# Very ample linear systems - intersections with multiplicity >1

On a degree $$n$$ Hirzebruch surface $$F_n$$, suppose we have a very ample linear system. It is known that its generic smooth irreducible members give a Lefschetz pencil on $$F_n$$. Let us take a member, $$G$$, in this pencil. And suppose we know that $$G$$ intersects the fiber, $$F$$, of $$F_n$$ $$m$$ times. Generically they intersect at $$m$$ points.

Is it possible to move $$G$$ in a way that $$G$$ and $$F$$ intersect at 1 point with multiplicity $$m$$? I don't want to change the genus of $$G$$ while moving. Or is it possible to find $$G'$$ in the pencil which intersects $$F$$ in the desired way?

I am not an algebro-geometer so, sorry if this question is trivial. I'd appreciate any suggestions. Thanks.

• Welcome new contributor. The Picard group of $F_n$ is generated by the divisor class of $S$ together with the divisor class of a "fiber", i.e., a rational curve of self-intersection $0$ having intersection number $1$ with $S$. Relative to these two generators, could you please tell us the coefficients of the divisor class $G$? As near as I can tell, you have not yet specified enough information to identify the divisor class of $G$. – Jason Starr Jan 23 '19 at 21:23
• Thanks. Let divisor class of a fiber be F. Then divisor class of G is 2S+5F in second degree Hirzebruch surface $F_2$. Then it intersects S once, but intersects F twice. I was trying to ask more general question. I have edited my question above. – apm Jan 24 '19 at 11:07
• In the case in your comment, the cohomology group $H^1(\mathbb{F}_n,\mathcal{O}(G-S))$ does vanish. Thus, the answer below by @abx settles the case in your comment. – Jason Starr Jan 24 '19 at 12:56

A partial answer (too long for a comment): suppose $$G$$ is sufficiently ample so that $$H^1(\mathbb{F}_{n},\mathcal{O}(G-S))=0$$. Then the restriction map $$H^0(\mathbb{F}_{n},\mathcal{O}(G))\rightarrow H^0(S,\mathcal{O}(G)_{|S})$$ is surjective. This shows that:
1) There exists $$G'\in \lvert G \rvert$$ such that $$G\cdot S=4s$$, for any $$s\in S$$;
2) There will be no such $$G'$$ in a general pencil $$P\subset \lvert G \rvert$$.
• "$4s$" means the point $s$ counted with multiplicity 4. Since $S$ is a rational curve, all effective divisors of degree 4 form a unique linear system; the surjectivity implies that any divisor in this system is cut out by some $G'\in\lvert G\rvert$. And yes, the same argument works with $F$ -- you need $H^1(\mathcal{O}(G-F))=0$. This works in your particular case, since $G-F\equiv 2S+4F\equiv -K$. – abx Jan 24 '19 at 14:34
• Thank you very much. One last question; F and S intersect at one point, say $p$. And from your answer we know that there is some $G’$ in the pencil such that $G' \dot F = 2f$ and also $G’ \dot S = s$, $f \in F, s \in S$. Does there exist a member $G''$ of my pencil for which $p, f$ and $s$ are the same point? I.e., $F, S, G’'$ intersect at the same point with different multiplicities? Thanks for your answers! – apm Jan 24 '19 at 15:04