Special Number Theory Seminar From Random Matrix Theory to L-Functions

Transcription

1 Special Number Theory Seminar From Random Matrix Theory to L-Functions Steven J. Miller Brown University December 23, sjmiller/math/talks/talks.html

2 Origins of Random Matrix Theory Classical Mechanics: 3 Body Problem Intractable. Heavy nuclei like Uranium (200+ protons / neutrons) even worse! Get some info by shooting high-energy neutrons into nucleus, see what comes out. Fundamental Equation: Hψ n = E n ψ n E n are the energy levels Approximate with finite matrix. 1

3 Origins (cont) Statistical Mechanics: for each configuration, calculate quantity (say pressure). Average over all configurations most configurations close to system average. Nuclear physics: choose matrix at random, calculate eigenvalues, average over matrices. Look at: Real Symmetric, Complex Hermitian, Classical Compact Groups. 2

4 Random Matrix Ensembles Real Symmetric Matrices: A = a 11 a 12 a 13 a 1N a 12. a 22. a 23. a 2N.... a 1N a 2N a 3N a N = AT Let p(x) be a probability density. p(x) 0 p(x)dx = 1. R Often assume p(x) has finite moments: k th -moment = x k p(x)dx. R Define Prob(A) = 1 i<j N p(a ij ). 3

5 Eigenvalue Distribution Key to Averaging: Trace(A k ) = N λ k i (A). i=1 By the Central Limit Theorem: N N Trace(A 2 ) = a ij a ji = i=1 N j=1 N i=1 j=1 N 2 1 N λ 2 i (A) N 2 i=1 a 2 ij Gives NAve(λ 2 i (A)) N 2 or λ i (A) N. 4

6 Eigenvalue Distribution (cont) δ(x x 0 ) is a unit point mass at x 0. To each A, attach a probability measure: Obtain: µ A,N (x) = 1 N N δ i=1 ( x λ ) i(a) 2 N k th -moment = x k µ A,N (x)dx = 1 N N i=1 = Trace(Ak ) 2 k N k 2 +1 λ k i (A) (2 N) k 5

7 Semi-Circle Law N N real symmetric matrices, entries i.i.d.r.v. from fixed p(x). Semi-Circle Law: Assume p has mean 0, variance 1, other moments finite. Then µ A,N (x) 2 π 1 x2 with probability 1 Trace formula converts sums over eigenvalues to sums over entries of A. Expected value of k th -moment of µ A,N (x) is R R Trace(A k ) 2 k N k 2 +1 i<j p(a ij )da ij 6

8 Proof: 2 nd -Moment Trace(A 2 ) = N i=1 N j=1 a ij a ji = N i=1 N j=1 a 2 ij. Substituting into expansion gives N 2 R N R i,j=1 Integration factors as a 2 ji p(a 11 )da 11 p(a NN )da NN a ij R a 2 ijp(a ij )da ij (k,l) (ij) k<l a kl R p(a kl )da kl = 1. Have N 2 summands, answer is 1 4. Key: Averaging Formula, Trace Lemma. 7

9 Random Matrix Theory: Semi-Circle Law Distribution of eigenvalues Gaussian, N=400, 500 matrices Matrices: Gaussian p(x) = 1 e x2 /2 2π 8

10 Random Matrix Theory: Semi-Circle Law 2500 The eigenvalues of the Cauchy distribution are NOT semicirular Cauchy Distr: Not-Semicircular (Infinite Variance) p(x) = 1 π(1+x 2 ) 9

11 GOE Conjecture GOE Conjecture: N N Real Symmetric, entries iidrv. As N, the probability density of the distance between two consecutive, normalized eigenvalues approaches π2 4 d 2 Ψ dt 2 (the GOE distr). Ψ(t) is (up to constants) the Fredholm determinant of the operator f t t K f, kernel K = 1 ( ) sin(ξ η) sin(ξ + η) + 2π ξ η ξ + η Only known if entries chosen from Gaussian. Consecutive spacings well approximated by Axe Bx2. 10

12 Uniform Distribution: p(x) = x 104 The local spacings of the central 3/5 of the eigenvalues of x300 uniform matrices, normalized in batches of : uniform on [ 1, 1] 11

13 Cauchy Distribution: p(x) = 1 π(1+x 2 ) The local spacings of the central 3/5 of the eigenvalues of x100 Cauchy matrices, normalized in batches of : Cauchy 3.5 x 104 The local spacings of the central 3/5 of the eigenvalues of x300 Cauchy matrices, normalized in batches of : Cauchy 12

