Equidistributions in arithmetic geometry

Transcription

1 Equidistributions in arithmetic geometry Edgar Costa Dartmouth College 14th January 2016 Dartmouth College 1 / 29 Edgar Costa Equidistributions in arithmetic geometry

2 Motivation: Randomness Principle Rigidity/Randomness Dichotomy [Sarnak] Given an arithmetic problem, either 1 rigid structure rigid solution, or 2 the answer is difficult to determine random behaviour 2 / 29 Edgar Costa Equidistributions in arithmetic geometry

3 Motivation: Randomness Principle Rigidity/Randomness Dichotomy [Sarnak] Given an arithmetic problem, either 1 rigid structure rigid solution, or 2 the answer is difficult to determine random behaviour Understanding and providing the probability law deep understanding of the phenomenon Real world applications 2 / 29 Edgar Costa Equidistributions in arithmetic geometry

4 Motivation: Problem p a prime number X an integral object, e.g.: a integer a polynomial with integer coefficients a curve or a surface defined by a polynomial equation with integer coefficients... 3 / 29 Edgar Costa Equidistributions in arithmetic geometry

5 Motivation: Problem p a prime number X an integral object, e.g.: a integer a polynomial with integer coefficients a curve or a surface defined by a polynomial equation with integer coefficients... We can consider X modulo p. 3 / 29 Edgar Costa Equidistributions in arithmetic geometry

6 Motivation: Problem p a prime number X an integral object, e.g.: a integer a polynomial with integer coefficients a curve or a surface defined by a polynomial equation with integer coefficients... We can consider X modulo p. Question Given X mod p What can we say about X? 3 / 29 Edgar Costa Equidistributions in arithmetic geometry

7 Motivation: Problem p a prime number X an integral object, e.g.: a integer a polynomial with integer coefficients a curve or a surface defined by a polynomial equation with integer coefficients... We can consider X modulo p. Question Given X mod p What can we say about X? What if we consider infinitely many primes? 3 / 29 Edgar Costa Equidistributions in arithmetic geometry

8 Motivation: Problem p a prime number X an integral object, e.g.: a integer a polynomial with integer coefficients a curve or a surface defined by a polynomial equation with integer coefficients... We can consider X modulo p. Question Given X mod p What can we say about X? What if we consider infinitely many primes? How does it behave as p? 3 / 29 Edgar Costa Equidistributions in arithmetic geometry

9 Overview 1 Polynomials in one variable 2 Elliptic curves 3 Quartic surfaces 4 / 29 Edgar Costa Equidistributions in arithmetic geometry

10 Counting roots of polynomials f (x) Z[x] an irreducible polynomial of degree d > 0 p a prime number Consider: N f (p) := # {x {0,..., p 1} : f (x) 0 mod p} = # {x F p : f (x) = 0} N f (p) {0, 1,..., d} Question How often does each value occur? 5 / 29 Edgar Costa Equidistributions in arithmetic geometry

11 Example: quadratic polynomials f (x) = ax 2 + bx + c, = b 2 4ac, the discriminant of f. 0 if is not a square modulo p Quadratic formula = N f (p) = 1 0 mod p 2 if is a square modulo p 6 / 29 Edgar Costa Equidistributions in arithmetic geometry

12 Example: quadratic polynomials f (x) = ax 2 + bx + c, = b 2 4ac, the discriminant of f. 0 if is not a square modulo p Quadratic formula = N f (p) = 1 0 mod p 2 if is a square modulo p Half of the numbers modulo p are squares. Hence, if isn t a square, then Prob( is a square modulo p) = 1/2 = Prob(N f (p) = 0) = Prob(N f (p) = 2) = / 29 Edgar Costa Equidistributions in arithmetic geometry

