Introduction Point counting on K3 surfaces and applications Problem Given a homogeneous polynomial f ∈ ❩ [ X 0 , . . . , X n ]. Compute Andreas-Stephan Elsenhans # { x ∈ P n ( ❋ p d ) | f ( x ) = 0 } Universit¨ at Paderborn for some prime p and several values of d . Providence RI October 2015 Variation Study the double cover W 2 = f ( X 0 , . . . , X n ) A variation of a Kedlaya – Harvey method for K3 surfaces of degree 2. instead of the variety f = 0. Joint work with J. Jahnel. A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 1 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 2 / 33 Naive point counting Optimized naive point counting Example Algorithms W 2 = 6 X 6 + 6 X 5 Y + 2 X 5 Z + 6 X 4 Y 2 + 5 X 4 Z 2 + 5 X 3 Y 3 Evaluate f at all points of P n and count zeroes. + X 2 Y 4 + 6 XY 5 + 5 XZ 5 + 3 Y 6 + 5 Z 6 Optimization 1 Compute roots of univariate polynomials f ( x 0 , . . . , x n − 1 , t ) Number of points over ❋ 7 , . . . , ❋ 7 10 : for all ( x 0 , . . . , x n − 1 ) ∈ P n − 1 ( ❋ p d ). 60 , 2 488 , 118 587 , 5 765 828 , 282 498 600 , Optimization 2 13 841 656 159 , 678 225 676 496 , 33 232 936 342 644 , Count Frobenius orbits of points instead of points. 1 628 413 665 268 026 , 79 792 266 679 604 918 . Complexity O ( p nd ) and O ( p ( n − 1) d ). Remark Equation has no monomial with Y and Z . A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 3 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 4 / 33
Point counting using ´ etale cohomology Point counting on cubic surfaces 27 Lines Lefschetz Trace Formula A smooth cubic surface has 27 lines. They generate the Picard group. Via 2 n the cycle map we get the ´ etale cohomology. � ( − 1) i Tr(Frob , H i # V ( ❋ p ) = et ( V , ◗ ℓ )) ´ Pic( V ) ∼ et ( V , ❩ ℓ (1)) ∼ = ❩ 7 , H 2 i =0 = Pic( V ) ⊗ ❩ ❩ ℓ ´ for a n -dimensional projective variety V with good reduction at p . Example Problem > p := NextPrime(31^31); Find explicit description of ´ etale cohomology. > p; 17069174130723235958610643029059314756044734489 Example > rr<x,y,z,w> := PolynomialRing(GF(p),4); Let E be an elliptic curve. > gl := x^3 + 2*y^3 + 3*z^3 + 5*w^3 - 7*(x+y+z+w)^3; Then the ℓ n torsion of E give an explicit description of H 1 et ( E , ❩ /ℓ n ❩ ). > time NumberOfPointsOnCubicSurface(gl); ´ 2913567055049513379297281477203956430954204417435461480 Remark 74332970578622743757660955298550825611 This is the starting point of the Schoof algorithm. Time: 1.180 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 5 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 6 / 33 Congruences on the number of points Generalization Lemma Remark Let V be a cubic surface over ❋ p . Then # V ( ❋ p ) ≡ 1 mod p . The approach above can be used to show that any hypersurface S in P n ( ❋ p ) of degree at most n satisfies # S ( ❋ p f ) ≡ 1 mod p . Proof Let f ( X , Y , Z , W ) be the corresponding cubic form. f ( x , y , z , w ) p − 1 mod p is 0 or 1 by Fermat’s theorem. Thus, Resulting Algorithm To count the number of points on the hypersurface f = 0 in P n ( ❋ p d ) modulo p , we need all the terms of f p − 1 having only exponents divisible by p − 1. p ) ≡ p 4 − 1 − f ( x , y , z , w ) p − 1 mod p � ( p − 1)# V ( ❋ x , y , z , w =0 ,..., p − 1 Variation To treat W 2 = f ( X 0 , . . . , X n ), we have to inspect the quadratic character. Any monomial of f ( x , y , z , w ) p − 1 has degree 3( p − 1). Thus, one of the x =0 ,..., p − 1 x e ≡ 0 mod p for e = 0 , . . . , p − 2 Modulo p this is given by the p − 1 2 -th power. exponents is < p − 1. As � p − 1 the sum above is zero. We get 2 . For the point count modulo p we have to sum 1 + f ( p − 1)# V ( ❋ p ) ≡ − 1 mod p . Interpretation We have a p -adic approximation of # V ( ❋ p ) with precision 1. A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 7 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 8 / 33
