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, F = m P bi i i multiple fibers, R1f O = L ⊕ T , for L invertible on B and T torsion. b ∈ ∗ X Supp(T ) ⇔ h0(O ) ≥ 0 ⇔ h1(O ) ≥ 2 ⇔ F is an exceptional/wild fiber. Fb Fb b Theorem 1. With the above notation, ω = f∗(L−1 ⊗ ω ) ⊗O (� a P ), where X B X i i 0 ≤ a < m,a = m − 1 unless F is exceptional, and deg(L−1 ⊗ ω ) = i i i bi B 2p (B) − 2+ χ(O )+ �(T ), where �(T ) is its length as an O -module. a X B Proof. We have proved most of this: specifically, we have that ω = f∗(L−1 ⊗ X ω ) ⊗ O (� a P ) for 0 ≤ a < m. We have a Leray spectral sequence Epq = B X i i i 2 Hp(B,Rqf O ) =⇒ Hp+q(X, O ). The smaller order terms give us a short ∗ X X exact sequence 0 → H0(O ) → H1(O ) → H0(R1f O ) → H2(O ) = 0 B X ∗ X B (1) 0 → H2(O ) → H1(R1f O ) → 0 X ∗ X Using this, we see that χ(O ) = h0(O ) − h1(O )+ h2(O ) X X X X = h0(O ) − h1(O ) − h0(L ⊕ T )+ h1(L ⊕ T ) B B (2) = χ(O ) − χ(L) − h0(T ) B = −deg L − �(T ) by Riemann-Roch, so deg L = −χ(O ) − �(T ). Since deg ω = 2p (B) − 2, we X B a have deg (L−1 ⊗ ω ) = 2p (B) − 2 + χ(O ) + �(T ). It remains to show that B a X a = m − 1 if F is not exceptional. If fact, we can prove something stronger: i i bi let α be the order of O (P ) ⊗O in Pic(P ). Then we claim that i X i Pi i (1) α divides m and a + 1, i i i (2) h0(P , O ) ≥ 2 and h0(P , O ) = 1, and i (αi+1)Pi i αiPi (3) h0(P ,nP ) is a nondecreasing function of n. i i Assuming this, if a < m − 1, then α < m , so m P is exceptional by (b) and i i i i i i (c), since then h0(O ) ≥ 2. miPi We now prove the claim. If m > n ≥ 1, then O → O → 0 gives mP nP H1(P, O ) → H1(P, O ) → 0, implying that n �→ h1(P, O ) is nondecreas mP nP nP ing. But by Riemann-Roch and the definition of canonical type, χ(O ) = 0, so nP 1 2 LECTURES: ABHINAV KUMAR h0 = h1 is also nondecreasing. Now, by the definition of α , O (α P ) ⊗O ∼= ∼ i X i i Pi O , implying that O (−n P ) ⊗ O = O as well. We thus obtain an exact Pi X i i Pi Pi sequence 0 → O (−α P ) ⊗ O = O → O → O → 0, inducing a X i i Pi Pi (αi+1)Pi αiPi long exact sequence (3) 0 → k ∼= H0(O ) → H0(O ) → H0(O ) → ··· Pi (αi+1)Pi αiPi and h0(O ) ≥ 2. But for 1 ≤ j < α ,L = O (−jP ) ⊗ O is an (αi+1)Pi i j X i Pi invertible O -module whose degree in each component of P equals 0. Since L �∼= O ,HP0(i L ) = 0, and 0 → L → O → O → 0 givie s H0(O ) ∼= Hj0 (O P)i . Sincej H0(O ) ∼= k forj P ico(cj+t,1 )HPi 0(O j)P i ∼= ··· ∼= H0(O ()j +∼=1) Pki as jPi P 2P αP well. Finally, (4) (O (P ) ⊗O )mi ∼= O (F ) ⊗O ∼= O X i Pi X bi Pi Pi This is proved as follows, Since the fiber is cut out by a rational function f, H0(O (F ) ⊗O ) �= 0. Via the exact sequence X bi Pi (5) 0 → O → O (F ) 1/f→ �→1/f O (F ) ⊗O → 0 X X bi X bi Fbi we get a global section of O (F ) ⊗O . But this also has degree 0 along the X bi Fbi components. So it must be trivial, but what we proved for icoct. We also have ∼ O ((a + 1)P ) ⊗O = ω ⊗O (P ) ⊗O (6) X i i Pi X X i Pi ∼ ∼ = ω = O Pi Pi implying that α |a + 1 as desired. � i i Corollary 1. K2 = 0. Corollary 2. If h1(O ) ≤ 1, then either a +1 = m or a + α +1 = m . X i i i i i Proof. Exercise. � Remark. Raynaud showed that m /α is a power of p = char(k) (or is 1 if i i 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) ω = f∗(L−1 ⊗ ω ) ⊗O ( a P ), 0 ≤ a < m X B X i i i i If