13 Example: quadratic polynomials f (x) = ax 2 + bx + c, = b 2 4ac, the discriminant of f. 0 if is not a square modulo p Quadratic formula = N f (p) = 1 0 mod p 2 if is a square modulo p Half of the numbers modulo p are squares. Hence, if isn t a square, then Prob( is a square modulo p) = 1/2 = Prob(N f (p) = 0) = Prob(N f (p) = 2) = if p 2, 3 mod 5 For example, if = 5 and p > 2, then N f (p) = 1 if p = 5 2 if p 1, 4 mod 5 6 / 29 Edgar Costa Equidistributions in arithmetic geometry

14 Example: cubic polynomials In general one cannot find explicit formulas for N f (p), but we can still determine their average distribution! 7 / 29 Edgar Costa Equidistributions in arithmetic geometry

15 Example: cubic polynomials In general one cannot find explicit formulas for N f (p), but we can still determine their average distribution! f (x) = x 3 2 = ( x 3 2 ) ( x 3 2e 2πi/3 ) ( x 3 2e 4πi/3 ) 1/3 if x = 0 Prob (N f (p) = x) = 1/2 if x = 1 1/6 if x = 3. f (x) = x 3 x 2 2x + 1 = (x α 1 ) (x α 2 ) (x α 3 ) { 2/3 if x = 0 Prob (N f (p) = x) = 1/3 if x = 3. 7 / 29 Edgar Costa Equidistributions in arithmetic geometry

16 The Chebotarëv density theorem f (x) = (x α 1 )... (x α d ), α i C G := Aut(Q(α 1,..., α d )/Q) = Gal(f /Q) G S d, as it acts on the roots α 1,..., α d by permutations. 8 / 29 Edgar Costa Equidistributions in arithmetic geometry

17 The Chebotarëv density theorem f (x) = (x α 1 )... (x α d ), α i C G := Aut(Q(α 1,..., α d )/Q) = Gal(f /Q) G S d, as it acts on the roots α 1,..., α d by permutations. Theorem (Chebotarëv, early 1920s) For i = 0,..., d, we have where Prob(N f (p) = i) = Prob(g G : g fixes i roots), Prob(N f (p) = i) := lim N #{p prime, p N, N f (p) = i}. #{p prime, p N} 8 / 29 Edgar Costa Equidistributions in arithmetic geometry

18 Example: Cubic polynomials, again f (x) = x 3 2 = ( x 3 2 ) ( x 3 2e 2πi/3 ) ( x 3 2e 4πi/3 ) 1/3 if x = 0 Prob (N f (p) = x) = 1/2 if x = 1 and G = S 3. 1/6 if x = 3 S 3 = id, (1 2), (1 3), (2 3), ( ), ( ) 9 / 29 Edgar Costa Equidistributions in arithmetic geometry

19 Example: Cubic polynomials, again f (x) = x 3 2 = ( x 3 2 ) ( x 3 2e 2πi/3 ) ( x 3 2e 4πi/3 ) 1/3 if x = 0 Prob (N f (p) = x) = 1/2 if x = 1 and G = S 3. 1/6 if x = 3 S 3 = id, (1 2), (1 3), (2 3), ( ), ( ) f (x) = x 3 x 2 2x + 1 = (x α 1 ) (x α 2 ) (x α 3 ) { 2/3 if x = 0 Prob (N f (p) = x) = and G = Z/3Z. 1/3 if x = 3 9 / 29 Edgar Costa Equidistributions in arithmetic geometry

20 Prime powers We may also define N f (p e ) = # {x F p e : f (x) = 0} Theorem (Chebotarëv continued) Prob ( N f (p) = c 1, N f ( p 2 ) = c 2, ) Prob ( g G : g fixes c 1 roots, g 2 fixes c 2 roots,... ) 10 / 29 Edgar Costa Equidistributions in arithmetic geometry

21 Prime powers We may also define N f (p e ) = # {x F p e : f (x) = 0} Theorem (Chebotarëv continued) Prob ( N f (p) = c 1, N f ( p 2 ) = c 2, ) Prob ( g G : g fixes c 1 roots, g 2 fixes c 2 roots,... ) Let f (x) = x 3 2, then G = S 3 and: Prob ( ( N f (p) = N f p 2 ) = 0 ) = 1/3 Prob ( ( N f (p) = N f p 2 ) = 3 ) = 1/6 Prob ( ( N f (p) = 1, N f p 2 ) = 3 ) = 1/2 10 / 29 Edgar Costa Equidistributions in arithmetic geometry

