Erdős and arithmetic progressions

Erdős and arithmetic progressions W. T. Gowers University of Cambridge July 2, 2013 W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 1 / 22

Two famous Erdős conjectures Let A be a set of positive integers such that n A n 1 =. Then A contains arbitrarily long arithmetic progressions. Let ɛ 1, ɛ 2, ɛ 3,... be a sequence taking values in the set { 1, 1}. Then for every constant C there exist positive integers n and d such that n m=1 ɛ md C. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 2 / 22

Two famous Erdős conjectures Let A be a set of positive integers such that n A n 1 =. Then A contains arbitrarily long arithmetic progressions. Let ɛ 1, ɛ 2, ɛ 3,... be a sequence taking values in the set { 1, 1}. Then for every constant C there exist positive integers n and d such that n m=1 ɛ md C. A third famous Erdős conjecture about arithmetic progressions... For every m there exist distinct positive integers r 1,..., r k m and integers a 1, a 2,..., a k such that for every n N there exists i such that n a i mod r i. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 2 / 22

Interpreting the sum-of-reciprocals conjecture Given a set A, define the density function δ A by δ A (n) = n 1 A {1, 2,..., n} W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 3 / 22

Interpreting the sum-of-reciprocals conjecture Given a set A, define the density function δ A by δ A (n) = n 1 A {1, 2,..., n} The sum-of-reciprocals conjecture is roughly equivalent to the statement δ A (n) is significantly less than (log n) 1. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 3 / 22

Interpreting the conjecture continued Sufficient to prove For every k there exists n 0 such that for every n n 0, if A is any subset of {1,..., n} of cardinality at least n/ log n(log log n) 2, then A contains an arithmetic progression of length k. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 4 / 22

Motivation for the conjecture If we take A to be the set of primes, then n A n 1 = (reproved by Erdős when he was a boy). W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 5 / 22

Motivation for the conjecture If we take A to be the set of primes, then n A n 1 = (reproved by Erdős when he was a boy). So it would yield a purely combinatorial proof of the assertion that the primes contain arbitrarily long arithmetic progressions W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 5 / 22

Motivation for the conjecture If we take A to be the set of primes, then n A n 1 = (reproved by Erdős when he was a boy). So it would yield a purely combinatorial proof of the assertion that the primes contain arbitrarily long arithmetic progressions This statement is now the Green-Tao theorem. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 5 / 22

Motivation for the conjecture If we take A to be the set of primes, then n A n 1 = (reproved by Erdős when he was a boy). So it would yield a purely combinatorial proof of the assertion that the primes contain arbitrarily long arithmetic progressions This statement is now the Green-Tao theorem. But much motivation remains! W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 5 / 22

What do we know? Erdős-Turán conjecture (1936). For every k, if A is a set of integers containing no arithmetic progression of length k, then δ A (n) 0. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 6 / 22

What do we know? Erdős-Turán conjecture (1936). For every k, if A is a set of integers containing no arithmetic progression of length k, then δ A (n) 0. In 1975 this became Szemerédi s theorem. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 6 / 22

What do we know? Erdős-Turán conjecture (1936). For every k, if A is a set of integers containing no arithmetic progression of length k, then δ A (n) 0. In 1975 this became Szemerédi s theorem. Szemerédi s proof did not give useful quantitative information. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 6 / 22

What do we know? Erdős-Turán conjecture (1936). For every k, if A is a set of integers containing no arithmetic progression of length k, then δ A (n) 0. In 1975 this became Szemerédi s theorem. Szemerédi s proof did not give useful quantitative information. Alternative proof in 1977 by Furstenberg. Very important and influential, but again not quantitative. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 6 / 22

Progressions of length 3 The sum-of-reciprocals conjecture not known in first non-trivial case. Roth (1953) obtained δ A (n) C/ log log n. Heath-Brown (1987) and Szemerédi (1990) obtained δ A (n) (log n) c. Bourgain (1999) obtained δ A (n) C(log log n/ log n) 1/2. Bourgain (2008) obtained δ A (n) (log log n) 2 /(log n) 2/3. Sanders (2011) obtained δ A (n) (log n) 3/4+o(1). Sanders (2011) obtained δ A (n) C(log log n) 5 / log n. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 7 / 22

