Transcendental lattices of complex algebraic surfaces

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Ichiro Shimada

Hiroshima University

November 25, 2009, Tohoku

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Transcendental lattices of complex algebraic surfaces Introduction

LetAut(C) be the automorphism group of the complex number fieldC.

For a schemeV →SpecC and an elementσ ∈Aut(C), we define a schemeV^σ →SpecCby the following Cartesian diagram:

V^σ −→ V

↓ ¤ ↓

SpecC −→

σ^∗ SpecC.

Two schemesV and V⁰ overCare said to be conjugate ifV⁰ is isomorphic toV^σ overCfor someσ ∈Aut(C).

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Conjugate complex varieties can never be distinguished by any algebraic methods (they are isomorphic overQ),

but they can be non-homeomorphic in the classical complex topology.

The first example was given by Serre in 1964.

Other examples have been constructed by:

Abelson (1974),

Grothendieck’s dessins d’enfants (1984),

Artal Bartolo, Carmona Ruber, and Cogolludo Agust (2004), Easton and Vakil (2007), F. Charles (2009).

We will construct such examples by means of

transcendental latticesof complex algebraic surfaces.

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Example (S.- and Arima)

Consider two smooth irreducible surfacesS_± 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. ThenS₊ and S₋ are not homeomorphic.

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§ Definition of the transcendental lattice

§ The transcendental lattice is topological

§ The discriminant form is algebraic

§ Fully-rigged surfaces

§ Maximizing curves

§ Arithmetic of fully-riggedK3 surfaces

§ Construction of the explicit example

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Transcendental lattices of complex algebraic surfaces Definition of the transcendental lattice

By alattice, we mean a freeZ-module Lof finite rank with a non-degenerate symmetric bilinear form

L×L→Z.

A latticeL is naturally embedded into the dual lattice L^∨:=Hom(L,Z).

Thediscriminant group ofL is the finite abelian group D_L:=L^∨/L.

A latticeL is calledunimodular if D_L is trivial.

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LetX be a smooth projective surface overC. Then H²(X) :=H²(X,Z)/(the torsion) is regarded as a unimodular lattice by the cup-product.

TheN´eron-Severi lattice

NS(X) :=H²(X)∩H^1,1(X)

of classes of algebraic curves onX is a sublattice of signature sgn(NS(X)) = (1, ρ−1).

Thetranscendental latticeofX is defined to be the orthogonal complement ofNS(X) in H²(X):

T(X) :=NS(X)^⊥.

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Proposition (Shioda)

T(X) is a birational invariant of algebraic surfaces.

Proof.

Suppose thatX andX⁰ are birational. There exists a smooth projective surfaceX⁰⁰ with birational morphisms

X⁰⁰→X and X⁰⁰ →X⁰.

Every birational morphism between smooth projective surfaces is a composite of blowing-ups at points.

A blowing-up at a point does not change the transcendental lattice.

Hence, for a surface S (possibly singular and possibly open), the transcendental latticeT(S) is well-defined.

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Transcendental lattices of complex algebraic surfaces The transcendental lattice is topological

LetX be a smooth projective surface overC,

and letC₁, . . . ,C_n⊂X be irreducible curves. Suppose that [C₁], . . . ,[C_n]∈NS(X) span NS(X)⊗Q overQ.

We consider the open surface

S :=X \(C₁∪ · · · ∪C_n).

By definition, we haveT(S) =T(X).

Consider the intersection pairing

ι : H₂(S)×H₂(S)→Z.

We put

H₂(S)^⊥:={x ∈H₂(S) | ι(x,y) = 0 for all y ∈H₂(S)}.

ThenH₂(S)/H₂(S)^⊥ becomes a lattice.

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Proposition

The latticeT(S) =T(X) is isomorphic toH₂(S)/H₂(S)^⊥. Proof.

We putC :=C₁∪ · · · ∪C_n. Consider the diagram:

H₂(S) −→^j^∗ H₂(X)

↓ o ↓ o

H²(X,C) −→ H²(X) −→^r^C H²(C) =L

H²(C_i), wherej :S ,→X is the inclusion. From this, we see that Imj_∗ =T(X). SinceT(X) is non-degenerate, we have Kerj_∗ =H₂(S)^⊥.

