Rational Curves On K3 Surfaces

Jun Li

Department of MathematicsStanford University

Conference in honor of Peter Li

Overview of the talk

I The problem: existence of rational curves on a K3 surface

I The conjecture: infinitely many rational curves on an algebraic K3

I The technique: lifting from a special K3 to a general K3

I The new approach: of Bogomolov-Hassett-Tschinkel, using thejumping of Picard ranks

I Theorem (L-Liedtke): When ρ(X ) is odd, then X contains infinitelymany rational curves.

The Problem

Problem: Find rational curves in a complex K3 surfaces.

K3 surfaces: smooth, compact complex surface S satisfying

I π1(S) = 1;I KS

∼= OS ;I this is ∧2TCX ∼= OS ;I By Yau’s celebrated theorem on Calabi conjecture, Complex K3

surfaces are Ricci flat Kahler metrics, and vice versa.

Examples:

I degree four hypersurfaces: S = (p4 = 0) ⊂ CP3;

I resolutions of a torus quotient by involution: S = (T 2C/〈τ〉)desing.

Rational curves: image of a u : CP1 → S

I non-constant;

I generic one-one onto image;

I holomorphic;

I distinct rational curves if they have different images.

Distinct rational curves: if two have distinct images.

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?

The existence of (at least) one rational curve

Theorem (Bogomolov-Mumford) The existence of L 6∼= OS producesrational curves.

Outline

I L 6∼= OS produces s 6= 0 ∈ H0(L⊗n), C = s−1(0) a curve in S ;

I c1(L)2 = 0 or = −2 are ok; (elliptic fibration; (−2)-curves);

I c1(L)2 > 0 implies S is algebraic (projective).

For S projective,

I find a K3 X0 so that it has two smooth rationals C1 and C2 so thatthey intersects transversally; let L0 = OX0 (C1 + C2);

I find a deformation (Xt , Lt) of (X0, L0), with (X1, L1) = (S , L);

I show that C1 ∪ C2 ⊂ X0 deforms to rationals Ct ⊂ Xt , for general t;(this argument will be revisited)

I the limit of rational curves are union of rational curves, so X1 = Scontains rational curves.

Theorem (Lefschetz) A simply connected compact complex manifold Shas a line bundle L 6∼= OS if (the Hodge classes)

H1,1(S ,C) ∩ H2(S ,Z) 6= 0.

Answer to Q1: S has a rational curve if H1,1(S ,C) ∩ H2(S ,Z) 6= 0.

Remarks We have three different proofs of this existence theorem

I Bogomolov-Mumford: find the existence of a special K3, usingdeformation to prove the existence for general K3, plus limit ofrational curves is a union rational curves;

I Mori: using reduction to char= p technique, not relying ondeformation of K3 surfaces;

I Yau-Zaslow: using moduli of sheaves on K3, plus beautiful geometryof Hyperekahler manifolds; not relying on deformation of K3surfaces;

I the theorem holds for any field k = k .

The existence of more than one rational curves

Corollary: The space H1,1(S ,C) ∩ H2(S ,Q) 6= 0 is spanned by classesgenerated by rational curves.

Answer to Q2: The number of rational curves is

# ≥ dimH1,1(S ,C) ∩ H2(S ,Q).

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 P1-bundle);

b. S K3 implies there is a non-zero (2, 0)-form on S ;

c. ϕ∗(Ω) = 0 from (b), 6= 0 from (a), a contradiction.

The Conjecture (on Q3)

Conjecture: Any smooth complex K3 surface S contains infinitely manyrational curves.

This is motivated by Lang’s conjecture:

Lang Conjecture: Let X be a general type complex manifold. Then theunion of the images of holomorphic u : C→ X lies in a finite union ofproper subvarieties of X .

Key to the existence of rational curves:A class α 6= 0 ∈ H2(S ,Z) is Hodge (i.e. ∈ H1,1(S ,C) ∩ H2(S ,Q)) isnecessary and sufficient for the existence of a union of rational curves Ci

so that∑

[Ci ] = α.

Example: Say we can have a family St , t ∈ disk,

I α ∈ H1,1(S0,C) ∩ H2(S0,Q) so that S0 has C0∼= CP1 ⊂ S0 with

[C0] = α;

I in case α 6∈ H1,1(St ,C) for general t, then CP1 ∼= C0 → S0 can notbe extended to holomorphic ut : CP1 → St .

Preparations

I We will consider polarized K3 surfaces (S ,H), c1(H) > 0;

I we can group them according to H2 = 2d :

M2d = (S ,H) | H2 = 2d.

