Some Problems in Graph Ramsey Theory by Andrey Vadim Grinshpun Submitted to the Department of Mathematics in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics at the MASSACHUSETTS INSTITUTE OF TECHNOLOGY February 2015 c Andrey Vadim Grinshpun, MMXV. All rights reserved. ○ The author hereby grants to MIT permission to reproduce and to distribute publicly paper and electronic copies of this thesis document in whole or in part in any medium now known or hereafter created.
Author...... Department of Mathematics December 12, 2014
Certified by...... Jacob Fox Associate Professor Thesis Supervisor
Accepted by...... Peter Shor Chairman, Applied Mathematics Committee, Department of Mathematics 2 Some Problems in Graph Ramsey Theory by Andrey Vadim Grinshpun
Submitted to the Department of Mathematics on December 12, 2014, in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics
Abstract
A graph 퐺 is 푟-Ramsey minimal with respect to a graph 퐻 if every 푟-coloring of the edges of 퐺 yields a monochromatic copy of 퐻, but the same is not true for any proper subgraph of 퐺. The study of the properties of graphs that are Ramsey minimal with respect to some 퐻 and similar problems is known as graph Ramsey theory; we study several problems in this area. Burr, Erdős, and Lovász introduced 푠(퐻), the minimum over all 퐺 that are 2- Ramsey minimal for 퐻 of 훿(퐺), the minimum degree of 퐺. We find the values of 푠(퐻) for several classes of graphs 퐻, most notably for all 3-connected bipartite graphs which proves many cases of a conjecture due to Szabó, Zumstein, and Zürcher. One natural question when studying graph Ramsey theory is what happens when, rather than considering all 2-colorings of a graph 퐺, we restrict to a subset of the possible 2-colorings. Erdős and Hajnal conjectured that, for any fixed color pattern 퐶, there is some 휀 > 0 so that every 2-coloring of the edges of a 퐾푛, the complete graph on 푛 vertices, which doesn’t contain a copy of 퐶 contains a monochromatic clique on 푛휀 vertices. Hajnal generalized this conjecture to more than 2 colors and asked in particular about the case when the number of colors is 3 and 퐶 is a rainbow triangle (a 퐾3 where each edge is a different color); we prove Hajnal’s conjecture for rainbow triangles. One may also wonder what would happen if we wish to cover all of the vertices with monochromatic copies of graphs. Let = 퐹1, 퐹2,... be a sequence of graphs ℱ { } such that 퐹푛 is a graph on 푛 vertices with maximum degree at most ∆. If each 퐹푛 is bipartite, then the vertices of any 2-edge-colored complete graph can be partitioned into at most 2퐶Δ vertex disjoint monochromatic copies of graphs from , where 퐶 is an absolute constant. This result is best possible, up to the constant 퐶ℱ.
Thesis Supervisor: Jacob Fox Title: Associate Professor
3 4 Acknowledgments
I would like to thank my advisor, Jacob Fox, for his support and advice. He has set an example of an exemplary researcher and has dedicated much time to giving me guidance. I also thank my coauthors Jacob Fox, János Pach, Gábor Sárközy, Tibor Szabó, Yury Person, Anita Liebenau, Troy Retter, Rik Sengupta, and Raj Raina for working with me on projects now contained within this thesis. I would like to thank Steven Rudich for encouraging me to pursue mathematical research as a high school student and undergraduate and the Cornell REU for my first formalized opportunity to pursue research. I would like to thank Tom Leighton and Akamai for funding the Akamai fellowship at MIT, which supported me financially during my first year of graduate school, and the NPSC fellowship for supporting me financially from my second year onward. Finally, I am indebted to my family and especially my fiancée for support and advice.
5 6 Contents
1 Introduction 9 1.1 Chapter 2: Minimum Degrees of Minimal Ramsey Graphs ...... 10 1.2 Chapter 3: The Erdős-Hajnal Conjecture for Rainbow Triangles . . . 11 1.3 Chapter 4: Packing Vertex-Disjoint Monochromatic Copies of Sparse Graphs ...... 12
2 Minimum Degrees of Minimal Ramsey Graphs 15 2.1 Introduction ...... 15 2.2 Preliminaries ...... 18 2.3 A large class of Ramsey-simple graphs ...... 19 2.4 Stronger lower bounds for graphs with hanging edges ...... 23 2.5 Upper bounds for complete bipartite graphs with hanging edges . . . 30 2.6 Upper bounds for graphs with hanging edges ...... 36 2.7 The complete graph with an added vertex ...... 45 2.8 Concluding Remarks ...... 55 2.9 Sparse random graphs are Ramsey simple ...... 55 2.10 Random Caley graphs with a pendant edge ...... 58
3 The Erdős-Hajnal Conjecture for Rainbow Triangles 65 3.1 Introduction ...... 65 3.2 Lexicographic product colorings ...... 73 3.3 Simple bounds for three colors ...... 75 3.4 A weighted Ramsey theorem ...... 77
7 3.5 Tight lower bound for three colors ...... 80 3.6 Upper bound for many colors ...... 88 3.7 Weak lower bound for many colors ...... 91 3.8 Lower bound for many colors ...... 93 3.8.1 Discrepancy lemma in edge-weighted graphs ...... 93 3.8.2 Proof of lower bound for many colors ...... 94 3.9 Proof of Lemma 3.4.1 ...... 112 3.10 Proof of Lemma 3.5.3 ...... 114 3.11 Proof of Lemma 3.8.1 ...... 119 3.12 Proof of Lemma 3.8.3 ...... 126
4 Packing Vertex-Disjoint Monochromatic Copies of Sparse Graphs 131 4.1 Introduction ...... 131 4.2 Notation ...... 135 4.3 Regularity and blow-up lemmas ...... 136 4.4 Proofs of Theorems 4.1.2, 4.1.3, and 4.1.4 ...... 138 4.4.1 Proof of Lemma 4.4.3 ...... 140 4.4.2 Applying Lemma 4.4.3 ...... 142 4.5 Concluding Remarks ...... 147 4.6 Lower Bound ...... 148
8 Chapter 1
Introduction
A graph 퐺 is 퐻-Ramsey with 푟 colors, denoted by 퐺 푟 퐻, if any 푟-coloring of the → edges of 퐺 contains a monochromatic copy of 퐻. If 푟 = 2 we simply write 퐺 퐻. → The fact that for every graph 퐻 there is a graph 퐺 such that 퐺 is 퐻-Ramsey was first proved by Ramsey [79] in 1930 and rediscovered independently by Erdősand Szekeres a few years later [36]. Ramsey theory is currently one of the most active areas of combinatorics with connections to number theory, geometry, analysis, logic, and computer science.
A fundamental problem in graph Ramsey theory is to understand the graphs 퐺 satisfying 퐺 is 퐾푘-Ramsey, where 퐾푘 denotes the complete graph on 푘 vertices. The Ramsey number 푟(퐻) is the minimum number of vertices of a graph 퐺 which is 퐻- Ramsey. The most famous question in this area is that of estimating the Ramsey number 푟(퐾푘). Classical results of Erdős [31] and Erdős and Szekeres [36] show that 푘/2 2푘 2 푟(퐾푘) 2 . While there have been several improvements on these bounds ≤ ≤ (see, for example, [21]), despite much attention, the constant factors in the above exponents remain the same. Given these difficulties, the field has naturally stretched in different directions. In this thesis we prove new results in three of these directions. It should be noted that the text of this thesis, including the abstract and introduction, may closely follow or be directly taken from various papers of the author including [41, 42, 44, 51, 52].
9 1.1 Chapter 2: Minimum Degrees of Minimal Ram- sey Graphs
The first such direction is to understand, for a fixed graph 퐻, properties of the collection of graphs that are Ramsey for 퐻. Clearly, for a fixed graph 퐻, the collection of graphs that are Ramsey for it is upwards closed. That is, if 퐺 is Ramsey for 퐻 and 퐺 is a subgraph of 퐺′, then 퐺′ is also Ramsey for 퐻. Therefore, to understand the collection of graphs that are Ramsey for 퐻, it is sufficient to understand the collection (퐻) of graphs that are minimal subject to being Ramsey for 퐻; these graphs are ℳ called Ramsey minimal for 퐻. For a graph 퐺, let 훿(퐺) denote the minimum degree of the vertices of 퐺. Our interest in Chapter 2 lies in 푠(퐻), which is the minimum of 훿(퐺) over all graphs 퐺 (퐻). This parameter was first introduced and studied by Burr, Erdős, and ∈ ℳ Lovász in 1976 [12]. A simple upper bound is 푠(퐻) 푟(퐻) 1. Indeed, one may ≤ − take any Ramsey-minimal graph on 푟(퐻) vertices, and the minimum degree of this graph is at most 푟(퐻) 1. Since 푟(퐾푡) is exponential in 푡, the result of Burr, Erdős, − 2 and Lovász [12] that 푠(퐾푡) = (푡 1) may be surprising. − Fox and Lin [43] observed the simple lower bound 푠(퐻) 2훿(퐻) 1 which holds ≥ − for every graph 퐻. We say a graph 퐻 is Ramsey simple if this lower bound is tight, that is if 푠(퐻) = 2훿(퐻) 1. In recent years, the study of 푠(퐻) has received increased − 2 attention. Fox and Lin [43] present an alternative proof that 푠(퐾푡) = (푡 1) and also − show that 퐾푠,푡 is Ramsey simple, where 퐾푠,푡 is a complete bipartite graph with parts of size 푠 and 푡. Szabó, Zumstein, and Zürcher [88] conjectured that every bipartite graph without isolated vertices is Ramsey simple. They prove this conjecture for a variety of bipartite graphs including trees, even cycles, and bipartite graphs where every vertex in one of the parts has degree 훿(퐻). They also prove the conjecture for connected bipartite graphs with parts 퐴 of size 푎 and 퐵 of size 푏 with 푏 푎 in which 퐴 contains a minimum degree vertex. It is worth ≥ noting that they also address the case of isolated vertices.
In Chapter 2 we prove this conjecture for all 3-connected bipartite graphs, as
10 well as for several other graphs. We also find the first examples of graphs that are
Ramsey simple but are not bipartite and the first examples of connected graphs 퐻, 퐻′ satisfying 퐻 퐻′ with 푠(퐻) > 푠(퐻′), a rather surprising property. ⊆
1.2 Chapter 3: The Erdős-Hajnal Conjecture for Rain- bow Triangles
In Chapter 3 we pursue a different variant when studying which graphs satisfy 퐺 퐻, → namely by considering what happens when, rather than considering all colorings of
퐺, we restrict which colorings are allowed. Erdős and Hajnal [35] famously conjecture that, for any 2-coloring 퐶 of a complete graph, there is a 휀 = 휀(퐶) > 0 such that, for every 푛, every 2-coloring of the edges of a complete graph on 푛 vertices satisfying that
the 2-coloring contains no copy of 퐶 does contain a monochromatic 퐾푛휀 . In other words, if we only considered those colorings which don’t contain 퐶, Ramsey numbers would grow polynomially rather than exponentially. There are now several partial results on the Erdős-Hajnal conjecture. Erdős and
Hajnal [35] proved that, for each fixed coloring 퐶, there is 휖 = 휖(퐶) > 0 such that
every coloring of a 퐾푛 which does not contain a copy of 퐶 has a monochromatic √ clique with 푒휖 log 푛 vertices. Fox and Sudakov [45], strengthening an earlier result of Erdős and Hajnal, proved that for each fixed coloring 퐶 there is 휖 = 휖(퐶) > 0 such
that every coloring of a 퐾푛 which does not contain a copy of 퐶 has either a balanced complete bipartite graph in the first color or a clique of order 푛휖 in the second color. Erdős and Hajnal also proposed studying a multicolor generalization of their con-
jecture. It states that for every fixed 푘-coloring of the edges 퐶 of a complete graph, there is an 휖 = 휖(퐶) > 0 such that every 푘-coloring of the edges of the complete graph on 푛 vertices without a copy of 퐶 contains a clique of order 푛휖 which only uses 푘 1 √ − colors. They proved a weaker estimate, replacing 푛휖 by 푒휖 log 푛. Note that the case of two colors is what is typically referred to as the Erdős-Hajnal conjecture. Hajnal [59] conjectured that the following special case of the multicolor generaliza-
11 tion of the Erdős-Hajnal conjecture holds. There is 휖 > 0 such that every 3-coloring of the edges of the complete graph on 푛 vertices without a rainbow triangle (that is, a triangle with all its edges different colors), contains a set of order 푛휖 which uses at most two colors. We prove Hajnal’s conjecture, and further determine a tight bound on the order of the largest guaranteed 2-colored set in any such coloring; this size is Θ(푛1/3 log2 푛). Indeed, we actually find asymptotically tight bounds for every 푟 푠 1 for the largest size of a clique that uses at most 푠 colors in an 푟-coloring of ≥ ≥ a 퐾푛 that contains no rainbow triangles.
1.3 Chapter 4: Packing Vertex-Disjoint Monochro- matic Copies of Sparse Graphs
In Chapter 4 we pursue another variant of Ramsey theory where, rather than taking
a coloring of a graph 퐺 and finding a single monochromatic copy of some graph 퐻, we instead wish to partition all of the vertices of 퐺 into monochromatic copies of graphs. An area that has attracted much interest is the study of Ramsey numbers for bounded degree graphs. In 1975, Burr and Erdős [11] raised the problem that every
graph 퐺 with 푛 vertices and maximum degree ∆ has a linear Ramsey number, so 푟(퐺) 퐶(∆)푛, for some constant 퐶(∆) depending only on ∆. This was proved ≤ by Chvátal, Rödl, Szemerédi and Trotter [20] in one of the earliest applications of Szemerédi’s celebrated Regularity Lemma [89]. Because the proof uses the Regularity
Lemma, the bound on 퐶(∆) is quite weak; it is of tower type in ∆. Recently, Conlon [22] and, independently, Fox and Sudakov [45] have shown how
푂(Δ) to prove the bound of 퐶퐵(∆) 2 in the bipartite case. For the non-bipartite ≤ graph case, the current best bound is due to Conlon, Fox, and Sudakov [25] 퐶(∆) ≤ 2푂(Δ log Δ). Graham, Rödl, and Ruciński [49] proved that there are bipartite graphs with 푛 vertices and maximum degree ∆ for which the Ramsey number is at least 2Ω(Δ)푛,
12 meaning that the upper bound of Conlon, Fox, and Sudakov is best possible, up to the constant in the exponent. It is a natural question (initiated by András Gyárfás) to ask how many monochro- matic members from a bounded-degree graph family are needed to partition the vertex set of a 2-edge-colored complete graph. In Chapter 4 we study this problem and re-
lated questions. Given = 퐹1, 퐹2,... a sequence of graphs, we say it is a proper ℱ { } graph sequence if 퐹푛 is a graph on 푛 vertices. We say it has some graph property if every graph of has that property (e.g. is bipartite if 퐹푛 is bipartite for every 푛). ℱ ℱ We prove, for bipartite proper graph sequences with maximum degree at most ℱ ∆, that the vertices of any 2-edge-colored complete graph may be partitioned into 2푂(Δ) monochromatic copies of graphs from , and that this is best possible up to ℱ the constant in the exponent. We further prove that for proper graph sequences ℱ with maximum degree at most ∆, that the vertices of any 2-edge-colored complete graph may be partitioned into 2푂(Δ log Δ) monochromatic copies of graphs from . ℱ Finally, we generalize this to arrangeable graph sequences by showing that, for proper
graph sequences with arrangeability 푎 and chromatic number 푘 satisfying that ℱ the maximum degree of 퐹푛 is at most √푛/ log 푛, the vertices of any 2-edge-colored complete graph may be partitioned into 2푂(푎4푘2) monochromatic copies of graphs from . ℱ
13 14 Chapter 2
Minimum Degrees of Minimal Ramsey Graphs
2.1 Introduction
For graphs 퐺 and 퐻 we write 퐺 퐻 and say 퐺 is Ramsey for 퐻 if in any 2-coloring → of the edges of 퐺 there exists a monochromatic copy of 퐻. Ramsey’s theorem [79] states that for any 퐻 there is a 퐺 that is Ramsey for 퐻. Clearly, for a fixed graph 퐻, the collection of graphs that are Ramsey for it is upwards closed. That is, if 퐺 is Ramsey for 퐻 and 퐺 is a subgraph of 퐺′, then 퐺′ is also Ramsey for 퐻. Therefore, to understand the collection of graphs that are Ramsey for 퐻, it is sufficient to understand the collection (퐻) of graphs that are ℳ minimal subject to being Ramsey for 퐻; these graphs are called Ramsey minimal for 퐻. Many fundamental problems in graph theory concern the study of (퐻). The ℳ most famous of these, one of the driving motivations for the field of Ramsey theory, is to compute or estimate the Ramsey number 푟(퐻) for various 퐻, where 푟(퐻) is the smallest number of vertices of any graph in (퐻). Of particular interest is 푟(퐾푡), the ℳ Ramsey number of the complete graph on 푡 vertices. Classical results of Erdős and 푡/2 2푡 Szekeres [36] and Erdős [31] imply that 2 푟(퐾푡) 2 for 푡 2. Despite much ≤ ≤ ≥ interest (see, e.g., [21]), there have been no improvements in the constant factors in
15 the above exponents.
