Dept of Math., SCU+USTC

2015 s s Joint work with Xiaosheng Wu Dept of Math., SCU+USTC April, 2015

Outlineµ s 1 Background 2 A conjecture 3 Bohr 4 Main result

1. Background s Given a subset S = {s 1 < s 2 < } of natural numbers N = {1, 2, }, we may consider it as a big set as following: 1 S is called syndetic, if there is L N such that {i, i + 1,, i + L 1} S for any i {1, 2, }; 2 S has positive upper Banach density, if BD (S) = lim sup N + is larger than zero. sup M N 1 S {M, M + 1,, M + N 1} N 3 S satisfies Erdös condition, if i=1 1 s i = +. For example: Prime numbers P.

1. Backgroud s A basic qustion: µdoes a big subset of natural numbers contain a good linear structure? egµare there any arbitrarily long arithmetic progressions in Prime numbers set P? In 1939 van der Corput prove the case k = 3 for prime set P. Szemerédi Theorem(1975 Cominatorial method): A subset of natural numbers with positive upper Banach density contains arithmetic progressions of arbitrary finite length.

1. Background s In fact, Szemerédi proved that for any δ > 0 and k 3, there is a minimal natural number N(δ, k) such that If N N(δ, k) and A {1, 2,, N} with A δn, then A contains an arithmetic progression of length k. (In 1953, Roth proved the case k=3.) In 1977, Furstenberg gave a ergodic theory s proof for Szemerédi Theorem.

1. Background s Gowers(1998,2001) gave a new proof for Szemerédi Theorem, in particular he proved N(δ, k) 2 2δ c k, ck = 2 2k+9. For k = 3, Roth Theorem(1953), Heath-Brown(1987), Szemerédi(1990). Bourgain(1999) proved: N(δ, 3) 2 Cδ 2 log( 1 δ ). Bourgain(2003), Green(2005) ect..

1. Background s In 2008, Green and Tao proved (Szemerédi.s theorem in the primes). Let A be any subset of the prime numbers of positive relative upper density: if A {1,2,,N} lim sup N + P {1,2,,N} > 0, then A contains infinitely many arithmetic progressions of length k for all k. Green,Tao,Ziegler... Erdös conjectureµ Suppose that A satisfies Erdös condition, then A contains arbitrarily long arithmetic progressions.

1. Background: our question s Now our question isµ Given k 1. If S is a /big set0of natural numbers, then all common differences n of arithmetic progressions a, a + n,a + 2n,, a + kn with length k + 1 appeared in the subset S forms a set C k (S) := {n N : S (S n) (S 2n) (S kn) }. How about the structure of the set C k (S)? For exampleµfor prime numbers P, Is C k (P) syndetic? Recall that C k (P) is the set formed by all common differences of arithmetic progressions with length k + 1 appeared in P.

2. A conjecture s In fact, we have the following cojectureµ Conjecture Given k 1. If S is a /big set0of natural numbers, then C k (S) := {n N : S (S n) (S 2n) (S kn) } is an almost Nil k Bohr 0 set?

2. A conjecture s Where a subset A of natural numbers is called Nil k Bohr 0 set, if there exists a k-step nilsystem (X, T), x 0 X and an open neighborhood U of x 0 such that Remark A {n N : T n x 0 U}. Nil k Bohr 0 set is also called Bohr 0 set; Bohr 0 set is a classic notion; Nil k Bohr 0 sets were introduced by Host and Kra recently; Each Nil k Bohr 0 set is syndetic.

k-nilsystem s Definition Let G be a k-step nilpotent Lie group and Γ a discrete cocompact subgroup of G. The compact manifold X = G/Γ is called a k-step nilmanifold. The group G acts on X by left translations and we write this action as (g, x) gx. The Haar measure µ of X is the unique probability measure on X invariant under this action. Let τ G and T be the transformation x τx of X, i.e the rotation induced by τ G. Then (X, T, µ) is called a k-step nilsystem.

k-nilsystem s Remark Rotation on torus is 1-step nilsystem. Katznelson(2001) proved: A subset A of natural numbers is Bohr 0 set if and only if a 1, a 2,, a m R and ɛ > 0 such that A m {n N : na i (mod Z) ( ɛ, ɛ)}. i=1 In later, one of Huang+Shao+Ye results is to give a similar characterization of Nil k Bohr 0 set by generalized polynomials of degree k.

