bound on the genus of a fiber of the Albanese map of a surface with $h^1({\mathcal O})=1$?

2010-12-12T20:12:39Z

This is maybe more an open problem than a question, since I have seriously thought about it and asked several people working on algebraic surfaces with no success. I hope somebody here can suggest an approach different from the standard arguments in surface theory.

BACKGROUND: let $X$ be a smooth minimal complex projective surface of general type. An irrational pencil is a morphism with connected fibers $f\colon X\to B$, with $B$ a smooth curve of genus $b>0$.

For $b>1$, $X$ has at most finitely many pencils of genus $b$, having such a pencil is a topological property and it is possible to bound explicitly the genus of a general fiber of $f$ in terms of $K^2_X$ (Arakelov' theorem).

For $b=1$, namely for elliptic pencils, things are very different in general: a surface can have infinitely many such pencils, the genus of the general fibers of these pencils can be unbounded, and it is possible that a surface with an elliptic pencil deforms to a surface without elliptic pencils.

However, if $h^1({\mathcal O}_X)=1$ , then the Albanese map $a\colon X\to Alb(X)$ is an elliptic pencil, and for fixed $K^2$ the genus of a general fiber of $a$ is bounded, since the moduli space of surfaces with fixed $K^2$ is quasiprojective.

QUESTION: can one give a bound for the genus of the general fiber of the Albanese pencil of a minimal surface of general type $X$ with $h^1({\mathcal O}_X)=1$ in terms of $K^2_X$? Such a bound would be very interesting in the fine classification of surfaces of general type.

bound on the genus of a fiber of the Albanese map of a surface with $h^1({\mathcal O})=1$? - MathOverflow

bound on the genus of a fiber of the Albanese map of a surface with $h^1({\mathcal O})=1$?