then D 1 D n = D 1 D n.

Lecture 8. Intersection theory and ampleness: revisited. In this lecture, X will denote a proper irreducible variety over k = k, chark = 0, unless otherwise stated. We will indicate the dimension of X by n. As usual by the term variety we mean an irreducible and reduced scheme over k. We wish to define an itersection product n Div(X) Z (D 1, D 2,..., D n ) D 1 D 2 D n. In analogous way to the two-dimensional case, we would like for the intersection product to feature the following 3 basic properties: (1) the intesection product is symmetric and multilinear, i.e., it is linear in every component and it does not depend on the order in which divisors appear; (2) the intesection product only depends on the linear equivalence class of the divisors involved, i.e., if D i D i, i {1,..., n} then D 1 D n = D 1 D n; (3) if D 1,..., D n are all smooth divisor and all intersections are transverse, then D 1 D n = D 1 D n. Remember also that in the case of surfaces we showed that if C X is a curve in X and D a Cartier divisor on X, then D C = deg C (i O X (D)), where i: C X is the closed embedding of the curve in X. We would like to have an analogous property: given any irreducible subvariety V X then the intersection product of D 1,..., D dim V Div(X), (4) D 1 D dim V [V ] = (D 1 V D dim V V ) V. In particular, if H D, D, D 1,..., D n 1 Div(X), then (4 ) D D 1 D n 1 = D 1 D n 1 [H] = (D 1 H D n 1 H ) H. As we did before, we can simply compute intersection numbers by defining the intersection form in the algebraic realm in two different ways. Let D 1,..., D n Div(X) be Cartier divisors and assume that X is projective. Then we know very well at this point, that each of the D i can be written as a difference of very ample divisors D i Hi 1 H2 i, Hj i very ample i, j. Then, we can define the intersection number D 1 D n in the following way, by using very ample divisors and property (3): D 1 D n = Hi 2 Hj 1. I 1,...,n( 1) I From this definition properties (1)(-4) (and (4 ) as well) follow at once. What one needs to verify, though, is that the definition of the interesection product is independent of the choice of the veri ample divisors H j i. 1 j I

2 Exercise 1. Let X be a projective irreducible variety. Verify that the definition of the intersection product just given is independent of the choice of the very ample divisors H j i. (Hint: proceed as in the analogous check that we did in Lecture 5 for the 2-dimensional case) Alternatively, we can define the intersection number in terms of Euler characteristic. Remember that in case of surfaces, we showed that: D 1 D 2 = χ(x, O X ) χ(x, O X ( D 1 )) χ(x, O X ( D 2 )) + χ(x, O X ( D 1 D 2 )). Let H 1,..., H n be very ample divisors. Then by property 4) we know that H 1 H n = (H 2 H1 H n n ) H1. Using this property and induction, then we can compute inductively the right hand side term as (H 2 H1 H n n ) H1 = H 1 H n = I {2,...,n}( 1) I χ(h 1, O ( H1 H i H1 )). Using the short exact sequence 0 O X ( H 1 ) O X O H1 0 and for each I {2,..., n} taking the sequence obtained by tensoring each term with O X ( H i), we see immediately that H 1 H n = I {2,...,n}( 1) I (χ(x, O X ( H i )) χ(x, O X ( H 1 D i ))). Hence, we can try to adopt the following formula as the definition of the intersection product: D 1 D n = I {1,...,n}( 1) I χ(x, O X ( D i )). Exercise 2. Let X be a projective variety. Then show that the definition of the intersection product in terms of Euler characteristic is well defined and respects all the needed properties (1)-(4 ) stated above. (Hint: to prove linearity, you may want to prove the following fact: Let E, F, D 1,..., D dim X be Cartier divisors. Then I {2,...,n}( 1) I (χ(x, D i ) χ(x, E D i ) χ(x, F D i ) + χ(x, E F D i ) = 0. I suggest to try and prove this statement by induction on the dimension of X, using the usual reduction about very ample divisors.) The definition of intersection product that we just gave in terms of Euler characteristic does not actually require the variety X to be projective. Actaully, it does not even requiere X to be a variety but merely a scheme. In fact, the Euler characteristic is a well defined quantity on every scheme and moreover, even though we used the assumption on the projectivity of X in the previous exercise to check that such definition of intersection product makes it linear, it is possible to see that that it is actually unnecessary, cf. [D, Chap. 1]. Exercise 3. Let X be a projective variety. Let H be a very ample divisor on X. Then H n = deg H (X), where deg H (X) indicates the degree of X with respect to the embedding of X given by H.

3 Definition 0.1. Let X be as above. Let D be a Cartier divisor. Then D is said to be numerically equivalent to 0 if D C = 0, C curve in X. We will denote a divisor D numerically equivalent to 0 by D 0. We will denote by Num(X) Div(X) the subgroup given by all divisors numerically equivalent to 0 Lemma 0.2. Let X be as above and let D Num(X). Then D D 1 D n 1 = 0, D i Div(X), i = 1,..., n 1. Proof. By the usual arguments, it is enough to prove the statement when the D i are all very ample and effective on X. But then D Dn 1 is also a numerically trivial divisor and D D 1 D dim X 1 = D D 1 [D dim X 1 ] and the right hand side vanishes by induction on the dimension of X. The onedimensional case is trivial, by the definition of degree. From the definition of numerical equivalence and of intersection product, it follows that the latter is well defined on those classes in N 1 (X) that contain a representative which is a Cartier divisor. Exercise 4. Let X be a proper algebraic variety. (1) Assume that any Weil divisor D Div(X) is Cartier, then show that the intersection product is a symmetric multilinear form defined over N 1 (X). (2) Assume that for any Weil divisor D Div(X) there exists m N >0 s.t. md is Cartier. Can you still define the intersection product on N 1 (X)? If yes, how? Definition 0.3. Let X be as above. Then the Néron-Severi group of X, denoted N 1 (X), is the group of divisors modulo linear equivalence N 1 (X) := Div(X)/Num(X). Theorem 0.4. Let X be as above. Then N 1 (X) is a finitely generated torsion-free abelian group. The torsion-freeness of N 1 (X) is an immediate consequence of the definitions. In fact, if D is a Cartier divisor on X s.t. nd 0 for some n N >0, then nd C = 0 for any irreducible curve C X. This implies, by linearity of the intersection product, that D C = 0, proving the claim. Definition 0.5. Let X be as above. The Picard number of X, denoted by ρ(x), is the rank of the Néron-Severi group of X, ρ(x) := rank N 1 (X). Remark 0.6 (TBC). When X is a proper smooth complex variety X, then we can explain the Theorem 0.4 in terms of complex geometric data. In fact, in the euclidean topology, we have the following short exact sequence 0 Z O X an OX an 0.

