Green-Tao’s theorem on k prime tuples. Wang Yonghui Department of Mathematics Capital Normal University Beijing 10037 P.R. China [email protected]

1

What Green-Tao proved?

Furstenberg [1] proved the multiple ergodic theorem:

0-0

Theorem 1 Let (X, X , µ, T ) be a measure preserving probability system, for any fixed postive integer k, and any set E ∈ X with µ (E) > 0, we have N  1 X  lim inf µ E ∩ T −n E ∩ · · · ∩ T −(k−1)n E > 0. N →∞ N n=1 Therefore, there exists an integer n such that   µ E ∩ T −n E ∩ · · · ∩ T −(k−1)n E > 0. Green-Tao provided more explicit settings and self-contained proof, without assuming any assumption of ergodic theory. Denote by ZN = Z/N Z, and for a function f : ZN → C, the average of f is defined as: N 1 X f (n) . E (f ) := N n=1 They proved that [2, Theorem 3.5]

0-1

Theorem 2 (Green-Tao’s Ergodic Theorem) Let ν : ZN → R+ be a k-pseudorandom measure (20). Suppose f is a non-negative function satisfying that 0 ≤ f (x) ≤ ν (x) for all x ∈ ZN and E (f ) ≥ δ for some δ > 0. Then we have for any fixed k ≥ 3, E (f (x) f (x + r) · · · f (x + (k − 1) r) | x, r ∈ ZN ) ≥ c (k, δ) , for sufficiently large N. Remark 1. Taking f = 1E the characteristic function of E, and ν = 1, we immediately get a quantitative version of Theorem 1.  Remark 2. If one can prove that, for any fixed positive integers k, r, E (f (x) f (x + r) · · · f (x + (k − 1) r) | x ∈ ZN ) ≥ c (k, δ) . Then the k-tuple primes conjecture follows, which includes the twin prime conjecture.  0-2

By choosing suitable ν and f related to the primes, they closed a famous arithmetic conjecture on k-tuple primes with the following theorem: Theorem 3 (Green-Tao’s arithmetic theorem) The set of prime numbers contain infinitely many arithmetic progressions of length k for all k. i.e. for any fixed k, there are infinitely many pairs (n, r) such that the series n, n + r, n + (k − 1) r are all primes. Indeed, their results holds for any subset of primes A, satifying that {A ∩ [1, N ]} > 0. lim N →∞ {primes ∩ [1, N ]} History of Green-Tao Theorem: 1. Hardy-Littlewood (1923), k-tuple conjecture; 2. van der Corput (1939), k = 3;

0-3

3. Szemer´edi (1975), A = N. Proof of Green-Tao’s arithmetic Theorem 3. It is sufficient to choose the appropriate function f, ν related to prime numbers, and prove that ν is a k-pseudorandom measure. Let  −1 −k−5 ˜ (n) if εk N ≤ n ≤ 2εk N , k 2 Λ f (n) = (1) 0 otherwise. ˜ (n) is the modified von Mangoldt function where Λ  Φ(W ) if W n + 1 is a prime, ˜ (n) = W log (W n + 1) Λ 0 otherwise. Q where W = p≤log log N p. This “W -trick” removes the small primes from consideration, and makes the pseudorandom to be possible (more uniformly distributed among the arithmetic progressions). Hence, there exits a k-pseudorandom measure related to f : ( Φ(W ) Λ2R (W n+1) if εk N ≤ n ≤ 2εk N , W log R ν (n) := (2) 1 otherwise. 0-4

where R = R (N, k) and ΛR (n) =

X

µ (d) log

d|n d≤R

R . d

˜ (n) , ΛR (n) are modified from the usual von Mangoldt function Here, Λ  X n log p if n = pr , Λ (n) = µ (d) log = 0 otherwise. d d|n

which enable of complex analysis by relating the generating P the tool function of Λ (n) n−s to Riemann zeta-functions. Hence, by Green-Tao theorem 2, we have E (f (x) f (x + r) · · · f (x + (k − 1) r) | x, r ∈ ZN ) ≥ ck .

The contribution from r = 0 is O( N1 logk N ) = o (1) . Therefore, there must exists integer r, n 6= 0 such that W n+1, W (n + r)+1, · · · , W (n + (k − 1) r) 1 are all primes. The theorem follows then.  Further, we expect the multiple convergence: 0-5

Conjecture 4 Let f be as in formula (1), then we have E (f (x) f (x + r) · · · f (x + (k − 1) r) | x, r ∈ ZN ) ∼ c (k) , as N → ∞. And then, we have c (k) N 2 # {(n, r) ∈ Pk |n ≤ N } ∼ , as N → ∞. k log N where (n, r) ∈ Pk means that n+r, n+2r, · · · , n+(k − 1) r are all primes. Understanding-1. This indicates that primes are “probablity” independent distribution (in arithmetic progression). As we know, a single prime in [1, N ] has the “probablity” log1k N , the k independent primes occurs as the probability log1k N . The conjectured term then follows since the number of the choice of pair (n, r) is of order N 2 . 