This impasse has naturally led to the study of other problems related to (퐻). ℳ For example, the size Ramsey number 푟ˆ(퐻) is the minimum number of edges of any graph in (퐻). This parameter was introduced by Erdős et al. [32]. While this ℳ parameter is still far from understood, there are now several beautiful results about size Ramsey numbers of sparse graphs (see, e.g., [5], [64], [80]). Another related problem asks which graphs 퐻 are Ramsey-infinite, that is for which graphs 퐻 is the family (퐻) infinite (see, e.g., the book [50]). ℳ For a graph 퐺, let 훿(퐺) denote the minimum degree of the vertices of 퐺. Our interest in this chapter lies in 푠(퐻), which is the minimum of 훿(퐺) over all graphs 퐺 (퐻). This parameter was first introduced and studied by Burr, Erdős, and ∈ ℳ Lovász in 1976 [12]. A simple upper bound is 푠(퐻) 푟(퐻) 1. Indeed, one may ≤ − take any Ramsey-minimal graph on 푟(퐻) vertices, and the minimum degree of this
graph is at most 푟(퐻) 1. Since 푟(퐾푡) is exponential in 푡, the result of Burr, Erdős, − 2 and Lovász [12] that 푠(퐾푡) = (푡 1) may be surprising. − Fox and Lin [43] observed the simple lower bound 푠(퐻) 2훿(퐻) 1 which holds ≥ − for every graph 퐻. To see this, assume for contradiction there is some minimal Ramsey graph 퐺 for 퐻 with a vertex 푣 of degree at most 2훿(퐻) 2. By minimality, − there must be some red-blue coloring of the edges of 퐺 푣 without a monochromatic − copy of 퐻; we may extend this to an edge-coloring of 퐺 by coloring at most 훿(퐻) 1 − of the edges incident to 푣 blue and at most 훿(퐻) 1 of the edges incident to 푣 red. − As 퐺 is Ramsey for 퐻, it follows that this coloring must have a monochromatic copy of 퐻 containing 푣. However, 푣 has degree less than 훿(퐻) in any monochromatic subgraph, contradicting that 푣 is in a monochromatic copy of 퐻, and completing the proof. We say a graph 퐻 is Ramsey simple if this lower bound is tight, that is if 푠(퐻) = 2훿(퐻) 1. − In recent years, the study of 푠(퐻) has received increased attention. Fox and Lin 2 [43] present an alternative proof that 푠(퐾푡) = (푡 1) and also show that 퐾푠,푡 is − Ramsey simple. Szabó, Zumstein, and Zürcher [88] conjectured that every bipartite graph without isolated vertices is Ramsey simple.
16 Conjecture 2.1.1. [88] If 퐻 is bipartite with no isolated vertices, then 퐻 is Ramsey simple.
They prove this conjecture for a variety of bipartite graphs including trees, even cycles, and bipartite graphs where every vertex in one of the parts has degree 훿(퐻). They also prove the conjecture for connected bipartite graphs with parts 퐴 of size 푎 and 퐵 of size 푏 with 푏 푎 in which 퐴 contains a minimum degree vertex. It ≥ is worth noting that they also address the case of isolated vertices: they show that for any graph 퐻 on 푛 vertices (not necessarily bipartite), if we denote the graph obtained from 퐻 by adding 푡 isolated vertices by 퐻 + 푡퐾1, then 푠(퐻 + 푡퐾1) = 푠(퐻) if 푡 푟(퐻) 푛, and 푠(퐻 + 푡퐾1) = 0 if 푡 > 푟(퐻) 푛. ≤ − − In this chapter we prove Conjecture 2.1.1 for all 3-connected bipartite graphs.
Theorem 2.1.2. If 퐻 is bipartite and 3-connected, then 푠(퐻) = 2훿(퐻) 1. − In the process of proving the above result, we prove the following theorem which gives the first examples of Ramsey simple graphs that are not bipartite.
Theorem 2.1.3. If 퐻 is 3-connected and has some vertex 푣 of degree 훿(퐻) so that the neighbors of 푣 are contained in an independent set of size 2훿(퐻) 1, then 퐻 is − Ramsey simple.
In a fairly wide range of values of 푝, the binomial random graph 퐺(푛, 푝) satisfies the conditions of the above theorem, and hence we have the following corollary.
Corollary 2.1.4. If 푛−1 log 푛 푝 푛−2/3, then 퐺(푛, 푝) is almost surely Ramsey ≪ ≪ simple.
The lower bound on 푝 originates from the need to be 3-connected (for 푝 푛−1 log 푛 ≪ there will almost surely be isolated vertices). The upper bound on 푝 guarantees that the expected number of triangles is 표(푛) and hence the neighborhood of a typical vertex is an independent set. Further, one expects that there is a minimum degree vertex 푣 whose neighborhood is an independent set, and can be extended to an inde- pendent set twice larger. The details of this proof are given in Section 2.9. Note also
17 that in this range 퐺(푛, 푝) is almost surely not bipartite, so these graphs do not fall under the assumptions of Theorem 2.1.2, and are examples of Ramsey simple graphs that are not bipartite.
There has also been much interest in the value of 푠(퐾푡 퐾2), where 퐻 퐾2 denotes · · the collection of graphs obtained by adding a new vertex 푣 to 퐻, picking a vertex 푢 of 퐻, and connecting 푣 to 푢. It was shown [88] that 푠(퐾푡 퐾2) 푡 1, and they · ≥ − 2 conjecture that 푠(퐾푡 퐾2) = 푠(퐾푡) = (푡 1) , for 푡 sufficiently large. The motivation · − for this conjecture is that, for 푡 sufficiently large, it intuitively may be the case that any graph which is Ramsey for 퐾푡 is also Ramsey for 퐾푡 퐾2. This conjecture was · disproved in [42], where it is shown that 푠(퐾푡 퐾2) = 푡 1. In this chapter, we · − generalize the lower bound of [88] that 푠(퐾푡 퐾2) 푡 1 to graphs other than 퐾푡, · ≥ − and we find upper bounds on 푠(퐻 퐾2) for many graphs 퐻. Most notably, we find · that 푠(퐾푡,푡 퐾2) = 1, where 퐾푡,푡 is the complete bipartite graph with parts of size 푡. · This is strong support for Conjecture 2.1.1, as 퐾푡,푡 퐾2 was thought to be the best · candidate for a counterexample. Theorem 2.1.2 and Theorem 2.1.3 use a powerful tool originating from [12] and generalized in [13] which requires the graphs to be 3-connected. In Section 2.3, we present the tools necessary to prove Theorems 2.1.2 and 2.1.3 and then present their proofs. We defer the proof that Corollary 2.1.4 follows from Theorem 2.1.3 and some basic facts about 퐺(푛, 푝) to Section 2.9. In Section 2.4, we prove lower bounds for
푠(퐻 퐾2) for many graphs 퐻. In Section 2.5, we show that 푠(퐾푡,푡 퐾2) = 1, and · · in Section 2.6 we give an upper bound for 푠(퐻 퐾2) for many graphs 퐻. Finally, · in Section 2.7, we give the first examples of connected graphs 퐻 퐻′ with 푠(퐻) > ⊆ 푠(퐻′).
2.2 Preliminaries
In this section, we introduce notation and tools that we use to prove bounds on 푠(퐻). Unless stated otherwise, all our colourings are red-blue colourings of the edges of a graph. We call two edge-disjoint graphs 푅, 퐵 on the same vertex set 푉 a colour
18 pattern on 푉 . For a graph 퐻, a colour pattern is called 퐻-free if neither 푅 nor 퐵 contains 퐻 as a subgraph. Let 퐺 be a graph that contains 푅 퐵 as a subgraph, ∪ where 푅, 퐵 is a colour pattern. We say a colouring 푐 of 퐺 extends (or has) colour pattern 푅 퐵 if 푅 and 퐵 are both monochromatic with different colours. For a graph ∪ 퐻, we call a colouring 푐 퐻-free if there is no monochromatic copy of 퐻 in 푐. Given a graph 퐺 which contains some 퐺0 as an induced subgraph, we say the pair (퐺, 퐺0) is 퐻-robust if any graph which is obtained from 퐺 by adding some vertices 푆 to 퐺 and adding edges within 푆 푉 (퐺0) satisfies that any copy of 퐻 is either contained ∪ entirely within 푆 푉 (퐺0) or is contained entirely within 퐺. ∪ Burr, Erdős, and Lovász [12] introduced a powerful tool in determining 푠(퐾푡). It states that, given any 퐾푡-free colour pattern 푅, 퐵, there is some graph 퐺 푅 퐵 ⊇ ∪ which is not Ramsey for 퐾푡, but any 퐾푡-free coloring of 퐺 extends the colour pattern 푅 퐵. For a graph 퐻 and a colour pattern 푅, 퐵, we call a graph B = B(퐻, 푅, 퐵) ∪ a BEL gadget for 퐻 (with colour pattern 푅 퐵) if B contains 푅 퐵 as an induced ∪ ∪ subgraph, B 9 퐻 and any 퐻-free colouring of B has colour pattern 푅 퐵. Burr, ∪ Nešetřil and Rödl [13] extended the proofs in [12] in the following way.
Lemma 2.2.1. [13] For any 3-connected graph 퐻 and any 퐻-free colour pattern 푅 퐵, there exists a graph B = B(퐻, 푅, 퐵) that is a BEL gadget for 퐻 with colour ∪ pattern 푅, 퐵 so that (B, 푅 퐵) are 퐻-robust. Furthermore, if 퐻 and 푅 퐵 are ∪ ∪ bipartite, then so is B.
Let us say that BEL gadgets exist for 퐻 if for any 퐻-free colour pattern 푅, 퐵 there is a BEL gadget for 퐻 with colour pattern 푅 퐵. ∪
2.3 A large class of Ramsey-simple graphs
In this section, we prove Theorem 2.1.2 and Theorem 2.1.3. In the following, we give
a sufficient condition for a graph 퐻 to be Ramsey-simple.
Lemma 2.3.1. Let 퐻 be a graph and suppose 퐻 has no isolated vertices and BEL gadgets exist for 퐻. If there is an 퐻-free graph 퐺 with an independent set 푆 of size
19 2훿(퐻) 1 so that any graph obtained from 퐺 by adding a vertex 푣 and connecting 푣 − to 훿(퐻) vertices of 푆 contains a copy of 퐻, then 퐻 is Ramsey simple.
Proof. Take 퐺0 to be a red copy 푅 of 퐺 with distinguished set 푆 and a blue copy 퐵 of 퐺 with the same distinguished set 푆 (note this creates no conflicts since 푆 is an independent set). Note that 푅, 퐵 is a colour pattern that is 퐻-free, since both 푅 and 퐵 consist of a copy of 퐺 along with isolated vertices. Since there exist BEL gadgets for 퐻, we may create a graph B so that B 9 퐻, but any 퐻-free coloring of B extends the colour pattern 푅 퐵. Now, add a vertex 푣 to B and add edges ∪ from 푣 to all of 푆. Call the resulting graph 퐺′. The degree of 푣 in 퐺′ is 2훿(퐻) 1, − and the graph obtained by removing 푣 is not Ramsey for 퐻. In any two-coloring of the edges containing 푣, at least 훿(퐻) of those edges must have the same colour, say, without loss of generality, red. Then, by assumption on 퐺, 푣 along with these 훿(퐻) neighbours in 푆 and the red copy of 퐺 contain a monochromatic copy of 퐻. Hence, 퐺′ is Ramsey for 퐻. Any Ramsey-minimal subgraph of 퐺′ must contain 푣, and so 푠(퐻) 2훿(퐻) 1, i.e., 퐻 is Ramsey simple. ≤ − In the next lemma, we show how to construct 퐺 under certain assumptions on 퐻.
Lemma 2.3.2. Let 퐻 be a graph and suppose 퐻 has no isolated vertices and BEL gadgets exist for 퐻. If there is a vertex 푢 of degree 훿(퐻) in 퐻 whose neighbourhood is contained in an independent set of size 2훿(퐻) 1, then 퐻 is Ramsey simple. − Proof. Let 푛 be the number of vertices of 퐻. Consider a graph 퐺 on 푛 1 vertices that − is complete except for an independent set 푆 on 2훿(퐻) 1 vertices; that is, the graph − consists of an independent set on 2훿(퐻) 1 vertices and a clique on 푛 1 (2훿(퐻) 1) − − − − vertices, and there is a complete bipartite graph between them. Notice that 퐻 푢 − is a subgraph of 퐺 for any 퐻 satisfying the assumptions of the theorem. Adding a vertex 푣 to 퐺 and connecting 푣 to any 훿(퐻) vertices of 푆 creates a copy of 퐻 where 푣 acts as a copy of 푢. The graph 퐺 is 퐻-free since it has only has 푛 1 vertices. − Therefore, we can apply Lemma 2.3.1 to conclude that 퐻 is Ramsey simple.
Note that Theorem 2.1.3 is a corollary of Lemma 2.3.2 and Lemma 2.2.1. We now
20 apply Lemma 2.2.1 to show that every bipartite 3-connected graph is Ramsey simple (Theorem 2.1.2). We again prove a slightly stronger result.
Theorem 2.3.3. Suppose 퐻 is a connected bipartite graph with at least two vertices. If BEL gadgets exist for 퐻, then 퐻 is Ramsey simple.
Proof. Let 퐴, 퐵 be a bipartition of 퐻 with 퐴 퐵 , and take 푎 = 퐴 , 푏 = 퐵 , | | ≤ | | | | | | so 푎 푏. Let 푛 = 푎 + 푏 = 푉 (퐻) . Let 훿 = 훿(퐻). If 퐵 contains only vertices of ≤ | | degree 훿, if 퐴 contains a vertex of degree 훿, or if 푎 = 푏, then it was proved in [88] that 퐻 is Ramsey simple, as desired. So we may assume that 퐵 contains some vertex of degree larger than 훿, that 퐴 contains no vertex of degree 훿, and that 푏 > 푎; this also means that 퐵 must contain some vertex 푢 of degree 훿. Under these assumptions we will show that there is a graph 퐺 satisfying the assumptions of Lemma 2.3.1, thus completing the proof. That is, 퐺 is 퐻-free and has an independent set 푆 of size 2훿 1 − so that adding a vertex 푣 to 퐺 and connecting 푣 to any 훿 vertices of 푆 creates a copy of 퐻. If 푏 2훿, then we may take 퐺 to be a complete bipartite graph with both parts ≥ of size 푏 1. We will take 푆 to be any 2훿 1 vertices from one of the parts. 퐺 will − − not contain a copy of 퐻, as neither of its parts has size at least 푏. Adding a vertex 푣 and connecting it to any 훿 vertices of 푆 creates a copy of 퐻 with 푣 serving as a copy of 푢. Therefore, we may assume that 푎 < 푏 < 2훿. In particular, this means that any two vertices in 퐵 have a common neighbour and any two vertices in 퐴 have a common neighbour. Now, we will instead consider the graph 퐺 obtained as follows. Take an independent set 푆 on 2훿 1 vertices. For any set 푆′ of 훿 vertices of 푆, add a copy of − 퐻 푢 where 푆′ is 푁(푢). Formally, 퐺 will have vertex set 푆 (︀(︀푆)︀ [푛 훿 1])︀. Enu- − ∪ 훿 × − − merate the vertices of 퐻 as 푢1, . . . , 푢푛 so that 푢 = 푢푛 and 푁(푢) = 푢푛−훿, . . . , 푢푛−1 . { } ′ For each set 푆 푆 of size 훿, fix some ordering 푣푆 ,푛−훿, 푣푆 ,푛−훿+1, . . . , 푣푆 ,푛−1 of the ver- ⊆ ′ ′ ′ tices of 푆′. The edges of 퐺 that are not incident to 푆 are those pairs of vertices of the ′ ′ ′ form (푆 , 푘1), (푆 , 푘2) where 푆 is a set of 훿 vertices from 푆 and 푢푘 , 푢푘 is an edge { } { 1 2 } ′ of 퐻. The edges that are incident to 푆 are those pairs of the form 푣푆 ,푘 , (푆 , 푘2) { ′ 1 } 21 where 푢푘 , 푢푘 is an edge of 퐻. Note that 퐺 is bipartite. { 1 2 } The subgraph of 퐺 consisting of those edges incident to 푆 is bipartite with one part being 푆. In the other part, all of the vertices have degree at most 훿 (since the vertex (푆′, 푘) can be adjacent only to vertices in 푆′). Therefore, if this subgraph contains a copy of 퐻, one of the parts of 퐻 must contain only vertices of degree 훿, contradicting the assumption.