14 Poisson Distribution: p(n) = λn n! e λ 3.5 x 104 The local spacings of the central 3/5 of the eigenvalues of x300 Poisson matrices with lambda=5 normalized in batches of : Poisson, λ = x 104 The local spacings of the central 3/5 of the eigenvalues of x300 sign matrices, normalized in batches of : Poisson, λ = 20 13

15 Fat Thin Families Need a family F A T enough to do averaging. Need a family THIN enough so that everything isn t averaged out. Real Symmetric Matrices have N(N+1) 2 independent entries. Examples of thin sub-families: Band Matrices Random Graphs Special Matrices (Toeplitz) 14

16 Random Graphs Degree of a vertex = number of edges leaving the vertex. Adjacency matrix: a ij = number edges from Vertex i to Vertex j. A = These are Real Symmetric Matrices. 15

17 McKay s Law (Kesten Measure) Density of States for d-regular graphs f(x) = { d 2π(d 2 x 2 ) 4(d 1) x 2 x 2 d 1 0 otherwise d = 3. 16

18 McKay s Law (Kesten Measure) d = 6. Idea of proof: Trace lemma, combinatorics and counting. Fat Thin: fat enough to average, thin enough to get something different than Semi-circle. 17

19 d-regular and GOE Regular, 2000 Vertices Graph courtesy of D. Jakobson, S. D. Miller, Z. Rudnick, R. Rivin 18

20 Riemann Zeta Function: ζ(s) Riemann Zeta-Function: ζ(s) = n 1 n s = p ( 1 1 p s ) 1, Re(s) > 1. Functional Equation: ( s ξ(s) = Γ π 2) s 2ζ(s) = ξ(1 s). Riemann Hypothesis: All non-trivial zeros have Re(s) = 1 2 ; ie, on the critical line. Spacings between zeros same as spacings between eigenvalues of Complex Hermitian matrices. 19

21 Contour Integration ζ (s) ζ(s) = d log ζ(s) ds = d log ( 1 p s) ds = p = p p log p p s 1 p s log p p s + Good(s). Contour Integration: ζ (s) x s ζ(s) s ds vs p x ρ ζ(ρ)=0 x ρ ρ vs p x log p log p. ( ) s x ds p s 20

22 L-Functions L-functions: Re(s) > s 0 : L(s, f) = n=1 a n (f) n s = p L p (p s, f) 1. Functional equation: s 1 s. GRH: All L-functions (after normalization) have their non-trivial zeros on the critical line. 21

23 Measures of Spacings: n-level Correlations {α j } be an increasing sequence of numbers, B R n 1 a compact box. Define the n-level correlation by lim N #{ ( α j1 α j 2,..., α j n 1 α j n N ) B, j i j k } Instead of using a box, can use a smooth test function. Results: 1. Normalized spacings of ζ(s) starting at (Odlyzko) 2. Pair and triple correlations of ζ(s) (Montgomery, Hejhal) 3. n-level correlations for all automorphic cupsidal L-functions (Rudnick-Sarnak) 4. n-level correlations for the classical compact groups (Katz-Sarnak) 5. insensitive to any finite set of zeros 22

24 Measures of Spacings: n-level Density and Families Let φ(x) = i φ i(x i ), φ i even Schwartz functions whose Fourier Transforms are compactly supported. D n,f (φ) = j 1,...,jn distinct ( ) ( ) φ 1 L f γ (j 1) φ n L f γ (j n) f f 1. individual zeros contribute in limit 2. most of contribution is from low zeros 3. average over similar curves (family) To any geometric family, Katz-Sarnak predict the n- level density depends only on a symmetry group attached to the family. 23

25 Correspondences Similarities b/w Nuclear Physics and L-Functions Zeros Energy Levels Support Neutron Energy Conjecture: Zeros near central point in a family of L-functions behave like eigenvalues near 1 of a classical compact group (Unitary, Symplectic, Orthogonal). 1 F f F D n,f (φ) = 1 F f F j 1,...,jn j i ±j k i ( ) log N f φ i 2π γ(j i) f φ(x)w n,g(f) (x)dx = φ(u) W n,g(f) (u)du. 24

26 Some Number Theory Results Orthogonal: Iwaniec-Luo-Sarnak: 1-level density for H ± k (N), N square-free; Hughes-Miller: n-level density for H ± k (N), N square-free; Dueñez-Miller: 1, 2-level {φ sym 2 f : f H k (1)}, φ even Maass; Miller: 1, 2-level for one-parameter families of elliptic curves. Symplectic: Rubinstein: n-level densities for L(s, χ d ); Dueñez-Miller: 1-level for {φ f : f H k (1)}, φ even Maass. Unitary: Miller, Hughes-Rudnick: Families of Primitive Dirichlet Characters. 25

