transcendental obstructions to rational points on general ... · transcendental obstructions to...

Transcendental obstructions to rational points ongeneral K3 surfaces

Anthony Varilly-AlvaradoRice University

(joint work with Brendan Hassett and Patrick Varilly)

Ramification in Algebra and Geometry at Emory,Emory University, May 2011

Anthony Varilly-Alvarado Transcendental obstructions on K3s

Fix a number field k , and let Ωk be the set of places of k .Let S be a class of nice (smooth, projective, geometrically integral)k-varieties. For X ∈ S, we have the embedding

φ : X (k) →∏

v∈Ωk

X (kv ) = X (A)

Definition

S satisfies the Hasse principle if for all X ∈ S,

X (A) 6= ∅ =⇒ X (k) 6= ∅.

Definition

S satisfies weak approximation if, for all X ∈ S, the image of φ isdense for the product of the v -adic topologies.


Manin (1970): the Brauer group Br X of X can be used toconstruct an intermediate “obstruction set”:

X (k) ⊆ X (A)Br ⊆ X (A).

In fact, X (A)Br already contains the closure of X (k) for the adelictopology:

X (k) ⊆ X (A)Br ⊆ X (A).

This set may be used to explain the failure of the Hasse principleand weak approximation on many kinds of varieties.


Three kinds of Brauer elements

Constant elements Br0 X := im(Br k → Br X )No obstructions: X (A)A = X (A) for all A ∈ Br0 X

Algebraic elements Br1 X := ker(Br X → Br Xk)If X (A) 6= ∅ then there is an isomorphism

Br1 X

Br0 X∼−→ H1

(Gal(k/k),Pic Xk

)(Hochschild-Serre spectral sequence)If Pic Xk = Z then Br1 X gives no obstructions

Transcendental elements Br X \ Br1 X .“geometric: they survive base-change to an algebraic closure”


Where might we find transcendental classes?

For curves and surfaces of negative Kodaira dimension we have

Br X = Br1 X ,

i.e., these varieties have no transcendental Brauer classes.

Thus, if one is interested in transcendental classes, it is reasonableto start by looking at surfaces of Kodaira dimension 0. Within thisclass, we will consider K3 surfaces.

Definition

A K3 surface is a nice surface with

ωX∼= OX and h1

(X ,OX

)= 0.


Examples of K3 surfaces

Double covers of P2 ramified along a smooth sextic planecurve:

w 2 − f (x , y , z) = 0 ⊆ P(1, 1, 1, 3) = Proj k[x , y , z ,w ],

where f (x , y , z) ∈ k[x , y , z ]6.

Smooth quartic surfaces in P3.

Smooth complete intersections of 3 quadrics in P5.


Transcendental elements: basic questions

Theorem (Harari, 1996)

There exist infinitely many explicit conic bundles V over P2 with atranscendental Brauer-Manin obstruction to the Hasse principle.

Question

Are there nice algebraic surfaces that fail to satisfy the Hasseprinciple on account of a transcendental Brauer-Maninobstruction? Can we write down an example?


Transcendental elements: basic questions

Many authors have constructed explicit transcendental Brauerclasses, including Artin–Mumford (1969),Colliot-Thelene–Ojanguren (1989), Harari (1996), Wittenberg(2004), Harari–Skorobogatov (2005),Skorobogatov–Swinnerton-Dyer (2005), Ieronymou (2009),Ieronymou–Skorobogatov–Zarhin (2009), Preu (2010).

This body of work includes examples of transcendentalBrauer-Manin obstructions to weak approximation on K3 surfaces.In all cases, the K3 surfaces considered are endowed with anelliptic fibration, which is used in an essential way to constructtranscendental Brauer classes.


Question

Can we construct an explicit K3 surface X with Pic Xk∼= Z with a

transcendental obstruction to weak approximation?

For such a surface

Br1 X/Br0 X∼−→ H1

(Gal(k/k),Pic Xk

)= 0, i.e., there are no

algebraic Brauer-Manin obstructions to weak approximation.

there are no elliptic fibrations: we have to bring some freshgeometric insight to the table.


Weak Approximation

Theorem (Hassett, Varilly, V-A; 2010)

Let X be the K3 surface of degree 2 given by

w2 = det

0BB@2(2x + 3y + z) 3x + 3y 3x + 4y 3y2 + 2z2

3x + 3y 2(z) 3z 4y2

3x + 4y 3z 2(x + 3z) 4x2 + 5xy + 5y2

3y2 + 2z2 4y2 4x2 + 5xy + 5y2 2(2x3 + 3x2z + 3xz2 + 3z3)

1CCA

in PQ(1, 1, 1, 3). Then Pic XQ∼= Z, and there is a transcendental

Brauer-Manin obstruction to weak approximation on X . Theobstruction arises from a quaternion Azumaya algebraA ∈ im (Br X → Br κ(X )). Explicitly, if Mi denotes the i-thleading principal minor of the above matrix, then

A =

(−M2

M21

,− M3

M1M2

).