′ ′ Hence, any copy of 퐻 in 퐺 must contain some edge (푆 , 푘1), (푆 , 푘2) not incident { } to 푆. Note that there are only 푛 1 vertices in 푆′ or of the form (푆′, 푘). Therefore, − a copy of 퐻 must contain some other vertex. However, a copy of 퐻 cannot contain any vertex of 푆 other than those in 푆′. ′ Indeed, such a vertex would share no common neighbours with both (푆 , 푘1) and ′ ′ ′ (푆 , 푘2). However, as (푆 , 푘1) and (푆 , 푘2) are adjacent, they must be in different parts of the bipartition, contradicting the assumption that any vertex of the copy of 퐻 must have a common neighbour with each vertex in the same part.
′′ ′′ Therefore, a copy of 퐻 must contain a vertex of the form (푆 , ℓ1) with 푆 distinct from 푆′. However, since 푆′′ and 푆′ are distinct sets of size 훿, they can intersect in at most 훿 1 vertices. Since the only vertices from 푆 in a copy of 퐻 must be contained in − ′ ′′ ′′ 푆 , and the only neighbours of (푆 , ℓ1) in 푆 are in 푆 , we must have that all neighbours ′′ ′ ′′ in 푆 of (푆 , ℓ1) used by this copy of 퐻 must be contained in 푆 푆 . Since this is ∩ ′′ at most 훿 1 vertices, the vertex (푆 , ℓ1) must have degree at least 훿 and so must − ′′ have another neighbour in the copy of 퐻. In 퐺, the only neighbours of (푆 , ℓ1) not ′′ ′′ contained in 푆 are of the form (푆 , ℓ2), and so 퐻 must contain some vertex (푆 , ℓ2) ′′ as a neighbour of (푆 , ℓ1). In particular, in this copy of 퐻, the vertices of the form (푆′, 푘) contain vertices from both parts of 퐻, as do vertices of the form (푆′′, 푘). In order to have that any two vertices from the same part share a common neighbour, we must have that 푆′ 푆′′ contains vertices from both parts of 퐻. However, 퐺 is ∩ bipartite and 푆′ 푆′′ is contained in one of the bipartitions, contradicting that a copy ∩ of the connected graph 퐻 can have vertices from both parts in 푆′ 푆′′. Therefore, ∩ 퐺 has no copy of 퐻. By construction, adding a vertex to 퐺 and connecting it to any 훿 vertices of 푆 creates a copy of 퐻. Therefore, 퐺 has the desired properties, and so,
22 by Lemma 2.3.1, 퐻 is Ramsey simple.
2.4 Stronger lower bounds for graphs with hanging edges
For a collection of graphs, we say that a graph 퐺 is -Ramsey, and denote this by ℋ ℋ 퐺 , if in any two colouring of the edges of 퐺 there exists a monochromatic copy → ℋ of some 퐻 in . If 퐺 is minimal with this property, we call 퐺 -Ramsey minimal. ℋ ℋ We denote by = ( ) the class of -Ramsey minimal graphs. Note this class ℳ ℳ ℋ ℋ does not need to relate by inclusion to any of the classes (퐻) for 퐻 due to ℳ ∈ ℋ the minimality assumption. We also set 푠( ) := min퐺∈ℳ(ℋ) 훿(퐺), as before. ℋ Given a graph 퐻 = (푉, 퐸), we denote by 퐻 퐾2 the collection of graphs obtained by · picking some vertex 푣 of 퐻, adding some new vertex 푤, and adding an edge between
푣 and 푤. This is a slight abuse of notation as we have already defined 퐾푘 퐾2 to be · a single graph, but a graph 퐺 is Ramsey for the previous definition of 퐾푘 퐾2 if and · only if it is Ramsey for the new definition, so the notation is consistent. Note thatif
퐻 is vertex transitive, then 퐻 퐾2 contains, up to isomorphism, only one graph. · The following proof closely follows the ideas of the proof that 푠(퐾푡 퐾2) 푡 1 · ≥ − in [88].
For a graph 퐻, let
(퐻) := 퐶 퐻[푁(푥)] : 푥 푉 (퐻), 퐶 is a connected component of 퐻[푁(푥)] ℱ { ⊆ ∈ }
denote the collection of all connected graphs that appear in the neighbourhood of any
vertex 푥 of 퐻. We will prove the following theorem:
Theorem 2.4.1. Let 퐻 = (푉, 퐸) be a graph on 푛 vertices, and assume 퐻 has the following properties:
1. 퐻 is connected.
2. 퐻 has minimum degree at least two.
23 3. There is a colouring of 퐾훼(퐻) that is 퐶-free for every 퐶 (퐻). ∈ ℱ
Then 푠(퐻 퐾2) 훿(퐻). · ≥ Remark 2.4.2. Note that Condition (3) trivially fails, e.g., when 퐻 is bipartite, or more generally, if there exists a vertex 푥 of 퐻 so that 퐻[푁(푥)] contains an isolated vertex. Note also, that when 퐻 is the complete graph, then Condition (3) is trivially satisfied. In Section 2.10, we give a large class of non-trivial examples ofsparse
(vertex-transitive) graphs 퐻 that fulfill all conditions of the theorem, meaning that the single graph 퐻 퐾2 satisfies 푠(퐻 퐾2) 훿(퐻). · · ≥ ′ ′ Proof. Let 퐺 be an 퐻 퐾2-Ramsey minimal graph. We want to show that 훿(퐺 ) · ≥ 훿(퐻). Assume the opposite and remove some vertex from 퐺′ of degree 훿(퐺′) < 훿(퐻). This leaves some graph 퐺 = (푉, 퐸). By minimality of 퐺′, there is a two colouring 휒 of 퐸(퐺) such that there is no monochromatic copy of 퐹 for any 퐹 퐻 퐾2. Call the ∈ · two colours red and blue.
We say a vertex of 퐺 is critical under some colouring 휓 if it is contained in both a red and a blue copy of 퐻. We will show below that we can convert 휒 to a colouring
휓 of 퐺 with no monochromatic 퐹 퐻 퐾2 and without critical vertices. Let us first ∈ · show how the existence of such a colouring implies Theorem 2.4.1.
Claim 2.4.3. If there is a colouring 휓 of 퐺 with no monochromatic 퐹 퐻 퐾2 ∈ · and with no critical vertices, then there is a colouring of 퐺′ with no monochromatic
퐻 퐾2. · Proof. If 푣 is the vertex we removed from 퐺′ of degree less than 훿(퐻), we define the colouring 휓′ of 퐺′ as follows: 휓′ agrees with 휓 on 퐺, and an edge 푣푤 is coloured blue if 푤 is contained under 휓 in a red copy of 퐻, and otherwise the edge is coloured red. ′ ′ Assume 퐺 has a monochromatic copy 퐹 of some 퐹 퐻 퐾2. By choice of 휓, ∈ · this 퐹 ′ must use 푣. Since 푑(푣) < 훿(퐻), 푣 must be the hanging vertex. Suppose 퐹 ′ were red. Then 퐹 ′ 푣 is red copy of 퐻, so the pending edge 푣푤 must be coloured − blue in 휓′, a contradiction. On the other hand, if 퐹 ′ were blue, then 푣푤 would be blue and by definition of 휓′ 푤 is also contained in a red copy of 퐻 under 휓. Then 푤 would be critical in 퐺 under 휓, a contradiction.
24 It remains to show the existence of a colouring 휓 of 퐸(퐺) without a monochromatic
퐹 퐻 퐾2 and no critical vertices, completing the proof. Let us first study the ∈ · structure of 휒, the given colouring of 퐺.
Lemma 2.4.4. Let 휒 be any colouring of 퐺 without a monochromatic copy of any
퐹 퐻 퐾2. ∈ ·
(푖) For any red (blue) copy 퐻1 of 퐻 in 퐺, all edges between 푉 (퐻1) and 푉 (퐺) 푉 (퐻1) ∖ are blue (red).
(푖푖) Given 퐻1, 퐻2 two monochromatic copies of 퐻 in 퐺 of the same colour, either
푉 (퐻1) = 푉 (퐻2) or 푉 (퐻1) 푉 (퐻2) = . ∩ ∅
(푖푖푖) Let 퐻1 be a red and 퐻2 be a blue copy of 퐻 in 퐺. Then there are no edges in
퐺 between 푉 (퐻1) 푉 (퐻2) and 푉 (퐻2) 푉 (퐻1). ∖ ∖
(푖푣) For any critical vertex 푣, if it is contained in a red copy 퐻1 and a blue copy 퐻2
of 퐻, 푣 is not adjacent in 퐺 to any vertex 푤 not in 푉 (퐻1) 푉 (퐻2). ∪
Proof. By assumption, there is no monochromatic 퐹 퐻 퐾2, so (푖), (푖푖푖) and (푖푣) ∈ · are immediate. (푖푖) follows by connectivity of 퐻.
We would like to keep track of the positions of monochromatic copies of 퐻. There- fore, for a red-blue-colouring 휓 of 퐺, we set
red(휓) := 푉 ′ 푉 (퐺): There is a red copy of 퐻 in 퐺 such that 푉 (퐻) = 푉 ′ , and 풱 { ⊆ } blue(휓) := 푉 ′ 푉 (퐺): There is a blue copy of 퐻 in 퐺 such that 푉 (퐻) = 푉 ′ . 풱 { ⊆ }
In other words, red(휓) ( blue(휓)) is the collection of 푉 ′ so that 푉 ′ is the vertex set 풱 풱 of a red (blue) copy of 퐻. We are now ready to describe the recolouring algorithm. The main motivation
behind the definition of 휓 below is that we have strong control over the kinds of struc- tures that contain edges incident to critical vertices, and so have much leeway when recolouring said edges. Indeed, we will use this structure to remove all monochromatic
copies of 퐻 that contain critical vertices.
25 ′ 퐼. Let red(휒) = 푉0, 푉1,... be the hosts of red copies of 퐻 under 휒. Define 휒 풱 { } to agree with 휒 on any edge that is not internal to any 푉푖, and to be red on any
edge internal to some 푉푖.
′ ′ 퐼퐼. Let blue(휒 ) = 푊0, 푊1,... be the hosts of blue copies of 퐻 under 휒 . Define 풱 { } ′′ 휒 to be blue on any edge that is internal to some 푊푖 but not internal to any ′ 푉푖, and to agree with 휒 on all other edges.
Note that by Lemma 2.4.4 (푖푖), the elements of red(휒) are pairwise disjoint, and by 풱 ′ 2.4.4 (푖) all edges between any two 푉푖 are blue. Now, 휒 only colours edges red inside ′′ the 푉푖. Further, 휒 only changes edges not inside the 푉푖 to blue. We therefore have
red(휒) = red(휒′′) =: red. 풱 풱 풱
′ Further note, since 퐺 has no monochromatic 퐻 퐾2 under 휒 and 휒 only coloured · ′ edges red inside the 푉푖, 퐺 has no monochromatic 퐻 퐾2 under 휒 . Therefore, again · ′ by Lemma 2.4.4 (푖) and (푖푖), the elements 푊푗 blue(휒 ) are pairwise disjoint, and ∈ 풱 ′′ all edges between any two 푊푗 are red. Now, since 휒 recolours only edges inside ′ 푊푗 blue(휒 ) to blue, the new hosts of blue 퐻 are the same as before. That is, ∈ 풱
blue(휒′) = blue(휒′′) =: blue. 풱 풱 풱
′′ For each 푉푖 containing a critical vertex under 휒 , choose some critical 푣푖 푉푖. For ∈ each 푊푗 containing a critical vertex, if possible, choose some critical 푤푗 푊푗 such ∈ that for every 푖, 푤푗 = 푣푖. If this is not possible, choose any critical vertex 푤푗 푊푗. ̸ ∈ Take 퐴 to be the set of 푣푖 and 퐵 the set of 푤푗. We will now describe the final recolouring step. Since the 푉푖 are pairwise disjoint, as are the 푊푗, the sets of the form
푉푖 푊푗 are pairwise disjoint. So we may colour their internal edges independently of ∩ each other.
퐼푉. Our final colouring 휓 will be obtained by recolouring some edges of 휒′′. First,
recolour the edges internal to any (푉푖 푊푗) (퐴 퐵) so that it contains no ∩ ∖ ∪ 26 monochromatic copy of any connected component of 퐻[푁(푣)] for any 푣 푉 (퐻) ∈ (we will show that this is possible). 휓 will also recolour edges incident to some ′′ 푣푖, 푤푗. Given any 푣푖, it is a critical vertex under 휒 , so take 푗 to be the unique ′ ′ index such that 푣푖 푊푗. For any vertex 푣 푉푖 with 푣푖푣 퐸, colour the edge ∈ ∈ ∈ ′ ′ ′ 푣푖푣 blue if 푣 = 푤푗. If 푣 = 푤푗, colour the edge arbitrarily. ̸ ′′ Given any 푤푗, it is a critical vertex under 휒 , so take 푖 to be the unique index ′ ′ such that 푤푗 푉푖. For any vertex 푤 푊푗 with 푤푗, 푤 퐸, colour the edge ∈ ∈ ∈ ′ ′ ′ 푤푗푤 red if 푤 = 푣푖, and colour it arbitrarily if 푤 = 푣푖. (We will later check ̸ that this is well-defined, i.e. that we haven’t coloured any edges twice, except those coloured arbitrarily.)
휓 will agree with 휒′′ on all other edges.
We now begin proving that 휓 is well-defined and has the desired properties. We first note some properties of 휒′′.
Observation 2.4.5. 휒′′ above satisfies
′′ (푖) 휒 has no monochromatic 퐻 퐾2; ·
(푖푖) if an edge 푒 is internal to 푉푖 and not internal to any 푊푗, then 푒 has colour red;
(푖푖푖) if an edge 푒 is internal to 푊푖 and not internal to any 푉푗, then 푒 has colour blue.
′ Proof. We noted above already that 퐺 contains no monochromatic 퐻 퐾2 under 휒 . · Now, 휒′′ only recolours edges inside blue to blue, so (푖) follows. (푖푖) and (푖푖푖) are 풱 immediate from the colouring procedure.
Note that each property listed by the lemma above is symmetric with respect to the colours.
Lemma 2.4.6. 휓 is well-defined.
Proof. Let us first note that for each 푖 and 푗, 푉푖 푊푗 훼(퐻). To see this, note any | ∩ | ≤ ′′ edge internal to 푉푖 푊푗 is coloured red by 휒 , so since there is a blue copy of 퐻 on ∩ 푊푗 by definition, we must have that in this copy 푉푖 푊푗 forms an independent set, ∩ 27 so 푉푖 푊푗 훼(퐻). Therefore, by assumption, 푉푖 푊푗 and so 푉푖 푊푗 (퐴 퐵) | ∩ | ≤ ∩ ∩ ∖ ∪ may be coloured so there is no monochromatic copy of any connected component of
퐻[푁(푣)] for any 푣 푉 . ∈ We now check that we do not ask for 휓 to colour an edge both red and blue; note
if an edge does not contain a 푣푖 or 푤푗 then it is only coloured once by 휓. Otherwise,
assume it has the form 푣푖, 푣 with 푣 푉푖 퐵 and so was coloured blue. But then { } ∈ ∖ for it to be coloured red we must have 푣푖 = 푤푗 for some 푗 and 푣 푊푗 퐴. But if ∈ ∖ 푣푖 = 푤푗 we must have that 푤푗 could not have been chosen such that 푤푗 = 푣푖, so we ̸ must have 푉푖 푊푗 = 1, which is impossible since 푣, 푣푖 푉푖 푊푗. | ∩ | ∈ ∩ We collect some immediate facts about the colouring 휓. Observation 2.4.7.
