Cubic fourfolds and odd-torsion Brauer Manin obstructions on K3 surfaces

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1 Cubic fourfolds and odd-torsion Brauer Manin obstructions on K3 surfaces Anthony Várilly-Alvarado Rice University Arithmetic Geometry, Number Theory and Computation August 22nd, 2018

2 We report on results of a collaboration Everything today is joint work with Jen Berg.

3 Brauer Manin recap X /k a nice variety over a number field. H Br(X ) := H 2 et(x,g m ) tors gives rise to an obstruction set X (k) X (A) H X (A) := v X (k v ) that contains X (k) (for product topology). Hasse principle: X (A) }{{} = X (k) }{{} locally soluble globally soluble Weak Approximation: locally soluble + X (k) = X (A).

4 Brauer Manin recap Brauer Manin obstruction to Hasse Principle: X (A) but X (A) H = for some H Br(X ). Brauer Manin obstruction to Weak Approximation: X (A) \ X (A) H for some H Br(X ).

5 K3 surfaces K3 surface: nice surface X with ω X O X and H 1 (X,O X ) = 0. Example A smooth double cover X P 2, branched along a plane sextic: w 2 = f 6 (x,y,z) This is a degree 2 K3 surface. Conjecture (Skorobogatov, 2009) If X /k is a locally soluble K3 surface over a number field, then X (k) = X (A) Br(X ). Informally: the Brauer Manin obstruction accounts for failures of the Hasse Principle and Weak Approximation on K3 surfaces.

6 A brief history of K3 obstructions X /k a K3 surface over a number field. H = Br(X )[2] can obstruct Weak Approximation: Wittenberg, 2004 Ieronymou, Hassett, V.-A., Elsenhans, Jahnel, Mckinnie, Sawon, Tanimoto, V.-A., H = Br(X )[2] can obstruct the Hasse principle: Martin Bright, Hassett, V.-A., H = Br(X )[3] or Br(X )[5] can obstruct the Weak Approximation: Preu, Ieronymou Skorobogatov, 2015.

7 What about odd torsion and the Hasse Principle? Question (Ieronymou Skorobogatov, 2015) Is there a locally soluble K3 surface X over a number field with X (A) Br(X ) odd =? Theorem (Corn Nakahara, 2017) The degree 2 K3 surface X /Q w 2 = 3x 6 +97y z 6 is locally soluble. The cyclic algebra ( A := Q( 3 28,ζ 3 )/Q(ζ 3 ), w ) 3x 3 w + BrQ(ζ 3 )(X ) 3x 3 extends to an element of Br(X Q(ζ3 )) that gives rise to a Brauer-Manin obstruction to the Hasse Principle on X.

8 What about odd torsion in the Brauer group? The class A in Corn Nakahara is algebraic. Recall there is a filtration Br 0 (X ) Br 1 (X ) Br(X ). }{{}}{{} im(brk BrX ) ker(br(x ) Br(X )) constant classes algebraic classes = X (A) Br 0(X ) = X (A), so we often consider the quotients Br 1 (X )/Br 0 (X ) and Br(X )/Br 1 (X ) when computing Brauer Manin obstructions.

9 Can odd transcendental classes obstruct HP on a K3? Theorem (Berg, V.-A., 2018) There is a degree 2 K3 surface X /Q, together with an A Br(X )[3], such that X (A) and X (A) A =. There is no algebraic Brauer Manin obstruction to the Hasse Principle on X. For last claim: Pic(X ) Z, and hence Br 1 (X )/Br 0 (X ) = 0.

10 Can odd transcendental classes obstruct HP on a K3? How can we find transcendental 3-torsion in Br(X )? X /C: a complex projective K3 surface with NS(X ) = Zh, h 2 = 2d. Let T X := NS(X ) H 2 (X,Z) = U 3 E 8 ( 1) 2 =: Λ K3 be the transcendental lattice of X. Br(X ) TX Q/Z, so there is a one-to-one correspondence { A BrX of order n} 1 1 {surjections T X Z/nZ} Hence, to A as above, we may associate T A T X : T A = ker(a : T X Z/nZ).

11 Sublattices of index 3 in T X Theorem (Mckinnie, Sawon, Tanimoto, V.-A., 2017) Let X be a complex projective K3 surface with PicX = Zh, h 2 = 2, and let A (BrX )[3]. Then either 1. there is a unique primitive embedding T A Λ K3. This gives a degree 18 K3 surface Y associated to the pair (X, A ); OR 2. T A ( 1) = H 2,T H 4 (Y,Z), where Y is a cubic fourfold of discriminant 18 (H is the hyperplane class); OR, 3. T A ( 1) is a lattice with discriminant group Z/2Z (Z/3Z) 2.