27 Main Tools Explicit Formula: Relates sums over zeros to sums over primes. Averaging Formulas: Petersson formula in ILS, Orthogonality of characters in Rubinstein, Hughes-Rudnick, Miller. Control of conductors: Monotone. 26

28 1-Level Densities The Fourier Transforms for the 1-level densities are W 1,SO(even) (u) = δ 0 (u) η(u) W 1,SO (u) = δ 0 (u) W 1,SO(odd) (u) = δ 0 (u) 1 2 η(u) + 1 W 1,Symplectic (u) = δ 0 (u) 1 2 η(u) W 1,Unitary (u) = δ 0 (u) where δ 0 (u) is the Dirac Delta functional and η(u) = { 1 if u < if u = 1 0 if u > 1 27

29 Dirichlet Characters: m Prime (Z/mZ) is cyclic of order m 1 with generator g. Let ζ m 1 = e 2πi/(m 1). Principal character χ 0 : { 1 (k, m) = 1 χ 0 (k) = 0 (k, m) > 1. The m 2 primitive characters are determined (by multiplicativity) by action on g. As each χ : (Z/mZ) C, for each χ there exists an l such that χ(g) = ζm 1. l Thus { ζm 1 la k g a (m) χ l (k) = 0 (k, m) > 0 28

30 Dirichlet L-Functions Let χ be a primitive character mod m. Let c(m, χ) = m 1 k=0 χ(k)e 2πik/m. c(m, χ) is a Gauss sum of modulus m. L(s, χ) = (1 χ(p)p s ) 1 p ( s + ɛ ) Λ(s, χ) = π 1 2 (s+ɛ) Γ m 1 2 (s+ɛ) L(s, χ), 2 where { 0 if χ( 1) = 1 ɛ = 1 if χ( 1) = 1 Λ(s, χ) = ( i) ɛc(m, χ) Λ(1 s, χ). m 29

31 Explicit Formula Let φ be an even Schwartz function with compact support ( σ, σ). Let χ be a non-trivial primitive Dirichlet character of conductor m. = φ (γ log(m π ) ) 2π φ(y)dy log p log(m/π) φ ( log p log(m/π) p log p log(m/π) φ ( log p 2 log(m/π) p ( 1 ) +O. log m ) [χ(p) + χ(p)]p 1 2 ) [χ 2 (p) + χ 2 (p)]p 1 30

32 Expansion {χ 0 } {χ l } 1 l m 2 are all the characters mod m. Consider the family of primitive characters mod a prime m (m 2 characters): 1 m 2 φ(y)dy χ χ 0 1 m 2 χ χ 0 p ( 1 ) + O. log m p log p log(m/π) φ ( log p log p log(m/π) φ log(m/π) ( log p 2 log(m/π) ) [χ(p) + χ(p)]p 1 2 ) [χ 2 (p) + χ 2 (p)]p 1 Note can pass Character Sum through Test Function. 31

33 Character Sums χ(k) = χ { m 1 k 1(m) 0 otherwise For any prime p m { m 1 1 χ(p) = 1 χ χ 0 p 1(m) otherwise Substitute into 1 m 2 χ χ 0 p log p log(m/π) φ ( log p ) [χ(p)+ χ(p)]p 1 2 log(m/π) 32

34 First Sum 2 m m 1 m 2 1 m 1 m 1 m m σ p m σ k m σ k 1 m mσ/2. m σ p log p log(m/π) φ ( log p ) p 1 2 log(m/π) m σ p 1(m) p k k 1 2 log p log(m/π) φ ( log p ) p 1 2 log(m/π) m σ p 1(m) m σ + 1 m k 1(m) k m+1 m σ No contribution if σ < 2. k p 1 2 k 1 2 k

35 1 m 2 χ χ 0 p Second Sum log p log(m/π) φ ( log p ) χ 2 (p) + χ 2 (p) 2. log(m/π) p [χ 2 (p) + χ 2 (p)] = 2 χ χ 0 ( ) Up to O we find that 1 log m { 2(m 2) p ±1(m) p ±1(m) 1 m 2 1 m 2 m σ/2 p m σ/2 k log(mσ/2 ) m 2 log m m p 1 + 2m 2 m 2 k m + log m m m σ/2 k m σ/2 k 1(m) k m+1 m σ/2 p ±1(m) k 1 + k m + log m m. p 1 m σ/2 k 1(m) k m 1 m σ/2 k k 1 k 1 + O ( ) 1 m 34