The red degree of 푣푖 in 푉푖 under 휓 is at most one.
The blue degree of 푤푗 in 푊푗 under 휓 is at most one.
Any edge in 푉푖 that is not internal to any 푊푗 and is blue under 휓 is incident to 푣푖.
Any edge in 푊푗 that is not internal to any 푉푖 and is red under 휓 is incident to 푤푗.
Lemma 2.4.8. If 푣 was a critical vertex under 휒′′, then 푣 is not contained in any monochromatic copy of 퐻 under 휓.
Proof. Let 퐻1 be a monochromatic red copy of 퐻 in 퐺 under 휓 (the case for blue is ′′ symmetric) and assume it contains a vertex critical under 휒 . If 푉 (퐻1) = 푉푖 for some
푖, then the red-degree of 푣푖 in 푉푖 is at most 1, giving a contradiction since 훿(퐻) > 1.
Therefore, 퐻1 needs to use some edge 푒 = 푣푤 that has been recoloured by 휓 to red. But 휓 only recoloured edges that were incident to some critical vertices, that is
푣 푉푖 푊푗 and 푤 푉푖 푊푗 for some 푉푖 red, 푊푗 blue. Since 푉 (퐻1) = 푉푖, ∈ ∩ ∈ ∪ ∈ 풱 ∈ 풱 ̸ we may assume that 푒 is an edge leaving 푉푖, i.e. 푤 푊푗 푉푖. Hence 푒 is not ∈ ∖ contained in 푉푖 푊푗, so by the definition of 휓, 푒 is incident to 푤푗, the critical vertex ∩ we chose for 푊푗. That is, 푣 = 푤푗. Note that there are no red edges between 푊푗 푉푖 ∖ and 푉푖 excluding those incident to 푤푗. We consider now the neighbourhood of 푤푗 in 푉 (퐻1) 푊푗. Note that property (3) of Theorem 2.4.1 implies in particular that ∩ 28 there are no isolated vertices in 퐻1[푁(푤푗)]. Therefore some connected component 퐶
of 퐻1[푁(푤푗)] containing at least one edge is contained in 푊푗. But the only red edges ′ of 푊푗 are either incident to 푤푗 or contained in some 푉푖 푊푗 (where 푖 and 푖 may or ′ ∩ may not coincide). Therefore by connectivity, we must have that 퐶 is contained in
some (푉푖 푊푗) 푤푗 . ′ ∩ ∖ { } By construction (푉푖 푊푗) (퐴 퐵) = (푉푖 푊푗) 푣푖 , 푤푗 contains no red copy ′ ∩ ∖ ∪ ′ ∩ ∖ { ′ } of 퐶. But 푣푖 is blue to all vertices of 푉푖 except possibly 푤푗, so by connectivity (푉푖 ′ ′ ′ ∩ 푊푗) 푤푗 contains no red copy of 퐶, but this is a contradiction as the neighbourhood ∖{ } of 푤푗 must have a connected component in a set of this form.
Lemma 2.4.9. 휓 contains no critical vertices.
Proof. Let a vertex 푣 be given, and assume 푣 is critical under 휓. Then 푣 must be
contained in 휓 in some red copy 퐻1 of 퐻 and in some blue copy 퐻2 of 퐻. By
Lemma 2.4.8, neither 퐻1 nor 퐻2 may contain any vertices that were critical under 휒′′. However, any recoloured edge of 휓 is incident to a critical vertex of 휒′′; therefore, ′′ ′′ the colourings of 퐻1 and 퐻2 agree with those of 휒 , but then 푣 is critical under 휒 , a contradiction.
Lemma 2.4.10. 휓 contains no monochromatic copy of 퐻 퐾2. ·
Proof. We already know that under 휒′′, 퐺 does not contain a monochromatic copy
of any 퐹 퐻 퐾2. Suppose 퐹1 is a monochromatic copy of some 퐹 퐻 퐾2 ∈ · ∈ · ′′ under 휓. Then 퐹1 must use some edge 푒 = 푣푤, where 푣 is critical under 휒 , since 휓 only recoloured such edges. By Lemma 2.4.8, 푒 must be the pending edge and 푣
the pending vertex. Thus, 푤 is contained in a monochromatic copy 퐻1 of 퐻 under 휓. Again by Lemma 2.4.8, 푤 cannot be a critical vertex under 휒′′. Also, since 휓 only recoloured edges which contained a critical vertex under 휒′′, and none of these vertices are contained in a monochromatic copy of 퐻 under 휓, none of the edges internal to
퐻1 were recoloured by 휓. Therefore, 퐻1 was already a monochromatic copy of 퐻 in ′′ 휒 . But since 휓 recoloured 푒, it needs to be internal to some 푉푖 or some 푊푗, none of ′′ which are equal to 푉 (퐻1). But that means 푤 is critical under 휒 , a contradiction.
29 Now, Claim 2.4.3, Lemma 2.4.9, and Lemma 2.4.10 prove Theorem 2.4.1.
2.5 Upper bounds for complete bipartite graphs with hanging edges
In this section we will show that, for every 푡 2, we have 푠(퐾푡,푡 퐾2) = 1. Since ≥ · 퐾푡,푡 퐾2 is not 3-connected, we cannot simply apply Lemma 2.2.1 to create BEL · gadgets for it. We will instead use BEL gadgets for 퐾푡,푡 to construct a weaker version
of BEL gadgets for 퐾푡,푡 퐾2. However, first we must show that BEL gadgets do exists · for 퐾2,2, as 퐾2,2 is not 3-connected. Since the proof is almost identical for 퐾2,푡, we prove the more general version here. Throughout this and the next section, we call a
graph 푆 = 푆(퐻, 푒, 푓) a negative (positive) signal sender if 푆 9 퐻 and in any 퐻-free colouring of 푆, the edges 푒 and 푓 receive a different (the same) colour. The two edges 푒 and 푓 of 푆 are called signal edges.
Lemma 2.5.1. For 푡 2, let 푅, 퐵 be a 퐾2,푡-free colour pattern. Then there exists ≥ a graph B = B(퐾2,푡, 푅, 퐵) that is a BEL gadget for 퐾2,푡 so that (B, 푅 퐵) is 퐾2,푡- ∪ robust. Furthermore, if 푅 퐵 is bipartite, then so is B. ∪
Proof. Let 푠 = 6(푡 1) + 1. We first show that the graph 퐾3,푠 with one edge removed − is a negative signal sender for 퐾2,푡 in which the two signal edges are adjacent. It is
known [43] that 퐾3,푠 is Ramsey minimal for 퐾2,푡. Take a copy of 퐾3,푠−1 and name the three vertices 푎, 푏, 푐 from the part of size three. Add a vertex 푣 to the graph and connect it to both 푎 and 푏. Call this graph 푆−. We claim that 푆− is a negative signal sender for 퐾2,푡 with signal edges 푣푎 and 푣푏. To see this, assume there is some colouring of 푆− in which 푣푎 and 푣푏 have the same colour, say red, and there is no monochromatic copy of 퐾2,푡. Then we may add an edge from 푣 to 푐 and colour it blue. This graph is a copy of 퐾3,푠 and so must have a monochromatic copy of
퐾2,푡. This copy must use the added edge and therefore 푣, but 푣 has blue-degree 1, a contradiction.
30 − Furthermore, there is a 퐾2,푡-free colouring of 푆 (in which 푣푎 and 푣푏 necessarily have different colours), since 퐾3,푠 is Ramsey-minimal. Once the existence of a negative signal sender 푆− has been established in which the signal edges are adjacent, it follows along the lines in [13] that BEL-gadgets for 퐾2,2 exist, with the modification that, rather than using 3-connectivity, we use that the graph 푆− above is bipartite and therefore has girth at least 4. The general approach is to glue several copies of 푆− along their signal edges to obtain signal senders (both positive and negative) in which the signal edges are arbitrarily far apart. Due to the similarity, we omit this argument.
Next, we show the existence of a “weak” BEL gadget for 퐾푡,푡 퐾2 conditioned on · 푠(퐾푡,푡 퐾2) > 1. We need this weak version to construct “strong” signal senders in · Lemma 2.5.3.
Lemma 2.5.2. For any 푡 2, let 푅, 퐵 be a colour pattern. There is a graph B̃︀ with ≥ an induced copy of 푅 퐵 so that (B̃︀, 푅 퐵) is 퐾푡,푡-robust and so that the following ∪ ∪ hold.
(1) Any 퐾푡,푡-free colouring in which 푅 is red and 퐵 is blue extends to a 퐾푡,푡-free colouring of B̃︀.
(2) Any 퐾푡,푡 퐾2-free colouring of B̃︀ has colour pattern 푅 퐵. · ∪ Furthermore, if 푅 퐵 is bipartite, then so is B̃︀. ∪
Proof. Let B = B(퐾푡,푡, 푅, 퐵) be a BEL gadget for 퐾푡,푡. B exists by Lemma 2.2.1 for 푡 3, and by Lemma 2.5.1 for 푡 = 2. If 푅 퐵 is bipartite, we may assume that B is ≥ ∪ bipartite. Note that B satisfies Property (1) and satisfies Property (2) with 퐾푡,푡 퐾2 · replaced by 퐾푡,푡. We now modify B to create the desired weak signal sender. To do this, if B is bipartite with parts 퐴, 퐵, then for every set 푆 of 푡 vertices contained either entirely within 퐴 or entirely within 퐵, we add a new set of 푡 + 1 vertices 푉푆 and add a complete bipartite graph between 푉푆 and 푆. If B is not bipartite, we take every set
푆 of 푡 vertices and add a set 푉푆 as above. Note that the degree of each of the vertices
31 2t 1 −
t 1 −
f
Figure 2-1: The graph 퐺0 in the proof of Lemma 2.5.3 with special signal edge 푓. The white circles are all sets of 푡 1 vertices, and thick lines indicate that vertices between those sets are pairwise connected.− The blue edges are all edges in 퐵 and the red edges are all edges in 푅.
of 푉푆 is 푡, so any colouring of B without a monochromatic copy of 퐾푡,푡 extends to
a colouring of the modified graph without a monochromatic copy of 퐾푡,푡 by giving every vertex added this way degree 푡 1 in red and degree 1 in blue. However, if − there is a monochromatic copy of 퐾푡,푡 in B, without loss of generality in colour red, then, picking one of the parts of 푡 vertices, call it 푆, from the monochromatic copy,
either one of the edges from 푆 to 푉푆 is red and we have a red 퐾푡,푡 퐾2, or all of the · edges are blue and the complete bipartite graph between 푆 and 푉푆 contains a blue
퐾푡,푡 퐾2. Note also that we maintain robustness when adding the various 푉푆. · We now construct a version of a signal sender which we call “strong” (negative or positive) signal sender. The reason for this name is that we have control of the colours of the edges incident to one of the signal edges.
− − Lemma 2.5.3. For 푡 2 there is a bipartite graph 푆 = 푆 (퐾푡,푡, 푒, 푓) with two ≥ − independent edges 푒, 푓 so that any 퐾푡,푡 퐾2-free colouring of 푆 satisfies that 푒 and · − 푓 have different colours, and there isa 퐾푡,푡-free colouring of 푆 so that every edge incident to 푓 has a different colour from 푓.
Proof. We first describe a colour pattern 푅, 퐵 and then apply the previous lemma; we
32 will then add some vertices and edges to 푅 퐵 to obtain a graph 퐺0; an illustration ∪ of 퐺0 with colour pattern 푅 퐵 can be found in Figure 2-1. ∪ Let 퐴0, 퐴1, . . . , 퐴2푡−1, 퐵0, 퐵1, . . . , 퐵2푡−1 be disjoint sets of 푡 1 elements each. For − each 1 푖 2푡 1, add a complete bipartite graph between 퐵0 and 퐴푖 , between ≤ ≤ − 퐴0 and 퐵푖, and between 퐴푖 and 퐵푖. Furthermore, add a new edge 푓, say between new vertices and , and add all edges between and ⋃︀ , and add all 푣푎 푣푏 푣푏 1≤푖≤2푡−1 퐴푖 ⋃︀ edges between 푣푎 and 퐵푖. This is the graph 퐺0. All edges between 퐴0 퐵0 1≤푖≤2푡−1 ∪ ⋃︀ ⋃︀ and 퐴푖 퐵푖 form the subgraph 퐵. All edges between 퐴푖 and 1≤푖≤2푡−1 ∪ 1≤푖≤2푡−1 ⋃︀ form the subgraph . 1≤푖≤2푡−1 퐵푖 푅 Now, apply Lemma 2.5.2 to obtain a graph B̃︀ that contains 푅 퐵 as an induced ∪ subgraph such that Property (1) and (2) of the lemma hold. The graph 푆− is obtained
by adding to B̃︀ the vertices 푣푎, 푣푏 and the edges incident to them, as described above. Let 푒 be an arbitrary edge in 푅.
By construction, there exists a 퐾푡,푡-free colouring of B̃︀ that has the colour pattern 푅 퐵. Without loss of generality, we may assume that 푅 is red and 퐵 is blue in ∪ − this colouring. We extend this to a 퐾푡,푡-free colouring of 푆 in the following way.
Colour 푓 blue and colour all other edges incident to 푣푎 and 푣푏 red. By construction, 푓 has a different colour than all edges adjacent to it. It is easy to see that there isno
monochromatic 퐾푡,푡 in this colouring of 퐺0, and, therefore, by robustness there is no
monochromatic 퐾푡,푡.
− It remains to prove that any 퐾푡,푡 퐾2-free colouring of 푆 satisfies that 푒 and 푓 · − have different colours. Let 푐 be a 퐾푡,푡 퐾2-free colouring of 푆 . By construction, the · colouring 푐 has colour pattern 푅, 퐵, say without loss of generality that 푅 is red and 퐵 is blue. Assume for a contradiction that 푓 is red. If there exists 1 푖 2푡 1 ≤ ≤ − such that all the edges 푣푎푥 and 푣푏푦 for 푥 퐵푖, 푦 퐴푖 are red, then 퐴푖 퐵푖 푣푎, 푣푏 ∈ ∈ ∪ ∪ { } forms a red 퐾푡,푡. If 푓 is adjacent to one more red edge, this forms a red 퐾푡,푡 퐾2, a · contradiction. If 푣푏 has blue degree at least 푡, then it and its neighbors along with 퐵0
and any other blue edge out of 퐵0 form a monochromatic 퐾푡,푡 퐾2. The symmetric · statement holds for 푣푎. However, in any colouring in which 푣푏 and 푣푎 both have
blue degree less than 푡, there must be some 푖 so that the edges from 푣푎, 푣푏 are { } 33 monochromatic in red to 퐴푖 퐵푖; then 퐴푖 퐵푖 along with 푣푎 and 푣푏 and one more ∪ ∪ red edge incident to 푣푎 form a red copy of 퐾푡,푡 퐾2, a contradiction. · As in the case of Lemma 2.5.1 and Lemma 2.5.2, the above lemma will imply the existence a stronger version of BEL gadgets for 퐾푡,푡 퐾2, and we omit the proof. · Lemma 2.5.4. For any 푡 2, let 푅, 퐵 be a colour pattern. There is a graph B̃︀ with ≥ an induced copy of 푅 퐵 so that (B̃︀, 푅 퐵) is 퐾푡,푡-robust and so that the following ∪ ∪ hold.
(1) Any 퐾푡,푡 퐾2-free colouring in which 푅 is red and 퐵 is blue and no 퐾푡,푡 in 푅 · is incident or contains any edges of 퐵 and no 퐾푡,푡 in 퐵 is incident or contains
any edges of 푅 extends to a 퐾푡,푡 퐾2-free colouring of B̃︀ in which no vertex · of 푅 퐵 is contained in a monochromatic copy of a 퐾푡,푡, except those vertices ∪ contained in a 퐾푡,푡 within 푅 or within 퐵.