4 By taking the associated long exact sequence in cohomology we see that... H 1 (X an, Z) H 1 (X an, O X an) H 1 (X an, O X an) c 1... H 2 (X an, Z) Why do we care about intersection products? Well, first of all, they provide a synthetic way to frame certain enumerative problems, such as counting the cardinality of the interesections of certain subvarieties of a fixed variety. We have seen that they also provide a generalization of the conept of degree of a variety. Let us give one more piece of motivation. Let X be a projective scheme over a noetherian ring A and let O X (1) be a very ample invertible sheaf. Then, recall that the Hilbert polynomial of X with respect to O X (1) is defined as the unique polynomial such that h 0 (X, O X (m)) = P (m), m 0. Moreover, the degree of this polynomial is equal to the dimension of X and the coefficient of the leading term is of the form deg O X (1) (X) dim X!, where deg OX (1)(X) is the degree of X with respect to the embedding given by O X (1). Via the intersection product just defined, we can extend this result to the more general case of all divisors on a projective variety X if we just consider the Euler characteristic rather than h 0 (X, D). Theorem 0.7 (Asymptotic Riemman-Roch). Let X be a normal projective variety. Let D be a Cartier divisor on X and E be a Weil divisor also on X. Then the Euler characteristic of O X (md+e) is a polynomial of degree dim X in m with rational coefficients. More precisely, it has the following (asymptotic) form: χ(x, O X (md + E)) = Dn n! mn + O(m n 1 ). Proof. We work by induction on dim X. The case dim X = 1 is nothing but Riemann-roch theorem for curves which we have already seen. Hence we can prove the inductive step and assume that the theorem has been proven for any normal projective variety of dimension dim X 1. Let H be a very ample divisor on X s.t. H + D is also very ample. Let H H, G H + D sufficiently general members of the corresponding complete linear systems given by theorem 0.8. In particular both this irreducible codimension 1 subvarieties are normal. Let us consider the two folliwng short exact sequences: 0 O X (md + E) O X (md + E + H) O H (md + E + H) 0, and 0 O X ((m 1)D + E) O X (md + E + H) O G (md + E + H) 0. Comparing Euler characteristics: χ(x, O X (md + E + H)) = χ(x, O X (md + E)) + χ(h, O H (md + E + H)) = = χ(x, O X ((m 1)D + E)) + χ(g, O G (md + E + H)).

5 Hence, χ(x, O X (md + E)) χ(x, O X ((m 1)D + E)) = = χ(g, O G (md + E + H)) χ(h, O H (md + E + H)) = = Dn 1 (G H) n 1! m n 1 + O(m n 2 ) = = Dn n 1! mn 1 + O(m n 2 ). The final formula now follows from the proof of [Har, Prop. I.7.3]. Theorem 0.8. Let X be a normal quasi projective variety and let H H be the general member of globally generated (finite dimensional) linear system on X. Then H is also normal. Exercise 5. (1) Is it possible to prove the Asymptotic Riemann-Roch formula in the case where X is just an irreducible projective variety? What if X is just a projective scheme? (2) Is it possible to identify the second term in the asymptotic Riemann-Roch formula above? That is, if we write χ(x, O X (md + E)) = Dn n! mn + b n 1 m n 1 + O(m n 2 ), what is b n 1? Try to give an answer in terms of intersection numbers. (Hint: use Riemann-Roch for surfaces to get a first inductive step and then try to argue as in the proof of the asymptotic Riemann-Roch formula). (3) Prove the following generalised form of Asymptotic Riemann-Roch. Let X be an irreducible projective variety. Let D be a Cartier divisor and F be a coherent sheaf on X. Then the Euler characteristic of F(mD) = F O X (md) is a polynomial of degree dim X in m with rational coefficients. More precisely, it has the following (asymptotic) form: χ(x, F(mD)) = rank(f) Dn n! mn + O(m n 1 ). Exercise 6. Let X d be a degree d hypersurface in P n. Compute P t (m) = χ(x d, O Xd (mt)). Exercise 7. Let F 1 be the blow-up of P 2 at one point. Recall that F 1 has a natural fibration π : F 1 P 1. Let F indicate a fibre of π, F P 1. Let E denote the exceptional curve E P 1, E 2 = 1 for the blow-up. (1) Show that all effective divisors on F 1 are linearly equivalent to ae + bf for some choice of a, b N 2. (2) Compute the asymptotic Riemann-Roch formula for divisors of the above form. What can you say about the growth-rate of the higher cohomology groups of such divisors? References [D] O. Debarre, Higher-Dimensional Algebraic Geometry, Springer. [Har] R. Hartshorne, Algebraic Geometry. Springer, GTM 52, 1997. 2 5