3.Reasons of ConjectureµHigher s In 1968, W.Veech provedµ Theorem (Veech) Let S N be a syndetic set. Then S S := {s 1 s 2 : s 1, s 2 S, s 1 > s 2 } (That is, the common differences of arithmetic progressions with length 2 appeared in S) is an /almost0bohr 0 set, More precise, there exists a set M with BD (M) = 0 such that (S S) M is a Bohr 0 set.

Bohr Problem s Problem (Bohr ) Let S N be a syndetic set. Is S S a Bohr 0 set? Furstenberg(1981), Glasner(1998,2004), Weiss(2000), Kantznelson(2001), Bergelson+ Furstenberg+Weiss(2006), Host+Kra(2011), Huang+Ye(2002,2011)

Furstenberg s Problem (Higher order ) Let S N be a syndetic set. How about the structure of the set {n N : S (S n) (S 2n) (S kn) } Similarly, we may consider the case that S has positive upper Banach density?

Furstenberg s Using Furstenberg, the above is changed to the following dynamical : Problem (Higher order ) Given a minimal system (X, T) and an non-empty open set U of X. How about the structure of the set {n N : U T n U T kn U }? RemarkµA dynamical system means a pair (X, T), where X is a compact metric space and T : X X is a homeomorphism. (X, T) is called minimal if for each x X, orb(x, T) = {T n x : n Z} is dense in X.

Furstenberg topological case s Proposition (Topological case) Let S N be a syndetic set. Then there exist a minimal dynamical system(x, T) and a non-empty open subset U of X such that {n N : U T n U T kn U } {n N : S (S n) (S kn) }. For any minimal dynamical system(x, T) and non-empty open subset U of X, we can find a syndetic S N such that {n N : S (S n) (S kn) } {n N : U T n U T kn U }.

Furstenberg measurable case s Proposition (measurable case) Let S N have positive upper Banach density. Then we can find a measure-preserving system (X, B, µ, T) and A B µ(a) = BD (S) > 0 such that {n N : µ(a T n A T kn A) > 0} {n N : S (S n) (S kn) }. Let (X, B, µ, T) be a measure-preserving system and A B, µ(a) > 0. Then there exists S N, BD (S) µ(a) such that {n N : S (S n) (S kn) } {n N : µ(a T n A T kn A) > 0}.

Measure-preserving system s RemarkµA measure-preserving system(x, B, µ, T) means (X, B, µ) is a probability space, T : X X is a measure-preserving transformation, that is, T 1 A B and µ(t 1 A) = µ(a) for any A B.

Furstenberg s Higher order Problem: Let S N have positive upper Banach density. How about the structure of the set C k (S) := {n N : S (S n) (S 2n) (S kn) }? change to the following ergodic theory s : Higher order Problem: Given a measure-preserving system (X, B, µ, T) and A B, µ(a) > 0. How about the structure of the Higher order return set {n N : µ(a T n A T kn A) > 0}? Furstenberg tell us: Szemeredi Theorem The above set is non-empty.

Multiple ergodic s In 1977, Furstenberg proved thatµ Theorem (Furstenberg) Given a measure-preserving system (X, B, µ, T) and A B, µ(a) > 0. Then 1 lim inf N + N N µ(a T n A T kn A) > 0. n=1 Particularly, there exists n N such that µ(a T n A T kn A) > 0. Szemeredi Theorem.

Multiple ergodic s For learning the structure of the the Higher order return time set {n N : µ(a T n A T kn A) > 0}, we need to know more information about the. For example, Is 1 lim N + N N µ(a T n A T kn A) n=1 existºect.

