Hyperkähler manifolds

Transcription

1 Hyperkähler manifolds Kieran G. O Grady May 20

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3 Contents I K3 surfaces 5 1 Examples The definition A list of examples Exercises to Chapter Curves and linear systems on K3 surfaces Adjunction for curves on a K Riemann-Roch and linear systems Ample divisors and projective embeddings Periods Introduction Topology of K3 surfaces Periods of marked K3 s Infinitesimal Torelli Noether-Lefschetz loci and K3 surfaces up to deformation Isomorphisms and isometries The Kähler cone Exercises to Chapter Global Torelli Introduction The plan Picard-Lefschetz Hausdorffization of the moduli space of marked K3 s Calabi-Yau metrics Geometry of twistor conics Proof of Global Torelli

4 4 CONTENTS

5 Part I K3 surfaces 5

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7 Chapter 1 Examples 1.1 The definition A K3 surface is a compact (connected) Kählerian surface with vanishing first Betti number (i.e. regular) and trivial canonical bundle. A smooth quartic surface X = V (F ) := {[X] P 3 F (X) = 0}, F C[X 0,..., X 3 ] 4 is a K3 surface. In fact by adjunction K X 0, and by Lefschetz Hyperplane Section Theorem (LHT) b 0 (X) = 1 and b 1 (X) = 0. One obtains similar examples by considering hypersurfaces D 1,..., D n 2 P n with transversal intersection X = D 1... D n 2. Since the intersection is transversal it is smooth of dimension 2. By the LHT X is connected and regular. Now suppose that in addition d d n 2 = n + 1. By adjunction X has trivial canonical bundle and hence it is a K3 surface. If n > 5 then necessarily d i = 1 for some i. Thus we get only 3 distinct families of examples: quartics in P 3, intersections of a quadric and a cubic in P 4 and the intersection of 3 quadrics in P 5. By the classification of compact complex surfaces [1] a Kählerian surface with trivial canonical bundle is either a K3 surface or a complex torus. 1.2 A list of examples Example (Double covers). We recall how one constructs double covers of a (complex) manifold Y. Let L be a line-bundle on Y, and assume that there exists a non-zero σ H 0 (L 2 ). Thus σ : Y L 2 is an embedding, where L 2 is the total space of L 2. Let m: L L 2 be the square map (between total spaces of the line-bundles). We define X := m 1 σ(y ) - this is an analytic subset of the total space of L. Identifying Y with σ(y ) and restricting m to X we get a double cover π : X Y. The cover is trivial if and only if L is trivial. The branch divisor of π is B := div(σ); X is smoth if and only if B is smooth. If there is a unique square-root of L 2 = O Y (B) then the double cover is uniquely determined by the choice of B. Now suppose that B is smooth. Let R Div(X) be the reduced divisor with support π 1 (B). We notice that O X (R) = π L. 7

8 8 CHAPTER 1. EXAMPLES In fact the inclusion X L defines a tautological section of π L: since it vanishes exactly on R, with multiplicity 1, we get the above isomorphism. Thus the adjunction formula for π gives K X = π K Y (R) = π (K Y L). (1.2.1) Another useful formula is obtained by decomposing the push-forward π O X into the ±1 eigensheaves for the action of the covering involution of π: one gets π O X = OY L 1, where O Y is the (+1)-eigensheaf and L 1 is the ( 1) eigensheaf. We claim that H p (O X ) = H p (π O X ) = H p (O Y ) H p (L 1 ). (1.2.2) One way to prove the first isomorphism of (1.2.2) is to choose an open Stein covering U := {U i } i I of Y ; thus H p (π O X ) is the p-th cohomology of the complex of Cech cochains associated to U. Moreover π 1 U := {π 1 U i } i I is an open Stein covering of X (this is where the hypothesis that π is a finite cover comes in); thus H p (O X ) is the p-th cohomology of the complex of Cech cochains associated to π 1 U. Comparing the two complexes of Chech cochains we get the first isomorphism of (1.2.2). Now let Y be a compact regular projective surface such that there exists a (non-zero) smooth curve B 2K Y. Let π : X Y be the double cover ramified over B, corresponding to the square root of O Y (B) given by K Y. Then X is a K3 surface: in fact K X is trivial by (1.2.1), and h 1 (O X ) = 0 by (1.2.2). One can take Y = P 2 : then X is a double plane ramified over a sextic. Another choice is Y = P 1 P 1 : then X is a double quadric ramified over a curve of type (4, 4). Example (Elliptic K3 s). Let X be a K3 surface. An elliptic fibration of X is a regular non-constant map f : X T to a smooth (connected) curve T such that the fiber over a non-critcical value t is a smooth curve of genus 1 (an elliptic curve). A K3 is elliptic if it has an elliptic fibration (it might have more than one). Notice that if X is an elliptic K3, the base T of the elliptic fibration must be P 1, because we have an injection f : H 0 (Ω 1 T ) H0 (Ω 1 X ). Conversely suppose that X is a K3, and f : X P 1 is a regular dominant map with connected fibers; a generic fiber of f is an elliptic fiber by adjunction. Elliptic K3 s appear already in the examples above. If a smooth quartic contains a line L, then the pencil H L, where H is the hyperplane section, is a pencil whose generic member is an elliptic curve, and thus defines an elliptic fibration. Another example is as follows. Let F 4 := P(O P 1(4) O P 1) be the rational ruled surface with a section Σ := P(O P 1(4)) of self-intersection 4. Then K F4 (2Σ + 6F ) where F is a fiber of the ruling F 4 P 1. Let B 2K F4 = 4Σ + 12F be smooth. The double cover π : X F 4 ramified over B is a K3 (argue as above) and the composition X F 4 P 1 is an elliptic fibration f : X P 1. We claim that there exists B = Σ + B 0 where B 0 3Σ + 12F is a smooth curve which does not intersect Σ (notice that Σ (3Σ + 12F ) = 0). To see why let H P 5 be a hyperplane and C H be a rational normal quartic curve, choose p (P 5 \ H) and let C be the cone over C with vertex p. Let ρ: C C be the blow-up of p; then C = F 4 and ρ O C (1) = O F4 (Σ + 4F ). Let V P 5 be a cubic hypersurface intersecting transversely C (in particular V does not contain p); then B 0 = ρ V will do. With this choice of B the elliptic fibration f has a section namely π 1 Σ. It is a fact that the elliptic fibration of a general quartic surface containing a line does not admit a section, thus the two examples are different.

