Roth s Theorem on Arithmetic Progressions

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1 September 25, 2014

2 The Theorema of Szemerédi and Roth For Λ N the (upper asymptotic) density of Λ is the number σ(λ) := lim sup N Λ [1, N] N [0, 1]

3 The Theorema of Szemerédi and Roth For Λ N the (upper asymptotic) density of Λ is the number σ(λ) := lim sup N Λ [1, N] N [0, 1] Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression.

4 The Theorema of Szemerédi and Roth For Λ N the (upper asymptotic) density of Λ is the number σ(λ) := lim sup N Λ [1, N] N [0, 1] Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Today we ll focus on the first non-trivial case of this result - known as Roth s Theorem.

5 The Theorema of Szemerédi and Roth For Λ N the (upper asymptotic) density of Λ is the number σ(λ) := lim sup N Λ [1, N] N [0, 1] Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Today we ll focus on the first non-trivial case of this result - known as Roth s Theorem. Theorem (Roth, 1953) If Λ N has positive density, then Λ contains an arithmetic progression of length 3.

6 Arithmetic progressions in the primes Theorem (Green-Tao, 2004) For any k N, the set of primes P contains infinitely many k-term arithmetic progressions.

7 Arithmetic progressions in the primes Theorem (Green-Tao, 2004) For any k N, the set of primes P contains infinitely many k-term arithmetic progressions. Note, by the Prime Number Theorem σ(p) := lim sup N 0 P [1, N] N lim sup N 0 1 log N = 0.

8 A big conjecture Conjecture (Erdős-Turán) Suppose Λ N is such that x Λ 1 x =. Then for any k N, Λ contains infinitely many k-term arithmetic progressions.

9 A big conjecture Conjecture (Erdős-Turán) Suppose Λ N is such that x Λ 1 x =. Then for any k N, Λ contains infinitely many k-term arithmetic progressions. Erdős offered $3000 for a resolution of this conjecture.

10 A big conjecture Conjecture (Erdős-Turán) Suppose Λ N is such that x Λ 1 x =. Then for any k N, Λ contains infinitely many k-term arithmetic progressions. Erdős offered $3000 for a resolution of this conjecture. Even the k = 3 case remains open!

11 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression.

12 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth s original proof involved Fourier analysis (circle method and density increments ).

13 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth s original proof involved Fourier analysis (circle method and density increments ). Szemerédi s proof was purely combinatorial. A tour de force, it introduced a number of techniques of central importance (Energy increments, Szemerédi s Regularity Lemma).

14 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth s original proof involved Fourier analysis (circle method and density increments ). Szemerédi s proof was purely combinatorial. A tour de force, it introduced a number of techniques of central importance (Energy increments, Szemerédi s Regularity Lemma). In 1977 Furstenberg reproved Szemerédi s Theorem, introducing the important Ergodic Theory approach. This lead to later multi-dimensional generalisations.

15 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth combined his Fourier analytic method with some of Szemerédi s ideas to prove the k = 4 case. His arguments provided precise quantitative statements.

16 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth combined his Fourier analytic method with some of Szemerédi s ideas to prove the k = 4 case. His arguments provided precise quantitative statements. In the late 1990s and early 2000s Gowers developed higher order Fourier analysis and the theory of Gowers Uniformity Norms to give another proof of Szemerédi s theorem. This gave much more precise quantitative information.

17 A brief history Theorem (Szemerédi, 1975) If Λ N has positive density, then for each k N, Λ contains a k-term arithmetic progression. Roth combined his Fourier analytic method with some of Szemerédi s ideas to prove the k = 4 case. His arguments provided precise quantitative statements. In the late 1990s and early 2000s Gowers developed higher order Fourier analysis and the theory of Gowers Uniformity Norms to give another proof of Szemerédi s theorem. This gave much more precise quantitative information. There have been numerous other important new proofs and developments...

18 Structure vrs. Randomness Although there are many proofs using very different techniques they all share some characteristics.

19 Structure vrs. Randomness Although there are many proofs using very different techniques they all share some characteristics. In all the arguments notions of Structure and Randomness play a significant rôle.