0-6

Understanding-2. We can regard Green-Tao’s Theorem as the following prime Diophantine equations:  pk − pk−1 = r    ··· (A) p2 − p1 = r    p2 − p1 = r, and their result states that the number of solution (p1 , . . . , pk , r) for 2 pi , r ≤ N is greater than c (k) logNk N . As far as we know, before their ingenune method, one can only solve the type of prime Diophantine equations by circle method and sieve method. 

2

Ergodic Theorems for proving Theorem

Szemer´edy has proved Theorem 2 when ν = νconst = 1 the constant function.

0-7

Theorem 5 (Szemer´ edy) Let f : ZN → R be a nonnegative function, satisfying that 0 ≤ f (x) ≤ 1, for all x ∈ ZN , and E (f (x) |x ∈ ZN ) ≥ δ > 0. Then E (f (x + r) f (x + 2r) · · · f (x + (k − 1) r) |x, r ∈ ZN ) ≥ c (k, δ) . for sufficiently large N. Proof. Szemer´edy first proved it by Ergodic theory. Furstenberg, Gowers, Tao also give different version of proof. see history in the explanation which follows proposition 2.3 of [2].  As a consequence, Szemer´edy immediately generate Corollary 6 For fixed 0 < δ ≤ 1, if A v N is any set such that A ∩ {1, . . . , N } is of cardinality of at least δN. Then A contains infinitely arithmetic progression of length k. 0-8

For detecting prime numbers, the function f in formula 1 is of order log N which canot be bounded by constant measures. Insteadly, f is bounded by a k-pseudorandom measure ν, Green-Tao develop a strategy that: 1. Generate a σ-algebra B to mollify f by the conditional expectation function fU ⊥ (x) = E (f |B) (x) = E (f (y) |y ∈ B (x)) , where B (x) is the unique atom in B which contains x. It is obvious that fU ⊥ (x) = constant = E (f (x) |B0 ) = E (f (x) |x ∈ ZN ) ≤ 1 for B0 = {∅, ZN } . More subtle σ-algebra B0 , . . . , BK will be generated iteratively in the step 2, and every time, reserving the condition (i)

fU ⊥ (x) = fU ⊥ (x) = E (f |Bi ) (x) ≤ 1 + o (1) , i = 1, . . . , K. 0-9

2. After a finite K iterative steps, fU = f − fU ⊥ will be very small in the Gowers’ uniform norms (see (3)), in another words kf − fU ⊥ kU k−1 = o (1) . K depends only on k, independent of N. Remark. This method can be understood as an algorithm to classify the scores of students. At first step, everyone is in one class (atom), and after finite iterative steps independent of N , we can have more subtle classification (atoms of a σ-algebra). And, we can regard that all the students in the same atoms has the same quality. It is reasonable, since on every atoms, the locally averaging score fU ⊥ approxiate the original score f in Gowers uniformity norms. The problem is, whether the classification is unique or consistent, i.e. for two such permissible σ-algebra, B1 , B2 , we must have B1 v B2 or B1 ⊇ B2 . F Now Green-Tao’s Theorem 2 can be immediately obtained by Szemer´edy theorem and the following Generalized von-Neumann theorem. Theorem 7 (Generalized von Neumann Theorem) Let ν be a kpseudorandom measure, and fi (x) ≤ ν (x) + 1 for all x ∈ ZN .Then we 0-10

have  E (f0 (x + r) f1 (x + 2r) · · · fk−1 (x + (k − 1) r) |x, r ∈ ZN ) = O

min

0≤j≤k−1

Proof. By variable change and introducing the Gowers’ Uniformity norms, we are able to explore the linear forms condition to the left side of the formula, i.e. fj can be understood independent in some sense, and then the right upper bound is controled by the least one. This can be obtained by using Cauchy inequality k − 2 times and the bound fi (x) ≤ ν (x) + 1 for i = 1, . . . , k − 1. Therefore, we obtain the result in combinational form of f0 (x) and ν (x) . The part of ν (x) will contribue 1 + o (1) by the linear form condition (the 1th/2 condion of k-pseudorandom). And then the final remaining part will take the form of f0 as 1/2k−1

  kf0 kU k−1 = E 

Y

 f0 (x + ωh) |x ∈ ZN , h ∈ Zk−1 N

.

