Content. CIMPA-ICTP Research School, Nesin Mathematics Village The Riemann zeta function Michel Waldschmidt

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1 Artin L functions, Artin primitive roots Conjecture and applications CIMPA-ICTP Research School, Nesin Mathematics Village 7 The Riemann zeta function Michel Waldschmidt Université Pierre et Marie Curie (Paris 6) Institut de Mathématiques de Jussieu /74 Content The ring of arithmetic functions. Dirichlet series. Distribution of prime numbers : elementary results. Abel s Partial Summation Formula PSF. Abscissa of convergence. Euler Product. Analytic continuation of the Riemann zeta function (I). The Prime Number Theorem. Analytic continuation of the Riemann zeta function (II). Special values of the Riemann zeta function. Functional Equation. Riemann memoir. Explicit formulae. Hadamard product Expansion. The Riemann Hypothesis. Diophantine problems /74 The fundamental Theorem of arithmetic n = Y p p vp(n)!(n) =, (n) = v p (n). p n p n Divisor function (n) = (n) =d(n) =. d n Arithmetic functions f : N! C Multiplicative function : f (ab) =f (a)f (b) if gcd(a, b) =. Example : for k C, k(n) = P d n d k. Completely multiplicative function : f (ab) =f (a)f (b) for all a, b. Example : for k C, j k = p k : n 7! n k. A multiplicative function is characterised by its values at prime powers : f (n) = Y f (p vp(n) ). p A completely multiplicative function is characterised by its values at prime numbers : f (n) = Y p f (p) vp(n). 3/74 4/74

2 Dirichlet series Product of Dirichlet series D(f, s) = n f (n)n s. Main example : Euler (736), Riemann (859) (s) = n n s Convolution : D(f, s)d(g, s) =D(f? g, s). f? g(n) = ab=n f (a)g(b). Euler : () = /6, (4) = 4 /9, (n) n Q (n ), () = /, ( ) = /, ( n) =(n ), ( n ) Q (n ). 5/74 6/74 The ring A of arithmetic functions The ring of arithmetic functions is a commutative factorial ring A, the unity of which is the Kronecker delta function with ( for n =, (n) = for n >. The units are the arithmetic functions f such that f () 6=. Exercise: check that the inverse g of a unit f is defined recursively by g() = f (), g(n) = f () f (n/d)g(d) (n > ). d n,d<n Examples of arithmetic functions The constant function = j : n 7!. The Kronecker delta function. The Möbius function : µ. Euler totient function : '. The function j : j(n) =n for all n. The characteristic function of the squares : apple. The lambda function : =( ). The von Mangoldt function : ( log p if n = p m with p prime and m, (n) = if n is not a prime power. Iwaniec Chap. I 7/74 8/74

3 Exercises Examples of Dirichlet series D(f, s) = n f (n)n s. µ? =, µ? j = ', µ?apple=, µ?apple=, µ? =!, j k? = k,? =, µ? log =. Möbius inversion formula : g =? f () f = µ? g. 9/74 D(, s) = D(, s) = (s) D(j k, s) = (s k) D(apple, s) = (s) D(µ, s) = (s) D(, s) = (s) D( k, s) = (s k) (s) (s ) D(', s) = D( µ, s) = (s) (s) (s) D(, s) = (s) (s) D(!, s) = (s) (s) D(log, s) = (s) D(, s) = (s) (s) More exercises : Patterson p. exercise.3 / 74 The Prime Number Theorem. For x > let (x) be the number of prime numbers apple x : (x) = papplex. Distribution of prime numbers : elementary results. (x) = log p, (x) = log p = (n) papplex p m applex napplex () = 4, () = 5, () = 68, ( ) = 9. Conjecture of Gauss (79) and Adrien-Marie Legendre (798), proved by Hadamard and de la Vallée Poussin (896) : PNT. Forx!, (x) x log x This is equivalent to p n n log n. Better approximation of (x) : integral logarithm Li(x) = Z x dt log t M(x) = napplex µ(n). Theorem (elementary PNT) (P.L. Tchebyshev) : 86 s cf. course of Francesco Pappalardi (x) x, (x) x, (x) x for x!. log x The Prime Number Theorem (x) x/ log x is equivalent (x) x and to (x) x and to M(x) =o(x) as x!. / 74 / 74