36 Results Theorem [Hughes-Rudnick 2002] F N all primitive characters with prime conductor N. If supp( φ) ( 2, 2), as N agrees with Unitary. Theorem [Miller 2002] F N all primitive characters with conductor odd square-free integer in [N, 2N]. If supp( φ) < ( 2, 2), as N agrees with Unitary. 35

37 Elliptic Curves E : y 2 +a 1 xy+a 3 y = x 3 +a 2 x 2 +a 4 x+a 6, a i Q Often can write E : y 2 = x 3 + Ax + B. Let N p be the number of solns mod p: N p = x(p) [ 1 + ( x 3 )] + Ax + B p = p+ x(p) ( x 3 ) + Ax + B p Local data: a E (p) = p N p. More generally, let a i = a i (T ) Z[T ]. 36

38 Elliptic Curves (cont) L(E, s) = n=1 a E (n) n s = p L p (E, s). By GRH: All zeros on the critical line. Rational solutions: E(Q) = Z r T. Birch and Swinnerton-Dyer Conjecture: Geometric rank r equals analytic rank (order of vanishing at central point). 37

39 Elliptic Curves Conductors grow rapidly. Results for small support, where Orthogonal densities indistinguishable. Study 1 and 2-Level Densities. D n,e (φ) = j 1,...,jn distinct ( φ 1 L E γ (j ) 1) E ( φ n L E γ (j ) n) E D n,f (φ) = 1 F E F D n,e (φ). 38

40 2-Level Densities c(g) = { 0 if G = SO(even) 1 2 if G = O 1 if G = SO(odd) For G = SO(even), O or SO(odd): = φ 1 (u 1 ) φ 2 (u 2 )Ŵ2,G(u)du 1 du 2 [ φ1 (0) + 1 ][ 2 φ 1(0) f2 (0) + 1 2(0)] 2 φ + 2 u φ 1 (u) φ 2 (u)du 2 φ 1 φ 2 (0) φ 1 (0)φ 2 (0) + c(g)φ 1 (0)φ 2 (0). 39

41 Comments on Previous Results explicit formula relating zeros and Fourier coeffs; averaging formulas for the family; conductors easy to control (constant, monotone) Elliptic curve E t : discriminant (t), conductor N Et = C(t) is C(t) = p (t) p f p(t) Conj: Distribution of Low Zeros agrees with Orthogonal Densities. 40

42 1-Level Expansion D 1,F (φ) = = 1 F 1 F 2 F E F E F E F ( log N E φ 2π j φ(0) + φ i (0) p γ(j) E ) ( log p 1 log N E p φ log p log N E ( 2 log p 1 F log N E F p E p φ 2 ( ) log log N E + O log N E ) 2 log p log N E a E (p) ) a 2 E(p) Want to move 1 F A r,f (p) = E F t mod p, leads us to study a r t (p), r = 1 or 2. 41

43 2-Level Expansion Need to evaluate terms like 1 F E F 2 i=1 1 p r i i ( ) log p g i i log N E a r i E (p i). Analogue of Petersson / Orthogonality: If p 1,..., p n are distinct primes, t mod p 1 p n a r1 t (p 1 ) a r n t (p n ) = A r1,f(p 1 ) A rn,f(p n ). 42

44 Input For many families (1) : A 1,F (p) = rp + O(1) (2) : A 2,F (p) = p 2 + O(p 3/2 ) Rational Elliptic Surfaces (Rosen and Silverman): If rank r over Q(T ): 1 A 1,F(p) log p = r X p lim X p X Surfaces with j(t ) non-constant (Michel): ( A 2,F (p) = p 2 + O p 3/2). 43

45 DEFINITIONS D n,f (φ) = 1 F E F j 1,...,jn j i ±j k i ( ) log N φ E i 2π γ(j i) E D (r) n,f (φ): n-level density with contribution of r zeros at central point removed. F N : Rational one-parameter family, t [N, 2N], conductors monotone. 44

46 ASSUMPTIONS 1-parameter family of Ell Curves, rank r over Q(T ), rational surface. Assume GRH; j(t ) non-constant; Sq-Free Sieve if (T ) has irr poly factor of deg 4. Pass to positive percent sub-seq where conductors polynomial of degree m. φ i even Schwartz, support ( σ i, σ i ): ( ) σ 1 < min 12, 2 3m for 1-level σ 1 + σ 2 < 1 3m for 2-level. 45

