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

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

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 on is a “topological” invariant: it is unchanged under deformations. Given an algebraic family of line bundles on — in other words, a scheme and a line bundle on which restricts on the fibers to — the Euler characteristics 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 — 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:

*Proof:* This isn’t a surprising formula: we already know that the Euler characteristic of a line bundle is a quadratic function on . Moreover, the Serre duality theorem tells us that

It follows that the two sides of the equation (1) are quadratic functions on which satisfy the desired symmetry and which agree at or ; our goal is to show from here that they are equal.

More precisely, given a quadratic function on an abelian group , define the associated bilinear form

When on , we know that 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 which vanishes at zero and whose associated bilinear form vanishes; it follows that the difference is itself a linear map (which in particular factors through the torsion subgroup of ). Since it satisfies the symmetry

it must vanish.

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) on the surface satisfying . In this case, the Riemann-Roch formula

shows that for , the Euler characteristic is positive and in fact tends to as . Using the Serre duality theorem, we find however (denoting by lower-case the dimension of the upper case ).

so that at least one of tends to as through some subsequence.

It follows that there are two cases to consider.

- Let’s suppose that with . In this case, for , is an
*effective*divisor: it is represented by the union of a finite number of (possibly singular) curves. If is a very ample divisor on , thenbecause restricted to the curves in is very ample as well. In particular, . For , so .

- Now suppose as . Dually, we find that is effective, so we are forced to have (where is the same very ample divisor). This implies that for all .

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

Proposition 2Suppose . Then exactly one of the two possibilities happens:

- and with (while for ).
- and with (while for ).

The **Hodge index theorem** is a consequence of the above analysis. Namely, let 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

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

given by the intersection pairing. Tensoring with , we get a -vector space 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 on , clearly , and we have a decomposition

Theorem 3 (Hodge index theorem)Notation as above, the intersection pairing is negative definite on : that is, if , then .

*Proof:* Since the intersection pairing is nondegenerate on , it suffices to show that for each such . But if , we have seen that either or in the above two-case analysis.

**3. The complex analytic viewpoint**

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

which can be described via

The first Chern classes take their values in . Conversely, every class in comes from a holomorphic line bundle: to see this, use the exponential sequence

The Picard group of is the group , and the Chern class map is the coboundary

Conversely, given an element in , it is in the image of if and only if it maps to zero under

in other words (since these classes are real), if it belongs to .

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 be a complex projective surface with the class of an ample line bundle. Then the intersection form is negative definite on

It follows that a line bundle is numerically equivalent to zero if and only if is zero.

Corollary 5One has a natural inclusion

In particular, the group is a finitely generated abelian group (a form of the “theorem of the base”). This is true in characteristic as well, but that requires different methods.

## Leave a Reply