The minimum number of monochromatic 4-term progressions
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1 The minimum number of monochromatic 4-term progressions Rutgers University May 28, 2009
2 Monochromatic 3-term progressions Monochromatic 4-term progressions 1 Introduction Monochromatic 3-term progressions Monochromatic 4-term progressions 2 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions 3 A simple counting argument with a slight tweak. 4 Thomason s construction(s) Giraud s lower bound 5
3 The discrete Fourier transform Monochromatic 3-term progressions Monochromatic 4-term progressions Let us recall the definition of the discrete Fourier transform.
4 The discrete Fourier transform Monochromatic 3-term progressions Monochromatic 4-term progressions Let us recall the definition of the discrete Fourier transform. Fourier transform: f (t) := E x Zp f (x)ω tx
5 The discrete Fourier transform Monochromatic 3-term progressions Monochromatic 4-term progressions Let us recall the definition of the discrete Fourier transform. Fourier transform: f (t) := E x Zp f (x)ω tx Fourier inversion: f (x) = t c Z p f (t)ω tx
6 The discrete Fourier transform Monochromatic 3-term progressions Monochromatic 4-term progressions Let us recall the definition of the discrete Fourier transform. Fourier transform: f (t) := E x Zp f (x)ω tx Fourier inversion: f (x) = t Z c p f (t)ω tx Parseval s identity: E x Zp f (x) 2 = t Z c p f (t) 2
7 The discrete Fourier transform Monochromatic 3-term progressions Monochromatic 4-term progressions Let us recall the definition of the discrete Fourier transform. Fourier transform: f (t) := E x Zp f (x)ω tx Fourier inversion: f (x) = t Z c p f (t)ω tx Parseval s identity: E x Zp f (x) 2 = t Z c p f (t) 2 Note that 1 A (0) = α whenever A Z p of density α.
8 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions Fact If Z p is 2-colored and one of the color classes has density α, then there are precisely 1 2 (α3 + (1 α) 3 )p 2 monochromatic 3-term progressions.
9 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions Fact If Z p is 2-colored and one of the color classes has density α, then there are precisely 1 2 (α3 + (1 α) 3 )p 2 monochromatic 3-term progressions. As a trivial consequence we have:
10 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions Fact If Z p is 2-colored and one of the color classes has density α, then there are precisely 1 2 (α3 + (1 α) 3 )p 2 monochromatic 3-term progressions. As a trivial consequence we have: Fact If Z p is 2-colored, then there are at least 1 8 p2 monochromatic 3-term progressions.
11 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions The number of monochromatic 3-term progression equals
12 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions The number of monochromatic 3-term progression equals E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d)+e x,d Zp 1 A C (x)1 A C (x+d)1 A C (x+2d)
13 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions The number of monochromatic 3-term progression equals E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d)+e x,d Zp 1 A C (x)1 A C (x+d)1 A C (x+2d) = t c Zp 1 A (t) 2 1 A ( 2t) + t czp 1 A C (t) 2 1 A C ( 2t)
14 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions The number of monochromatic 3-term progression equals E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d)+e x,d Zp 1 A C (x)1 A C (x+d)1 A C (x+2d) = t c Zp 1 A (t) 2 1 A ( 2t) + t czp 1 A C (t) 2 1 A C ( 2t) = α 3 + (1 α) 3
15 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 3-term progressions The number of monochromatic 3-term progression equals E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d)+e x,d Zp 1 A C (x)1 A C (x+d)1 A C (x+2d) = t c Zp 1 A (t) 2 1 A ( 2t) + t czp 1 A C (t) 2 1 A C ( 2t) since 1 A (t) = 1 A C (t) for t 0. = α 3 + (1 α) 3
16 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 4-term progressions Question Is there a simple formula for 4-term progressions?
17 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 4-term progressions Question Is there a simple formula for 4-term progressions? No.
18 Monochromatic 3-term progressions Monochromatic 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-term progressions in dense sets. quadratic Fourier analysis
19 Monochromatic 3-term progressions Monochromatic 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-term progressions in dense sets. quadratic Fourier analysis Because we are using 2 colors only, the coloring problem is closely related to density problems such as Szemerédi s theorem for longer progressions.