22 1 Polynomials in one variable 2 Elliptic curves 3 Quartic surfaces 11 / 29 Edgar Costa Equidistributions in arithmetic geometry

23 Elliptic curves An elliptic curve is a smooth plane algebraic curve defined by y 2 = x 3 + ax + b over the complex numbers C this is a torus: 12 / 29 Edgar Costa Equidistributions in arithmetic geometry

24 Elliptic curves An elliptic curve is a smooth plane algebraic curve defined by y 2 = x 3 + ax + b over the complex numbers C this is a torus: There is a natural group structure! If P, Q, and R are colinear, then P + Q + R = 0 12 / 29 Edgar Costa Equidistributions in arithmetic geometry

25 Elliptic curves An elliptic curve is a smooth plane algebraic curve defined by y 2 = x 3 + ax + b over the complex numbers C this is a torus: There is a natural group structure! If P, Q, and R are colinear, then P + Q + R = 0 Applications: cryptography integer factorization pseudorandom numbers, / 29 Edgar Costa Equidistributions in arithmetic geometry

26 Counting points on elliptic curves Given an elliptic curve over Q X : y 2 = x 3 + ax + b, a, b Z We can consider its reduction modulo p (ignoring bad primes and p = 2). 13 / 29 Edgar Costa Equidistributions in arithmetic geometry

27 Counting points on elliptic curves Given an elliptic curve over Q X : y 2 = x 3 + ax + b, a, b Z We can consider its reduction modulo p (ignoring bad primes and p = 2). As before, consider: N X (p) := #X(F p ) = { (x, y) (F p ) 2 : y 2 = f (x) } + 1 One cannot hope to write N X (p) as an explicit function of p. 13 / 29 Edgar Costa Equidistributions in arithmetic geometry

28 Counting points on elliptic curves Given an elliptic curve over Q X : y 2 = x 3 + ax + b, a, b Z We can consider its reduction modulo p (ignoring bad primes and p = 2). As before, consider: N X (p) := #X(F p ) = { (x, y) (F p ) 2 : y 2 = f (x) } + 1 One cannot hope to write N X (p) as an explicit function of p. Instead, we look for statistical properties of N X (p). 13 / 29 Edgar Costa Equidistributions in arithmetic geometry

29 Hasse s bound Theorem (Hasse, 1930s) p + 1 N X (p) 2 p. 14 / 29 Edgar Costa Equidistributions in arithmetic geometry

30 Hasse s bound Theorem (Hasse, 1930s) p + 1 N X (p) 2 p. In other words, N x (p) = p + 1 pλ p, λ p [ 2, 2] What can we say about the error term, λ p, as p? 14 / 29 Edgar Costa Equidistributions in arithmetic geometry

31 Weil s theorem Theorem (Hasse, 1930s) N x (p) = p + 1 pλ p, λ p [ 2, 2]. 15 / 29 Edgar Costa Equidistributions in arithmetic geometry

32 Weil s theorem Theorem (Hasse, 1930s) N x (p) = p + 1 pλ p, λ p [ 2, 2]. Taking λ p = 2 cos θ p, with θ p [0, π] we can rewrite N X (p) = p + 1 p(α p + α p ), α p = e iθp. 15 / 29 Edgar Costa Equidistributions in arithmetic geometry

33 Weil s theorem Theorem (Hasse, 1930s) N x (p) = p + 1 pλ p, λ p [ 2, 2]. Taking λ p = 2 cos θ p, with θ p [0, π] we can rewrite N X (p) = p + 1 p(α p + α p ), α p = e iθp. Theorem (Weil, 1940s) N X (p e ) = p e + 1 p e ( α e p + α p e ) = p e + 1 p e 2 cos (e θ p ) We may thus focus our attention on p α p S 1 or p θ p [0, π] or p 2 cos θ p [ 2, 2] 15 / 29 Edgar Costa Equidistributions in arithmetic geometry

