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 19 LECTURES: ABHINAV KUMAR Corollary 1. If D is an indecomposable curve of canonical type (icoct), then ωD ∼= OD, where ωD is the dualizing sheaf of D. Proof. By Serre duality, h1(ωD) = h0(OD) = 0. We have the short exact sequence (1) 0 → OX (K) → OX (K + D) → ωD → 0 so χ(ωD) = χ(OX (K +D))− χ(OX (K)) = 1 ((K +D) D) = 0 by Riemann-Roch 2 ·(using D2 = 0 and D K = 0). Thus, h0(ωD) = 1. Since ωD has degree 0 along · the Ei, (2) deg Ei (OD ⊗ OX (K + D) ⊗ OEi ) = (K + D) · Ei = 0 It follows from the proposition last time that ωD ∼�= OD. Corollary 2. If D = niEi is an icoct, D� an effective divisor on X s.t. D� Ei = 0 for all i, then D� = nD + D�� where n ≥ 0, D�� an effective divisor · disjoint from D. Proof. Let n be the largest integer s.t. D� − nD ≥ 0, and let D�� = D� − nD, L = OD(D��). ∃ an exact sequence (3) 0 → OX (D�� − D) → OX (D��) → OD(D��) = L → 0 Let s ∈ H0(X, OX (D��)) be s.t. div X (s) = D��. Since D�� − D = D − (n + 1)D is not effective, s doesn’t come from H0(OX (D�� −D)), so its image in H0(OD(D��)) is nonzero. But deg (L ) = D�� Ei = (D� − nD) Ei = 0 = ==|Ei · · ⇒ L ∼OD ⇒ s(x) = 0 � ∀ x ∈ D, so that the support of D�� must be disjoint from that of D. � Theorem 1. Let X be a minimal surface with K2 = 0 and K C ≥ 0 for all curves ·on X. If D is an icoct on X, ∃ an elliptic or quasielliptic fibration f : X B→ on X obtained from the Stein factorization of φ|nD| : X → P(H0(OX (nD))∨) for some n > 0. Proof. Idea: use D and K to get an elliptic/quasielliptic fibration. Then show that the fiber must be a multiple of D. 1������ � 2 LECTURES: ABHINAV KUMAR Case 1: pg = 0. or n ≥ 0, we have the exact sequence (4) 0 → OX (nK + (n − 1)D) → OX (nK + nD) → OD → 0 obtained by tensoring 0 → OX (−D) → OX → OD → 0 by n(K + D) and using OX (nK + nD) ⊗ OD = ω⊗n = since D is an icoct. We claim ∼D ∼OD that (5) H2(OX (nK + (n − 1)D)) = H0(−(n − 1)(K + D))0 mfor n ≥ 2. To see this, note that if Δ ∈either Δ = 0 (K + D) for m > 0, then n H = −D H < 0 for an ample · divisor H, giving a contradiction, or Δ > 0 with a similar contradiction. = = ⇒ mK ∼ −mD ⇒ K · Also, H2(OD) = 2 since D has support of dimension 1, implying that H2(OX (nK + nD)) = 0, and H1(OD) = H0(ωD) = H0(OD) =� 0 gives H1(OX (nK + nD)) = 0. We know from Riemann-Roch that �1 (6) χ(OX (nK + nD)) = χ(OX ) + 2(nK + nD)(nK + nD − K) = χ(OX ) = 1 − q (since pg = 0). Noether’s formula states that (7) 12 − 12q = 12 − 12q − 12p2g = K + 2 − 2b1 + b2 with b1 = 2q since the irregularity Δ = 0 because pg = 0. So (8) 10 − 8q = b2 ≥ 1 =⇒ q ≤ 1 =⇒ χ(OX ) = 0, 1 and χ(OX (nK + nD)) = 0 or 1 for n ≥ 2. Since H1(OX (nK + nD)) = 0 and H2(OX (nK + nD)) = 0, we must have H0(OX (nK + nD)) = 0 for n ≥ 2. So ∃ Dn ∈ |nK + nD|. As before, we see that Dn = 0. We claim that Dn is of canonical type. Letting D = �niEi, we find that (9) Dn · Ei = n(K · Ei) + n(D · Ei) = 0 ��This implies that Dn = aD + kjFj for some a ≥ 0, kj > 0 integers, Fj distinct irreducible curves that don’t intersect D. Now K Fj ≥ 0, and · by our hypothesis ( kj Fj ) K ≥ 0. But it equals K nK + nD − D = 0, · · so K Fj = 0 for all j. Finally, · (10) Dn · Fj = n(K Fj ) + n(D Fj ) = 0 · · so Dn is of canonical type. Now, Dn can’t be a multiple of D for all n, For then Dn = mD = ⇒nK ∼ λnD for some integer λn for each n ≥ 2 = K = 3K 2K is a ⇒ · multiple of D, say λ D = K. If λ < 0, this contradicts K H ≥ 0. If· · λ ≥ 0, then |K| = |λD| = ∅ which contradicts pg = 0. So ∃ a curve of� ALGEBRAIC SURFACES, LECTURE 19 3 canonical type D� on X s.t. removing the multiple of D and decomposing to get an icoct, we get something disjoint from D. So let D� be an icoct, disjoint from D. Then 1) 0 → OX (2K + D + D�) → OX (2K + 2D + 2D�) → OD ⊕ OD� → 0 (using ω2D ∼= OD, ωD� ∼= OD� ). As before, we can show that H ( 2 OX (2K +D+D�)) = 0, and therefore H (OX (2K +2D+2D�)) = 0. So χ(OX (2K + 2D+2D�)) = χ(OX ) = 0 or 1, while h1(OX (2K +2D+2 D�)) 1 1 0 ≥ 2 (because h (OD), h (OD�) ≥ 1) implies that h (OX (2K + 2D + 2D�)) ≥ 0. Now, take 2)Δ ∈ |2K + 2D + 2D�| , Δ > 0, Δ2= 0, dim |Δ| ≥ 1 Since D, D� are of canonical type, so is Δ (easy exercise). We now claim that |Δ| is composed with a pencil (i.e. it gives a map to a curve). To see this, let C be the fixed part of |Δ|, then since Δ is of canonical type, we get (Δ − C)2 ≤ 0 (the self-intersection of a divisor supported on a curve of canonical type is ≤ 0). So the rational map 3) φΔ : X → φ | Δ(X) = B| | |⊂ |Δ| is defined everywhere (else would have (Δ−C)2 > 0. Use ˜C1·C2 = C1 ·˜C2+ m1m2 for a single blowup at p if C1, C2 pass through p with multiplicity m , m and apply to two elements of Δ� C with zero intersection after (1(1(11 …