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Let S be a surface and L be a line bundle on S. For any zero-dimensional closed subschemes x of S, there is natural map from global sections of L to the global sections of L restricting to x (which is a (r+1)-dimension vector space. ). For any positive integers r, the line bundle L on a surface S is called r-very ample if for any length r+1 zero-dimensional closed subschemes x of S, this map is surjective. For example, very ampleness implies 1-very ampleness. If L and K are both very ample then tensoring l copies of L and k copies of K is (k+l)-very ample.

My question is, for every positive integer r, can you find an algebraic K3 surface with Picard number 1 such that the primitive ample line bundle is r-very ample? Moreover, can you find infinitely many line bundles of this kind whose self-intersection numbers are all different?

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  • $\begingroup$ I don't understand something: why do you expect the self intersection numbers of these line bundles to be different ? $\endgroup$
    – David Lehavi
    Commented Oct 22, 2009 at 13:52

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In Beauville's "Counting rational curves on K3 surfaces" is implictly assumed the existence of algebraic K3 surfaces with Pic of rank one and generated by a curve of genus g.

How do we show the existence of such K3 surfaces ? Edit: See Ferreti's first comment below for an answer.

Using the argument pointed out by Ferreti in his first comment and taking care to avoid the difficulty pointed out in his second comments, we can reduce the existence of the sought K3 surfaces to the following statement:

There exists infinitely many integers k for which (2k)(2k +3) is squarefree.

Start with a smooth quartic S in P(3) and let H be a hyperplane section. For a fixed r and k>>0, the restriction of kH to S is r-very ample.

Suppose S contains a line L. Then the linear system |E|=|H-L| defines a fibration by elliptic curves on S. Thus kH + E is also r-very ample.

Let SS be a family of K3 surfaces that deforms S in such a way that the class of O(k) is preserved, and for a generic member of the family every element in H1,1\cap H^2(Z) not proportional to O(k) becomes non-rational. Thus the generic element has Pic = Z. Since r-very ampleness is an open condition ( the points in the relative Hilb^r(SS) where it does not hold is closed) we obtain a K3 surface with Pic = Z and a r-very ample line-bundle of self-intersection 4k^2 + 6k = 2k(2k +3). If this number is squarefree then the line bundle is primitive.

After googling a bit I found general results about squarefree values of polynomials which seems to ensure the existence of infinitely many integers k for which 2k(2k +3) is squarefree.

Edit: I would like to know if it is necessary to impose the number theoretical condition to obtain primitiviness.

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    $\begingroup$ First you can show the existence of algebraic K3 surfaces such that the generic hyperplane section is a smooth curve of genus g. This is done in Beauville's book on algebraic surfaces Then you study deformations using the local Torelli theorem. Deformations which keep a given cohomology class of type (1,1) are locally an hypersurface in the deformation space. So you can keep the class of your curve of type (1,1) and kill all other cohomology classes, if any. Use exponential sequence and Riemann-Roch to check that this is still the class of a curve, so the Picard group has now rank 1. $\endgroup$
    – Andrea Ferretti
    Commented Oct 21, 2009 at 13:54
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    $\begingroup$ There is a slight problem, which I should have pointed out before if comments could be larger. In Beauville's construction you can always take the genus g curve to be primitive, and it will remain primitive when you take the deformation. But your construction does not yield a primitive vector, since O(k) = k O(1). $\endgroup$
    – Andrea Ferretti
    Commented Oct 21, 2009 at 16:35
  • $\begingroup$ More simply, the only nontrivial common factor between 2k and 2k+3 can be 3. So if the product is a square, either both are squares (in which case 2k+1, so no solution here) or else 2k = 3 a^2 2k +3 = 3 b^2 But then 6k and 6k + 9 are both squares, which has no solution again. So I would say that number is never a square. $\endgroup$
    – Andrea Ferretti
    Commented Oct 22, 2009 at 12:42
  • $\begingroup$ I hope it's readable; line breaks were lost :-( $\endgroup$
    – Andrea Ferretti
    Commented Oct 22, 2009 at 12:43
  • $\begingroup$ @Andrea: I agree that it is never a square, but I don't see how it implies the primitiviness. If the primitive curve has self-intersection N then its multiples will have self-intersection N k^2. $\endgroup$
    – Jorge Vitório Pereira
    Commented Oct 22, 2009 at 14:45

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