4 Elementary results on prime numbers. For n, For n and p prime, n apple v p (n!) = m n apple 4 n. n n. p m Elementary results (continued) For apple y < x, (x) (y) apple log y ( (x) (y)). For n, Consequence : (n) (n) apple 4logn. (x) apple C x, (x) C x log x (x) log x apple (x) apple (x) log x +loglogx + x (log x) Hindry p / 74 4 / 74 Exercises von Mangoldt function Deduce and papplex p papplex log p p =logx + O() =loglogx + C + O(/ log x) where (n) = (s) (s) = (n) n s n ( log p if n = p m with p prime and m, if n is not a prime power. 5 / 74 6 / 74

5 Exercise Check lim (s) s! where is Euler constant : = s = lim N! N log N. Abel summation. n A =, A n = a m, a n = A n A n (n ) m= N N a n b n = A n (b n b n+ )+A N b N. n= n= Hint: Z N N =+ ([t] t) dt t (telescopic sum) 7 / 74 8 / 74 Abel s Partial Summation Formula PSF. Let a n be a sequence of complex numbers and A(x) = P napplex a n. Let f be a function of class C. Then y<napplex a n f (n) =A(x)f (x) Z x y A(t)f (t)dt. Hindry p. 9 3 Abel summation : an example. Take a n =for all n, so that A(t) =btc. Then N n=m+ f (n) = Take f (t) =/t. Then where N n= Z N M is Euler constant = Z f (t)dt + n =logn + Z N M (t btc)f (t)dt + O(/N) {t} dt t where {t} = t btc. 9 / 74 / 74

6 Abel summation for. Take a n = n s for all n ; assume Re(s) >. An approximation of the series is Z n n s t s dt = s It is a fact that (s ) (s)! as s! +. We will see that if we remove this singularity of at s =, the di erence (s) s becomes an entire function (analytic in C). Analytic continuation of (s) to Re(s) > ( s ) (s) = s s +3 s 4 s + +(n ) s (n) s + log( u) = u + 3 u u = log. 4 lim ( s ) (s) = log s! + lim (s ) (s) = s! + u n n / 74 P Hardy and Wright : lim s! ann s = P a nn if the series in RHS converges. / 74 Analytic continuation of (s) (continued) The function (s) /(s ) extends to an analytic function in Re(s) >. n s = s (s) =s n (s) =s Z t s n Z t s Z n Z s dt, t s dt + s dt = s Z t =[t]. n= Z t s dt = s +. [t]t s dt ([t] t)t s dt. Abscissa of convergence Let f (s) = P n a nn s be a Dirichlet series converging at s. Then for any C it converges uniformly on the sector n o s C Re(s s ), s s apple CRe(s s ) A Dirichlet series f (s) = P n a nn s has an absissa of convergence R [ {, +}, such that the series converges for Re(s) > and diverges for Re(s) <. The sum of the series is analytic in the half plane Re(s) > and its derivatives are given by f (k) (s) = n a n ( log n) k n s. Hindry p. 39 Hindry p / 74 4 / 74

7 Absolute abscissa of convergence The absolute abscissa of convergence a is defined in the same way. One can prove apple a apple +. Infinite products Definition of the convergence of a product. Remark. If z n is any sequence of complex numbers and if one at least of the z n vanishes, then the sequence For the Riemann zeta function, = a =. For the Dirichlet series converges to. (z z z z n ) n we have =and a =. ( s ) (s) = n ( ) n n s, Riemann Hypothesis : the abscissa of convergence of the Dirichlet series D(µ, s) = (s) is /. 5 / 74 Remark. lim NY N! n= =. n + Hindry p / 74 Convergent infinite products Convergent infinite products Definitions. An infinite product Q n ( + u n) is convergent (resp. absolutely convergent) ifthereexistsn such that u n < for n n and the series P n n log( + u n ) is convergent (resp. absolutely convergent). An infinite product of functions Q n ( + u n(x)) is uniformly convergent on a compact K if there exists n such that u P n (z) < for n n and all z K and such that the series n n log( + u n (z)) is uniformly convergent on K. A product Q n ( + u n) is absolutely convergent if and only if the series P n u n is convergent A convergent infinite product is zero if and only if one of the factors is zero. If the series P n log( + u n(z)) is uniformly convergent on every compact subset of an open set U, then the infinite product Q n ( + u n(z)) is holomorphic on U. 7 / 74 8 / 74

