2.6 Numerical equivalence
For this section, assume that \(X\) is an \(n\)-dimensional projective variety. We have a bilinear pairing
We set \(Z_1^{\operatorname{\mathit{num}}} X = (\operatorname{\mathrm{Div}}X)^{\perp }\). So \(Z_1^{\operatorname{\mathit{num}}} X = \{ Z\in Z_1 X : \deg D\cdot Z = 0 \} \) for all \(D\in \operatorname{\mathrm{Div}}X\). We set \(\operatorname{\mathrm{Div}}_{\operatorname{\mathit{num}}} X = (Z_1 X)^{\perp }\). So \(D\in \operatorname{\mathrm{Div}}_{\operatorname{\mathit{num}}} X\) if \(\deg D\cdot Z = 0\) for all \(Z\in Z_1 X\). It is clear that \(Z_1^{\operatorname{\mathit{num}}} X\) (resp. \(\operatorname{\mathrm{Div}}_{\operatorname{\mathit{num}}} X\)) is a subgroup of \(Z_1 X\) (resp. \(\operatorname{\mathrm{Div}}X\)). A \(1\)-cycle in \(Z_1^{\operatorname{\mathit{num}}} X\) (resp. \(\operatorname{\mathrm{Div}}_{\operatorname{\mathit{num}}} X\)) is said to be numercially equivalent to \(0\).
We set \(N_1 X = Z_1 X/Z_1^{\operatorname{\mathit{num}}} X\), and we set \(N^1 X = \operatorname{\mathrm{Div}}X/\operatorname{\mathrm{Div}}_{\operatorname{\mathit{num}}} X\). Then the pairing 2.2 gives rise to a pairing
The group \(N^1 X\) is called the Néron-Severi group of \(X\) and is often written as \(\operatorname{\mathrm{NS}}X\).
Suppose the base field \(k\) is \(\mathbb {C}\). Then \(N^1 X\) is a finitely generated abelian group. In fact, when \(X\) is smooth, the map \(D\mapsto c_1(\mathcal{O}_D)\) induces an injection from \(N^1 X\) into the quotient of \(H^2(X,\mathbb {Z})\) by its torsion subgroup.
This follows from that fact that, as we already saw, that \(c_1(\mathcal{O}_X(D))\) determines the intersection number of \(D\) with any curve. There’s another proof of the finite generation in Hartshorne, Exercise V.1.8 on page 368.
Note that the natural map \(\operatorname{\mathrm{Div}}X\to N^1 X\) factors as \(\operatorname{\mathrm{Div}}X\to \operatorname{\mathrm{Pic}}X \to H^2(X,\mathbb {Z}) \to N^1(X)\). In particular, if \(D\) is linearly equivalent to \(0\), then \(D\) is numerically equivalent to \(0\). We say that \(D\) is homologically equivalent to \(0\) if \(D\) is in the kernel of the map \(\operatorname{\mathrm{Div}}X\to H^2(X,\mathbb {Z})\). So, we get a sequence of inclusions
The rank of the finite abelian group \(N^1 X\) is called the Picard rank \(\rho X\) of \(X\). Note that, the pairing 2.3 is perfect by definition (because we modded out by the kernel) and \(N_1 X\) and \(N^1 X\) are both torsion free. In particular the map \(N_1 X\to \operatorname{\mathrm{Hom}}(N^1 X,\mathbb {Z})\) is injective. So, as abelian groups, we have \(N_1 X\cong \mathbb {Z}^{\rho X}\cong N^1 X\).
For Cartier divisors \(D\) and \(D'\), let’s write \(D\equiv D'\) if \(D\) is numerically equivalent to \(D'\), and \(D\sim D'\) if \(D\) is linearly equivalent to \(D'\). Then, by what we’ve just said, \(D\sim D'\Rightarrow D\equiv D'\).
Suppose \(V\) is a \(k\)-dimensional subvariety in \(X\) and
are Cartier divisors with \(D_i\equiv D_i'\) for all \(i\). Then \(\int _V D_1\ldots D_k = \int _V D_1'\ldots D_k'\).
\(\int _V D_1\ldots D_k\) is symmetric and \(\mathbb {Z}\)-linear in each of the factors. So it suffices to show that, if \(D_1\equiv 0\), then \(\int _V D_1\ldots D_k = 0\). But then assume that \(D_2\cdots D_{k}\cdot V\) is a \(1\)-cycle. So \(\int _V D_1\cdots D_k = \deg D_1\cdot (D_2\cdots D_k\cdot V) = 0\).
Using linearity, we this gives us a pairing
factoring \(\int _V D_1\ldots D_k\).
Suppose \(D\) and \(D'\) are Cartier divisor with \(D\equiv D'\) and \(D\) ample. Then \(D'\) is ample as well.
Direct using Nakai-Moishezon and Theorem 2.5.