ω∈{0,1}k−1

(3) 0-11

kfj kU k−

Using elementary method (although Fourier analyisis can also be interviewed), kf kU k−1 is proved to be a norm for k ≥ 3 by Gowers, and hence named as Gowers uniformity norms.  Proof of Green-Tao’s Ergodic Theorem. Let f = fU + fU ⊥ where the subscript U denotes the Gowers’ uniformity. fU ⊥ = 1 + o (1), and kfU kU k−1 = o (1) . Then by Szemer´edy theorem E (fU ⊥ (x + r) fU ⊥ (x + 2r) · · · fU ⊥ (x + (k − 1) r) |x, r ∈ ZN ) ≥ c (k, δ) for sufficiently large N. Applying the generalized von Neumann theorem we thus see that E (f0 (x + r) f1 (x + 2r) · · · fk−1 (x + (k − 1) r) |x, r ∈ ZN ) = o (1) , whenever each fj (x) is equal to fU or fU ⊥ , with at least one fj equals to fU . Adding these two estimates together we obtain E (f (x + r) f (x + 2r) · · · f (x + (k − 1) r) |x, r ∈ ZN ) ≥ c (k, δ) − o (1) .  0-12

2.1

The strategy of getting fU ⊥ , fU

Let 0 < ε < 1 be a positive small number. We need to construct the σ-algebra B0 , . . . , BK iteratively, such that Target I. The k-pseudorandom measure ν reserves uniform-distribution with respect to BK as K goes large. Explicit speaking, there exist a BK measurable set ΩK such that, (Prop 8.1 and in particular Prop 8.2), E ((ν + 1) 1ΩK ) = Oε (1) ;

k(1 − 1ΩK ) E ((ν − 1) |BK )kL∞ (ZN ) = Oε (1) .

Here, ΩK is added for technical reason (6). As a consequence, for 0 ≤ f ≤ ν, we immediately get kfU ⊥ kL∞ (ZN ) := k(1 − 1ΩK ) E (f |BK )kL∞ (ZN ) ≤ 1 + Oε (1) . for every iterative step. Target II. Energy increment happens as K goes large, that is, for FK+1 = (1 − 1ΩK ) (f − E (f |BK )) ,

0-13



if kFK+1 kU k−1 > ε

1/2k

, then we have

2


1 − 1Ω E (f |BK+1 ) L2 (Z K+1

2

k

−2 +1 +2 ε. ≥ k(1 − 1 ) E (f |B )k 2 Ω K K L (ZN ) N) (4)

2
Since 1 − 1ΩK+1 E (f |BK+1 ) L2 (Z ) are always bounded by 1 + o (1) , N

k

after finite step K ≤ K0 = [22 /ε + 2], we must have k

kfU kU k−1 = kFK kU k−1 ≤ ε1/2 .

 Both Target I and Target II are attacked by introducing the Dual Function DF (14) into the iterative process. Let B0 , . . . , BK and F1 , F2 , . . . , FK+1

0-14

be defined as B0 = {∅, ZN } , Ω0 = ∅ F1 = (1 − 1Ω0 ) (f − E (f |B0 )) = f − E (f ) , ··· BK = B0 ∨ Bε,η (DF1 ) ∨ · · · ∨ Bε,η (DFK ) FK+1 = (1 − 1ΩK ) (f − E (f |BK )) ··· where ΩK+1 ⊇ ΩK is a small measurable set in BK+1 . Each Bε,η (DFj ) can be generated by atoms of DFj−1 ([ε (n + α) , ε (n + α + 1)]) for some α in Prop7.2., with an immediate consequence that kDFj − E (DFj |BK )kL∞ ≤ ε.

(5)

Hence BK is generated by OK,ε (1) atoms by boundness of DFj (16). Proof of Target I (??), Prop 7.3. Let Ω be generated by the all 1/2 small atoms A satisfied with E ((ν + 1) 1 ) ≤ η .It is apparent that A
E ((ν + 1) 1Ω ) = Oε η 1/2 . For the second equation in (??), it suffices 0-15

to prove that   E ((ν − 1) 1A ) E (ν − 1|A) = = O η 1/2 E (1A )

(6)

for all atoms E ((ν + 1) 1A ) > η 1/2 . Since E ((ν − 1) 1A ) + 2E (1A ) = E ((ν + 1) 1A ) > η 1/2 , it will sufficient to prove that E ((ν − 1) 1A ) = O (η) . The triangle inequality is applied for this key formula. Explicit speaking, a continuous function ΨA is introduced so that E ((ν − 1) (1A − ΨA (DF1 , . . . , DFK ))) = O (η) .

(7)

This is obtained by an argument of pigeonhole principle and choosing 1A − Ψj (DFj ) to be a continous characteristic function of very small interval (Prop7.2). 0-16

Now, we need the last involved formula that, for any continuous function Ψ, E ((ν − 1) Ψ (DF1 , . . . , DFK )) = O (η) , which is induced by (18) from correlation condition and by (21) from linear form condition of ν and the definition of Dual function. See (17) of Prop 6.2.  k Proof of Target II (4) (Prop 8.2). If kFK+1 kU k−1 > ε1/2 , by definition of Dual function (15), hFK+1 , DFK+1 i = h(1 − 1ΩK ) (f − E (f |BK )) , DFK+1 i =

2k−1 kFK+1 kU k−1

> ε1/2 .