What is the right bound probably? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 8 / 22

What is the right bound probably? Best known lower bound is due to Behrend (1946). Crucial observations are that A d-dimensional sphere contains no three points in arithmetic progression. Some d-dimensional spheres contain many points. Subsets of Z d can be isomorphic to subsets of Z. This gives a bound of the form exp( c log n). So the extremal examples are not random. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 8 / 22

Is the apparent barrier at (log n) 1 significant? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 9 / 22

Is the apparent barrier at (log n) 1 significant? Meshulam (1995) observed that Roth s argument works very cleanly in F n 3 and gives an upper bound of C/n. Very hard to improve on this bound a favourite problem of many people. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 9 / 22

Is the apparent barrier at (log n) 1 significant? Meshulam (1995) observed that Roth s argument works very cleanly in F n 3 and gives an upper bound of C/n. Very hard to improve on this bound a favourite problem of many people. Bateman and Katz (2012) did improve it to n 1 ɛ for a small ɛ > 0. However, their proof hard to combine with Sanders s. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 9 / 22

Another piece of evidence Schoen and Shkredov (2011) proved the following. Theorem Let A be a subset of {1, 2,..., n} of density exp( c(log n) 1/6 ɛ ). Then A contains distinct elements x 1, x 2, x 3, x 4, x 5, y such that x 1 + x 2 + x 3 + x 4 + x 5 = 5y. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 10 / 22

Another piece of evidence Schoen and Shkredov (2011) proved the following. Theorem Let A be a subset of {1, 2,..., n} of density exp( c(log n) 1/6 ɛ ). Then A contains distinct elements x 1, x 2, x 3, x 4, x 5, y such that x 1 + x 2 + x 3 + x 4 + x 5 = 5y. Thus, for a closely related problem a Behrend-type bound is correct. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 10 / 22

Another piece of evidence Schoen and Shkredov (2011) proved the following. Theorem Let A be a subset of {1, 2,..., n} of density exp( c(log n) 1/6 ɛ ). Then A contains distinct elements x 1, x 2, x 3, x 4, x 5, y such that x 1 + x 2 + x 3 + x 4 + x 5 = 5y. Thus, for a closely related problem a Behrend-type bound is correct. However, convolutions of three or more sets are very different from convolutions of two sets. Therefore, it is hard to say how strong this evidence is. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 10 / 22

Longer progressions W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 11 / 22

Longer progressions To obtain a progression of length k a density of C(log log n) 1/22k+9 suffices (WTG 2001). W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 11 / 22

Longer progressions To obtain a progression of length k a density of C(log log n) 1/22k+9 suffices (WTG 2001). When k = 4, Green and Tao have improved this to exp( c log log n). In a finite-field model they have obtained (log n) c. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 11 / 22

Erdős s discrepancy problem W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 12 / 22

Erdős s discrepancy problem A homogeneous arithmetic progression is a set of the form {d, 2d,..., md}. If we colour the integers from 1 to n red or blue, the discrepancy of the colouring is the maximum over all HAPs of the difference between the number of blue elements and the number of red elements. How small can this discrepancy be? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 12 / 22

Erdős s discrepancy problem A homogeneous arithmetic progression is a set of the form {d, 2d,..., md}. If we colour the integers from 1 to n red or blue, the discrepancy of the colouring is the maximum over all HAPs of the difference between the number of blue elements and the number of red elements. How small can this discrepancy be? Erdős conjectured that it tends to infinity with n. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 12 / 22

Again, random examples are not best A random red/blue colouring gives n 1/2+o(1). But we can do much better. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 13 / 22

Again, random examples are not best A random red/blue colouring gives n 1/2+o(1). But we can do much better. Observe that the sequence 1, 1, 0, 1, 1, 0,... is completely multiplicative. (It is a multiplicative character mod 3.) Therefore, its discrepancy is 1. Same is true of the dilation by 3 : that is, the sequence 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0,... Adding together all dilations by powers of 3, we get a growth rate of log 3 n for the discrepancy. (Borwein, Choi 2006) W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 13 / 22