(2) Any 퐾푡,푡 퐾2-free colouring of B̃︀ has colour pattern 푅, 퐵. · Furthermore, if 푅 퐵 is bipartite, then so is B̃︀. ∪ We are now ready to prove the main theorem of this section.
Theorem 2.5.5. For 푡 2, 푠(퐾푡,푡 퐾2) = 1. ≥ ·
Proof. We first describe a graph 퐺0 together with a colour pattern 푅 퐵 퐺0 and ∪ ⊆ then apply Lemma 2.5.4 to force this colour pattern. An illustration of the graph 퐺0 with colour pattern 푅 퐵 can be found in Figure 2-2. ∪ (︀푈)︀ Let 푈 be a set of size 2푡 1, and let 푈푇 : 푇 be a collection of disjoint sets − { ∈ 푡 } of size 2푡, indexed by the 푡-subsets of 푈. Form a graph 퐹 on the union of those sets in the following way. The subgraph 퐹 [푈] forms a clique 퐾2푡−1, and each subgraph (︀푈)︀ (︀푈)︀ 퐹 [푈푇 ] for 푇 forms a clique 퐾2푡. Furthermore, for each 푇 , we choose a ∈ 푡 ∈ 푡 subset 푆푇 of size 푡 1 in 푈푇 and add all edges between this set and 푇 ; we also add, − for each 푇 , a new vertex and connect it to one vertex of 푆푇 . Take two copies, 퐹 and 퐹 ′, of the above graph. Add a vertex 푣 and add all edges ′ between 푣 and 푈 푈 . This graph is 퐺0. The colour pattern 푅 퐵 we define as ∪ ∪ 34 v
S = K2t 1 ∼ − Kt+1 Kt 1 t −
Figure 2-2: The graph 퐺0 in the proof of Theorem 2.5.5 for 푡 = 2 with colour pattern 푅 퐵 where 퐵 consists of all blue edges and 푅 consists of all red edges. ∪ follows:
(︂ )︂ (︂ )︂ 푈 ⋃︁ 푈푇 푅 = 퐹, 2 ∪ 2 ⊆ 푇 ∈ 푈 ( 푡 ) (︂푈 ′)︂ ⋃︁ (︂푈 ′ )︂ 퐵 = 푇 퐹 ′. 2 ∪ 2 ⊆ 푇 ∈ 푈′ ( 푡 )
We claim there is a 퐾푡,푡 퐾2-free colouring of the edges of 퐺0 that extends the · colour pattern 푅 퐵 so that any copy of 퐾푡,푡 in 푅 is not incident to any edge of ∪ 퐵 and any copy of 퐾푡,푡 in 퐵 is not incident to any edge in 푅. To see this, colour 푅 red and 퐵 blue. Further, colour all remaining edges of 퐹 blue and all remaining edges of 퐹 ′ red. Finally, colour all of the edges from 푣 to 푈 red and from 푣 to 푈 ′ blue. The only monochromatic copies of 퐾푡,푡 are contained within one of the 푈푇 or ′ or are contained in along with or along with ′. The edges touching those 푈푇 푣 푈 푣 푈 monochromatic copies are either contained themselves in a monochromatic copy of
퐾푡,푡 or are not contained in 푅 퐵. ∪
We now show that any 퐾푡,푡 퐾2-free colouring of the edges of 퐺0 that extends the · colour pattern 푅 퐵 must satisfy that 푣 is contained in both a red and a blue copy of ∪ 퐾푡,푡. In any such colouring, all of the edges of 푅 must have the same colour, without loss of generality red. Then, since the colouring has no monochromatic 퐾푡,푡 퐾2, all of · the edges leaving any 푈푇 must be blue. Therefore, any 푡 vertices of 푈 form one of the
35 parts in a blue 퐾푡,푡−1. If 푣 had blue degree at least 푡 to 푈, then 푣 along with 푡 of its blue neighbours would be contained in a blue 퐾푡,푡 퐾2, contradicting our assumption. · Therefore, 푣 must have red degree at least 푡 to 푈. In this case, 푣 is contained in a red 퐾푡,푡 with the vertices of 푈. By symmetry, 푣 is contained in a blue 퐾푡,푡 with the vertices of 푈 ′.
Now, applying Lemma 2.5.4 to 퐺0 with 푅, 퐵 and adding a vertex 푤 to B and connecting 푤 only to 푣 gives the desired result.
2.6 Upper bounds for graphs with hanging edges
In this section, we generalize the methods of the previous section. Throughout this section, let 퐻 be a graph that is sufficiently connected, which we define to be a graph that is either 3-connected or isomorphic to the complete bipartite graph 퐾2,푡 ′ with 푡 2. Further, let 퐻 퐻 퐾2. We call a vertex 푤 a distinguished vertex ≥ ∈ · of 퐻 if attaching a pendant edge to 퐻 at 푤 yields a copy of 퐻′. Clearly, if 퐻 is vertex-transitive, any vertex in it is distinguished.
′ Let 훿2(퐺) be the second-smallest degree in 퐺. Note that 훿2(퐻 ) 훿(퐻) + 1. ≤ ′ ′ ′ In this section we show that 푠(퐻 ) 훿2(퐻 ), and if 퐻 is bipartite then 푠(퐻 ) = 1. ≤ Since 퐻′ is not 3-connected, we cannot directly apply Lemma 2.2.1 to construct BEL gadgets for it. By applying Lemma 2.2.1 to 퐻, we will get a weaker version of BEL gadgets for 퐻′. First, however, we need an even simpler lemma.
′ ′ Lemma 2.6.1. If 퐻 is a 2-connected graph and 퐻 퐻 퐾2, then either 푠(퐻 ) = 1 ∈ · or there is a graph 퐹 with a vertex 푢 satisfying that
(1) there is an 퐻′-free colouring of 퐹 in which 푢 is not a distinguished vertex of any monochromatic copy of 퐻, and
(2) in any 퐻′-free colouring of 퐹 , 푢 is incident to edges of both colours.
Furthermore, if 퐻 is bipartite, so is 퐹 .
Proof. Let 푡 be the number of vertices of 퐻. Let 퐹̃︀ be a minimal Ramsey graph for 퐻′ (if 퐻 is bipartite, we may take 퐹̃︀ to be bipartite as well) and obtain a graph 퐹 ′
36 by removing an edge 푒 = 푤0, 푤1 from 퐹̃︀, and adding a pendant edge to both 푤0 { } ′ ′ and 푤1. By minimality, the graph 퐹̃︀ 푒 is not Ramsey for 퐻 . Therefore, if 퐹 is − Ramsey for 퐻′, then one of the pendant edges is necessary for being Ramsey, and thus 푠(퐻′) = 1. Otherwise, 퐹 ′ is not Ramsey for 퐻′. If 퐹 ′ satisfies Property (1) and
(2) with 푢 = 푤0 then we are done. We now split the argument into two cases, based on which of the two properties
퐹 ′ fails to possess. In both cases we conclude that one of the following holds:
1. 푠(퐻) = 1,
2. There is a graph 퐹 (different from 퐹 ′) with the properties desired by the lemma,
3. There is a graph 퐹 ′′ with a special property (푃 ).
After this, we will show that property (푃 ) implies the existence of 퐹 , as desired.
′′ The Property (푃 ) is the following. 퐹 contains 푡 1 vertices 푢 = 푣0, 푣1, 푣2, . . . , 푣푡−2 − that are at pairwise distance at least 푡 so that
′ ′′ (푎) there is an 퐻 -free colouring 휒 of 퐹 in which all edges containing 푢 = 푣0 are
red and, for every 0 푖 푡 2, 푣푖 is not a distinguished vertex of any red copy ≤ ≤ − of 퐻; and
(푏) in any colouring of 퐹 ′′ in which 푢 is incident to edges of only one colour, all of
푣1, . . . , 푣푡−2 are contained as distinguished vertices in a monochromatic copy of 퐻.
′ ′ Assume first that in any 퐻 -free colouring of 퐹 , the vertex 푤0 is a distinguished vertex in a monochromatic copy of 퐻. It is possible that in every 퐻′-free colouring ′ of 퐹 , the vertex 푤0 is a distinguished vertex in a monochromatic copy of 퐻 in both colours. In this case, 푠(퐻) = 1 (just add a pendant edge to 푤0). Otherwise, there is an 퐻-free colouring in which 푤0 is a distinguished vertex in a monochromatic copy of 퐻 in only one colour. In the latter case, the graph 퐹 ′′ may be obtained by taking
푡 2 copies of 퐹 taking 푣1, . . . , 푣푡−2 to be the copies of 푤0, and taking 푢 to be an − isolated vertex.
37 For the second case, assume that there is a colouring of 퐹 ′ without a monochro- ′ matic copy of 퐻 in which 푤0 is not a distinguished vertex in any monochromatic copy of 퐻 and that there is a colouring of 퐹 ′ without a monochromatic copy of 퐻′
in which 푤0 is incident to edges of only one colour. Note that if we add the edge 푒 to 퐹 ′ then 퐹 ′ is Ramsey for 퐻′ and so, in particular, if we take the colouring without a ′ monochromatic copy of 퐻 in which 푤0 is incident to edges of only one colour, say 퐶0, ′ and colour 푒 in the other colour, say 퐶1, we must create a copy of 퐻 ; this copy must
contain 푒 and be of colour 퐶1. But the degree of 푤0 in colour 퐶1 in this colouring is
1, and so 푤0 cannot be contained in a monochromatic copy of 퐻 in colour 퐶1. There- fore, the edge 푒 must be a pendant edge in a monochromatic copy of 퐻′. Note further
that in this colouring 푤0 is not a distinguished vertex in a monochromatic copy of ′ ′ 퐻, since in 퐹 we added a pendant edge to 푤0 and this would create a copy of 퐻 in ′ ′ the colouring of 퐹 , so in the colouring of 퐹 we must have that 푤1 is a distinguished vertex in a monochromatic copy of 퐻. Note that, since we added a pendant edge to
푤1, it cannot be a distinguished vertex in monochromatic copies in both colours, for otherwise there would be a copy of 퐻′ (in the colour of the pendant edge).
Now, obtain 퐹 ′′ as follows. Take 푉 (퐻) (푡 1) copies of 퐹 ′, call them 퐹 ′, 퐹 ′,..., | | − 0 1 ′ . For each copy ′, associate the copy of in ′ with the copy of in 퐹|푉 (퐻)|(푡−1) 퐹푖 푤1 퐹푖 푤0 ′ . Take to be the copy of from ′ and take to be the copy of from 퐹푖+1 푢 푤0 퐹0 푣푖 푤0 ′ . 퐹푖*|푉 (퐻)|
The distance from 푢 to any 푣푖 and between any two 푣푖, 푣푗 is at least 푉 (퐻) by | | construction. There is a colouring without a monochromatic copy of 퐻′ in which 푢 is
incident to edges of only one colour and each 푣푖 is not contained as the distinguished vertex in a monochromatic copy of 퐻 in that same colour; to see this, colour each ′ independently without a monochromatic copy of ′ so that its copy of is 퐹푖 퐻 푤0
monochromatic in colour 퐶0. The connectivity condition on 퐻 guarantees that any monochromatic copy of must be internal to some copy of ′, and we do not create 퐻 퐹푖 any monochromatic copies of 퐻′ since we have said that in any colouring of 퐹 ′ in ′ which there is no monochromatic copy of 퐻 and in which 푤0 is incident to edges of ′ only one colour, 푤1 must not be contained in a monochromatic copy of 퐻 in the other
38 colour. We also see by induction that each 푣푖 must be contained as a distinguished
vertex in a monochromatic copy of 퐻; indeed, we see this for each copy of 푤0 in each of the ′, as by induction the vertex in ′ is contained as a distinguished vertex 퐹푖 푤1 퐹푖−1 in a monochromatic copy of , and so in ′ must be incident to only edges of the 퐻 푤0 퐹푖 other colour in ′, and so the copy of in ′ is also contained as a distinguished 퐹푖 푤1 퐹푖 vertex in a monochromatic copy of 퐻.
We now show how to obtain a graph 퐹 with the desired properties from 퐹 ′′ that has Property . To ′′, add two isolated vertices and ′ and put a copy of on (푃 ) 퐹 푣 푣1 퐻
the vertex set 푢, 푣, 푣1, . . . , 푣푡−2 in which 푣1 is a distinguished vertex and so that 푢푣 forms an edge. This is possible since 퐻 is 2-connected. Finally, add an edge between ′ and . This is the graph . 푣1 푣1 퐹 To see that there is an 퐻′-free colouring of 퐹 in which 푢 is not contained in a monochromatic copy of 퐻 as a distinguished vertex, let 휒 be the colouring of 퐹 ′′
from (푎). Now colour all edges of the copy of 퐻 on vertex set 푢, 푣, 푣1, . . . , 푣푡−2 red, except for the edge 푢푣 which we colour blue. Note that we have not created any new monochromatic copies of 퐻 since 퐻 is 2-connected and the distance in 퐹 ′′ between
any 푣푖 and 푣푗 (0 푖 < 푗 푡 2) is large. Since none of 푢, 푣1, . . . , 푣푡−2 is contained in ≤ ≤ − a red copy of 퐻 as a distinguished vertex, we have also not created a monochromatic copy of 퐻′. Thus, Property (1) follows.
Finally, let 휒′ be a colouring of 퐹 in which 푢 is incident to edges of only one
colour, say red. By Property (푏), each 푣푖, 1 푖 푡 2, is a distinguished vertex in a ≤ ≤ − ′′ monochromatic copy of 퐻 in 퐹 . Suppose, one of those copies, say 퐻푖 that “hangs” at
vertex 푣푖 is red. Then, either it forms a red copy with another edge containing 푣푖 (and
there is nothing to prove), or all other edges not in 퐻푖 that are incident to 푣푖 must
be blue. But then, on the shortest path between 푢 and 푣푖 that misses 푣 (which exists
by 2-connectivity of 퐻) there is some 푣푗 which is incident to edges of both colours, and since it is the distinguished vertex of a monochromatic copy of 퐻 in 퐹 ′′, there is a monochromatic 퐻′. So suppose that all the monochromatic copies of 퐻 in 퐹 ′′, of which the vertices 푣1, . . . , 푣푡−2 are distinguished, are blue. Again, either we find a blue copy of ′, or all edges between the , including the edge ′ are coloured red. 퐻 푣푖 푣1푣1
39 By assumption, all edges containing 푢 are red, including the edge 푢푣, yielding a red ′ ′ copy of 퐻 on the vertex set 푢, 푣, 푣 , 푣1, . . . , 푣푡−2 . { 1 } We now prove a (weak) generalization of Lemma 2.5.2 by using BEL gadgets for
퐻.
′ Lemma 2.6.2. Let 퐻 be a graph that is sufficiently connected and let 퐻 퐻 퐾2. ∈ · Either 푠(퐻′) = 1 or the following holds. Given an 퐻-free colour pattern 푅, 퐵 there is a graph 퐺 with an induced copy of 푅 퐵 so that (퐺, 푅 퐵) is 퐻-robust and: ∪ ∪
(1) There exists an 퐻′-free colouring of 퐺 that extends the colour pattern 푅, 퐵 in which none of the vertices of 푅 퐵 are distinguished vertices of a monochromatic ∪ copy of 퐻.
(2) Any 퐻′-free colouring of 퐺 has the colour pattern 푅, 퐵.
Furthermore, if 푅 퐵 and 퐻 are bipartite, then so is 퐺. ∪
Proof. Let B = B(퐻, 푅, 퐵) be a BEL gadget for 퐻 which exists by Lemma 2.2.1 when
퐻 is 3-connected, and by Lemma 2.5.1 when 퐻 = 퐾2,푡 for some 푡 2. Furthermore, ≥ if 퐻 is bipartite and if 푅 퐵 is bipartite, we may assume that B is bipartite. Note ∪ that B satisfies Property (1) and Property (2) if we replace in the properties 퐻′ by 퐻. For every vertex 푢 of B, add a copy of the graph 퐹 given by the previous lemma (which exists unless 푠(퐻′) = 1) and identify the distinguished vertex of 퐹 with 푢; this is the graph 퐺. Consider the colouring in which 푅 is red and 퐵 is blue. Since B is a BEL gadget for 퐻, this colouring extends to an 퐻-free colouring of B. This colouring extends to an 퐻′-free colouring of 퐺 in which no vertex of B is a distinguished vertex in a monochromatic copy of 퐻. This follows from Property (1) in Lemma 2.6.1, since any copy of 퐹 meets B only in its distinguished vertex 푢, and since 퐻 is connected. Therefore, property (1) holds.
