the minimum number of monochromatic 4-term progressions · since c1 a(t) = 1d c (t) for t 6= 0....

IntroductionAn analytic upper bound

A combinatorial lower boundAn analogue in the world of graphs

Remarks

The minimum number of monochromatic 4-termprogressions

Julia Wolf

Rutgers University

May 28, 2009

Julia Wolf The minimum number of monochromatic 4-term progressions

Remarks

Monochromatic 3-term progressionsMonochromatic 4-term progressions

1 IntroductionMonochromatic 3-term progressionsMonochromatic 4-term progressions

2 An analytic upper boundA crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions

3 A combinatorial lower boundA simple counting argument...... with a slight tweak.

4 An analogue in the world of graphsThomason’s construction(s)Giraud’s lower bound

5 Remarks

Remarks

The discrete Fourier transform

Let us recall the definition of the discrete Fourier transform.

Fourier transform: f (t) := Ex∈Zp f (x)ωtx

Fourier inversion: f (x) =∑

t∈cZpf (t)ω−tx

Parseval’s identity: Ex∈Zp |f (x)|2 =∑

t∈cZp|f (t)|2

Note that 1A(0) = α whenever A ⊆ Zp of density α.

Remarks

t∈cZpf (t)ω−tx

t∈cZp|f (t)|2

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t∈cZpf (t)ω−tx

t∈cZp|f (t)|2

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t∈cZpf (t)ω−tx

t∈cZp|f (t)|2

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t∈cZpf (t)ω−tx

t∈cZp|f (t)|2

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Counting monochromatic 3-term progressions

If Zp is 2-colored and one of the color classes has density α, thenthere are precisely 1

2(α3 + (1− α)3)p2 monochromatic 3-termprogressions.

As a trivial consequence we have:

If Zp is 2-colored, then there are at least 18p2 monochromatic

3-term progressions.

Remarks

The number of monochromatic 3-term progression equals

Ex ,d∈Zp 1A(x)1A(x+d)1A(x+2d)+Ex ,d∈Zp 1AC (x)1AC (x+d)1AC (x+2d)

=∑t∈cZp

1A(t)21A(−2t) +∑t∈cZp

1AC (t)21AC (−2t)

= α3 + (1− α)3

since 1A(t) = −1AC (t) for t 6= 0.

Remarks

=∑t∈cZp

1A(t)21A(−2t) +∑t∈cZp

1AC (t)21AC (−2t)

= α3 + (1− α)3

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=∑t∈cZp

1A(t)21A(−2t) +∑t∈cZp

1AC (t)21AC (−2t)

= α3 + (1− α)3

Remarks

=∑t∈cZp

1A(t)21A(−2t) +∑t∈cZp

1AC (t)21AC (−2t)

= α3 + (1− α)3

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=∑t∈cZp

1A(t)21A(−2t) +∑t∈cZp

1AC (t)21AC (−2t)

= α3 + (1− α)3

Remarks

Question

Is there a simple formula for 4-term progressions?

The Fourier transform is not sufficient for counting 4-termprogressions in dense sets. → quadratic Fourier analysis

Because we are using 2 colors only, the coloring problem isclosely related to density problems such as Szemeredi’stheorem for longer progressions.

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Question

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Question

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Question

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Theorem

There exists a 2-coloring of Zp with fewer than

(1− 1

monochromatic 4-term progressions.

Any 2-coloring of Zp contains at least

Remarks

Theorem

There exists a 2-coloring of Zp with fewer than

(1− 1

Any 2-coloring of Zp contains at least

Remarks

A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions

Counting 3-term arithmetic progressions in dense sets

Definition

We say a set A ⊆ G is uniform if the largest non-trivial Fouriercoefficient of its characteristic function is small.

If a subset A of G of density α is uniform, then it contains theexpected number α3 of 3-term progressions.