12 Cubic fourfolds of discriminant 18 Hence, lattice theory suggests: Y cubic fourfold of discriminant 18 (X, A ), where X is a K3 surface of degree 2 and A Br(X )[3]. A cubic fourfold of discriminant 18 is a smooth cubic fourfold Y P 5, together with a rank two saturated lattice H 2,T H 2,2 (Y ) H 4 (Y,Z) of discriminant 18, where H is the hyperplane class, T H T 6 18 H 2 Hassett, 2000: such fourfolds exist.

13 What is the surface T? Theorem (Addington, Hassett, Tschinkel, V.-A., 2016) A general cubic fourfold Y of discriminant 18 contains an elliptic ruled surface T of degree 6. The linear system of quadrics in P 5 containing T is 3-dimensional. Let Q 0,Q 1,Q 2 C[x 0,...,x 5 ] 2 be a basis for H 0 (T,I T (2)) Bl T (Y ) π Y P(H 0 (T,I T (2)) ) = P 2 x = [x 0,...,x 5 ] [Q 0 ( x),q 1 ( x),q 2 ( x)]

14 Fibers of π Key insight: General fiber of π: Bl T (Y ) P 2 is a del Pezzo surface of degree 6. Geometrically, a dp6 is a blow-up of P 2 at 3 non-colinear points. The locus of fibers that are geometrically blow-ups of P 2 at three distinct colinear points has image a sextic curve under π: C : f (x,y,z) = 0. The extension K := C(P 2 )( f ) splits the two triples of pairwise skew exceptional curves in the generic fiber S of π. K is the function field of our K3 surface X of degree 2!

15 Brauer elements of order 3 Let S K be the (dp6) generic fiber of π base extended to K. S K contains two triples of pairwise skew exceptional curves: {E 1,E 2,E 3 } and {L E 1 E 2,L E 2 E 3,L E 1 E 3 }. These can be blown-down, respectively, with the morphisms associated to the linear systems 3L and 3(2L E 1 E 2 E 3 ).

16 Brauer elements of order 3 We get φ 3L : S K X 1 and φ 3(2L E1 E 2 E 3 ) : S K X 2 X 1 and X 2 are Severi-Brauer varieties of dimension 2 over K. Hence X 1 and X 2 give rise to two classes A 1 and A 2 (BrK)[3]. Proposition (Corn, 2005) A 1 A 2 = Id in BrK. Combining work of Corn, Kollár, AHTVA, and Kuznetsov, we can spread these classes to our K3 surface X. We have recovered a pair (X, { Id,A 1,A 2 } )!

17 Brauer elements of order 3: alternatively... In each smooth fiber S of π, there are two families of twisted cubics, each one two dimensional, parametrized by P(H 0 (S,O S (L))) = P 2 and P(H 0 (S,O S (2L E 1 E 2 E 3 ))) = P 2. These two P 2 s come together over the locus f (x,y,z) = 0 of the base P 2 of π. Let H be the relative Hilbert scheme that parametrizes twisted cubics in the fibers of π. The Stein factorization of H P 2 is H X P 2 where X is our K3 surface, and H is an étale P 2 bundle over X.

18 Brauer elements of order 3: alternatively... Hence H gives rise to an element A Br(X [3]). Notes: 1. The covering involution ι: X X sending w w induces ι : Br(X ) Br(X ) sending A A 1, so H actually encodes the group A. 2. If (Y,T ) are defined over a field k (e.g., a number field), then so are H and X.

19 Computing the obstruction Theorem (Wedderburn) Every central simple algebra of degree 3 over a field is cyclic. In principle, can write down cyclic representatives for A 1 and A 2. Challenge Do it.

20 Computing the obstruction { X (A) A = (P v ) X (A) : v } inv v A (P v ) = 0. inv v A (P v ) = 0 the fiber in H above P v X (k v ) is a split P 2 (i.e., is isomorphic to P 2 over k v ) the fiber has a k v -rational point. From now on, X is defined over Q. To produce an example with X (A) A =, we rig (X,A ) so that inv v A (P v ) = { 0 if v 3, 1/3 or 2/3 if v = 3.

21 Lang Nishimura to the rescue Lemma (Lang Nishimura) Let Z 1 and Z 2 be birational nice Q v -varieties. Then Z 1 (Q v ) Z 2 (Q v ). The fiber in H above P v X (Q v ) is birational to the dp6 fiber of π above the image of P v in P 2. Apply Lang-Nishimura: to have inv 3 A (P) 0 for all P X (Q 3 ), it suffices to have Bl T (Y )(Q 3 ) =. Applying Lang-Nishimura again, it suffices to have Y (Q 3 ) =.

