Pretentiousness in analytic number theory. Andrew Granville A report on joint work with K. Soundararajan, and Antal Balog

Pretentiousness in analytic number theory Andrew Granville A report on joint work with K. Soundararajan, and Antal Balog

The number of primes up to x Gauss, Christmas eve 1849: As a boy of 15 or 16, I determined that, at around x, the primes occur with density 1 ln x.

The number of primes up to x Gauss, Réveillon de Noël 1849: Quand j étais un garçon de 15 ou 16 ans, j ai determiné que, autour x, les nombres premiers ont la densité 1 ln x. #{premiers x} [x] n=2 1 ln n

The number of primes up to x Gauss, Christmas eve 1849: As a boy of 15 or 16, I determined that, at around x, the primes occur with density 1 ln x. #{primes x} [x] n=2 x 2 1 ln n dt ln t = Li(x)

The number of primes up to x Gauss, Christmas eve 1849: As a boy of 15 or 16, I determined that, at around x, the primes occur with density 1 ln x. #{primes x} [x] n=2 x 2 x ln x 1 ln n dt ln t = Li(x)

Ici Li(x) := x 2 dt ln t x π(x) = #{premiers x} Excès: [Li(x) π(x)] 10 8 5761455 753 10 9 50847534 1700 10 10 455052511 3103 10 11 4118054813 11587 10 12 37607912018 38262 10 13 346065536839 108970 10 14 3204941750802 314889 10 15 29844570422669 1052618 10 16 279238341033925 3214631 10 17 2623557157654233 7956588 10 18 24739954287740860 21949554 10 19 234057667276344607 99877774 10 20 2220819602560918840 222744643 10 21 21127269486018731928 597394253 10 22 201467286689315906290 1932355207 10 23 1925320391606803968923 7250186214

Here Li(x) := x 2 dt ln t x π(x) = #{primes x} Overcount: [Li(x) π(x)] 10 8 5761455 753 10 9 50847534 1700 10 10 455052511 3103 10 11 4118054813 11587 10 12 37607912018 38262 10 13 346065536839 108970 10 14 3204941750802 314889 10 15 29844570422669 1052618 10 16 279238341033925 3214631 10 17 2623557157654233 7956588 10 18 24739954287740860 21949554 10 19 234057667276344607 99877774 10 20 2220819602560918840 222744643 10 21 21127269486018731928 597394253 10 22 201467286689315906290 1932355207 10 23 1925320391606803968923 7250186214 Guess: 0 < Li(x) π(x) < π(x).

x π(x) = #{premiers x} Excès: [Li(x) π(x)] 10 8 5761455 753 10 9 50847534 1700 10 10 455052511 3103 10 11 4118054813 11587 10 12 37607912018 38262 10 13 346065536839 108970 10 14 3204941750802 314889 10 15 29844570422669 1052618 10 16 279238341033925 3214631 10 17 2623557157654233 7956588 10 18 24739954287740860 21949554 10 19 234057667276344607 99877774 10 20 2220819602560918840 222744643 10 21 21127269486018731928 597394253 10 22 201467286689315906290 1932355207 10 23 1925320391606803968923 7250186214 On devins: 0 < x 2 dt ln t π(x) < π(x). L hypothèse de Riemann : x dt ln t π(x) x ln x. 2 State belief in PNT

Théorème des nombres premiers π(x) x log x.

Prime Number Theorem: Proof π(x) x log x. Define ζ(s) := 1 n s = ( 1 1 ) 1 p s, Re(s) > 1 n 1 p prime (s 1)ζ(s) has analytic continuation to all of C

Prime Number Theorem: Proof π(x) x log x. Define ζ(s) := 1 n s = ( 1 1 ) 1 p s, Re(s) > 1 n 1 p prime (s 1)ζ(s) has analytic continuation to all of C Riemann Hypothesis: ζ(σ + it) 0 if σ > 1 2 x dt ln t π(x) x ln x. PNT : π(x) x log x 2 ζ(1 + it) 0, t R Proved in 1896 by Hadamard and de la Vallée Poussin.

