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In Beauville's "Complex Algebraic Surfaces", given an elliptic surface $f : X \to C$ with a generic fiber $E$. Then either $\text{Alb}(X) \cong \text{Jac}(C)$ or there is an exact sequence of abelian varieties

$0 \to F \to \text{Alb}(X) \to \text{Jac}(C) \to 0$

with $F$ being isogenous to $E$.

Assume that $X$ is a properly elliptic that is in the second case. Does anyone know an example where $F$ is isomorphic to $E$ (and $X$ is not a product), and an example where $F$ is not isomorphic to $E$?

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This is a hint rather than a solution: consider (twisted) products. –  Donu Arapura Oct 22 '10 at 18:09
    
ops, I forgot I had examples in the case of Kodaira dimension $-\infty$, I just wonder about the properly elliptic case. –  Tuan Oct 23 '10 at 0:16

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up vote 8 down vote accepted

Donu Arapura is right:

If you want $F$ isomorphic to $E$ you can do the following: Consider the product $E \times P^1$. You find $F \simeq Alb(X) \simeq E$.

If you want $F$ and $E$ to be non isomorphic, one can do as follows. Let $c$ be a 2-torsion point of $E$. Form the quotient $X$ of $E \times E$ by the involution defined by $(x, y) \mapsto (-x, y + c)$.

The first projection induced an elliptic fibration $X \to \mathbb P^1$. A generic fibre is $E$. However, the albanese of $X$ is $E'$ = $E$/{0, $c$}.

The statement about the albanese follows by considering the elliptic fibration $X \to E'$ given by the second projection, and the statement of Beauville.

So let us continue this discussion with examples of Kodaira dimension one:

Take an elliptic curve $E$ and a curve $C$ with $g(C) \ge 2$. Then $E \times C \to C$ is an example with $E$ and $F$ being isomorphic.

Now let $C$ be an hyperelliptic curve with involution $i \colon C \to C$. Let $E$ be elliptic, and let $b \in E$ be a 2-torsion point. We define $X$ to be the quotient of $E \times C$ by the involution given by $(x, y) \mapsto (x + b, i(y))$. As before, the fibration $X \to \mathbb P^1$ gives an example for $E$ not isomorphic to $F$.

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so I suppose that when $X$ is not a product $E\times C$, it's not expected that $E$ and $F$ are isomorphic then. –  Tuan Oct 25 '10 at 17:48

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