the minimum number of monochromatic 4-term progressions · since c1 a(t) = 1d c (t) for t 6= 0....
TRANSCRIPT
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
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
Fact
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:
Fact
If Zp is 2-colored, then there are at least 18p2 monochromatic
3-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
Fact
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:
Fact
If Zp is 2-colored, then there are at least 18p2 monochromatic
3-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
Fact
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:
Fact
If Zp is 2-colored, then there are at least 18p2 monochromatic
3-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 3-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Question
Is there a simple formula for 4-term progressions?
No.
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Question
Is there a simple formula for 4-term progressions?
No.
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Question
Is there a simple formula for 4-term progressions?
No.
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Question
Is there a simple formula for 4-term progressions?
No.
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Theorem
There exists a 2-coloring of Zp with fewer than
1
16
(1− 1
2025
)p2
monochromatic 4-term progressions.
Any 2-coloring of Zp contains at least
1
32p2
monochromatic 4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Monochromatic 3-term progressionsMonochromatic 4-term progressions
Counting monochromatic 4-term progressions
Theorem
There exists a 2-coloring of Zp with fewer than
1
16
(1− 1
2025
)p2
monochromatic 4-term progressions.
Any 2-coloring of Zp contains at least
1
32p2
monochromatic 4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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.
Fact
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)
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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.
Fact
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)
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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.
Fact
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)
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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.
Fact
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)
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
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.
Fact
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)
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
The same is true for any linear configuration defined by a singlelinear equation. However:
Fact
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
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
The same is true for any linear configuration defined by a singlelinear equation. However:
Fact
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
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
The same is true for any linear configuration defined by a singlelinear equation. However:
Fact
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
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
The same is true for any linear configuration defined by a singlelinear equation. However:
Fact
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
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
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?
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
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?
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Counting 4-term progressions in dense sets
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?
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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
4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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
4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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
4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
Gowers’s construction
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
4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
The number of monochromatic 4-term progressions equals
Ex ,d1A(x)1A(x + d)1A(x + 2d)1A(x + 3d)
+Ex ,d1AC (x)1AC (x + d)1AC (x + 2d)1AC (x + 3d)
= 1− 4α + 6α2 + 4p4(A)
−Ex ,d1A(x)1A(x+d)1A(x+2d)−Ex ,d1A(x+d)1A(x+2d)1A(x+3d)
−Ex ,d1A(x)1A(x +d)1A(x +3d)−Ex ,d1A(x)1A(x +2d)1A(x +3d).
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
The number of monochromatic 4-term progressions equals
Ex ,d1A(x)1A(x + d)1A(x + 2d)1A(x + 3d)
+Ex ,d1AC (x)1AC (x + d)1AC (x + 2d)1AC (x + 3d)
= 1− 4α + 6α2 + 4p4(A)
−Ex ,d1A(x)1A(x+d)1A(x+2d)−Ex ,d1A(x+d)1A(x+2d)1A(x+3d)
−Ex ,d1A(x)1A(x +d)1A(x +3d)−Ex ,d1A(x)1A(x +2d)1A(x +3d).
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
The number of monochromatic 4-term progressions equals
Ex ,d1A(x)1A(x + d)1A(x + 2d)1A(x + 3d)
+Ex ,d1AC (x)1AC (x + d)1AC (x + 2d)1AC (x + 3d)
= 1− 4α + 6α2 + 4p4(A)
−Ex ,d1A(x)1A(x+d)1A(x+2d)−Ex ,d1A(x+d)1A(x+2d)1A(x+3d)
−Ex ,d1A(x)1A(x +d)1A(x +3d)−Ex ,d1A(x)1A(x +2d)1A(x +3d).
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
The number of monochromatic 4-term progressions equals
Ex ,d1A(x)1A(x + d)1A(x + 2d)1A(x + 3d)
+Ex ,d1AC (x)1AC (x + d)1AC (x + 2d)1AC (x + 3d)
= 1− 4α + 6α2 + 4p4(A)
−Ex ,d1A(x)1A(x+d)1A(x+2d)−Ex ,d1A(x+d)1A(x+2d)1A(x+3d)
−Ex ,d1A(x)1A(x +d)1A(x +3d)−Ex ,d1A(x)1A(x +2d)1A(x +3d).