8 Euler Product (s) 6= for Re(s) > The infinite product Q p ( p s ) is uniformly convergent on any compact included in the half plane Re(s) >. It defines an analytic function in this half plane and we have (s) = Y p p s For Re(s) = >, (s) Y papple = p s n6n ( ) n s apple n> n s = n> n For >, let N ( ) be the set of positive integers, the prime factors of which are apple. Then Y papple p s = Y papple m p ms = nn ( ) n s. Hindry p. 37 log( u) = m u m m p ms log( p s )= m m! (s) =exp. mp ms p m 9 / 74 3 / 74 log (s) for Re(s) > log (s) = p (s) (s) = p m m mp ms log p p ms Recall the von Mangoldt function ( log p if n = p m with p prime and m, (n) = if n is not a prime power. and the formula (s) (s) = (n) n s n Euler products for multiplicative functions Proposition : an arithmetic function f is multiplicative (resp. completely multiplicative) if and only if D(f, s) = Y! f (p )p s, p (resp. D(f, s) = Y p = = f (p)p s! = Y p f (p)p s ) Hindry p / 74 3 / 74

9 The Prime Number Theorem Theorem. The integral is convergent. Z ( (t) Corollary : (x) x as x!. t) dt t Hindry p. 48 Laplace Transform Let h(t) be a bounded piecewise continuous function. Then the integral F (s) = Z h(u)e su du is convergent and defines a holomorphic function on the half plane Re(s) >. If this function F can be analytically continued to a holomorphic function on an open set containing the closed half plane Re(s), the the integral converges for s =and F () = Z h(u)du. Hindry p / / 74 Consequence Take h(u) = (e u )e u F (s) = Z can be written as the sum of. Then the function h(u)e su du = (s +) (s +) (s +) Z s (t) t s+ t dt and a holomorphic function on Re(s) /. The key point in the proof of the PNT is the following result of Hadamard and de la Vallée Poussin Theorem. The function (s) has no zero on the line Re(s) =. No zero of (s) on the line Re(s) = Trigonometric formula : 4cosx + cos(x)+3=(+cosx). Consequence : ( + it) 4 ( +it) ( ) 3 for and t R. Hindry p / / 74

10 Euler Gamma function Euler Gamma function The integral (s) = Z e t t s defines an analytic function on the half plane Re(s) > which satisfies the functional equation dt (s) = apple s e s t s + s Z e t t s dt = s (s +). (s +)=s (s). The Gamma function can be analytically continued to a meromorphic function in the complex plane C with a simple pole at any integer apple. The residue at s = k (k ) is ( ) k /k!. (s) = (s + n +) s(s +) (s + n) Remark. From () = we deduce (n + ) = n! for n. 37 / / 74 (/) = p Hence (/) = Z 4 (/) = e t t / dt = = = Z Z e x Z Z / apple e r = 4 Z e x dx. y dxdy e r rdrd Euler Gamma function Exercises: s (s) ( s) = sin s s + = s (s) = lim n! n s n! s(s +) (s + n) (s) = se s Y n= + s e s/n. n (s). 39 / 74 4 / 74

11 Analytic continuation of (s) (continued) For Re(s) >, (s) = (s) Z e t e t = n e nt, dt ts t, Analytic continuation Lemma. Let f C (R > ) be a fast decreasing function at infinity. Then the function defined for Re(s) > by the integral L(f, s) = Z f (t)t s dt (s) t has an analytic continuation to C. Special values : Under the assumptions of the lemma, for n we have Z e nt t s dt t = (s) n s L(f, n) =( ) n f (n) (). Colmez p. 4 Colmez p. 4 4 / 74 4 / 74 Bernoulli numbers The function f (t) = t e t satisfies the hypotheses of the Lemma. Define (B n ) n by f (t) = n B n t n n! B =, B =, B = 6, B 4 = 3, B 6 = 4, B 3 = B 5 = B 7 = =. 43 / / 74

