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 18 LECTURES: ABHINAV KUMAR Let X be as from last time, i.e. equipped with maps f : X B, g : X P1 .→ →Assume char(k) =� 2, 3 and let S = {c ∈ P1|Fc is multiple}. If c ∈ P1 � S, fc : Fc�→ B is an ´etale morphism. Then we have the map fc ∗ : Pic 0(B) → Pic 0(Fc�), and Pic 0(F �) acts canonically on F �. Thus, we get an action B × F �F �forc c c c each c ∈ P1 � S, and thus actions → (1) σ0 : B × g−1(P1 × S) → g−1(P1 � S), σ : B × X → X Explicitly, if b ∈ B, x ∈ F �⊂ X with c ∈ P1 � S, then b X = y, wherec · f∗OB(b − b0) ⊗ OF �(s) = OF �(y). Here b0 is a fixed base point on B, which c c acts as the zero element of the elliptic curve B. Apply the norm NFc�|B to get OB(nb − nb0 + f(x)) − OB (f(y)) where n = deg fc = Fb Fc�. We thus obtain· commutative diagrams b X ��X f(2) f B ��B tnb (where tnb is translation by nb) and σ B × X ��X (3) idB ×f f B × B (b,b�)�→nb+b� ��B Let B0 = Fb0 and An = Ker nB : B B a group subscheme of B. We see that →the fibers of f are invariant under the action of An on X. In particular, An acts on B0. Denote this by α : An Aut (B0), where Aut (B0) is the group scheme→of automorphisms of B0. The action of B on X gives τ : B × B0 X, which→1� �� �2 LECTURES: ABHINAV KUMAR completes the diagram τ (4) B × B�0 �������� ��X f nB ◦pr1 B Note that we can’t use b0 for an arbitrary element of B0, since we already used it for a base point of B0. So replace it by b ∈ B and b� ∈ B0. On can check that τ(b, x) = τ (b�, x�) ⇔ σ(b − b�, x) = x�. Thus, X is isomorphic to the quotient of B × B0 by the action of An given by a (b, b�) = (b + a, α(a)(b�)) for· a ∈ An, b ∈ B, b� ∈ B0. We can substitute the curve B/Ker (α) for B to get the following theorem: Theorem 1. Every hyperelliptic surface X has the form X = B1 × B0/A, where B0, B1 are elliptic curves, A is a finite group subscheme of B1, and A acts on the product B1 × B0 by a(b, b�) = (b + a, α(a)(b�)) for a ∈ A, b ∈ B1, b� ∈ B0, and α : A Aut (B0) an injective homomorphism. The two elliptic fibrations of X→ are given by f : B1 × B0/A → B1/A = B, g : B1 × B0/A → = P1(5) B0/α(A) ∼We can classify these, using the structure of a group of automorphisms of an elliptic curve Aut (B0) = B0 � Aut (B0, 0) (the group of translations and the group of automorphisms fixing 0 respectively). Explicitly, we have that ⎧ ⎨ Z/2Z j(B0) = 0, 1728� 2 3(6) Aut (B0, 0) ∼Z/4Z j(B0) = 1728, i.e. B0 ∼= x= ⎩ = {y2 3 − x}Z/6Z j(B0) = 0, i.e. B0 ∼= x= {y − 1} Now α(A) can’t be a subgroup of translations, else B0/α(A) would be an elliptic curve, not P1 . Let α ∈ A be s.t. α(a) generates the cyclic group α(A) in Aut (B0)/B0 = Aut (B0, 0). It is easy to see that α(a) must have a fixed∼point. Choose that point to be the zero point of B0. Now α(A) is abelian, so is a direct product A0 × Z/nZ. A0 is a subgroup of translations of B0 and thus a finite subgroup scheme of B0. Since A0 and α(A) commute, we must have A0 ⊂ {b� ∈ B0|α(a)(b�) = b�}. We thus have the following possibilities: (a) n = 2 = the fixed points are Z/2Z × Z/2Z (b) n = 3 = ⇒ the fixed points are Z/3Z (c) n = 4 = ⇒ the fixed points are Z/2Z⇒(d) n = 6 = the fixed points are {0}⇒ We thus obtain the following classification (Bagnera-de Franchis): (a1) (B1×B0)/(Z/2Z), with the generator a of Z/2Z ⊂ B1[2] acting on B1 ×B0 by a(b1, b0) = (b1 + a, −b0).� ALGEBRAIC SURFACES, LECTURE 18 3 (a2) (B1 × B0)/(Z/2Z)2, with the generators a and g of (Z/2Z)2 ⊂ (B1[2])2 acting by a(b1, b0) = (b1 + a, −b0), g(b1, b0) = (b1 + g, b0 + c) for c ∈ B0[2]. (b1) (B1 ×B0)/(Z/3Z), with the generator a of Z/3Z = B1[3] (s.t. α(a) = ω ∈Aut (B0, 0) an automorphism of order 3 [only when j(B0) = 0]) acting on B1 × B0 by a(b1, b0) = (b1 + a, ω(b0)). (b2) (B1 × B0)/(Z/3Z)2, with the generators a and g of (Z/3Z)2 = (B1[3])2 acting by a(b1, b0) = (b1 +a, ω(b0)), g(b1, b0) = (b1 +g, b0 +c) for c ∈ B0[3], is fixed by ω, i.e. ω(c) = c. (c1) (B1 × B0)/(Z/4Z), with the generator a of Z/4Z ⊂ B1[4] (s.t. α(a) = i ∈ Aut (B0, 0) an automorphism of order 4 [only when j(B0) = 1728]) acting on B1 × B0 by a(b1, b0) = (b1 + a, i(b0)). (c2) (B1 ×B0)/(Z/4Z ×Z/2Z), with the generators a and g of Z/4Z ×Z/2Z = B1[4] × B1[2] acting by a(b1, b0) = (b1 + a, i(b0)), g(b1, b0) = (b1 + g, b0 + c) for c ∈ B0[2]. (d) (B1×B0)/(Z/6Z), with the generator a of Z/6Z = B1[6] acting on B1 ×B0 by a(b1, b0) = (b1 + a, −ω(b0)). 1. Classification (contd.) Our first goal is to prove the following theorem: Theorem 2. Let X be a minimal surface. Then (a) ∃ an integral curve C on X s.t. K · C < 0 ⇔ κ(X) = −∞ ⇔ pg = p0 = 0 p12 = 0.⇔(b) K · C = 0 for all integral curves C on X (i.e. K ≡ 0) ⇔ κ(X) = 0 ⇔4K ∼ 0 or 6K ∼ 0 12K ∼ 0.⇔(c) K2 = 0, K C ≥ 0 for all integral curves C on X, and ∃ an integral curve · C� with K · C� > 0 ⇔ κ(X) = 1 ⇔ K2 = 0, |4K| or |6K| contains a strictly positive divisor ⇔ K2 = 0, |12K| has a strictly positive divisor. (d) K2 > 0, K · C ≥ 0 for all integral curves C on X ⇔ κ(X) = 2, in which case |2K| = ∅. We already showed that the 4 classes (given by the first clause) are exhaustive and mutually exclusive. We also proved …