when an extension of the deformation to ${R'}$ exists.

These facts are classical; the key point in these arguments is that deformations of an affine smooth algebra are always trivial (this follows from the nilpotent lifting definition of smoothness), but that they have automorphisms given by vector fields. So, first-order deformations form a gerbe banded by the action of vector fields: this means that the gerbe is classified by an obstruction class in ${H^2}$ , and if that vanishes (i.e., the gerbe is trivial) isomorphism classes are given by torsors for the sheaf of vector fields. That’s a very loose summary; I’ve heard that using derived algebraic geometry there is a better way to think of the obstructions in ${H^2}$ , but I don’t understand it will enough to say anything here.

2. Deforming as an abstract variety

So, our first step in showing that the deformation problem for abelian varieties is unobstructed is the following.

Proposition 11 Let ${\overline{X}/R}$ be an abelian scheme for ${R \in \mathrm{Art}_k}$ , and let ${R' \twoheadrightarrow R}$ be a surjection whose kernel ${I \subset R'}$ satisfies ${\mathfrak{m}_{R'} I = 0}$ . Then there exists a smooth scheme ${\overline{X}'/R'}$ which restricts to ${\overline{X}}$ on ${\mathrm{Spec} R}$ .

In other words, if we are given an abelian scheme ${\overline{X}}$ over ${\mathrm{Spec} R}$ and a small extension ${\mathrm{Spec} R \hookrightarrow \mathrm{Spec} R'}$ , we can deform ${\overline{X}}$ as an abstract scheme. We will later show that you can deform the abelian structure as well.

Proof: Let ${X}$ be the underlying variety, i.e. ${X = \overline{X} \times_{\mathrm{Spec} R} \mathrm{Spec} k}$ . By the above theory, there is an obstruction

$\displaystyle o \in H^2(X, T_X\otimes_k I)$

which vanishes if and only if ${\overline{X}}$ can be deformed as a variety (not as an abelian scheme) to ${\mathrm{Spec} R'}$ . The strategy is now to use naturality to argue that it must be zero. For simplicity, we assume ${I}$ is one-dimensional and so drop it from the notation.

Suppose first that the characteristic of ${k}$ is not ${2}$ . In this case, we can argue as follows: there is an inverse map ${\overline{X} \rightarrow \overline{X}}$ of ${R}$ -schemes, which restricts to the inverse map ${i: X \rightarrow X}$ . This is an isomorphism, and we find that it must preserve the obstruction. That is,

$\displaystyle i^* o = o \in H^2(X, T_X).$

But the claim is that ${i^*}$ acts by ${-1}$ on ${H^2(X, T_X)}$ , which implies that ${2o = 0}$ or ${o = 0}$ ; so if the characteristic is not ${2}$ , the obstruction must vanish.

We can see this claim as follows. First, ${T_X \simeq T_{0, X} \otimes \mathcal{O}_X}$ : any abelian variety is parallelizable. The cohomology ring ${H^\bullet(X, \mathcal{O}_X)}$ acquires the structure of a graded, connected Hopf algebra because of the group law on ${X}$ , and it is in fact a theorem (in the theory of abelian varieties) that

$\displaystyle H^\bullet(X, \mathcal{O}_X) \simeq \bigwedge^\bullet H^1(X, \mathcal{O}_X)$

is an exterior algebra. It now follows formally that the antipode on this Hopf algebra must be given by multiplication by ${-1}$ in degree ${1}$ (since elements in degree ${1}$ are primitive), and the antipode must thus be the identity in degree ${2}$ . In particular, the action

$\displaystyle i^*: H^2(X, \mathcal{O}_X) \rightarrow H^2(X, \mathcal{O}_X)$

is the identity; so since ${i}$ acts by ${-1}$ on the tangent space at zero, we find that