47 MAIN RESULT Theorem (M ): Under previous conditions, as N, n = 1, 2: where D (r) n,f N (φ) G = φ(x)w G (x)dx, { O if half odd SO(even) if all even SO(odd) if all odd 1 and 2-level densities confirm Katz-Sarnak, B-SD predictions for small support. 46

48 Excess Rank One-parameter family, rank r over Q(T ), RMT = 50% rank r, r+1. For many families, observe Percent with rank r = 32% Percent with rank r+1 = 48% Percent with rank r+2 = 18% Percent with rank r+3 = 2% Problem: small data sets, sub-families, convergence rate log(conductor)? 47

49 Data on Excess Rank y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a6 Family: a 1 : 0 to 10, rest 10 to 10. Percent with rank 0 = 28.60% Percent with rank 1 = 47.56% Percent with rank 2 = 20.97% Percent with rank 3 = 2.79% Percent with rank 4 =.08% 14 Hours, 2,139,291 curves (2,971 singular, 248,478 distinct). 48

50 Data on Excess Rank y 2 + y = x 3 + T x Each data set 2000 curves from start. t-start Rk 0 Rk 1 Rk 2 Rk 3 Time (hrs) < < Last set has conductors of size 10 11, but on logarithmic scale still small. 49

51 Excess Rank Calculations Families with y 2 = f t (x); D(t) SqFree Family t Range Num t r r r + 1 r + 2 r (4t + 2) [2, 2002] (4t + 2) [2, 2002] t + 1 [2, 247] t 2 + 9t + 1 [2, 272] t(t 1) [2, 2002] (6t + 1)x 2 [2, 101] (6t + 1)x [2, 77] x 3 + 4(4t + 2)x, 4t + 2 Sq-Free, odd. 2. x 3 4(4t + 2)x, 4t + 2 Sq-Free, even. 3. x ( 3) 3 (9t + 1) 2, 9t + 1 Sq-Free, even. 4. x 3 + tx 2 (t + 3)x + 1, t 2 + 3t + 9 Sq-Free, odd. 5. x 3 + (t + 1)x 2 + tx, t(t 1) Sq-Free, rank x 3 + (6t + 1)x 2 + 1, 4(6t + 1) Sq-Free, rank x 3 (6t + 1) 2 x + (6t + 1) 2, (6t + 1)[4(6t + 1) 2 27] Sq-Free, rank 2. 50

52 Excess Rank Calculations Families with y 2 = f t (x); All D(t) Family t Range Num t r r r + 1 r + 2 r (4t + 2) [2, 2002] (4t + 2) [2, 2002] t + 1 [2, 247] t 2 + 9t + 1 [2, 272] t(t 1) [2, 2002] (6t + 1)x 2 [2, 101] (6t + 1)x [2, 77] x 3 + 4(4t + 2)x, 4t + 2 Sq-Free, odd. 2. x 3 4(4t + 2)x, 4t + 2 Sq-Free, even. 3. x ( 3) 3 (9t + 1) 2, 9t + 1 Sq-Free, even. 4. x 3 + tx 2 (t + 3)x + 1, t 2 + 3t + 9 Sq-Free, odd. 5. x 3 + (t + 1)x 2 + tx, t(t 1) Sq-Free, rank x 3 + (6t + 1)x 2 + 1, 4(6t + 1) Sq-Free, rank x 3 (6t + 1) 2 x + (6t + 1) 2, (6t + 1)[4(6t + 1) 2 27] Sq-Free, rank 2. 51

53 Orthogonal Random Matrix Model RMT: 2N eigenvalues, in pairs e ±iθ j, probability measure on [0, π] N : dɛ 0 (θ) j<k (cos θ k cos θ j ) 2 j dθ j Model: forced zeros independent (suggested by Function Field analogue) {( ) } g A 2N,2r = : g SO(2N 2r) I 2r 52

54 Orthogonal Random Matrix Models RMT: 2N eigenvalues, in pairs e ±iθ j, probability measure on [0, π] N : dɛ 0 (θ) j<k (cos θ k cos θ j ) 2 j dθ j Interaction Model: NOT SUGGESTED BY FUNC- TION FIELD Sub-ensemble of SO(2N) with the last 2n of the 2N eigenvalues equal +1: dε 2n (θ) j<k (cos θ k cos θ j ) 2 j (1 cos θ j ) 2n j dθ j, with 1 j, k N n. Independent Model: SUGGESTED BY FUNCTION FIELD {( ) } g A 2N,2n = : g SO(2N 2n) I 2n 53

