Does the line bundle have global sections?

Question

Let $X$ be a smooth projective surface over $\mathbb{C}$. Let $H$ be an ample divisor on $X$. Consider a line bundle $N$ such that $N\cdot H>0$ and $H^2(N)=0$. Is it possible that $H^0(N)=0$? I feel it is not possible because $N\cdot H>0$ should mean $N$ is effective.

Answer 1

Yes, it is possible. Let me start by cooking up an example and then I will address your last sentence.

Let $X$ be a quartic in $\mathbf P^3$ containing two disjoint lines $L_1$ and $L_2$. (Note that $X$ is a K3 surface.) Let $N$ be the line bundle associated to the divisor $2L_1-L_2$. Let $H$ be the restriction of the hyperplane class to $X$. Then:

• $H \cdot N =1$
• $H^0(N)=0$
• $H^2(N)=H^0(N^\ast)=0$.

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As for: "I feel it is not possible because $N⋅H>0$ should mean $N$ is effective.":

This inequality is certainly a necessary condition for a (nontrivial) line bundle to be effective, but it is (almost always) not sufficient.

There is an vector space $N^1(X)$ of numerical classes of line bundles. For a given line bundle $H$ the condition $N \cdot H >0$ defines an open half-space $H_{>0}$ inside $N^1(X)$. If $H$ is ample then the numerical class of any effective line bundle must be in $H_{>0}$. But in fact the set of numerical classes of effective line bundles is much smaller: it is a strictly convex cone inside $H_{>0}$, called the effective cone. In general the structure of this cone can be very complicated.

A good reference to learn more is Volume I of Positivity in Algebraic Geometry by Lazarsfeld.