20 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 4-term progressions Theorem There exists a 2-coloring of Z p with fewer than ( ) p monochromatic 4-term progressions.
21 Monochromatic 3-term progressions Monochromatic 4-term progressions Counting monochromatic 4-term progressions Theorem There exists a 2-coloring of Z p with fewer than ( ) p monochromatic 4-term progressions. Any 2-coloring of Z p contains at least 1 32 p2 monochromatic 4-term progressions.
22 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 Fourier coefficient of its characteristic function is small.
23 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 Fourier coefficient of its characteristic function is small. Fact If a subset A of G of density α is uniform, then it contains the expected number α 3 of 3-term progressions.
24 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 Fourier coefficient of its characteristic function is small. Fact If a subset A of G of density α is uniform, then it contains the expected number α 3 of 3-term progressions. E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d)
25 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 Fourier coefficient of its characteristic function is small. Fact If a subset A of G of density α is uniform, then it contains the expected number α 3 of 3-term progressions. E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d) = t c Zp 1 A (t) 2 1 A ( 2t)
26 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 Fourier coefficient of its characteristic function is small. Fact If a subset A of G of density α is uniform, then it contains the expected number α 3 of 3-term progressions. E x,d Zp 1 A (x)1 A (x+d)1 A (x+2d) = t c Zp 1 A (t) 2 1 A ( 2t) = α 3 +o(1)
27 Counting 4-term progressions in dense sets A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions The same is true for any linear configuration defined by a single linear equation. However:
28 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Counting 4-term progressions in dense sets The same is true for any linear configuration defined by a single linear equation. However: Fact Fourier analysis is not sufficient for counting longer progressions.
29 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Counting 4-term progressions in dense sets The same is true for any linear configuration defined by a single linear equation. However: Fact Fourier analysis is not sufficient for counting longer progressions. For example, the following set is uniform in the Fourier sense but contains many MORE than the expected number of 4-APs. A = {x Z p : x 2 small}
30 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Counting 4-term progressions in dense sets The same is true for any linear configuration defined by a single linear equation. However: Fact Fourier analysis is not sufficient for counting longer progressions. For example, the following set is uniform in the Fourier sense but contains many MORE than the expected number of 4-APs. A = {x Z p : x 2 small} x 2 3(x + d) 2 + 3(x + 2d) 2 (x + 3d) 2 = 0
31 Counting 4-term progressions in dense sets A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions This is the starting point of Gowers s analytic proof of Szemerédi s Theorem for longer progressions.
32 Counting 4-term progressions in dense sets A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions This is the starting point of Gowers s analytic proof of Szemerédi s Theorem for longer progressions. uniformity norms, inverse theorems
33 Counting 4-term progressions in dense sets A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions This is the starting point of Gowers s analytic proof of Szemerédi s Theorem for longer progressions. uniformity norms, inverse theorems Question Are there any subsets of Z p that are uniform but contain FEWER than the expected number of 4-term progressions?
34 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Yes. We will briefly sketch Gowers s construction:
35 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Yes. We will briefly sketch Gowers s construction: We take a small example and blow it up.
36 A crash course in quadratic Fourier analysis Gowers s construction Implications 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
37 A crash course in quadratic Fourier analysis Gowers s construction Implications 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 f (x)f (x + d)f (x + 2d)f (x + 3d) = 36. x,d
38 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Define a function F : Z p { 1, 0, 1} by setting F (x) = f (t) whenever x I t, where I t stands for the interval [(2t 1)m, 2tm] and m is a positive integer between p/(5 18) and p/(4 18).
39 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Define a function F : Z p { 1, 0, 1} by setting F (x) = f (t) whenever x I t, where I t stands for the interval [(2t 1)m, 2tm] and m is a positive integer between p/(5 18) and p/(4 18). The 4-AP counts of F and f are related via F (x)f (x + d)f (x + 2d)F (x + 3d) x,d = s x,d f (x)f (x + d)f (x + 2d)f (x + 3d), where s m 2 /9.