héorème des nombres premiers: Démonstration ζ(1 + it) 0, t R (Hadamard et de la Vallée Poussin, 1896) On utilise que (s 1)ζ(s) est analytique

héorème des nombres premiers: Démonstration ζ(1 + it) 0, t R (Hadamard et de la Vallée Poussin, 1896) On utilise que (s 1)ζ(s) est analytique Stratégie: ζ(1 + it) = 0 ζ(1 + + it) c r pour c 0, r 1 si est petite... ζ(1 + 2it) =, contradiction Preuve habituelle: On utilize l indentité de Mertens: pour σ > 1, ζ(σ) 3 ζ(σ + it) 4 ζ(σ + 2it) 1. Maintenant, ζ(1 + ) 1/ ainsi 3 c 4r ζ(1 + + 2it) 1. comme 0 +. Alors ζ(1+ +2it) doit être au moins c / 4r 3, et donc a une pole à s = 1, contradiction.

Prime Number Theorem: Proof ζ(1 + it) 0, t R (Hadamard and de la Vallée Poussin, 1896) Use that (s 1)ζ(s) is analytic Tactic: ζ(1 + it) = 0 ζ(1 + + it) c r for c 0, r 1 if is small... ζ(1 + 2it) =, Contradiction! Preuve prétentieuse: Take = 1/ log x = ζ(1 + + it) ( 1 1 ) 1 p p x 1+it c 1 (log x) r; Now 1 1/p 1+it 1 + 1/p so that c 1 (log x) r ( 1 + 1 ) 1 c 2 p log x, p x so r = 1.

Hence r = 1 and ( 1 1 ) 1 p p x 1+it 1 log x p x ( 1 + 1 ) 1 p and so 1/p 1+it must be close to 1/p for most p x; that is p it pretends to be 1.

Hence r = 1 and ( 1 1 ) 1 p p x 1+it 1 log x p x ( 1 + 1 ) 1 p and so 1/p 1+it must be close to 1/p for most p x; that is p it pretends to be 1. But then p 2it pretends to be ( 1) 2 = 1, and so ) 1 p x ( 1 1 p 1+2it p x ( 1 1 p) 1 log x, so ζ(1 + + 2it) 1/, and thus ζ has a pole at 1 + 2it, Contradiction! We would like a measure for pretentiousness : Perhaps above could be written: D(p it, 1; x) 1 which implies (by a -inequality) that D(p 2it, 1; x) 2D(p it, 1; x) 1.

The Prime Number Theorem π(x) x log x. More explicitly: For any ɛ > 0 there exists x ɛ such that if x x ɛ then π(x) x log x ɛ x log x.

The Prime Number Theorem π(x) x log x. More explicitly: For any ɛ > 0 there exists x ɛ such that if x x ɛ then π(x) x log x ɛ x log x. Let π(x; q, a) = #{p x : p a (mod q)}. Analogous proof gives, if (a, q) = 1 and x x ɛ,q then x π(x; q, a) φ(q) log x ɛ x φ(q) log x. How small can we take x ɛ,q, for a given ɛ? Unconditionally: x ɛ,q = e q Assuming GRH: x ɛ,q = q 2+ɛ We believe: x ɛ,q = q 1+ɛ

Les fonctions multiplicatives f : N... tel que f(mn) = f(m)f(n) quand (m, n) = 1. Souvent f : N U := {z C : z 1}, et f est totalement multiplicatif: f(mn) = f(m)f(n) m, n 1 Exemples: f(n) = 1 f(n) = n it pour un t R Les caractières de Dirichlet: χ : (Z/qZ) G U où G est l ensemble des mième racines d unité, une groupe finite. f(n) = χ(n)n it

Connection/Lien? The Mőbius function { ( 1) k if n = p µ(n) = 1 p 2... p k 0 if p 2 n is multiplicative and we have the useful identity { log p if n = p e µ(a) log b = 0 otherwise n=ab Summing this identity over all n x we obtain log p = µ(a) log b. p e x ab x