55 Random Matrix Models and One-Level Densities Fourier transform of 1-level density: ˆρ 0 (u) = δ(u) η(u). Fourier transform of 1-level density (Rank 2, Independent): ˆρ 2,Ind (u) = [δ(u) + 12 ] η(u) + 2. Fourier transform of 1-level density (Rank 2, Interaction): ˆρ 2,Int (u) = [δ(u) + 12 ] η(u) ( u 1)η(u). 54

56 Testing RMT Model For small support, 1-level densities for Elliptic Curves agree with ρ r,indep. Curve E, conductor N E, expect first zero iγ (1) E with γ(1) E log 1 N. E If r zeros at central point, if repulsion of zeros is of size r c log N, might detect in 1-level E density: ( 1 (j) γ φ E log N ) E. F N 2π E F N Corrections of size j φ (x 0 + c r ) φ(x 0 ) φ (x(x 0, c r )) c r. 55

57 Theoretical Distribution of First Normalized Zero First normalized eigenvalue: 230,400 from SO(6) with Haar Measure First normalized eigenvalue: 322,560 from SO(7) with Haar Measure 56

58 Rank 0 Curves: 1st Normalized Zero (Far left and right bins just for formatting) curves, log(cond) [3.2, 12.6]; mean = curves, log(cond) [12.6, 14.9]; mean =.88 57

59 Rank 2 Curves: 1st Normalized Zero curves, log(cond) [10, ]; mean = curves, log(cond) [16, 16.5]; mean =

60 Rank 2 Curves: [0, 0, 0, t 2, t 2 ] 1st Normalized Zero curves, log(cond) [7.8, 16.1]; mean = curves, log(cond) [16.2, 23.3]; mean =

61 Summary Similar behavior in different systems. Find correct scale. Average over similar elements. Need an Explicit Formula. Different statistics tell different stories. Evidence for B-SD, RMT interpretation of zeros Need more data. 60

62 Appendix Below are some additional details / topics of interest. 61

63 Appendix One: Dirichlet Characters Below is a sketch of the calculation for squarefree conductors (and not just prime conductors). 62

64 Dirichlet Characters: m Square-free Fix an r and let m 1,..., m r be distinct odd primes. m = m 1 m 2 m r M 1 = (m 1 1)(m 2 1) (m r 1) = φ(m) M 2 = (m 1 2)(m 2 2) (m r 2). M 2 is the number of primitive characters mod m, each of conductor m. A general primitive character mod m is given by χ(u) = χ l1 (u)χ l2 (u) χ lr (u). Let F = {χ : χ = χ l1 χ l2 χ lr }. 1 M 2 1 M 2 p p log p log(m/π) φ ( log p log p log(m/π) φ log(m/π) ( log p 2 log(m/π) ) p 1 2 [χ(p) + χ(p)] χ F ) p 1 χ F[χ 2 (p) + χ 2 (p)] 63

65 Characters Sums: m i 2 l i =1 χ li (p) = { m i 1 1 p 1(m i ) 1 otherwise Define Then χ(p) = χ F δ mi (p, 1) = = = m 1 2 l 1 =1 m i 2 r i=1 i=1 l i =1 { 1 p 1(m i ) 0 otherwise m r 2 l r =1 χ li (p) χ l1 (p) χ lr (p) ( ) r 1 + (m i 1)δ mi (p, 1). 64

66 Expansion Preliminaries: k(s) is an s-tuple (k 1, k 2,..., k s ) with k 1 < k 2 < < k s. This is just a subset of (1, 2,..., r), 2 r possible choices for k(s). δ k(s) (p, 1) = s δ mki (p, 1). i=1 If s = 0 we define δ k(0) (p, 1) = 1 p. Then = r ( i=1 ) 1 + (m i 1)δ mi (p, 1) r ( 1) r s δ k(s) (p, 1) s=0 k(s) s (m ki 1) i=1 65

67 First Sum: m σ p p M 2 (1 + r δ k(s) (p, 1) s=1 k(s) As m/m 2 3 r, s = 0 sum contributes S 1,0 = 1 mσ M 2 p s ) (m ki 1). i=1 p r m 1 2 σ 1, hence negligible for σ < 2. Now we study S 1,k(s) = 1 s (m ki 1) M 2 i=1 1 s (m ki 1) M 2 i=1 m σ p p 1 2δ k(s) (p, 1) m σ n 1(m k(s) ) 1 s 1 (m ki 1) M s 2 i=1 (m k i ) i=1 3 r m 1 2 σ 1. n 1 2 m σ n n

68 First Sum (cont): There are 2 r choices, yielding S 1 6 r m 1 2 σ 1, which is negligible as m goes to infinity for fixed r if σ < 2. Cannot let r go to infinity. If m is the product of the first r primes, log m = r log p k k=1 = p r log p r Therefore 6 r m log 6 m