9 1.2. A LIST OF EXAMPLES 9 Example (Kummer surfaces). Let T = C 2 /L be a two-dimensional compact torus, i.e. L is a subgroup of C 2 isomorphic to Z 4 and generating C 2 as real vector space; thus T is a compact Kähler surface. Let ι be the automorphism of T sending x to x. The quotient K(T ) := T/ ι is an analytic variety, smooth away from the fixed points of ι, i.e. the 2-torsion points T [2] := {x T 2x = 0} Let (u, v) be local coordinates centered at a 2-torsion point which are induced by affine coordinates on C 2. The action of ι on (u, v) is multiplication by 1, hence locally the quotient map T K(T ) is given by B r Br 2 V (xy z 2 ) (u, v) (u 2, v 2, 2uv) where B r C 2 and B r 2 C 3 are the balls of radii r and r 2 centered at the origin. Thus K(T ) has sixteen ordinary double points and is smooth elsewhere. Blowing up the singularities we get a complex manifold K(T ) with sixteen ( 2)-curves, i.e. copies of P 1 with self-intersection equal to 2, which are called nodal curves. This is the Kummer surface associated to T. (Classically the singular model K(T ) was called the Kummer surface.) We constructed K(T ) by first taking the quotient and then blowing up, but we will get the same result if we proceed in the opposite order. Let π : T T be the blow up of T [2] and ι: T T the involution lifting ι: then T / ι is isomorphic to K(T ). If T is projective then K(T ) is projective, in particular Kähler; since being Kähler is an open condition it follows that every small deformation of K(T ) is Kähler (but not projective for very generic T ). In fact K(T ) is Kähler for any T, see [19]. Let s prove that K(T ) is a K3 surface. Obviously it is connected. Next notice that pull-back defines isomorphisms π : H 1 (T ; Q) H 1 ( T ; Z), π : H 0 (Ω 2 T ) H 0 (Ω 2 T ). (1.2.3) Let ρ: T K(T ) be the quotient map: pull-back defines isomorphisms ρ : H 1 ( K(T ); Q) H 1 ( T ; Z) ι, ρ : H 0 (Ω 2 K(T ) ) H 0 (Ω 2 T ) ι. (1.2.4) The action of ι on π H p (T ) is identified with the action of ι on H p (T ); since ι acts on H p (T ) as multiplication by ( 1) p it follows that b 1 ( K(T )) = 0 and h 0 (Ω 2 K(T ) = 1. Let ) 0 σ H 0 (Ω 1 K(T ) ); we claim that σ is nowhere zero i.e. K(T ) has trivial canonical bundle. By (1.2.4) there exists τ H 0 (Ω 2 T ) such that ρ σ = τ and by (1.2.3) there exists τ H 0 (Ω 2 T ) such that π τ = τ. Since σ 0 we have τ 0 and hence τ is nowhere zero (T has trivial canonical bundle). It follows that σ is certainly non-zero outside the nodal classes. Thus div(σ) = 16 i=1 a i R i K K(T ) where R 1,..., R 16 are the nodal curves and a i 0 for each i. We must prove that a i = 0 for all i. Each R i is a smooth rational curve of self-intersection 2; thus adjunction gives that Thus a i = 0. 2 = R i (K K(T ) + R i ) = (a i + 1)R i R i = 2(a i + 1).

10 10 CHAPTER 1. EXAMPLES It is interesting to count the number of effective parameters for each of the examples given above. The parameter space for quartic surfaces has dimension 34, but the group of projectivities of P 3 (of dimension 15) acts on it, leaving at most 19 effective parameters for quartics up to isomorphism. By the Noether-Lefschetz Theorem [21] two very generic quartics have Picard group generated by the class of a hyperplane; it follows that they are isomorophic if and only if they are projectively equivalent and hence we may say that we have 19 effective parameters (warning: there do exist quartics which are isomorphic without being projectively equivalent see ). A similar computation gives 19 effective parameters for complete intersections and double covers of P 2. Of course Kummers have 4 effective parameters. All the other examples that we have given have 18 effective parameters. The first to introduce the name K3 was A. Weil [22], it was meant to recall Kummer, Klein Kodaira and also the K2 mountain. Usually the condition of being Kähler is omitted when defining a K3 but in fact every compact complex surface with b 1 = 0 and trivial canonical bundle admits a Kähler metric by a Theorem of Siu [17]. Weil did not discover K3 surfaces. Kummer surfaces had been studied by Klein (and Kummer?). The classical (Italian) algebraic geometers had studied surfaces embedded in a projective space P g and whose hyperplane sections are genus-g canonical curves (for example the complete intersections considered above) - these are the projective K3 s. Exercises to Chapter 1 Exercise Let X be a K3 surface. Since h 0 (T X ) = 0 the connected component of the identity in Aut(X) is trivial (it equals {1}). We may say that Aut(X) is a discrete group. The purpose of this exercise is to give an example of K3 surface with infinite automorphism group. Let (a 1, b 1 ) and (a 2, b 2 ) be couples of strictly positive integers such that 3 = a 1 + a 2 = b 1 + b 2. Let D i O P 2(a i ) O P 2(b i ). If D 1, D 2 are generic they intersect transversely. Let X := D 1 D 2 where D 1, D 2 are transverse. (1) Show that X is a K3 surface. (2) Let f 1 : X P 2 and f 2 : X P 2 be the restrictions to X of the two projections of P 2 P 2. Show that if D 1, D 2 are generic then both f 1 and f 2 are finite of degree 2. (3) Suppose that X is as in Item (2). Let τ i Aut(X) be the covering involution of f i. Thus τ 1, τ 2 define a homomorphism of groups ρ: Z Z Aut(X). (1.2.5) Prove that ρ is injective. (Hint: examine the action of τ 1, τ 2 on the subgroup of H 2 (X; Z) given by {c 1 (O P 2(x) O P 2(y)) (x, y) Z 2 }.)

11 Chapter 2 Curves and linear systems on K3 surfaces The main references for the present section are [11] and [14]. A curve is a compact analytic space of pure dimension 1 and without embedded points 1 ; it is integral if it is reduced and irreducible. Let X be a smooth surface; we may and will identify curves in X with effective divisors on X. 2.1 Adjunction for curves on a K3 Let X be a K3 surface. Suppose that C X is a curve. Then the dualizing sheaf ω C is locally-free, in fact adjunction gives that ω C = OX (C) C. (2.1.1) In particular we get the following key equality: p a (C) = C C (2.1.2) where p a (C) is the arithmetic genus of C. (Given divisors A, B on a compact (complex) surface we let A B be their intersection number.) The result below follows at once from (2.1.2). Claim Let X be a K3 surface and C X an integral curve. Then C C 2 and equality holds if and only if C is a smooth rational curve (i.e. isomorphic to P 1 ). Let us assume that C X is an integral curve. Let s analyze the linear system C := P(H 0 (O X (C)). Consider the exact sequence of coherent sheaves 0 O X O X (C) O C (ω C ) 0. (2.1.3) Here X has the classical topology and the sheaves are holomorphic sheaves, if X is projective we may also consider the Zariski topology and sheaves of regular (algebraic) functions. The exact cohomology sequence associated to (2.1.3) gives a surjection H 0 (X, O X (C)) H 0 (C, O C (ω C )). (2.1.4) Proposition Let X be a K3 surface and C X an integral curve. Then h 0 (X, O X (C)) = p a (C) = 2 + C C 2 (2.1.5) If C is not a smooth rational curve then the linear system C has no base-locus i.e. O X (C) is globally generated. 1 Let X be a smooth surface: an analytic space C X is a curve if and only if the ideal sheaf I C is locally principal. 11

