As in the previous two posts, let ${S/k}$ be a smooth, projective surface over an algebraically closed field ${k}$. In the previous posts, we set up an intersection theory for divisors, which was a symmetric bilinear form

$\displaystyle \mathrm{Pic}(S) \times \mathrm{Pic}(S) \rightarrow \mathbb{Z},$

that gave the “natural” answer for the intersection of two transversely intersecting curves. Specifically, we had

$\displaystyle \mathcal{L} . \mathcal{L}' = \chi(\mathcal{O}_S) - \chi(\mathcal{L}^{-1}) - \chi(\mathcal{L}'^{-1}) + \chi( \mathcal{L}^{-1} \otimes \mathcal{L}'^{-1});$

the bilinearity of this map had to do with the fact that the Euler characteristic was a quadratic function on the Picard group. The purpose of this post is to prove a few more general and classical facts about this intersection pairing. As usual, Hartshorne’s Algebraic geometry and Mumford’s Lectures on curves on an algebraic surface are very helpful sources for this material; I also found Abhinav Kumar’s lecture notes useful.

1. The Riemann-Roch theorem

The Euler characteristic of a line bundle ${\mathcal{L}}$ on ${S}$ is a “topological” invariant: it is unchanged under deformations. Given an algebraic family of line bundles ${\mathcal{L}_t}$ on ${S}$ — in other words, a scheme ${T}$ and a line bundle on ${S \times_{k} T}$ which restricts on the fibers to ${\mathcal{L}_t}$ — the Euler characteristics ${\chi(\mathcal{L}_t)}$ are constant. This is one of the parts of the semicontinuity theorem on the cohomology of a flat family of sheaves. Over the complex numbers, one can see this by observing that the Euler characteristic of a line bundle is the index of an elliptic operator — more specifically, the index of the Dolbeault complex associated to ${\mathcal{L}}$ — and can therefore be computed in purely topological terms via the Hirzebruch-Riemann-Roch formula.

In algebraic geometry, the fact that the Euler characteristic is a topological invariant is reflected in the following result, which computes it solely in terms of intersection numbers:

Theorem 1 Let ${\mathcal{L}}$ be a line bundle on ${S}$. Then

$\displaystyle \chi(\mathcal{L}) = \frac{1}{2} \mathcal{L}.( \mathcal{L} - K) + \chi(\mathcal{O}_S), \ \ \ \ \ (1)$

where ${K}$ is the canonical divisor on ${S}$.

Proof: This isn’t a surprising formula: we already know that the Euler characteristic of a line bundle is a quadratic function on ${\mathrm{Pic}(S)}$. Moreover, the Serre duality theorem tells us that

$\displaystyle \chi(\mathcal{L}) = \chi( K - \mathcal{L}).$

It follows that the two sides of the equation (1) are quadratic functions on ${\mathrm{Pic}(S)}$ which satisfy the desired symmetry and which agree at ${\mathcal{L} = \mathcal{O}_S}$ or ${\mathcal{L} = K_S}$; our goal is to show from here that they are equal.

More precisely, given a quadratic function ${f: A \rightarrow \mathbb{Q}}$ on an abelian group ${A}$, define the associated bilinear form

$\displaystyle B_f(x,y) = f(x+y) - f(x) - f(y) + f(0).$

When ${f = \chi}$ on ${\mathrm{Pic}(S)}$, we know that ${B_f}$ is the intersection pairing. It’s also easy to check that the right-hand-side of (1) is a quadratic function whose associated bilinear form is the intersection pairing. In particular, the difference of the two sides of (1) is a quadratic function on ${\mathrm{Pic}(S)}$ which vanishes at zero and whose associated bilinear form vanishes; it follows that the difference is itself a linear map ${ \ell: \mathrm{Pic}(S) \rightarrow \mathbb{Q}}$ (which in particular factors through the torsion subgroup of ${\mathrm{Pic}(S)}$). Since it satisfies the symmetry

$\displaystyle \ell( K - \mathcal{L}) = \ell(\mathcal{L}),$

it must vanish. $\Box$

The goal of the rest of the post is to describe some applications of the Riemann-Roch formula.

2. Application: the Hodge index theorem

The first example is to divisors (not necessarily effective) ${D}$ on the surface ${S}$ satisfying ${D^2 > 0}$. In this case, the Riemann-Roch formula

$\displaystyle \chi(nD) = \frac{1}{2}(nD).(nD - K) +\chi(\mathcal{O}_S)$

shows that for ${n \gg 0}$, the Euler characteristic ${\chi(nD)}$ is positive and in fact tends to ${\infty}$ as ${n \rightarrow \infty}$. Using the Serre duality theorem, we find however (denoting by lower-case $h$ the dimension of the upper case $H$).

$\displaystyle h^0(nD) + h^0(K - nD) \geq h^0(nD) + h^0(K - nD) - \dim h^1(nD) = \chi(nD),$

so that at least one of ${\dim H^0(nD), \dim H^0(K - nD)}$ tends to ${\infty}$ as ${n \rightarrow \infty}$ through some subsequence.

It follows that there are two cases to consider.

• Let’s suppose that ${h^0(nD) \stackrel{\mathrm{def}}{=} \dim H^0(nD) \rightarrow \infty}$ with ${n}$. In this case, for ${n \gg 0}$, ${nD}$ is an effective divisor: it is represented by the union of a finite number of (possibly singular) curves. If ${H}$ is a very ample divisor on ${S}$, then

$\displaystyle n D . H > 0$

because ${H}$ restricted to the curves in ${nD}$ is very ample as well. In particular, ${D.H > 0}$. For ${n \gg 0}$, ${(K - nD).H < 0}$ so ${h^0(K - nD) = 0}$.

