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In viewing the tags about ample bundle with no global sections I found an example below:

If $C$ is a curve of genus $2$, and $p,q,r$ are general points on $C$, then the bundle $L=\mathcal{O}_C(p+q-r)$ is ample, but has no global sections at all.

But I don't know how to verify this? How is 'general' used here?

(Also I think it is right if we give the divisor consisting of a single point on a irrational curve. )

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    About your last remark: the divisor consisting of a single point on a curve does have a global section. But if the curve is irrational, the space of sections is 1-dimensional, so it is not very ample. –  Apr 22 '14 at 13:05

1 Answers1

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Let $\sigma$ be the hyperelliptic involution on $C$. Suppose that $q\ne \sigma(p)$, then I claim that $H^0(C, L)=0$ for all $r$.

Indeed, the hypothesis on $q$ implies that $L(p+q):=H^0(C, O_C(p+q))=k$ (the base field). This is a basic property of the Weierstrass points on $C$. So if $H^0(C, L)\ne 0$ for some $r$, then the inclusion $L(p+q-r)\subseteq L(p+q)$ is an equality by comparing the vector dimensions. In other words, any global section $s\in L(p+q)$ vanishes at some point of $C$, hence $s=0$ (because $s$ is constant !) and $L(p+q)=0$, contradiction.

Note that if $q=\sigma(p)$, then $p+q-r=p+\sigma(p)-r \sim \sigma(r)$, so $H^0(C, L)\cong H^0(C, O_C(\sigma(r)))\ne 0$. This explains why we need $p, q, r$ be "general". This also shows that $H^0(C, L)\ne 0$ if and only if $q=\sigma(p)$.

  • Can you give a reference on the property of Weierstrass points or give more details? Thanks a lot! –  Apr 23 '14 at 04:02
  • If $L(p+q)$ has dimension bigger than $1$, then there is a non-constant rational function $f\in L(p+q)$. The morphism $f: C\to \mathbb P^1_k$ has degree two and $p+q=f^*(\infty)$. This implies that $q=\sigma(p)$ because $f$ is automatically the unique $2:1$ morphism from $C$ to $\mathbb P^1_k$. Weierestrass points are those $p$ such that $p=\sigma(p)$ in the particular case of hyperelliptic curves, more over, for any $p$, $p+\sigma(p)$ is equivalent to $2q$ for any Weierestrass point $q$. –  Apr 23 '14 at 09:56
  • An element of $L(p+q)$ is a rational function with $\operatorname{div} (f) \geq -p-q$ does it imply $p+q=f^*(\infty)$? (Sorry I am a beginner..) –  Apr 23 '14 at 15:06
  • Yes, in fact the negative (pole) part of div$(f)$ is exactly $-p-q$ because otherwise $f$ would be a rational function in $L(p)=k$ or $L(q)=k$. –  Apr 23 '14 at 15:13