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
The number of monochromatic 4-term progressions equals
Ex ,d1A(x)1A(x + d)1A(x + 2d)1A(x + 3d)
+Ex ,d1AC (x)1AC (x + d)1AC (x + 2d)1AC (x + 3d)
= 1− 4α + 6α2 + 4p4(A)
−Ex ,d1A(x)1A(x+d)1A(x+2d)−Ex ,d1A(x+d)1A(x+2d)1A(x+3d)
−Ex ,d1A(x)1A(x +d)1A(x +3d)−Ex ,d1A(x)1A(x +2d)1A(x +3d).
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
But remember that in a uniform set, all 3-term configurationsoccurred with the expected frequency α3.
Therefore, using Gowers’s set to induce the coloring, we have thatthe number of monochromatic 4-term progressions equals
1− 4α + 6α2 − 4α3 + 4p4(A),
with the number of 4-APs p4(A) in A being less than the expectedα4/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
But remember that in a uniform set, all 3-term configurationsoccurred with the expected frequency α3.Therefore, using Gowers’s set to induce the coloring, we have thatthe number of monochromatic 4-term progressions equals
1− 4α + 6α2 − 4α3 + 4p4(A),
with the number of 4-APs p4(A) in A being less than the expectedα4/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
But remember that in a uniform set, all 3-term configurationsoccurred with the expected frequency α3.Therefore, using Gowers’s set to induce the coloring, we have thatthe number of monochromatic 4-term progressions equals
1− 4α + 6α2 − 4α3 + 4p4(A),
with the number of 4-APs p4(A) in A being less than the expectedα4/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A crash course in quadratic Fourier analysisGowers’s constructionImplications for monochromatic 4-term progressions
What does this mean for monochromatic progressions?
But remember that in a uniform set, all 3-term configurationsoccurred with the expected frequency α3.Therefore, using Gowers’s set to induce the coloring, we have thatthe number of monochromatic 4-term progressions equals
1− 4α + 6α2 − 4α3 + 4p4(A),
with the number of 4-APs p4(A) in A being less than the expectedα4/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
An averaging argument using Van der Waerden’s Theorem yieldsat least
1
370p2
monochromatic 4-term progressions.
Cameron, Cilleruelo and Serra proved the following in 2005:
Theorem
Any 2-coloring of Zp with p a prime contains at least
1
33p2
monochromatic 4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
An averaging argument using Van der Waerden’s Theorem yieldsat least
1
370p2
monochromatic 4-term progressions.Cameron, Cilleruelo and Serra proved the following in 2005:
Theorem
Any 2-coloring of Zp with p a prime contains at least
1
33p2
monochromatic 4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
An averaging argument using Van der Waerden’s Theorem yieldsat least
1
370p2
monochromatic 4-term progressions.Cameron, Cilleruelo and Serra proved the following in 2005:
Theorem
Any 2-coloring of Zp with p a prime contains at least
1
33p2
monochromatic 4-term progressions.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
We’ll give a brief sketch of (a slight refinement of) their argument:
For i = 0, 1, 2, 3, 4, let ci denote the proportion of 4-termprogressions in Zp which have precisely i red elements.
Lemma
With the ci defined as above, we have that
4(c0 + c4) + (c1 + c3) = 4(1− 3α + 3α2)
for any coloring of Zp in which the red color class has density α.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
We’ll give a brief sketch of (a slight refinement of) their argument:For i = 0, 1, 2, 3, 4, let ci denote the proportion of 4-termprogressions in Zp which have precisely i red elements.
Lemma
With the ci defined as above, we have that
4(c0 + c4) + (c1 + c3) = 4(1− 3α + 3α2)
for any coloring of Zp in which the red color class has density α.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
We’ll give a brief sketch of (a slight refinement of) their argument:For i = 0, 1, 2, 3, 4, let ci denote the proportion of 4-termprogressions in Zp which have precisely i red elements.
Lemma
With the ci defined as above, we have that
4(c0 + c4) + (c1 + c3) = 4(1− 3α + 3α2)
for any coloring of Zp in which the red color class has density α.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
The preceding lemma together with the identity∑4
i=0 ci = 1 nowimplies that
(∗) c0 + c4 =1
3c2 + (1− 4α + 4α2).
