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.

  • $\begingroup$ 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 $G$. $\endgroup$ – Jason Starr Jan 23 at 21:23
  • $\begingroup$ 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. $\endgroup$ – apm Jan 24 at 11:07
  • $\begingroup$ 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. $\endgroup$ – Jason Starr Jan 24 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 $P\subset \lvert G \rvert$.

  • $\begingroup$ Dear abx, what did you mean by 4s, for any s in S? Does it mean an intersection point with multiplicity 4? Also why does the surjectivity of that map implies the existence of such G’, could you suggest a reference? $\endgroup$ – apm Jan 24 at 14:11
  • $\begingroup$ And, does the same argument work with the fiber class F, not with the section S? (I have edited the question, sorry for the confusion) Thanks! $\endgroup$ – apm Jan 24 at 14:22
  • $\begingroup$ "$4s$" 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 $G-F\equiv 2S+4F\equiv -K$. $\endgroup$ – abx Jan 24 at 14:34
  • $\begingroup$ 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! $\endgroup$ – apm Jan 24 at 15:04

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.