# Variation of the Albanese map

Let $S$ be an irregular surface of general type over $\mathbb{C}$ and $a \colon S \to A:=\textrm{Alb}(S)$ be its Albanese map. Let Def($S$) and Def($A$) be the bases of the Kuranishi family of $S$ and $A$, respectively. Then $a$ induces a map $f \colon \textrm{Def}(S) \to \textrm{Def}(A)$, whose differential is $f_* \colon H^1(S, T_S) \to H^1(A, T_A)$.

Since every small deformation of $S$ is again a surface of general type, if Def(S) is generically smooth then the image of $f_*$ is contained in the subspace of dimension $\frac{g(g+1)}{2}$ of $H^1(A, T_A)$ corresponding to the algebraic deformations of $A$ (here $g := \dim(A)$ ).

Is this still true when Def($S$) is not generically smooth, i.e. everywhere non-reduced? I suspect that the answer should be "yes", but I would like to see a rigorous proof (or a counterexample, if my guess is wrong).

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Ehm, I'm new in MathOverflow and unintentionally I deleted a comment trying to answer it. Sorry! Anyway, the answer was the following. Yes you are right, I've been sloppy in defining $f$ . Anyway, you can also define $f_*$ directly, considering the map induced in cohomology by the pullback, namely $H^1(S, T_S) \to H^1(S, a^*T_A)$, and then using the fact that $T_A$ is trivial and that $H^1(S, \mathcal{O}_S)=H^1(A, \mathcal{O}_A)$. Your translation seems to me correct; moreover, since S$is assumed to be a surface of general type, its first-order analytic and algebraic deformations coincide – Francesco Polizzi Jul 9 '10 at 15:01 @Francesco: you didn't delete my comment, I deleted it upon seeing that Angelo had given an answer ("projective" seems avoidable, and in general settings one shouldn't expect all first-order deformations to be projective). Actually, for the equality of first-order analytic and algebraic deformations of a smooth proper$\mathbf{C}\$-scheme, can't we just make an application of GAGA, and so not require mention of "general type", etc.? –  BCnrd Jul 9 '10 at 15:15
You are right, I wrote "first-order deformations" but I was actually thinking of the semiuniversal formal deformation, that can be actually non algebraizable (e.g. for K3 surfaces or complex tori). –  Francesco Polizzi Jul 9 '10 at 15:35

## 1 Answer

Yes, this is true. The point is that the Albanese is well defined in families, as a family of abelian varieties; that is, given a flat family of smooth projective varieties, there is a projective smooth family of Albanese varieties over the same base. This family is the dual of the family of Pic^0, which is well known to be smooth and projective.

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Thank you for the very clear answer! –  Francesco Polizzi Jul 9 '10 at 14:17
In char. 0 we'll often have non-projective algebraic deformations of general projective varieties, so may be worth noting that Angelo's nice answer still works in those cases, since the story of Pic and duality is OK with "projective" relaxed to "proper" (at the expense of more work in that story). –  BCnrd Jul 9 '10 at 15:19
Right, good remark. –  Angelo Jul 9 '10 at 19:03