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 have
where 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 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 1 The 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
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
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 3 Let 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.
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 4 If 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.