34 Histograms If one picks an elliptic curve and computes a histogram for the values N X (p) 1 p p = 2 Re α p = 2 cos θ p over a large range of primes, one always observes convergence to one of three limiting shapes! / 29 Edgar Costa Equidistributions in arithmetic geometry

35 Histograms If one picks an elliptic curve and computes a histogram for the values N X (p) 1 p p = 2 Re α p = 2 cos θ p over a large range of primes, one always observes convergence to one of three limiting shapes! One can confirm the conjectured convergence with high numerical accuracy: drew/g1satotatedistributions.html 16 / 29 Edgar Costa Equidistributions in arithmetic geometry

36 Classification of Elliptic curves Elliptic curves can be divided in two classes: special and ordinary 17 / 29 Edgar Costa Equidistributions in arithmetic geometry

37 Classification of Elliptic curves Elliptic curves can be divided in two classes: CM (special) non-cm (ordinary) 17 / 29 Edgar Costa Equidistributions in arithmetic geometry

38 Classification of Elliptic curves Elliptic curves can be divided in two classes: Consider the elliptic curve over C CM (special) non-cm (ordinary) X/C C/Λ and Λ = Zω 1 Zω 2 = 17 / 29 Edgar Costa Equidistributions in arithmetic geometry

39 Classification of Elliptic curves Elliptic curves can be divided in two classes: CM (special) non-cm (ordinary) Consider the elliptic curve over C X/C C/Λ and Λ = Zω 1 Zω 2 = non-cm the generic case, End(Λ) = Z CM Λ has extra symmetries, Z End(Λ) and ω 2 /ω 1 Q( d) for some d N. 17 / 29 Edgar Costa Equidistributions in arithmetic geometry

40 CM Elliptic curves Theorem (Deuring 1940s) If X is a CM elliptic curve then α p = e iθ are equidistributed with respect to the uniform measure on the semicircle, i.e., { e iθ C : Im(z) 0 } with µ = 1 2π dθ If the extra endomorphism is not defined over the base field one must take µ = 1 π dθ δ π/2 In both cases, the probability density function for 2 cos θ is {1, 2} 4π 4 z 2 = / 29 Edgar Costa Equidistributions in arithmetic geometry

41 non-cm Elliptic curves Conjecture (Sato Tate, early 1960s) If X does not have CM then α p = e iθ are equidistributed in the semi circle with respect to µ = 2 π sin2 θ dθ. The probability density function for 2cosθ is 4 z 2 2π = / 29 Edgar Costa Equidistributions in arithmetic geometry

42 non-cm Elliptic curves Conjecture (Sato Tate, early 1960s) If X does not have CM then α p = e iθ are equidistributed in the semi circle with respect to µ = 2 π sin2 θ dθ. The probability density function for 2cosθ is 4 z 2 2π = Theorem (Clozel, Harris, Taylor, et al., late 2000s; very hard!) The Sato Tate conjecture holds for K = Q (and more generally for K a totally real number field). 19 / 29 Edgar Costa Equidistributions in arithmetic geometry

43 Group-theoretic interpretation There is a simple group-theoretic descriptions for these measures! 20 / 29 Edgar Costa Equidistributions in arithmetic geometry

44 Group-theoretic interpretation There is a simple group-theoretic descriptions for these measures! There is compact Lie group associated to X called the Sato Tate group ST X. It can be interpreted as the Galois group of X. 20 / 29 Edgar Costa Equidistributions in arithmetic geometry

45 Group-theoretic interpretation There is a simple group-theoretic descriptions for these measures! There is compact Lie group associated to X called the Sato Tate group ST X. It can be interpreted as the Galois group of X. The pairs {α p, α p } are distributed like the eigenvalues of a matrix chosen at random from ST X with respect to its Haar measure. non-cm CM CM (with the δ) SU(2) U(1) N SU(2) (U(1)) / 29 Edgar Costa Equidistributions in arithmetic geometry

