K3 Surfaces and Lattice Theory

K3 Surfaces and Lattice Theory

Ichiro Shimada

Hiroshima University

2014 Aug Singapore

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Example

Consider two surfaces S+ and S− in C³ defined by w²(G(x,y)±√

5·H(x,y)) = 1, where

G(x,y) := −9x⁴−14x³y + 58x³−48x²y²−64x²y +10x²+ 108xy³−20xy²−44y⁵+ 10y⁴, H(x,y) := 5x⁴+ 10x³y −30x³+ 30x²y²+

+20x²y −40xy³+ 20y⁵.

Since S₊ and S₋ are conjugate by Gal(Q(√ 5)/Q), they can not be distinguished algebraically.

ButS₊ and S₋ are not homeomorphic (in the classical topology).

Many examples of non-homeomorphic conjugate complex varieties are known since Serre (1964).

Introduction

Definition

A smooth projective surfaceX is called a K3surface if

∃a nowhere vanishing holomorphic 2-form ω_X onX, and π₁(X) ={1}.

We consider the following geometric problems onK3 surfaces:

enumerate elliptic fibrations on a givenK3 surface, enumerate elliptic K3 surfaces up to some equivalence relation,

enumerate projective models of a given K3 surface, enumerate projective models of K3 surfaces,

determine the automorphism group of a givenK3 surface, . . . .

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Thanks to the theory of period mapping, some of these problems are reduced to computational problems in lattice theory, and the latter can often be solved by means of computer.

In this talk, we explain how to use lattice theory and computer in the study ofK3 surfaces.

We then demonstrate this method on the problems of constructing Zariski pairs of plane curves of degree 6.

Alattice is a free Z-moduleL of finite rank with a non-degenerate symmetric bilinear form

h i: L×L→Z.

LetL be a lattice of rank n. We choose a basise₁, . . . ,e_n ofL.

The latticeL is given by the Gram matrix G := (hei,eji)i,j=1,...,n.

O(L) is the group of all isometries ofL.

L is unimodular if detG =±1.

Thesignature sgn(L) is the signature L⊗R.

A lattice L is said to be hyperbolic if sgn(L) = (1,n−1), and is positive-definiteif sgn(L) = (n,0).

A lattice L is evenif v² ∈2Zfor all v ∈L.

A sublattice L⁰ of L isprimitive if L/L⁰ is torsion free.

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Lattices associated to a K 3 surface

K3 surfaces are diffeomorphic to each other.

Suppose thatX is a K3 surface. Then H²(X,Z) with the cup product is an even unimodular lattice of signature (3,19), and hence is isomorphic to

U^⊕3⊕E₈^−⊕2,

whereU is the hyperbolic plane with a Gram matrix µ 0 1

1 0

¶ ,

and E₈⁻ is the negative definite root lattice of type E₈.





















−2 0 0 1 0 0 0 0

0 −2 1 0 0 0 0 0

0 1 −2 1 0 0 0 0

1 0 1 −2 1 0 0 0

0 0 0 1 −2 1 0 0

0 0 0 0 1 −2 1 0

0 0 0 0 0 1 −2 1

0 0 0 0 0 0 1 −2





















The Gram matrix of E₈⁻

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TheN´eron-Severi lattice

SX :=H²(X,Z)∩H^1,1(X)

of cohomology classes of divisors on X is an even hyperbolic lattice of rank≤20. Moreover the sublattice S_X of H²(X,Z) is primitive.

Problem

Suppose that an even hyperbolic lattice of rank ≤20is given.

Is there a K3 surface X such that S ∼=S_X?

We have the following corollary of the surjectivity of the period map:

Theorem

Let S be a primitive hyperbolic sublattice of U^⊕3⊕E₈^−⊕2. Then∃ a K3 surface X such that S ∼=S_X.

Problem

Suppose that an even lattice L and an even unimodular lattice M are given. Can L be embedded into M primitively?

A lattice L is canonically embedded into its dual lattice L^∨ :=Hom(L,Z)

as a submodule of finite index. The finite abelian group D_L :=L^∨/L

is called thediscriminant group of L.

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The symm. bil. form onL extends to aQ-valued symm. bil.

form on L^∨, and it defines a finite quadratic form q_L: D_L→Q/2Z, ¯x 7→x² mod 2Z.

LetM be an even unimodular lattice containing L primitively with the orthogonal complementL^⊥. Then we have

(D_L,q_L)∼= (D_L^⊥,−q_L^⊥).

