Motivation Polynomials in one variable Elliptic curves K3 surfaces Equidistributions in arithmetic geometry Edgar Costa ICERM/Dartmouth College 10th December 2015 IST 1 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Motivation Question Given an “integral” object X , for example: an integer a one variable polynomial with integer coefficients an algebraic curves defined by one polynomial equation with integer coefficients a smooth surface defined over Q . . . I can consider its reduction modulo a prime p . What kind of geometric properties of X can we read of X mod p ? What if we consider infinitely many primes? How does X mod p behave when we take p → ∞ ? Does it behave as random as it should? 2 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Overview Polynomials in one variable 1 Elliptic curves 2 K3 surfaces 3 3 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces 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? 4 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Example: quadratic polynomials f ( x ) = ax 2 + bx + c ∆ = b 2 − 4 ac , 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 If ∆ isn’t a square, then Prob( N f ( p ) = 0) = Prob( N f ( p ) = 2) = 1 2 In this case, one can even give an explicit formula for N f ( p ), using the law of quadratic reciprocity. For example, if ∆ = 5 (for p > 2): 0 if p ≡ 2 , 3 mod 5 N f ( p ) = 1 if p = 5 2 if p ≡ 1 , 4 mod 5 5 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Example: cubic polynomials In general one cannot find explicit formulas for N f ( p ), but one can still determine their average distribution! √ √ √ f ( x ) = x 3 − 2 = � 3 � � 3 2 e 2 π i / 3 � � 3 2 e 4 π i / 3 � x − 2 x − x − 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 − 2 x + 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 . 6 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces The Chebotar¨ ev 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 (Chebotar¨ ev, early 1920s) For i = 0 , . . . , d, we have Prob( N f ( p ) = i ) = Prob( g ∈ G : g fixes i roots ) where, # { p prime , p ≤ N , N f ( p ) = i } Prob( N f ( p ) = i ) := lim . # { p prime , p ≤ N } N →∞ 7 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Example: Cubic polynomials, again √ √ √ f ( x ) = x 3 − 2 = � 3 � � 3 2 e 2 π i / 3 � � 3 2 e 4 π i / 3 � x − 2 x − x − 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) , (1 → 2 → 3 → 1) , (1 → 3 → 2 → 1) } f ( x ) = x 3 − x 2 − 2 x + 1 = ( x − α 1 ) ( x − α 2 ) ( x − α 3 ) � 2 / 3 if x = 0 Prob ( N f ( p ) = x ) = if x = 3 and G = Z / 3 Z . 1 / 3 8 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Prime powers We may also define N f ( p e ) = # { x ∈ F p e : f ( x ) = 0 } Theorem (Chebotar¨ ev continued) p 2 � � � � Prob N f ( p ) = c 1 , N f = c 2 , · · · || g ∈ G : g fixes c 1 roots , g 2 fixes c 2 roots , . . . � � Prob Let f ( x ) = x 3 − 2, then G = S 3 and: p 2 � � � � Prob N f ( p ) = N f = 0 = 1 / 3 p 2 � � � � Prob N f ( p ) = N f = 3 = 1 / 6 � � p 2 � � Prob N f ( p ) = 1 , N f = 3 = 1 / 2 9 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Polynomials in one variable 1 Elliptic curves 2 K3 surfaces 3 10 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Elliptic curves An elliptic curve over a field K is a smooth proper algebraic curve over K of genus 1. Taking K = C we get a torus: . These are projective algebraic curves defined by equations of the form y 2 = f ( x ) f ∈ K [ x ] , deg f = 3 , and no repeated roots There is a natural group structure ! If P , Q , and R are colinear, then P + Q + R = 0 . Applications: cryptography, integer factorization . . . 11 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Counting points on elliptic curves Given an elliptic curve over Q X : y 2 = f ( x ) , f ( x ) ∈ Z [ x ] We can consider its reduction modulo p (we will ignore the bad primes and p = 2). As before, consider: N X ( p e ) := # X ( F p e ) ( x , y ) ∈ ( F p e ) 2 : y 2 = f ( x ) � � = + 1 One cannot hope to write N X ( p e ) as an explicit function of p e . Instead, we will look for statistical properties of N X ( p e ). 12 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Hasse’s bound Theorem (Hasse, 1930s) For any positive integer e | p e + 1 − N X ( p e ) | ≤ 2 √ p e . In other words, N x ( p e ) = p e + 1 − √ p e λ p , λ p ∈ [ − 2 , 2] What can we say about the error term, λ p , as p → ∞ ? 13 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Weil’s theorem Theorem (Hasse, 1930s) N x ( p e ) = p e + 1 − √ p e γ 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 i θ p . Theorem (Weil, 1940s) N X ( p e ) = p e + 1 − √ p e � α e e � p + α p = 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] 14 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces 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! - 2 - 1 1 2 - 2 - 1 1 2 - 2 - 1 0 1 2 One can confirm the conjectured convergence with high numerical accuracy: http://math.mit.edu/ ∼ drew/g1SatoTateDistributions.html 15 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces Classification of Elliptic curves Elliptic curves can be divided in two classes: CM and non-CM Consider the elliptic curve over C X / C ≃ ≃ C / Λ and Λ = Z ω 1 ⊗ Z ω 2 = non-CM End(Λ) = Z , the generic case √ CM Z � End(Λ) and ω 2 /ω 1 ∈ Q ( − d ) for some d ∈ N . 16 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces CM Elliptic curves Theorem (Deuring 1940s) If X is a CM elliptic curve then α p = e i θ are equidistributed with respect to the uniform measure on the semicircle, i.e., with µ = 1 e i θ ∈ C : Im( z ) ≥ 0 � � 2 π d θ If the extra endomorphism is not defined over the base field one must take µ = 1 π d θ + 1 2 δ π/ 2 In both cases, the probability density function for t = 2 cos θ is { 1 , 2 } √ 4 − z 2 = 4 π - 2 - 1 1 2 17 / 27 Edgar Costa Equidistributions in arithmetic geometry
Motivation Polynomials in one variable Elliptic curves K3 surfaces non-CM Elliptic curves Conjecture (Sato–Tate, early 1960s) If X does not have CM then α p = e i θ are equidistributed in the semi π sin 2 θ d θ . circle with respect to µ = 2 The probability density function for t = 2 cos θ is √ 4 − t 2 = 2 π - 2 - 1 1 2 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). 18 / 27 Edgar Costa Equidistributions in arithmetic geometry
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