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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.

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  • $\begingroup$ 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
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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 $P\subset \lvert G \rvert$.

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  • $\begingroup$ 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

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