22 Construction of the cubic fourfold Thus, the hardest part of our task is to produce a cubic fourfold Y /Q of discriminant 18, such that Y (Q 3 ) =. Let P 5 := ProjQ[x 0,x 1,x 2,x 3,x 4,x 5 ], and define quadrics cut out by Q 1 := x 0 x 3 +x 2 x 3 x 0 x 4 +x 1 x 4 +3x 2 x 4 +5x 0 x 5 x 1 x 5 Q 2 := x 1 x 3 +5x 0 x 4 2x 2 x 4 2x 0 x 5 +5x 1 x 5 +x 2 x 5 Q 3 := 2x 2 x 3 x 0 x 4 2x 1 x 4 2x 2 x 4 +x 1 x 5 Each quadric contains the planes Π 1 = {x 0 = x 1 = x 2 = 0} and Π 2 = {x 3 = x 4 = x 5 = 0}.

23 Construction of the cubic fourfold The sextic elliptic surface T is obtained by saturating the ideal Q 1,Q 2,Q 3 with respect to I (Π 1 )I (Π 2 ). It is cut out by Q 1,Q 2,Q 3 and the two cubics C 1 := 2x3 3 +5x 3 2 x 4 +x 3 x x x 3 2 x 5 26x 3 x 4 x 5 11x4 2 x 5 +47x 3 x x 4x5 2 +5x 5 3, C 2 := 2x0 3 x 0 2 x 1 2x 0 x1 2 x x 0 2 x 2 +10x 0 x 1 x 2 +x1 2 x 2 11x 0 x2 2 18x 1x2 2 4x 2 3 Magical property: The plane cubics, {C 1 = 0} and {C 2 = 0} have no Z/3Z-points! Bhargava, Cremona, Fisher: 99.25% of plane cubics have Q 3 -points.

24 Construction of the cubic fourfold C 1 := 2x3 3 +5x 3 2 x 4 +x 3 x x x 3 2 x 5 26x 3 x 4 x 5 11x4 2 x 5 +47x 3 x x 4x5 2 +5x 5 3, C 2 := 2x0 3 x 0 2 x 1 2x 0 x1 2 x x 0 2 x 2 +10x 0 x 1 x 2 +x1 2 x 2 11x 0 x2 2 18x 1x2 2 4x 2 3 The cubic fourfold Y is defined by 3C 1 +C 2 +9 (suitable element of Q 1,Q 2,Q 3 ) which has the property that Y (Z/9Z) =!

25 The cubic fourfold The cubic fourfold Y is defined by 2x0 3 x 0 2 x 1 2x 0 x1 2 x x 0 2 x 2 +10x 0 x 1 x 2 +x1 2 x 2 11x 0 x2 2 18x 1x2 2 4x x 0 2 x 3 +18x 0 x 1 x 3 +9x1 2 x 3 +18x 0 x 2 x 3 +18x 1 x 2 x 3 +18x2 2 x 3 +9x 1 x3 2 +6x x0 2 x 4 +9x 0 x 1 x 4 +18x1 2 x 4 9x 0 x 2 x 4 +18x 1 x 2 x 4 +18x2 2 x 4 27x 0 x 3 x 4 +18x 2 x 3 x 4 +15x3 2 x 4 +27x 0 x4 2 36x 2x4 2 +3x 3x x4 3 90x 0 2 x 5 72x 0 x 1 x 5 45x1 2 x 5 18x 1 x 2 x 5 +36x 0 x 3 x 5 45x 1 x 3 x 5 +9x 2 x 3 x 5 60x3 2 x 5 54x 0 x 4 x 5 +27x 1 x 4 x 5 18x 2 x 4 x 5 78x 3 x 4 x 5 33x4 2 x 5 90x 0 x x 3x x 4x x 5 3 = 0.

26 The K3 surface The K3 surface X P(1,1,1,3) is given by w 2 = x x 5 y x 4 y x 3 y x 2 y xy y x 5 z x 4 yz x 3 y 2 z x 2 y 3 z xy 4 z y 5 z x 4 z x 3 yz x 2 y 2 z xy 3 z y 4 z x 3 z x 2 yz xy 2 z y 3 z x 2 z xyz y 2 z xz yz z 6. We show Pic(X ) Z. Hence Br 1 (X )/Br 0 (X ) = 0.

27 Primes of bad reduction of Y 3, 5, 29, 2851, , , , ,

28 Main Theorem Theorem (Berg, V.-A., 2018) There is a degree 2 K3 surface X /Q, together with an A Br(X )[3], such that X (A) and X (A) A =. There is no algebraic Brauer Manin obstruction to the Hasse Principle on X.

Transcendental obstructions on K3 surfaces

Transcendental obstructions on K3 surfaces Anthony Várilly-Alvarado February 28, 2012 1 Introduction Let X be a smooth projective geometrically integral variety over Q. We have an inclusion φ : X(Q) p,