We get by triangle inequality that  


 1 − 1ΩK+1 (f − E (f |BK )) , E (DFK+1 |BK+1 ) > ε1/2 −O η 1/2 −O (ε) , since the uniform distribution of ν w.r.t BK and BK+1 asserts that


1ΩK+1 − 1ΩK (f − E (f |BK )) , DFK+1

≤ kDFK+1 kL∞ E 1ΩK+1 − 1ΩK (ν + 1)  η 1/2 , 0-17

and DFK+1 is uniform distributed w.r.t BK+1 (5) implies that


 1 − 1ΩK+1 (f − E (f |BK )) , DFK+1 − E (DFK+1 |BK+1 )

≤ kDFK+1 − E (DFK+1 |BK+1 )kL∞ E 1 − 1ΩK+1 |f − E (f |BK )|

≤ εE 1 − 1ΩK+1 (ν + 1) = ε.
Further, 1 − 1ΩK+1 , E (f |BK ) and E (DFK+1 |BK+1 ) are all measurable in BK+1 , i.e. locally constant on the atoms of BK+1 . Hence by taking conditional expectation w.r.t BK+1 on the both side of the above inequality, we have  


 1/2 1/2 1 − 1ΩK+1 (E (f |BK+1 ) − E (f |BK )) , E (DFK+1 |BK+1 ) > ε −O η −O By Cauchy inequality and the boundness of Dual function (16) in Lem 6.1, we have


−2k +1 1/2

1 − 1Ω

ε . (8) (E (f |B ) − E (f |B )) > 2 K+1 K K+1 L2 (Z ) N

This implies (4) thanks to Pythagoras’s theorem, by the approximate 0-18

orthogonality



 1 − 1ΩK+1 E (f |BK ) , 1 − 1ΩK+1 (E (f |BK+1 ) − E (f |BK )) < η 1/2 . The left hand is majorized by



2E 1ΩK+1 − 1ΩK (f − E (f |BK )) ≤ 4E 1ΩK+1 − 1ΩK (ν + 1)
1/2 by the uniform distribution property of ν.  which is O η

3

Arithmetic Theorems for verifying k-pseudoran measure ν.

Now, we need prove ν in formula (2) to be a pseudorandom measure, i.e. it is a measure which satisfies linear forms condition and correlation conditon. Definition 8 (Linear forms condition) Let ν : ZN → R+ , we say ν satisfies the (m0 , t0 , L0 )-linear forms condition if
t E ν (ψ1 (x)) · · · ν (ψm (x)) | x ∈ ZN = 1 + oL0 ,m0 ,t0 (1) , 0-19

Pt

where ψi (x) = j=1 Lij xj + bi are arbitrary system of linear forms for x = (x1 , . . . , xt ) ∈ ZtN , satisfing that 1. m ≤ m0 , t ≤ t0 , and m0 , t0 will be determined w.r.t to k. q

2. Lij = pij ∈ Q with |pij | , |qij | ≤ L0 , and Lij xj is intepreted as elements ij of ZN in usual manner, bi are arbitrary elements in ZN ; 3. the t-tuples (Lij )1≤j≤t ∈ Qt are non-zero, and no t-tuple is a rational multiple of each other. Example: m, t, L
= 1 is the measure condition. For (m0 , t0 , L0 ) = (4, 3, 1) = 2k−1 , k, 1 , E (ν (x) ν (x + h1 ) ν (x + h2 ) ν (x + h1 + h2 ) | x, h1 , h2 ∈ ZN ) = 1 + o (1)

0-20

is used to prove that kν − 1kU k−1 = o (1) for k = 3. For (m0 , t0 , L0 ) = (12, 5, 2) ,      x−y x − y + h2 E ν ν ν (−y) ν (−y − h1 ) × 2 2     0 0 x−y x − y + h2 ×ν ν ν (−y 0 ) $ (−y 0 − h1 ) × 2 2 ×ν (x) ν (x + h1 ) ν (x + h2 ) ν (x + h1 + h2 ) | x, h1 , h2 , y, y 0 ∈ ZN ) = 1 + o (1) is used to prove von Neumann theorem for k = 3. Definition 9 (Correlation condition) Let ν : ZN → R+ , we say ν satisfies m0 -correlation condition if there exists τ = τm : ZN → R+ X E (ν (x + h1 ) ν (x + h2 ) · · · ν (x + hm ) | x ∈ ZN ) ≤ τ (hi − hj ) 1≤i
for every 1 ≤ m ≤ m0 and E (τ q ) = Om,q (1) for all 1 ≤ q < ∞. 0-21

Correlation condition is only used to prove the weakly-mixing Theorem of Dual function. Prop6.2 Definition 10 Let ν : ZN → R+ , we
say ν is a k-pseudorandom measure if it satisfies the k2k−1 , 3k − 4, k -linear forms condition and also the 2k−1 -correlation condition.

