Let be an abelian variety. In the previous post, we saw that the functor
sending a local artinian algebra with residue field to the set of isomorphism classes of deformations of over 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 and a surjection , it can be lifted to a deformation over . 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 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 be a smooth variety.
- Isomorphism classes of deformations over are parametrized by .
- More generally, given a deformation of over and given an extension
in , where , then isomorphism classes of extensions of this deformation to are either empty or parametrized by the vector space .
- Given an extension as above, there is an obstruction which is natural under isomorphisms in and in , such that vanishes precisely when an extension of the deformation to 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 , 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 , 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 be an abelian scheme for , and let be a surjection whose kernel satisfies . Then there exists a smooth scheme which restricts to on .
In other words, if we are given an abelian scheme over and a small extension , we can deform as an abstract scheme. We will later show that you can deform the abelian structure as well.
Proof: Let be the underlying variety, i.e. . By the above theory, there is an obstruction
which vanishes if and only if can be deformed as a variety (not as an abelian scheme) to . The strategy is now to use naturality to argue that it must be zero. For simplicity, we assume is one-dimensional and so drop it from the notation.
Suppose first that the characteristic of is not . In this case, we can argue as follows: there is an inverse map of -schemes, which restricts to the inverse map . This is an isomorphism, and we find that it must preserve the obstruction. That is,
But the claim is that acts by on , which implies that or ; so if the characteristic is not , the obstruction must vanish.
We can see this claim as follows. First, : any abelian variety is parallelizable. The cohomology ring acquires the structure of a graded, connected Hopf algebra because of the group law on , and it is in fact a theorem (in the theory of abelian varieties) that
is an exterior algebra. It now follows formally that the antipode on this Hopf algebra must be given by multiplication by in degree (since elements in degree are primitive), and the antipode must thus be the identity in degree . In particular, the action
is the identity; so since acts by on the tangent space at zero, we find that
In characteristic , this argument does not work, but there is a slightly more complicated naturality trick that does. The idea is to consider the abelian scheme . This has an obstruction which turns out to be the sum of the pull-backs of the obstructions to lifting . The abelian scheme has a shearing automorphism which must preserve ; this means that
In particular, we find that , so . In this case, too, the obstruction vanishes, and the scheme is liftable.
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 be a small extension in : that is, . Let be a smooth, proper scheme together with a section . Suppose given the structure of an abelian scheme on (for the given section as identity element). Then there is a unique extension of this abelian scheme structure to .
Proof: Since is an abelian scheme, there exists a map given by . The entire structure of abelian scheme is encoded by (together with the identity section). The goal is to lift to a map which satisfies the desired axioms. In other words, we want to deform the morphism . For simplicity, we will assume .
The observation is that deforming is the same as deforming its graph. Right now, is determined by its graph, which is a closed subscheme which projects isomorphically onto the first two factors. If we can deform to a flat subscheme , then will necessarilyproject isomorphically onto the first two factors and will thus define a deformation of the morphism .
Now, there is an analogous theory of deformations for a closed subscheme. Let be the special fiber, so that is an abelian variety over a field. There is an obstruction to deforming the subscheme , which is for the normal bundle of . However, the point is that this obstruction vanishes when restricted to either factor or the diagonal , because the constant map at zero or the identity map can clearly be deformed.
Using the Künneth theorem, we find that has to vanish, so we can deform to a map
The collection of such deformations (again, the collection of deformations of the graph of ) is parametrized by for the normal bundle of in . The condition that we want, which is that (i.e., is unital) determines uniquely the particular choice of deformation: in fact, each element of determines by restriction an element of . This restriction map is an isomorphism—this means that deforming is equivalent to deforming the zero section, or specifying an element of . We choose the one so that this element of is zero.
So, anyway, it follows from all this that there is a map
which sends the zero element of into the zero element of . From this, we want to claim that we have in fact an abelian scheme. The inverse map and the multiplication map are defined in terms of , and we just need to check that they satisfy the right identities. For instance, we need to check that satisfies associativity.
For this, it is convenient to formulate the above analysis in:
Lemma 13 Let be a small extension in . Let be smooth proper schemes (for ) together with sections such that the restrictions to are given the structure of abelian schemes. Then, given a morphism of abelian schemes , there is at most one deformation of to a map of pointed -schemes.
In other words, if we are given a homomorphism of the restrictions of to , then we get a at most one map preserving the basepoints of (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 is either empty or parametrized by of the normal bundle of the graph in . 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 : 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 on is associative. This is now straightforward, because the two morphisms and both deform the same morphism of abelian schemes over 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.
4. The local moduli space
We can now see that, given an abelian variety , the functor
is smooth. In fact, given a deformation and a surjection , which we may assume is a small extension, we first deform as an abstract scheme to a scheme over ; we’ve seen that the obstructions for that vanish. Then, we extend the zero section of to a section of (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 , which shows that is formally smooth.
By Schlessinger’s theorem, we find that is prorepresentable by for some . It remains to determine . For this, we have to study first-order deformations.
Proposition 14 Let be an abelian variety. Then there is a bijection (in fact, an isomorphism of vector spaces) between isomorphism classes of first-order deformations of as an abelian variety and isomorphism classes of first-order deformations of as an abstract variety.
Proof: In fact, we have seen that there is an equivalence between first-order deformations of as an abelian variety and first-order deformations of 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 (i.e. a vector field) to remedy the problem of choosing different sections.
In particular, we find that isomorphism classes of deformations of the abelian variety over the dual numbers are parametrized by , which has dimension if . This, together with Schlessinger’s theorem and the above analysis, finally proves:
Theorem 15 The deformation space of an abelian variety of dimension is .
In the next post, we’ll see that this deformation space has another interpretation (when the characteristic of is ): it is the deformation space of the -divisible group of .