I each M2d is smooth, of dimension 19;

I each M2d is defined over Z. (defined by equation with coefficientsin Z.)

I to show that (S ,H) contains infinitely many rational curves, itsuffices to show that

I for any N, there is a rational curve R ⊂ S so that [R] · H ≥ N.

I we define ρ(S) = dimH1,1(S ,C) ∩ H2(S ,Q), called the rank of thePicard group of S .

The existence of ∞ rational curves on complex algebraicK3 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 groupcontains infinitely many rational curves.

Theorem (Bogomolov-Hassett-Tschinkel, 10) A degree 2 polarizedcomplex K3 surface (S ,H) (i.e. H2 = 2) having ρ(S) = 1 containsinfinitely many rational curves.

Extending their method, we prove

Theorem (L–Liedtke, 11) A polarized complex K3 surface (S ,H) havingρ(S) = 1 contains infinitely many rational curves.

I ρ can run between 1 and 20;

I A K3 with ρ ≥ 5 is an elliptic K3, thus covered by an earlier theorem.

Technical approach

Extension Problem:

I a family of K3 surface St , t ∈ T (a parameter space);

I C0 ⊂ S0 a union of rational curves; let α = [C0] ∈ H2(S ,Z); (it is aHodge class);

I suppose α ∈ H1,1(St ,C) for all t ∈ T ;

We like to show that

I exists a family of curves Ct ⊂ St , such thatI Ct are union of rational curves;I C0 = C0.

Use moduli of genus 0 stable maps

I A genus zero stable map is a holomorphic map u : C → X such thatI C is a nodal, genus 0 complex curve (i.e. a tree of CP1’s);I for any Σ ∼= CP1 ⊂ C with u|Σ = const., Σ contains ≥ 3 nodes of C .

Moduli of genus 0 stable maps

I A genus zero stable map is a holomorphic map u : C → X such thatI C is a nodal, genus 0 complex curve (i.e. a tree of CP1’s);I for any Σ ∼= CP1 ⊂ C with u|Σ = const., Σ contains ≥ 3 nodes of C .

Let S be a K3 surface, α ∈ H2(S ,Z),

I M0(S , α) = [u : C → S ] | genus 0 stable maps with u∗([C ]) = αis an algebraic space;

I has exp.dimension −1.I (expected, as when α is not Hodge, then M0 = ∅.)

Extension Lemma

I St , t ∈ T , be a family of K3 surfaces;

I α ∈ H2(S0,Z) is Hodge for all t;

I [u0] ∈ M0(S0, α).

Extension Lemma (Ran, Bogomolov-Tschinkel, –): Suppose[u0] ∈ M0(S0, α) is isolated, then u0 extends to ut ∈ M0(St , α) forgeneral t ∈ T .

Extension Lemma (Ran, Bogomolov-Tschinkel, –): Suppose[u0] ∈ M0(S0, α) is isolated, then u0 extends to ut ∈ M0(St , α) forgeneral t ∈ T .

Outline of the proof

I let U be an open neighborhood of all deformations of S0; U is asmooth complex manifold of C dimension 20;

I let X → U be the tautological family of K3 surfaces;

I form M0(X , α), the moduli of genus zero stable morphismsu : C → Xz , for some z ∈ U.

I exp. dim M0(S0, α) = −1 plus dimU = 20 implies

I exp. dim M0(X , α) = 19.

Let [u0] ∈W ⊂ M0(X , α) be an irreducible component; dimW ≥ 19.

I Look at π : W → U, π(W ) ⊂ U has dimension at least 19;

I U(α) ⊂ U is the locus where α ∈ H2(Sz ,Z) remain Hodge;dimU(α) = 19;

I π(W ) ⊂ U(α);

I by assumption, T ⊂ U(α);

This implies that T ⊂ π(W ); thus u0 lifts to a family ut : Ct → St ,t ∈ T .

Extension criterion

Definition: We say a map [u] ∈ M0(S , α) rigid if [u] is an isolated pointin M0(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 classu∗[C ] ∈ H2(S ,Z) remains Hodge.

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

I find a family of K3 surfaces (Xt ,Ht), t ∈ T a parameter space, suchthat

I (S ,H) is a general member of (Xt ,Ht);I for a 0 ∈ T , we have C0 ⊂ X0 a union of rational curves, and an

irreducible R0 ⊂ C0 so that [R0] · H0 ≥ N;I [C0] = nH0;I we can represent C0 as the image of a rigid genus zero stable map

[u0].

