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 21 LECTURES: ABHINAV KUMAR From last time: f : X B is an elliptic/quasi-elliptic fibration, Fbi = miPi multiple fibers, R1f∗OX = → L ⊕ T , for L invertible on B and T torsion. b ∈Supp (T ) h0(OFb ) ≥ 0 h1(OFb ) ≥ 2 Fb is an exceptional/wild fiber. ⇔ ⇔ ⇔ � Theorem 1. With the above notation, ωX = f∗(L−1 ⊗ ωB) ⊗ OX ( aiPi), where 0 ≤ ai < m, ai = mi − 1 unless Fbi is exceptional, and deg (L−1 ⊗ ωB ) = 2pa(B) − 2 + χ(OX ) + �(T ), where �(T ) is its length as an OB -module. Proof. We have proved most of this: specifically, we have that ωX = f∗(L−1 ⊗ωB) ⊗ OX ( aiPi) for 0 ≤ ai < m. We have a Leray spectral sequence Epq = 2 Hp(B, Rqf∗OX ) = ⇒ Hp+q(X, OX ). The smaller order terms give us a short exact sequence (1) 0 → H0(OB ) → H1(OX ) → H0(R1f∗OX ) → H2(OB ) = 0 0 → H2(OX ) → H1(R1f∗OX ) → 0 Using this, we see that χ(OX ) = h0(OX ) − h1(OX ) + h2(OX ) = h0(OB) − h1(OB ) − h0(L ⊕ T ) + h1(L ⊕ T )(2) = χ(OB ) − χ(L) − h0(T ) = −deg L − �(T ) by Riemann-Roch, so deg L = −χ(OX ) − �(T ). Since deg ωB = 2pa(B) − 2, we have deg (L−1 ⊗ ωB ) = 2pa(B) − 2 + χ(OX ) + �(T ). It remains to show that ai = mi − 1 if Fbi is not exceptional. If fact, we can prove something stronger: let αi be the order of OX (Pi) ⊗ OPi in Pic (Pi). Then we claim that (1) αi divides mi and ai + 1, (2) h0(Pi, O(αi+1)Pi ) ≥ 2 and h0(Pi, OαiPi ) = 1, and (3) h0(Pi, nPi) is a nondecreasing function of n. Assuming this, if ai < mi − 1, then αi < mi, so miPi is exceptional by (b) and (c), since then h0(OmiPi ) ≥ 2. We now prove the claim. If m > n ≥ 1, then OmP → OnP → 0 gives H1(P, OmP ) → H1(P, OnP ) → 0, implying that n �→ h1(P, OnP ) is nondecreas-ing. But by Riemann-Roch and the definition of canonical type, χ(OnP ) = 0, so 1� 2 LECTURES: ABHINAV KUMAR h0 = h1 is also nondecreasing. Now, by the definition of αi, OX (αiPi) ⊗ OPi =∼OPi , implying that OX (−niPi) ⊗ OPi = OPi as well. We thus obtain an exact ∼sequence 0 → OX (−αiPi) ⊗ OPi = OPi → O(αi+1)Pi → OαiPi → 0, inducing a long exact sequence (3) = H0(OPi H0(O(αi+1)Pi0 → k ∼) → ) → H0(OαiPi ) → · · · and h0(O(αi+1)Pi ) ≥ 2. But for 1 ≤ j < αi, Lj = OX (−jPi) ⊗ OPi is an invertible OPi -module whose degree in each component of Pi equals 0. Since Lj � , H0(Lj ) = 0, and 0 → → OjPi → 0 gives H0(O(j+1)Pi =∼= OPi Lj → O(j+1)Pi ) ∼H0(OjPi ).= = ∼H0(OαP ) ∼k asSince H0(OP ) ∼k for P icoct, H0(O2P ) ∼= = · · · well. Finally, = OPi(4) (OX (Pi) ⊗ OPi )mi ∼= OX (Fbi ) ⊗ OPi ∼This is proved as follows, Since the fiber is cut out by a rational function f, H0(OX (Fbi ) ⊗ OPi ) = 0. Via the exact sequence �(5) ) 1/f �→1/f 00 → OX → OX (Fbi → OX (Fbi ) ⊗ OFbi → we get a global section of OX (Fbi ) ⊗ OFbi . But this also has degree 0 along the components. So it must be trivial, but what we proved for icoct. We also have OX ((ai + 1)Pi) ⊗ OPi ∼(6) = ωX ⊗ OX (Pi) ⊗ OPi = ωPi = OPi∼ ∼implying that αi|ai + 1 as desired. � Corollary 1. K2 = 0. Corollary 2. If h1(OX ) ≤ 1, then either ai + 1 = mi or ai + αi + 1 = mi. Proof. Exercise. � Remark. Raynaud showed that mi/αi is a power of p = char(k) (or is 1 if char(k) = 0). Therefore, there are no exceptional fibers in characteristic 0. 1. Classification (contd.) If f : X B is an elliptic/quasi-elliptic fibration, then → � (7) ωX = f∗(L−1 ⊗ ωB) ⊗ OX ( aiPi), 0 ≤ ai < mi If n ≥ 1 is a multiple of m1, . . . , mr, then (8) H0(X, ω⊗n) = H0(B, L−n ⊗ ωn ai(n/mi)bi))X B ⊗ OB( Now we recall the 4 classes of surfaces: (a) ∃ an integral curve C on X s.t. K C < 0.·ALGEBRAIC SURFACES, LECTURE 21 3 (b) K ≡ 0. (c) K2 = 0, K C ≥ 0 for all integral curves C, and ∃C� s.t. K C� > 0.· · (d) K2 > 0, and K C ≥ 0 for all integral curves C.· Lemma 1. If X is in (a), then κ(X) = −∞, i.e. pn = 0 for all n ≥ 1. If X is in (b), then κ(X) ≤ 0. If X has an elliptic or quasielliptic fibration f : X B,� ai →and if we let λ(f) = 2pa(B) − 2 + χ(OX ) + �(T ) + , then X is not in class mi (d) and • X is in (a) iff λ(f) < 0, in which case κ(X) = −∞, • X is in (b) iff λ(f) = 0, in which case κ(X) = 0, • X is in (c) iff λ(f) > 0, in which case κ(X) = 1. Proof. If K C < 0, then X is ruled, and κ(X) = We did this before, · ∞. and there is an easy way to see that pn = 0 for all n ≥ 1. For every divisor D ∈ Div (X), ∃nD s.t. |D + nK| = ∅ for n > nD. (Since (D + nK) · C = D C + n(K C) becomes negative eventually. Now C is effective. We claim that · · C2 ≥ 0, so by our useful lemma, 6 |D + nK| can’t have an effective divisor. If C2 < 0, then C K < 0 would imply that C was an exceptional curve of the first · kind, contradicting the minimality of X. Thus, C2 ≥ 0.) In particular, D = K gives |nK| = ∅ for large enough n, implying that |nK| = ∅ for all n (since pn < pmn). Next, assume K ≡ 0 (case (b)). If pn ≥ 2, then dim |nK| ≥ 1 = ⇒ ∃ a strictly positive divisor Δ > 0 in |nK|. Then Δ · H > 0 for a hypersurface section, contradicting nK H = 0 since K …