n ≥ 1 is a multiple of m ,...,m , then 1 r � (8) H0(X,ω⊗n) = H0(B,L−n ⊗ ωn ⊗O ( a (n/m )b )) X B B i i i 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. p = 0 for all n ≥ 1. If X is n in (b), then κ(X) ≤ 0. If X has an elliptic or quasielliptic fibration f : X → B, and if we let λ(f) = 2p (B) − 2+ χ(O )+ �(T )+ � ai , then X is not in class a X 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 p = 0 for all n ≥ 1. For every divisor n D ∈ Div(X), ∃n s.t. |D + nK| = ∅ for n > n . (Since (D + nK) · C = D D 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 p < p ). n mn Next, assume K ≡ 0 (case (b)). If p ≥ 2, then dim |nK| ≥ 1 =⇒ ∃ a n strictly positive divisor Δ > 0 in |nK|. Then Δ · H > 0 for a hypersurface section, contradicting nK · H = 0 since K ≡ 0. So p ≤ 1 for all n, implying n that κ(X) ≤ 0. Now assume X has an elliptic/quasielliptic fibration, and let M = f (ω ) = ∗ X L−1 ⊗ ω from last time. Then M has degree λ(f). Let H be a very ample B divisor on X. Then π = f| : H → B is some finite map of degree = H · F > 0. H Now n(K · H) = deg(ωn | ) = deg (π∗M) = (deg π)(deg M) = (H · F )λ . So X H H B f if λ < 0, then K · H < 0 and X is in (a). f Similarly, λ(f) = 0 =⇒ K · H = 0 for every irreducible hyperplane section H, and any curve C can be written, up to ∼, as the difference of 2 such. This implies that K · C = 0 ∀ C =⇒ K ≡ 0. Lastly, λ(f) > 0 =⇒ K · C > 0 for all horizontal irreducible C. For vertical C, K · C = 0 by the formula for K, implying that K · C ≥ 0 for all C integral, (K2) = 0 by the formula, implying that we are in class (c). � Let X be a minimal surface with K2 = 0,p ≤ 1 (in particular, every surface in g class (b) is of this form. Then Noether’s formula gives 10 − 8q +12p = b +2Δ. g 2 Since p ≤ 1, 0 ≤ Δ ≤ 2p ≤ 2, also Δ = 2(q − s) is even, we obtain the following g g possibilities. (1) b = 22,b = 0,χ(O ) = 2,q = 0,p = 1, Δ = 0. 2 1 X g (2) b = 14,b = 2,χ(O ) = 1,q = 1,p = 1, Δ = 0. 2 1 X g 4 LECTURES: ABHINAV KUMAR (3) b = 10,b = 0,χ(O ) = 1, and either q = 0,p = 0, Δ = 0 or q = 1,p = 2 1 X g g 1, Δ = 2. (4) b = 6,b = 4,χ(O ) = 0,q = 2,p = 1, Δ = 0. 2 1 X g (5) b = 2,b = 2,χ(O ) = 0, and either q = 1,p = 1, Δ = 0 or q = 2,p = 2 1 X g g 0, Δ = 2. Note. If X is in class (b) and p = 1, then K ∼ 0 (because K = 0,H0(K) �= 0 g imply that K ∼ 0.). Let’s deal with case 4 of class (b). Proposition 1. Let X be minimal in class (b), and b = 2,b = 2. Then s = 1, 2 1 Alb(X) is an elliptic curve, and X → Alb(X) gives an elliptic/quasielliptic fibration. Proof. Let’s see that the fibers of f are irreducible. If not, we would have ρ > 2(F,H, component of F ) and b ≥ ρ > 2, contradicting b = 2. Now, to see that 2 2 the fibers are not multiple, note that χ(O ) = 0 from the list. X (9) deg (L−1 ⊗ ω ) = 2p (B) − 2+ χ(O )+ �(T ) = �(T ) ≥ 0 B a X Since ω = f∗(L−1 ⊗ω )⊗O ( �a P ) ≡ 0, we see that �(T )·f−1(y)+ �a P ≡ X B X i i i i 0. But it is an effective divisor, implying that all the a = 0,�(T ) = 0 and thus i a = m − 1 ∀ i (there are no wild fibers since T = 0). So m = 1 ∀ i. Thus, we i i i have integral fibers, which is the case of a bielliptic surface. �

Topics in Algebraic Geometry- Algebraic Surfaces- Algebraic Surfaces, Lecture 21

Similar Topics in Algebraic Geometry- Algebraic Surfaces- Algebraic Surfaces, Lecture 21