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 classification from before: (a) ∃ an integral curve C on X s.t. K C < 0.· (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.· 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 K2 = 0, K C ≥ 0 for any integral curve · C. Thus, either 2K ∼ 0 or X has an elliptic/quasielliptic fibration. 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 fibration, 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 K2 = 0, pg = 0, we have that pg = h0(B, L−1 ⊗ ωB ) = 0 and deg (L−1 ⊗ ωB) = 2pa(B) − 2 + χ(OX ) + �(T ). But χ(OX ) ≥ 0 from the list of last time, so by Riemann-Roch, pa(B) = 1, χ(OX ) = 0, �(T ) = 0 or pa(B) = 0, χ(OX ) + �(T ) < 2. We analyze these two cases separately. Case 1: having no exceptional fibers implies that ai = mi − 1 for all i. If ∃ a multiple fiber miPi with mi ≥ 2, then (say m1 ≥ 2) ωX = f∗(L−1 ⊗ ωB ) ⊗ OX ( (mi − 1)Pi) (1) � X Bω2 = f∗(L−2 ⊗ ω2 ) ⊗ OX ( (2mi − 2)Pi) ⊗ f∗OB (b1) ⊗ OX ((m1 − 2)P1) i>1 Since deg (L−2 ⊗ω2 B ⊗OB (b1) ≥ 1 and B is an elliptic curve, p2 ≥ 1 and so p4, p6 > 0, proving the theorem. If f has no multiple fibers, then ωX = f∗(L−1 ⊗ ωB ), and deg (L−1 ⊗ωB) = 2pa(B)−2 +χ(OX )+�(T ) = 0 = ⇒ ωX ≡ 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 fibration g : X → P1 1, placing it in case 2 of our analysis.2 LECTURES: ABHINAV KUMAR Case 2: pa(B) = 0, i.e. B = P1, pg = 0, i.e. deg (L−1 ⊗ ωB) = −2 + χ(OX ) + �(T ) < 0. Now, a(2) λ(f) = −2 + χ(OX ) + i �(T ) + ≥ 0 mi and an easy check gives � na h0i (3) (nK) = 1 + n (−2 + χ(OX ) + �(T )) + mi Case 2A: �(T ) = 0, so ai = mi − 1 for all i. Also, −2 + �χ(� �OX ) + �(T ) < 0 gives χ(OX ) = 0, 1. If χ(OX ) �= 0, then we must have m ers at least 3 ultiple fib(because we need −2 + mi−1 i ≥ 0. m– ≥ 4 multiple fibers, mi ≥ 2. Then check |2K| = ∅, i.e. p2 > 0. – 3 multiple fibers, all mi ≥ 3. Then |3K| = ∅, so p3, p6 > 0. – 3 multiple fibers, m1 = 2, m2, m3 ≥ 4 = ⇒ |4K| = ∅. – 3 multiple fibers, m1 = 2, m2 = 3, m3 ≥ 6 =⇒ |6K| = ∅. Case 2B: �(T ) = 1, i.e. there is exactly one wild/exceptional fiber F . Now −2 + χ(OX ) + �(T ) < 0 =⇒ χ(OX ) = 0. Also, pg = 0 =⇒ q = 1. Applying the corollary from the previous lecture, we get a1 = m1 −1 or m1 1 α1, where α1 is a common divisor of m1 and −a−a1 + 1. Since −1 + i ≥ 0,m there iare at least 2 multiple fibers, so either – There exist at least 2 multiple curves with ai = mi − 1, so |2K| = ∅. – F has m1 = 3, a1 = 1, α1 = 1, m2 3 = 3K = ∅. ≥– ⇒ | |F has m1 = 4, a1 = 1, α1 = 2, m2 ≥ 4 =⇒ |4K| = ∅. – F has ma1 1 1 = β1α1, β1 ≥ 4 =⇒ ≥ =⇒ 22| K| = ∅. m1 – F has m1 = 2α1, a1 = α1 1, α1 3, m2 3 = 3K = ∅. – F − ≥ ≥ ⇒ | |has m1 = 3α1, a1 = 2α1 − 1, α1 ≥ 2, m2 ≥ 2 =⇒ |2K| = ∅.This concludes the proof. � So for X a minimal surface with elliptic/quasielliptic fibration f : X → B,�����������• X is in (a) ⇔ λ(f) < 0 ⇔ κ(X) = −∞ ⇔ pn = 1∀n ≥ 1 ⇔ p4 = p6 = 0 p12 = 0, ⇔ • X is in (b) ⇔ λ(f) = 0 ⇔ κ(X) = 0 ⇔ nK ∼ 0 for some n ≥ 1 ⇔ 4K ∼0 or 6K ∼ 0 12K ∼ 0,⇔ • X is in (c) ⇔ λ(f) > 0 ⇔ κ(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. K2 > 0, K C ≥ 0· for all curves C on X. Then |2K| =� ∅, and for sufficiently large n, the linear system |nK| is free of base points and defines a morphism φn = φ|nK| : X →P(H0(OX (nK))∨) s.t. • Xn = φn(X) is normal, with at most rational double points as singu-larities, i.e. desingularizing gives a fixed cycle Z (the smallest divisor3 ALGEBRAIC SURFACES, LECTURE 22-23 with support in the exceptional locus with Z Ei ≤ 0 ∀ i) which satisfies · pa(Z) = 0, Z2 = −2. • φn is an isomorphism away from the singular locus, i.e. (4) X � φn−1(Sing(Xn)) ∼ Xn � Sing(Xn)→ In this case, κ(X) = 2. Proof. Exercise. Use Riemann-Roch, the Hodge index theorem, and Nakai-Moishezon. 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 K2 > 0 and on the orthogonal complement of K, the form ( )· ·is negative definite, so there at most finitely such curves Ei. Show that they are rational (pa(Ei) = 0), and …