40 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Next multiply F by an appropriate sum of quadratic exponentials to make it uniform:
41 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Next multiply F by an appropriate sum of quadratic exponentials to make it uniform: G(x) = F (x)(ω x2 + ω 3x2 + ω 3x2 + ω x2 )
42 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Next multiply F by an appropriate sum of quadratic exponentials to make it uniform: G(x) = F (x)(ω x2 + ω 3x2 + ω 3x2 + ω x2 ) Finally, turn the function G into a set A Z p via the standard procedure of choosing an element x to lie in A Z p with probability (1 + G(x))/2.
43 Gowers s construction A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions Next multiply F by an appropriate sum of quadratic exponentials to make it uniform: G(x) = F (x)(ω x2 + ω 3x2 + ω 3x2 + ω x2 ) Finally, turn the function G into a set A Z p via the standard procedure of choosing an element x to lie in A Z p with probability (1 + G(x))/2. With high probability the resulting set A is uniform but contains at most 1/16(1 36/9(5 18) 2 )p 2 4-term progressions.
44 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? The number of monochromatic 4-term progressions equals E x,d 1 A (x)1 A (x + d)1 A (x + 2d)1 A (x + 3d)
45 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? The number of monochromatic 4-term progressions equals E x,d 1 A (x)1 A (x + d)1 A (x + 2d)1 A (x + 3d) +E x,d 1 A C (x)1 A C (x + d)1 A C (x + 2d)1 A C (x + 3d)
46 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? The number of monochromatic 4-term progressions equals E x,d 1 A (x)1 A (x + d)1 A (x + 2d)1 A (x + 3d) +E x,d 1 A C (x)1 A C (x + d)1 A C (x + 2d)1 A C (x + 3d) = 1 4α + 6α 2 + 4p 4 (A)
47 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? The number of monochromatic 4-term progressions equals E x,d 1 A (x)1 A (x + d)1 A (x + 2d)1 A (x + 3d) +E x,d 1 A C (x)1 A C (x + d)1 A C (x + 2d)1 A C (x + 3d) = 1 4α + 6α 2 + 4p 4 (A) E x,d 1 A (x)1 A (x+d)1 A (x+2d) E x,d 1 A (x+d)1 A (x+2d)1 A (x+3d)
48 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? The number of monochromatic 4-term progressions equals E x,d 1 A (x)1 A (x + d)1 A (x + 2d)1 A (x + 3d) +E x,d 1 A C (x)1 A C (x + d)1 A C (x + 2d)1 A C (x + 3d) = 1 4α + 6α 2 + 4p 4 (A) E x,d 1 A (x)1 A (x+d)1 A (x+2d) E x,d 1 A (x+d)1 A (x+2d)1 A (x+3d) E x,d 1 A (x)1 A (x +d)1 A (x +3d) E x,d 1 A (x)1 A (x +2d)1 A (x +3d).
49 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? But remember that in a uniform set, all 3-term configurations occurred with the expected frequency α 3.
50 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? But remember that in a uniform set, all 3-term configurations occurred with the expected frequency α 3. Therefore, using Gowers s set to induce the coloring, we have that the number of monochromatic 4-term progressions equals
51 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? But remember that in a uniform set, all 3-term configurations occurred with the expected frequency α 3. Therefore, using Gowers s set to induce the coloring, we have that the number of monochromatic 4-term progressions equals 1 4α + 6α 2 4α 3 + 4p 4 (A),
52 A crash course in quadratic Fourier analysis Gowers s construction Implications for monochromatic 4-term progressions What does this mean for monochromatic progressions? But remember that in a uniform set, all 3-term configurations occurred with the expected frequency α 3. Therefore, using Gowers s set to induce the coloring, we have that the number of monochromatic 4-term progressions equals 1 4α + 6α 2 4α 3 + 4p 4 (A), with the number of 4-APs p 4 (A) in A being less than the expected α 4 /2.