By Extension criterion:

I we can extend u0 to ut for general t;

I since t is general, there is an irreducible component Rt of image(ut)so that [Rt ] · Ht ≥ [R0] · H0 ≥ N;

I since (S ,H) is a general member, we can assume (S ,H) = (Xt ,Ht).

Theorem (Chen) Very general K3 (S ,H) ∈M2d contains infinitely manyrational curves.

Proof

I Form a family, so that (X0,H0) is a singular K3 surface, thusC0 ⊂ X0 can be constructed explicitly.

I Applying the lifting property. So for any N, all K3’s in a dense opensubset in M2d contains rational curves of degree ≥ N.

Theorem (Bogomolov-Hassett-Tschikel). Fix an r . Assume that every K3(X ,H) ∈M2d (of ρ(X ) = r) defined over a number field K containsinfinitely many rational curves in XQ, then every (X ,H) ∈M2d containsinfinitely many rational curves.

I Let X be any K3, not defined over a number field, then X is thegeneric fiber of X → S , S a variety over K ;

Example X = (πx41 + x4

2 + x43 + x4

4 = 0) ⊂ CP3.

I Switch π by t, we define X = (tx41 + x4

2 + x43 + x4

4 = 0) ⊂ P3 × A1,S = A1.

Theorem (Bogomolov-Hassett-Tschikel). Fix an r . Assume that every K3(X ,H) ∈M2d (of ρ(X ) = r) defined over a number field K containsinfinitely many rational curves in XQ, then every complex (X ,H) ∈M2d

contains infinitely many rational curves.

I Let X be any K3. Then X is the generic fiber of X → S , S a varietyover K ;

I By (Deligne), we can find ξ ∈ S , over L/K , s.t. PicX = PicXξ;

I By assumption, ∃C0 ⊂ Xξ, high degree; [C0 → Xξ] ∈ M0(Xξ) rigid;

I applying lifting criterion, [C0 → Xξ] lifts to generic fiber of X → B;thus lifts to C → X , of high degree.

When (S ,H) is defined over a number field, say Q, after deleting a finiteprime numbers, (S ,H) is the generic member of a family of K3’ overT = specZ−finite places:

Theorem (Bogomolov-Zarhin, Nygaard-Ogus). Let X be a K3 over anumber field K . Then

1. for all but finitely many places p of K , the reductions Xp aresmooth, not supersingular;

2. for all such places with char p ≥ 5, ρ((Xp)Fp) are even.

Proof

I Use Tate Conjecture for K3;

I Use Weil Conjecture for K3.

Theorem (Bogomolov-Hassett-Tschinkel). Let (X ,H) be a polarized K3in M2. Suppose ρ(X ) = 1, then X contains infinitely many rationalcurves.

Sketch

I we only need to prove the case where X is defined over K , sayK = Q;

I for any places p, ρ((Xp)Fp) ≥ 2 imply ∃Dp ⊂ (Xp)Fp

rational curves,not in ZH;

I if Dp · H ≤ N uniformly, implies ρ(XQ) ≥ 2, a contradiction;

Sketch

I places p, ∃Dp ⊂ (Xp)Fprational curves, not in ZH, Dp · H arbitrary

large;

I using H2 = 2, find D ′p ⊂ (Xp)Fprational curves, Dp + D ′p ∈ |npH|;

I Dp + D ′p is the image of a rigid genus zero stable map.

Applying extension criterion, we can find a rational curve of degree ≥ Nin the generic fiber, which is X .

Theorem (– Liedtke) Let (X ,H) be a polarized complex K3 surface suchthat ρ(X ) is odd. Then X contains infinitely many rational curves.

Outline of proof

I We only need to prove the Theorem for (X ,H) defined over anumber field K ;

I say K = Q, we get a family Xp for every prime p ∈ Z, X is thegeneric member of this family;

I ∀p, exists Dp ⊂ Xp, Dp 6∈ ZH,I we have supDp · H →∞;

I pick Cp ⊂ Xp union of rationals, Dp + Cp ∈ |npH|Difficulty: Dp + Cp may not be representable as the image of a rigidgenus zero stable map.

Solution: Suppose we can find a nodal rational curves R ⊂ X , of classkH for some k , then for some large m we can represent

Cp + Dp + mR,

which is a class in (n + mk)H, by a rigid genus zero stable map.

End of the proof: In general, X may not contain any nodal rational curvein |kH|. However, we know a small deformation of X in M2d containsnodal rational curves in |kH| (a Theorem of Chen). Using this, plus somefurther algebraic geometry argument, we can complete the proof.

Happy Birthday, Peter.