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Let $X$ be a smooth projective rational surface over an algebraically closed field of characteristic zero, and $D$ a divisor on $X$ such that $D$ is nef and $D^2 = 0$ with the following properties:

  • $h^0(X,hD) = 0$ for $1\leq h\leq h_0-1$;
  • $h^0(X,m(h_0D)) = 1$ for all $m \geq 1$;
  • $h^0(X,kD) = 0$ if $k$ is not a multiple of $h_0$.

Is there any example of a surface $X$ and a divisor $D$ with this behavior?

Thank you.

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2 Answers 2

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Yes, for $h_0=2$, I believe the following construction works.

Let $C$ degree 8 curve in $P^2$ with 16 nodes $p_1,\ldots,p_{16}$ as its singular locus. Let $X$ be the blow-up of $P^2$ along the 16 points, and let $D=4H-E_1-\ldots-E_{16}$. Then $2D$ is represented by an irreducible, smooth curve $\tilde C$ on $X$. Since $D^2=0$, $D$ is nef. If $C$ is generic among 16-nodal curves, one easily checks that the 16 points do not lie on a quartic curve, so $h^0(X,D)=0$. In this generic situation, there should be no relation between the points $p_1,\ldots,p_{16}$ on the curve $\tilde C$ and the hyperplane section $H$, so $$H^0(\tilde C,\mathcal O_{\tilde C}(mD))=0$$ for all $m\ge 1$. Then by the exact sequence $$ 0\to \mathcal O_{X}((m-2)D)\to \mathcal O_{X}(mD)\to \mathcal O_{\tilde C}(mD)\to 0 $$we get the three properties you want.

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  • $\begingroup$ To show that $D$ is nef shouldn't one prove that $D\cdot C\geq 0$ for all curves $C\subset X$? Also, how do you check that if $C$ is general among the curves of degree $8$ with $16$ nodes then there is no quartic through the nodes? $\endgroup$
    – Puzzled
    Commented Sep 13, 2021 at 23:02
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    $\begingroup$ Note that if $C, C'$ are distinct irreducible curves, then $C\cdot C'\ge 0$ by the definition of the intersection product on surfaces. Since $2D$ is represented by an irreducible curve, the only curve it could potentially intersect negatively would be itself; but then $D^2=0$ shows that this doesn`t happen, so $D$ is nef. $\endgroup$
    – Ennio Mori cone
    Commented Sep 14, 2021 at 21:05
  • $\begingroup$ One way would be to exhibit a single example of an degree 8 curve and 16 points with this property. I used a degree 8 curve obtained by a generic map $P^1\to P^2$ given by three degree 8 polynomials, computed the singular locus by hand, and saw that it was defined by forms of degree $\ge 7$. $\endgroup$
    – Ennio Mori cone
    Commented Sep 14, 2021 at 21:07
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An example is given by the canonical class $K_S$ of a bielliptic surface $S$, which is torsion of order $h_0=2, 3, 4$ or $6$. For the definition of a bielliptic surface see for instance Chapter VI of Beauville's book "Complex algebraic surfaces".

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  • $\begingroup$ I think the OP wanted an example with a rational surface - this makes it a little bit trickier. $\endgroup$
    – Ennio Mori cone
    Commented Oct 4, 2021 at 9:51
  • $\begingroup$ You are right, I had not read the question carefully. Thanks for pointing this out! $\endgroup$
    – rita
    Commented Oct 17, 2021 at 14:32

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