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Letσ be an element of Aut(C). Then

[C₁^σ], . . . ,[C_n^σ]∈NS(X^σ) span NS(X^σ)⊗QoverQ, because the intersection pairing onNS(X) is defined algebraically.

Since the latticeH₂(S)/H₂(S)^⊥ is defined topologically, we obtain the following:

Corollary

IfS^σ andS are homeomorphic, then T(S^σ) =T(X^σ) is isomorphic toT(S) =T(X).

Corollary

IfT(X) andT(X^σ) are not isomorphic, then there exists a Zariski open subsetS ⊂X such thatS and S^σ are not homeomorphic.

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The discriminant form of the transcendental lattice is algebraic

LetLbe a lattice. TheZ-valued symmetric bilinear form on L extends to

L^∨×L^∨→Q.

Hence, on the discriminant groupD_L:=L^∨/L ofL, we have a quadratic form

q_L:D_L→Q/Z, ¯x 7→x² modZ, which is called thediscriminant formof L.

A latticeL is said to beevenif x² ∈2Z for anyx∈L.

IfLis even, then q_L:D_L→Q/Z is refined to q_L:D_L→Q/2Z.

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The discriminant form of the transcendental lattice is algebraic

SinceH²(X) is unimodular and both ofT(X) andNS(X) are primitive inH²(X), we have the following:

Proposition

(D_T_(X₎,q_T(X₎)∼= (D_NS(X₎,−q_NS(X₎).

Since the N´eron-Severi lattice is defined algebraically, we obtain the following:

Corollary

For anyσ ∈Aut(C), we have

(D_T_(X₎,q_T(X₎)∼= (D_T_(Xσ),q_T(Xσ)).

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Transcendental lattices of complex algebraic surfaces Fully-rigged surfaces

Recall that:

Our aim is to construct conjugate open surfacesS andS^σ that are not homeomorphic.

For this, it is enough to construct conjugate smooth projective surfacesX andX^σ with non-isomorphic transcendental lattices T(X)6∼=T(X^σ).

ButT(X) andT(X^σ) have isomorphic discriminant forms.

Problem

To what extent does the discriminant form determine the lattice?

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Proposition

LetLand L⁰ be even lattices of the same rank.

IfLandL⁰ have isomorphic discriminant forms and

the same signature, thenLandL⁰ belong to the same genus.

Theorem (Eichler)

Suppose thatL andL⁰ are indefinite.

IfLandL⁰ belong to the same spinor-genus, thenL andL⁰ are isomorphic.

The difference between genus and spinor-genus is not big.

Hence we need to search forX such thatT(X) is definite.

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Definition (Katsura)

LetS be a surface with a smooth projective birational modelX. S isfully-rigged

⇐⇒ rank(NS(X)) =h^1,1(X)

⇐⇒ rank(T(S)) = 2p_g(X)

⇐⇒ T(S) is positive-definite.

Remark

For abelian surfaces orK3 surfaces, fully-rigged surfaces are called

“singular”.

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Transcendental lattices of complex algebraic surfaces Maximizing curves

Definition (Persson)

A reduced (possibly reducible) projective plane curveB⊂P² of even degree 2m is maximizingif the following hold:

B has only simple singularities (ADE-singularities), and the total Milnor number of B is 3m²−3m+ 1.

Equivalently,B ⊂P² is maximizing if and only if

the double cover Y_B →P² of P² branching along B has only RDPs, and

for the minimal resolutionX_B ofY_B, the classes of the exceptional divisors span a sublattice of NS(X_B) with rank h^1,1(X_B)−1.

In particular,X_B is fully-rigged.

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Persson (1982) found many examples of maximizing curves.

Example

The projective plane curve

B :xy(xⁿ+yⁿ+zⁿ)²−4xy((xy)ⁿ+ (yz)ⁿ+ (zx)ⁿ) = 0 has singular points of type

2n×D_n+2 + n×A_n−1 + A₁. It is maximizing.

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IfB⊂P² is of degree 6 and has only simple singularities, then the minimal resolutionX_B of the double cover of P² branching alongB is aK3 surface.

