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 18 LECTURES ABHINAV KUMAR Let X be as from last time i e equipped with maps f X B g X P1 Assume char k 2 3 and let S c P1 Fc is multiple If c P1 S fc Fc B is an e tale morphism Then we have the map fc Pic 0 B Pic 0 Fc and Pic 0 Fc acts canonically on Fc Thus we get an action B Fc Fc for each c P1 S and thus actions 1 0 B g 1 P1 S g 1 P1 S B X X Explicitly if b B x Fc X with c P1 S then b X y where f OB b b0 OFc s OFc y Here b0 is a xed base point on B which acts as the zero element of the elliptic curve B Apply the norm NFc B to get OB nb nb0 f x OB f y where n deg fc Fb Fc We thus obtain commutative diagrams X 2 f b X B f B tnb where tnb is translation by nb and B X 3 idB f B B X b b nb b f B Let B0 Fb0 and An Ker nB B B a group subscheme of B We see that the bers of f are invariant under the action of An on X In particular An acts on B0 Denote this by An Aut B0 where Aut B0 is the group scheme of automorphisms of B0 The action of B on X gives B B0 X which 1 2 LECTURES ABHINAV KUMAR completes the diagram X f nB pr1 B B 0 4 B Note that we can t use b0 for an arbitrary element of B0 since we already used it for a base point of B0 So replace it by b B and b B0 On can check that b x b x b b x x Thus X is isomorphic to the quotient of B B0 by the action of An given by a b b b a a b for a An b B b B0 We can substitute the curve B Ker for B to get the following theorem Theorem 1 Every hyperelliptic surface X has the form X B1 B0 A where B0 B1 are elliptic curves A is a nite group subscheme of B1 and A acts on the product B1 B0 by a b b b a a b for a A b B1 b B0 and A Aut B0 an injective homomorphism The two elliptic brations of X are given by 5 f B1 B0 A B1 A B g B1 B0 A B0 A P1 We can classify these using the structure of a group of automorphisms of an elliptic curve Aut B0 B0 Aut B0 0 the group of translations and the group of automorphisms xing 0 respectively Explicitly we have that Z 2Z j B0 0 1728 Z 4Z j B0 1728 i e B0 6 Aut B0 0 y 2 x3 x Z 6Z j B0 0 i e B0 y 2 x3 1 Now A can t be a subgroup of translations else B0 A would be an elliptic curve not P1 Let A be s t a generates the cyclic group A in Aut B0 B0 Aut B0 0 It is easy to see that a must have a xed point Choose that point to be the zero point of B0 Now A is abelian so is a direct product A0 Z nZ A0 is a subgroup of translations of B0 and thus a nite subgroup scheme of B0 Since A0 and A commute we must have A0 b B0 a b b We thus have the following possibilities a n 2 the xed points are Z 2Z Z 2Z b n 3 the xed points are Z 3Z c n 4 the xed points are Z 2Z d n 6 the xed points are 0 We thus obtain the following classi cation Bagnera de Franchis a1 B1 B0 Z 2Z with the generator a of Z 2Z B1 2 acting on B1 B0 by a b1 b0 b1 a b0 ALGEBRAIC SURFACES LECTURE 18 3 a2 B1 B0 Z 2Z 2 with the generators a and g of Z 2Z 2 B1 2 2 acting by a b1 b0 b1 a b0 g b1 b0 b1 g b0 c for c B0 2 b1 B1 B0 Z 3Z with the generator a of Z 3Z B1 3 s t a Aut B0 0 an automorphism of order 3 only when j B0 0 acting on B1 B0 by a b1 b0 b1 a b0 b2 B1 B0 Z 3Z 2 with the generators a and g of Z 3Z 2 B1 3 2 acting by a b1 b0 b1 a b0 g b1 b0 b1 g b0 c for c B0 3 is xed by i e c c c1 B1 B0 Z 4Z with the generator a of Z 4Z B1 4 s t a i Aut B0 0 an automorphism of order 4 only when j B0 1728 acting on B1 B0 by a b1 b0 b1 a i b0 c2 B1 B0 Z 4Z Z 2Z with the generators a and g of Z 4Z Z 2Z B1 4 B1 2 acting by a b1 b0 b1 a i b0 g b1 b0 b1 g b0 c for c B0 2 d B1 B0 Z 6Z with the generator a of Z 6Z B1 6 acting on B1 B0 by a b1 b0 b1 a b0 1 Classification contd Our rst goal is to prove the following theorem Theorem 2 Let X be a minimal surface Then a an integral curve C on X s t K C 0 X pg p0 0 p12 0 b K C 0 for all integral curves C on X i e K 0 X 0 4K 0 or 6K 0 12K 0 c K 2 0 K C 0 for all integral curves C on X and an integral curve C with K C 0 X 1 K 2 0 4K or 6K contains a strictly positive divisor K 2 0 12K has a strictly positive divisor d K 2 0 K C 0 for all integral curves C on X X 2 in which case 2K We already showed that the 4 classes given by the rst clause are exhaustive and mutually exclusive We also proved the equivalences in a As a pre liminary we need some results on elliptic and quasielliptic brations Recall that an e ective divisor D ri 1 ni Ei 0 is said to be of canonical type if Ki Ei D Ei 0 i if X B is an elliptic quasielliptic bration then every ber has this property If D is also connected and gcd n1 nr 1 then we say that D is an indecomposable curve or …