Rational Curves On K3 Surfaces

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1 Rational Curves On K3 Surfaces Jun Li Department of Mathematics Stanford University Conference in honor of Peter Li

2 Overview of the talk The problem: existence of rational curves on a K3 surface The conjecture: infinitely many rational curves on an algebraic K3 The technique: lifting from a special K3 to a general K3 The new approach: of Bogomolov-Hassett-Tschinkel, using the jumping of Picard ranks Theorem (L-Liedtke): When ρ(x ) is odd, then X contains infinitely many rational curves.

3 The Problem Problem: Find rational curves in a complex K3 surfaces. K3 surfaces: smooth, compact complex surface S satisfying π 1 (S) = {1}; K S = OS ; Examples: this is 2 T C X = O S ; By Yau s celebrated theorem on Calabi conjecture, Complex K3 surfaces are Ricci flat Kahler metrics, and vice versa. degree four hypersurfaces: S = (p 4 = 0) CP 3 ; resolutions of a torus quotient by involution: S = (T 2 C / τ )desing.

4 Rational curves: image of a u : CP 1 S non-constant; generic one-one onto image; holomorphic; distinct rational curves if they have different images. Distinct rational curves: if two have distinct images.

5 Main questions: given a compact, complex K3 surface S, Q1 : does S contain a rational curve? Q2 : if does, does it contain more than one rational curves? Q1 : if does, does it contain infinitely many rational curves? Q1 : if does, does it contain a continuous family of rational curves?

6 The existence of (at least) one rational curve Theorem (Bogomolov-Mumford) The existence of L = OS produces rational curves. Outline L = OS produces s 0 H 0 (L n ), C = s 1 (0) a curve in S; c 1 (L) 2 = 0 or = 2 are ok; (elliptic fibration; ( 2)-curves); c 1 (L) 2 > 0 implies S is algebraic (projective).

7 For S projective, find a K3 X 0 so that it has two smooth rationals C 1 and C 2 so that they intersects transversally; let L 0 = O X0 (C 1 + C 2 ); find a deformation (X t, L t ) of (X 0, L 0 ), with (X 1, L 1 ) = (S, L); show that C 1 C 2 X 0 deforms to rationals C t X t, for general t; (this argument will be revisited) the limit of rational curves are union of rational curves, so X 1 = S contains rational curves.

8 Theorem (Lefschetz) A simply connected compact complex manifold S has a line bundle L = OS if (the Hodge classes) H 1,1 (S, C) H 2 (S, Z) {0}. Answer to Q1: S has a rational curve if H 1,1 (S, C) H 2 (S, Z) {0}.

9 Remarks We have three different proofs of this existence theorem Bogomolov-Mumford: find the existence of a special K3, using deformation to prove the existence for general K3, plus limit of rational curves is a union rational curves; Mori: using reduction to char= p technique, not relying on deformation of K3 surfaces; Yau-Zaslow: using moduli of sheaves on K3, plus beautiful geometry of Hyperekahler manifolds; not relying on deformation of K3 surfaces; the theorem holds for any field k = k.

10 The existence of more than one rational curves Corollary: The space H 1,1 (S, C) H 2 (S, Q) {0} is spanned by classes generated by rational curves. Answer to Q2: The number of rational curves is # dim H 1,1 (S, C) H 2 (S, Q).

11 No family of rational curves Answer to Q4: There can be no continuous family of rational curves. Otherwise, a. we can find a birational morphism ϕ : X S, X is a ruled surface (generic P 1 -bundle); b. S K3 implies there is a non-zero (2, 0)-form on S; c. ϕ (Ω) = 0 from (b), 0 from (a), a contradiction.

12 The Conjecture (on Q3) Conjecture: Any smooth complex K3 surface S contains infinitely many rational curves. This is motivated by Lang s conjecture: Lang Conjecture: Let X be a general type complex manifold. Then the union of the images of holomorphic u : C X lies in a finite union of proper subvarieties of X.

13 Key to the existence of rational curves: A class α 0 H 2 (S, Z) is Hodge (i.e. H 1,1 (S, C) H 2 (S, Q)) is necessary and sufficient for the existence of a union of rational curves C i so that [C i ] = α. Example: Say we can have a family S t, t disk, α H 1,1 (S 0, C) H 2 (S 0, Q) so that S 0 has C 0 = CP 1 S 0 with [C 0 ] = α; in case α H 1,1 (S t, C) for general t, then CP 1 = C 0 S 0 can not be extended to holomorphic u t : CP 1 S t.