Multiple ergodic s The above question is generalized as the following:multiple ergodic µgiven a measure-preserving system (X, B, µ, T) and f 1, f 2,, f k L (µ). Is 1 lim N + N N f 1 (T n x)f 2 (T 2n x) f k (T kn x) n=1 convergent in L 2 (µ)? (L 2 -ergodic of order k) convergent for µ-a.e. x X? (Pointwise-ergodic of order k)

Multiple ergodic s Furstenberg+Katznelson+Ornstein (1982) proved multiple L 2 -ergodic Theorem for weakly mixing system. In later, we will see that in the study of Multiple ergodic, there will naturally appear characteristic factor, nilsystem and nilsequence ect.. Using the higher order return time set and nilsequence, we will prove thata higher order return times set contains a alost Nil Bohr 0 set, this is the reason of our conjecture.

Multiple L 2 -ergodic Theorem s L 2 -ergodic Theorem of order 1 = Von Neumann ergodic Theorem(1932). In 1984, Conze and Lesigne proved L 2 -ergodic Theorem of order 2. In 2001, Host and Kra proved L 2 -ergodic Theorem of order 3; Bergelson,Furstenberg+Weiss, Zhang ect. In 2005, Host and Kra proved Multiple L 2 -ergodic Theorem(2007,Ziegler). Tao(2008),Austin(2010),Walsh(2011)... Some survey on this topic: B.Kra (ICM2006), V.Bergelson (ICM2006), H.Furstenberg (ICM2010).

Multiple pointwise-ergodic Theorem s pointwise-ergodic Theorem of order 1 = Birkhoff ergodic Theorem(1931). In 1990, Bourgain proved pointwise-ergodic Theorem of order 2. Huang, Shao and Ye (2014) proved pointwise-ergodic Theorem of order k, k 3 for ergodic distal systems.

Characteristic factor and nilsystem s The Characteristic factor of a measure-preserving system (X, B, µ, T) was introduced by Furstenberg, Weiss in 1996, the key point lies inµ to show L 2 -ergodic Theorem of order k for (X, B, µ, T) to show L 2 -ergodic Theorem of order k for its k-characteristic factor (Z k, B k, µ k, T k ).

Characteristic factor and nilsystem s Using a kind of semi-norm (called Host-Kra- Gowers seminorm), Host and Kra(2005) provedµ Theorem (Host+Kra) k-characteristic factor (Z k, B k, µ k, T k ) of a measure preserving (X, B, µ, T) is an inverse limit of k-nilsystem. For k-nilsystem, Multiple ergodic Theorem had been investigated by Parry(1969), Lesigne(1989), Shah(1996), Leibman (2005), Ziegler (2005), See also Ratner equi-distribution Theorem(1991). This establisned the L 2 -ergodic Theorem of order k.

Nilsequence s Definition (Nilsequence) Let (X, T) be a k-nilsystem, X = G/Γ, f : X R is a continuous function. Then {f (T n x)} is called a basic k-nilsequence"a k-nilsequence is uniformly limit of basic nilsequences. Theorem (Bergelson, Host and Kra, 2005) Let (X, B, µ, T) be an ergodic a measure preserving, k N and A B, µ(a) > 0. Then (µ(a T n A T kn A) = F k (n) + a(n) where F k is a k-nilsequence, a(n) convergences to zero in uniform density, i.e. lim sup sup M+N 1 M Z n=m a(n) = 0. N +

A result s Using Bergelson-Host-Kra Theorem, Huang, Shao and Ye (Mem. Amer. Math. Soc.) prove the following resultµ Theorem Let (X, B, µ, T) be an ergodic measure-preserving system, k N and A B, µ(a) > 0. Then the set I = {n N : µ(a T n A... T kn A) > 0} is an almost Nil k Bohr 0 set, that is, there exists M N with BD (M) = 0 such that I M is a Nil k Bohr 0 set. Remark: almost Nil k Bohr 0 set is syndetic.

