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 D OD where D is the dualizing sheaf of D Proof By Serre duality h1 D h0 OD 0 We have the short exact sequence 1 0 OX K OX K D D 0 so D OX K D OX K 21 K D D 0 by Riemann Roch using D2 0 and D K 0 Thus h0 D 1 Since D 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 D OD Corollary 2 If D ni Ei is an icoct D an e ective divisor on X s t D Ei 0 for all i then D nD D where n 0 D an e ective 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 H 0 X OX D be s t div X s D Since D D D n 1 D is not e ective s doesn t come from H 0 OX D D so its image in H 0 OD D is nonzero But deg L Ei D Ei D nD Ei 0 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 K 2 0 and K C 0 for all curves on X If D is an icoct on X an elliptic or quasielliptic bration f X B on X obtained from the Stein factorization of nD X P H 0 OX nD for some n 0 Proof Idea use D and K to get an elliptic quasielliptic bration Then show that the ber 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 n using OX nK nD OD D OD since D is an icoct We claim that 5 6 H 2 OX nK n 1 D H 0 n 1 K D 0 for n 2 To see this note that if m K D for m 0 then n either 0 mK mD K H D H 0 for an ample divisor H giving a contradiction or 0 with a similar contradiction Also H 2 OD 2 since D has support of dimension 1 implying that H 2 OX nK nD 0 and H 1 OD H 0 D H 0 OD 0 gives 0 We know from Riemann Roch that H 1 OX nK nD 1 OX nK nD OX nK nD nK nD K 2 OX 1 q since pg 0 Noether s formula states that 7 12 12q 12 12q 12pg K 2 2 2b1 b2 with b1 2q since the irregularity 0 because pg 0 So 8 10 8q b2 1 q 1 OX 0 1 and OX nK nD 0 or 1 for n 2 Since H 1 OX nK nD 0 and H 2 OX nK nD 0 we must have H 0 OX nK nD 0 for 0 n 2 So Dn nK nD As before we see that Dn ni Ei we nd We claim that Dn is of canonical type Letting D that 9 10 Dn Ei n K Ei n D Ei 0 This implies that Dn aD kj Fj 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 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 n D for some integer n for each n 2 K 3K 2K is a multiple of D say D K If 0 this contradicts K H 0 If 0 then K D 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 11 0 OX 2K D D OX 2K 2D 2D OD OD 0 using D OD D OD As before we can show that H 2 OX 2K D D 0 and therefore H 2 OX 2K 2D 2D 0 So OX 2K 2D 2D OX 0 or 1 while h1 OX 2K 2D 2 D 2 because h1 OD h1 OD 1 implies that h0 OX 2K 2D 2D 0 Now take 12 2K 2D 2D 0 2 0 dim 1 Since D D are of canonical type so is easy exercise We now claim that is composed with a pencil i e it gives a map to a curve To see this let C be the xed part of then since is of canonical type we get C 2 0 the self intersection of a divisor supported on a curve of canonical type is 0 So the rational map 13 X X B is de ned everywhere else would have C 2 0 Use C1 C2 C 1 C 2 m1 m2 for a single blowup at p if C1 C2 pass through p with multiplicity m1 m2 and apply to two elements of C with zero intersection after the blowup Since dim 1 B can t be a point And it can t be a surface else we would have C H C 2 0 So is composed with a pencil and is a morphism Now D D 2K 2D 2D 0 and D C 0 and D C 0 write C as ki Ei Fi where F doesn t have any of the Ei as components This forces D C 0 Since D is connected it is contained in one of the bers and D2 0 We see that D is a rational multiple of one of the bers of the Stein factorization f X B B Since the gcd of the coe cients of D is 1 the ber must be a positive integral multiple of D It is easy to see that the genus of the ber is 1 implying that it is an elliptic quasielliptic bration Case 2 pg 0 As before it is enough to show that dim H 0 OX 2 for some divisor of canonical type We ll show that n 0 …