$\displaystyle i^* = -1: H^2(X, T_X) \rightarrow H^2(X, T_X).$

In characteristic ${2}$ , this argument does not work, but there is a slightly more complicated naturality trick that does. The idea is to consider the abelian scheme ${\overline{X} \times_{\mathrm{Spec} R} \overline{X}}$ . This has an obstruction ${o' \in H^2(X \times X, T_{X \times X})}$ which turns out to be the sum of the pull-backs ${p_1^* o + p_2^* o}$ of the obstructions to lifting ${X}$ . The abelian scheme ${\overline{X} \times_{\mathrm{Spec} R} \overline{X}}$ has a shearing automorphism ${(x,y) \mapsto (x+y, y)}$ which must preserve ${o}$ ; this means that

$\displaystyle p_1^* o + p_2^* o = p_1^* o + 2p_2^* o \in H^2(X \times X, T_X \oplus T_X).$

In particular, we find that ${p_2^* o = 0}$ , so ${o = 0}$ . In this case, too, the obstruction vanishes, and the scheme is liftable. $\Box$

The use of naturality tricks in showing that certain obstructions vanish seems to crop up quite frequently; the same sort of idea was used in the theorem of the cube. In dimension 1 (for smooth curves), the obstructions automatically vanish because the relevant cohomology groups are zero, so one can always lift deformations of curves (this is a reflection of the smoothness of the moduli of curves). For abelian varieties, the situation is a bit more subtle because the cohomology groups aren’t zero, but these kinds of tricks can still show that the obstructions vanish.

3. Extending the abelian structure

This is an incomplete answer to our goals though; we need to show that we can extend the lifting as an abelian scheme, not just as a scheme. The following result shows that we can do so: once we’ve lifted it as a scheme, we can get the abelian structure for free.

Proposition 12 Let ${0 \rightarrow I \rightarrow R' \rightarrow R \rightarrow 0}$ be a small extension in ${\mathrm{Art}_k}$ : that is, ${\mathfrak{m}_{R'} I =0}$ . Let ${X'/R'}$ be a smooth, proper scheme together with a section ${\mathrm{Spec} R' \rightarrow X'}$ . Suppose given the structure of an abelian scheme on ${X_0 = X' \times_{\mathrm{Spec} R'} \mathrm{Spec} R}$ (for the given section as identity element). Then there is a unique extension of this abelian scheme structure to ${X}$ .

Proof: Since ${X_0}$ is an abelian scheme, there exists a map ${j: X_0 \times_{\mathrm{Spec} R} X_0 \rightarrow X_0}$ given by ${(x,y) \mapsto x-y}$ . The entire structure of abelian scheme is encoded by ${j}$ (together with the identity section). The goal is to lift ${j}$ to a map ${j': X' \times_{\mathrm{Spec} R'} X' \rightarrow X'}$ which satisfies the desired axioms. In other words, we want to deform the morphism ${j}$ . For simplicity, we will assume ${\dim_k I = 1}$ .

The observation is that deforming ${j}$ is the same as deforming its graph. Right now, ${j}$ is determined by its graph, which is a closed subscheme ${\Gamma \subset X_0 \times_{ R} X_0 \times_{ R} X_0}$ which projects isomorphically onto the first two factors. If we can deform ${\Gamma}$ to a flat subscheme ${\Gamma' \subset X' \times_{R'} X' \times_{R'} X'}$ , then ${\Gamma'}$ will necessarilyproject isomorphically onto the first two factors and will thus define a deformation of the morphism ${j}$ .

Now, there is an analogous theory of deformations for a closed subscheme. Let ${X_s}$ be the special fiber, so that ${X_s}$ is an abelian variety over a field. There is an obstruction to deforming the subscheme ${\Gamma}$ , which is ${o \in H^1( X_s \times X_s, N_{\Gamma_s})}$ for ${N_{\Gamma_s}}$ the normal bundle of ${\Gamma_s = X_s \times X_s \subset X_s \times X_s \times X_s}$ . However, the point is that this obstruction vanishes when restricted to either factor ${X_s \times \left\{0\right\}}$ or the diagonal ${X_s \times X_s}$ , because the constant map at zero or the identity map can clearly be deformed.

