On the supersingular K 3 surface in characteristic 5 with Artin - PowerPoint PPT Presentation

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds On the supersingular K 3 surface in characteristic 5 with Artin invariant 1 Ichiro Shimada Hiroshima University May 28, 2014, Hakodate 1 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds Introduction A K 3 surface is called supersingular if its Picard number is 22. Let Y be a supersingular K 3 surface in characteristic p > 0. eron-Severi lattice, and put S ∨ Let S Y be its N´ Y := Hom ( S Y , Z ). → S ∨ The intersection form on S Y yields S Y ֒ Y . Artin proved that S ∨ Y / S Y ∼ = ( Z / p Z ) 2 σ , where σ is an integer such that 1 ≤ σ ≤ 10, which is called the Artin invariant of Y . Ogus and Rudakov-Shafarevich proved that a supersingular K 3 surface with Artin invariant 1 in characteristic p is unique up to isomorphisms. 2 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds We consider the supersingular K 3 surface X in characteristic 5 with Artin invariant 1. We work in characteristic 5. Let B F be the Fermat sextic curve (or the Hermitian curve) in P 2 : x 6 + y 6 + z 6 = 0 ( x ¯ x + y ¯ y + z ¯ z = 0) . Let π F : X → P 2 be the double cover of P 2 branched along B F : X : w 2 = x 6 + y 6 + z 6 . Then X is a supersingular K 3 surface in characteristic 5 with Artin invariant 1 3 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds Proof. Let P be an F 25 -rational point of B F , and ℓ P the tangent line to B F at P . Then ℓ P intersects B F at P with multiplicity 6, and hence π − 1 F ( ℓ P ) splits into two smooth rational curves. Since | B F ( F 25 ) | = 126, we obtain 252 smooth rational curves on X . Calculating the intersection numbers of these 252 smooth rational curves, we see that their classes span a lattice of rank 22 (hence X is supersingular) with discriminant − 25 (hence σ = 1). 4 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds In fact, the lattice S X is generated by appropriately chosen 22 curves among these 252 curves. Corollary Every class of S X is represented by a divisor defined over F 25 . Corollary Every projective model of X can be defined over F 25 . Remark Sch¨ utt proved the above results for supersingular K 3 surfaces of Artin invariant 1 in any characteristics. 5 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds 2 − 2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 3 − 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 7 6 7 6 1 0 − 2 1 1 1 0 0 0 0 0 1 1 1 1 0 0 1 1 1 0 0 7 6 7 6 7 1 0 1 − 2 1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 1 6 7 6 7 6 1 0 1 1 − 2 1 1 1 1 1 0 1 0 0 0 1 1 0 0 1 0 0 7 6 7 6 7 1 0 1 1 1 − 2 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 1 6 7 6 7 1 0 0 0 1 0 − 2 0 0 1 0 0 0 0 0 0 0 1 1 1 1 0 6 7 6 7 6 7 1 0 0 0 1 0 0 − 2 1 0 0 1 1 1 1 0 0 0 0 0 1 1 6 7 6 7 6 1 0 0 0 1 0 0 1 − 2 0 1 0 1 0 0 0 1 1 1 1 0 1 7 6 7 6 7 1 0 0 0 1 0 1 0 0 − 2 1 0 1 1 1 0 1 1 0 0 0 1 6 7 6 7 6 1 0 0 1 0 0 0 0 1 1 − 2 1 0 0 1 1 0 1 1 0 0 0 7 6 7 6 7 1 0 1 0 1 0 0 1 0 0 1 − 2 0 0 1 1 0 1 1 0 1 1 6 7 6 7 6 1 0 1 1 0 0 0 1 1 1 0 0 − 2 1 0 0 1 0 0 1 0 0 7 6 7 6 7 1 0 1 0 0 0 0 1 0 1 0 0 1 − 2 0 1 0 1 1 0 0 0 6 7 6 7 6 1 0 1 0 0 1 0 1 0 1 1 1 0 0 − 2 0 1 0 0 1 0 0 7 6 7 6 7 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 − 2 1 0 0 1 0 0 6 7 6 7 6 1 0 0 0 1 1 0 0 1 1 0 0 1 0 1 1 − 2 0 1 0 1 0 7 6 7 6 7 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 0 − 2 0 1 1 0 6 7 6 7 6 1 0 1 1 0 0 1 0 1 0 1 1 0 1 0 0 1 0 − 2 0 0 0 7 6 7 6 7 1 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 0 1 0 − 2 1 1 6 7 6 7 6 1 0 0 1 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0 1 − 2 0 7 4 5 1 0 0 1 0 1 0 1 1 1 0 1 0 0 0 0 0 0 0 1 0 − 2 6 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds Problem: Find distinct projective models of X (especially of degree 2) as many as possible. We put P 2 := { h ∈ S X | h is a polarization of degree 2 } , that is, h ∈ S X belongs to P 2 if and only if the line bundle L → X corresponding to h gives a double covering Φ |L| : X → P 2 . Let B h be the branch curve of Φ |L| : X → P 2 . For h , h ′ ∈ P 2 , we say h ∼ h ′ if there exists g ∈ Aut ( X ) such that g ∗ ( h ) = h ′ , or equivalently, there exists φ ∈ PGL 3 ( k ) such that φ ( B h ′ ) = B h . Problem: Describe P 2 / ∼ . 