Counting points over extensions Philosophy Setup q = p d with p odd, f ∈ ❩ [ X 0 , . . . , X n ]. The quadratic character as a power Result from cohomology f q − 1 ( x 0 , . . . , x n ) ∈ { 0 , 1 } for x 0 , . . . , x n ∈ ❋ The number of points is given by the alternating sum of the traces of q Frobenius on cohomology. q − 1 2 ( x 0 , . . . , x n ) ∈ { 0 , ± 1 } for x 0 , . . . , x n ∈ ❋ f q Can we convert the above formula to the trace of a matrix? The quadratic character as a norm N ( f p − 1 ( x 0 , . . . , x n )) = N ( f ( x 0 , . . . , x n )) p − 1 ∈ { 0 , 1 } for x 0 , . . . , x n ∈ ❋ Extending the base field should result in the trace of a power of the matrix. q p − 1 p − 1 2 ( x 0 , . . . , x n )) = N ( f ( x 0 , . . . , x n )) N ( f ∈ { 0 , ± 1 } for x 0 , . . . , x n ∈ ❋ 2 q Conclusion We get the numbers of points over extensions without further powering. We just have to take norms. A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 9 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 10 / 33 Hasse-Witt Matrix Hasse-Witt Matrix II Polynomials and linear maps g ∈ K [ X ], deg( g ) ≤ m ( p − 1). m g : K [ X ] → K [ X ] multiplication by g . Reminder B l := { X 0 , X p l , . . . , X mp l } Special monomials Remark The matrix A represents m g ( X pl − 1 ) with domain basis B l − 1 and co-domain B 0 := { X 0 , X 1 , . . . , X m } basis B l . B 1 := { X 0 , X p , . . . , X mp } B l := { X 0 , X p l , . . . , X mp l } Theorem The matrix A l represents m g ( X pl − 1 ) ··· g ( X p ) g ( X ) with domain basis B 0 and Let A be the matrix of m g with domain basis B 0 and co-domain basis B 1 . co-domain basis B l . I.e. we combine m g with an inclusion and a projection map. (Hasse-Witt Matrix) Proof The projections remove only those terms in g ( X p j ) · · · g ( X p ) g ( X ) that do Observation not contribute to the final result. Then Tr( A ) is the sum of all coefficients of monomials of g that have degree divisible by ( p − 1). A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 11 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 12 / 33
p -adic point counting I More p -adic precision – Using extrapolation Binomial formula Let X = ± 1 + pE with E ∈ ❩ p be given. Algorithm X = ± 1 + pE Given a variety V : f = 0 or C : w 2 = f . X 2 = 1 ± 2 pE + p 2 E 2 Compute g := f p − 1 or g := f p − 1 2 . X 3 = ± 1 + 3 pE ± 3 p 2 E 2 + p 3 E 3 Use the coefficients to build up the Hasse-Witt matrix A for g . X 4 = 1 ± 4 pE + 6 p 2 E 2 ± 4 p 3 E 3 + p 4 E 4 X 5 = ± 1 + 5 pE ± 10 p 2 E 2 + 10 p 3 E 3 ± 5 p 4 E 4 + p 5 E 5 Compute trace of the d -th power of A . Derive # V ( ❋ p d ) modulo p . Linear combinations Summary E.g. We can do the point count with a p -adic precision of one digit. 1 8(15 X − 10 X 3 + 3 X 5 ) = ± 1 + 5 2 p 3 E 3 ± 15 p 4 E 4 + 3 p 5 E 5 gives us ± 1 with a p -adic precision 3. A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 13 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 14 / 33 Outline of the point counting algorithm Example K3-surface V : w 2 = x 6 + y 6 + z 6 + ( x + 2 y + 3 z ) 6 Goal Counting solutions of w 2 = f ( x , y , z ) with x , y , z ∈ ❋ ∗ q , q = p d and Count point over ❋ p d for p = 23 , 29 , 31 , 37 and d = 1 , . . . , 10. f ∈ ❩ [ X , Y , Z ]: Number of points over ❋ p 10 : 1716155831334527151964160602, 176994576151110959542233115893, 671790528819083879907512196232, Algorithm 23122483666661170932546556282656 g k := f (2 k − 1) p − 1 ∈ ( ❩ / p n ❩ )[ X , Y , Z ] for k = 1 , . . . , n . 2 Build the Hasse-Witt matrices A k corresponding to g k . Benchmark Compute traces of powers of A k . q ) mod p 11 Computation of # V ( ❋ Each trace results in an approximation with p -adic precision 1. Highest inspected power f 378 of degree 2268 with 2575315 terms 6 Use the extrapolation method to get p -adic precision n . Matrices up to size 2080 × 2080 Time per prime: 2 minutes 1.6 GB memory usage A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 15 / 33 A.-S. Elsenhans (Universit¨ at Paderborn) Point Counting October 2015 16 / 33
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