Cameron, Cilleruelo and Serra immediately discarded the secondterm on the right-hand side, which is indeed equal to zero forα = 1/2.They then went on to observe that one only needs to color 7points in arithmetic progression before one is guaranteed to find amonochromatic 4-AP or one which is evenly colored.This together with (∗) yields a lower bound on c0 + c4.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
The preceding lemma together with the identity∑4
i=0 ci = 1 nowimplies that
(∗) c0 + c4 =1
3c2 + (1− 4α + 4α2).
Cameron, Cilleruelo and Serra immediately discarded the secondterm on the right-hand side, which is indeed equal to zero forα = 1/2.
They then went on to observe that one only needs to color 7points in arithmetic progression before one is guaranteed to find amonochromatic 4-AP or one which is evenly colored.This together with (∗) yields a lower bound on c0 + c4.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
The preceding lemma together with the identity∑4
i=0 ci = 1 nowimplies that
(∗) c0 + c4 =1
3c2 + (1− 4α + 4α2).
Cameron, Cilleruelo and Serra immediately discarded the secondterm on the right-hand side, which is indeed equal to zero forα = 1/2.They then went on to observe that one only needs to color 7points in arithmetic progression before one is guaranteed to find amonochromatic 4-AP or one which is evenly colored.
This together with (∗) yields a lower bound on c0 + c4.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
Counting to get a lower bound
The preceding lemma together with the identity∑4
i=0 ci = 1 nowimplies that
(∗) c0 + c4 =1
3c2 + (1− 4α + 4α2).
Cameron, Cilleruelo and Serra immediately discarded the secondterm on the right-hand side, which is indeed equal to zero forα = 1/2.They then went on to observe that one only needs to color 7points in arithmetic progression before one is guaranteed to find amonochromatic 4-AP or one which is evenly colored.This together with (∗) yields a lower bound on c0 + c4.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
A simple counting argument...... with a slight tweak.
The crucial observation
Any 3-term progression S of the form x , x + d , x + 2ddetermines a unique (unordered) pair of points (a, b) such thatthe five points and each of the quadruples a, x , x + d , x + 2dand x , x + d , x + 2d , b lie in arithmetic progression.
We shall call the pair (a, b) a frame pair.
It is straightforward to see that each frame pair belongs to aunique 3-term progression.
Note that in these statements we have used the assumptionthat p is prime.
Two 4-APs containing S have different color parities if andonly if the frame pair of S is bichromatic.
The total number of monochromatic pairs is at its minimumfor densities close to 1/2.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
An analogue in the world of graphs
Question
If the edges of Kn are 2-colored, what is the minimum number ofmonochromatic K4s?
It was conjectured by Erdos that the number ofmonochromatic K4s is always at least the number expected inthe random case.
It is not difficult to see that this is true for triangles.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
An analogue in the world of graphs
Question
If the edges of Kn are 2-colored, what is the minimum number ofmonochromatic K4s?
It was conjectured by Erdos that the number ofmonochromatic K4s is always at least the number expected inthe random case.
It is not difficult to see that this is true for triangles.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
An analogue in the world of graphs
Question
If the edges of Kn are 2-colored, what is the minimum number ofmonochromatic K4s?
It was conjectured by Erdos that the number ofmonochromatic K4s is always at least the number expected inthe random case.
It is not difficult to see that this is true for triangles.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A disproof of Erdos’ s conjecture
Thomason disproved Erdos’s conjecture in 1989.
Theorem
There exists a 2-coloring of Kn for which the minimum number ofmonochromatic K4s is at most
1
33n4.
The initial construction is rather obscure, using a quadraticform over a finite field.
It was subsequently simplified and rephrased.
Now many examples are known computationally, but atheoretical framework is lacking.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A disproof of Erdos’ s conjecture
Thomason disproved Erdos’s conjecture in 1989.
Theorem
There exists a 2-coloring of Kn for which the minimum number ofmonochromatic K4s is at most
1
33n4.
The initial construction is rather obscure, using a quadraticform over a finite field.
It was subsequently simplified and rephrased.
Now many examples are known computationally, but atheoretical framework is lacking.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A disproof of Erdos’ s conjecture
Thomason disproved Erdos’s conjecture in 1989.
