Ramsey theory. Andrés Eduardo Caicedo. Graduate Student Seminar, October 19, Department of Mathematics Boise State University

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1 Andrés Eduardo Department of Mathematics Boise State University Graduate Student Seminar, October 19, 2011

2 Thanks to the NSF for partial support through grant DMS My work is mostly in set theory, but I am also interested in finite combinatorics. can be seen as part of both these fields.

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4 Whereas the entropy theorems of probability theory and mathematical physics imply that, in a large universe, disorder is probable, certain combinatorial theorems show that complete disorder is impossible. T.S. Motzkin (1967)

5 Let s begin with an example. The following question was asked by Neil Hindman in 1979; it is still open. Suppose that the positive integers are partitioned into finitely many pieces, N = A 1 A 2 A n. Must there be integers x, y such that x, y, x + y, and xy all belong to the same A i? We usually think of the A i as colors, so we are coloring the positive integers, and we look for a monochromatic set of the form x, y, x + y, xy. To get a feeling of where the difficulty lies, let s think about the case of 2 colors.

6 Let s show that there must be a monochromatic set among the first 48 integers. A 1 A 2

7 Let s show that there must be a monochromatic set among the first 48 integers. A 1 1 A 2

8 Let s show that there must be a monochromatic set among the first 48 integers. A 1 1 A 2 2

9 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2

10 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2 3, 5

11 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2 3, 5

12 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2 3, 5 7

13 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2 3, 5 7

14 Let s show that there must be a monochromatic set among the first 48 integers. A A 2 2 3, 5 7 9, 11, 24

15 Let s show that there must be a monochromatic set among the first 48 integers. A , 14, 18 A 2 2 3, 5 7 9, 11, 24

16 Let s show that there must be a monochromatic set among the first 48 integers. A , 14, 18 A 2 2 3, 5 7 9, 11, 24 16, 19, 48

17 Let s show that there must be a monochromatic set among the first 48 integers. A , 14, 18 A 2 2 3, 5 7 9, 11, 24 16, 19, 48

18 If we require x y, Ronald Graham (with computer assistance) showed that one needs to consider all numbers up to 252 to find a monochromatic set. This involves an analysis of 27 cases. For 3 colors, if we require x y, all we know is that one needs to analyze all numbers up to at least 11,706,659.

19 Hindman was led to this problem after he proved the following result: Given a subset A N, let F S(A) be the set of sums of finite non-empty subsets of A, and let F P (A) be the corresponding set of products. For example, if A = {1, 2, 7}, then F S(A) = {1, 2, 3, 7, 8, 9, 10} and F P (A) = {1, 2, 7, 14}.

20 Theorem (Hindman) If N is colored using finitely many colors, then there is an infinite subset A N such that F S(A) is monochromatic, and there is an infinite subset B N such that F P (B) is monochromatic.

21 The typical framework of a problem in is as follows: Given a certain size, a large structure is partitioned (colored) into a small number of pieces. Must it be the case that there is a monochromatic substructure of the given size?

22 Ramsey s theorem, the result from which the field takes its name, is a generalization of the following statement: Given a number n, let K n denote the complete graph on n-vertices, and let K n denote the graph on n-vertices with no edges. Theorem (Ramsey) For any n, m there is an M so large that any graph on M vertices either contains a copy of K n, or a copy of K m. The smallest such M is the Ramsey number R(n, m). Clearly, R(n, m) = R(m, n), R(1, m) = 1, and R(2, m) = m.

23 Frank Plumpton Ramsey (22 Feb., Jan., 1930). Philosopher, Economist, Mathematician.

24 Statements of the form R(n, m) M are usually stated in the following way: In any party of at least M people, either there are n that know each other, or else there are m that do not know one another. We can also think of them as follows: Color each edge of K M red or blue. Then there is either a red copy of K n, or a blue copy of K m. Let s see some examples.

25 R(3, 3) = 6. That it is at least 6 follows from considering the following graph:

26 R(3, 3) = 6. That it is at most 6 follows from noticing that in any coloring of K 6, there must be 3 edges from the same vertex with the same color:

27 R(4, 3) = 9. That it is at least 9 follows from considering the following graph:

28 R(4, 3) = 9. That it is at most 10 follows from considering the following: To see that it is at most 9 is a bit more involved.

29 R(4, 4) = 18. That it is at least 18 follows from considering the following graph:

30 It seem very likely that the nice regularity of the previous example is an accident, an instance of what Richard K. Guy calls The strong law of small numbers.

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32 R(3, 5) = 14, R(3, 6) = 18, R(3, 7) = 23, R(3, 8) = 28, R(3, 9) = R(3, 10) 43. Theorem (Ajtai-Komlós-Szemerédi, Kim) R(3, t) grows like t 2 log t.

33 R(4, 5) = 25, 35 R(4, 6) 41. That R(4, 5) 25 was proved by Kalbfleisch in That R(4, 5) 25 was proved by McKay and Radziszowski in Again, computers were used in an essential fashion.

