# Presented by: Ryan ODonnell Carnegie Melloni by D. H. J. Polymath.

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presented by: Ryan ODonnell Carnegie Melloni by D. H. J. Polymath

If A {0,1,2} n has density Ω(1), then A contains a (combinatorial) line. DHJ(3): point:1 1 2 1 2 1 0 0 0 1 0 1 1 0 1 2 0 1 2 2 line: 1 0 2 1 1 2 1 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 2 1 1 2 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 2 1 1 2 1 1 0 2 2 1 0 0 2 1 1 2 1 1 { },, 1 0 2 1 1 2 1 1 0 1 0 0 1 1 1 1

[Furstenberg-Katznelson91] : DHJ(k) is true k. (k = 3): δ > 0, n 0 (δ) s.t. n n 0 (δ), A {0,1,2} n has density δ A contains a line proof: Used ergodic theory. No effective bound on n 0 (δ).

Importance in combinatorics: Implies Szemerédis Theorem. 1 0 2 1 1 2 1 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 2 1 1 2 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 2 1 1 2 1 1 0 2 2 1 0 0 2 1 1 2 1 1

Why I was interested: 0 1 2 0 0, 1, 2, 1 0 1 2 2 A {0,1,2} n, no trips w/ all cols 0 0 0 1 1 0, 0, 1, 0, 1 0 1 0 0 1 A {0,1} n, no trips w/ all cols see [O-Wu09]

[D.H.J. Polymath]: δ > 0, n 2 O(1/δ 3 ), A {0,1,2} n has density δ A contains a line proof: Elementary probability/combinatorics. also the general k case, with Ackermann-type bounds

Proof sketch

DHJ(2): A {0,1} n dens. δ A contains line line: 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1 1 0 1 1 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 { }, 1 0 1 1 1 1 1 1 0 1 0 0 1 1 1 1 contrapositive: If A has no comparable pairs i.e., A is an antichain then A is very sparse. Sperners Theorem

DHJ(2): A {0,1} n dens. δ A contains line 1234567n 00011111 equal-slices distribution

DHJ(2): A {0,1} n dens. δ A contains line equal-slices distribution 5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 1 7 1 6 1 3 1 4 1 2 0 n 0 5 0 (each Hamming wt. equally likely)

DHJ(2): A {0,1} n dens. δ A contains line 5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 1234567n equal-slices distribution DHJ(2): A {0,1} n eq-slices dens. δ A contains line

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 1234567n DHJ(2): A {0,1} n eq-slices dens. δ A contains line 00000011

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 1234567n DHJ(2): A {0,1} n eq-slices dens. δ A contains line 0000001100000011 0 0 0 0 0 1 1 0 0000001100000011

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 1234567n DHJ(2): A {0,1} n eq-slices dens. δ A contains line 0000001100000011 0 0 0 0 0 1 1 0 00000011

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 DHJ(2): A {0,1} n eq-slices dens. δ A contains line 0000001100000011 0 0 0 0 0 1 1 0 00000011 ( Pr[degen] = ) random eq-slices line

DHJ(2): A {0,1} n eq-slices dens. δ eq-slice-line in A w.p. δ 2 Pr[ x, y A ] = E xy x y A A & Pr = E x x A 2 Pr E x x A 2 = δ 2 (independent)

DHJ(2): A {0,1} n eq-slices dens. δ eq-slice-line in A w.p. δ 2

DHJ(2): A {0,1} n eq-slices dens. δ n 1/δ 2 A contains a line

DHJ(2): A {0,1} n eq-slices dens. δ Distribution Hackery 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1unif rand. coords unif on eq-slices on 1 0 0 1 1 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1

[ eq-slices ( A ) ] DHJ(2): A {0,1} n eq-slices dens. δ Distribution Hackery 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1unif 1 0 0 1 1 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1 easy: d TV ( unif, unif ) = o(1)unif (A) δ = E [ eq-slices A y y rand. coords

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 0000001100000011 0 0 0 0 0 1 1 0 00000011 1234567n DHJ(2): A {0,1} n eq-slices dens. δ eq-slice-line in A w.p. δ 2

DHJ(3): A {0,1,2} n eq-slices dens. δ eq-slice-line in A w.p. δ 3 ???