Conversely, if R is an even lattice such that (D_L,q_L)∼= (D_R,−q_R),

then there exist an even unimodular latticeM and a primitive embeddingL ,→M such that L^⊥ ∼=R.

Problem

Suppose that s₊,s₋∈Z_≥0 and a finite quadratic form (D,q) are given. Can we determine whether∃ an even lattice L such thatsgn(L) = (s+,s−) and (DL,qL)∼= (D,q)?

Theorem YES.

Corollary

We can determine whether a given even hyperbolic lattice of rank≤20 is a N´eron–Severi lattice of a K3 surface X or not.

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Polarized K 3 surfaces

Forv ∈S_X, letL_v →X be the corresponding line bundle.

Definition

Ford ∈Z_>0, a vector h∈S_X of h² =d is a polarization of degree d if |L_h| 6=∅ and has no fixed-components.

Leth be a polarization of degree d. Then |L_h| defines Φh :X →P^1+d/2. We denote by

X −→^φh Y_h −→^ψh P^1+d/2

the Stein factorization of Φ_h. The normal surface Y_h is the projective modelof (X,h), and has only rational double points as its singularities.

Example

A plane curve B is a simple sextic if B is of degree 6 and has only simple singularities (ADE-singularities; ordinary nodes, ordinary cusps, tacnodes, . . . ).

LetB be a simple sextic, and Y_B →P² the double covering branched along B. The minimal resolution X_B of Y_B is a K3 surface.

We denote by

Φ_B :X_B →Y_B →P²

the composite of the min. resol. and the double covering, and byh_B ∈S_XB the class of the pull-back of a line. Then h_B is a polarization of degree 2, and Y_B is its projective model.

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Problem

Suppose that h∈S_X with h² >0 is given. Is h a polarization?

If so, what is the ADE -type of Sing(Y_h)?

We consider the second question first.

Proposition

The ADE -type ofSingY_h is equal to the ADE -type of the root system{r ∈S_X| hh,ri= 0,hr,ri=−2}.

The sublattice {x ∈SX | hh,xi= 0} is negative-definite.

Problem

Given a positive-definite lattice L and an integer d. Calculate the set{r ∈L| hr,ri=d}.

Suppose that we are given a triple [Q, λ,c], where Q is a positive-definiten×n symmetric matrix with entries inQ,

λ is a column vector of lengthn with entries in Q, c ∈Q.

ForQT := [Q, λ,c], we defineFQT :Rⁿ→R by F_QT(v) :=v Q^tv + 2vλ+c.

We have an algorithm to calculate the finite set E(QT) := {v ∈Zⁿ | F_QT(v)≤0}

by induction on n, and hence we can determine the ADE-type of SingYh.

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Criterion for a polarization

LetL be an even hyperbolic lattice. Let PL be one of the two connected components of{x ∈L⊗R|x² >0}.

We put

RL :={r ∈L | r² =−2}.

Eachr ∈ R_L defines a reflections_r into the hyperplane (r)^⊥ :={x ∈ PL| hx,ri= 0}:

s_r :x 7→x +hx,rir,

The closure inP_L of each connected component of P_L\[

r∈RL(r)^⊥

is a standard fundamental domain of the action onP_L of W(L) :=hsr |r ∈ RL i.

LetP(X)⊂S_X ⊗R be the positive cone that contains an ample class (e.g., the class of a hyperplane section). We put

N(X) := {x ∈ P(X)| hx,[C]i ≥0for any curve C on X }.

Proposition

This N(X) is a standard fundamental domain of the action of W(SX) on P(X).

It is obvious that, if h is a polarization, thenh ∈N(X).

Proposition

Let h∈S_X be a vector with h² = 2. Then h is a polarization if and only if h∈N(X) and 6 ∃ e ∈SX with e² = 0 and he,hi= 1.

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Problem

Suppose that h∈S_X with h² >0 is given. Does h belong to N(X)?

Since N(X) is bounded by (r)^⊥, this problem is reduced to the following:

Problem

Suppose that we are given vectors h,h₀ ∈ P_L. Then, for a negative integer d, calculate the set

{r ∈S_X | hr,hi>0, hr,h₀i<0, hr,ri=−2}.

There is an algorithm for this task, and hence we can determine whether a given h∈SX with h² = 2 is a polarization or not.

Zariski pairs

For a simple sextic B,

RB : the ADE-type ofSingB (or of SingYB),

degsB the list of degrees of irreducible components of B.