14 Preparations We will consider polarized K3 surfaces (S, H), c 1 (H) > 0; we can group them according to H 2 = 2d: M 2d = {(S, H) H 2 = 2d}. each M 2d is smooth, of dimension 19; each M 2d is defined over Z. (defined by equation with coefficients in Z.) to show that (S, H) contains infinitely many rational curves, it suffices to show that for any N, there is a rational curve R S so that [R] H N. we define ρ(s) = dim H 1,1 (S, C) H 2 (S, Q), called the rank of the Picard group of S.

15 The existence of rational curves on complex algebraic K3 surfaces Theorem (Chen, 99): Very general polarized complex K3 surfaces (i.e. (S, H)) contains infinitely many rational curves. Theorem: A complex elliptic K3 surface contains infinitely many rational curves. Theorem: A complex elliptic K3 surface with infinite automorphism group contains infinitely many rational curves.

16 Theorem (Bogomolov-Hassett-Tschinkel, 10) A degree 2 polarized complex K3 surface (S, H) (i.e. H 2 = 2) having ρ(s) = 1 contains infinitely many rational curves. Extending their method, we prove Theorem (L Liedtke, 11) A polarized complex K3 surface (S, H) having ρ(s) = 1 contains infinitely many rational curves. ρ can run between 1 and 20; A K3 with ρ 5 is an elliptic K3, thus covered by an earlier theorem.

17 Technical approach Extension Problem: a family of K3 surface S t, t T (a parameter space); C 0 S 0 a union of rational curves; let α = [C 0 ] H 2 (S, Z); (it is a Hodge class); suppose α H 1,1 (S t, C) for all t T ; We like to show that exists a family of curves C t S t, such that Ct are union of rational curves; C0 = C 0.

18 Use moduli of genus 0 stable maps A genus zero stable map is a holomorphic map u : C X such that C is a nodal, genus 0 complex curve (i.e. a tree of CP 1 s); for any Σ = CP 1 C with u Σ = const., Σ contains 3 nodes of C.

19 Moduli of genus 0 stable maps A genus zero stable map is a holomorphic map u : C X such that C is a nodal, genus 0 complex curve (i.e. a tree of CP 1 s); for any Σ = CP 1 C with u Σ = const., Σ contains 3 nodes of C. Let S be a K3 surface, α H 2 (S, Z), M 0 (S, α) = {[u : C S] genus 0 stable maps with u ([C]) = α} is an algebraic space; has exp.dimension 1. (expected, as when α is not Hodge, then M0 =.)

20 Extension Lemma S t, t T, be a family of K3 surfaces; α H 2 (S 0, Z) is Hodge for all t; [u 0 ] M 0 (S 0, α). Extension Lemma (Ran, Bogomolov-Tschinkel, ): Suppose [u 0 ] M 0 (S 0, α) is isolated, then u 0 extends to u t M 0 (S t, α) for general t T.

21 Extension Lemma (Ran, Bogomolov-Tschinkel, ): Suppose [u 0 ] M 0 (S 0, α) is isolated, then u 0 extends to u t M 0 (S t, α) for general t T. Outline of the proof let U be an open neighborhood of all deformations of S 0 ; U is a smooth complex manifold of C dimension 20; let X U be the tautological family of K3 surfaces; form M 0 (X, α), the moduli of genus zero stable morphisms u : C X z, for some z U. exp. dim M 0 (S 0, α) = 1 plus dim U = 20 implies exp. dim M 0 (X, α) = 19. Let [u 0 ] W M 0 (X, α) be an irreducible component; dim W 19.

22

23 Look at π : W U, π(w ) U has dimension at least 19; U(α) U is the locus where α H 2 (S z, Z) remain Hodge; dim U(α) = 19; π(w ) U(α); by assumption, T U(α); This implies that T π(w ); thus u 0 lifts to a family u t : C t S t, t T.

24 Extension criterion Definition: We say a map [u] M 0 (S, α) rigid if [u] is an isolated point in M 0 (S, α). Extension principle: In case (a genus zero stable map) u : C S is rigid, then u extends to nearby K3 surfaces as long as the class u [C] H 2 (S, Z) remains Hodge.