20 Example: (Pseudo)-Random sets Select each x N to be a member of Λ independently at random with probability 0 p 1.

21 Example: (Pseudo)-Random sets Select each x N to be a member of Λ independently at random with probability 0 p 1. Then σ(λ) = p almost surely.

22 Example: (Pseudo)-Random sets Select each x N to be a member of Λ independently at random with probability 0 p 1. Then σ(λ) = p almost surely. Each progression of length k has probability p k of lying in Λ. From this one deduces Λ contains arithmetic progressions of all lengths almost surely.

23 Example: (Pseudo)-Random sets Select each x N to be a member of Λ independently at random with probability 0 p 1. Then σ(λ) = p almost surely. Each progression of length k has probability p k of lying in Λ. From this one deduces Λ contains arithmetic progressions of all lengths almost surely. The same is true if one considers pseudo-random sets - that is, Λ is deterministic but certain correlations are negligible.

24 Example: Linearly structured sets Consider an almost-periodic set Λ := {x N: {xα} δ} 0 < δ < 1, α R.

25 Example: Linearly structured sets Consider an almost-periodic set Λ := {x N: {xα} δ} 0 < δ < 1, α R. If α R \ Q, then by Weyl s Equidistribution Theorem σ(λ) = δ.

26 Example: Linearly structured sets Consider an almost-periodic set Λ := {x N: {xα} δ} 0 < δ < 1, α R. If α R \ Q, then by Weyl s Equidistribution Theorem σ(λ) = δ. Observe Λ is almost-periodic - there is a strong correlation between the event x Λ and x + r Λ due to the identity {α(x + r)} {αx} = {αr} mod 1

27 Example: Linearly structured sets Consider an almost-periodic set Λ := {x N: {xα} δ} 0 < δ < 1, α R. If α R \ Q, then by Weyl s Equidistribution Theorem σ(λ) = δ. Observe Λ is almost-periodic - there is a strong correlation between the event x Λ and x + r Λ due to the identity {α(x + r)} {αx} = {αr} mod 1 By Dirichlet s Theorem one may deduce Λ contains k-term arithmetic progressions for all k.

28 Quadratically structured sets Now consider Λ := {x N: {x 2 α} δ} 0 < δ < 1, α R.

29 Quadratically structured sets Now consider Λ := {x N: {x 2 α} δ} 0 < δ < 1, α R. By Weyl s Polynomial Equidistribution theorem, if α R \ Q, then σ(λ) = δ.

30 Quadratically structured sets Now consider Λ := {x N: {x 2 α} δ} 0 < δ < 1, α R. By Weyl s Polynomial Equidistribution theorem, if α R \ Q, then σ(λ) = δ. This set is not linearly structured. However, it is quadratically structured in the sense that the event x Λ, x + r Λ and x + 2r Λ are highly correlated due to the identity {α(x + 2r) 2 } 2{α(x + r) 2 } + {αx 2 } = 2{αr 2 } mod 1

31 Quadratically structured sets Now consider Λ := {x N: {x 2 α} δ} 0 < δ < 1, α R. By Weyl s Polynomial Equidistribution theorem, if α R \ Q, then σ(λ) = δ. This set is not linearly structured. However, it is quadratically structured in the sense that the event x Λ, x + r Λ and x + 2r Λ are highly correlated due to the identity {α(x + 2r) 2 } 2{α(x + r) 2 } + {αx 2 } = 2{αr 2 } mod 1 This information can be used to prove the existence of k-length arithmetic progressions in Λ.

32 Structure vrs Randomness Broadly speaking, the proofs of Szemerédi s Theorem rely on showing the following two facts:

33 Structure vrs Randomness Broadly speaking, the proofs of Szemerédi s Theorem rely on showing the following two facts: If a set behaves randomly, then it contain many arithmetic progressions.

34 Structure vrs Randomness Broadly speaking, the proofs of Szemerédi s Theorem rely on showing the following two facts: If a set behaves randomly, then it contain many arithmetic progressions. If a set is not random, then it has a sizeable intersection with a structured set.

35 Structure vrs Randomness Broadly speaking, the proofs of Szemerédi s Theorem rely on showing the following two facts: If a set behaves randomly, then it contain many arithmetic progressions. If a set is not random, then it has a sizeable intersection with a structured set. The analysis is then restricted to this structured set. Furthermore, it is typically shown one cannot pass to a structured subset an indefinite number of times which leads to the proof.

