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 22 23 LECTURES ABHINAV KUMAR 1 Classification contd Recall the classi cation from before a an integral curve C on X s t K C 0 b K 0 c K 2 0 K C 0 for all integral curves C and C s t K C 0 d K 2 0 and K C 0 for all integral curves C Theorem 1 Let X be a minimal surface in class b or c Then p4 0 or p6 0 So if X belongs to class b then 4K 0 or 6K 0 and if X belongs to class c then 4K or 6K contains a strictly positive divisor Proof If X is in class b or c then K 2 0 K C 0 for any integral curve C Thus either 2K 0 or X has an elliptic quasielliptic bration If 2K 0 then of course K 0 implying that X is in class b and p2 0 p4 p6 0 as well Otherwise let f X B be the stated bration and assume that pg 0 if pg p1 0 then pn 0 for all n 2 and the theorem holds Now for X minimal with K 2 0 pg 0 we have that pg h0 B L 1 B 0 and deg L 1 B 2pa B 2 OX T But OX 0 from the list of last time so by Riemann Roch pa B 1 OX 0 T 0 or pa B 0 OX T 2 We analyze these two cases separately Case 1 having no exceptional bers implies that ai mi 1 for all i If a multiple ber mi Pi with mi 2 then say m1 2 X f L 1 B OX mi 1 Pi 1 2 X f L 2 B2 OX 2mi 2 Pi f OB b1 OX m1 2 P1 i 1 Since deg L 2 B2 OB b1 1 and B is an elliptic curve p2 1 and so p4 p6 0 proving the theorem If f has no multiple bers then X f L 1 B and deg L 1 B 2pa B 2 OX T 0 X 0 and thus X is in class b It also must be case 5 from last time thus giving a bielliptic surface and another elliptic bration g X P1 placing it in case 2 of our analysis 1 2 LECTURES ABHINAV KUMAR Case 2 pa B 0 i e B P1 pg 0 i e deg L 1 B 2 OX T 0 Now ai 0 f 2 OX T 2 mi and an easy check gives nai 0 h nK 1 n 2 OX T 3 mi Case 2A T 0 so ai mi 1 for all i Also 2 OX T 0 gives 0 then we must have at least 3 multiple bers OX 0 1 If OX mi 1 0 because we need 2 mi 4 multiple bers mi 2 Then check 2K i e p2 0 so p3 p6 0 3 multiple bers all mi 3 Then 3K 3 multiple bers m1 2 m2 m3 4 4K 3 multiple bers m1 2 m2 3 m3 6 6K Case 2B T 1 i e there is exactly one wild exceptional ber F Now 2 OX T 0 OX 0 Also pg 0 q 1 Applying the corollary from the previous lecture we get a1 m1 1 or m 1 1 1 ai 0 where 1 is a common divisor of m1 and a1 1 Since 1 mi there are at least 2 multiple bers so either There exist at least 2 multiple curves with ai mi 1 so 2K F has m1 3 a1 1 1 1 m2 3 3K F has m1 4 a1 1 1 2 m2 4 4K a1 1 2 K F has m1 1 1 1 4 m 2 1 F has m1 2 1 a1 1 1 1 3 m2 3 3K F has m1 3 1 a1 2 1 1 1 2 m2 2 2K This concludes the proof So for X a minimal surface with elliptic quasielliptic bration f X B X is in a f 0 X pn 1 n 1 p4 p6 0 p12 0 X is in b f 0 X 0 nK 0 for some n 1 4K 0 or 6K 0 12K 0 X is in c f 0 X 1 nK has a strictly positive divisor for some n 1 12K has a strictly positive divisor Theorem 2 Let X be a minimal surface in class d i e K 2 0 K C 0 for all curves C on X Then 2K and for su ciently large n the linear system nK is free of base points and de nes a morphism n nK X P H 0 OX nK s t Xn n X is normal with at most rational double points as singu larities i e desingularizing gives a xed cycle Z the smallest divisor ALGEBRAIC SURFACES LECTURE 22 23 3 with support in the exceptional locus with Z Ei 0 i which satis es pa Z 0 Z 2 2 n is an isomorphism away from the singular locus i e 4 X 1 n Sing Xn Xn Sing Xn In this case X 2 Proof Exercise Use Riemann Roch the Hodge index theorem and NakaiMoishezon The point is that if K C 0 for all integral curves C on X then K is ample and we re through So the problem comes from curves Ei s t K Ei 0 But K 2 0 and on the orthogonal complement of K the form is negative de nite so there at most nitely such curves Ei Show that they are rational pa Ei 0 and satisfy the criteria of rational double points There are some things left to prove in classi cation one is that every minimal surface with X 0 b2 6 is Abelian see Bombieri Mumford Remark For a surface of general type nK is base point free for n 4 and for n 5 nK is an isomorphism away from the union of nitely many rational curves To review we have shown that if X is a minimal surface and f X B is an elliptic quasi elliptic bration then X is in a i f 0 X 4K 6K X is in b i f 0 X 0 4K 0 or 6K 0 X is in c i f 0 X 1 4K or 6K contain strictly positive divisors and K 2 0 X is in class d X 2 and in this case 2K Since a d are mutually disjoint and exhaustive we get the following the orem Theorem 3 Let X be a minimal surface Then C 0 on X s t K C 0 X p4 p6 0 p12 0 X is ruled K 0 X 0 4K 0 or 6K 0 12K 0 X is …