Assume $C$ is a smooth projective curve and $L$ is a line bundle on it. We say $L$ is $p$-very ample if for any effective divisor $D$ of degree $p+1$, the evaluation map $H^0(C,L)\to H^0(C,L\otimes\mathcal{O}_D)$ is surjective. Kemeny wrote in his paper "The extremal secant conjecture for curves of arbitrary gonality" that it is "rather straightforward" that if $L$ is not $p+1$-very ample then the Koszul cohomology $K_{p,2}(C;L)\not=0$. But I cannot see this, at least from the algebraic definition of the Koszul cohomology.

Is there any geometric explanation for Koszul cohomology, in which $p+1$-very ampleness might be involved?

  • 1
    $\begingroup$ You will have to do some digging yourself, but there is a paper of Green and Lazarsfeld in which they show how to construct non-zero Koszul cohomology classes from points in special position w.r.t a line bundle. 'not p+1 very ample' means that there are p+1 points in special position relative to L. The construction of the class is explicit and imho, geometric. Happy hunting. $\endgroup$
    – meh
    Nov 17 '19 at 17:47
  • $\begingroup$ Hi ! This is a result from Koh-Stillman's paper "Linear Syzygies and Line Bundles on an Algebraic Curve", see Prop. 3.6. You can also look at Chapter 4.4 of Aprodu-Nagel's book "Koszul Cohomology and Algebraic Geometry", in particular Theorem 4.36. $\endgroup$
    – mkemeny
    Nov 24 '19 at 4:00
  • $\begingroup$ @mkemeny But those propositions induce that $L$ does not satisfy $N_p$(in fact $N_{p-1}$ in my situation). This doesn't imply the nonvanishment of $K_{p,2}$, right? $\endgroup$
    – Li Li
    Dec 12 '19 at 23:57
  • $\begingroup$ These two conditions are basically the same in our setting, see the top of page 8 of arxiv.org/pdf/1408.4164.pdf $\endgroup$
    – mkemeny
    Dec 14 '19 at 0:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.