36 An equivalent formulation For A N, let r 3 (A) be the largest cardinality of a subset of A containing no 3-term arithmetic progression. Then Roth s theorem is equivalent to the statement r 3 ([1, N]) N 0 as N

37 An equivalent formulation For A N, let r 3 (A) be the largest cardinality of a subset of A containing no 3-term arithmetic progression. Then Roth s theorem is equivalent to the statement r 3 ([1, N]) N 0 as N Suppose Λ has positive density but contains no 3-term arithmetic progressions. Then r 3 ([1, N]) Λ [1, N]

38 An equivalent formulation For A N, let r 3 (A) be the largest cardinality of a subset of A containing no 3-term arithmetic progression. Then Roth s theorem is equivalent to the statement r 3 ([1, N]) N 0 as N Suppose Λ has positive density but contains no 3-term arithmetic progressions. Then r 3 ([1, N]) N Λ [1, N] N

39 An equivalent formulation For A N, let r 3 (A) be the largest cardinality of a subset of A containing no 3-term arithmetic progression. Then Roth s theorem is equivalent to the statement r 3 ([1, N]) N 0 as N Suppose Λ has positive density but contains no 3-term arithmetic progressions. Then lim sup N r 3 ([1, N]) N lim sup N Λ [1, N] N = σ(λ) > 0

40 Things are simplified considerably if we consider the analogous problem in the setting of F n p..

41 Things are simplified considerably if we consider the analogous problem in the setting of F n p. Let r 3 (F n p) denote the cardinality of the largest subset of F n p containing no 3-term arithmetic progression..

42 Things are simplified considerably if we consider the analogous problem in the setting of F n p. Let r 3 (F n p) denote the cardinality of the largest subset of F n p containing no 3-term arithmetic progression. Theorem (Meshulam, 1995) For any odd prime p and any dimension n one has r 3 (F n p) F n p < 3 n Henceforth fix p an odd prime, n N and let Z := F n p.

43 Step 1 If a set contains no 3-term arithmetic progression, then it is not random.

44 Random sets and arithmetic progressions For A Z define 3 (A) := P x,r Z (x, x + r, x + 2r A) = {(x, r) Z 2 : x, x + r, x + 2r A} Z 2

45 Random sets and arithmetic progressions For A Z define 3 (A) := P x,r Z (x, x + r, x + 2r A) = {(x, r) Z 2 : x, x + r, x + 2r A} Z 2 Heuristically, if A is randomly distributed, then the events x A, x + r A and x + 2r A should be in some sense independent and then one would expect 3 (A) P x,r Z (x A)P x,r Z (x + r A)P x,r Z (x + 2r A) = P Z (A) 3.

46 Random sets and arithmetic progressions For A Z define 3 (A) := P x,r Z (x, x + r, x + 2r A) = {(x, r) Z 2 : x, x + r, x + 2r A} Z 2 Heuristically, if A is randomly distributed, then the events x A, x + r A and x + 2r A should be in some sense independent and then one would expect 3 (A) P x,r Z (x A)P x,r Z (x + r A)P x,r Z (x + 2r A) = P Z (A) 3. Thus if A is dense in Z, then we expect the number of arithmetic progressions in A to be large.

47 Random sets and arithmetic progressions If A contains no arithmetic progressions of length 3, then 3 (A) = P Z (A) Z which conflicts with the heuristic 3 (A) P Z (A) 3 for Z large.

48 Random sets and arithmetic progressions If A contains no arithmetic progressions of length 3, then 3 (A) = P Z (A) Z which conflicts with the heuristic 3 (A) P Z (A) 3 for Z large. The proof of Meshulam s Theorem proceeds by making this heuristic precise.

49 Random sets and arithmetic progressions If A contains no arithmetic progressions of length 3, then 3 (A) = P Z (A) Z which conflicts with the heuristic 3 (A) P Z (A) 3 for Z large. The proof of Meshulam s Theorem proceeds by making this heuristic precise. One needs a measurement of the randomness of A: this realised as the linear bias of A, a quantity defined via Fourier analysis.

