Let be a smooth, projective surface over an algebraically closed field and let be curves (subschemes pure of codimension one) on . In the previous post, we discussed what a good theory of intersections would look like. We wanted to be able to define the intersection in such a manner that:

- If intersect transversely, then .
- The intersection product is
**additive**. That is, given curves , we havewhere is treated as an effective Cartier divisor.

- The intersection product is invariant under linear equivalence and descends to a pairing on the Picard group.

**1. Definition of the intersection product**

In the previous post, we saw that any intersection theory as above was necessarily unique, and suggested that the Euler characteristic formula

would be a good definition: i.e., that the failure of

in general was due to two factors: the existence of nilpotents in the (scheme-theoretic as opposed to set-theoretic) intersection and higher homotopy groups in the (derived as opposed to scheme-theoretic) intersection . The main goal of this post is to prove that (1)does give a good theory. That is, we would like to prove:

Theorem 1The definition of in (1) satisfies the conditions desired of an intersection product.

*Proof:* This is now relatively “formal” to prove, except for additivity. If intersect transversely, then

that is, there are no higher homotopy groups in the “derived” intersection. Moreover, the intersection is only a finite collection of reduced points, so the above Euler characteristic formula counts them. To prove this, observe that if , then there exist such that cuts out , cuts out (scheme-theoretically), and . This means

because form a *regular sequence.* This shows that the “derived” intersection is the ordinary set-theoretic one and proves the first claim.

Now let’s check that the intersection depends only on the linear equivalence class of (and by the evident symmetry in the definition, in ). We have an exact sequence

which gives an exact triangle (in the derived category)

This clearly only depends on the line bundle associated to . This proves that the intersection pairing depends only on the linear equivalence class.

**2. Proof of additivity**

The additivity of the intersection product is the “non-formal” part of the above proof, and it relies on the fact that we are working with surfaces. Let’s first obtain another formula for the intersection number. We saw in the previous section that

and, using the exact sequence

we conclude the following useful formula:

Let be an abelian group. We say that a function

is **polynomial of degree ** if, for any map , the composite

is given by a polynomial of (total) degree in its arguments. If is **quadratic** (i.e., polynomial of degree ), then the difference

defines a **symmetric bilinear form** on . In order to prove additivity of the intersection numbers, it thus suffices to prove that the Euler characteristic is a quadratic function on the Picard group of a surface. More generally:

Theorem 2 (Snapper)Let be a projective variety of dimension . Then the Euler characteristic

is polynomial of degree .

*Proof:* The proof is induction on . Without loss of generality, assume irreducible.

Let be a homomorphism. By possibly expanding , we can assume that a very ample line bundle is in the image. By an appropriate change of basis of , we may also assume that each of the coordinate vectors maps to a very ample line bundle on .

Now if is very ample, we have a global nonzero section whose zero locus is of dimension . The exact sequence

shows that

is a polynomial function of degree , by induction on the dimension. In particular, for each , we find that is a polynomial function on of degree . The next lemma now shows that is a polynomial function of degree .

Lemma 3Let be a function. Suppose for each basis element , is polynomial of degree . Then is polynomial of degree .

*Proof:* First, observe that the assertion is straightforward when . We will use this case below.

The hypothesis implies that is polynomial of degree for any . By induction on , we may assume that is polynomial of degree . Then

is polynomial of degree . Since the restriction of to any line has degree , we are done.

Over the complex numbers, Snapper’s theorem has a simple proof via Hirzebruch-Riemann-Roch, which states that for any vector bundle on a complex manifold , the holomorphic Euler characteristic can be computed topologically:

The Chern character of a line bundle is , which varies **polynomially** in since is additive in and since we are working on a finite-dimensional space. Thus the integral that computes is polynomial in , and the polynomial is of degree at most the (complex) dimension.

**3. Examples**

Let’s now try to make this a bit more concrete. Let be curves. Suppose have no common irreducible component. This implies that the derived tensor product is the ordinary tensor product. In fact, near a point , is cut out by an equation and is cut out by an equation . Since

is an artinian local ring, it follows that must be a regular sequence. In this case, we have

where we do not need derived tensor products to make the intersection, but we do need nilpotents (and schemes).

Let’s now make a calculation that works even to calculate the self-intersection of a curve. We have , and as we saw that implied (see (2))

If is smooth, this is equivalently the **degree** of the line bundle on : in fact, this is a consequence of the Riemann-Roch theorem for curves. So we conclude:

Proposition 4If is a smooth curve, then .

In particular, if is a smooth curve, then the **self-intersection** can be calculated as the degree of . The adjunction formula gives that

is the normal bundle of in , so we find that the self-intersection of a curve in is the degree of its normal bundle.

January 28, 2013 at 1:00 pm

There’s something weird about saying . Should the word length be in there (I always confuse these terms, maybe not length, but something like it)?

January 28, 2013 at 3:56 pm

Yes, in fact I should have said the length (or -dimension). That was a typo which I’ve now fixed; thanks for pointing it out.