12 12 CHAPTER 2. CURVES AND LINEAR SYSTEMS ON K3 SURFACES Proof. Equation (2.1.5) follows from surjectivity of (2.1.4). Now suppose that C is not a smooth rational curve. Then p a (C) > 0. Since the dualizing sheaf of an integral Gorenstein curve of arithmetic genus at least 1 is globally generated [8] it follows that O C (C) is globally generated. Thus O X (C) is globally generated as well. Let C be as in Proposition and suppose that C is not a smooth rational curve. By the proposition we have a morphism X ϕ C C = P g, g := p a (C). Now suppose that C is not hyperelliptic (in particular g 3). Then the canonical map of C is an embedding and hence the restriction of ϕ C to C is an embedding by surjectivity of (2.1.4). In particular ϕ C is birational onto its image. Thus we see that C is very well-behaved: we will get back to this in Section Riemann-Roch and linear systems Let us write out the Riemann-Roch formula for a holomorphic vector-bundle F on X - (actually Atiyah-Singer s index Theorem if X is not projective): χ(f ) = ch(f ) Td(X) = rk(f ) c 2(X) + c 1(F ) 2 c 2 (F ). (2.2.1) 12 2 X Here rk(f ) is the rank of F and to simplify notation we have dropped the integration sign in the last expression. Letting F = O X we get c 2 (X) 2 = χ(o X ) = i.e. χ top (X) = 24. (2.2.2) 12 X We will return to the above equality later on. Now assume that L is a (holomorphic) line-bundle: we rewrite Riemann-Roch as χ(l) = 2 + c 1(L) 2. (2.2.3) 2 By Serre duality h 2 (L) = h 0 (L 1 ) and hence (2.2.3) gives that In particular Equation (2.2.5) gives the converse of Claim h 0 (L) + h 0 (L 1 ) 2 + c 1(L) 2. (2.2.4) 2 If c 1 (L) 2 2 then h 0 (L) > 0 or h 0 (L 1 ) > 0. (2.2.5) Remark Suppose that h 0 (L) > 0 and L is not trvial, i.e. L = O X (C) where C > 0 and C 0. How do we decide whether Inequality (2.2.4) is an equality? By Serre duality h 2 (L) = h 0 (O X ( C) = 0 and hence h 0 (O X (C)) = 2 + C C + h 1 (O X (C)). (2.2.6) 2 By Serre duality h 1 (O X (C)) = h 1 (O X ( C)). On the other hand the exact sequence 0 O X ( C) O X O C 0 (2.2.7) gives that h 1 (O X ( C)) = h 0 (O C ) 1. Thus h 1 (O X ( C)) vanishes if and only if h 0 (O C ) = 0. Before stating our next result we recall a classical definition and a classical theorem. Let X be a compact complex manifold and D a complete linear system without fixed components i.e. such that the base-scheme Bs( D ) := E (2.2.8) E D

13 2.2. RIEMANN-ROCH AND LINEAR SYSTEMS 13 has codimension at least 2. Then D is composed with a pencil if there exist a smooth (integral) curve and a dominant rational map f : X T (2.2.9) such that for each E D we have E = f A where A Div(T ). Theorem (Bertini s second Theorem). Let X be a compact complex manifold and D a complete linear system without fixed components. Then either the generic divisor in D is a prime divisor or else D is composed with a pencil. A comment regarding Bertini s Theorem. Since D is without fixed components we have a rational (meromorphic) map f : X D with image Y (the closure of the image of the open subset U X of regular points of f) a closed irreducible subset of D of dimension at least 1. If dim Y = 1 then D is composed with a pencil; the point is to prove that if dim Y > 1 then the generic divisor in D is a prime divisor. For a proof when dim X = 2 (the only case that we will need) see [7]. We will also need the following elementary observation. Proposition Let X be a compact complex manifold such that h 0 (Ω 1 X ) = 0. Suppose that D is a complete linear system on X without fixed components and composed with a pencil. Then there exists a prime divisor A such that D na for some n and moreover h 0 (O X (D) = n + 1. Proof. We may assume that D is composed with the pencil given by (2.2.9). Let f : X T be a resolution of indeterminacies of f (e.g. the closure of the graph of f in X T ) and π : X X the projection. Let X ρ T ϑ T be the Stein factorization of f. There exists B Div(T ) such that π D f B ρ (ϑ B). Pull-back gives an injection ρ H 0 (Ω 1 T ) H 0 (Ω 1 X; by our hypothesis we get that h 0 (Ω 1 T ) = 0 and hence T = P 1. Thus O T (ϑ B) = O P 1(n) for some n > 0. It follows that D na for some n. Let A 1,..., A n A be generic, in particular pairwise distinct (this makes sense because O X (A) f O P 1(1)); by hypothesis each A i is irreducible and hence h 0 (O Ai ) = 1. The exact cohomology sequence associated to the exact sequence 0 O X O X (D) O A1... O An 0 gives that h 0 (O X (D) (n+1). On the other hand pull-back of sections gives an injection f : H 0 (O P 1(n)) H 0 (O X (D)) and hence h 0 (O X (D) n + 1. Proposition Let X be a K3 surface and D a linear system on X without fixed components. Then D has no base-locus and moreover one of the following holds: (1) There exists an integral C D. (2) There exists an elliptic pencil ϕ C : X C = P 1 such that D = {C C g C i C }. (2.2.10) Proof. By Bertini s second Theorem Theorem either there exists an integral C D or D is composed with a pencil. If the former holds then D has no base-locus by Proposition and of course Item (1) holds. Now suppose that D is composed with a pencil. Since X is regular Theorem gives that D nc where C is an irreducible curve and h 0 (O X (D) = n + 1. Inequality (2.2.4) gives n + 1 = h 0 (O X (nc)) 2 + C C n 2. 2 Since D has no fixed components we have C C 0; it follows that C C = 0. By adjunction C has arithmetic genus 1. Since C is irreducible Proposition gives that h 0 (O X (C)) = 2 and C has no base-locus. Thus ϕ C : X C = P 1 is an elliptic pencil such that (2.2.10) holds. Moreover D = nc has no base-locus because C has no base-locus.