50 Fourier Analysis in F n p Given a function f : Z C one may define its Fourier transform ˆf : Z C as follows: ˆf (ξ) := 1 2πi ξ, x /p f (x)χ ξ (x), χ ξ (x) := e Z x Z where ξ, x = n j=1 ξ j.x j Z.

51 Fourier Analysis in F n p Given a function f : Z C one may define its Fourier transform ˆf : Z C as follows: ˆf (ξ) := 1 2πi ξ, x /p f (x)χ ξ (x), χ ξ (x) := e Z x Z where ξ, x = n j=1 ξ j.x j Z. Thus, for a set A Z and ξ 0 it follows ˆ1 A (ξ) measures the correlation between 1 A and the oscillatory function χ ξ.

52 Fourier Analysis in F n p Given a function f : Z C one may define its Fourier transform ˆf : Z C as follows: ˆf (ξ) := 1 2πi ξ, x /p f (x)χ ξ (x), χ ξ (x) := e Z x Z where ξ, x = n j=1 ξ j.x j Z. Thus, for a set A Z and ξ 0 it follows ˆ1 A (ξ) measures the correlation between 1 A and the oscillatory function χ ξ. Define the linear bias A u of a set A Z to be A u := sup ˆ1 A (ξ) ξ Z\{0}

53 Fourier Analysis in F n p Given a function f : Z C one may define its Fourier transform ˆf : Z C as follows: ˆf (ξ) := 1 2πi ξ, x /p f (x)χ ξ (x), χ ξ (x) := e Z x Z where ξ, x = n j=1 ξ j.x j Z. Thus, for a set A Z and ξ 0 it follows ˆ1 A (ξ) measures the correlation between 1 A and the oscillatory function χ ξ. Define the linear bias A u of a set A Z to be A u := sup ˆ1 A (ξ) ξ Z\{0} If A has large linear bias, it is strongly correlation with a linear phase function.

54 Lack of progressions implies non-uniformity The earlier heuristic 3 (A) P Z (A) 3 is made rigorous by the following inequality: 3 (A) P Z (A) 3 A u P Z (A).

55 Lack of progressions implies non-uniformity The earlier heuristic 3 (A) P Z (A) 3 is made rigorous by the following inequality: 3 (A) P Z (A) 3 A u P Z (A). Recall, if A contains no 3-term progression then 3 (A) = P Z (A)/ Z. Plugging this into the above inequality one obtains:

56 Lack of progressions implies non-uniformity The earlier heuristic 3 (A) P Z (A) 3 is made rigorous by the following inequality: 3 (A) P Z (A) 3 A u P Z (A). Recall, if A contains no 3-term progression then 3 (A) = P Z (A)/ Z. Plugging this into the above inequality one obtains: Proposition (Lack of progressions implies non-uniformity) If A contains no 3-term arithmetic progression, then A u P Z (A) 2 1 Z.

57 Step 2 If a set is not random, then it has a sizeable intersection with a structured set.

58 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u.

59 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u. By definition there exists ξ Z \ {0} such that 1 A u = 1 A (x)e( x, ξ ) Z, e(t) := e2πit/p x Z

60 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u. By definition there exists ξ Z \ {0} such that 1 A u = (1 A (x) P Z (A)) e( x, ξ ), e(t) := e 2πit/p Z }{{} x Z f (x):=

61 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u. By definition there exists ξ Z \ {0} such that 1 A u = (1 A (x) P Z (A)) e( x, ξ ), e(t) := e 2πit/p Z }{{} x Z f (x):= Furthermore, we can find θ {0,..., p 1} such that A u = R 1 f (x)e( x, ξ + θ) Z x Z

62 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u. By definition there exists ξ Z \ {0} such that 1 A u = (1 A (x) P Z (A)) e( x, ξ ), e(t) := e 2πit/p Z }{{} x Z f (x):= Furthermore, we can find θ {0,..., p 1} such that A u = R 1 f (x)e( x, ξ + θ) Z x Z

