MIT OpenCourseWare http://ocw.mit.edu 18.727 Topics in Algebraic Geometry: Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.ALGEBRAIC SURFACES, LECTURE 2 LECTURES: ABHINAV KUMAR Remark. In the definition of (L, M) we wrote M = OX (A − B) where A and B are irreducible curves. We can think of this as a moving lemma. 1. Linear Equivalence, Algebraic Equivalence, numerical equivalence of divisors Two divisors C and D are linearly equivalent on X there is an f ∈ k(X)⇔ s.t. C = D + (f ). This is the same as saying there is a sheaf isomorphism OX (C) ∼= OX (D), 1 �→ f. Two divisors C and D are algebraically equivalent if OX (C) is algebraically equivalent to OX (D). We say two line bundles L1 and L2 on X are algebraically equivalent if there is a connected scheme T , two closed points t1, t2 ∈ T and a line bundle L on X × T such that LX×{t1} ∼= L1 =and LX×{t2} ∼L2, with the obvious abuse of notation. Alternately, two divisors C and D are alg. equivalent if there is a divisor E on X × T , flat on T , s.t. E| = C and E| = D. We say C ∼alg D.t1 t2 We say C is numerically equivalent to D, C ≡ D, if C E = D E for every · · divisor E on X. We have an intersection pairing Div X × Div X → Z which factors through Pic X × Pic X → Z, which shows that linear equivalence = ⇒ num equivalence. In fact, lin equivalence = alg equivalence (map to P1 defined by (f)) and ⇒alg equivalence = numerical equivalence (χ() is locally constant for a flat ⇒morphism, T connected). Notation. Pic (X) is the space of divisors modulo linear equivalence, Pic τ (X) is the set of divisor classes numerically equivalent to 0, Pic 0(X) ⊂ Pic τ (X) ⊂Pic (X) is the space of divisor classes algebraically equivalent to 0. Num(X) = Pic (X)/Pic τ (X) and NS(X) = Pic (X)/Pic 0(X). 1.1. Adjunction Formula. Let C be a curve on X with ideal sheaf I. (1) O → I/I2 → ΩX/k ⊗ OC → ΩC/k → 0 with dual exact sequence (2) 0 → TC → TX ⊗ OC → NC/X = (I/I2)∗ → 0 12 LECTURES: ABHINAV KUMAR Taking ∧2 gives ωX ⊗ OC = OX (−C)|C ⊗ ΩC or KC = (KX + C)|C so deg KC = 2g(C) − 2 = C.(C + K) (genus formula). Note: C2 = deg (OX (C) ⊗ OC ) by definition. I/I2 is the conormal bundle, and is ∼O(−C) ⊗ OC , while NC/X is= the normal bundle ∼= O(C) ⊗ OC . Theorem 1 (Riemann-Roch). χ(L) = χ(OX ) + 12 (L2 − L ωX ).· Proof. L−1 · L ⊗ ω−1 = χ(OX ) − χ(L) − χ(ωX ⊗ L−1) + χ(ωX ). By Serre duality, X χ(OX ) = χ(ωX ) and χ(ωX ⊗L−1) = χ(L). So we get that the RHS is 2(χ(OX ) −χ(L)) and thus the desired formula. � As a consequence of the generalized Grothendieck-Riemann-Roch, we get Theorem 2 (Noether’s Formula). χ(OX ) = 1 (c12 +c2) = 1 (K2 +c2) where c1, c212 12 are the Chern classes of TX , K is the class of ωX , c2 = b0 −b1 +b2 −b3 +b4 = e(X) is the Euler characteristic of X. See [Borel-Serre], [Grothendieck: Chern classes], [Igusa: Betti and Picard numbers], [SGA 4.5], [Hartshorne]. Remark. If H is ample on X, then for any curve C on X, we have C H > 0 1 · (equals n (degree of C in embedding by nH) for larger n).· 1.2. Hodge Index Theorem. Lemma 1. Let D1, D2 be two divisors on X s.t. h0(X, D2) = 0� . Then h0(X, D1) ≤h0(X, D1 + D2). aProof. Let a = 0 � ∈ H0(X, D2). Then H0(X, D1) → H0(X, D1) ⊗k H0(X, D2) →H0(X, D1 + D2) is injective. � Proposition 1. If D is a divisor on X with D2 > 0 and H is a hyperplane section of X, then exactly one of the following holds: (a) (D H) > 0 and· h0(nD) → ∞ as n → ∞. (b) (D · H) < 0 and h0(nD) → ∞ as n → −∞. Proof. Since D2 > 0, as n → ±∞ we have 1 1 (3) h0(nD) + h0(K − nD) ≥ 2 n 2D2 − 2 n(D · K) + χ(OX ) → ∞ We can’t have h0(nD) and h0(K − nD) both going to ∞ as n → ∞ or n → −∞ (otherwise h0(nD) = 0 gives � h0(K − nD) ≤ h0(K), a contradiction). Similarly, h0(nD) can’t go to ∞ both as n → ∞ and as n → −∞. Similarly for h0(K−nD). Finally, note that h0(nD) �= 0 implies (nD · H) > 0 and so D · H > 0. � Corollary 1. If D is a divisor on X and H is a hyperplane section on X s.t. (D · H) = 0 then D2 ≤ 0 and D2 = 0 ⇔ D ≡ 0. Proof. Only the last statement is left to be proven. If D �≡ 0 but D2 = 0, then ∃E on X s.t. D.E = 0. Let � E� = (H2)E − (E · H)H, and get D · E� = (H2)D · E = 0 �and H E� = 0. Thus, replacing E with E�, we can assume H E = 0. Next, let · ·3 ALGEBRAIC SURFACES, LECTURE 2 D� = nD + E, so D� H = 0 and D�2 = 2nD E + E2 . Taking n >> 0 if D E > 0· · · and n << 0 if D E < 0, we get D�2 > 0 and D� H = 0, contradicting the above · · proposition. � Theorem 3. (HIT): Let NumX = Pic X/Pic τ X. Then we get a pairing NumX×NumX → Z. Let M = NumX ⊗Z R. This is a finite dimensional vector space over R of dimension ρ (the Picard number) and signature (1, ρ − 1). Proof. Embed this in �-adic cohomology H2(X, Q�(1)) which is finite dimen-sional, and C.D equals C ∪ D under (4) H2(X, Q�(1)) × H2(X, Q�(1)) → H4(X, Q�(2)) ∼= Q� The map NumX � C → [C] ∈ H2 is an injective map. The intersection numbers define a symmetric bilinear nondegenerate form on M(= NumX ⊗Z R). Let h be the class in M of a hyperplane section on X. We can complete to a basis for M, say h = H1, h2, . . . , hρ s.t. (h, hi) = 0 for i ≥ 2, (hi, hj ) = 0 for i =� j. By the above, ( ) has signature (1, ρ − 1) in this basis. Therefore, if E is any divisor on ·, ·X s.t. E2 > 0, then for every divisor D on X s.t. D E = 0, we have D2 …