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 17 LECTURES: ABHINAV KUMAR 1. K3 Surfaces (contd.) Remark. Note that K3 surfaces can only be elliptic over P1: on a K3 surface, however, one can have many different elliptic fibrations, though not every K3 surface has one. 2. Enriques Surfaces Recall that such surfaces have κ(X) = 0, KX ≡ 0, b2 = 10, b1 = 0, χ(OX ) = 1. A classical Enriques surface has pg = 0, q = 0, Δ = 0, while a non-classical En-riques surface has pg = 1, q = 1, Δ = 2 (which can only happen in characteristic 2). We will discuss only classical Enriques surfaces. Proposition 1. For an Enriques surface, ωX �= OX , but ω2 X = OX .∼ ∼Proof. Since pg = 0, ωX = OX . By Riemann-Roch, χ(OX (−K)) = χ(OX ) + ∼1 �2 (−K)(−2K) = χ(OX ) = 1, so h0(OX (−K)) + h0(OX (2K)) ≥ 1. Since KX �� OX = ⇒ KX �� OX , h0(OX (−K)) = 0 (since −K ≡ 0), and so h0(OX (2K)) ≥ 1. Since 2K ≡ 0, 2K = 0, i.e. ω2 = So the order of K in Pic (X) is 2. Note X ∼OX . that Pic (X) = NS (X), because Pic 0(X) = 0 since q = 0, Δ = 0 for classical Enriques surfaces. � Proposition 2. Pic τ (X) = Z/2Z, where the former object is the space of divi-sors numerically equivalent to zero modulo linear (or algebraic) equivalence, or similarly the torsion part of NS . Proof. Let L ≡ 0. By Riemann-Roch, χ(L) = χ(OX )+ 1 L (L−K) = χ(OX ) = 1 2 ·Thus, h0(L) = 0 or � h2(L) = h0(K − L) = 0. But both � L and K − L are ≡ 0, so either L ∼or ω ⊗ L−1 ∼= ω.= OX = OX , i.e. L ∼� Proposition 3. Let X be an Enriques surface. Suppose char(k) = 2� . Then ∃ an ´etale covering X� of degree 2 of X which is a K3 surface, and the fundamental group of X�/X is Z/2Z. Proof. KX is a 2-torsion divisor class. Let (fij ) ∈ Z1({Ui}, O∗ ) be a cocycle X representing K. in Pic (X) = H1(X, O∗ ). Since 2K ∼ 0, (f2 ) is a coboundary, X ij 12 LECTURES: ABHINAV KUMAR so we can write is as fij 2 = ggji on Ui ∩ Uj , giX ). Now π : X� →∈ Γ(Ui, O∗ X defined locally by zi 2 = gi on Ui given by zzji = fij . This is ´etale since char(k) = 2. �ωX� = π∗(ωX ) = ⇒ κ(X�) = 0 as well. Since χ(OX� ) = 2χ(OX ) = 2, X� is a K3 surface from the classification theorem. � Remark. Over C, in terms of line bundles, take X� = {s ∈ L α(S⊗2) = 1}, where | ∼ωX = L = O(K) is a line bundle equipped with an isomorphism α : L⊗2 → OX . The map L ⊃ X� � s → (x, x) ∈ X� ×X L defines a nowhere vanishing section of π∗L which is trivial, implying that π∗L = KX� is trivial. This implies that χ(OX� ) = 2, and thus X� is K3. Proposition 4. Let X� be a K3 surface and i a fixed-point-free involution s.t. it gives rise to an ´etale connected covering X� X. If char(K) = 2, then X is an → �Enriques surface. Proof. ωX� = π∗(ωX ), and since ωX� ≡ OX� , ωX ≡ 0, κ(X) = 0, and χ(OX ) = 1 χ(OX� ) = 1. By classification, X is an Enriques surface. �2 Thus, Enriques surfaces are quotients of K3 surfaces by fixed-point free invo-lutions. Example. The smooth complete intersection of 3 quadrics in P5 is a K3 surface. Let fi = Qi(x0, x1, x2) + Q�i(x3, x4, x5) for i = 1, 2, 3, where Qi, Q�i are homo-geneous quadratic forms; the fi cut out X�, a K3 surface. Now, let σ : P5 →P5, σ(x0 : : x5) = (x0 : x1 : x2 : −x3 : −x4 : −x5) be an involution. Note· · · that σ(X�) = X�. Generically, the 3 conics Qi = 0 in P2 (respectively the conics Q�i = 0) have no points in common, implying that σ� = σX� has no fixed points in X�, giving us an Enriques surface as above. Theorem 1. Every Enriques surface is elliptic (or quasielliptic). Proof. Exercise. � 3. Bielliptic surfaces This is the fourth class of surfaces with κ(X) = 0 : b2 = 2, χ(OX ) = 0, b1 = 2, KX ≡ 0. There are two cases: (1) pg = 0, q = 1, Δ = 0: the classical, bielliptic/hyperelliptic surface. (2) pg = 1, q = 2, Δ = 2, which only happens in positive characteristic. In either case, b1 = 2 = s = b2 = 1 = dim Alb (X), so the Albanese variety is ⇒ 2 an elliptic curve. Theorem 2. The map f : X Alb (X) has all fibers either smooth elliptic →curves, or all rational curves, each having one singular point which is an ordinary cusp. The latter case happens only in characteristic 2 or 3.� � 3 ALGEBRAIC SURFACES, LECTURE 17 Proof. Let B = Alb (X), b ∈ B a closed point, F = Fb = f−1(b). Then F 2 = 0, F K = 0 = pa(f) = 1 = f : X B is an elliptic or quasi-elliptic · ⇒ ⇒ →fibration (the latter only in characteristic 2 or 3). All the fibers of f are irreducible (if we had a reducible fiber F = aiEi, then the classes of F, Ei, and H (the hyperplane section) would give 3 independent classes in NS (X), implying that b2 ≥ ρ ≥ 3 by the Igusa-Severi inequality, a contradiction). Similarly, one can show that there are no multiple fibers, implying that all fibers are integral. If the general fiber is smooth (or any closed fiber is smooth), then f∗ω, ω ∈ F 0(B, ω�)Bis a regular 1-form on X, vanishes exactly where f is not smooth, implying that it is a global section of Ω1 whose zero locus is either empty or of pure X/k codimension 2. A result of Grothendieck shows that the degree of the zero locus is c2(Ω1 ) = c2 = 2 − 2b1 + b2 = 0, implying that f∗ω is everywhere nonzero X/kand f is smooth. � Remark. If all fibers of the Albanese map are smooth, call it a hyperellip-tic/bielliptic surface. If all fibers of the Albanese map are singular, call it a quasihyperelliptic/quasibielliptic surface. Next, we find a second elliptic fibration. Theorem 3. Let X be as above, f : X B = Alb (X) a hyperelliptic or→quasihyperelliptic fibration. Then ∃ another elliptic fibration g : X → P1 . Proof. (Idea) Find an indecomposable curve C of …