53 Counting to get a lower bound A simple counting argument with a slight tweak. An averaging argument using Van der Waerden s Theorem yields at least p2 monochromatic 4-term progressions.
54 Counting to get a lower bound A simple counting argument with a slight tweak. An averaging argument using Van der Waerden s Theorem yields at least p2 monochromatic 4-term progressions. Cameron, Cilleruelo and Serra proved the following in 2005:
55 A simple counting argument with a slight tweak. Counting to get a lower bound An averaging argument using Van der Waerden s Theorem yields at least p2 monochromatic 4-term progressions. Cameron, Cilleruelo and Serra proved the following in 2005: Theorem Any 2-coloring of Z p with p a prime contains at least 1 33 p2 monochromatic 4-term progressions.
56 Counting to get a lower bound A simple counting argument with a slight tweak. We ll give a brief sketch of (a slight refinement of) their argument:
57 Counting to get a lower bound A simple counting argument with a slight tweak. We ll give a brief sketch of (a slight refinement of) their argument: For i = 0, 1, 2, 3, 4, let c i denote the proportion of 4-term progressions in Z p which have precisely i red elements.
58 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 c i denote the proportion of 4-term progressions in Z p which have precisely i red elements. Lemma With the c i defined as above, we have that 4(c 0 + c 4 ) + (c 1 + c 3 ) = 4(1 3α + 3α 2 ) for any coloring of Z p in which the red color class has density α.
59 Counting to get a lower bound A simple counting argument with a slight tweak. The preceding lemma together with the identity 4 i=0 c i = 1 now implies that ( ) c 0 + c 4 = 1 3 c 2 + (1 4α + 4α 2 ).
60 Counting to get a lower bound A simple counting argument with a slight tweak. The preceding lemma together with the identity 4 i=0 c i = 1 now implies that ( ) c 0 + c 4 = 1 3 c 2 + (1 4α + 4α 2 ). Cameron, Cilleruelo and Serra immediately discarded the second term on the right-hand side, which is indeed equal to zero for α = 1/2.
61 Counting to get a lower bound A simple counting argument with a slight tweak. The preceding lemma together with the identity 4 i=0 c i = 1 now implies that ( ) c 0 + c 4 = 1 3 c 2 + (1 4α + 4α 2 ). Cameron, Cilleruelo and Serra immediately discarded the second term on the right-hand side, which is indeed equal to zero for α = 1/2. They then went on to observe that one only needs to color 7 points in arithmetic progression before one is guaranteed to find a monochromatic 4-AP or one which is evenly colored.
62 Counting to get a lower bound A simple counting argument with a slight tweak. The preceding lemma together with the identity 4 i=0 c i = 1 now implies that ( ) c 0 + c 4 = 1 3 c 2 + (1 4α + 4α 2 ). Cameron, Cilleruelo and Serra immediately discarded the second term on the right-hand side, which is indeed equal to zero for α = 1/2. They then went on to observe that one only needs to color 7 points in arithmetic progression before one is guaranteed to find a monochromatic 4-AP or one which is evenly colored. This together with ( ) yields a lower bound on c 0 + c 4.
63 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and x, x + d, x + 2d, b lie in arithmetic progression.
64 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and x, x + d, x + 2d, b lie in arithmetic progression. We shall call the pair (a, b) a frame pair.
65 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and 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 a unique 3-term progression.
66 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and 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 a unique 3-term progression. Note that in these statements we have used the assumption that p is prime.
67 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and 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 a unique 3-term progression. Note that in these statements we have used the assumption that p is prime. Two 4-APs containing S have different color parities if and only if the frame pair of S is bichromatic.
68 The crucial observation A simple counting argument with a slight tweak. Any 3-term progression S of the form x, x + d, x + 2d determines a unique (unordered) pair of points (a, b) such that the five points and each of the quadruples a, x, x + d, x + 2d and 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 a unique 3-term progression. Note that in these statements we have used the assumption that p is prime. Two 4-APs containing S have different color parities if and only if the frame pair of S is bichromatic. The total number of monochromatic pairs is at its minimum for densities close to 1/2.