Pr[ x, y A ] = E xy x y A A & Pr = E x y x A 2 Pr E xy x A 2 = δ 2 (independent)

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 1 0 DHJ(3): A {0,1,2} n eq-slices dens. δ ? 00000011 1234567n

5 0 n 0 2 0 4 1 1 1 3 1 6 1 7 1 1 0 1 1 0 1 1 0 00000011 0 0 0 0 0 1 1 0 1234567n DHJ(3): A {0,1,2} n eq-slices dens. δ 00022211 2 0 2 2 0 1 1 0 eq-slices pt in {0,1,2} n eq-slices line over {0,1} n if in A as a line, & in A as a point, done!

DHJ(3): A {0,1,2} n eq-slices dens. δ by distrib. hackery: A {0,1} n eq-slices dens. δ also {0,1,2} n {0,1} n A A

DHJ(3): A {0,1,2} n eq-slices dens. δ by distrib. hackery: A {0,1} n eq-slices dens. δ also {0,1,2} n {0,1} n A A L L = { x {0,1,2} n : x 21, x 20 A } δ 2

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n {0,1} n A A L δ 2 L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n {0,1} n A A L δ 2

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n {0,1} n A A L δ 2 if L A, done! δ

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n {0,1} n A A L δ 2 δ idea: A has dens. in

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n idea: A has dens. in A δ + δ 3

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n A δ + δ 3 Suppose, somehow, that

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log n A δ + δ 3 Suppose, somehow, that

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log n A δ + δ 3 Suppose, somehow, that Repeat.

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log n A δ + δ 3 Suppose, somehow, that Repeat. δ 2

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log n A δ + δ 3 Suppose, somehow, that Repeat. δ 2

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log log log log n A δ + 2δ 3 Suppose, somehow, that Repeat.

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log log log log log log log log log n A δ + 3δ 3 Suppose, somehow, that Repeat.

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} log (3/δ 3 ) n A > 2/3 Suppose, somehow, that Repeat. A contains line trivially. Density Increment Argument

L = { x {0,1,2} n : x 21, x 20 A } {0,1,2} n A δ + δ 3 Suppose, somehow, that

L = { x {0,1,2} n : x 21, x 20 A } Suppose, somehow, that L = { x : x 21 A } { x : x 20 A } = { x : x 21 A } (harmless cheat) pretend idea: is a 12-insensitive set 12-insensitive set

= { x : x 21 A } 12-insensitive set

= { x : x 21 A } 12-insensitive set monkeying with 1s and 2s does not affect presence in the set Closer to an isomorphic copy of

key thm: A dense 12-insensitive set can be almost completely partitioned into copies of

key thm: A dense 12-insensitive set can be almost completely partitioned into copies of

A has dens. δ + δ 3 inside A has dens. δ + δ 3 inside some

key thm: A dense 12-insensitive set can be almost completely partitioned into copies of 2 1 1 0 1 1 1 0 1 0 2 1 1 2 1 2 2 2 1 1 1 2 lemma 1: Dense 12-insensitive set contains a copy of

key thm: A dense 12-insensitive set can be almost completely partitioned into copies of lemma 1: Dense 12-insensitive set contains a copy of lemma 2: Dense 12-insensitive set contains a copy of {0, 1, 2} i.e., a line

lemma 2: Dense 12-insensitive set contains a copy of {0, 1, 2} i.e., a line lemma 2: Dense 12-insensitive set contains a copy of {0, 1, 2} i.e., a line

lemma 2: Dense 12-insensitive set contains a copy of {0, 1, 2} i.e., a line (i) Distrib. hack: make it dense in {0,1} n also. (ii) Apply DHJ(2). 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1 A 1 0 1 1 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 A 1 0 1 1 1 1 1 1 0 2 2 1 0 0 2 1 1 2 1 1 A (iii) … by 12-insensitivity.

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