3.1

Proof of linear forms condition

Let ν be as in formula (2), (

Φ(W ) Λ2R (W n+1) W log R

if εk N ≤ n ≤ 2εk N , 1 otherwise. Q 1 1/(k2k+1 ) k with R = N , εk = 1/2 (k + 4)! = 2k (k+4)! , W = p≤w(N ) p and ν (n) =

ΛR (n) =

X d|n d≤R

R µ (d) log . d

0-22

Hence if x, x+r, . . . , x+(k − 1) r mod N lies in [εk N, 2εk N ] , we must have r ≤ εk N, and then x, x + r, . . . , x + (k − 1) r is a genuine arithmetic progression. Such restriction costs two pages of elementary method, to transform the linear forms condition in the following regular form, Pt Proposition 11 (Prop. 9.5) Let θi = W ψi +1, ψi (x) = j=1 Lij xj + p bi be linear forms with integer cofficients |Lij | ≤ w (N )/2. Assuming t that the t-tuples (Lij )j=1 are never zero, and that no two t-tuples are rational multiple of each other. Then we have  m
W log R 2 2 E ΛR (θ1 (x)) · · · ΛR (θm (x)) | x ∈ B = (1 + om,t (1)) φ (W ) (9) Qt in which B is a product j=1 Ij @ Rt of t intervals Ij , each of length at least R10m . This proposition is attributed to Goldston-Yildirim. Firstly, elementary number theory and the analytic number theory method is applied to rewrite the summation in (9) as an Euler product, in order to transform 0-23

it into a multiple integration in terms of Riemann ζ-functions. And then the analytic method is applied to estmate such multiple integral. Proof of The Euler Product: By definition,
2 2 E ΛR (θ1 (x)) · · · ΛR (θm (x)) | x ∈ B   m Y  = E i=1

=

X di ,d0i ≤R di ,d0i |θi (x)

 R R  µ (di ) µ (d0i ) log log 0 | x ∈ B di di 

X

m Y

d1 ,...,dm ,d01 ,...,d0m ≤R

i=1

µ (di ) µ (d0i ) log

R R log 0 di di

! E

m Y

!

1di ,d0i |θi (x) | x ∈ B .

i=1

We may assume that di , d0i are all square-free because of the presence of the Mobius functions. Write D := [d1Q , . . . , dm , d01 , . . . , d0m ] , thus D ≤ m R2m . It is apparent that the expression i=1 1di ,d0i |θi (x) is periodic with period D in each components of x, and then can be safely defined on

0-24

ZtD , E

m Y i=1

! 1

di ,d0i |θi (x)

|x∈B

=E

m Y

! 1

di ,d0i |θi (x)

|x∈

ZtD

+Om,t R

−8m




,

i=1

since B is a product of intervals of length at least R10m . Hence, it suffices to prove that ! ! m m Y X Y R R 0 1di ,d0i |θi (x) | x ∈ ZtD µ (di ) µ (di ) log log 0 E di di i=1 d1 ,...,dm ,d01 ,...,d0m ≤R i=1  m W log R = (1 + om,t (1)) . φ (W ) By Chinese remainder theorem and the square-free nature of di , d0i ,

0-25

we have E

m Y



! 1di ,d0i |θi (x) | x ∈

ZtD

=

i=1

Y

E

 Y

1θi (x)≡0(mod p) | x ∈ Ztp 

i:p|di d0i

p|D

 =

Y

E

Y

1θi (x)≡0(mod p) | x ∈ Ztp  .

i:p|di d0i

p

Here, we notice that Y



1θi (x)≡0(mod p) ,

i:p|di d0i

Y i

1θi (x)≡0(mod p) di d0i ≡0(mod p)

have different meaning. If p - di d0i , the i-term of the first product will Q take can write p instead Q 1, whereas the second will be zero. Hence we of p|D . Let’s see the simple case firstly, if p1 , p01 , p2 are coprime to each other, then    
E 1p1 ,p01 |θ1 (x) | x ∈ Ztp1 p01 = E 1p1 |θ1 (x) | x ∈ Ztp1 E 1p01 |θ1 (x) | x ∈ Ztp01 , 0-26

and E 1p1 |θ1 (x) 1p2 |θ2 (x) | x ∈

Ztp1 p2




= E 1p1 |θ1 (s) | s ∈

Ztp1




E 1p2 |θ2 (t) | t ∈

by taking x = p2 P2 s+p1 P1 t, in which p2 P2 ≡ 1 (mod P1 ) , p1 P1 ≡ 1 (mod P2 ) . Hence,   ! m Y Y Y t E 1di ,d0i |θi (x) | x ∈ ZD = E  1p|θi (x) | x ∈ ZtD  p i:p|di d0i

i=1

 =

Y p

E

 Y

1p|θi (x) | x ∈ Ztp 

i:p|di d0i

by the similar argument. That is, the restriction condition in the left side is the form of intersection, but the restriction in the right side is the form of union. For abbreviation, we write the right hand side as   Y Y Y t E 1θi (x)≡0(mod p) | x ∈ Zp  = ωXd1 ,...,dm (p) ∪Xd0 ,...,d0 (p) (p) p

i:p|di d0i

p

0-27

1

m

Ztp2




.

where Xd1 ,...,dm (p) := {1 ≤ i ≤ m : p|di } and ! ωX (p) := E

Y

1θi (x)≡0(mod p) | x ∈ Ztp

.

