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Sorry for the vague title. This question is about the Albanese map from the variety $M$ of canonically polarized varieties to the set of abelian varieties. (The variety $M$ is not of finite type...)

Let $A$ be an abelian variety of dimension $g$. For which integers $N$ does there exist an $N$-dimensional subvariety $X$ of $M$ which maps to $A$ under the Albanese map?

In words, when does the fibre of $A$ under the Albanese map contain an $N$-dimensional variety? (In such a subvariety I want the Hilbert polynomial of the varieties to be constant.)

The question is even interesting for $g=0$. In this case, I am asking for big families of canonically polarized varieties with zero Albanese. I am sure one can find families of canonically polarized surfaces with trivial Albanese, but I don`t know of any explicit examples.

Can anyone provide a smart construction? I only know of the following example (and some of its generalizations).

Example Starting from an abelian surface $A$, one can consider double covers $X_D\to A$ ramified over precisely one smooth ample divisor $D$ on $A$. Varying the ample divisor gives a positive-dimensional family $(X_D)$ of canonically polarized surfaces with Albanese $A$. The hilbert polynomial of these guys is all different though, but you can get it to be constant by sticking with ample divisors $D$ on $A$ with the same self-intersection.

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1 Answer

up vote 8 down vote accepted

Smooth hypersurfaces of fixed degree $d$ in $\mathbb P^n$ ($n\ge 3$) are simply connected, so they have trivial Albanese, and are of general type for $d>n+1$. The Hilbert polynomial is determined by $d$.

To get families of polarized varieties with fixed Albanese, just take products $X\times Y$, where $X$ varies in a family as above and $Y$ is a fixed variety of general type.

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This is nice. Thank you. I assume that one takes the variety Y to have Albanese A, am I right? But why does such a can pold variety exist? –  Fabiano Rug Jun 4 '13 at 21:42
To construct $Y$ with Albanese variety $A$ for instance you can take a cover as described in your answer (in this case $\dim Y=g$), or you can take a complete intersection of very ample divisors inside $A$. Of course these are not all the possibilities, just the simplest ones. –  rita Jun 5 '13 at 6:49
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