Ex ,d∈Zp 1A(x)1A(x+d)1A(x+2d) =∑t∈cZp

1A(t)21A(−2t) = α3+o(1)

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Definition

1A(t)21A(−2t) = α3+o(1)

Remarks

Definition

Ex ,d∈Zp 1A(x)1A(x+d)1A(x+2d)

=∑t∈cZp

1A(t)21A(−2t) = α3+o(1)

Remarks

Definition

1A(t)21A(−2t)

= α3+o(1)

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Definition

1A(t)21A(−2t) = α3+o(1)

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Counting 4-term progressions in dense sets

The same is true for any linear configuration defined by a singlelinear equation. However:

Fourier analysis is not sufficient for counting longer progressions.

For example, the following set is uniform in the Fourier sense butcontains many MORE than the expected number of 4-APs.

A = {x ∈ Zp : |x2| small}

x2 − 3(x + d)2 + 3(x + 2d)2 − (x + 3d)2 = 0

Remarks

x2 − 3(x + d)2 + 3(x + 2d)2 − (x + 3d)2 = 0

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x2 − 3(x + d)2 + 3(x + 2d)2 − (x + 3d)2 = 0

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x2 − 3(x + d)2 + 3(x + 2d)2 − (x + 3d)2 = 0

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This is the starting point of Gowers’s analytic proof of Szemeredi’sTheorem for longer progressions.

→ uniformity norms, inverse theorems

Question

Are there any subsets of Zp that are uniform but contain FEWERthan the expected number of 4-term progressions?

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Question

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Question

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Gowers’s construction

Yes. We will briefly sketch Gowers’s construction:

We take a small example and ”blow it up”.

On the interval [1, 18], let f take successive values

−1,−1,−1, 1,−1,−1, 1,−1,−1,−1,−1, 1, 1, 1,−1, 1,−1,−1

so that ∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d) = −36.

Remarks

−1,−1,−1, 1,−1,−1, 1,−1,−1,−1,−1, 1, 1, 1,−1, 1,−1,−1

so that ∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d) = −36.

Remarks

−1,−1,−1, 1,−1,−1, 1,−1,−1,−1,−1, 1, 1, 1,−1, 1,−1,−1

so that ∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d) = −36.

Remarks

−1,−1,−1, 1,−1,−1, 1,−1,−1,−1,−1, 1, 1, 1,−1, 1,−1,−1

so that ∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d) = −36.

Remarks

Define a function F : Zp → {−1, 0, 1} by setting F (x) = f (t)whenever x ∈ It , where It stands for the interval[(2t − 1)m, 2tm] and m is a positive integer betweenp/(5× 18) and p/(4× 18).

The 4-AP counts of F and f are related via∑x ,d

F (x)F (x + d)F (x + 2d)F (x + 3d)

= s∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d),

where s ≥ m2/9.

Remarks

Define a function F : Zp → {−1, 0, 1} by setting F (x) = f (t)whenever x ∈ It , where It stands for the interval[(2t − 1)m, 2tm] and m is a positive integer betweenp/(5× 18) and p/(4× 18).

The 4-AP counts of F and f are related via∑x ,d

F (x)F (x + d)F (x + 2d)F (x + 3d)

= s∑x ,d

f (x)f (x + d)f (x + 2d)f (x + 3d),

where s ≥ m2/9.

Remarks

Next multiply F by an appropriate sum of quadraticexponentials to make it uniform:

G (x) = F (x)(ωx2+ ω3x2

+ ω−3x2+ ω−x2

Finally, turn the function G into a set A ⊆ Zp via thestandard procedure of choosing an element x to lie in A ⊆ Zp

with probability (1 + G (x))/2.

With high probability the resulting set A is uniform but contains atmost

1/16(1− 36/9(5× 18)2)p2

Remarks

G (x) = F (x)(ωx2+ ω3x2

+ ω−3x2+ ω−x2

1/16(1− 36/9(5× 18)2)p2

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G (x) = F (x)(ωx2+ ω3x2

+ ω−3x2+ ω−x2

1/16(1− 36/9(5× 18)2)p2

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G (x) = F (x)(ωx2+ ω3x2

+ ω−3x2+ ω−x2

1/16(1− 36/9(5× 18)2)p2

Remarks

What does this mean for monochromatic progressions?

The number of monochromatic 4-term progressions equals