12 (s) as an integral (s) = (s) Z t s e t dt Values at negative integers The Riemann zeta function (s) has a meromorphic continuation to C which is analytic in C \{}, with a simple pole at s =with residue. For n apositiveinteger, Poles : For (s) : s =,,,... residue ( ) n /n! at s = n. Z t s For e t dt : s =,,,,... residue B n+/(n +)! at s = n. Hence the poles cancel except for s =. ( n) =( ) n B n+ n + In particular ( n) Q for n and ( n) =for n. () =, ( ) =, ( 3) =, ( 5) =, ( 7) = / / 74 Values at positive even integers Theorem. Let n be a positive integer. Then (n) = B (i ) n n (n)! In particular (n)/ n is a rational number. Examples : () = /6 (The Basel problem). (4) = 4 /9, (6) = 6 /945, (8) = 8 /945. The Basel problem, first posed by Pietro Mengoli in 644, was solved by Leonhard Euler in 735, when he was 8 only. 47 / 74 Proof of () = /6, followingeuler The sum of the inverses of the roots of a polynomial f with f () = is f () : for +a z + a z + + a n z n =( z) ( n z) we have + + n = a. Write sin x x = x 3! + x 4 5! Set z = x. The zeroes of the function sin p z z p = z 3! + z 5! x 6 7! + z 3 7! + are, 4, 9,...hence the sum of the inverses of these numbers is n = 6 n 48 / 74

13 Remark Completing Euler s proof Let C. The functions f (z) =+a z + a z + sin x x = Y n x. n and e z f (z) =+(a + )z + sin x x = x 6 + =) n n = 6 have the same zeroes, say / i. The sum P i i cannot be at the same time a and a. problem Evaluating (). Fourteen proofs compiled by Robin Chapman. 49 / 74 5 / 74 Another proof (Calabi) Another proof (Calabi) x y = n x n y n. P. Cartier. Fonctions polylogarithmes, nombres polyzêtas et groupes pro-unipotents. Sém. Bourbaki no. 885 Astérisque 8 (), Z Z Z x n dx = n + dxdy x y = n (n +) Z Z dxdy x y = x = sin u cos v, Z y = sin v cos u, appleuapple /, applevapple /, u+vapple / dudv = 8 5 / 74 5 / 74

14 Completing Calabi s proof of () = /6 From one deduces n (n +) = 8 n = (n) + n n n n () = n (n) = 4 n n (n +) n = 4 3 (n +) = 6 n The function cotg( z) Proposition : for z C \ Z, cotg z = z + n= n= z + n +. z n z z n = (n)z n. n= cotg z = i ei z + i = i + e i z e i z = i + z n= (i z) n B n n! Colmez p / / 74 Functional equation An entire function (analytic in C) isdefinedby (s) =s(s ) s/ (s/) (s). () = () =. Non trivial zeroes Denote by Z the multiset of zeroes (counting multiplicities) of (s) in the critical strip < Re(s) < and by Z + the multiset of zeroes (counting multiplicities) of (s) in the critical strip < Re(s) < with positive imaginary part. Then Z + = Z [ { Z}. Theorem (Riemann) : (s) = ( s). Iwaniec Chap / / 74

15 Hadamard product expansion Explicit formula : s( s) s/ (s/) (s) = e Bs Y Z s e s/ Explicit formula for the logarithmic derivative with B = Z ( ) = + log(4 ) = (s) (s) = log (s/) (s/) s s + Z s We can write e Y Bs Z s e s/ = Y Z + s s. Patterson p.33-34, Iwaniec Chap / / 74 Poisson formula For f L (R) let ˆf be its Fourier transform : bf (y) = Z + f (x)e i xy dx. Assume that the function x 7! P nz f (x + n) is continuous with bounded variation on [, ] ; then f (n) = f b (m). mz nz Recall van der Corput Formula (first course by Francesco Pappalardi). Poisson formula Corollary. The theta series (u) = nz e un satisfies the functional equation, for u > : For Re(s) >, (/u) = p u (u). Z (s) =s(s ) ( (u) )u s/ du. u Colmez p / 74 6 / 74