i∈X

Therefore, we need to express m Y

X d1 ,...,dm ,d01 ,...,d0m ≤R

µ (di ) µ (d0i ) log

i=1

R R log 0 di di

! Y

ωXd1 ,...,dm (p) ∪Xd0 ,...,d0 1

p

(10) X

m Y

d1 ,...,dm ,d01 ,...,d0m ∈Z+

i=1

= ×

Y p

µ (di ) µ (d0i )

ωXd1 ,...,dm (p) ∪Xd0 ,...,d0 1

m (p)



R log di

  +

R log 0 di

 ! × +

(p)

as an Euler Product. Here, µ and ωX are all multiplicative functions, whereas (log x)+ = log (max (1, x)) is an additive function. Using the 0-28

m (p)

(p)

Mellin transform 1 (log x)+ = 2πi

Z Γ1

xz dz for x > 0, z2

where Γ1 is taken to be the vertical line with the real part σ = log1 R , we can express the logarithms in terms of multiplicative functions. Hence, (10) can be rewritten as 

1 2πi

m zj +zj0 Y R 0 ··· F (z, z 0 ) dz dz j j 2 z 02 z Γ1 Γ1 j j j=1

m Z

Z

0-29

0 where z := (z1 , . . . , zm ) and z 0 := (z10 , . . . , zm ), and  
m X Y Y µ (dj ) µ d0j 0   F (z, z ) := ωXd1 ,...,dm (p) ∪Xd0 ,...,d0 (p) (p 0 0z zj m 1 j dj dj p j=1 d1 ,...,dm ,d01 ,...,d0m ∈Z+

=

Y

|X|+|X 0 |

X

p X,X 0 v{1,...,m}

=

Y

Ep (z, z 0 ) ,

p

(−1) P

P

j∈X zj +

0 j∈X 0 zj

ωX∪X 0 (p)

say.

p

By elementary discussing (Lem 10.1), it can be seen that  0, if p ≤ w (N )  p−1 , p > w (N ) and |X| = 1 ωX (p) =  ≤ p−2 , p > w (N ) and |X| ≥ 2,

0-30

which implies 0

Ep (z, z ) = 1 − 1p>w(N )

m  X

p

−1−zj

+p

−1−zj0

−p

−1−zj −zj0



(11)

j=1

X

+ 1p>w(N )

X,X 0 v{1,...,m} |X∪X 0 |≥2

p

P

P

O 1/p j∈X

zj +

2




j∈X 0

zj0

.

Ep (z, z 0 ) then can be majorized by m Y

1 − 1p>w(N ) p


−1−zj

1 − 1p>w(N ) p

j=1

0-31

−1−zj0



1 − 1p>w(N ) p

−1−zj −zj0

−1

.

(1)

(2)

(3)

Therefore, we factorized Ep = Ep Ep Ep , where Ep(1)

Ep (z, z 0 )

0

(z, z ) := Q m

j=1

Ep(2) (z, z 0 ) := Ep(3) (z, z 0 ) :=

m Y j=1 m Y

1 − 1p>w(N ) p−1−zj

1 − 1p




−1 −1−zj


−1−zj

1−p

−1−zj0

−1−zj0



−1−zj0

−1 

1 − 1p>w(N ) p

1 − 1p
1−p

−1−zj −zj0

−1

1 − 1p>w(N ) p−

1 − 1p
.

j=1

Q

Writing Gj = the main term

(j)

p Ep , and thus F = G1 G2 G3 , in which G3 contributes

0

G3 (z, z ) =

m Y j=1

ζ 1 + zj +

zj0

ζ (1 + zj ) ζ 1 +

0-32


zj0


.

The other two terms G1 , G2 is permissible by (11) and Lem 10.3, that is G1 (0, 0) = 1 + om (1) , m G2 (0, 0) = (W/φ (W )) kG1 kC m (Dm ) ≤ Om (1) ,

(12)

σ

kG2 k

m) C m ( Dσ

for σ =

≤ Om,w(N ) (1) .

1 6m ,

Dσm

where
 0 0 := zj , zj : −σ < Re (zj ) , Re zj < 100, j = 1, . . . , m

and kGkC k (Dσm ) :=

  am  a0m a01 

∂ a1

∂ ∂ ∂

sup ··· G · · ·

0 0 0

∂z1 ∂zm ∂z1 ∂zm a1 +···+am +a1 +···+a0m ≤k

m L∞ (Dσ

0-33

Lemma 1 (Goldston, Yildirim) Suppose that    1/3 kGkC m (Dm ) ≤ exp Om,σ log R , σ

Then Z

1 2m

(2πi)

Z ···

Γ1

m Y

G (z, z 0 )

Γ1

= G (0, . . . , 0) logm R +

ζ 1 + zj +

zj0

zj +zj0



0

j=1 m X

ζ (1 + zj ) ζ 1 + zj 

R 0 dz dz j j zj2 zj02 



Om,σ kGkC m (Dσm ) logm−j R + Om,σ e−δ

√

j=1

for some δ = δ (m) > 0. Prop 9.5 and then the linear forms condition follows immediately from this lemma and (12). The correlation condition also use this lemma and a similar tragedy of generating Euler Products.