14 14 CHAPTER 2. CURVES AND LINEAR SYSTEMS ON K3 SURFACES 2.3 Ample divisors and projective embeddings We will prove the following result. Theorem Let X be a K3 surface and L an ample line-bundle on X. Then one of the following holds: (a) There exist a smooth rational curve R X, an elliptic fibration f : X P 1 and an integer g 3 such that L = O X (R + ge) where E is a fiber of f. We have In particular R is the base-locus of L. L = {R + f A A O P 1(g) }. (b) L is globally generated and ϕ L defines a double cover X ϕ L (X) L. (2.3.1) Let C L be a smooth curve; then C is hyperelliptic (of genus at least 2) and the restriction to C of the covering involution of (2.3.1) is equal to the hyperelliptic involution. (c) L is globally generated and ϕ L : X L is an embedding with image a surface whose smooth hyperplane sections are canonically embedded curves. In all cases L n is globally generated for n 2 and very ample for n 3. Before proving Theorem let s give examples of Items (1), (2) and (3). In Example we have given examples of K3 s X which have an elliptic fibration f : X P 1 with a section R. Let F be a fiber of f. Let g 3. One checks easily that R + gf is ample on X; then L := O X (R + gf ) is an example of Item (a). Examples of Item (b) are given by the double covers of Example Examples of Item (c) are given by complete intersections. Proof of Theorem By Kodaira vanishing 0 = h 1 (L) = h 2 (L) and hence (see (2.2.3)) we have h 0 (L) = χ(l) = 2 + c 1(L) 2. (2.3.2) 2 Since L is ample we have L L > 0; it follows that h 0 (L) 3. Thus we may write L = O X (M +F ) where M (the moving part of L ) is non-empty and has no fixed components, F is the fixed component of L i.e. L = {C + F C M }. (2.3.3) We apply Proposition to M. Thus either Item (1) or Item (2) of Proposition holds. Suppose that Item (2) holds and hence M gc where C X is an integral curve with C C = 0. Write F = i I a if i where each a i > 0 and the F i s are pairwise distinct prime divisors. We claim that there exists i 0 such that M F i0 0. In fact if M F i = 0 for all i I then c 1 (L) 2 = F F 0 (the inequality holds by Hodge index), that is absurd. Let R := F i0 ; since C has no fixed components we have C R > 0. We have g + 1 = h 0 (O X (gc)) = h 0 (O X (gc + R)) 2 + (gc + R) (gc + R) 2 = 2 + g(c R) + R R 2 (The last inequality holds because by Claim we have R R 2.) Thus Let In both cases we have D := 2 + g 1. (2.3.4) C R = 1, R R = 2. (2.3.5) { M if Item (1) of Proposition holds, M + R if Item (2) of Proposition holds. h 0 (O X (D)) = 2 + D D. (2.3.6) 2

15 2.3. AMPLE DIVISORS AND PROJECTIVE EMBEDDINGS 15 In fact if Item (1) holds then the above equality follows from Proposition 2.1.2, if Item (2) holds it follows from (2.3.4) (all inequalities are necessarily equalities). On the other hand h 0 (O X (D)) = h 0 (L); by (2.3.2) we get that D D = c 1 (L) 2 = (D + F ) (D + F ) (2.3.7) where F := { F if Item (1) of Proposition holds, (F R) if Item (2) of Proposition holds. Notice that in both cases F 0. Let s prove that F = 0. We have (D + F ) (D + F ) D D = (D + F ) F + D F = L F + D F. (2.3.8) Since L is ample and F 0 it will suffice to show that D F 0. If Item (1) of Proposition holds then D F 0 because D has no fixed components. Now suppose that Item (2) of Proposition holds and hence D (gc + R). First let s prove that g 2. In fact since c 1 (L) 2 > 0 we get that D D > 0 by (2.3.7); thus (2.3.5) gives that g 2. It follows that (gc + R) is nef. In fact let Z X be an irreducible curve: then C Z 0 because C has no base-locus, and R Z 0 unless Z = R, thus (gc + R) Z 0 if Z R, on the other hand (gc + R) R = (g 2) which is non-negative because g 2. This finishes the proof that F = 0. Moreover if Item (2) holds we get that g 3. We have proved that either Item (a) of the theorem holds or else L contains a smooth curve C. Before going on with the proof of the remaining statements we notice the following. Let τ H 0 (L) such that and C := div(τ). For n 1 we have the complex 0 H 0 (L (n 1) ) sn H 0 (L n ) rn H 0 ((L C ) n ) 0 (2.3.9) where s n is multiplication by τ H 0 (L) and r n is restriction. We claim that (2.3.9) is exact. The issue is surjectivity of r n : we have coker r n H 1 (L (n 1) ) and H 1 (L (n 1) ) = 0 for n = 1 by definition of K3 and for n > 1 by Kodaira vanishing, thus (2.3.9) is indeed exact. Now suppose that L contains a smooth non-hyperelliptic curve C of genus g. We claim that for each n 1 the natural map S n H 0 (L) mn H 0 (L n ) (2.3.10) is surjective. The proof is by induction on n. If n = 1 it is trivially true. In order to prove the inductive step suppose that n 2 and σ H 0 (L n ). By Noether s Theorem the canonical curve ϕ L (C) is projectively normal and hence there exist α 1,..., α g H 0 (L C ) and P C[x 1,..., x g ] n such that r n (σ) = P (α 1,..., α g ). By Exact Sequence (2.3.9) there exist α 1,..., α g H 0 (L) such that r 1 ( α i ) = α i for 1 i g.thus σ P ( α 1,..., α g ) s n (H 0 (L (n 1) )). By inductive hypothesis H 0 (L (n 1) ) is in the image of (2.3.10); since s n is multiplication by τ H 0 (L) it follows that σ is in the image of (2.3.10) as well. We have proved that (2.3.10) is surjective. Since there exists n such that L n is very ample it follows that L is very ample as well. This proves that either (a) or (c) holds or else every smooth C L is hyperelliptic. Since the canonical map of a hyperelliptic curve has degree 2 onto its image it follows that ϕ L has degree 2 onto its image. We have proved that one of (a), (b), (c) holds. It remains to prove the statements on global generation and very ampleness of powers L n. First we prove that L n is globally generated for n 2. If Item (b) or (c) there is nothing to prove because L is already globally generated. Suppose that Item (a) holds. Let s prove that L 2 is globally generated. Since the base-locus of L is the rational curve R and L 2 R = OR (1) it suffices to prove surjectivity of the map H 0 (L 2 ) H 0 (L 2 R ) induced by the exact sequence 0 L 2 O X ( R) L 2 L 2 R 0, (2.3.11) Now L 2 O X ( R) = O X (2gC + R). One proves that h 1 (O X (2gC + R)) = 0 as follows. By Serre duality h 1 (O X (2gC + R)) = h 1 (O X ( 2gC R)). Let C 1,..., C 2g C be pairwise distinct; then

16 16 CHAPTER 2. CURVES AND LINEAR SYSTEMS ON K3 SURFACES C C 2g + R is a reduced connected curve and hence h 0 (O C1+...+C 2g+R)) = 1: arguing as in Remark we get that h 1 (O X ( 2gC R)) = 0. This proves that L 2 is globally generated. It follows that there exists a smooth C L 2. For every n 2 the restriction L n C is globally generated; arguing as in the proof of exactness of (2.3.9) we get surjectivity of H 0 (L n ) H 0 (L n C ) for n 2. Let n 2: since L n C is globally generated get that L n is globally generated as well. It remains to prove that L n is very ample for n 3. In all cases there exists a smooth C L 3. The multiplication map S N 3 H 0 (L C ) H 0 ((L C ) 3 ) H 0 ((L C ) N ) is surjective for N 3. Arguing as above we get that S N 3 H 0 (L) H 0 (L 3 ) H 0 (L N ) is surjective for N 3. By hypothesis L N is ample for large N; it follows that L n is very ample for n 3.