25 Lifting to a general K3 surface Given an (S, H) and an integer N, to construct a rational R S of [R] H N, we will find a family of K3 surfaces (X t, H t ), t T a parameter space, such that (S, H) is a general member of (Xt, H t); for a 0 T, we have C0 X 0 a union of rational curves, and an irreducible R 0 C 0 so that [R 0] H 0 N; [C0] = nh 0; we can represent C0 as the image of a rigid genus zero stable map [u 0]. By Extension criterion: we can extend u 0 to u t for general t; since t is general, there is an irreducible component R t of image(u t ) so that [R t ] H t [R 0 ] H 0 N; since (S, H) is a general member, we can assume (S, H) = (X t, H t ).

26

27 Theorem (Chen) Very general K3 (S, H) M 2d contains infinitely many rational curves. Proof Form a family, so that (X 0, H 0 ) is a singular K3 surface, thus C 0 X 0 can be constructed explicitly. Applying the lifting property. So for any N, all K3 s in a dense open subset in M 2d contains rational curves of degree N.

28 Theorem (Bogomolov-Hassett-Tschikel). Fix an r. Assume that every K3 (X, H) M 2d (of ρ(x ) = r) defined over a number field K contains infinitely many rational curves in X Q, then every (X, H) M 2d contains infinitely many rational curves. Let X be any K3, not defined over a number field, then X is the generic fiber of X S, S a variety over K; Example X = (πx x x x 4 4 = 0) CP3. Switch π by t, we define X = (tx x x x 4 4 = 0) P3 A 1, S = A 1.

29 Theorem (Bogomolov-Hassett-Tschikel). Fix an r. Assume that every K3 (X, H) M 2d (of ρ(x ) = r) defined over a number field K contains infinitely many rational curves in X Q, then every complex (X, H) M 2d contains infinitely many rational curves. Let X be any K3. Then X is the generic fiber of X S, S a variety over K; By (Deligne), we can find ξ S, over L/K, s.t. PicX = PicX ξ ; By assumption, C 0 X ξ, high degree; [C 0 X ξ ] M 0 (X ξ ) rigid; applying lifting criterion, [C 0 X ξ ] lifts to generic fiber of X B; thus lifts to C X, of high degree.

30 When (S, H) is defined over a number field, say Q, after deleting a finite prime numbers, (S, H) is the generic member of a family of K3 over T = specz finite places:

31 Theorem (Bogomolov-Zarhin, Nygaard-Ogus). Let X be a K3 over a number field K. Then 1. for all but finitely many places p of K, the reductions X p are smooth, not supersingular; 2. for all such places with char p 5, ρ((x p ) F p ) are even. Proof Use Tate Conjecture for K3; Use Weil Conjecture for K3.

32 Theorem (Bogomolov-Hassett-Tschinkel). Let (X, H) be a polarized K3 in M 2. Suppose ρ(x ) = 1, then X contains infinitely many rational curves. Sketch we only need to prove the case where X is defined over K, say K = Q; for any places p, ρ((x p ) F p ) 2 imply D p (X p ) F p rational curves, not in ZH; if D p H N uniformly, implies ρ(x Q ) 2, a contradiction;

33 Sketch places p, D p (X p ) F p rational curves, not in ZH, D p H arbitrary large; using H 2 = 2, find D p (X p ) F p rational curves, D p + D p n p H ; D p + D p is the image of a rigid genus zero stable map. Applying extension criterion, we can find a rational curve of degree N in the generic fiber, which is X.

34 Theorem ( Liedtke) Let (X, H) be a polarized complex K3 surface such that ρ(x ) is odd. Then X contains infinitely many rational curves. Outline of proof We only need to prove the Theorem for (X, H) defined over a number field K; say K = Q, we get a family X p for every prime p Z, X is the generic member of this family; p, exists D p X p, D p ZH, we have sup Dp H ; pick C p X p union of rationals, D p + C p n p H Difficulty: D p + C p may not be representable as the image of a rigid genus zero stable map.

35 Solution: Suppose we can find a nodal rational curves R X, of class kh for some k, then for some large m we can represent C p + D p + mr, which is a class in (n + mk)h, by a rigid genus zero stable map.

36 End of the proof: In general, X may not contain any nodal rational curve in kh. However, we know a small deformation of X in M 2d contains nodal rational curves in kh (a Theorem of Chen). Using this, plus some further algebraic geometry argument, we can complete the proof.

37 Happy Birthday, Peter.

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