34 The two implementations required 3.2 years and 6 years of cpu time on Sun Microsystems computers (mostly Sparcstation SLC). This was achieved without undue delay by employing a large number of computers (up to 110 at once). McKay-Radziszowski (1993)

35 43 R(5, 5) R(6, 6) 165. Suppose aliens invade the Earth and threaten to destroy it in a year if human beings do not find R(5, 5). It is (probably) possible to save the Earth by putting together the world ss best minds and computers. If, however, the invaders were to demand R(6, 6), the human beings might as well attempt a preemptive strike without even trying to ponder the problem. P. Erdős (1993)

36 In general, we know that 2 e 2 n2n/2 (1 + o(1)) < R(n, n) < (n 1) c for an appropriate constant c > 0. log(n 1) log log(n 1) Ronald Graham will reward with $100 a proof that lim R(n, n)1/n n exists, and with $250 a computation of this limit. ( ) 2n 1 n 1

37 Let K Q be the complete graph with set of vertices Q. Theorem (Galvin) Color the edges of K Q using finitely many colors. Then there is a copy of Q that uses at most 2 colors.

38 Let tan(z) = n=0 T n n! zn be the Taylor series expansion of tan(z) about z = 0. Define the sequence (t n 1 n) of odd tangent numbers by t k = T 2k 1, so t 1 = 1, t 2 = 2, t 3 = 16,... Given a set X, let [X] k be the collection of subsets of X of size k. For example, the set of edges of the complete graph K X is [X] 2. Theorem (Devlin) Suppose that [Q] k is colored with finitely many colors. Then there is a copy of Q that uses at most t k many colors, and this is best possible.

39 Indeed one can view as the study of generalizations and repeated applications of the pigeonhole principle. Tao-Vu (2006)

40 The pigeonhole principle If sufficiently many objects are distributed over not too many classes, then at least one class contains many of these objects. The principle was first stated by Dirichlet in 1834, who called it the Schubfachprinzip.

41 Dirichlet used the principle to prove a result on Diophantine approximation. Given a real α, let α denote the closest distance from α to an integer. Theorem (Dirichlet) For any real α and n N there is a positive integer m n such that mα 1 n + 1. If α is irrational, this implies that there are infinitely many fractions n/m such that α n < 1 m m 2. These fractions are known as the convergents of α and are closely related to its continued fraction.

42 For example: π = Its convergents are 3, , , = 3, 22 7, , = 15 3, , ,....

43 Note that π 3 1 < 1, π 22 7 = < 1 49 = , π = < = ,...

44 Thus for any irrational α, we can find infinitely many fractions n m with α n < 1 m m 2. On the other hand, the Thue-Siegel-Roth theorem states that if α is irrational and algebraic, then the equation α n < 1 m m 2+ɛ has only finitely many solutions for any ɛ > 0. One of the key ingredients of the proof is the construction of what is called an auxiliary function. Curiously, one of the main steps of the construction is an application of the pigeonhole principle.

45 (Off-topic) To complete the picture: π is transcendental, i.e., it is not algebraic. Then, it is conceivable that there is some θ > 2 such that π n < 1 m m θ holds infinitely often. The best result we currently know is that, if it exists, any such θ must be less than

46 has many applications in number theory, analysis, and algebra. For example, beyond the pigeonhole principle, the earliest recorded Ramsey theoretic result is due to Hilbert. He used it as a lemma towards a proof of the irreducibility of certain polynomials.

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48 An n-dimensional affine cube is a set Q Z for which there exist a, x 1,..., x n N such that Q = {a + i F x i F {1,..., n}}. Theorem (Hilbert) For every pair of positive integers r, n, there exists an M such that if {1,..., M} is colored with at most r colors, then there is a monochromatic n-dimensional affine cube.

49 Let me close by mentioning one of the most famous results in. Theorem (Van der Waerden) For any n, r there is an N such that if {1, 2,..., N} is split into r pieces, then one of the pieces contains an arithmetic progression of size at least n. Call W (n, r) the smallest possible N.

50 Not much is known in terms of bounds. The original proof is a double induction and gives an Ackermannian bound.

51 Ackermannian bounds are as bad as they get. For example, the diagram above shows that W (3, 3) 7( )(2 3 7( ) + 1) Actually, W (3, 3) = 27.

52 Van der Waerden s proof is from In 1988, Shelah found a new proof that gives much better bounds (still huge). For example, even for two colors, the bounds obtained by Shelah are given by towers of towers of powers of 2. In general, he obtains bounds that are towers of towers of towers of powers of 2.

53 The best known bounds are due to Timothy Gowers, Fields medalist in In work from the early 2000s, Gowers established that W (n, 2) n+9.

54 Ronald Graham conjectures that W (n, 2) 2 n2. He offers $1000 for a proof of the conjecture.

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