Transcendental obstructions to rational points on general K3 - - PowerPoint PPT Presentation

Transcendental obstructions to rational points on general K3 surfaces

Anthony V´ arilly-Alvarado Rice University (joint work with Brendan Hassett and Patrick Varilly) Ramification in Algebra and Geometry at Emory, Emory University, May 2011

Anthony V´ arilly-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) = ∅ = ⇒ X(k) = ∅. Definition S satisfies weak approximation if, for all X ∈ S, the image of φ is dense for the product of the v-adic topologies.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Manin (1970): the Brauer group Br X of X can be used to construct 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 adelic topology: X(k) ⊆ X(A)Br ⊆ X(A). This set may be used to explain the failure of the Hasse principle and weak approximation on many kinds of varieties.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

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) = ∅ 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”

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

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 reasonable to start by looking at surfaces of Kodaira dimension 0. Within this class, we will consider K3 surfaces. Definition A K3 surface is a nice surface with ωX ∼ = OX and h1 X, OX

• = 0.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Examples of K3 surfaces

Double covers of P2 ramified along a smooth sextic plane curve: {w2 − 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.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Transcendental elements: basic questions

Theorem (Harari, 1996) There exist infinitely many explicit conic bundles V over P2 with a transcendental Brauer-Manin obstruction to the Hasse principle. Question Are there nice algebraic surfaces that fail to satisfy the Hasse principle on account of a transcendental Brauer-Manin

• bstruction? Can we write down an example?

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Transcendental elements: basic questions

Many authors have constructed explicit transcendental Brauer classes, including Artin–Mumford (1969), Colliot-Th´ el` ene–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 transcendental Brauer-Manin obstructions to weak approximation on K3 surfaces. In all cases, the K3 surfaces considered are endowed with an elliptic fibration, which is used in an essential way to construct transcendental Brauer classes.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

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 fresh geometric insight to the table.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Weak Approximation

Theorem (Hassett, Varilly, V-A; 2010) Let X be the K3 surface of degree 2 given by

w2 = det B B @ 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) 1 C C A

in PQ(1, 1, 1, 3). Then Pic XQ ∼ = Z, and there is a transcendental Brauer-Manin obstruction to weak approximation on X. The

• bstruction arises from a quaternion Azumaya algebra

A ∈ im (Br X ֒ → Br κ(X)). Explicitly, if Mi denotes the i-th leading principal minor of the above matrix, then A =

• −M2

M2

1

, − M3 M1M2

• .

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

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Hasse principle

Consider the following polynomials in Q[x, y, z]: A := 62x2 − 16xz + 16y2 + 16z2 B := −5x2 − 2y2 − 2z2 C := −4x2 − 6xz − 4y2 − 2yz − 3z2 D := 48x2 − 16xz + 30y2 + 48yz + 32z2 E := −2x2 − y2 − 4yz − 6z2 F := 16x2 + 48xz + 32y2 + 62z2

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Hasse principle

Theorem (Hassett, V-A; 2011) Let X be the K3 surface of degree 2 given by w2 = det   2A B C B 2D E C E 2F   in PQ(1, 1, 1, 3). Then Pic XQ ∼ = Z, and there is a transcendental Brauer-Manin obstruction to the Hasse principle on X. The

• bstruction arises from a quaternion Azumaya algebra

A ∈ im (Br X ֒ → Br κ(X)). Explicitly, if Mi denotes the i-th leading principal minor of the above matrix, then A =

• −M2

M2

1

, − M3 M1M2

• .

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

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-one correspondence {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).

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Theorem (van Geemen; 2005) Let X be a complex projective K3 surface of degree 2 with Pic X ∼ = Z, and let α ∈ (Br X)[2]. Then one of the following three things must happen:

1 There is a unique primitive embedding Tα ֒

→ ΛK3. This gives a degree 8 K3 surface Y associated to the pair (X, α).

2 Tα(−1) ∼

= h2, P⊥ ⊆ H4(Z, Z), where Z is a cubic fourfold with a plane P (h is the hyperplane class).

3 Tα(−1) ∼

= h2

1, h1h2, h2 2⊥ ⊆ 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 under the two projections P2 × P2 → P2). Idea: “go backwards” and work over any field k.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

The pair (X, A) in the K3 counter-example to weak approximation is naturally associated to a cubic fourfold Y ⊆ P5

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

2X 2

1 Y1 + 3X 2 1 Y2 + X 2 1 Y3 + 3X1X2Y1 + 3X1X2Y2 + 3X1X3Y1

+ 4X1X3Y2 + 3X1Y 2

2 + 2X1Y 2 3 + X 2 2 Y3 + 3X2X3Y3

+ 4X2Y 2

2 + X 2 3 Y1 + 3X 2 3 Y3 + 4X3Y 2 1 + 5X3Y1Y2

+ 5X3Y 2

2 + 2Y 3 1 + 3Y 2 1 Y3 + 3Y1Y 2 3 + 3Y 3 3 = 0.

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

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Theorem Let Y be a cubic fourfold smooth over a field k. Suppose that Y contains a plane P, and let Y denote the blow up of Y along P, q : Y → P2 the corresponding quadric surface bundle, and r : W → P2 its relative variety of lines. Assume that there exists no plane P′ ⊂ Y¯

k such that P′ meets P along a line.

Then the Stein factorization r : W

π1

→ X

φ

→ P2 consists of a smooth P1-bundle followed by a degree-two cover of P2, which is a K3 surface. Note: The case k = C goes back at least to Voisin (1985).

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

The pair (X, A) in the K3 counter-example to the Hasse principle is naturally associated to a double cover of P2 × P2 = Proj Q[x0, x1, x2] × Proj Q[y0, y1, y2] ramified along the divisor of type (2, 2) given by 62x2

0y2 0 + 16x2 1y2 0 − 16x0x2y2 0 + 16x2 2y2 0 − 5x2 0y0y1 − 2x2 1y0y1

− 2x2

2y0y1 + 48x2 0y2 1 + 30x2 1y2 1 − 16x0x2y2 1 + 48x1x2y2 1

+ 32x2

2y2 1 − 4x2 0y0y2 − 4x2 1y0y2 − 6x0x2y0y2 − 2x1x2y0y2

− 3x2

2y0y2 − 2x2 0y1y2 − x2 1y1y2 − 4x1x2y1y2 − 6x2 2y1y2

+ 16x2

0y2 2 + 32x2 1y2 2 + 48x0x2y2 2 + 62x2 2y2 2 = 0.

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Let W ⊆ P2 × P2 be a smooth type (2, 2) divisor. The projections πi : W → P2 define conic bundle structures ramified along smooth sextic curves C1, C2 (respectively). Let φi : Xi → P2 be a double cover ramified along Ci. Then W ×P2 Xi → Xi is a smooth P1-bundle for the ´ etale topology. This is our element A ∈ (Br X)[2].

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Caveats

The constructions above don’t necessarily yield K3 surfaces for which Pic Xk ∼ = Z. We use ideas of van Luijk, Elsenhans and Jahnel to construct surfaces for which we can prove this is the case. This requires some intensive point counts over finite fields. In the Hasse principle counter-example, we need the primes of bad reduction of X. A Groebner basis computation over Z shows these primes divide

1575961389009482560437286583534678113874684452575330515905422247848 6447126409097867530208903148583982831806707465087640413725725692102 0660524389357878901956308029862880978409927736101780460331885931216 3834734316641126951561545616484413034372280131483926551238024753488 7685949966462337614350040948449859415529952750002506898422776271778 4930382467087815253432944143520

(366 digits!)

Anthony V´ arilly-Alvarado Transcendental obstructions on K3s