We say that B and B⁰ are of the same config type and write B ∼cfgB⁰ if

RB =RB⁰, degsB =degsB⁰,

their intersection patterns of irred. comps are same.

Example

Zariski showed the existence of a pair [B,B⁰] such that R_B =R_B⁰ = 6A₂, degsB =degsB⁰ = [6], and π1(P² \B)∼=Z/(2)∗Z/(3),

π₁(P² \B⁰)∼=Z/(2)×Z/(3).

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For a simple sextic B with

ΦB :XB →YB →P²,

let E_B be the set of exceptional curves of X_B →Y_B, and let Σ_B :=h[E]|E ∈ E_Bi ⊕ hh_Bi ⊂ H²(X_B,Z), whereh_B is the class of the pull-back of a line. We have

B ∼_cfg B⁰ ⇒ Σ_B ∼= ΣB⁰. We denote the primitive closure of Σ_B by

Σ_B ⊂ S_XB ⊂ H²(X_B,Z).

After the partial results by Urabe, Yang (1996) made the complete list configuration type of simple sextics by classifying all such Σ .

We writeB ∼_emb B⁰ if there exists a homeomorphism ψ : (P²,B) →∼ (P²,B⁰).

We have B ∼_emb B⁰ =⇒ B ∼_cfgB⁰.

# of config types= 11159<# of emb-top types=?

Definition

AZariski pair is a pair [B,B⁰] of projective plane curves of the same degree with only simple singularities such that

B ∼cfgB⁰ but B6∼_embB⁰.

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We consider the finite abelian group G(B) := Σ_B/Σ_B. We put

Θ_B := (Σ_B ⊂H²(X_B,Z))^⊥. Theorem

If B ∼_embB⁰, thenΘ_B ∼= ΘB⁰.

In fact, Θ_B is a topological invariant of the open surface U_B := Φ⁻¹_B (P²\B) ⊂ X_B,

because we have Θ_B ∼= H²(U_B,Z)/Ker, where

Ker:={v ∈H²(U_B) | hv,xi= 0for allx ∈H²(U_B)}.

Since the discriminant groups of ΣB and ΘB are isomorphic, we have:

Corollary

If B ∼_cfgB⁰ but |G(B)| 6=|G(B⁰)|, then B6∼_embB⁰. This corollary produces many examples of Zariski pairs.

Example

In Zariski’s example [B,B⁰] with R_B =R_B⁰ = 6A₂, degsB =degsB⁰ = [6] and

π1(P²\B)∼=Z/(2)∗Z/(3), π1(P²\B⁰)∼=Z/(2)×Z/(3), we have G(B)∼=Z/3Z and G(B⁰) = 0.

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Singular K 3 surfaces

Definition

AK3 surface X is called singularif rank(S_X) = 20.

Theorem (Shioda and Inose) The map

X 7→T(X) := (S_X ⊂H²(X,Z))^⊥

is a bijection from the set of isom. classes of singular K3 surfaces to the set of isom. classes of oriented positive-definite even lattices of rank2.

Theorem (Shioda and Inose)

Every singular K3 surface X is defined over Q.

Theorem (S. and Sch¨utt)

Let X and X⁰ be singular K3 surfaces over Q such that q_T(X)∼=q_T(X⁰₎.

Then there∃σ∈Gal(Q/Q) such that X⁰ ∼=X^σ.

If B is a simple sextic with total Milnor numberµ(B) = 19, then X_B is a singular K3 surface with Θ_B ∼=T(X_B).

Corollary

Let B be a sextic withµ(B) = 19 defined over Q.

If ∃ an even pos-def lattice T⁰ of rank 2with q_T⁰ ∼=q_T(XB) and T⁰ 6∼=T(X_B),

then ∃σ ∈Gal(Q/Q) such that B6∼_embB^σ.

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The first example revisited

Consider the config type of sexticsB =L+Q, where degL = 1, degQ = 5,

L and Q are tangent at one point with multiplicity 5 (A₉-singularity), and

Q has oneA₁₀-singular point.

Such sextics are projectively isomorphic to z·(G(x,y,z)±√

5·H(x,y,z)) = 0,

whereG(x,y,z) and H(x,y,z) are homogenizations of the polynoms in the 1st slide with L={z = 0}.

The genus corresponding to (D_ΣB,−q_ΣB) and signature (2,0) (that is, the genus containingT(X_B)) consists of

· 2 1 1 28

¸

(for +√ 5),

· 8 3 3 8

¸

(for−√ 5).