Completely multiplicative functions The resulting sequence, 1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, 1, 1, -1, -1, 1, -1, -1,... is completely multiplicative. It is 1 or -1 according as the last non-zero ternary digit is 1 or 2. Partial sum up to m is the number of 1s in the ternary expansion of m. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 14 / 22

Completely multiplicative functions The resulting sequence, 1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, 1, 1, -1, -1, 1, -1, -1,... is completely multiplicative. It is 1 or -1 according as the last non-zero ternary digit is 1 or 2. Partial sum up to m is the number of 1s in the ternary expansion of m. Discrepancy conjecture not even known for completely multiplicative functions! W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 14 / 22

What makes the problem hard? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 15 / 22

What makes the problem hard? The extremal examples are not random. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 15 / 22

What makes the problem hard? The extremal examples are not random. The example 1, -1, 0, 1, -1, 0, 1, -1, 0,... makes it hard to turn the problem into an analytic one. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 15 / 22

What makes the problem hard? The extremal examples are not random. The example 1, -1, 0, 1, -1, 0, 1, -1, 0,... makes it hard to turn the problem into an analytic one. There are several examples (discovered by computer) of sequences of length 1124 and discrepancy 2. (Polymath5 2010) W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 15 / 22

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Interesting variants Replace ±1 by complex numbers of modulus 1. Replace ±1 by unit vectors in a Hilbert space. Replace ±1 by non-zero elements mod p for arbitrarily large prime p. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 18 / 22

Completely multiplicative sequences again W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 19 / 22

Completely multiplicative sequences again An observation of Terence Tao. Suppose that there exists an infinite ±1 sequence of discrepancy at most C. Then there exists a completely multiplicative sequence z 1, z 2,... of complex numbers of modulus 1 such that the averages N 1 N n=1 n i=1 z i 2 are bounded above by a constant that depends on C only. Proof is a short argument using Fourier analysis. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 19 / 22

Analytic versions after all An observation of Moses Charikar (Polymath5 2010). If we can find non-negative coefficients c m,d and b n such that c m,d = 1, b n =, and m,d n c m,d (x d + x 2d + + x md ) 2 n m,d is positive semidefinite, then Erdős s conjecture is true. b n x 2 n W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 20 / 22

Why is this approach promising? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 21 / 22

Why is this approach promising? The existence of such a quadratic form is equivalent to the Hilbert-spaces version of the conjecture. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 21 / 22

Why is this approach promising? The existence of such a quadratic form is equivalent to the Hilbert-spaces version of the conjecture. One can use semidefinite programming to work out optimal coefficients for fairly large values of n and try to get insight into what they might look like in general. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 21 / 22

Why is this approach promising? The existence of such a quadratic form is equivalent to the Hilbert-spaces version of the conjecture. One can use semidefinite programming to work out optimal coefficients for fairly large values of n and try to get insight into what they might look like in general. Shows that an analytic version may exist. If this approach works, it yields weights b n such that every sequence (x n ) with n b nx 2 n = has unbounded discrepancy. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 21 / 22

Why is this approach promising? The existence of such a quadratic form is equivalent to the Hilbert-spaces version of the conjecture. One can use semidefinite programming to work out optimal coefficients for fairly large values of n and try to get insight into what they might look like in general. Shows that an analytic version may exist. If this approach works, it yields weights b n such that every sequence (x n ) with n b nx 2 n = has unbounded discrepancy. Interesting problem: what are those weights? W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 21 / 22

Another analytic generalization The following conjecture easily implies Erdős s conjecture and is again an analytic statement. If A = (a ij ) is any infinite matrix with 1s on the diagonal, then for every C there exist homogeneous arithmetic progressions P and Q such that i P j Q a ij C. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 22 / 22

Another analytic generalization The following conjecture easily implies Erdős s conjecture and is again an analytic statement. If A = (a ij ) is any infinite matrix with 1s on the diagonal, then for every C there exist homogeneous arithmetic progressions P and Q such that i P j Q a ij C. To recover EDP, set a ij = ɛ i ɛ j. W. T. Gowers (University of Cambridge) Erdős and arithmetic progressions July 2, 2013 22 / 22