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

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.

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 does X = S? hmmm? (15 character lower limit) – roy smith Dec 12 2010 at 20:39 Yes, thanks! I've fixed it. – rita Dec 12 2010 at 20:47 I GUESS in some sense you can make it. For example, you can consider the canonical map of the surface, which will work if $K_S^2$ is not so small. Assume that the fiber of the Albanese map is $F$, then consider whether $\mathscr{O}(K_S-F)$ has global section or not. If it has a global section, then after computing some intersection number, maybe you can have a bound of $g(F)$. – Michael Zhang Mar 28 2011 at 4:42 If $K_S-F$ is effective, than of course I have a bound since $K_S(K_S-F)\ge 0$ and $K_SF=2g(F)-2$. And by the same argument, I would have a bound if I could determine explicitly an $m$ such that $mK_S-F$ is effective, but that's precisely what I don't know how to do. – rita Mar 29 2011 at 18:26