This is the second post in a series on deformations of abelian varieties. In the previous post, I described the basic outline of the goals and strategies, as well as a weak version of Schlessinger’s criterion useful in showing prorepresentability of smooth moduli problems without infinitesimal automorphisms. In this post, we’ll see that the deformation problem for abelian varieties satisfies the second condition above: that abelian varieties are rigid. The material here is very classical; I learned it from Oort’s article (from a summer school in the 1970s) and Katz’s article. Most of the material in this post comes from chapter 6 of Mumford’s GIT book, which is surprisingly readable without knowledge of any other parts of it.

Let ${R}$ be an artin ring, and let ${X/\mathrm{Spec} R}$ be an abelian scheme. Consider a morphism of ${R}$-group schemes

$\displaystyle f: X \rightarrow X$

inducing the identity on the special fiber. We would like to show that it is the identity, as in the next proposition:

Proposition 7 Such a morphism ${f}$ is necessarily the identity: that is, an infinitesimal deformation of an abelian variety has no nontrivial infinitesimal automorphisms.

This will imply prorepresentability of the deformation functor, using the general form of Schlessinger’s theorem.

1. A general rigidity lemma

To prove this, consider the difference ${f-1: X \rightarrow X}$. This is a morphism of schemes ${X \rightarrow X}$ which restricts to a constant over the special fiber. Now we have the following useful rigidity lemma, which is extremely handy in generalizing results for abelian varieties to abelian schemes.

Lemma 8 (Rigidity lemma) Let ${S}$ be the ${\mathrm{Spec}}$ of an artin local ring with closed point ${s \in S}$. Let ${X, Y}$ be ${S}$-schemes with ${X}$ proper and flat; suppose that ${h^0(X_s) = 1}$. Suppose ${f: X \rightarrow Y}$ is such that ${f(X)}$ is a single point (set-theoretically). Then ${f}$ factors through ${S}$.

Proof: The observation is that, set-theoretically, the morphism ${f}$ of spaces has to factor

$\displaystyle X \rightarrow S \stackrel{}{\rightarrow} Y,$

because the image of ${f}$ is a point. To describe ${f}$, though, we need to do more: we need the map of sheaves of rings, since a morphism of schemes is a morphism of locally ringed spaces.

To avoid confusion, we will write ${|X|}$ for the underlying space of ${X}$. Let ${F: |X| \rightarrow |Y|}$ be the underlying map of topological spaces of ${f: X \rightarrow Y}$. Then, since the image of ${F}$ is one point, we find that there is a (unique) map ${T: |S| \rightarrow |Y|}$ of topological spaces such that ${F}$ factors ${|X| \rightarrow |S| \stackrel{T}{\rightarrow} |Y|}$. Our goal is to make ${T}$ into a morphism of schemes and make the diagram commute in the category of schemes.

The description of ${f}$ comes from ${F}$ together with a map of sheaves of rings ${\mathcal{O}_Y \rightarrow F_*(\mathcal{O}_X)}$. In order to factor ${f}$ through ${S}$, and upgrade ${T}$ to a map of schemes, we need to define a map of sheaves ${\phi :\mathcal{O}_Y \rightarrow T_*(\mathcal{O}_S)}$ such that the composite

$\displaystyle \mathcal{O}_Y \stackrel{\phi}{\rightarrow} T_*(\mathcal{O}_S) \rightarrow F_*(\mathcal{O}_X),$

is the structure map of ${f}$. (Here ${T_*(\mathcal{O}_S) \rightarrow F_* (\mathcal{O}_X)}$ is the push-forward of structure map on sheaves of rings from ${X \rightarrow S}$; it is already determined.)

The whole point, though, is that ${T_*(\mathcal{O}_S) \rightarrow F_*(\mathcal{O}_X)}$ is an isomorphism; in fact, let ${p: X \rightarrow S}$ be the map making ${X}$ into an ${S}$-scheme. Then

$\displaystyle \mathcal{O}_S \stackrel{\rightarrow}{\simeq }p_*(\mathcal{O}_X)$

by the assumption ${h^0(X_s) = 1}$ and cohomology and base change. It follows that the pushforward ${T_*(\mathcal{O}_S) \rightarrow F_*(\mathcal{O}_X)}$ is an isomorphism too, and there is a unique way of defining a map ${\phi: \mathcal{O}_Y \rightarrow T_*(\mathcal{O}_S)}$ of sheaves of rings to factor ${f}$ through ${S}$.

This almost upgrades ${T}$ to a map of schemes; the only last thing to check is that ${T}$ is a map of locally ringed spaces (rather than simply ringed spaces). In other words, we need to show that ${\mathcal{O}_{Y, T(s)} \rightarrow \mathcal{O}_{S, s}}$ is a local homomorphism of local rings. But this follows because we have a composite map of rings

$\displaystyle \mathcal{O}_{Y, T(s) } \rightarrow \mathcal{O}_{S, s} \rightarrow \mathcal{O}_{X, x}$

(for any ${x \in X}$), which is a local homomorphism because ${f}$ is a map of schemes. $\Box$

Suppose ${X}$ admits a section ${t: S \rightarrow X}$. Then it follows in the above proposition that ${f}$ is actually equal to the composite ${X \rightarrow S \stackrel{t}{\rightarrow} X \rightarrow Y}$, by staring at the commutative diagram

which implies that the map ${S \rightarrow Y}$ comes from ${S \stackrel{t}{\rightarrow} X \rightarrow Y}$.

This corollary was stated for artinian rings, but it quickly yields a powerful generalization.