17 Chapter 3 Periods 3.1 Introduction Let X be a K3 surface. By definition there exists a nowhere vanishing holomorphic 2-form σ whose cohomology class generates H 2,0 (X). The periods of σ are the integrals σ where γ is any (piecewise γ smooth) 2-cycle; these integrals determine the line H 2,0 (X) H 2 (X; Z) C. Clearly the periods are determined up to a common scalar factor. More intrinsically: we are interested in the integral Hodge structure on H 2 (X; C) determined by the Hodge decomposition and the integral lattice H 2 (X; Z) equipped with the symmetric bilinear form given by cup-product (there is no torsion in the cohomology of a K3, we will prove it later on). Periods are a key tool in the theory of K3 surfaces. We will give the proof that periods determine locally the isomorphism class of a K3 (the so-called local Torelli Theorem). We will also use the period map to prove (following Kodaira) that any two K3 s are deformation equivalent. We will introduce the (weak and strong) Global Torelli Theorem for K3 s, the proof will be given in Chapter 4. Remark The set of integrals of a non-zero holomorphic 2-form σ determine the integral Hodge structure on H 2 (X; C). In fact the integrals determine H 0,2 (X) and H 0,2 (X) = H 2,0 (X), H 1,1 (X) = (H 2,0 (X) H 0,2 (X)). 3.2 Topology of K3 surfaces For us a lattice is a finitely generated free abelian group L equipped with a symmetric bilinear form L L Z which we denote (, ) L ; we let q L be the associated quadratic form. Recall: a lattice is even if q L (α) is even for all α L. Let (L 1, (, ) L1 ), (L 2, (, ) L2 ) be lattices; an isomorphism of groups f : L 1 L 2 is an isometry of lattices if (α, β) L1 = (f(α), f(β)) L2 for all α, β L 1. Let X be a K3 surface. Let Tors < H 2 (X; Z) be the torsion subgroup. We let H 2 (X; Z) := H 2 Z(X; Z)/ Tors. (Soon we will show that there is no torsion in H 2 (X; Z), see Corollary ) Thus H 2 (X; Z) < H 2 (X; C) by the universal coefficient Theorem. Let [X] H 4 (X; Z) be the fundamental class determined by the (complex) orientation of X. The intersection-form H 2 (X; C) H 2 (X; C) C (α, β) [X] α β equips H 2 (X; Z) with a structure of lattice. One determines the isometry class of H 2 (X; Z) as follows. Claim Let X be a K3 surface. Then b 2 (X) = 22 and h 1,1 (X) =

18 18 CHAPTER 3. PERIODS Proof. By (2.2.2) the Euler characteristic of X is 24; by definition we have b 1 (X) = 0 and hence b 3 (X) = 0 by Poincarè duality; it follows that b 2 (X) = 22. The Hodge decomposition gives that 22 = b 2 (X) = h 2,0 (X)+h 1,1 (X)+h 0,2 (X). By definition h 2,0 (X) = 1 and hence Hodge simmetry gives that h 0,2 (X) = 1; it follows that h 1,1 (X) = 20. The signature of the intersection form on H 2 (X; R) is (3, 19): in fact the maximum dimension of a subspace of H 2 (X; R) on which the intersection-form is positive-definite equals 2h 2,0 (X) + 1 = 3 by the Hodge index Theorem. By Poincaré duality the intersection form on H 2 (X; Z) is unimodular. Lastly we claim that the intersection form on H 2 (X; Z) is even. By Poincarè duality it suffices to show that the intersection form on H 2 (X; Z)/ Tors is even. Let w 2 (X) H 2 (X; Z/(2)) be the 2-nd Stiefel-Whitney class of X; one has γ γ + w 2 (X), γ 0 (mod 2), γ H 2 (X; Z) (3.2.1) (represent γ by an immersed smooth oriented surface f : Σ X and compute w 2 (X), γ by splitting f T X = T Σ N Σ/X ). On the other hand we have the general formula (see Problem 14-B of [12]) w 2 (X) = [c 1 (X)] H 2 (X; Z/(2)). (3.2.2) By definition c 1 (X) = 0; thus Equations (3.2.1) and (3.2.2) give that the intersection form on H 2 (X; Z)/ Tors is even. By the classification of unimodular indefinite lattices [15] we get that the intersection form on H 2 (X; Z) is isometric to the lattice Λ := H H H E 8 ( 1) E 8 ( 1) (3.2.3) where H is the hyperbolic lattice i.e. the unique even unimodular lattice of rank 2 and signature (1, 1) and E 8 ( 1) is the unique even unimodular negative-definite lattice of rank Periods of marked K3 s Let X be a K3 surface. Let σ H 2,0 (X) be non-zero: a pointwise computation gives that σ σ = 0, σ σ > 0. (3.3.1) [X] It follows that the periods of X are not an arbitrary set of 22 numbers. Now let L be a lattice such that (, ) L is non-degenerate of signature (3, b 3). (3.3.2) We will define the period domain Ω L associated to L - the definition is motivated by (3.3.1). Fater that we will show how to map a K3 with a chosen isometry H 2 (X; Z) Λ (here Λ is as in (3.2.3)) to a point of Ω Λ. For a commutative ring R let L R := L Z R (usually R = Q, R or C). The scalar product (, ) L on L extends to an R-valued bilinear symmetric non-degenerate form on L R ; given α, β L R we denote their product by (α, β) L or simply (α, β) if there is no risk of confusion. We let q L be the associated quadratic form on L R. Let Ω L P(L C ) defined by [X] Ω L := {[α] P(L C ) q L (α) = 0, q L (α + α) > 0}. (3.3.3) Thus Ω L is an open subset (in the classical topology) of a smooth quadric of dimension (b 2) and hence it is naturally a complex manifold: it is the period domain associated to L. Up to isomorphism Ω L depends only on the rank b of L - on the other hand O(L) acts naturally on Ω L and the different realizations of Ω L correspond to different group actions. Let Λ be the lattice given by (3.2.3): then Ω Λ is the period domain for K3 surfaces. Thus Ω Λ is an open subset (in the classical topology) of a 20-dimensional smooth quadric. Let X be a K3 surface: in Section 3.2 we proved that there exists an isometry f : H 2 (X; Z) Λ.

