3.3 The Hodge Index Theorem
In this section, \(X\) is a nonsingular projective surface. For \(\mathcal{F}\) a coherent sheaf (resp. for \(D\) a divisor) on \(X\) and \(i\in \mathbb {Z}\), we write \(h^i(\mathcal{F})\) (resp. \(h^i(D)\)) for \(\dim \operatorname{\mathrm{H}}^i(X,\mathcal{F})\) (resp. \(\dim \operatorname{\mathrm{H}}^i(X,\mathcal{O}_X(nD))\)). Write \(K\) for the canonical bundle of \(X\), i.e., \(K=\omega _X\).
Suppose \(H,D\in \operatorname{\mathrm{Div}}X\) with \(H\) very ample and \(D\cdot H {\gt} K\cdot H\). Then \(h^2(D) = 0\).
By Serre duality, we have \(h^2(D) = h^0(K-D)\), and \(h^0(K-D)\neq 0\) if and only if \(K-D\sim E\) for some effective divisor \(E\). So assume \(h^2(K-D)\neq 0\). So
Suppose \(H,D\in \operatorname{\mathrm{Div}}X\) with \(H\) very ample, \(D^2 {\gt} 0\) and \(D\cdot H\neq 0\). Then
\(D\cdot H {\gt} 0 \Leftrightarrow \lim _{n\to \infty } h^0(nD) = \infty \).
\(D\cdot H {\lt} 0 \Leftrightarrow \lim _{n\to -\infty } h^0(nD) = \infty \).
It suffices to prove the first statement. Suppose then that \(D\cdot H {\gt} 0\). Then, by Lemma 3.11, \(h^2(nD) = 0\) for \(n\gg 0\). Therefore, by the Riemann-Roch
It follow that \(\lim _{n\to \infty } h^0(nD) = \infty \).
On the other hand, suppose \(\lim _{n\to \infty } h^0(nD) = \infty \). Then there exists an \(n{\gt}0\) and an effective divisor \(E\) such that \(nD\sim E\). But then \(nD\cdot H = E\cdot H \geq 0\). So \(D\cdot H \geq 0\) as well, and, since we are assuming that \(D\cdot H\neq 0\), we must have \(D\cdot H {\gt} 0\).
Suppose \(D,H\in \operatorname{\mathrm{Div}}X\) with \(H\) very ample, and suppose that \(D\cdot H = 0\). Suppose further that \(D\) is not numerically trivial. Then \(D^2 {\lt} 0\).
Suppose to get a contradition that \(D^2\geq 0\).
If \(D^2 {\gt} 0\), then set \(H(n) = D + nH\). Then \(H(n)\) is ample for \(n\gg 0\), and \(D\cdot H(n) = D^2 {\gt} 0\). It follows from Corollary 3.12 that, \(\lim _{k\to \infty } h^0(kD) = \infty \). But then there exists \(k\gg 0\) and \(E\sim kD\) such that \(E{\gt}0\). So \(E\cdot H{\gt}0\). But this implies that \(D\cdot H {\gt} 0\), which is a contradiction.
On the other hand, suppose \(D^2 = 0\). Since \(D\not\equiv 0\), there exists a curve \(C\) such that \(D\cdot C \neq 0\). Set \(E := (H^2) C - (C\cdot H) H\). Then \(D\cdot E = (H^2) D\cdot C \neq 0\), while \(E\cdot H = 0\). So set \(D' = nD + E\) for \(n\in \mathbb {Z}\). We get that \(D'\cdot H = 0\), while \((D')^2 = 2n (D\cdot E) + E^2\). So by taking a suitable \(n\in \mathbb {Z}\), we can guarantee that \((D')^2 {\gt} 0\) and \(D'\cdot H = 0\). But this contradicts what we have just proved.
Suppose \(X\) is a surface and \(E\in \operatorname{\mathrm{Div}}X\) is a divisor with \(E^2 {\gt} 0\). Then, for every \(D\in \operatorname{\mathrm{Div}}X\) such that \(E\cdot D = 0\), we have \(D^2\leq 0\) with \(D^2=0\) if and only if \(D\) is numerically trivial.
By the Hodge Index theorem for ample divivisors (Corollary 3.14), we know that the intersection pairing on \(\operatorname{\operatorname{\mathrm{NS}}_{\mathbb {R}}}X\) has signature \((1, \rho -1)\), where \(\rho = \dim \operatorname{\operatorname{\mathrm{NS}}_{\mathbb {R}}}\) is the Picard rank of \(X\). The result then follows from the theory of quadratic forms and the invariance of the signature, i.e., from Sylvester’s Law of Inertia [ 8 ] .