To see that property (2) holds, let 휒 be an 퐻′-free colouring of 퐺. Then by property (2) in Lemma 2.6.1, every vertex 푢 of B is incident to edges of both colours
40 that do not lie inside B. Therefore, the colouring must be 퐻-free on B. And thus, by the property of a BEL gadget, 휒 must extend the colour pattern 푅 퐵. ∪
Using the above lemma, we will construct signal senders in which we have some control over the structure of edges incident to the signal edges, similar to the graphs in Lemma 2.5.3.
′ Lemma 2.6.3. Let 퐻 be a graph that is sufficiently connected, let 퐻 퐻 퐾2 and ∈ · assume that 푠(퐻′) = 1. Take 푡 to be the number of vertices of 퐻 Then there is a graph ̸ 푆− = 푆−(퐻′, 푒, 푓) with two independent edges 푒, 푓 so that any 퐻′-free colouring of 푆− satisfies that 푒 and 푓 have different colours, and there is an 퐻′-free colouring of 푆− so that every edge incident to 푓 has a different colour from 푓, and so that any vertex in 푒 or 푓 is not the distinguished vertex in a monochromatic copy of 퐻. Furthermore, if 퐻 is triangle-free, then 푒 and 푓 are not contained in any triangles.
Proof. Let 휒 be the chromatic number of 퐻 and let 휒 be the set of all 휒-colorings of 퐻. Define 휎 to be the size of the smallest color class over all 휒-colorings of 퐵; i.e., 휎 = min min 휒−1(푖) . 휒∈휒 푖∈[휒] | | If 휒 > 2, let be any (휒 1)-uniform linear hypergraph having girth greater than 푡 ℋ − and chromatic number at least 3. If 휒 = 2, we take 푉 ( ) = 푣1 and 퐸( ) = 푣1 . ℋ { } ℋ {{ }} In either case, set 푁 = 푉 ( ) and 푣1, 푣2, . . . , 푣푁 = 푉 ( ). | ℋ | { } ℋ From we build a graph 퐺0 on 푉 ( ) by joining two vertices if and only if they ℋ ℋ are both contained in some hyperedge of . Throughout this construction we will ℋ keep the underlying structure of in mind. At this point, each hyperedge in ℋ ℋ corresponds to a clique on (휒 1) vertices. − We now build a graph 퐺1 by blowing up each vertex 푣푖 푉 (퐺0) into an inde- ∈ pendent set 푉푖 of size 2푡(푡 1) vertices; that is 푉 (퐺1) = 푉1 푉2 푉푁 with − ⊔ ⊔ · · · ⊔ 푉푖 = 2푡(푡 1) and 푥푦 퐸(퐺1) if and only if there exists a pair of integers 푖, 푗 [푁] | | − ∈ ∈ such that 푥 푉푖, 푦 푉푗, and 푣푖푣푗 퐸(퐺0). Each hyperedge of now corresponds ∈ ∈ ∈ ℋ to a complete (휒 1)-partite graph and the intersection of two hyperedges is either − the empty set or some independent set 푉푖.
41 To 푉 (퐺1) add an independent set 휃 of 휎 1 vertices and join every vertex in 휃 − to every vertex in 푉 (퐺1). Add, for every vertex 푣 from 푉 (퐺1), a new vertex 푤 and make 푣푤 an edge. Add two more vertices and an edge 푒 between them. Let 퐺2 be the graph obtained by this procedure. Take 퐵 = 퐺2.
If 퐻 has an edge 푟 not contained in any 퐾3, take 푥1 and 푥2 to be the endpoints of 푟; note such an edge exists if 퐻 is bipartite. Otherwise, take 푟 to be any edge of
퐻 and 푥1, 푥2 to be its endpoints. Define 퐻˜ to be the (labeled) graph on 푡 1 vertices − ′ obtained from 퐻 by removing the vertices 푥1 and 푥2. We obtain 퐺3 from 퐺2 by inserting the edges of 2푡 vertex-disjoint (labeled) copies of 퐻˜ into each 푉푖. Take 푅 to be the edges we just added, that is the edges of 퐺3 that are not in 퐺2. Let B be the graph obtained by applying Lemma 2.6.2 to the colour pattern 푅, 퐵 (we will show later that 푅, 퐵 has the necessary properties). The vertex set of 푆− is obtained by adding two vertices ′ and ′ to . We add the edge ′ ′ and, 푥1 푥2 푉 (B) 푓 := 푥1푥2 for each and for each copy of ˜ that we added to , we connect ′ to the neighbors 푖 퐻 푉푖 푥1 of in that copy, and ′ to the neighbors of in that copy. 푥1 푥2 푥2 We will now check that the colour pattern 푅, 퐵 does not contain a monochromatic 퐻. Note that 푅 is the disjoint union of connected components that contain at most 푡 1 vertices, so 푅 cannot contain a copy of 퐻. Note also that 퐵, excluding vertices − of degree 1, is a 휒-chromatic graph where one of the parts has size 휎 1, so 퐵 cannot − contain a monochromatic copy of 퐻 by definition of 휎. Indeed, we argue that we may extend to colour the edges incident to ′ ′ 푅, 퐵 푥1, 푥2 without creating a monochromatic copy of 퐻. To do this, colour all of 푅 red, colour all of 퐵 blue, colour 푓 blue, and colour all of the edges incident to 푓 red. Clearly, we do not create a blue copy of 퐻. Note that, in the case 퐻 is 3-connected, any red copy of using ′ and/or ′ must remain connected after removing ′ and ′ , 퐻 푥1 푥2 푥1 푥2 and so vertices of the copy, excluding ′ and ′ , must be contained entirely within 푥1 푥2 the vertex set of one of the copies of 퐻˜ inside one of the 푉푖. If we fix any copy of ˜ and consider the graph induced by this copy along with ′ and ′ , this graph is 퐻 푥1 푥2 isomorphic to 퐻′ in which 푟 is coloured blue rather than red. This graph has no red
42 copy of 퐻, as the vertex of degree 1 cannot be used in a copy of 퐻, and then the number of red edges remaining after this vertex is omitted is fewer than the number
of edges in 퐻. Finally, we wish to show that if the edges of 퐵 are coloured blue, the edges of 푅 are coloured red, and 푓 is coloured red, then there must be a monochromatic copy of
퐻. Recall that the induced graph on 푉푖 consists of 2푡(푡 1) vertex-disjoint red copies − ′ ′ of 퐻˜ . For each such copy of 퐻˜ in 푉푖, if the bipartite graph between 퐻˜ and 푥 , 푥 { 1 2} is colored entirely red, then this forms a red copy of 퐻′. If this is not the case, for
each 푉푖 and all the copies of 퐻˜ in 푉푖, there is at least one blue edge in the bipartite ′ ′ ′ ′ graph between 퐻˜ and 푥 , 푥 . Hence the bipartite graph on 푉푖 푥 , 푥 contains at { 1 2} ∪ { 1 2} ′ least 2(푡 1) blue edges, which implies that either the bipartite graph 푉푖 푥 has − ∪ { 1} ′ at least has 푡 1 blue edges or the bipartite graph 푉푖 푥 has at least 푡 1 blue − ∪ { 1} − edges. In the case 휒 = 2, then the graph 퐵 along with the 푡 1 blue edges contain − a blue 퐾휎,푡−1, which must contain a blue copy of 퐻, and there are other blue edges incident to either part, so taking one of those forms a blue 퐻′, as desired. The remaining case is when 휒 > 2. In this case, we define a 2-colouring of ; we ℋ ′ colour vertex 푣푖 with colour 1 if 푥 has at least 푡 1 blue edges to 푉푖, and otherwise 1 − ′ 푥 has at least 푡 1 blue edges to 푉푖 and we colour vertex 푣푖 with colour 2. Because 2 − has chromatic number at least 3, there must exist a monochromatic edge under ℋ this colouring. Therefore, there is some 푗 [2] and some edge 푠 of the hypergraph ∈ ′ so that 푥 has blue-degree at least 푡 1 to each of the 푉푖 with 푖 푗. However, this 푗 − ∈ forms a complete blue multipartite graph with one part of size 휎 and the remaining parts of size 푡 1, which must contain a blue 퐻′. − As discussed around Lemma 2.5.2, the above lemma immediately gives the follow-
ing version of BEL gadgets for 퐻′.
′ ′ Lemma 2.6.4. If 퐻 is sufficiently connected and 퐻 퐻 퐾2 with 푠(퐻 ) = 1, let ∈ · ̸ 푅, 퐵 be a colour pattern. There is a graph B̃︀ with an induced copy of 푅 퐵 so that ∪ (B̃︀, 푅 퐵) is 퐻-robust and so that the following hold. ∪ (1) Any 퐻′-free colouring in which 푅 is red and 퐵 is blue and no 퐻 in 푅 is incident
43 or contains any edges of 퐵 and no 퐻 in 퐵 is incident or contains any edges of 푅 extends to a 퐻′-free colouring of B̃︀ in which no vertex of 푅 퐵 is contained ∪ as a distinguished vertex in a monochromatic copy of a 퐻, except those vertices contained as a distinguished vertex in a copy of 퐻 within 푅 or within 퐵.
(2) Any 퐻′-free colouring of B̃︀ has colour pattern 푅, 퐵.
′ ′ The previous lemma will allow us to show the upper bound 푠(퐻 ) 훿2(퐻 ). ≤
′ ′ Theorem 2.6.5. If 퐻 is sufficiently connected and 퐻 퐻 퐾2, we have 푠(퐻 ) ∈ · ≤ ′ 훿2(퐻 ).
Proof. If 푠(퐻) = 1, we are done, and otherwise we may apply Lemma 2.6.4. Take 푡 to be the number of vertices of 퐻. Take 퐵 to be 푡 1 vertex-disjoint cliques, each on − 푡 vertices. Apply Lemma 2.6.4 to this colour pattern (푅 is empty) to get a graph B.
Pick vertices 푣1, . . . , 푣푡−1, one from each of the cliques from 퐵. Pick a vertex 푣푡 = 푣1 ̸ from the first clique. Add edges to form a clique on 푣1, . . . , 푣푡−1, and add one more ′ edge between 푣1 and 푣푡. Observe that this is not Ramsey for 퐻 , as we may colour all of 퐵 blue and all of the edges added to 퐵 red; in blue there is no copy of 퐻′ by construction, and in red we have a connected component on 푡 vertices, but one of those vertices has degree 1 so there is no red 퐻 in this part of the graph. We will ′ now add one more vertex 푣 of degree 훿2(퐻 ) and show that with this vertex the graph ′ ′ is Ramsey for 퐻 . We first consider the case where there is a vertex ofdegree 훿2(퐻 ) that is not incident to the vertex of degree 1 in 퐻′. Add a vertex 푣 and connect it to . Otherwise, if the vertex of degree ′ is incident to the vertex of 푣1, . . . , 푣훿2(퐻′) 훿2(퐻 ) degree , then connect to and to . In either case, if is coloured 1 푣 푣1, . . . , 푣훿2(퐻′)−1 푣푡 퐵 blue, then if 푣 has any outgoing blue edges it is contained in a blue 퐻′ as the pendant vertex, and, otherwise, all of its outgoing edges are red and it is contained in a red
퐻′.
44 2.7 The complete graph with an added vertex
We define 퐻푡,푑 to be the graph on 푡 + 1 vertices that contains a 퐾푡 and in which the remaining vertex (not in the 퐾푡) has degree 푑, with its neighbors being any 푑 vertices of the 퐾푡.
2 Note 퐻푑,푑 is isomorphic to 퐾푑+1, for which 푠(퐾푑+1) is known to be 푑 [12]. For
푑 = 1, it was recently shown that 푠(퐻푡,1) = 푡 1 [42]. For 푑 = 0, it was found − 2 푠(퐻푡,0) = 푠(퐾푡) = (푡 1) [88]. A natural question that arises is how 푠(퐻푡,푑) behaves − when 푑 is between 1 and 푡. We now state the main result of this section.
Theorem 2.7.1. For all 1 < 푑 < 푡 we have
2 푠(퐻푡,푑) = 푑 .
The proof of this theorem is presented in two parts. In the first part, we prove
2 that 푠(퐻푡,푑) 푑 for all values of 푑. The second part expands on the ideas in [12] ≥ and [42] and deals with the upper bound on 푠(퐻푡,푑) for 푑 2: we construct a graph ≥ 2 퐺 with a vertex 푣 of degree 푑 that is Ramsey for 퐻푡,푑 such that 퐺 푣 퐻푡,푑. It − ̸→ 2 2 follows from this that 푠(퐻푡,푑) 푑 , and so we obtain 푠(퐻푡,푑) = 푑 for all 1 < 푑 < 푡. ≤ We now begin with the first part of our proof, which closely follows the ideas of[12].
Lemma 2.7.2. Let 퐻 be a graph such that, for all 푣 푉 (퐻), the neighborhood of 푣 ∈ 2 contains a copy of 퐾푑. Then 푠(퐻) 푑 . ≥
Proof. Suppose there exists 퐹 (퐻) and some 푣 푉 (퐹 ) with deg (푣) < 푑2. Since ∈ ℳ ∈ 퐹 is minimal, we can 2-color the edges of 퐹 푣 so that there is no monochromatic − copy of 퐻. Consider any such 2-coloring of 퐹 푣. In this coloring, let 푆 denote the − neighborhood of 푣 and let 푇1, . . . , 푇푘 be a maximal set of vertex-disjoint red copies of 2 2 퐾푑 in 푆. Since deg (푣) < 푑 , we must have 푆 < 푑 , and so 푘 푑 1. Now we color | | ≤ − all the edges connecting 푣 to 푇1, . . . , 푇푘 blue, and all other edges incident to 푣 red. We claim that no monochromatic copy of 퐻 arises in such a coloring. Note that such a copy would need to use 푣. We will now show that there is no red 푑-clique in the
45 red neighborhood of 푣 and that there is no blue 푑-clique in the blue neighborhood of 푣, thus showing that 푣 cannot be contained in any monochromatic copy of 퐻.
Any red 푑-clique in 푆 must intersect one of 푇1, . . . , 푇푘 and therefore would have a blue edge from 푣. On the other hand, suppose there exists a blue 푑-clique in the blue neighborhood of 푣, which is precisely 푇1 푇푘. Since 푘 푑 1, by the pigeonhole ∪ · · · ∪ ≤ − principle, at least two vertices of this blue 푑-clique must be contained in the same 푇푖. These two vertices, however, are connected by a red edge, a contradiction. It follows that such an 퐹 (퐻) cannot exist, and hence 푠(퐻) 푑2. ∈ ℳ ≥
Since the neighborhood of each vertex in 퐻푡,푑 contains a copy of 퐾푑, we have the following corollary.
2 Corollary 2.7.3. For all values of 푑 we have 푠(퐻푡,푑) 푑 . ≥ This completes the first part of our proof, establishing a lower bound on the value of 푠(퐻푡,푑). For the upper bound, we wish to construct an 퐻-minimal graph with vertex of 2 degree exactly 푑 for 푑 2. To that end, we wish to show that 퐻푡,푑 has BEL gadgets. ≥ Theorem 2.2.1 implies this in the case 푑 3, but not when 푑 = 2; the majority of ≥ the work in this section is proving that 퐻푡,2 has BEL gadgets.
Theorem 2.7.4. For all 2 푑 푡, the graph 퐻푡,푑 has BEL gadgets. ≤ ≤ We postpone the proof of this theorem to the end of the section; let us first see why it implies the desired upper bound on 푠(퐻푡,푑).