Connection? The Mőbius function { ( 1) k if n = p µ(n) = 1 p 2... p k 0 if p 2 n is multiplicative and we have the useful identity { log p if n = p e µ(a) log b = 0 otherwise n=ab Summing this identity over all n x we obtain log p = µ(a) log b. p e x ab x Fix a, sum over b; easy since b B log b = log B! accurate estimates using Stirling s formula. Leaves us with a sum like x µ(a) (log(x/a) 1), a a x

The prime number theorem is equivalent to x a x µ(a) a (log(x/a) 1),

The prime number theorem is equivalent to x a x µ(a) a (log(x/a) 1), We can evaluate this if we can show that µ(n) = o(n). n N In fact this is equivalent to the PNT.

The prime number theorem is equivalent to x a x µ(a) a (log(x/a) 1), We can evaluate this if we can show that µ(n) = o(n). n N In fact this is equivalent to the PNT. Let λ(n) = ( 1) #{prime powers pe n}. Prime number theorem is also equivalent to λ(n) = o(n). which leads us to: n N

The Prime Number Theorem (version 2) For any ɛ > 0 there exists x ɛ such that if x x ɛ then λ(n) ɛx. n x Moreover if x x ɛ,q and (a, q) = 1 then λ(n) ɛ x n x q. n a (mod q)

Mean values of multiplicative functions PNT is n N λ(n) = o(n), which is 1 λ(n) = 0. N lim N n N

Mean values of multiplicative functions PNT is n N lim N λ(n) = o(n), which is 1 λ(n) = 0. N lim N n N Now λ(n) = 1 or 1 n, so half λ(n) are 1, half 1; ie for g = 1 or 1, PNT is equivalent to 1 #{n N : λ(n) = g} N exists and equals 1 2. For totally mult f : N G, finite group, does lim N 1 N exist, for each g G? #{n N : f(n) = g}

Mean values of multiplicative functions Can we get upper and lower bounds on 1 lim #{n N : f(n) = g} N N that depend on only on G?

Mean values of multiplicative functions Can we get upper and lower bounds on 1 lim #{n N : f(n) = g} N N that depend on only on G? E.g. G = { 1, 1}. f(n) = 1 n = limit = 1.

Mean values of multiplicative functions Can we get upper and lower bounds on 1 lim #{n N : f(n) = g} N N that depend on only on G? E.g. G = { 1, 1}. f(n) = 1 n = limit = 1. Can we get limit = 1? (f(2) = f(3) = 1 = f(6) = +1) Hall (1984): No! Heath-Brown and Montgomery conjectured actual limit. That is, what is the minimum possible proportion of f(n)-values that equal 1 when G = { 1, 1}? Sound/Granville (2001): MinProp = 1 is 17.15% 1 π2 6 log(1 + e) log e 1 + e + 2 n=1 1 n 2 1 (1 + e) n

Mean values of multiplicative functions For totally mult f : N G, finite group, 1 lim #{n N : f(n) = g} N N exists, for each g G, so that 1 ( ) f(n). N exists. lim N n N

Mean values of multiplicative functions For totally mult f : N G, finite group, 1 lim #{n N : f(n) = g} N N exists, for each g G, so that 1 ( ) f(n). N lim N n N exists. What about when f : N U?

Mean values of multiplicative functions For totally mult f : N G, finite group, 1 lim #{n N : f(n) = g} N N exists, for each g G, so that 1 ( ) f(n). N lim N n N exists. What about when f : N U? Example: f(n) = n it has mean value 1 n it 1 N u it dt = 1 N 1+it N N N 1 + it = N it 1 + it, n N 0 which equals 1/ 1 + t 2 in absolute value, but varies in angle with N. That is,

(+) lim N exists but ( ) does not. lim N 1 N 1 N f(n). f(n). n N n N

(+) lim N exists but ( ) lim N 1 N 1 N n N n N f(n) f(n). does not. In fact (+) does exist for all totally multiplicative f : N U, but (*) does not necessarily exist..

(+) lim N exists but ( ) lim N 1 N 1 N n N n N f(n) f(n). does not. In fact (+) does exist for all totally multiplicative f : N U, but (*) does not necessarily exist. Are there other examples where (*) fails? Halasz (1975) No! If the mean value of f is large in absolute value then f(n) pretends to be n it for some small real t..