69 Second Sum Expansions: m i 2 l i =1 χ 2 l i (p) = { m i 1 1 p ±1(m i ) 1 otherwise = = = χ 2 (p) χ F m 1 2 l 1 =1 m i 2 r i=1 i=1 l i =1 m r 2 l r =1 χ 2 l i (p) χ 2 l 1 (p) χ 2 l r (p) r ( ) 1 + (m i 1)δ mi (p, 1) + (m i 1)δ mi (p, 1) 68

70 Second Sum Bounds: Handle similarly as before. Say p 1 mod m k1,..., m ka p 1 mod m ka +1,..., m kb How small can p be? +1 congruences imply p m k1 m ka congruences imply p m ka+1 m kb 1. Since the product of these two lower bounds is greater than b i=1 (m k i 1), at least one must be greater than ( b i=1 (m 2 k i 1))1. There are 3 r pairs, yielding Second Sum = r S 2,k(s),j(s) 9 r m 1 2. s=0 k(s) j(s) 69

71 Summary: Agrees with Unitary for σ < 2. We proved: Lemma: m square-free odd integer with r = r(m) factors; m = r i=1 m i; M 2 = r i=1 (m i 2). Consider the family F m of primitive characters mod m. Then First Sum 1 M 2 2 r m 1 2 σ Second Sum 1 M 2 3 r m

72 Dirichlet Characters: m [N, 2N] Square-free F N all primitive characters with conductor odd squarefree integer in [N, 2N]. At least N/ log 2 N primes in the interval. At least N N log 2 N = N 2 log 2 N primitive characters: M N 2 log 2 N 1 M log2 N N 2. 71

73 Bounds S 1,m 1 M 2r(m) m 1 2 σ S 2,m 1 M 3r(m) m r(m) = τ(m), the number of divisors of m, and 3 r(m) τ 2 (m). While it is possible to prove τ l (n) x(log x) 2l 1 n x the crude bound τ(n) c(ɛ)n ɛ yields the same region of convergence. 72

74 First Sum Bound S 1 = 2N m=n m squaref ree 2N m=n S 1,m 1 M 2r(m) m 1 2 σ 1 M N 1 2 σ 2N m=n τ(m) 1 M N 1 2 σ c(ɛ)n 1+ɛ No contribution if σ < 2. log2 N N 2 N 1 2 σ c(ɛ)n 1+ɛ c(ɛ)n 1 2 σ+ɛ 1 log 2 N. Second sum handled similarly. 73

75 Appendix Two: Sketch of Proof for Elliptic Curve Families We give a quick sketch of the main ingredients. The greatest difficulty is the oscillatory behavior of the conductors. Localizing them to log N r +O(1) is too crude the O(1) factor is enough to ruin the results. 74

76 Sieving 2N S(t) = N k/2 µ(d) S(t) t=n D(t)sqfree d=1 D(t) 0(d 2 ) t [N,2N] = log l N µ(d) S(t) + N k/2 µ(d) S(t). d=1 D(t) 0(d 2 ) t [N,2N] d log l N D(t) 0(d 2 ) t [N,2N] Handle first by progressions. Handle second by Cauchy-Schwartz: The number of t in the second sum (by Sq- Free Sieve Conj) is o(n): 75

77 Sieving (cont) log l N µ(d) S(t) d=1 D(t) 0(d 2 ) t [N,2N] t i (d) roots of D(t) 0 mod d 2. [ ] t i (d), t i (d) + d 2,..., t i (d) + N d 2 d 2. If (d, p 1 p 2 ) = 1, go through complete set of residue classes N/d2 p 1 p 2 times. 76

78 Partial Summation ã d,i,p (t ) = a t(d,i,t ) (p), G d,i,p (u) is related to the test functions, d and i from progressions. Applying Partial Summation S(d, i, r, p) = [N/d 2 ] t =0 ã r d,i,p(t )G d,i,p (t ) = ( [N/d 2 ] ( ) ) A r,f (p) + O p R G d,i,p ([N/d 2 ]) p [N/d 2 ] 1 u=0 ( u ( ) )( ) p A r,f(p)+o p R G d,i,p (u) G d,i,p (u+1) 77

79 Difficult Piece: Fourth Sum I [N/d 2 ] 1 u=0 ( ) O(P R ) G d,i,p (u) G d,i,p (u + 1) Taylor G d,i,p (u) G d,i,p (u+1) gives P RN d 2 1 P r log N. 1 F i,d gives O( P R P r log N ). Problem is in summing over the primes, as we no longer have 1 F. 78