16 The Riemann Memoir (859). On the number of primes less than a given magnitude (9p.) I The P function (s) defined by the Dirichlet series n n s has an analytic continuation to the whole complex plane where it is holomorphic except a simple pole at s =with residue. I The following functional equation holds s s/ (s/) (s) = (s )/ ( s). The Riemann Memoir (859) (continued). I The following product formula holds s(s ) s/ (s/) (s) = Y I The following prime number formula holds [ (x) = napplex [ (n) =x x log( ) s log x. I The Riemann zeta function (s) has simple zeroes at s =, 4,, 6,... which are called the trivial zeroes, and infinitely many non-trivial zeroes in the critical strip of the form = + i with apple apple and R. 6 / 74 I The Riemann Hypothesis. Every non-trivial zero of (s) is on the critical line Re(s) =/ : = + i. Iwaniec Chap. 5 6 / 74 The Riemann Hypothesis. The complex zeroes of the Riemann zeta function (s) in the critical strip < Re(s) < lie on the critical line Re(s) =/ : s C, < Re(s) < and (s) = =) Re(s) =/. Equivalent statement: as x!. Asymptotic expansion : Li(x) ' x log x (x) =Li(x)+O(x / log x) n n! (log x) n ' x log x + Riemann hypothesis for the Möbius function : M(x) =O(x /+ ) x (log x) + 63 / 74 Notes by Riemann Non trivial zeroes : = = = = / 74

17 Non trivial zeroes of (s) The asymptotic formula for N(T ) Hardy (94) : infinitely many non-trivial zeroes of (s) are on the critical line. Norman Levinson (974) : (s) are on the critical line. /3 of the non-trivial zeroes of Brian Conrey (989) improved this further to two-fifths (precisely 4.77 %) are real. The present record seems to belong to S. Feng, who proved that at least 4.8% of the zeroes of the Riemann zeta function (s) are on the critical line. Let N(T ) be the number of zeroes = + i of (s) in the rectangle < <, < apple T. Then for T. N(T )= T log T e + O(log T ) Iwaniec Chap / / 74 Zero free region for (s) De la Vallée Poussin (896) : > for an absolute constant c >. c log( + t ) (x) =x + O x exp( c p log x). Korobov and Vinogradov (957) > c(log t) /3, t (x) =x + O exp( c(log x) 3/5 (log log x) /5 ), x 3 Diophantine problem Conjecture. The numbers are algebraically independent. Apéry (978) : (3) 6 Q, (3), (5),..., (n +),... Rivoal () : infinitely many (n + )are irrational ; the numbers (n +)span a Q vector space of infinite dimension. Zudilin (4) : At least one of the 4 numbers (5), (7), (9), () is irrational. Iwaniec Chap. 67 / / 74

18 Multizeta values (MZV) (s) is a period For s integer, Z (s) = >t >t >t s> dt dt s dt s t t s t s Euler (s,...,s k )= n >n > >n k for s,...,s k positive integers with s. n s ns k k Induction Z t >t >t s> dt t dt s t s dt s t s = n t n n s 69 / 74 7 / 74 MZV are periods Proof. We have Z t Z (, ) = dt 3 t 3 = n >t >t >t 3 > t n n, and Z hence Z >t >t >t 3 > dt t dt t next dt t dt t Z t t m dt = m, dt 3 = t 3 dt 3 t 3 t n dt t = t m m, m>n m>n m n = (, ) Linear relations among MZV As a consequence, multiple zeta values satisfy a lot of independent linear relations with integer coe cients. Example Product of series : Product of integrals : Hence () = (, ) + (4) () = (, ) + 4 (3, ) (4) = 4 (3, ). 7 / 74 7 / 74

19 Conjecture Rohrlich Lang Any algebraic dependence relation among the numbers ( ) / (a) with a Q lies in the ideal generated by the standard relations : () Translation : (a +)=a (a), () Reflexion : (a) ( a) = sin( a). (3) Multiplication : for any positive integer n, we have ny a + k =( ) (n )/ n na+(/) (na). n k= (Universal odd distribution). 73 / 74 Artin L functions, Artin primitive roots Conjecture and applications CIMPA-ICTP Research School, Nesin Mathematics Village 7 The Riemann zeta function Michel Waldschmidt Université Pierre et Marie Curie (Paris 6) Institut de Mathématiques de Jussieu 74 / 74