Intersection pairing on the Picard group of a surface ($\mathrm{Pic}(X)\times \mathrm{Pic}(X)\to \mathbb{Z}$)

On a smooth projective surface, two line bundles are represented by divisors $D$ and $E$, and their intersection is the degree of the zero-cycle obtained after moving them into proper position. It is a symmetric bilinear form on the Picard group, and $D\cdot E$ equals the degree of ${\mathcal{O}}_X(D)$ restricted to $E$ when $E$ is an integral curve not contained in $D$.

For a smooth projective surface and an ample divisor $A$, the intersection pairing on the Picard group of a surface is negative definite on the orthogonal complement of $[A]$ in $N^1(X{)}_{\mathbb{R}}$. Its signature is $(1,\rho -1)$. This is an algebraic statement valid in arbitrary characteristic, distinct from its analytic formulation for compact Kähler surfaces.

Let a surjective morphism from a smooth projective surface to a projective surface contract an irreducible curve $C$. Pull back an ample divisor $H$ from the target to obtain $L$. The projection formula for algebraic cycles gives $L\cdot C=0$ and $L^2=\mathrm{deg}(f)H^2>0$. The Hodge index theorem for algebraic surfaces makes the intersection form negative definite on the orthogonal complement of this positive-square class. Since an ample divisor on the source meets $C$ positively, its numerical class is nonzero, and $C^2<0$.

If $D$ is nonzero numerically with $D^2=0$ and $D\cdot A>0$ for ample $A$, then $D\cdot E=0$ implies $E^2\le 0$, with equality only when $[E]$ is proportional to $[D]$. To see this, use the Hodge index theorem for algebraic surfaces to split off the positive $A$ direction and diagonalize the remaining negative definite form. Orthogonality to a nonzero isotropic vector then leaves a negative semidefinite form whose radical is precisely that vector's span.

For the Hirzebruch surface ${\mathbb{F}}_n$, the negative section of a Hirzebruch surface $S$ and a fiber class of a Hirzebruch surface $F$ freely generate $\mathrm{Pic}({\mathbb{F}}_n)$ and satisfy