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 de nition 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 at on T s t E t1 C and E t2 D We say C alg D 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 de ned by f and alg equivalence numerical equivalence is locally constant for a at morphism T connected Notation Pic X is the space of divisors modulo linear equivalence Pic X is the set of divisor classes numerically equivalent to 0 Pic 0 X Pic X Pic X is the space of divisor classes algebraically equivalent to 0 Num X Pic X Pic X and N S X Pic X Pic 0 X 1 1 Adjunction Formula Let C be a curve on X with ideal sheaf I 1 O I I 2 X k OC C k 0 with dual exact sequence 2 0 TC TX OC NC X I I 2 0 1 2 LECTURES ABHINAV KUMAR Taking 2 gives X OC OX C C C or KC KX C C so deg KC 2g C 2 C C K genus formula Note C 2 deg OX C OC by de nition I I 2 is the conormal bundle and is O C OC while NC X is the normal bundle O C O C Theorem 1 Riemann Roch L OX 21 L2 L X 1 Proof L 1 L X OX L X L 1 X By Serre duality OX X and X L 1 L So we get that the RHS is 2 OX L and thus the desired formula As a consequence of the generalized Grothendieck Riemann Roch we get 1 1 c21 c2 12 K 2 c2 where c1 c2 Theorem 2 Noether s Formula OX 12 are the Chern classes of TX K is the class of X 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 equals n1 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 a Proof Let a 0 H 0 X D2 Then H 0 X D1 H 0 X D1 k H 0 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 n2 D2 n D K OX 2 2 0 0 We can t have h nD and h K nD both going to as n or n otherwise h0 nD 0 gives h0 K nD h0 K a contradiction Similarly 0 h 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 0 Let E H 2 E E H H and get D E H 2 D E 0 on X s t D E and H E 0 Thus replacing E with E we can assume H E 0 Next let ALGEBRAIC SURFACES LECTURE 2 3 D nD E so D H 0 and D 2 2nD E E 2 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 X Then we get a pairing NumX N umX Z Let M NumX Z R This is a nite dimensional vector space over R of dimension the Picard number and signature 1 1 Proof Embed this in adic cohomology H 2 X Q 1 which is nite dimen sional and C D equals C D under 4 H 2 X Q 1 H 2 X Q 1 H 4 X Q 2 Q The map NumX C C H 2 is an injective map The intersection numbers de ne 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 h s t h hi 0 for i 2 hi hj 0 for i j By the above has signature 1 1 in this basis Therefore if E is any divisor on X s t E 2 0 then for every divisor D on X s t D E 0 we have D2 0 1 3 Nakai Moishezon Let X k be a proper nonsingular surface over k Then L is ample for L L 0 and for every curve C on X L OX C 0 Note we de ne the intersection number of L M to be the coe cient of n1 n2 in Ln1 Mn2 check that this is bilinear etc and that it coincides with the previous de nition Proof Sketch when X is projective is easy For the converse Ln as n Riemann Roch or by defn Replace …