presented by: Ryan O’Donnell

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-Katznelson’91]: DHJ(k) is true ∀ k.

(k = 3): ∀ δ > 0, ∃ n0(δ) s.t. ∀ n ≥ n0(δ),

A ⊆ {0,1,2}n has density ≥ δ

⇒ A contains a line

proof: Used ergodic theory.

No effective bound on n0(δ).

Importance in combinatorics:

Implies Szemerédi’s 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-Wu’09]

[D.H.J. Polymath]:

∀ δ > 0, ∀ n ≥ 2↑↑ O(1/δ3),

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.

“Sperner’s Theorem”

DHJ(2): A ⊆ {0,1}n dens. δ ⇒ A contains line

12 345 6 7n

0 0 0 1 1 1 1 1

“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

1 2 3 4 5 6 7 n

“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

1 2 3 4 5 6 7 n

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

0 0 0 0 0 0 1 1

5

0

n

0

2

0

4

1

1

1

3

1

6

1

7

1

1 0 1 1 0 1 1 0

1 2 3 4 5 6 7 n

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

0 0 0 0 0 0 1 10 0 0 0 0 0 1 1

0 0 0 0 0 1 1 0

0 0 0 0 0 0 1 10 0 0 0 0 0 1 1

5

0

n

0

2

0

4

1

1

1

3

1

6

1

7

1

1 0 1 1 0 1 1 0

1 2 3 4 5 6 7 n

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

0 0 0 0 0 0 1 10 0 0 0 0 0 1 1

0 0 0 0 0 1 1 0

0 0 0 0 0 0 1 1

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

0 0 0 0 0 0 1 10 0 0 0 0 0 1 1

0 0 0 0 0 1 1 0

0 0 0 0 0 0 1 1

( Pr[degen] = )

“random eq-slices line”

DHJ(2): A ⊆ {0,1}n eq-slices dens. δ

⇒ eq-slice-line in A w.p. ≥ δ2 −

Pr[ x, y ∈ A ]

= Ex y

x

y

∈ A

∈ A

&Pr

= Ex

x ∈ A

2Pr

≥ Ex

x ∈ A

2Pr

= δ2

(independent)

DHJ(2): A ⊆ {0,1}n eq-slices dens. δ

⇒ eq-slice-line in A w.p. ≥ δ2 −

DHJ(2): A ⊆ {0,1}n eq-slices dens. δ

n ≥ 1/δ2 ⇒ 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

unif

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

unif 1 0 0 1 1 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1

easy: dTV( unif, unif ) = o(1)unif

unif (A) ≈ δ = E [eq-slices Ay

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

0 0 0 0 0 0 1 10 0 0 0 0 0 1 1

0 0 0 0 0 1 1 0

0 0 0 0 0 0 1 1

1 2 3 4 5 6 7 n

DHJ(2): A ⊆ {0,1}n eq-slices dens. δ

⇒ eq-slice-line in A w.p. ≥ δ2 −

DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ

⇒ eq-slice-line in A w.p. ≥ δ3 ???

DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ

⇒ eq-slice-line in A w.p. ≥ δ3 ???

Pr[ x, y ∈ A ]

= Ex y

x

y

∈ A

∈ A

&Pr

= Ex y

x ∈ A

2Pr

≥ Ex y

x ∈ A

2Pr

= δ2

(independent)

5

0

n

0

2

0

4

1

1

1

3

1

6

1

7

1

1 0 1 1 0 1 1 00 0 0 0 0 1 1 0

DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ

?

0 0 0 0 0 0 1 1

1 2 3 4 5 6 7 n

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 0 1 1

0 0 0 0 0 1 1 0

1 2 3 4 5 6 7 n

DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ

0 0 0 2 2 2 1 1

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 : x2→1, x2→0 ∈ A }

≥ δ2

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

{0,1}n

A

A L ≥ δ2

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

{0,1}n

A

A L ≥ δ2

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

{0,1}n

A

A L ≥ δ2

if L ⋂ A ≠ ∅, done!

≥ δ

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

{0,1}n

A

A L ≥ δ2

≥ δ

idea: A has dens. ≥ in

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

idea: A has dens. ≥ in

A

≥ δ + δ3

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

A

≥ δ + δ3

Suppose, somehow, that

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log n

A

≥ δ + δ3

Suppose, somehow, that

Suppose, somehow, that

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log n

A

≥ δ + δ3

Suppose, somehow, that

Repeat.

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log n

A

≥ δ + δ3

Suppose, somehow, that

Repeat.

≥ δ2

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log n

A

≥ δ + δ3

Suppose, somehow, that

Repeat.

≥ δ2

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log log log log n

A

≥ δ + 2δ3

Suppose, somehow, that

Repeat.

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log log log log log log log log log n

A

≥ δ + 3δ3

Suppose, somehow, that

Repeat.

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}log (3/δ3)n

A

> 2/3

Suppose, somehow, that

Repeat.

A contains line trivially.

“Density

Increment

Argument”

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

{0,1,2}n

A

≥ δ + δ3

Suppose, somehow, that

L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }

Suppose, somehow, that

L = { x : x2→1 ∉ A } ⋃ { x : x2→0 ∉ A }

= { x : x2→1 ∉ A } (harmless cheat)pretend

idea: is a 12-insensitive set 12-insensitive set

= { x : x2→1 ∉ A }

= { x : x2→1 ∉ A }

12-insensitive set

12-insensitive set

= { x : x2→1 ∉ A }

12-insensitive set

monkeying with 1’s and 2’s

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 + δ3 inside

⇒ A has dens. ≳ δ + δ3 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.

Questions?