Note: X has rational points! e.g., [15 : 15 : 16 : 13 752]


Hasse principle

Consider the following polynomials in Q[x , y , z ]:

A := 62x2 − 16xz + 16y 2 + 16z2

B := −5x2 − 2y 2 − 2z2

C := −4x2 − 6xz − 4y 2 − 2yz − 3z2

D := 48x2 − 16xz + 30y 2 + 48yz + 32z2

E := −2x2 − y 2 − 4yz − 6z2

F := 16x2 + 48xz + 32y 2 + 62z2


Hasse principle

Theorem (Hassett, V-A; 2011)

Let X be the K3 surface of degree 2 given by

w 2 = det

2A B CB 2D EC E 2F

in PQ(1, 1, 1, 3). Then Pic XQ

∼= Z, and there is a transcendentalBrauer-Manin obstruction to the Hasse principle on X . Theobstruction arises from a quaternion Azumaya algebraA ∈ im (Br X → Br κ(X )). Explicitly, if Mi denotes the i-thleading principal minor of the above matrix, then

A =

(−M2

M21

,− M3

M1M2

).


Hodge theoretic motivation

How did we know where to look for these examples?Let X be a complex projective K3 surface. Let

TX := NS(X )⊥ ⊆ H2(X ,Z)

be the transcendental lattice of X . Write

ΛK3 = U3 ⊕ E8(−1)2

for the abstract K3 lattice.The exponential sequence shows there is a one-to-onecorrespondence

α ∈ Br X of exact order n 1−1←→ surjections TX → Z/nZ

Hence, to α as above, we may associate Tα ⊆ TX :

Tα = ker(α : TX → Z/nZ).


Theorem (van Geemen; 2005)

Let X be a complex projective K3 surface of degree 2 withPic X ∼= Z, and let α ∈ (Br X )[2]. Then one of the following threethings must happen:

1 There is a unique primitive embedding Tα → ΛK3. This givesa degree 8 K3 surface Y associated to the pair (X , α).

2 Tα(−1) ∼= 〈h2,P〉⊥ ⊆ H4(Z ,Z), where Z is a cubic fourfoldwith a plane P (h is the hyperplane class).

3 Tα(−1) ∼= 〈h21, h1h2, h

22〉⊥ ⊆ H4(W ,Z), where W is a double

cover of P2 × P2 ramified along a type (2, 2) divisor (h1, h2

are the pullbacks to W of the hyperplane classes of P2 underthe two projections P2 × P2 → P2).

Idea: “go backwards” and work over any field k.


The pair (X ,A) in the K 3 counter-example to weak approximationis naturally associated to a cubic fourfoldY ⊆ P5

Q := Proj Q[X1,X2,X3,Y1,Y2,Y3] given by

2X 21 Y1 + 3X 2

1 Y2 + X 21 Y3 + 3X1X2Y1 + 3X1X2Y2 + 3X1X3Y1

+ 4X1X3Y2 + 3X1Y 22 + 2X1Y 2

3 + X 22 Y3 + 3X2X3Y3

+ 4X2Y 22 + X 2

3 Y1 + 3X 23 Y3 + 4X3Y 2

1 + 5X3Y1Y2

+ 5X3Y 22 + 2Y 3

1 + 3Y 21 Y3 + 3Y1Y 2

3 + 3Y 33 = 0.

This fourfold contains the plane Y1 = Y2 = Y3 = 0.


Theorem

Let Y be a cubic fourfold smooth over a field k. Suppose that Ycontains a plane P, and let Y denote the blow up of Y along P,q : Y → P2 the corresponding quadric surface bundle, andr :W → P2 its relative variety of lines. Assume that there existsno plane P ′ ⊂ Yk such that P ′ meets P along a line.Then the Stein factorization

r :W π1→ Xφ→ P2

consists of a smooth P1-bundle followed by a degree-two cover ofP2, which is a K3 surface.

Note: The case k = C goes back at least to Voisin (1985).


The pair (X ,A) in the K 3 counter-example to the Hasse principleis naturally associated to a double cover ofP2 × P2 = Proj Q[x0, x1, x2]× Proj Q[y0, y1, y2] ramified along thedivisor of type (2, 2) given by

62x20 y 2

0 + 16x21 y 2

0 − 16x0x2y 20 + 16x2

2 y 20 − 5x2

0 y0y1 − 2x21 y0y1

− 2x22 y0y1 + 48x2

0 y 21 + 30x2

1 y 21 − 16x0x2y 2

1 + 48x1x2y 21

+ 32x22 y 2

1 − 4x20 y0y2 − 4x2

1 y0y2 − 6x0x2y0y2 − 2x1x2y0y2

− 3x22 y0y2 − 2x2

0 y1y2 − x21 y1y2 − 4x1x2y1y2 − 6x2

2 y1y2

+ 16x20 y 2

2 + 32x21 y 2

2 + 48x0x2y 22 + 62x2

2 y 22 = 0.


Let W ⊆ P2 × P2 be a smooth type (2, 2) divisor. The projectionsπi : W → P2 define conic bundle structures ramified along smoothsextic curves C1, C2 (respectively).

Let φi : Xi → P2 be a double cover ramified along Ci . ThenW ×P2 Xi → Xi is a smooth P1-bundle for the etale topology. Thisis our element A ∈ (Br X )[2].


Caveats

The constructions above don’t necessarily yield K3 surfacesfor which Pic Xk

∼= Z.We use ideas of van Luijk, Elsenhans and Jahnel to constructsurfaces for which we can prove this is the case. This requiressome intensive point counts over finite fields.In the Hasse principle counter-example, we need the primes ofbad reduction of X . A Groebner basis computation over Zshows these primes divide

1575961389009482560437286583534678113874684452575330515905422247848

6447126409097867530208903148583982831806707465087640413725725692102

0660524389357878901956308029862880978409927736101780460331885931216

3834734316641126951561545616484413034372280131483926551238024753488

7685949966462337614350040948449859415529952750002506898422776271778

4930382467087815253432944143520

(366 digits!)


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