0-34

log R



3.2

Proof of Lemma 1

Lemma 1 is completed by the contour integration and the zero free region of ζ-function. Let Γ0 and Γ2 be two further contours in addtion to Γ1 , definded by β + it, − ∞ < t < ∞, Γ0 (t) := − log (|t| + 2) Γ2 (t) := 1 + it, − ∞ < t < ∞. Then ζ (1 + z) , ζ (1 + z 0 ) has no zero on the right hand side of Γ0 (t) , because of the logrithmic zero free region of ζ-function. We first detect the case of m = 1, and then finish Lemma 1 by deduction. We want to show that Z Z 0 z+z 0 1 ζ (1 + z + z ) R 0 0 I := f (z, z ) dzdz (13) 2 0 ) z 2 z 02 ζ (1 + z) ζ (1 + z (2πi) Γ1 Γ1 Z  √  1 f (z, −z) dz ∂f −δ log R = f (0, 0) log R + 0 (0, 0) + + O e . m ∂z 2πi Γ1 ζ (1 + z) ζ (1 − z) z 4 0-35

We firstly shift the z 0 contour from Γ1 to Γ2 with no poles, and then shift z contour from Γ1 to Γ0 with only one simple pole at z = 0, since in ζ (1+z+z 0 ) the shifting ζ(1+z)ζ(1+z0 ) has no poles and has a simple zero while z = 0. Hence, we have I = I1 + I2 , where 0

z R f (0, z 0 ) 02 dz 0 , z Γ2 Z Z 0 z+z 0 1 ζ (1 + z + z ) R 0 0 dzdz . I2 = f (z, z ) 2 0 2 02 ζ (1 + z) ζ (1 + z ) z z (2πi) Γ2 Γ0

1 I1 = 2πi

Z

Observe that z0

R f (0, z ) 02 = z 0



∂f f (0, 0) + 0 (0, 0) z 0 + · · · ∂z

0

the residue of f (0, z )

0

Rz z 02



(1 + z 0 log R + · · · ) , 02 z

∂f is therefore f (0, 0) log R+ ∂z 0 (0, 0) , and there-

0-36

fore Z

z0

∂f 1 0 R 0 f (0, z ) (0, 0) + dz ∂z 0 2πi Γ0 z 02   √ ∂f −δ log R . = f (0, 0) log R + 0 (0, 0) + Om e ∂z

I1 = f (0, 0) log R +

To estimate I2 , we first interchange the order of the integral, and move the z 0 contour from Γ2 to Γ0 . In such shifting, there are two simple poles, at z 0 = −z and z 0 = 0. The residue at the first is Z 1 f (z, −z) dz . 4 2πi Γ0 ζ (1 + z) ζ (1 − z) z The residue at z 0 = 0 is 1 2πi

Rz f (z, 0) 2 dz z Γ0

Z

0-37



which is also bounded by Om e−δ

√

log R



. The remaining term of I2 is 0

0 z+z ) ζ (1 + z + z R 0 dzdz f (z, z 0 ) ζ (1 + z) ζ (1 + z 0 ) z 2 z 02 Γ0 Γ0   √ which is still bounded by Om e−δ log R . This concludes the proof of 13. The proof of Lemma for m = 1 is then completes by Z 1 dz G (z, −z) 2πi 4 Γ0 ζ (1 + z) ζ (1 − z) z Z 1 ≤ sup |G (z, −z)| ζ (1 + z) ζ (1 − z) z 4 dz Γ0

1 2πi

Z

Z

≤ sup |G (z, −z)|  kGkC 1 (Dσ1 ) . For the general case, we use the deduction, suppose that we have established the result for m ≥ 1 and wish to deduce it for m+1. Applying

0-38

0 formula (13) in the variables zm+1 , zm+1 , we get

I (G, m + 1) =

Z

1 2m+1

(2πi)

Z ···

Γ1

Γ1

G (z, z 0 )

m+1 Y j=1

zj0




0

Rzj +zj 0
dz dz j j ζ (1 + zj ) ζ 1 + zj0 zj2 zj02 ζ 1 + zj +

m Y

zj0




zj + R 0 0
2 · · · G (z , . . . , z , 0, z , . . . , z , 0) = 1 m 1 m 2m 0 ζ (1 + zj ) ζ 1 + zj zj z (2πi) Γ1 Γ1 j=1
Z Z m 0 zj +zj0 Y ζ 1 + z + z R 1 j j 0 0
2 02 dz · · · H (z , . . . , z , z , . . . , z ) + 1 m 1 m 2m 0 ζ (1 + zj ) ζ 1 + zj zj zj (2πi) Γ1 Γ1 j=1   √ + O e−δ log R  √  0 0 −δ log R = I (G (z1 , . . . , zm , 0, z1 , . . . , zm , 0) , m) log R + I (H, m) + O e .