7 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds The lattice S X is characterized as the unique even hyperbolic X / S X ∼ lattice of rank 22 with S ∨ = ( Z / 5 Z ) 2 . Therefore we can obtain a list of combinatorial data of these B h by lattice theoretic method, which was initiated by Yang. We try to find defining equations of these B h , and understand their relations. 8 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds Naive method. Projective models of the supersingular K3 surface with Artin invariant 1 in characteristic 5. J. Algebra 403 (2014), 273-299. Specialization from σ = 3 (joint work with Pho Duc Tai). Unirationality of certain supersingular K3 surfaces in characteristic 5. Manuscripta Math. 121 (2006), no. 4, 425–435. Ballico-Hefez curve (joint work with Hoang Thanh Hoai). On Ballico-Hefez curves and associated supersingular surfaces, to appear in Kodai Math. J. Borcherds’ method (joint work with T. Katsura and S. Kondo). On the supersingular K3 surface in characteristic 5 with Artin invariant , preprint, arXiv:1312.0687 9 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds Naive method Classification by relative degrees with respect to h F . We have the polarization h F ∈ P 2 that gives the Fermat double sextic plane model π F : X → P 2 : h F = [1 , 1 , 0 , . . . , 0] . We have Aut ( X , h F ) = PGU 3 ( F 25 ) . 2 , which is of order 756000. For a ∈ Z > 0 , we put P 2 ( a ) := { h ∈ P 2 | � h F , h � = a } . 10 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds For any a ∈ Z > 0 , the set V 2 ( a ) := { h ∈ S X | h 2 = 2 , � h F , h � = a } is finite. Then h ∈ V 2 ( a ) belongs to P 2 ( a ) if h is nef and not of the form 2 · f + z , with f 2 = 0 , z 2 = − 2 , � f , z � = 1 . The vector h ∈ V 2 is nef if and only if there are no vectors r ∈ S X such that r 2 = − 2 , � h F , r � > 0 , � h , r � < 0 . Thus we can calculate P 2 ( a ) for a given a ∈ Z > 0 . 11 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds We have calculated P 2 ( a ) for a ≤ 5. Their union consists of 146 , 945 , 851 vectors. From the defining ideals of the 22 lines on X F we have chosen as a basis of S X , we can calculate the defining equations of B h for each h , and hence we can determine whether h ∼ h ′ or not. Under ∼ , they are decomposed into 65 equivalence classes. 12 / 22

Introduction Problem Naive method Specialization Ballico-Hefez Borcherds 0: Sing = 0: N = 13051: h = [1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0] : x 6 + y 6 + 1 1: Sing = 6 A 1 : N = 5607000: h = [0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 1 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 1] : x 6 + 3 x 5 y + x 4 y 2 + 2 x 3 y 3 + y 6 + 3 x 4 + 3 x 2 y 2 + xy 3 + 3 xy + 2 y 2 + 4 2: Sing = 7 A 1 : N = 6678000: h = [0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0] : x 6 + 2 x 4 y 2 + x 2 y 4 + x 2 y 3 + 2 y 5 + x 4 + 2 y 4 + 2 x 2 y + 2 y 3 + 3 y 2 + 3 y + 2 3: Sing = 3 A 1 + 2 A 2 : N = 2268000: h = [0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0] : x 6 + 3 x 3 y 3 + y 6 + 3 x 3 y + 2 y 2 + 2 4: Sing = 8 A 1 : N = 2457000: h = [0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0] : x 6 + 3 x 4 y 2 + x 2 y 4 + 4 x 2 y 3 + 4 y 5 + x 4 + 2 x 2 y 2 + 3 y 4 + 2 x 2 y + 4 x 2 + y 2 + 4 y 5: Sing = 8 A 1 : N = 2268000: h = [0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 0 , 0] : x 4 y 2 + x 2 y 4 + 2 x 4 + 4 x 2 y 2 + y 4 + x 2 + 4 y 2 + 4 6: Sing = 6 A 1 + A 2 : N = 1512000: h = [0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 1 , 0 , 1 , 1 , 0 , 1 , 0 , 0] : x 6 + 4 x 4 y 2 + 2 x 2 y 4 + 2 x 2 y + y 3 + 4 7: Sing = 6 A 1 + A 2 : N = 4914000: h = [0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 1 , 0 , 1 , 0 , 1] : √ √ √ √ √ 2 x 3 y 3 + “ ” x 2 y 4 + x 4 + “ ” x 3 y + “ ” x 2 y 2 + xy 3 + “ ” y 4 + 1 + 3 2 2 + 2 2 1 + 4 2 2 + 2 2 √ √ 2 x 2 + “ ” 1 + 3 2 xy 13 / 22