Lemma 2.7.5. For all 2 푑 푡 there exists a graph 퐹 ′ with vertex 푣 of degree 푑2 ≤ ≤ ′ ′ so that 퐹 퐻푡,푑 but 퐹 푣 퐻푡,푑. → − ̸→
2 Proof. If 푑 = 푡 then 푠(퐻푡,푑) = 푑 by [12], which immediately implies the lemma; we will henceforth assume 푑 < 푡.
The graph 퐻푡,푑 has BEL gadgets by Theorem 2.7.4. This means that, for any graph 퐺 and 2-coloring 휓 of 퐺 without a monochromatic copy of 퐻, there exists a graph 퐹 퐻푡,푑 with an induced copy of 퐺 such that every 2-coloring of 퐹 without ̸→ 46 a monochromatic copy of 퐻푡,푑 agrees with 휓 on the copy of 퐺, up to permutation of colors. We describe our graph 퐺 together with its coloring 휓 for our BEL gadget as follows:
1. 퐺 contains 푑 disjoint red copies 푇1, . . . , 푇푑 of 퐾푡,
2. For each distinct pair 푖 and 푗, there is a complete blue bipartite graph between
푇푖 and 푇푗, and
3. For each way there is to choose a 푑-tuple 푇 = (푡1, . . . , 푡푑) 푇1 푇푑 by ∈ × · · · × 푇 푇 taking one vertex from each 푇푖, we add a set of 푡 푑 vertices 푆푇 = 푣 , . . . , 푣 ; − { 1 푡−푑} we add blue edges between all pairs of vertices in 푆푇 so that 푆푇 becomes a blue clique, and add more blue edges so that there is a complete blue bipartite graph
between and . For distinct -tuples and ′, and are disjoint. 푆푇 푇 푑 푇 푇 푆푇 푆푇 ′
An example of this 퐺 with coloring 휓 is shown in Figure 2-3. We first claim that this
coloring 휓 contains no monochromatic copy of 퐻푡,푑. The connected components in
red are all copies of 퐾푡, so there is no red copy of 퐻푡,푑. We also claim there is no blue
copy of 퐻푡,푑. If we omit the vertices that are contained in the various 푆푇 , the blue
graph is 푑-partite and so contains no 퐾푡, as 푑 < 푡. Therefore, any blue copy of 퐻푡,푑
must use some vertex 푤 in some 푆푇 as part of a blue 퐾푡. Note that the blue degree of
푤 is 푡 1, and therefore this blue 퐾푡 must consist precisely of 푤 and its neighborhood. − However, any vertex that is not 푤 or contained in the blue neighborhood of 푤 has degree at most 푑 1 to the neighborhood of 푤 by construction, and so cannot be the − vertex of degree 푑 in 퐻푡,푑. Therefore, there is no blue copy of 퐻푡,푑.
Consider a graph 퐹 퐻푡,푑 with an induced copy of 퐺 such that any 2-coloring ̸→ of 퐹 without a monochromatic copy of 퐻푡,푑 restricts to the coloring 휓 on the induced copy of 퐺, up to permutation of the colors; this exists by Theorem 2.7.4. We now ′ modify 퐹 to 퐹 by adding a vertex 푣, and adding 푑 edges from 푣 to each 푇푖 in the induced copy of 퐺. The vertex 푣 clearly has degree 푑2. We claim that this modified ′ ′ graph 퐹 is Ramsey for 퐻푡,푑. Consider any 2-coloring of 퐹 . In this 2-coloring, if ′ there is a monochromatic copy of 퐻푡,푑 in the subgraph 퐹 = 퐹 푣, then we are done. − 47 Otherwise suppose the 2-coloring does not yield a monochromatic copy of 퐻푡,푑 in 퐹 . Then the induced graph 퐺 must have coloring 휓, up to permutation of colors. Let us
assume without loss of generality that each 푇푖 forms a red clique and the remaining edges are blue.
If 푣 had red degree 푑 to some 푇푖, then 푣 together with 푇푖 would be a red copy of
퐻푡,푑. Thus, at least one edge from 푣 to each copy of 푇푖 must be colored blue. Choose one vertex 푡푖 from each 푇푖 so that 푣 has a blue edge to 푡푖 and take 푇 = (푡1, . . . , 푡푑).
Then these vertices 푡푖 together with 푆푇 forms a blue 퐾푡, and adding 푣 creates a blue
퐻푡,푑.
This immediately gives the desired upper bound on 푠(퐻푡,푑).
2 Corollary 2.7.6. For every 2 푑 푡, we have 푠(퐻푡,푑) 푑 . ≤ ≤ ≤
Proof. By the previous lemma, there is a graph 퐹 ′ with a vertex 푣 of degree 푑2 which ′ ′′ is Ramsey for 퐻푡,푑 so that 퐹 푣 is not Ramsey for 퐻푡,푑. Take 퐹 to be a subgraph − of 퐹 ′ which is minimal subject to the constraint that 퐹 ′′ is Ramsey for 퐹 . 퐹 ′′ must ′′ 2 contain 푣, and so 푠(퐻푡,푑) 훿(퐹 ) 푑 , as desired. ≤ ≤
푆푇
푣푥
푣푦 푣푧
Figure 2-3: Example of 퐺 with the coloring 휓 for 푡 = 5 and 푑 = 3. Here, only one set 푆푇 is shown, corresponding to the triple 푇 = (푣푥, 푣푦, 푣푧). The dashed blue edges represent complete blue bipartite graphs. When we add the external vertex 푣, we will 2 connect it to three vertices from each copy of 퐾5, making its degree 푑 = 9.
48 We now prove that 퐻푡,2 has BEL gadgets. Note that, for 푡 = 2, the graph 퐻2,2 is isomorphic to 퐾3, for which it is known that BEL gadgets exist [12]. Henceforth, we will assume that 푡 3. The ideas behind the proof of BEL gadgets for 퐻푡,2 stems ≥ from a strategy in [42]. We now introduce the main tool that we will need.
휖 Definition 2.7.7. Write 퐹 퐻 to mean that, for every 푆 푉 (퐹 ) such that 푆 → ⊆ | | ≥ 휖 푉 (퐹 ) , the subgraph of 퐹 induced by 푆 is Ramsey for 퐻 (i.e. 퐹 [푆] 퐻). | | → The following lemma, which is a strengthening of a theorem in [76], is proven in [42].
Lemma 2.7.8. For any graph 퐻 and every 휖 > 0 and 푡 > 2, if 휔(퐻) < 푡 then there 휖 exists a graph 퐹 that is 퐾푡-free such that 퐹 퐻. →
We are now ready to construct a graph 퐺0 so that, for every coloring of 퐺0 without
a monochromatic copy of 퐻푡,2, a particular copy of some (arbitrary) graph 푅0 is forced
to be monochromatic. Furthermore, there is a coloring of 퐺0 where 푅0 is red, all of
the edges leaving 푅0 are blue, there is no red 퐻푡,2, and there is no blue 퐾푡. The proof of this lemma closely follows the arguments in [42].
Lemma 2.7.9. Let 푅0 be a graph that has no copy of 퐻푡,2. Then there exists a graph
퐺0 with an induced copy of 푅0 and the following properties:
1. There is a 2-coloring of 퐺0 without a red copy of 퐻푡,2 and without a blue copy
of 퐾푡 in which the edges of 푅0 are red, and all of the edges incident to, but not
contained in, 푅0 are blue, and
2. Every 2-coloring of 퐺0 without a monochromatic copy of 퐻푡,2 results in 푅0 being monochromatic.
−푛−푡2 Proof. Take 휖 = 2 , where 푛 is the number of vertices in 푅0. Let 퐹1, 퐹2, . . . , 퐹푡−2
be copies of the graph as defined in Lemma 2.7.8 when applied to 퐻 = 퐻푡−1,1. We
claim that the graph 퐺0 := 퐹1 퐹2 퐹푡−2 푅0 satisfies both desired conditions ··· (see Figure 2-4).
49 To see the first property, color all the edges internal to anyof 퐹1, 퐹2, . . . , 퐹푡−2, 푅0 red and the remaining edges blue. There can be no monochromatic red copy of 퐻푡,2, since each 퐹푖 is 퐾푡-free and 푅0 is 퐻푡,2-free. Furthermore, there is no blue 퐾푡, since the graph induced by the blue edges is (푡 1)-chromatic. −
To see the second property, we consider some 2-coloring 휓 of 퐺0 so that 퐺0 does not have a monochromatic copy of 퐻푡,2. We show that this forces 푅0 to be monochromatic. For a subset 푆 of the vertices and some vertex 푣 푆, define the ̸∈ color pattern 푐푣 with respect to 푆 to be the function with domain 푆 that maps a vertex 푤 푆 to the color of the edge (푣, 푤). This method was utilized in [42]. ∈
For a vertex 푣 퐹1, consider its color pattern 푐푣 with respect to 푉 (푅0). There ∈ 푛 −푛 are 2 possible color patterns, so at least a 2 fraction of the vertices in 퐹1 have the same color pattern with respect to 푉 (푅0). Call the set of these vertices 푆1. Then −푛 푆1 2 푉 (퐹1) 휖 푉 (퐹1) , so there must exist a monochromatic copy 퐻1 | | ≥ · | | ≥ · | | isomorphic to 퐻푡−1,1 in 푆1. Without loss of generality, suppose 퐻1 is monochromatic in red. We claim that all the edges going from 푆1 to 푅0 (and in particular from
퐻1 to 푅0) are blue. Indeed, since all vertices 푣 푆1 have the same color pattern ∈ with respect to 푅0, then for a fixed vertex 푖 푅0 the edges (푖, 푣) have the same ∈ color for all 푣 푆1. If that color is red, then 푖 along with all the vertices of 퐻1 ∈ would form a monochromatic red copy of 퐻푡,2, which contradicts our definition of
휓. We now proceed inductively. Suppose we have identified red copies of 퐻푡−1,1 labeled 퐻1, . . . , 퐻푘−1 in 퐹1, . . . , 퐹푘−1 with vertex sets 푉1, . . . , 푉푘−1 respectively, and that all edges between these copies as well as to 푅0 are blue. In 퐹푘, at least a −푛−푡(푘−1) 2 > 휖 fraction of the vertices 푆푘 have the same color pattern with respect to
푉 (푅0) 푉 (퐻1) 푉 (퐻푘−1). Since 푆푘 > 휖 푉 (퐹푘) , we have 퐹 [푆푘] 퐻푡−1,1. Find ∪ ∪· · ·∪ | | ·| | → a monochromatic copy of 퐻푡−1,1 and call it 퐻푘. Suppose 퐻푘 is blue. Then, as in the case before, all the edges between 퐻푘 and 푅0, as well as to 퐻1, . . . , 퐻푘−1, would have to be red, otherwise there would be a monochromatic blue copy of 퐻푡,2. But if all these edges are red, then any vertex of 퐻푘 along with 퐻1 forms a monochromatic copy of 퐻푡,2, a contradiction. Thus, 퐻푘 must be red, and consequently all edges between
퐻푘 and 퐻1, . . . , 퐻푘−1, 푅0 must be blue, completing the inductive step. After applying
50 this argument 푡 2 times, we have a collection (퐻1, . . . , 퐻푡−2) of red copies of 퐻푡−1,1 − with complete bipartite blue graphs between any two of them. Now, suppose some
edge in 푅0 was blue. Then this edge, along with one vertex in each of 퐻1, . . . , 퐻푡−2
and one other arbitrary vertex in 퐻1 forms a monochromatic blue copy of 퐻푡,2. Thus,
all the edges in 푅0 must be colored red, as required.
퐹1 퐹2 퐹3
푅0
Figure 2-4: Construction of the gadget graph 퐺0 for 푡 = 5 and 푑 = 2. The dashed lines represent complete bipartite graphs.
We now introduce a lemma which is a stronger version of an idea first introduced in [12] known as a positive signal sender.
Lemma 2.7.10. There is a graph 퐺 with two independent edges 푒 and 푓 so that, in any 2-coloring of 퐺 without a monochromatic copy of 퐻푡,2, both edges 푒 and 푓 must have the same color. Furthermore, there is a 2-coloring of 퐺 with no red 퐻푡,2 and no
blue 퐾푡 in which both edges 푒 and 푓 are red, and in which all of the edges incident to either of 푒 or 푓 are blue. Furthermore, there are no edges incident to both 푒 and 푓.
Proof. This follows by taking 푅0 in the previous lemma to be two disjoint edges, 푒 and 푓.
51 We now take the above lemma and use it to prove a slight strengthening of itself.
Lemma 2.7.11. There is a graph 퐺 with two independent edges 푒 and 푓 so that in any 2-coloring of 퐺 without a monochromatic copy of 퐻푡,2 both edges 푒 and 푓 must have the same color. Furthermore, there is a 2-coloring of 퐺 with no red 퐻푡,2 and no blue 퐾푡 in which both edges 푒 and 푓 are red, and in which all of the edges incident to either of 푒 or 푓 are blue. Furthermore, any path between a vertex of 푒 and a vertex of 푓 has length at least 3.
Proof. Lemma 2.7.10 gave us a graph that satisfied all of these constraints except for the last one. Take two copies 퐺′, 퐺′′ of this graph from Lemma 2.7.10, with distinguished pairs of edges (푒′, 푓 ′) and (푒′′, 푓 ′′), respectively. Identify 푓 ′ with 푒′′ and take 푒 = 푒′ and 푓 = 푓 ′′, and call the resulting (combined) graph 퐺. By construction, any path between a vertex of 푒 and a vertex of 푓 has length at least 3. Also by construction, in any 2-coloring of 퐺 without a monochromatic copy of 퐻푡,2, we must have that 푒 = 푒′ and 푓 ′ have the same color, and 푓 ′ = 푒′′ and 푓 ′′ = 푓 have the same color, and so 푒 and 푓 have the same color. Finally, if we color 푒, 푓 ′, and 푓 all red, then we may extend this to colorings of 퐺′ and 퐺′′ so that neither 퐺′ nor 퐺′′ contains a red
퐻푡,2 or a blue 퐾푡 so that all edges incident to either of 푒 or 푓 are blue. This coloring contains no red 퐻푡,2, as every connected component in red is contained entirely within ′ ′′ at least one of 퐺 and 퐺 , and neither one of these graphs has a red copy of 퐻푡,2.
There is no blue copy of 퐾푡, as every blue triangle is contained either entirely within ′ ′′ 퐺 or entirely within 퐺 , and neither one contains a blue copy of 퐾푡.
The next lemma uses these so-called strong positive signal senders to construct a weaker version of BEL gadgets for 퐻푡,2. It is weaker because it does not guarantee that we can agree with a given coloring 휓 of a graph up to permutation of colors; it only guarantees that in a monochromatic 퐻푡,2-free coloring of the graph, the edges that are red in 휓 all end up with one color 훼1 and the edges that are blue in 휓 all end up with one color 훼2. The two colors 훼1 and 훼2 may be the same. After proving this lemma, we will then show that the existence of this weaker version of BEL gadgets implies the full strength of the BEL theorem, completing the proof.
52 Lemma 2.7.12. Given edge-disjoint graphs 퐺0 and 퐺1 on the same vertex set that
are both 퐻푡,2-free, there is a graph 퐺 with an induced copy of 퐺0 퐺1 so that there ∪ is a 2-coloring of 퐺 without a monochromatic copy of 퐻푡,2 in which 퐺0 is red and 퐺1
is blue. Furthermore, in any 2-coloring of 퐺 without a monochromatic 퐻푡,2, all the
edges in 퐺0 have the same color and all the edges in 퐺1 have the same color.
Proof. Take 퐹 to be a copy of the graph given by Lemma 2.7.11.
Form a graph 퐺 as follows. Start with 퐺0 퐺1 on the same vertex set. Add two ∪ edges 푒0 and 푒1 independent from both 퐺0 and 퐺1. For any edge 푓0 in 퐺0, we add
a copy of 퐹 with 푒0 and 푓0 as the distinguished edges. For any edge 푓1 in 퐺1, we
add a copy of 퐹 with 푒1 and 푓1 as the distinguished edges. By construction, in any
2-coloring of 퐺 without a monochromatic 퐻푡,2, all of the edges in 퐺0 have the same
color and all of the edges in 퐺1 have the same color.
Consider coloring all edges of 퐺0 as well as 푒0 red and all edges of 퐺1 as well as
푒1 blue. By construction of 퐹 , we may extend this coloring to a coloring of 퐺 in
which every copy of 퐹 attached to two edges in 퐺0 contains no blue 퐾푡 and no red
퐻푡,2 and in which all of the edges of 퐹 that are incident to the two edges are blue.