Theorem
There exists a 2-coloring of Kn for which the minimum number ofmonochromatic K4s is at most
1
33n4.
The initial construction is rather obscure, using a quadraticform over a finite field.
It was subsequently simplified and rephrased.
Now many examples are known computationally, but atheoretical framework is lacking.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A disproof of Erdos’ s conjecture
Thomason disproved Erdos’s conjecture in 1989.
Theorem
There exists a 2-coloring of Kn for which the minimum number ofmonochromatic K4s is at most
1
33n4.
The initial construction is rather obscure, using a quadraticform over a finite field.
It was subsequently simplified and rephrased.
Now many examples are known computationally, but atheoretical framework is lacking.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A disproof of Erdos’ s conjecture
Thomason disproved Erdos’s conjecture in 1989.
Theorem
There exists a 2-coloring of Kn for which the minimum number ofmonochromatic K4s is at most
1
33n4.
The initial construction is rather obscure, using a quadraticform over a finite field.
It was subsequently simplified and rephrased.
Now many examples are known computationally, but atheoretical framework is lacking.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A combinatorial lower bound
A simple but very weak lower bound can be given via Ramsey’sTheorem.
Giraud had proved the following lower bound in 1979:
Theorem
Given a 2-coloring of Kn the minimum number of monochromaticK4s is at least
1
46n4.
The proof proceeds via ingenious combinatorial counting and anoptimization.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A combinatorial lower bound
A simple but very weak lower bound can be given via Ramsey’sTheorem.Giraud had proved the following lower bound in 1979:
Theorem
Given a 2-coloring of Kn the minimum number of monochromaticK4s is at least
1
46n4.
The proof proceeds via ingenious combinatorial counting and anoptimization.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A combinatorial lower bound
A simple but very weak lower bound can be given via Ramsey’sTheorem.Giraud had proved the following lower bound in 1979:
Theorem
Given a 2-coloring of Kn the minimum number of monochromaticK4s is at least
1
46n4.
The proof proceeds via ingenious combinatorial counting and anoptimization.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Thomason’s construction(s)Giraud’s lower bound
A combinatorial lower bound
A simple but very weak lower bound can be given via Ramsey’sTheorem.Giraud had proved the following lower bound in 1979:
Theorem
Given a 2-coloring of Kn the minimum number of monochromaticK4s is at least
1
46n4.
The proof proceeds via ingenious combinatorial counting and anoptimization.
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Open problems
A wide gap between upper and lower bounds persists for theproblem in both sets and graphs.
It would be interesting to make the analogy between the twomore precise.
For the integers 1, 2, . . . , n instead of Zp, the problem isunsolved even for 3-term progressions. → Schur triples
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Open problems
A wide gap between upper and lower bounds persists for theproblem in both sets and graphs.
It would be interesting to make the analogy between the twomore precise.
For the integers 1, 2, . . . , n instead of Zp, the problem isunsolved even for 3-term progressions. → Schur triples
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Open problems
A wide gap between upper and lower bounds persists for theproblem in both sets and graphs.
It would be interesting to make the analogy between the twomore precise.
For the integers 1, 2, . . . , n instead of Zp, the problem isunsolved even for 3-term progressions. → Schur triples
Julia Wolf The minimum number of monochromatic 4-term progressions
IntroductionAn analytic upper bound
A combinatorial lower boundAn analogue in the world of graphs
Remarks
Bibliography
P. Cameron, J. Cilleruelo, and O. Serra. On monochromaticsolutions of equations in groups, 2005. Available athttp://www.maths.qmw.ac.uk/∼pjc/papers.html.
G. Giraud. Sur le probleme de Goodman pour les quadrangleset la majoration des nombres de Ramsey, J. Combin. TheorySer. B, 27(3):237253, 1979.
W.T. Gowers. Two examples in additive combinatorics, 2005.Unpublished.
B.J. Green. Montreal lecture notes on quadratic Fourieranalysis, 2006. Available athttp://arxiv.org/abs/math/0604089.
A.G. Thomason. Graph products and monochromaticmultiplicities, Combinatorica, 17(1):125134, 1997.
Julia Wolf The minimum number of monochromatic 4-term progressions