• Now suppose ${h^0(K - nD) \rightarrow \infty}$ as ${n \rightarrow \infty}$. Dually, we find that ${K - nD}$ is effective, so we are forced to have ${D.H < 0}$ (where ${H}$ is the same very ample divisor). This implies that ${h^0(nD) = 0}$ for all ${n > 0}$.

In particular, the above two cases are mutually exclusive. We get:

Proposition 2 Suppose ${D^2 > 0}$. Then exactly one of the two possibilities happens:

1. ${D.H > 0}$ and ${h^0(nD) \rightarrow \infty}$ with ${n}$ (while ${h^0(K - nD) =0}$ for ${n \gg 0}$).
2. ${D. H < 0}$ and ${h^0(K - nD) \rightarrow \infty}$ with ${n}$ (while ${h^0(nD) = 0}$ for ${n \gg 0}$).

The Hodge index theorem is a consequence of the above analysis. Namely, let ${\mathrm{Pic}^n(S) \subset \mathrm{Pic}(S)}$ denote the subgroup of line bundles that are numerically equivalent to zero: that is, they have zero intersection multiplicity with all the other line bundles. Then let

$\displaystyle \mathrm{Num}(S) \stackrel{\mathrm{def}}{=} \mathrm{Pic}(S)/\mathrm{Pic}^n(S);$

this is the group of divisors (or line bundles) modulo the relation of numerical equivalence. One has a nondegenerate bilinear pairing

$\displaystyle \mathrm{Num}(S) \times \mathrm{Num}(S) \rightarrow \mathbb{Z}$

given by the intersection pairing. Tensoring with ${\mathbb{Q}}$, we get a ${\mathbb{Q}}$-vector space ${\mathrm{Num}(S) \otimes_{\mathbb{Z}} \mathbb{Q}}$ with a nondegenerate symmetric bilinear form; it is a nontrivial theorem that this vector space is finite-dimensional. See below for a proof in the complex case.

Given an ample divisor ${H}$ on ${S}$, clearly ${H.H > 0}$, and we have a decomposition

$\displaystyle \mathrm{Num}(S) \otimes_{\mathbb{Z}} \mathbb{Q} \simeq \mathbb{Q} \left\{H\right\} \oplus H^{\perp} .$

Theorem 3 (Hodge index theorem) Notation as above, the intersection pairing is negative definite on ${H^{\perp}}$: that is, if ${D \in H^{\perp} \setminus \left\{0\right\}}$, then ${D^2 < 0}$.

Proof: Since the intersection pairing is nondegenerate on ${H^{\perp}}$, it suffices to show that ${D^2 \leq 0}$ for each such ${D \in H^{\perp}}$. But if ${D^2 > 0}$, we have seen that either ${D.H > 0}$ or ${D.H < 0}$ in the above two-case analysis. $\Box$

3. The complex analytic viewpoint

Let’s try to motivate the above analysis from the complex analytic viewpoint. That is, suppose ${S}$ is a complex projective manifold of (complex) dimension ${2}$: there is then a pairing

$\displaystyle \mathrm{Pic}(S) \times \mathrm{Pic}(S) \rightarrow \mathbb{Z}$

which can be described via

$\displaystyle \mathcal{L}.\mathcal{L}' = \int_{S} c_1(\mathcal{L}) \wedge c_1(\mathcal{L}').$

The first Chern classes ${c_1(\mathcal{L})}$ take their values in ${H^{1,1}(S; \mathbb{C}) \subset H^2(S; \mathbb{C})}$. Conversely, every class in ${H^{1,1}(S; \mathbb{C}) \cap H^2(S; \mathbb{Z})}$ comes from a holomorphic line bundle: to see this, use the exponential sequence

$\displaystyle 0 \rightarrow \mathbb{Z} \stackrel{2\pi i}{\rightarrow} \mathcal{O}_S \stackrel{\exp}{\rightarrow} \mathcal{O}_{S}^{\times} \rightarrow 0.$

The Picard group of ${S}$ is the group ${H^1(S, \mathcal{O}_{S}^{\times})}$, and the Chern class map is the coboundary

$\displaystyle c_1: H^1(S, \mathcal{O}_S^{\times}) \rightarrow H^2(S; \mathbb{Z}).$

Conversely, given an element in ${H^2(S; \mathbb{Z})}$, it is in the image of ${c_1}$ if and only if it maps to zero under

$\displaystyle H^2(S; \mathbb{Z}) \stackrel{2 \pi i }{\rightarrow} H^2(S; \mathbb{C}) \rightarrow H^2(S; \mathcal{O}_S);$

in other words (since these classes are real), if it belongs to ${H^{1,1}(S; \mathbb{C}) \cap H^{2}(S; \mathbb{Z})}$.

Now the Hodge index theorem as above follows from the next formulation of it, which is a theorem in Kähler geometry:

Theorem 4 (Hodge) Let ${S}$ be a complex projective surface with ${\omega \in H^2(S; \mathbb{Z})}$ the class of an ample line bundle. Then the intersection form is negative definite on

$\displaystyle \omega^{\perp} \subset H^{1,1}(S; \mathbb{C}) \cap H^2(S; \mathbb{R}).$

It follows that a line bundle is numerically equivalent to zero if and only if ${c_1}$ is zero.

Corollary 5 One has a natural inclusion

$\displaystyle \mathrm{Num}(S) \otimes_{\mathbb{Z}} \mathbb{C} \subset H^{1,1}(S; \mathbb{C}).$

In particular, the group ${\mathrm{Num}(S)}$ is a finitely generated abelian group (a form of the “theorem of the base”). This is true in characteristic ${p}$ as well, but that requires different methods.