46 1 Polynomials in one variable 2 Elliptic curves 3 Quartic surfaces 21 / 29 Edgar Costa Equidistributions in arithmetic geometry

47 K3 surfaces K3 surfaces are a 2-dimensional analog of elliptic curves. 22 / 29 Edgar Costa Equidistributions in arithmetic geometry

48 K3 surfaces K3 surfaces are a 2-dimensional analog of elliptic curves. For simplicity we will focus on smooth quartic surfaces in P 3 X : f (x, y, z, w) = 0, f Z[x], deg f = 4 22 / 29 Edgar Costa Equidistributions in arithmetic geometry

49 K3 surfaces K3 surfaces are a 2-dimensional analog of elliptic curves. For simplicity we will focus on smooth quartic surfaces in P 3 X : f (x, y, z, w) = 0, f Z[x], deg f = 4 In this case N X (p e ) are associated to a orthogonal matrix! 22 / 29 Edgar Costa Equidistributions in arithmetic geometry

50 K3 surfaces K3 surfaces are a 2-dimensional analog of elliptic curves. For simplicity we will focus on smooth quartic surfaces in P 3 X : f (x, y, z, w) = 0, f Z[x], deg f = 4 In this case N X (p e ) are associated to a orthogonal matrix! Instead, we study other geometric invariant. 22 / 29 Edgar Costa Equidistributions in arithmetic geometry

51 Picard lattice We will be studying a lattice associated to X and X mod p. 23 / 29 Edgar Costa Equidistributions in arithmetic geometry

52 Picard lattice We will be studying a lattice associated to X and X mod p. Pic = Picard lattice of {curves on }/ ρ( ) = rk Pic ρ(x) is known as the geometric Picard number 23 / 29 Edgar Costa Equidistributions in arithmetic geometry

53 Picard lattice We will be studying a lattice associated to X and X mod p. Pic = Picard lattice of {curves on }/ ρ( ) = rk Pic ρ(x) is known as the geometric Picard number X Pic X ρ(x) [1,..., 20] X p := X mod p Pic X p ρ(x p ) [2, 4,... 22]??? 23 / 29 Edgar Costa Equidistributions in arithmetic geometry

54 Picard lattice We will be studying a lattice associated to X and X mod p. Pic = Picard lattice of {curves on }/ ρ( ) = rk Pic ρ(x) is known as the geometric Picard number X Pic X ρ(x) [1,..., 20] X p := X mod p Pic X p ρ(x p ) [2, 4,... 22]??? Theorem (Charles 2011) We have min q ρ(x q ) = ρ(x p ) for infinitely many p. 23 / 29 Edgar Costa Equidistributions in arithmetic geometry

55 Problem X Pic X ρ(x) [1,..., 20] X p := X mod p Pic X p ρ(x p ) [2, 4,... 22]??? Theorem (Charles 2011) We have min q ρ(x q ) = ρ(x p ) for infinitely many p. What can we say about the following: Π jump (X) := { p : min q ρ(x q ) < ρ(x p ) } 24 / 29 Edgar Costa Equidistributions in arithmetic geometry

56 Problem X Pic X ρ(x) [1,..., 20] X p := X mod p Pic X p ρ(x p ) [2, 4,... 22]??? Theorem (Charles 2011) We have min q ρ(x q ) = ρ(x p ) for infinitely many p. What can we say about the following: Π jump (X) := { p : min q ρ(x q ) < ρ(x p ) } γ(x, B) := # {p B : p Π jump(x)} # {p B} as B 24 / 29 Edgar Costa Equidistributions in arithmetic geometry

57 Problem What can we say about the following: Π jump (X) := { p : min q ρ(x q ) < ρ(x p ) } γ(x, B) := # {p B : p Π jump(x)} # {p B} as B Information about Π jump (X) Geometric statements How often an elliptic curve has p + 1 points modulo p? How often two elliptic curves have the same number of points modulo p? Does X have infinitely many rational curves? / 29 Edgar Costa Equidistributions in arithmetic geometry