69 Thomason s construction(s) Giraud s lower bound Question If the edges of K n are 2-colored, what is the minimum number of monochromatic K 4 s?
70 Thomason s construction(s) Giraud s lower bound Question If the edges of K n are 2-colored, what is the minimum number of monochromatic K 4 s? It was conjectured by Erdős that the number of monochromatic K 4 s is always at least the number expected in the random case.
71 Thomason s construction(s) Giraud s lower bound Question If the edges of K n are 2-colored, what is the minimum number of monochromatic K 4 s? It was conjectured by Erdős that the number of monochromatic K 4 s is always at least the number expected in the random case. It is not difficult to see that this is true for triangles.
72 A disproof of Erdős s conjecture Thomason s construction(s) Giraud s lower bound Thomason disproved Erdős s conjecture in 1989.
73 Thomason s construction(s) Giraud s lower bound A disproof of Erdős s conjecture Thomason disproved Erdős s conjecture in Theorem There exists a 2-coloring of K n for which the minimum number of monochromatic K 4 s is at most 1 33 n4.
74 Thomason s construction(s) Giraud s lower bound A disproof of Erdős s conjecture Thomason disproved Erdős s conjecture in Theorem There exists a 2-coloring of K n for which the minimum number of monochromatic K 4 s is at most 1 33 n4. The initial construction is rather obscure, using a quadratic form over a finite field.
75 Thomason s construction(s) Giraud s lower bound A disproof of Erdős s conjecture Thomason disproved Erdős s conjecture in Theorem There exists a 2-coloring of K n for which the minimum number of monochromatic K 4 s is at most 1 33 n4. The initial construction is rather obscure, using a quadratic form over a finite field. It was subsequently simplified and rephrased.
76 Thomason s construction(s) Giraud s lower bound A disproof of Erdős s conjecture Thomason disproved Erdős s conjecture in Theorem There exists a 2-coloring of K n for which the minimum number of monochromatic K 4 s is at most 1 33 n4. The initial construction is rather obscure, using a quadratic form over a finite field. It was subsequently simplified and rephrased. Now many examples are known computationally, but a theoretical framework is lacking.
77 Thomason s construction(s) Giraud s lower bound A simple but very weak lower bound can be given via Ramsey s Theorem.
78 Thomason s construction(s) Giraud s lower bound A simple but very weak lower bound can be given via Ramsey s Theorem. Giraud had proved the following lower bound in 1979:
79 Thomason s construction(s) Giraud s lower bound A simple but very weak lower bound can be given via Ramsey s Theorem. Giraud had proved the following lower bound in 1979: Theorem Given a 2-coloring of K n the minimum number of monochromatic K 4 s is at least 1 46 n4.
80 Thomason s construction(s) Giraud s lower bound A simple but very weak lower bound can be given via Ramsey s Theorem. Giraud had proved the following lower bound in 1979: Theorem Given a 2-coloring of K n the minimum number of monochromatic K 4 s is at least 1 46 n4. The proof proceeds via ingenious combinatorial counting and an optimization.
81 Open problems A wide gap between upper and lower bounds persists for the problem in both sets and graphs.
82 Open problems A wide gap between upper and lower bounds persists for the problem in both sets and graphs. It would be interesting to make the analogy between the two more precise.
83 Open problems A wide gap between upper and lower bounds persists for the problem in both sets and graphs. It would be interesting to make the analogy between the two more precise. For the integers 1, 2,..., n instead of Z p, the problem is unsolved even for 3-term progressions. Schur triples
84 Bibliography P. Cameron, J. Cilleruelo, and O. Serra. On monochromatic solutions of equations in groups, Available at pjc/papers.html. G. Giraud. Sur le problème de Goodman pour les quadrangles et la majoration des nombres de Ramsey, J. Combin. Theory Ser. B, 27(3):237253, W.T. Gowers. Two examples in additive combinatorics, Unpublished. B.J. Green. Montreal lecture notes on quadratic Fourier analysis, Available at A.G. Thomason. Graph products and monochromatic multiplicities, Combinatorica, 17(1):125134, 1997.
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