Mean values of multiplicative functions Halasz (1975) If mean value of f is large in absolute value then f(n) pretends to be n it for some small real t.

Mean values of multiplicative functions Halasz (1975) If mean value of f is large in absolute value then f(n) pretends to be n it for some small real t. Pretends means that 1 Re(f(p)/p it ) p p N is bounded. Distance function for pairs of multiplicative functions: If f and g are two multiplicative functions with values inside or on the unit circle define D(f, g; x) 2 := 1 Re(f(p)g(p)), p p x satisfies the triangle inequality D(f, g; x) + D(F, G; x) D(fF, gg; x).

Large character sums, I Non-principal character χ (mod q). Want ( ) χ(n) = o(x). for small x. n x

Large character sums, I Non-principal character χ (mod q). Want ( ) χ(n) = o(x). n x for small x. Burgess (1962): Proved for x > q 1/4+o(1). Want to show this for x q ɛ.

Large character sums, I Non-principal character χ (mod q). Want ( ) χ(n) = o(x). n x for small x. Burgess (1962): Proved for x > q 1/4+o(1). Want to show this for x q ɛ. Halasz (1975) If mean value of f is large in absolute value then f(n) pretends to be n it for some small real t. If (*) fails then χ(n) pretends to be n it (by Halasz s theorem); hence χ 2 (n) pretends to be n 2it, so we expect that (*) fails with χ replaced by χ 2.

Large character sums, II How large can max χ χ(n) be? Periodicity = q. n x

Large character sums, II How large can max χ χ(n) be? Periodicity = q. n x Pólya-Vinogradov (1919): q log q

Large character sums, II How large can max χ χ(n) be? Periodicity = q. n x Pólya-Vinogradov (1919): q log q Montgomery- Vaughan (1977): GRH = c q log log q

Large character sums, II How large can max χ χ(n) be? Periodicity = q. n x Pólya-Vinogradov (1919): q log q Montgomery- Vaughan (1977): GRH = c q log log q Paley (1932): χ = (./q), order 2, with c q log log q Pf: Primes q s.t. (n/q) pretends to be 1 n (log q) c. If primes q s.t. χ(n) pretends to be 1 n q c then x s.t. n x χ(n) > c q log q Don t believe exist, but...

Sound-G: If χ (mod q) has odd order g > 1 then χ(n) q(log q) 1 δg+o(1), n x where δ g = 1 2 sin(π/g) (1 ) > 0 (1 δ π/g 3 11/12).

Sound-G: If χ (mod q) has odd order g > 1 then χ(n) q(log q) 1 δg+o(1), n x where δ g = 1 sin(π/g) 2 (1 ) > 0 (1 δ π/g 3 11/12). Assuming GRH χ(n) q(log log q) 1 δg+o(1). n x

Sound-G: If χ (mod q) has odd order g > 1 then χ(n) q(log q) 1 δg+o(1), n x where δ g = 1 sin(π/g) 2 (1 ) > 0 (1 δ π/g 3 11/12). Assuming GRH χ(n) q(log log q) 1 δg+o(1). n x MV need RH for L(s, χψ) not L(s, χ) (χ pretends to be ψ, so χψ pretends to be 1). If large character sum for χ, then χ pretends to be ψ (mod m), with m small, χ( 1)ψ( 1) = 1, so χ 3 pretends to be ψ 3 (mod m) with χ 3 ( 1)ψ 3 ( 1) = ( 1) 3 = 1 = a large character sum for χ 3. E.g. Large character sum for character of Order 6 (mod q) implies large character sum for character of Order 6 (mod q)

Mult fns in arithmetic progressions Halasz: n N f(n) large = f(n) is n it -pretentious. 1 N If f(n) = n it, or if f(n) = χ(n) then 1 f(n) N n N n a (mod q) is large. Also if f(n) = χ(n)n it, or anything f which is χ(n)n it -pretentious. Any others? BSG: No! If mean value of f is large in an arithm prog mod q then f(n) is χ(n)n it -pretentious for some Dirichlet character χ mod q and some small real t.

