Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are abelian varieties unobstructed? That is, given an abelian scheme $X \to \mathrm{Spec}R $ for $R$ a local artinian ring and $\mathrm{Spec} R \to \mathrm{Spec} S $ a nilpotent thickening, can we extend $X$ to an abelian scheme over $\mathrm {Spec }S $?

This assertion was made somewhat quickly in an article of Katz I was trying to read, at the bottom of p. 144. Mumford shows in his GIT book that it's enough to deform $X$ together with the identity section $\mathrm{Spec} R \to \mathrm{Spec } X$, and the abelian structure follows. I don't see how to get that, though.

share|improve this question
5  
This follows from a theorem of Grothendieck. See, for example, Theorem 2.2.1 of the article by F. Oort "Finite group schemes, local moduli for abelian varieties ...", Compositio Math. 1971. –  ulrich Jun 15 '12 at 3:38
    
One idea: use limit theorems and reduction mod primes to reduce to the case of char p residue field. Now the Abelian scheme is projective: for an ample on the closed fiber, a sufficiently high $\text{p}^{\text{th}}$ power will extend to your deformation over $R$. Take a general complete intersection curve $C$ of sufficiently ample divisors (on the dual) to embed into $\text{Jac}(C)$. Since $C$ deforms over $S$, also $\text{Jac}(C)$ deforms. Now you have the simpler problem of showing that for some deformation of $C$, also $X$ deforms as a subvariety of $\text{Jac}(C)$ (sheafy arguments). –  Jason Starr Jun 15 '12 at 12:58
    
Thanks! Oort's article is great. –  Akhil Mathew Jun 15 '12 at 14:42
    
Why does a general curve which is a complete intersection of very ample divisors let you imbed an abelian variety into the Jacobian? –  Akhil Mathew Jun 15 '12 at 14:42
    
@Akhil -- The question is whether the pullback map from Picard schemes of the (dual) Abelian variety to the Picard scheme of the curve is a closed immersion. First, the pullback map on $H^1(\mathcal{O})$ is injective using Serre vanishing, Serre duality and the usual (Koszul-type) resolution of the structure sheaf of a complete intersection subvariety. So the pullback map is unramified, hence has finite, 'etale kernel. Since elements in the kernel are torsion, vanishing of the kernel reduces to a problem about surjectivity of pushforward on $\pi_1$, which follows by Bertini's theorem. –  Jason Starr Jun 15 '12 at 15:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.