19 3.3. PERIODS OF MARKED K3 S 19 Such an isometry is a marking of X, and the couple (X, f) is a marked K3. Let (X, f) be a marked K3. Abusing notation we let f : H 2 (X; C) Λ C be the isometry obtained by extending scalars (recall that H 2 (X; Z) H 2 (X; C)). By (3.3.1) f(h 2,0 (X)) Ω Λ ; we associate to the marked K3 surface (X, f) its period point P (X, f) := f(h 2,0 (X)) Ω Λ. Definition Let (X 1, f 1 ) and (X 2, f 2 ) be marked K3 s: an isomorphism ϕ: X 1 isomorphism of marked K3 s if the following diagram is commutative: X2 is an (3.3.4) H 2 (X 2 ; Z) H 2 (ϕ) H 2 (X 1 ; Z) f 2 f 1 As is easily checked isomorphism of marked K3 s is an equivalence relation. Λ Remark Let ϕ: X 1 X2 be an isomorphism of the marked K3 surfaces (X 1, f 1 ) and (X 2, f 2 ). Since ϕ H 2,0 (X 2 ) = H 2,0 (X 1 ) we have P (X, f 1 ) = P (X, f 2 ). Thus isomorphic marked K3 s have the same period point. Now let s go back to the period domain Ω L associated to a lattice L such that (3.3.2) holds. We will give a different realization of Ω L : it will turn out to be very convenient when proving the Global Torelli Theorem. Definition Let Gr + (2, L R ) Gr(2, L R ) be the set of 2-dimensional subspaces of L R on which q L is positive definite. Let Gr or + (2, L R ) be the set of couples (V, τ) where V Gr + (2, Λ R ) and τ an orientation of V. Thus Gr or + (2, L R ) is naturally a smooth manifold so that the forgetful map Gr or + (2, L R ) Gr + (2, L R ) is a smooth double covering. We define a diffeomorphism Ω L Gr or + (2, L R ) [α] (V α, τ α ) (3.3.5) as follows. We let V α := {xα + xα x C} (3.3.6) be the set of real vectors in the subspace [α] + [α]. We have a well defined R-linear isomorphism [α] V α ; we let τ α be the orientation corresponding to the complex orientation of [α]: an oriented basis of V α is {R(α), I(α)}. Claim The period domain Ω L has the homotopy type of S 2, in particular it is simply connected. Proof. Let s prove that Gr + (2, L R ) has the homotopy type of the real projective plane P 2 R. Write L R = P N where P N, q L P > 0 and q L N < 0. Thus dim P = 3 and dim N = 19. Let π : L R P be projection. Let V Gr + (2, L R ); then V N = {0} and hence π(v ) Gr(2, P ). Thus π defines a map f : Gr + (2, L R ) Gr(2, P ) which is clearly continuous. Let i: Gr(2, P ) Gr + (2, L R ) be inclusion. We have f i = Id Gr(2,P ). One checks easily that i f is homotopic to the identity. This proves that Gr + (2, L R ) is homotopy equivalent to Gr(2, P ) = P 2 R. A similar argument shows that Gror + (2, L R ) is homotopy equivalent to the universal cover of P 2 R i.e. S2.

20 20 CHAPTER 3. PERIODS 3.4 Infinitesimal Torelli A family of K3 surfaces consists of a proper flat map π : X B of analytic spaces such that each fiber X b := f 1 (b) is a K3 surface - B is the base of the family 1. In the present subsection we will consider exclusively families with smooth base: in that case flatness amounts to π being a submersive map of complex manifolds. Suppose that the local system R 2 π Z is trivial, e.g. if B is simply connected. Choose a trivialization of F : R 2 π Z B Z 22 ; it defines an isomorphism f b : H 2 (X b ; Z) Λ for each b B. If f b is an isometry for each b we say that F defines a marking of the family π : X B. Notice that if R 2 π Z is trivial there do exist markings of π. Given a marking F of π we have the period map B Pπ Ω b P (X b, f b ). (Of course P π depends on the trivialization chosen, our notation is somewhat imprecise.) A fundamental result of Griffiths [21] (valid for arbitrary families of Kähler manifolds) asserts that the period map is holomorphic and computes its differential as follows. Let σ be a non-zero holomorphic 2-form on X b and α := f b (σ). Then P π (b) = [α]. Since Ω is an open subset of the quadric defined by q Λ we have a canonical isomorphism T [α] Ω = Hom(α, α /Cα) where perpendicularity is with respect to q Λ. Let Θ Xb be the (holomorphic) tangent bundle of X b. Contraction with σ defines an isomorphism and hence an isomorphism for every q. Let L σ : H q (Θ X ) T b B Θ = Ω 1 H q (Ω 1 X) (3.4.1) κ H 1 (X b ; Θ Xb ) be the Kodaira-Spencer map associated to the family X. Griffiths formula asserts that for v T b B where we make the identification dp π (v), α f b (L σ (κ(v))) (mod [σ]), (3.4.2) H 1 (Ω 1 ) = σ /Cσ via the Hodge decomposition. The infinitesimal Torelli Theorem follows from the above formula and the result below. Proposition Let X be a K3 surface. Then X has a universal unobstructed deformation space of dimension 20. Proof. Let s prove that 0 = h 0 (Θ X ) = h 2 (Θ X ). (3.4.3) By definition of K3 we have h 0 (Ω 1 X ) = 0 and hence h0 (Θ X ) = 0 by (3.4.1). Serre duality gives that h 2 (Ω 1 X ) = h0 (Θ X ) and hence h 2 (Ω 1 X ) = 0; by (3.4.1) we get that h2 (Θ X ) vanishes as well. By Kodaira-Spencer s general results on deformations of compact complex manifolds it follows that Def(X) is smooth with tangent space isomorphic to H 1 (Θ X ); by (3.4.1) and Claim we get that dim Def(X) = 20. Let X be a K3 and π : X U a representative of Def(X), i.e. we are given 0 U and an isomorphism X 0 X which induce an isomorphism of germs (B, 0) Def(X) (recall that Def(X) is a germ of analytic space). By shrinking U we may assume that R 2 f Z is trivial and hence a marking F of π exist. The result below could be named Local surjectivity and Infinitesimal Torelli, it often goes under the name of Local Torelli (but this name should go to a different statement, see??). 1 We emphasize that X and B are allowed to be singular or non-reduced, e.g. B = Spec(C[t]/(t 2 ))

