Maximal dimension of linear system of curves of fixed genus on a surface - MathOverflow

Maximal dimension of linear system of curves of fixed genus on a surface

2010-11-11T19:42:07Z

To a (projective smooth) algebraic surface $S$ over an algebraically closed field and a divisor $D$ of $S$, we can associate $n= \dim |D|$ and $g$, the genus of a generic member of $|D|$. I would like to fix $g$ and vary $S,D$ so as to make $n$ as large as possible. Is $n$ unbounded and, if not, what is the optimal bound? I am mainly interested in the case of positive characteristic but an answer over the complex numbers would be welcome too.

Answer by Francesco Polizzi for Maximal dimension of linear system of curves of fixed genus on a surface

2010-11-11T21:26:26Z

The following example shows that $n$ can be unbounded, for any value of $g$, even if $S$ is fixed.

Let $C$ be a smooth curve of genus $g$, and let $\mathcal{E}$ be a normalized rank $2$ vector bundle on $E$. Let $\mathfrak{e}$ be the divisor on $C$ such that $\bigwedge^2 \mathcal{E}=\mathcal{O}_C(\mathfrak{e})$, and consider the projective bundle

$S:=\mathbb{P}(\mathcal{E}) \stackrel{\pi}{\longrightarrow} C$.

Let $C_0$ be a section such that $C_0^2 = \deg \mathfrak{e}$, and let $f$ be the class of a fiber of $\pi$.

Assume now that $\mathfrak{b}$ is any divisor on $C$ having the following properties:

$\mathfrak{b}$ is nonspecial;
$|\mathfrak{b}|$ and $|\mathfrak{b}+ \mathfrak{e}|$ have no base points.

Notice that these conditions are satisfied as soon as $\mathfrak{b}$ is a general divisor of sufficiently high degree.

By [Hartshorne, Algebraic geometry, Ex. 2.11 p. 385], there exists a section $D$ linearly equivalent to $C_0 + \mathfrak{b} f$, and moreover $|D|$ is base-point free. By Bertini's theorem it follows that the general element of $|D|$ is a smooth curve of genus $g$, and the dimension of $|D|$ clearly goes to infinity when $\deg \mathfrak{b}$ goes to infinity.

Answer by rita for Maximal dimension of linear system of curves of fixed genus on a surface

2010-11-11T21:38:04Z

$n$ is unbounded, as it is shown by the following example.

EDIT: my example did not work, as pointed out by quim in the comments. The example he suggests however works: take $S$ the blowup of $P^2$ at a point $x$ and $D$ the strict transform of a curve of degree $d$ with a singular point of multiplicity $d-1$ at $x$. The general $D$ is smooth and $|D|$ has dimension $2d$.\ Of course, a similar construction can be used to construct examples with $g>0$.

On the other hand, if the Kodaira dimension of $S$ is $\ge 0$ and $D$ is irreducible, then $n=\dim|D|$ is bounded by $g$ by the following argument.

Up to blowing up $S$ we may assume that the general $D$ is smooth. Let $m>0$ be such that $mK_S\ge 0$. If $n>0$, then $D$ is not in the fixed part of $|mK_S|$, hence $K_SD\ge 0$. Hence by the adjunction formula $D|_D$ is a divisor of $D$ of degree $\le 2g-2$ and therefore it satisfies $2\dim|D|_D|\le \deg D$ (this is Clifford's theorem if $D$ is special and it is trivially true otherwise).
So we have $\dim |D|\le \dim|D|_D|+1\le D^2/2+1\le (D^2+K_SD)/2+1=g$.

I, hence $K_S|_D$ is effective and $D|_D$ is special. Then Clifford's theorem and the adjunction formula give $2(n-1)\le D^2\le D^2+K_SD=2g-2$.