Corollary 9 (Rigidity lemma; strengthened version) Let ${f: X \rightarrow Y}$ be a morphism of separated schemes over a connected scheme ${S}$. Suppose ${X}$ is proper and flat over ${S}$ and that there is a section ${t: S \rightarrow X}$. Suppose there exists a point ${s \in S}$ such that ${f(X_s)}$ is set-theoretically a point and that for all ${s' \in S}$, ${h^0(X_{s'}) = 1}$. Then ${f}$ factors as ${X \rightarrow S \stackrel{q}{\rightarrow} Y}$ where ${q}$ is the map ${S \stackrel{t}{\rightarrow} X \stackrel{f}{\rightarrow} Y}$.

In other words, if ${f}$ is constant on one fiber, it is constant on all fibers, and a little more.

Proof: Without loss of generality, we can assume ${S}$ irreducible.

Consider the morphism ${f}$ and the other morphism ${\overline{f}: X \rightarrow S \stackrel{t}{\rightarrow} X \rightarrow Y}$. We want to show that they are equal. The previous version of the rigidity lemma and the comments following it imply that the two morphisms are equal when restricted to ${X \times_S S_0}$, for ${S_0}$ any nilpotent thickening of ${\mathrm{Spec} k(s) \hookrightarrow S}$.

Now, there is a maximal closed subscheme ${X' \subset X}$ such that ${f, \overline{f}}$ become equal on ${X'}$: this is the equalizer of ${f, \overline{f}}$, which can be represented by a suitable fibered product ${X \times_{X \times_Y X} X}$. What we have seen is that ${X \times_S S_0 \subset X'}$ (as closed subschemes) for any nilpotent thickening of ${\mathrm{Spec} k(s) \hookrightarrow S}$. Using the Krull intersection theorem, it follows that ${X'}$ must contain a neighborhood ${N}$ of the fiber ${X_s}$, which (by properness) we can assume to be of the form ${X \times_S U}$ for ${U \subset X}$ open containing ${s}$. From this, we want to conclude that ${X' = X}$.

So far, we find that ${f = \overline{f}}$ when restricted to ${X \times_S U}$ for ${U \subset S}$ an open subset containing ${s}$. Since ${U}$ is dense in ${S}$, it follows (by flatness) that ${X \times_S U}$ is dense in ${X}$, which means that the subscheme ${X'}$ must set-theoretically contain all of ${X}$. Using the artinian version of the rigidity lemma again, it follows that ${X'}$ must scheme-theoretically contain all of ${X}$. $\Box$

One of the intriguing things to me about this proof is how fundamentally scheme-theoretic it is; although the result is a nontrivial statement even in the category of varieties, the proof requires the use of nilpotents and infinitesimal neighborhoods to get at the global structure. It reminds me of the proof of the theorem of the cube. This is a statement about varieties and can be proved using only varieties (as in ch. 2 of Mumford’s book on abelian varieties), but the proof using a tiny bit of deformation theory is much simpler.

2. The deformation functor preserves products

In any event, we find from this that deformations of abelian varieties have no nontrivial (infinitesimal) automorphisms: in fact, we find that any endomorphism of an abelian scheme over a connected base which is the identity on one fiber is the identity. It follows that, given an abelian variety ${X/k}$, the functor

$\displaystyle F_1: \mathrm{Art}_k \rightarrow \textbf{Gpd}$

sending an abelian variety to its groupoid of infinitesimal deformations over an artin ring and infinitesimal isomorphisms, is actually discrete. This is intuitively at least unsurprising: an abelian variety is complex analytically a complex torus and isomorphisms morphisms between these are given by isomorphisms of lattices, which are very discrete.

Now, ${F_1}$ is product-preserving along surjections: this follows from the fact that if ${A', A'' \twoheadrightarrow A}$ are surjections in ${\mathrm{Art}_k}$, then ${\mathrm{Spec} A' \times_A A''}$ is obtained by “gluing” ${\mathrm{Spec} A', \mathrm{Spec} A''}$ along ${\mathrm{Spec} A}$. In other words, let for a ring ${R}$, ${\mathbf{Sch}_R^f}$ be the category of schemes flat over ${R}$. Then we have an equivalence of categories

$\displaystyle \mathbf{Sch}^f_{A' \times_A A''} \simeq \mathbf{Sch}_{A'}^f \times_{\mathbf{Sch}^f_A} \mathbf{Sch}_{A''}^f.$

In other words, to give a scheme flat over ${\mathrm{Spec} A' \times_A A''}$ is equivalent to giving a scheme flat over ${\mathrm{Spec} A'}$, a scheme flat over ${\mathrm{Spec} A''}$, together with an isomorphism between the restrictions to ${\mathrm{Spec} A}$ (in other words, this is the 2-categorical fiber product rather than the 1-categorical one).

In particular, to give a deformation of an abelian variety over ${ A' \times_A A''}$ is the same as giving a deformation over ${A' }$, one over ${A''}$, and an isomorphism on the restrictions to ${A}$. Since there are no infinitesimal automorphisms, though, we can say a little more: an isomorphism class of deformations over ${A' \times_A A''}$ is the same as giving an isomorphism class of deformations over ${A' }$, one over ${A''}$, which restrict to the same isomorphism class on ${A}$.

Corollary 10 The functor ${\mathrm{Art}_k \rightarrow \mathbf{Sets}}$ which sends an artinian ring ${R}$ to the set of isomorphism classes of deformations of an abelian variety ${X/k}$ over ${R}$ is product-preserving.

Using the general version of Schlessinger’s criterion (which we did not prove yesterday), it now follows that this functor is actually prorepresentable. In the next post, we’ll see that the functor above is formally smooth, so we can apply the weaker version of Schlessinger’s criterion proved last post to show that is prorepresentable by ${W(k)[[t_1, \dots, t_{g^2}]]}$.