Using the Künneth theorem, we find that ${o}$ has to vanish, so we can deform ${j}$ to a map

$\displaystyle j': X' \times_{R'} X' \rightarrow X'.$

The collection of such deformations (again, the collection of deformations of the graph of ${j}$ ) is parametrized by ${H^0(X_s \times X_s, N_{\Gamma_s})}$ for ${N_{X_s}}$ the normal bundle of ${\Gamma_s}$ in ${X_s \times X_s \times X_s}$ . The condition that we want, which is that ${j'|_{\mathrm{Spec} R' \times_{\mathrm{Spec} R'} \mathrm{Spec} R'} = \mathrm{Id}}$ (i.e., ${j'}$ is unital) determines uniquely the particular choice of deformation: in fact, each element of ${H^0(X_s \times X_s, N_{\Gamma_s})}$ determines by restriction an element of ${(N_{\Gamma_s})_{(0, 0)}}$ . This restriction map is an isomorphism—this means that deforming ${j}$ is equivalent to deforming the zero section, or specifying an element of ${(N_{\Gamma_s})_{(0, 0)}}$ . We choose the one so that this element of ${(N_{\Gamma_s})_{(0, 0)}}$ is zero.

So, anyway, it follows from all this that there is a map

$\displaystyle j' : X' \times_{R'} X' \rightarrow X'$

which sends the zero element of ${X' \times_{R'} X'}$ into the zero element of ${X'}$ . From this, we want to claim that we have in fact an abelian scheme. The inverse map ${X' \rightarrow X'}$ and the multiplication map ${m': X' \times_{R'} X' \rightarrow X'}$ are defined in terms of ${j'}$ , and we just need to check that they satisfy the right identities. For instance, we need to check that ${m'}$ satisfies associativity.

For this, it is convenient to formulate the above analysis in:

Lemma 13 Let ${R' \twoheadrightarrow R}$ be a small extension in ${\mathrm{Art}_k}$ . Let ${X'/R', Y'/R'}$ be smooth proper schemes (for ${R' \in \mathrm{Art}_k}$ ) together with sections ${\mathrm{Spec} R' \rightarrow X, \mathrm{Spec} R' \rightarrow Y}$ such that the restrictions to ${R}$ are given the structure of abelian schemes. Then, given a morphism of abelian schemes ${f_0: X \times_{R'} R \rightarrow Y \times_{R'} R}$ , there is at most one deformation of ${f}$ to a map ${X \rightarrow Y}$ of pointed ${R'}$ -schemes.

In other words, if we are given a homomorphism of the restrictions of ${X', Y'}$ to ${R}$ , then we get a at most one map ${X' \rightarrow Y'}$ preserving the basepoints of ${X', Y'}$ (i.e., the sections). This is proved using the same obstruction-theoretic argument as before: that is, we argue that the collection of deformations of ${f}$ is either empty or parametrized by ${H^0}$ of the normal bundle of the graph in ${X_s \times Y_s}$ . But a global section of this normal bundle is determined by restriction to one point. Note that there may well be obstructions to finding a map ${f}$ : for instance, there are non-isomorphic first-order deformations of an abelian variety.

With this lemma in mind, we can complete the proof of the proposition. Let’s say we want to show that the multiplication defined by the deformation ${m'}$ on ${X'}$ is associative. This is now straightforward, because the two morphisms ${m'(m'(\cdot, \cdot), \cdot)}$ and ${m'(\cdot, m'(\cdot, \cdot))}$ both deform the same morphism of abelian schemes over ${R}$ and are both pointed. This means that the two morphisms have to be equal by the above lemma. The other axioms are checked in the same way. $\Box$