21 3.5. NOETHER-LEFSCHETZ LOCI AND K3 SURFACES UP TO DEFORMATION 21 Theorem (Andreotti - Weil). Keep notation and hypotheses as above. Then P π : U Ω defines an isomorphism of an open neighborhood of 0 onto an open neighborhood of P (X 0, f 0 ) in Ω. Proof. By the implicit function Theorem it suffices to prove that the differential dp π (0): T 0 U T P (X0,f 0)Ω is an isomorphism. Since π : X U is a representative of Def(X) the Kodaira-Spencer map for U identifies the tangent space to U at 0 with H 1 (X; Θ). Thus dp π (0) is an isomorphism by Isomorphism (3.4.1) (for q = 1) and Formula (3.4.2). We spell out the meaning of the injectivity part of the above theorem. First a definition. Definition Let X, X be K3 surfaces. A map ϕ: H 2 (X; C) H 2 (X ; C) is an integral Hodge isometry if the following hold: (1) ϕ is an isometry with respect to the intersection forms, (2) ϕ is integral i.e. ϕ(h 2 (X; Z) = H 2 (X ; Z), (3) ϕ is a morphism of Hodge structures i.e. ϕ(h p,q (X; Z) = H p,q (X ; Z) for all p, q. Remark Suppose that (1) and (2) hold. Then in order that (3) hold it suffices that ϕ(h 2,0 (X; Z)) = H 2,0 (X ; Z): this follows from Remark Let U be as above; by Theorem we may assume, after shrinking U, that P π is an isomorphism of U onto an open subset of Ω. For b U let γ b : H 2 (X 0 ; Z) H 2 (X b ; Z) be the isomorphism given by parallel transport by the Gauss-Manin connection (equivalently: determined by a trivialization of R 2 π Z). Clearly γ b is an isometry. Let b 1, b 2 U; Theorem states (beyond local surjectivity of the period map) that if γ b2 γ 1 b 1 : H 2 (X b1 ) H 2 (X b2 ) is an integral Hodge isometry then X b1 is isomorphic to X b2. We emphasize that Theorem does not prove that if ϕ: H 2 (X b1 ) H 2 (X b2 ) is an arbitrary Hodge isometry then X b1 is isomorphic to X b2. The latter statement is the content of the celebrated (weak) Global Torelli for K3 surfaces - we will prove it in Chapter Noether-Lefschetz loci and K3 surfaces up to deformation Let X be a K3 surface. Let H 1,1 Z (X) < H 2 (X; Z) be the quotient of H 1,1 Z modulo the torsion subgroup. Let S X := H 1,1 Z (X) = c 1 (Pic(X))/ Tors be the lattice of algebraic cycles (the quadratic form is the restriction of the intersection-form on H 2 (X; Z)). The transcendental lattice T X < H 2 (X; Z) is the smallest lattice L < H 2 (X; Z) such that H 2,0 (X) L C and H 2 (X; Z)/L is torsion-free. It is clear that the Hodge structure on T X C has no non-trivial sub Hodge structures (we stipulate that the integral Hodge substructure is trivial if and only if it is trivial after tensorization by Q); it follows that T X SX. If the restriction of the intersection-form on S X is non-degenerate (e.g. if X is projective) then T X := SX. By Lefschetz (1, 1)-Theorem S X is completely determined by the periods of X: if σ is a symplectic form on X, then Let S < Λ be a sublattice; we let Ω S Ω be given by S X = H 2 (X; Z) σ. (3.5.1) Ω S := {[α] Ω α S}.

22 22 CHAPTER 3. PERIODS Let (X, f) be a marked K3; by (3.5.1) we have P (X, f) Ω S if and only if f 1 (S) S X. For v Λ we set Ω v := Ω Zv. We may tie in the computation of effective parameters for the examples given in Section 1.2 with the period map as follows. Let X 0 P 3 be a quartic surface and f 0 : H 2 (X 0 ; Z) Λ be a marking of X 0. Let h 0 := c 1 (O X0 (1)) and v := f(h 0 ). Then P (X 0, f 0 ) Ω v because h 0 H 1,1 (X 0 ; Z). Let π : X U be a representative of Def(X 0 ) with a marking F which is equal to f 0 on the fiber X 0 ; we assume that U is small, in particular P π : U Ω is an embedding. Let t Pπ 1 (Ω v ) and h t := ft 1 (v). Thus h t H 1,1 (X t ; Z) and hence there exists a divisor H t on X t such that h t = c 1 (O Xt (H t )). By semicontinuity arguments we get that H t is very ample (recall: U is arbitrarily small). It follows that ϕ t : X t H t embeds X t as a quartic surface in P 3. Viceversa if Y is a quartic surface close to X 0 then it is isomorphic to X t for some t Pπ 1 (Ω v ). Thus we see that periods of quartics fill out a 19-dimensional subset of Ω - this confirms the computations of Section 1.2. Noether-Lefschetz loci are used in the proof that K3 s form a single deformation class. First we recall the relevant definition. Let X 1 and X 2 be two compact complex manifolds. They are directly deformation equivalent if there exist a complex manifold X, a proper submersive map f : X B of complex manifolds where B is connected, and b 1, b 2 B such that X bi = Xi for i = 1, 2. Two manifolds X 1 and X 2 are deformation equivalent if there exists a chain of directly deformation equivalent manifolds starting with X 1 and ending with X 2 i.e. deformation equivalence is the equivalence relation generated by direct deformation equivalence. By Ehresmann s Theorem [?] two deformation equivalent manifolds are diffeomorphic. Theorem (Kodaira [?]). There is a single class of deformations of K3 surfaces i.e. two arbitrary K3 surfaces are deformation equivalent. Kodaira proved that any K3 may be deformed into an elliptic one, then he proved that elliptic K3 s form a connected family. We will give a similar proof: we will show that any K3 may be deformed to a double cover of P 2 ramified over a smooth sextic. First we will go through a series of preliminaries. Given k Z let P 2k Λ be the set of non-zero primitive vectors of square 2k (i.e. vectors v such that q Λ (v) = 2k). Lemma Let w P 0 and k Z. There exists a sequence {w n } with w n P 2k such that lim n [w n ] [w] (here [w n ] and [w] are the points in P(Λ R ) determined by w n and w respectively). Proof. Since Λ is unimodular and w is primitive there exists w such that (w, w ) Λ = 1. The sublattice generated by w and w is isomorphic to H. The lattice H is unimodular. By the classification of unimodular even lattices we get that H = H 2 E 8 ( 1) 2 ; thus we have an orthogonal decomposition Λ = H (H 2 E 8 ( 1) 2 ) (3.5.2) with w belonging to the first copy of H. Let {u, u } be a standard basis of one of the copies of H appearing in the second addend of (3.5.2), i.e. u, u P 0 and (u, u ) Λ = 1. Let w n := (nw + u + ku ); then w n P 2k and lim n [w n ] [w]. Let Q Λ P(Λ C ) be the cone of isotropic lines in Λ C. We let Q Λ (Q) and Q Λ (R) be the sets of isotropic lines spanned by a vector with rational or real coordinates respectively. Proposition Let k Z. Then Ω v is dense in Ω. v P 2k Proof. Let [α] Ω; then α Q Λ (R) = [V α ] Q Λ (R) where V α is as in (3.3.6). Since the restriction of (, ) to V α is of type (1, 19) there exists [v] α Q Λ (R).