A result and s Using Furstenberg (measurable case), we have Theorem (Huang,Shao+Ye, Mem. Amer. Math. Soc.) Let S N have positive upper Banach density. Then {n N : S (S n) (S 2n) (S kn) } is an almost Nil k Bohr 0 set, particularly it is a syndetic set. RemarkµSince a syndetic set has positive positive upper Banach density. Hnece the above Theorem also holds for syndetic set.

Nil k Bohr 0 set and generalized polynomials s Proposition (Huang,Shao+Ye, Mem. Amer. Math. Soc.) Let k N. Then a subset A of natural numbers is Nil k Bohr 0 set if and only if there exist (special) generalized polynomials P 1, P 2,, P m of degree k and ɛ > 0 such that A m {n N : P i (n)(mod Z) ( ɛ, ɛ)}. i=1

Example of generalized polynomials s SGP 1 = {an : a R}, SGP 2 = {an 2, bn cn, f en 2 : a, b, c, e, f R}. a 1 a 2 n 2 + b 1 b 2 b 3 n + c 1 n 2 + c 2 n GP 2, where a i, b i, c i R. SGP 3 = {an 3, an 2 bn, an bn 2, an bn cn : a, b, c R} 2 i=1 SGP i.

Remarks on generalized polynomials s Generalized polynomials have been studied extensively, For example: Bergelson and Leibman(2007, Acta Math) investigate the equi-distribution of Generalized polynomials, which is a generalization of A.Weil s result on the equi-distribution mod1 of polynomials having at least a non-zero irration coefficient. Green and Tao(2012, Ann Math) proved that Generalized polynomials is asymptotic orthogonal to Möbius function.

A on prime numbers s Problem For any positive integer k 1, whether the set of the commom difference of arithmetic progressions of length k + 1 in primes is an almost Nil k Bohr 0 set? (or a syndetic?)

4. Maillet s conjecture s Conjecture (Maillet,1905) Every even number is the difference of two primes. Actually, before Maillet s conjecture, there were two stronger forms of the conjecture. Conjecture (Kronecker, 1901) Every even number can be expressed in infinitely many ways as the difference of two primes. Conjecture (Polignac, 1849) Every even number can be written in infinitely many ways as the difference of two consecutive primes.

4. Zhang s result s It is easy to see that The twin prime conjecture is a special case of Kronecker s Conjecture. Recently, Zhang made a breakthrough and proved that Theorem There exists an even number not more than 7 10 7 which can be expressed in infinitely many ways as the difference of two primes. Soon after, Maynard and Tao reduced the limit of such even number to not more than 600. The best known result now is not more than 246.

4. Zhang s result s Let k be a positive integer, we say a given k-tuple of integers H = {h 1, h 2,, h k } is admissible if for any prime p, h i s never occupy all of the residue classes modulo p. Theorem (Zhang s Theorem) Let H = {h 1, h 2,, h k } be an k-tuple of integers. If H is admissible and k > C for some given constant C > 0, then there are infinitely many integers n such that at least two of the numbers n + h 1, n + h 2,, n + h k will be prime. Zhang proved that C = 3.5 10 6, then obtained there are infinitely many couple of primes with difference not more than 7 10 7. To obtain Maynard and Tao s result, they proved a much smaller value C = 105.

4.Pintz s result s Based on the GPY sieve method s result, Pintz proved Theorem Let ɛ > 0 be arbitrary. The interval [x, x + x ɛ ] contains even numbers which can be written as the difference of primes if x > x 0 (ɛ).

4.Our main results s Theorem (Main result) Let D 0 ={p q: p > q and p, q are primes}, then D 0 is syndetic. Actually, we prove a stronger form as Theorem (Main result) Let D={d: d can be expressed in infinitely many ways as the difference of two primes }, then D is syndetic. That is, There is L N such that D [m, m + L] for any m N. The proof of Theorem is based on Zhang s Theorem. However, our method doesn t give the exact value of L.

s Œ[! Thank you for the attention! Email:wenh@mail.ustc.edu.cn