58 Numerical experiments for a generic K3, ρ(x) = 1 ρ(x) is very hard to compute ρ(x p ) only now computationally feasible for large p [C.-Harvey] 26 / 29 Edgar Costa Equidistributions in arithmetic geometry

59 Numerical experiments for a generic K3, ρ(x) = 1 ρ(x) is very hard to compute ρ(x p ) only now computationally feasible for large p [C.-Harvey] 26 / 29 Edgar Costa Equidistributions in arithmetic geometry

60 Numerical experiments for a generic K3, ρ(x) = 1 ρ(x) is very hard to compute ρ(x p ) only now computationally feasible for large p [C.-Harvey] γ(x, B) c X B, B 26 / 29 Edgar Costa Equidistributions in arithmetic geometry

61 Numerical experiments for a generic K3, ρ(x) = 1 ρ(x) is very hard to compute ρ(x p ) only now computationally feasible for large p [C.-Harvey] γ(x, B) c X B, B = Prob(p Π jump (X)) 1/ p 26 / 29 Edgar Costa Equidistributions in arithmetic geometry

62 Numerical experiments for a generic K3, ρ(x) = 1 ρ(x) is very hard to compute ρ(x p ) only now computationally feasible for large p [C.-Harvey] γ(x, B) c X B, B = Prob(p Π jump (X)) 1/ p Similar behaviour observed in other examples with ρ(x) odd. In this case, data equidistribution in O(21)! 26 / 29 Edgar Costa Equidistributions in arithmetic geometry

63 Numerical experiments for ρ(x) = 2 No obvious trend / 29 Edgar Costa Equidistributions in arithmetic geometry

64 Numerical experiments for ρ(x) = 2 No obvious trend... Could it be related to some integer being a square modulo p? 27 / 29 Edgar Costa Equidistributions in arithmetic geometry

65 Numerical experiments for ρ(x) = 2 No obvious trend... Could it be related to some integer being a square modulo p? Similar behaviour observed in other examples with ρ(x) even. 27 / 29 Edgar Costa Equidistributions in arithmetic geometry

66 Numerical experiments for ρ(x) = 2 No obvious trend... Could it be related to some integer being a square modulo p? Similar behaviour observed in other examples with ρ(x) even. Data equidistribution in O(20)! 1 CPU year per example. 27 / 29 Edgar Costa Equidistributions in arithmetic geometry

67 Numerical experiments Theoretical Results In most cases we can explain the 1/2! 28 / 29 Edgar Costa Equidistributions in arithmetic geometry

68 Numerical experiments Theoretical Results In most cases we can explain the 1/2! Theorem ([C.] and [C.-Elsenhans-Jahnel]) Assume ρ(x) is even and ρ(x) = min q ρ(x q ), there is a d X Z such that: {p > 2 : d X is not a square modulo p} Π jump (X). In general, d X is not a square. Corollary If d X is not a square: lim inf B γ(x, B) 1/2 X has infinitely many rational curves. 28 / 29 Edgar Costa Equidistributions in arithmetic geometry

69 Numerical experiments for ρ(x) = 2, again What if we ignore {p : d X is not a square modulo p} Π jump (X)? 29 / 29 Edgar Costa Equidistributions in arithmetic geometry

70 Numerical experiments for ρ(x) = 2, again What if we ignore {p : d X is not a square modulo p} Π jump (X)? γ (X, B) c/ B, B 1 γ( ) 1 γ( ) 1 γ( ) { 1 if dx is not a square modulo p Prob(p Π jump (X)) = 1 p otherwise 29 / 29 Edgar Costa Equidistributions in arithmetic geometry

71 Summary Computing zeta functions of K3 surfaces via p-adic cohomology Experimental data for Π jump (X) Results regarding Π jump (X) New class of examples of K3 surfaces with infinitely many rational curves 29 / 29 Edgar Costa Equidistributions in arithmetic geometry

72 Thank you! 29 / 29 Edgar Costa Equidistributions in arithmetic geometry