Pretentiousness is repulsive Can f be pretentious two ways?

Pretentiousness is repulsive Can f be pretentious two ways? Can f(n) ψ(n)n it and f(n) χ(n)n iu for most n x? If so then χ(n)n iu ψ(n)n it, so that (χψ)(n) n i(u t) for most n x, which is impossible as χψ has small conductor. Why repulsive?

Exponential sums If n x f(n)e2iπnα is large then α is close to some rational a/b with b small (MV); f(n) is ψ(n)n it where ψ (mod b)-pretentious, t small.

Exponential sums If n x f(n)e2iπnα is large then α is close to some rational a/b with b small (MV); f(n) is ψ(n)n it where ψ (mod b)-pretentious, t small. Multiplicative f : N { 1, 1}. What proportion of a, b, c N : a + b = c satisfy f(a) = f(b) = f(c) = 1? At least 1 2 % (which is really (17.15%)3 ). If f 1, f 2, f 3 are totally mult fns in U, s.t. f 1 (a)f 2 (b)f 3 (c) ɛ N 2 2 a,b,c N a+b=c then f j (n) is ψ j (n)n it j-pretentious with ψ 1 ψ 2 ψ 3 = 1.

PNT in aps Gallagher (1970) Let λ be Liouville s function. Given ɛ > 0 there exists A > 1 such that λ(n) ɛ x n x q n a (mod q) for all (a, q) = 1 and all q x 1/A

PNT in aps Gallagher (1970) Let λ be Liouville s function. Given ɛ > 0 there exists A > 1 such that λ(n) ɛ x n x q n x n a (mod q) n a (mod q) for all (a, q) = 1 and all q x 1/A, except perhaps q that are multiples of some exceptional modulus r. If such a modulus r exists then there is a character ψ (mod r) such that λ(n) ψ(a) λ(n) ɛ x q n x n 1 (mod q) whenever (a, q) = 1 and r divides q, with q x 1/A. If so then λ(n) is ψ(n)n it pretentious for small t R.

PNT in aps If this last case occurs, that is if λ(n) is ψ(n)n it pretentious, it would contradict GRH. In fact since λ(n) is real-valued one can deduce from this that t = 0, ψ must be a real-valued character, there is a zero of the L(s, ψ) lying very close to s = 1. This exceptional zero is known as a Siegel zero. Given x, modulus r, character ψ and the zero are all unique the zeros of Dirichlet L-functions repel one another, a concept believed to lie deep in the theory of zeta-functions.

Mult fns in aps BSG (2008) Let f : N U totally mult function. Given ɛ > 0 there exists A > 1 such that f(n) ɛ x n x q n x n a (mod q) n a (mod q) for all (a, q) = 1 and all q x 1/A, except perhaps q that are multiples of some exceptional modulus r. If such a modulus r exists then there is a character ψ (mod r) such that f(n) ψ(a) f(n) ɛ x q n x n 1 (mod q) whenever (a, q) = 1 and r divides q, with q x 1/A. If so then f(n) is ψ(n)n it pretentious for small t R.

Primes in aps Suppose λ(n) is ψ(n)n it pretentious Given x, the modulus r, character ψ, and small real t, are all unique, which is a consequence of the fact that the zeros of Dirichlet L-functions repel one another, a concept believed to lie deep in the theory of zeta-functions. Mult fns in aps Suppose f(n) is ψ(n)n it pretentious Given x, the modulus r, character ψ, and small real t, are all unique, which is a consequence of the fact that pretentiousness is repulsive, which is not deep. How deep is our proof? Conclusions?

How deep? Original proof used deep results on distn of primes. But how can such a combinatorial result need such depth?

How deep? Original proof used deep results on distn of primes. But how can such a combinatorial result need such depth? Selberg (in his elt pf of PNT for arith progs): log x log p + log p 1 log p 2 p x p a (mod q) p 1 p 2 x p 1 p 2 a (mod q) 2x log x φ(q) for (a, q) = 1, for x e q (too large for us).