63 Non-uniformity implies density increment Proposition For A Z there exists an affine hyperplane H in Z such that P H (A) P Z (A) A u. By definition there exists ξ Z \ {0} such that 1 A u = (1 A (x) P Z (A)) e( x, ξ ), e(t) := e 2πit/p Z }{{} x Z f (x):= Furthermore, we can find θ {0,..., p 1} such that A u = R 1 f (x)e( x, ξ + θ) = R 1 f (x)e( x, ξ + θ) Z Z x Z x Z

64 Non-uniformity implies density increment Thus, A u = R 1 Z f (x)(e( x, ξ + θ) + 1). x Z

65 Non-uniformity implies density increment Thus, A u = R 1 Z f (x)(e( x, ξ + θ) + 1). x Z Let Z := ξ := {y Z : ξ, y = 0}. Note the function y e( x, ξ + θ) is constant on any translate of Z.

66 Non-uniformity implies density increment Thus, A u = R 1 Z f (x)(e( x, ξ + θ) + 1). x Z Let Z := ξ := {y Z : ξ, y = 0}. Note the function y e( x, ξ + θ) is constant on any translate of Z. Hence A u = 1 Z Z x Z x 0 Z f (x + x 0 )R ( e( x + x 0, ξ + θ) + 1 )

67 Non-uniformity implies density increment Thus, A u = R 1 Z f (x)(e( x, ξ + θ) + 1). x Z Let Z := ξ := {y Z : ξ, y = 0}. Note the function y e( x, ξ + θ) is constant on any translate of Z. Hence A u = 1 Z Z = 1 Z x Z x 0 Z f (x + x 0 )R ( e( x + x 0, ξ + θ) + 1 ) (P Z +x 0 (A) P Z (A))R ( e( x 0, ξ + θ) + 1 ) x 0 Z

68 Non-uniformity implies density increment Thus, A u = R 1 Z f (x)(e( x, ξ + θ) + 1). x Z Let Z := ξ := {y Z : ξ, y = 0}. Note the function y e( x, ξ + θ) is constant on any translate of Z. Hence A u = 1 Z Z = 1 Z x Z x 0 Z f (x + x 0 )R ( e( x + x 0, ξ + θ) + 1 ) (P Z +x 0 (A) P Z (A))R ( e( x 0, ξ + θ) + 1 ) x 0 Z Thus, there must exist x 0 Z such that for H = Z + x 0 one has 2(P H (A) P Z (A)) (P H (A) P Z (A))R ( e( x 0, ξ +θ)+1 ) A u

69 Bringing the steps together Proof of Meshulam s Theorem Theorem (Meshulam, 1995) For p an odd prime and n any dimension, if Z := F n p, then r 3 (Z) Z < 3 n.

70 Proof of Meshulam s Theorem We proceed by induction on n, noting the result is trivial for 1 n 3 and therefore we assume n > 3.

71 Proof of Meshulam s Theorem We proceed by induction on n, noting the result is trivial for 1 n 3 and therefore we assume n > 3. Suppose there exists A Z with density P Z (A) 3/n containing no 3-term arithmetic progression.

72 Proof of Meshulam s Theorem We proceed by induction on n, noting the result is trivial for 1 n 3 and therefore we assume n > 3. Suppose there exists A Z with density P Z (A) 3/n containing no 3-term arithmetic progression. By Steps 1 and 2 we can find an affine hyperplane H Z such that P H (A) P Z (A) A u P Z (A) + 1 ( P Z (A) 2 1 ). 2 Z

73 Proof of Meshulam s Theorem We proceed by induction on n, noting the result is trivial for 1 n 3 and therefore we assume n > 3. Suppose there exists A Z with density P Z (A) 3/n containing no 3-term arithmetic progression. By Steps 1 and 2 we can find an affine hyperplane H Z such that P H (A) P Z (A) A u P Z (A) + 1 ( P Z (A) 2 1 ). 2 Z Applying the hypothesis on P Z (A) we have P H (A) 3 n + 1 ( 9 2 n 2 1 ) 3 Z n 1 and so A H contains a 3-term progression by the induction hypothesis.

74 Thanks for listening!

Roth s Theorem on 3-term Arithmetic Progressions

Roth s Theorem on 3-term Arithmetic Progressions Mustazee Rahman 1 Introduction This article is a discussion about the proof of a classical theorem of Roth s regarding the existence of three term arithmetic