23 3.6. ISOMORPHISMS AND ISOMETRIES 23 Since Q Λ (Q) is not empty, it is dense in Q Λ (R). Thus there exists a sequence {[v n ]} converging to [v], where v n P 0. By Lemma we may replace the vectors v n by elements u n P 2k. Thus every point of v Ω is a limit of points in Ω un. Since α v Ω we get that α n Ω un. Proof of Theorem Let X 0 be a K3 surface. Let π : X U be a representative of Def(X 0 ). By shrinking U we may assume that R 2 f Z is trivial and hence a marking F of π exist. By Theorem we may also assume that P π is an embedding of U as an open subset of P (X 0, f 0 ) Ω. Let k Z. By Theorem and Proposition X 0 may be deformed to a K3 surface X t which has primitive h t H 1,1 (X t ) of square 2k. Applying again Theorem one can ensure that h 1,1 (X t ) = 1. Now we assume that k = 1 and let L t be a (holomorphic) line-bundle such that c 1 (L t ) = h t. By Riemann-Roch and Serre duality we have h 0 (L t ) + h 0 (L 1 t ) 3. Without loss of generality we may assume that h 0 (L t ) > 0; thus h 0 (L t ) 3. An easy argument (use Remark 2.2.1) shows that h 0 (L t ) = 3. Since h t = c 1 (L t ) generates H 1,1 (X t ) the linear system L t does not have fixed components. By Proposition we get that L t has no base-points. Thus the map ϕ Lt X t L t = P 2 is a regular double cover ramified over a smooth sextic. This proves that X is a deformation of a double plane ramified over a smooth sextic: since K3 surfaces of this type are all deformation equivalent, we conclude that two K3 surfaces are deformation equivalent. Corollary Let X be a K3 surface. Then X is simply connected, in particular H 2 (X; Z) has no torsion. Proof. By Theorem X is diffeomorphic to a smooth quartic in P 3. By the LHT a smooth quartic is simply connected; it follows that X is simply connected. Since H 1 (X; Z) = 0 the universal coefficient Theorem gives that H 2 (X; Z) = Hom(H 2 (X; Z), Z) and hence H 2 (X; Z) has no torsion. 3.6 Isomorphisms and isometries Let f : X Y be an isomorphism between K3 surfaces. Then f : H 2 (Y ) H 2 (X) is an integral Hodge isometry. Thus we have a map Iso(X, Y ) HIsom(H 2 (Y ), H 2 (X)) (3.6.1) between the set of isomorphisms from X to Y and the set of integral Hodge isometries from H 2 (Y ) to H 2 (X). Proposition Map (3.6.1) is injective. Proof. If Iso(X, Y ) = there is nothing to prove. If Iso(X, Y ) we choose a reference isomorphism ψ 0 and we write every isomorphism as ψ 0 ϕ where ϕ Aut(X). Thus we are reduced to proving that (3.6.1) is injective for X = Y. Then (3.6.1) is a homomorphism of groups and hence it suffices to prove that if H 2 (ϕ) = Id H 2 (X) then ϕ = Id X. Let π : X U be a representative of Def(X) with 0 U the reference point i.e. X 0 = X. By the functorial property of Def(X) there exist an open 0 V U, a map m: V U and an isomorphism Φ: X V X (here X V := π 1 V ) fitting into the commutative diagram Φ X (3.6.2) X V π V V m π U

24 24 CHAPTER 3. PERIODS and such that Φ(0) = ϕ 0. Since H 2 (ϕ) = Id H2 (X 0) we have P π m = P πv. By Local Torelli we get that m is the inclusion of V : thus we may assume that V = U and m = Id U. This means that Φ(t) is an automorphism ϕ t : X t X t : in other words ϕ exetnds to an automorphism of all K3 surfaces close to X. By the density result Proposition there exists a codimension-1 hypersurface D U such that for all t D there exists h t H 1,1 Z (X t) with q Xt (h t ) = 2. For a very general t we have (X t) = Zh t and hence ±h t is ample: it follows that there is an open D 0 D such that X t for H 1,1 Z t D 0 is a double covering of the plane branched over a sextic B t and moreover up to isomorphism we get an open subset of such double covers. If H 1,1 Z (X t) = Zh t then necessarily ϕ t (h t ) = h t it follows that ϕ t (h t ) = h t for all t D 0. By the openness result the sextic B t has no non-trivial automorphisms: it follows that ϕ t is the covering involution, that is absurd because the covering involution does not act as the identity on H 2 (X t ). 3.7 The Kähler cone A cone in a finite-dimensional real vector space V is a subset C V which is stable under multiplication by any strictly positive real number (thus C need not contain 0). Let X be a K3 surface. We let C X := {z H 1,1 R (X) (z, z) X > 0}. (3.7.1) Thus C X is a cone: it has two connected components because the intersection form restricted to H 1,1 R (X) has signature (1, h 1,1 X 1). The Kähler cone is the cone K X H 1,1 R (X) of Kähler classes. We have K X C X : thus there is a distinguished connected component C + X of C X containing K X, it is the positive cone. We have C X = C + X ( C + X ). Let X := {δ H 1,1 Z (X) (δ, δ) X = 2}, + X := {δ X δ = c 1 (O X (R)), R Eff(X)}. (3.7.2) Let δ X : by Riemann-Roch we have that either δ or δ is effective: thus X = + X ( + X ). (3.7.3) Let δ X ; then δ C X is not empty and the collection of all δ C X is a locally finite collection of closed subsets of C X. Thus the complement C X \ δ (3.7.4) δ X is an open subset of C X. A connected component of (3.7.4) is an open chamber of C X. Let ω H 1,1 R (X) be a Kähler class: then (ω, δ) X > 0 for every δ + X : by (3.7.3) it follows that K X belongs to a single open chamber of C X. Proposition Let X be a K3 surface. The Kähler cone K X is an open chamber of C X. Proof. Let A X be the open chamber of C X containing K X. Let α A X and ω K X. By [2] it suffices to prove that (α, c 1 (O X (R))) X > 0 for every irreducible curve R X of (strictly) negative self-intersection and moreover (α, ω) X > 0. (This is an analogue of the Nakai-Moishezon ampleness criterion: it was generalized to arbitrary compact Kähler manifolds by D ly and Paun [5].) Since R R < 0 we have R R = 2 and hence δ := c 1 (O X (R)) + X. We have (ω, δ) X > 0 and since α, ω belong to the same open chamber it follows that (α, c 1 (O X (R))) X > 0. Now let s prove that (α, ω) X > 0. If α Rω then α is a positive multiple of ω because it belongs to C + X and hence (α, ω) X > 0. Thus we may suppose that α, ω are linearly independent. The restriction of q X to Rα Rω has signature (1, 1) by the Hodge index Theorem. Since α, ω both belong to C + X it follows that they both belong to the same connected component of {xα + yω q X (xα + yω) > 0}: it follows that (α, ω) X > 0. Given only the Hodge structure we cannot pin down which one of the open chambers is the Kähler cone because the group of integral Hodge self-isometries acts transitively on the family of chambers of C X. Since multiplication by ( 1) exchanges C + X and ( C+ X ) it suffices to check that integral Hodge