80 Fourth Sum: II If exactly one of the r j s is non-zero, then = [N/d 2 ] 1 u=0 [N/d 2 ] 1 u=0 G d,i,p(u) G d,i,p (u + 1) ( ) ( ) g log p log p g log C(t i (d) + ud 2 ) log C(t i (d) + (u + 1)d 2 ) If conductors monotone, for fixed i, d and p, small independent of N (bounded variation). If two of the r j s are non-zero: a 1 a 2 b 1 b 2 = a 1 a 2 b 1 a 2 + b 1 a 2 b 1 b 2 a 1 a 2 b 1 a 2 + b 1 a 2 b 1 b 2 = a 2 a 1 b 1 + b 1 a 2 b 2 79

81 Handling the Conductors: I y 2 + a 1 (t)xy + a 3 (t)y = x 3 + a 2 (t)x 2 + a 4 (t)x + a 6 (t) C(t) = p f p(t) p (t) D 1 (t) = primitive irred poly factors (t), c 4 (t) share D 2 (t) = remaining primitive irred poly factors of (t) D(t) = D 1 (t)d 2 (t) D(t) sq-free, C(t) like D 2 1(t)D 2 (t) except for a finite set of bad primes. 80

82 Handling the Conductors: II y 2 +a 1 (t)xy+a 3 (t)y = x 3 +a 2 (t)x 2 +a 4 (t)x+a 6 (t) Let P be the product of the bad primes. Tate s Algorithm gives f p (t), depend only on a i (t) mod powers of p. Apply Tate s Algorithm to E t1. Get f p (t 1 ) for p P. For m large, p P, f p (τ) = f p (P m t + t 1 ) = f p (t 1 ), and order of p dividing D(P m t + t 1 ) is independent of t. D Get integers st C(τ) = c 1 2(τ) bad c 1 sq-free. D 2 (τ) c, D(τ) 2 81

83 Appendix III: Numerically Approximating Ranks We give a quick sketch of how to compute values of L-functions at the central point. If the conductor is of size N E, approximately NE log N E Fourier coefficients a E (p) are needed. 82

84 Numerically Approximating Ranks: Preliminaries Cusp form f, level N, weight 2: Define f( 1/Nz) = ɛnz 2 f(z) f(i/y N) = ɛy 2 f(iy/ N). i L(f, s) = (2π) s Γ(s) 1 ( iz) s f(z) dz z Λ(f, s) = (2π) s N s/2 Γ(s)L(f, s) = Get 0 0 f(iy/ N)y s 1 dy. Λ(f, s) = ɛλ(f, 2 s), ɛ = ±1. To each E corresponds an f, write 0 = and use transformations. 83

85 Algorithm for L r (s, E): I Λ(E, s) = = = f(iy/ N)y s 1 dy f(iy/ N)y s 1 dy + f(iy/ N)(y s 1 + ɛy 1 s )dy. Differentiate k times with respect to s: Λ (k) (E, s) = At s = 1, 1 Λ (k) (E, 1) = (1 + ɛ( 1) k ) 1 f(iy/ N)y s 1 dy f(iy/ N)(log y) k (y s 1 + ɛ( 1) k y 1 s )dy. 1 f(iy/ N)(log y) k dy. Trivially zero for half of k; let r be analytic rank. 84

86 Algorithm for L r (s, E): II Λ (r) (E, 1) = 2 = 2 Integrating by parts 1 f(iy/ N)(log y) r dy a n e 2πny/ N (log y) r dy. n=1 1 Λ (r) (E, 1) = N π n=1 a n n 1 e 2πny/ N (log y) r 1dy y. We obtain L (r) (E, 1) = 2r! n=1 ( ) a n 2πn n G r, N where G r (x) = 1 e xy (log y) r 1dy (r 1)! 1 y. 85

87 Expansion of G r (x) ( G r (x) = P r log 1 ) + x n=1 ( 1) n r n r n! xn P r (t) is a polynomial of degree r, P r (t) = Q r (t γ). Q 1 (t) = t; Q 2 (t) = 1 2 t2 + π2 12 ; Q 3 (t) = 1 6 t3 + π2 12 t ζ(3) 3 ; Q 4 (t) = 1 24 t4 + π2 24 t2 ζ(3) 3 t + π4 160 ; Q 5 (t) = t5 + π2 72 t3 ζ(3) 6 t2 + π4 160 t ζ(5) 5 ζ(3)π2. 36 For r = 0, Λ(E, 1) = N π n=1 a n N n e 2πny/. Need about N or N log N terms. 86

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