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 Ω k 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 ) ֒ → X ( k v ) = X ( A ) v ∈ Ω k 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 Br 0 X := im(Br k → Br X ) No obstructions: X ( A ) A = X ( A ) for all A ∈ Br 0 X Algebraic elements Br 1 X := ker(Br X → Br X k ) If X ( A ) � = ∅ then there is an isomorphism → H 1 � � Br 1 X ∼ − Gal( k / k ) , Pic X k Br 0 X (Hochschild-Serre spectral sequence) If Pic X k = Z then Br 1 X gives no obstructions Transcendental elements Br X \ Br 1 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 = Br 1 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 h 1 � � ω X ∼ and X , O X = 0 . = O X Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Examples of K3 surfaces Double covers of P 2 ramified along a smooth sextic plane curve: { 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 P 3 . Smooth complete intersections of 3 quadrics in P 5 . Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

Transcendental elements: basic questions Theorem (Harari, 1996) There exist infinitely many explicit conic bundles V over P 2 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 obstruction? 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 X k ∼ = Z with a transcendental obstruction to weak approximation? For such a surface → H 1 � � ∼ Br 1 X / Br 0 X − Gal( k / k ) , Pic X k = 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 3 y 2 + 2 z 2 0 2(2 x + 3 y + z ) 3 x + 3 y 3 x + 4 y 1 4 y 2 w 2 = det 3 x + 3 y 2( z ) 3 z B C 4 x 2 + 5 xy + 5 y 2 B C 3 x + 4 y 3 z 2( x + 3 z ) @ A 3 y 2 + 2 z 2 4 x 2 + 5 xy + 5 y 2 2(2 x 3 + 3 x 2 z + 3 xz 2 + 3 z 3 ) 4 y 2 in P Q (1 , 1 , 1 , 3) . Then Pic X Q ∼ = Z , and there is a transcendental Brauer-Manin obstruction to weak approximation on X. The obstruction arises from a quaternion Azumaya algebra A ∈ im (Br X ֒ → Br κ ( X )) . Explicitly, if M i denotes the i-th leading principal minor of the above matrix, then � � − M 2 , − M 3 A = . M 2 M 1 M 2 1 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 := 62 x 2 − 16 xz + 16 y 2 + 16 z 2 B := − 5 x 2 − 2 y 2 − 2 z 2 C := − 4 x 2 − 6 xz − 4 y 2 − 2 yz − 3 z 2 D := 48 x 2 − 16 xz + 30 y 2 + 48 yz + 32 z 2 E := − 2 x 2 − y 2 − 4 yz − 6 z 2 F := 16 x 2 + 48 xz + 32 y 2 + 62 z 2 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   2 A B C w 2 = det   B 2 D E C E 2 F in P Q (1 , 1 , 1 , 3) . Then Pic X Q ∼ = Z , and there is a transcendental Brauer-Manin obstruction to the Hasse principle on X. The obstruction arises from a quaternion Azumaya algebra A ∈ im (Br X ֒ → Br κ ( X )) . Explicitly, if M i denotes the i-th leading principal minor of the above matrix, then � � − M 2 , − M 3 A = . M 2 M 1 M 2 1 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 T X := NS( X ) ⊥ ⊆ H 2 ( X , Z ) be the transcendental lattice of X . Write Λ K 3 = U 3 ⊕ E 8 ( − 1) 2 for the abstract K3 lattice. The exponential sequence shows there is a one-to-one correspondence 1 − 1 { α ∈ Br X of exact order n } ← → { surjections T X → Z / n Z } Hence, to α as above, we may associate T α ⊆ T X : T α = ker( α : T X → Z / n Z ) . 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 T α ֒ → Λ K 3 . This gives a degree 8 K3 surface Y associated to the pair ( X , α ) . = � h 2 , P � ⊥ ⊆ H 4 ( Z , Z ) , where Z is a cubic fourfold 2 T α ( − 1) ∼ with a plane P (h is the hyperplane class). 3 T α ( − 1) ∼ 2 � ⊥ ⊆ H 4 ( W , Z ) , where W is a double = � h 2 1 , h 1 h 2 , h 2 cover of P 2 × P 2 ramified along a type (2 , 2) divisor (h 1 , h 2 are the pullbacks to W of the hyperplane classes of P 2 under the two projections P 2 × P 2 → P 2 ). Idea: “go backwards” and work over any field k . Anthony V´ arilly-Alvarado Transcendental obstructions on K3s

The pair ( X , A ) in the K 3 counter-example to weak approximation is naturally associated to a cubic fourfold Y ⊆ P 5 Q := Proj Q [ X 1 , X 2 , X 3 , Y 1 , Y 2 , Y 3 ] given by 2 X 2 1 Y 1 + 3 X 2 1 Y 2 + X 2 1 Y 3 + 3 X 1 X 2 Y 1 + 3 X 1 X 2 Y 2 + 3 X 1 X 3 Y 1 + 4 X 1 X 3 Y 2 + 3 X 1 Y 2 2 + 2 X 1 Y 2 3 + X 2 2 Y 3 + 3 X 2 X 3 Y 3 + 4 X 2 Y 2 2 + X 2 3 Y 1 + 3 X 2 3 Y 3 + 4 X 3 Y 2 1 + 5 X 3 Y 1 Y 2 + 5 X 3 Y 2 2 + 2 Y 3 1 + 3 Y 2 1 Y 3 + 3 Y 1 Y 2 3 + 3 Y 3 3 = 0 . This fourfold contains the plane { Y 1 = Y 2 = Y 3 = 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, Y → P 2 the corresponding quadric surface bundle, and q : � r : W → P 2 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 π 1 φ → P 2 r : W → X consists of a smooth P 1 -bundle followed by a degree-two cover of P 2 , 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 K 3 counter-example to the Hasse principle is naturally associated to a double cover of P 2 × P 2 = Proj Q [ x 0 , x 1 , x 2 ] × Proj Q [ y 0 , y 1 , y 2 ] ramified along the divisor of type (2 , 2) given by 62 x 2 0 y 2 0 + 16 x 2 1 y 2 0 − 16 x 0 x 2 y 2 0 + 16 x 2 2 y 2 0 − 5 x 2 0 y 0 y 1 − 2 x 2 1 y 0 y 1 − 2 x 2 2 y 0 y 1 + 48 x 2 0 y 2 1 + 30 x 2 1 y 2 1 − 16 x 0 x 2 y 2 1 + 48 x 1 x 2 y 2 1 + 32 x 2 2 y 2 1 − 4 x 2 0 y 0 y 2 − 4 x 2 1 y 0 y 2 − 6 x 0 x 2 y 0 y 2 − 2 x 1 x 2 y 0 y 2 − 3 x 2 2 y 0 y 2 − 2 x 2 0 y 1 y 2 − x 2 1 y 1 y 2 − 4 x 1 x 2 y 1 y 2 − 6 x 2 2 y 1 y 2 + 16 x 2 0 y 2 2 + 32 x 2 1 y 2 2 + 48 x 0 x 2 y 2 2 + 62 x 2 2 y 2 2 = 0 . Anthony V´ arilly-Alvarado Transcendental obstructions on K3s