log R

Z

Z

0-39

ζ 1 + zj +

where

∂G 0 (z1 , . . . , zm , 0, z10 , . . . , zm , 0) H := 0 ∂zm+1 Z 1 dzm+1 0 + G (z1 , . . . , zm , zm+1 , z10 , . . . , zm , −zm+1 ) 4 2πi Γ0 ζ (1 + zm+1 ) ζ (1 − zm+1 ) zm+ 0 (z1 , . . . , zm , z10 , . . . , zm )

Using the inductive hypothesis and kHkC j  kGkC j+1 , we therefore obtain m+1

m

I (G, m + 1) = G (0, . . . , 0) (log R) + H (0, . . . , 0) (log R) m+1     √ X m+1−j + Om kGkC j (Dσm+1 ) (log R) + O e−δ log R j=1 m+1

= G (0, . . . , 0) (log R)

+

m+1 X



m+1−j

Om kGkC j (Dσm+1 ) (log R)

j=1

which completes our proof.

0-40





√ −δ log

+O e

4 4.1

Index Dual Function

Define the dual function of F to be    DF (x) := E  

Y



  fω (x + ωh) |h ∈ Zk−1 N .  κ−1

(14)

ω∈{0,1} ω6=0k−1

Then we immediately get by the definition that (6.4) hF, DF i =

2k−1 kF kU k−1

.

(15)

Proof. Lemma 2 (Lemma 6.1 of [2], Boundness of Dual function) If |F | < ν + 1, then dual function is bounded |DF | ≤ 22

k−1

−1

0-41

+ o (1)

(16)

by the linear form conditon of ν. 

Proposition 12 (Propostion 6.2 in [2], Weakly mixing of Dual function k−1 k−1 k let Φ : I → R be a continous function in a Set I = [−22 , 22 ], and
compact subset E of C I k . For k-pseudorandom measure ν, if |Fi | < ν + 1 Then we have hν − 1, Φ (DF1 , . . . , DFK )i = oK,E (1) .

(17)

Proof. This is a vital role in proving Target I. By Weierstrass approximation theorem, it is sufficient to prove for Φ = P the polynomial functions. Further, since the formula (21) kν − 1kU k−1 = o (1) holds by applying linear form conditions. Hence it is sufficient to prove that, D Y E f, DFj = OK (1) (18) 0-42

for all f with kf kU k−1 ≤ 1.This formula follows by applying GowersCauchy-Schwarz inequality first, and the correlation condition (2nd/2 condition of k-pseudorandom, used only here in Tao’s paper) in the final step. 

4.2

Argument of Pigeonhole Principle (Prop 7.2)

Proof of (7) Prop 7.2. For any bounded function G(= DF ), we have Z

1

X


E 1G(x)∈[ε(n+α−η),ε(n+α+η)] (ν − 1) dα

0 n∈Z Z 1X

1 X 1G(x)∈[ε(n+α−η),ε(n+α+η)] (ν − 1) (m) dα = 0 n∈Z N m∈Z N Z 1X 1 X = (ν − 1) (m) 1G(x)∈[ε(n+α−η),ε(n+α+η)] (m) dα N 0 m∈ZN

n∈Z

= 2ηE (ν − 1) ,

0-43

(19)

since X

 1G(x)∈[ε(n+α−η),ε(n+α+η)] (m) =

n∈Z

G(m) − η}, { + η}], 1, α + n0 ∈ [{ G(m) ε ε 0, otherwise.

By pigeonhole principle, there exists a α ∈ [0, 1] such that X
E 1G(x)∈[ε(n+α−η),ε(n+α+η)] (ν − 1) = O (η) . n∈Z

Therefore, we only need to choose P 1A − ΨA to be continous and has its support in the support of the n∈Z 1G(x)∈[ε(n+α−η),ε(n+α+η)] . 

4.3

k-pseudorandom measure

The function ν : ZN → R+ is a measure if E (ν) = 1 + o (1) , as N → ∞. A measure ν is k-pseudorandom if it satisfies 0-44

(20)

• linear form condition • correllation conditon Linear form condition induce that — Generalized von-Neumann theorem; — The boundness of Dual function, (see application in (5) and (8)) — the smallness of Gower Norm of ν − 1,i.e. kν − 1kU d = o (1) ,

(21)

and then the weakly-mixing wrt Dual function Theorem 12. The correlation condition only has its application in the weaklymixing Dual function Theorem 12.

0-45

References [1] H. Furstenberg, Ergodic behavior of diagonal measures and a theorem of Szemer´edi on arithmetic progressions. J. d’Analyse Math., 31 (1977), 204–256. MR0498471 (58:16583). [2] B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions. To appear, Ann. Math.arXiv:math.NT/0404188 v4 2 Aug 2005. [3] E. Szemer´edi, On sets of integers containing no k elements in arithmetic progression, Acta Arith. 27 (1975), 299–345.

0-46