4. The local moduli space

We can now see that, given an abelian variety ${X/k}$ , the functor

$\displaystyle F: \mathrm{Art}_k \rightarrow \mathbf{Sets}, F(R) = \left\{\text{defos of$

is smooth. In fact, given a deformation ${X_0/R}$ and a surjection ${R' \twoheadrightarrow R}$ , which we may assume is a small extension, we first deform ${X_0}$ as an abstract scheme to a scheme ${X'}$ over ${R'}$ ; we’ve seen that the obstructions for that vanish. Then, we extend the zero section of ${X_0}$ to a section of ${X'}$ (the obstructions to this vanish, since we’re mapping out of something zero-dimensional). As we’ve seen, this gives us the structure of an abelian scheme on ${X'}$ , which shows that ${F}$ is formally smooth.

By Schlessinger’s theorem, we find that ${F}$ is prorepresentable by ${W(k)[[t_1, \dots, t_n]]}$ for some ${n}$ . It remains to determine ${n}$ . For this, we have to study first-order deformations.

Proposition 14 Let ${X/k}$ be an abelian variety. Then there is a bijection (in fact, an isomorphism of vector spaces) between isomorphism classes of first-order deformations of ${X}$ as an abelian variety and isomorphism classes of first-order deformations of ${X}$ as an abstract variety.

Proof: In fact, we have seen that there is an equivalence between first-order deformations of ${X}$ as an abelian variety and first-order deformations of ${X}$ as an abstract variety together with a section. But two different choices of a section give isomorphic deformations as abelian varieties: one chooses an infinitesimal automorphism of the deformation of ${X}$ (i.e. a vector field) to remedy the problem of choosing different sections. $\Box$

In particular, we find that isomorphism classes of deformations of the abelian variety ${X/k}$ over the dual numbers ${k[\epsilon]/\epsilon^2}$ are parametrized by ${H^1(X, T_X) \simeq H^1(X, \mathcal{O}_X) \otimes T_{X, 0}}$ , which has dimension ${g^2}$ if ${g = \dim X}$ . This, together with Schlessinger’s theorem and the above analysis, finally proves:

Theorem 15 The deformation space of an abelian variety ${X/k}$ of dimension ${g}$ is ${W(k)[[t_1, \dots, t_{g^2}]]}$ .

In the next post, we’ll see that this deformation space has another interpretation (when the characteristic of ${k}$ is ${p}$ ): it is the deformation space of the ${p}$ -divisible group of ${X}$ .

Jason Starr Says:

June 21, 2012 at 12:43 pm

This is a nice proof. It is also a nice exercise to see where the proof breaks down in closely related examples, e.g., deformations of (the total space of) an Abelian scheme over a smooth, projective curve. Of course such a scheme has the same automorphism acting fiber-by-fiber as inversion, yet the scheme can easily be obstructed.

Akhil Mathew Says:

June 22, 2012 at 8:41 am
Hmm, that’s an interesting point. I don’t really know anything about deformation theory on anything other than a local artinian ring though.

1. Jason Starr Says:
  
  June 22, 2012 at 9:23 am
  The base of the deformation is a small extension of local Artinian rings. However, the smooth projective scheme being deformed is the total space of an Abelian scheme over a smooth, projective curve, rather than an Abelian variety. Most of the proof in the case of an Abelian variety carries over to this setting as well. Yet, there are many examples of Abelian schemes over a curve such that deformations of the total space are obstructed. The question is, what breaks down in the proof?
  
  It is also interesting to consider such Abelian schemes for which the generic fiber has even more automorphisms than usual, e.g., the Eisenstein integers sit in the endomorphism algebra. Because there are “more” characters, the invariant part of the relevant cohomology groups is “smaller”.
  
  1. Akhil Mathew Says:
    
    June 22, 2012 at 6:01 pm
    Ah, I see I misunderstood. That’s pretty interesting! Presumably the fact that the inversion map on $H^2$ acts by $-1$ is what breaks down in the proof (it was sort of surprising to me how these sorts of tricks enabled one to show vanishing of various obstructions).

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