# Vanishing of a section of a vector bundle at more than e/n points

This is from the beginning of the first section of the paper "Second Chern class and Riemann-Roch for vector bundles on resolutions of surfaces singularities" by J. Wahl.

$$C$$ is a smooth projective curve over $$\mathbb{C}$$. $$E$$ is a vector bundle of degree $$e$$ and rank $$n$$ (I believe $$n$$ should be at least $$2$$). If $$E$$ is semi-stable, then $$h^0(E) \leq e+n$$; otherwise one can find a section of $$E$$ vanishing at more than $$\frac{e}{n}$$ points, producing a line subbundle of $$E$$ of slope $$> \frac{e}{n}$$.

So I think it means that if $$h^0(E) > e+n$$, then there is a divisor $$D$$ of degree $$\frac{e}{n}$$ such that $$h^0(E \otimes \mathcal{O}(-D)) > 0$$. I don't know how to show it. It seems that Riemann-Roch is not enough.

• Let $p$ be the smallest integer $>\frac{e}{n}$; then $np-e\leq n$. If $x$ is a point of $C$, $h^0(E(-x))\geq h^0(E)-n$. So for any effective divisor $D$ of degree $p$, $h^0(E(-D))> e+n-np\geq 1$.
– abx
Dec 7 '20 at 14:40

One chooses the smallest integer $$d$$ such that $$e-nd<0$$. Note that $$nd \le e+n$$.
Then if $$D$$ is effective of degree $$d$$, $$h^0(E(-D))=0$$ since $$E(-D)$$ is semistable with negative degree.
Since $$h^0(E)\le h^0(E(-D))+nd$$ (by using the exact sequence $$0\to \mathcal O(-D)\to \mathcal O \to \mathcal O_D\to 0$$ tensorized by $$E$$), the result follows from the inequality $$nd\le e+n$$ observed above.