In the paper

Yang, Jin-Gen

Sextic curves with simple singularities

Tohoku Math. J. (2) 48 (1996), no. 2, 203–227,

Yang classified all sextic curves with only simple singularities by means of Torelli theorem for complexK3 surfaces.

His method also gives the transcendental lattices of the fully-rigged K3 surfacesX_B obtained as the double plane sextics.

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Transcendental lattices of complex algebraic surfaces Arithmetic of fully-rigged (singular)K3 surfaces

(We use the terminology “fully-riggedK3 surfaces” rather than the traditional “singularK3 surfaces”.)

LetX be a fully-riggedK3 surface; that is,X is aK3 surface with the Picard number 20. Then the transcendental latticeT(X) is a positive-definite even lattice of rank 2.

The Hodge decomposition

T(X)⊗C=H^2,0(X)⊕H^0,2(X) induces an orientation onT(X). We denote by

T˜(X) the oriented transcendental lattice ofX. By Torelli theorem, we have

T˜(X)∼= ˜T(X⁰) =⇒ X ∼=X⁰.

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Construction by Shioda and Inose.

Every fully-riggedK3 surface X is obtained as a certain double cover of the Kummer surface

Km(E ×E⁰),

whereE andE⁰ are elliptic curves withCM by some orders of Q(p

−|discT(X)|).

Theorem (Shioda and Inose)

(1) For any positive-definite oriented even lattice ˜T of rank 2, there exists a fully-riggedK3 surface X such that ˜T(X)∼= ˜T. (2) Every fully-riggedK3 surface is defined over a number field.

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The class field theory of imaginary quadratic fields tells us how the Galois group acts on thej-invariants of elliptic curves with CM.

Using this, S.- and Sch¨utt (2007) proved the following:

Theorem

LetX andX⁰ be fully-riggedK3 surfaces defined overQ.

If

(D_T(X₎,q_T(X₎)∼= (D_T(X⁰₎,q_T(X⁰₎) (that is, ifT(X) andT(X⁰) are in the same genus), thenX andX⁰ are conjugate.

Therefore, if the genus contains more than one isomorphism class, then we can construct non-homeomorphic conjugate surfaces as Zariski open subsets of fully-riggedK3 surfaces.

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Transcendental lattices of complex algebraic surfaces Constructing explicit examples

From Yang’s table, we know that there exists a sextic curve B =L+Q,

whereQ is a quintic curve with one A₁₀-singular point, andLis a line intersecting Q at only one smooth point ofQ.

HenceB hasA₉+A₁₀.

The N´eron-Severi latticeNS(X_B) is an overlattice of R_A₉_+A₁₀⊕ hhi

with index 2,

whereR_A₉_+A₁₀ is the negative-definite root lattice of typeA₉+A₁₀, andh is the vector [O_P2(1)] withh²= 2.

(The extension comes from the fact thatB is reducible.)

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The genus of even positive-definite lattices of rank 2 corresponding to the discriminant form

(D_NS(X_B₎,−q_NS(X_B₎) ∼= (Z/55Z,[2/55] mod 2) consists of two isomorphism classes:

· 2 1 1 28

¸ ,

· 8 3 3 8

¸ .

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On the other hand, the maximizing sexticB is defined by the normal form

B_± : z·(G(x,y,z)±√

5H(x,y,z)) = 0, where

G = −9x⁴z−14x³yz+ 58x³z²−48x²y²z−64x²yz²+ 10x²z³+ 108xy³z −20xy²z²−44y⁵+ 10y⁴z, H = 5x⁴z + 10x³yz−30x³z²+ 30x²y²z+

20x²yz²−40xy³z+ 20y⁵.

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Hence the ´etale double coversS_± of the complementsP²\B_± are conjugate but non-homeomorphic. Indeed, we have

T(S₊)∼=

· 2 1 1 28

¸

and T(S₋)∼=

· 8 3 3 8

¸ .

Remark

There is another possibility T(S₊)∼=

· 8 3 3 8

¸

and T(S₋)∼=

· 2 1 1 28

¸ . The verification of the fact that the first one is the case needs a careful topological calculation.

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Transcendental lattices of complex algebraic surfaces

Transcendental lattices of complex algebraic surfaces

Thank you!