Symmetrically, in this coloring every copy of 퐹 attached to two edges in 퐺1 contains
no red 퐾푡 and no blue 퐻푡,2 and satisfies that all of the edges of 퐹 that are incident to the two edges are red.
We claim there is no blue 퐻푡,2. By symmetry it will follow that there is also no
red 퐻푡,2. First, observe that if we pick any two edges (푒, 푓) to which a copy of 퐹 is attached, the vertices of any triangle in 퐺 are either contained entirely in 퐹 or entirely in the graph 퐺′ obtained by removing the vertices of 퐹 except 푒 and 푓; this follows immediately from the construction. Note further that any triangle that is not
contained entirely in 퐺′ must use some vertex 푤 that belongs to 퐹 but not to 퐺′; since there is no vertex in 퐹 that has as a neighbor both a vertex of 푒 and a vertex of 푓, such a triangle may not use both a vertex of 푒 and a vertex of 푓; in particular, this means that all of the edges used by the triangle are contained in 퐹 (note that there are no edges between 푒 and 푓 that are not contained in 퐹 , by the way we
constructed 퐺0 and 퐺1). Therefore, any copy of 퐾푡 must be contained entirely in the
53 ′ edges of 퐹 or in entirely in 퐺 . Since there is no blue 퐾푡 in the copies of 퐹 attached
to edges from 퐺0, any blue copy of 퐻푡,2 must have its copy of 퐾푡 contained entirely
in 퐺1 or entirely in some copy of 퐹 attached to an edge of 퐺1. If we take a blue 퐾푡 contained in some copy of 퐹 attached an edge 푒 and some edge 푓 in 퐺1, then, since all of the edges incident to both 푒 and 푓 are red, if we take the connected component corresponding to the blue subgraph of 퐺 containing this copy of 퐾푡, we see that it is contained entirely in this copy of 퐹 . But by assumption this copy of 퐹 has no 퐻푡,2, and so this blue 퐾푡 is not contained in any copy of 퐻푡,2. Therefore, any blue copy of 퐻푡,2 must have its 퐾푡 contained in 퐺1. By assumption, 퐺1 contains no copy of
퐻푡,2, so this copy must have some vertex outside of 퐺1 that has blue degree at least
2 to this copy of 퐾푡. Such a vertex cannot be contained in the copies of 퐹 attached to an edge of 퐺1, as these are completely red to 퐺1. Therefore, this vertex must be contained in some copy of 퐹 attached to an edge 푒 and an edge 푓 of 퐺0. But neither 푒 nor 푓 may be edges of the blue clique, since they are both red, and so this vertex must have a blue neighbor in 푒 and a blue neighbor in 푓, but this contradicts our assumptions on 퐹 , concluding the proof.
If a graph 퐻 satisfies the conclusions of the above lemma, we say ithas weak BEL
gadgets. We now prove that this is enough to get strong BEL gadgets for 퐻푡,2, thus completing the proof of the upper bound.
Lemma 2.7.13. If 퐻 is connected and has weak BEL gadgets, then 퐻 has BEL gadgets.
Proof. Consider a graph 퐺 with a given 2-coloring 휓. Let 퐺 be composed of the graphs ′ and ′ , where ′ is the graph induced by the blue edges of and ′ is 퐺0 퐺1 퐺0 퐺 퐺1 the graph induced by the red edges of 퐺. Take 푡 to be the number of vertices in 퐻. Define a graph by taking ′ , adding to it some set of vertices, and adding 퐺0 퐺0 푆 푡 edges to so it forms a copy of with one edge removed. Define by taking ′ , 푆 퐻 퐺1 퐺1 adding to it 푆, and adding to 푆 the edge that was removed from 퐻. We will show that this resulting graph can be made a strong BEL gadget for 퐻. Note that neither
퐺0 nor 퐺1 contains a copy of 퐻; the connected components are either connected
54 components of 퐺0 or 퐺1, or are in 푆. Note further that in any 2-coloring of 퐺0 퐺1 ∪ in which all of the edges in 퐺0 have the same color and all of the edges of 퐺1 have
the same color, if 퐺0 and 퐺1 have the same color then there is a monochromatic copy
of 퐻, namely on vertex set 푆. Now, taking a weak BEL gadget for 퐺0 and 퐺1 yields the desired strong BEL gadget for ′ and ′ . 퐺0 퐺1
2.8 Concluding Remarks
We have shown that all 3-connected bipartite graphs are Ramsey simple. However, Conjecture 2.1.1 remains open.
Conjecture 2.8.1. [88] If 퐻 is bipartite with no isolated vertices, then 퐻 is Ramsey simple.
We have also demonstrated the first class of graphs that is not bipartite butis Ramsey simple, namely those graphs that have BEL, that don’t have isolated vertices,
and that contain a vertex of minimum degree 훿 whose neighbourhood is contained in an independent set of size 2훿 1. One may hope to use similar techniques to those − found in the proof of Theorem 2.1.2 to get a corresponding result for triangle-free graphs. This leads us to the following conjecture.
Conjecture 2.8.2. Every connected triangle-free graph without isolated vertices is Ramsey simple.
Based on the techniques used here, it may be significantly easier to prove the conjecture just for 3-connected graphs.
Conjecture 2.8.3. Every 3-connected triangle-free graph is Ramsey simple.
2.9 Sparse random graphs are Ramsey simple
We prove Corollary 2.1.4 that the random graph 퐺(푛, 푝) with 푛−1 log 푛 푝 푛−2/3 ≪ ≪ is Ramsey simple with high probability. This follows from showing that 퐺(푛, 푝) satisfies the conditions of Theorem 2.1.3 with high probability.
55 Proof of Corollary 2.1.4. Let 푛−1 log 푛 푝 푛−2/3 and assume that the vertex set ≪ ≪ of 퐺(푛, 푝) is [푛].
We will use the following basic facts about 퐺(푛, 푝), for 푝 in the aforementioned range.
(1) For every 푑, 푃 [훿(퐺(푛, 푝)) = 푑] = 표(1). (2) 푃 [훿(퐺(푛, 푝)) 푝푛] = 1 표(1). ≤ − (3) With high probability, there exists a unique vertex in 퐺(푛, 푝) of minimum degree. (4) With high probability, 퐺(푛, 푝) is 3-connected. (5) For any 푑, we have 푃 [훿(퐺(푛 1, 푝)) 푑 1] 푃 [훿(퐺(푛, 푝)) 푑] 표(1). − ≥ − ≥ ≥ − Facts (1), (2), and (3) follow from Theorem 3.9(i) and its proof in [8], and (4) follows from Theorem 7.7 in [8]. We now prove fact (5).
Consider the vertex 푛 [푛]. Since with high probability there is a unique vertex ∈ of minimum degree by fact (3), the probability that 푛 is this vertex is 표(1). Then, in the distribution obtained by taking 퐺(푛, 푝) and removing the vertex 푛, every vertex has degree at least its degree in the 퐺(푛, 푝) minus 1. Since with high probability the minimum degree vertex of 퐺(푛, 푝) was in [푛 1], we have that the minimum degree − of this 퐺(푛 1, 푝) is at least 훿(퐺(푛, 푝)) 1. This completes the proof of (5). − − We wish to show that, with high probability, 퐺 = 퐺(푛, 푝) contains a minimum degree vertex, call its degree 훿, whose neighborhood is contained in an independent set of size 2훿 1. Combining this with fact (4) and applying Theorem 2.1.3 immediately − implies Corollary 2.1.4.
For any 0 < 휀 < 1, pick 푑 minimal so that 푃 [훿(퐺(푛, 푝)) 푑] 1 휀. We will let ≤ ≥ − 휖 very slowly tend to 0. By (2) we must have that 푑 푝푛. By (1) we must have that ≤
1 휀 푃 [훿(퐺(푛, 푝)) 푑] 1 휀 + 표(1). − ≤ ≤ ≤ −
For every set 푆 [푛] with 푆 푑 and every vertex 푣 [푛] 푆, take 푇푆,푣 to be some ⊆ | | ≤ ∈ ∖ set containing 푆 of size 2 푆 1 so that 푇푆,푣 does not contain 푣. For any such 푆 and | | − 푣, let 퐴푆,푣 be the event that 푣 is the unique vertex of minimum degree in 퐺(푛, 푝) and that its neighbourhood is 푆. Let 퐵푆,푣 be the event that 푇푆,푣 is not an independent
56 set. Note that the probability that there is some vertex 푣 of minimum degree whose neighbourhood is contained in an independent set of size 2 deg(푣) 1 is at least the − probability that there exists a set 푆 of size at most 푝푛 and a vertex 푣 not in 푆 so that 퐴푆,푣 holds and 퐵푆,푣 does not hold. We will show that this is true with high probability.
Note that the events 퐴푆,푣 are disjoint. With probability at least 1 휖 표(1), which − − is 1 표(1) as 휖 tends to 0, the minimum degree is at most 푑 and there is a unique − vertex of minimum degree. Equivalently, exactly one of the events 퐴푆,푣 occurs. We next show that the conditional probability 푃 [퐵푆,푣 퐴푆,푣] is 표(1), which completes the | proof.
We consider the distribution 퐺(푛, 푝) 퐴푆,푣. For convenience, we will assume that | 푣 = 푛. Then, in this distribution, the neighbours of 푣 are 푆, and the remaining vertices form the distribution 퐺(푛 1, 푝) conditioned on the event that all vertices in − 푆 have degree at least 훿 and all vertices in [푛 1] 푆 have degree at least 훿 + 1; call − ∖ this event 퐶푆. Hence, 푃 [퐵푆,푣 퐴푆,푣] = 푃 [퐵푆,푣 퐶푆], where the first probability is taken | | with respect to 퐺(푛, 푝) and the second probability is taken with respect to 퐺(푛 1, 푝). −
By definition, we have 푃 [퐵푆,푣 퐶푆] = 푃 [퐵푆,푣 퐶푆]/푃 [퐶푆] 푃 [퐵푆,푣]/푃 [퐶푆]. The | ∧ ≤ expected number of edges in 푇푆,푣 in 퐺(푛 1, 푝) is − (︂2 푆 1)︂ (︂2푑)︂ | | − 푝 푝 4푑2푝 4푝3푛2 = 표(1), 2 ≤ 2 ≤ ≤ where the last inequality uses 푑 푝푛, and the last equality uses 푝 = 표(푛−2/3). Since ≤ the probability that there is an edge in 푇푆,푣 is at most the expected number of edges in 푇푆,푣, we have 푃 [퐵푆,푣] = 표(1).
We next give a lower bound for 푃 [퐶푆]. Note that 퐶푆 holds if 훿(퐺(푛 1, 푝)) 훿+1. − ≥ Hence,
푃 [퐶푆] 푃 [훿(퐺(푛 1, 푝)) 훿 + 1] 푃 [훿(퐺(푛, 푝)) 훿 + 2] 표(1) ≥ − ≥ ≥ ≥ − 푃 [훿(퐺(푛, 푝)) 푑 + 2] 표(1) 푃 [훿(퐺(푛, 푝)) 푑 + 1] 표(1) 휖 표(1) ≥ ≥ − ≥ ≥ − ≥ − 57 where the second inequality is by fact (5), the third inequality uses 훿 푑, the fourth ≤ inequality follows from fact (2), and the last inequality holds by the choice of 푑. Putting this together, we have
푃퐺(푛,푝)[퐵푆,푣 퐴푆,푣] = 푃퐺(푛−1,푝)[퐵푆,푣 퐶푆] | | ≤
푃퐺(푛−1,푝)[퐵푆,푣]/푃퐺(푛−1,푝)[퐶푆] = 표(1)/(휖 표(1)) = 표(1), − where the last inequality is by taking 휖 to tend to 0 slower than 푃퐺(푛,푝)[퐵푆,푣] tends to 0. This completes the proof.
2.10 Random Caley graphs with a pendant edge
We give a non-trivial example of a vertex-transitive graph 퐻 such that 퐻 퐾2 is not · Ramsey simple.
Let 퐺 be a group, and 푆 be a subset of elements of 퐺, called the set of gen- erators. The (undirected) Caley graph 푋(퐺, 푆) has vertex set 퐺 and edge set 푔, 푔푠 : 푔 퐺, 푠 푆 . Clearly, 푋(퐺, 푆) is vertex-transitive and the degree of a {{ } ∈ ∈ } vertex is 푆 푆−1 . For 0 푝 1, we denote by 퐻 푋(퐺, 푝) a random Caley graph | ∪ | ≤ ≤ ∼ 퐻 = 푋(퐺, 푆) where 푆 is chosen by including every element of 퐺 with probability 푝. √︁ Theorem 2.10.1. Let 퐺 be a group on 푛 elements, let 푝 0 such that 푝 ln 푛 , → ≫ 푛 and let 퐻 푋(퐺, 푝). Then a.a.s. 푠(퐻 퐾2) 훿(퐻) > 1. ∼ · ≥ Proof. We prove that 퐻 satisfies the conditions of Theorem 2.4.1 a.a.s. Let 푆 denote the set of generators of 퐻 chosen by including every element of 퐺 with probability 푝. It is clear that 훿(퐻) 2 a.a.s. since a.a.s. 훿(퐻) 푆 푝푛/2 1. ≥ ≥ | | ≥ ≫ For condition (1) and (3) of Theorem 2.4.1, let 퐶 be a large constant to be determined later. We pick the set 푆 in 퐶 rounds. Let 푞 [0, 1] be such that ∈ 푝 = 1 (1 푞)퐶 . Note that 푞 = (1 + 표(1))푝/퐶 if 푝 0. For 1 푖 퐶, pick − − → ≤ ≤ a set 푅푖 of elements each with probability 푞 of being in the set, choices for different 푖 being independent. Let 푆 = 푅1 ... 푅퐶 and note that this is an equivalent way of ∪ ∪ picking the set 푆 of elements each with probability 푝 of being in the set.
58 For an element 푥 퐺 and a subset 퐴 퐺, we use the shorthand notation ∈ ⊆ 푥.퐴 := 푥푎 : 푎 퐴 . { ∈ }
Lemma 2.10.2. Let 푥 퐺 be a vertex of 퐻, and let 푅1, . . . , 푅퐶 be as above. Then ∈ with probability tending to one, the set 푥.푅1 is connected.
Before we prove the lemma, let us show how it implies that 퐻 is connected a.a.s., i.e. condition (1) of Theorem 2.4.1 holds.
Let 푥 퐺 be an arbitrary element of 퐺. We claim that a.a.s. 푥.푅1 is a dominating ∈ set of 퐻, i.e. for every 푦 퐺, there is an edge between 푦 and 푥.푅1. By definition, ∈ there is an edge between 푦 and 푥.푅1 if there exists a generator 푠 푆 and an element ∈ 푟1 푅1 such that 푦푠 = 푥푟1. For fixed 푥 and 푦, there are 푅1 possible values 푠 퐺 ∈ | | ∈ such that 푦푠 = 푥푟1. With probability tending to one, 푥.푅1 푛푞/2. Therefore, | | ≥ choosing 푅2, . . . , 푅퐶 , the probability that there exists some 푦 퐺 such that there is ∈ no edge between 푥.푅1 and 푦 is at most
퐶 1 2 (퐶−1)푛푞/2 − − 푛푞 +ln 푛 푛(1 푞) + Pr( 푥.푅1 < 푛푞/2) 푒 2 + 표(1) = 표(1), − | | ≤ √︁ since 푞 ln 푛 . Together with Lemma 2.10.2 it follows that a.a.s. the graph 퐻 is ≫ 푛 connected.
To prove that 푥.푅1 is connected, we need to show that a.a.s. for every nontrivial partition 퐴 ˙ 퐵 of 푅1, there is an edge between 푥.퐴 and 푥.퐵. That is, there exists ∪ 푎 퐴, 푏 퐵 and 푦 푆 such that 푏 = 푎푦. We show first that with high probability ∈ ∈ ∈ −1 the set 푌퐴,퐵 := 푎 푏 : 푎 퐴, 푏 퐵 of distinct such values 푦 퐺 is large enough { ∈ ∈ } ∈ for any partition of 푅1, and then use the random choices of 푅2, . . . , 푅퐶 to show that at least one solution survives with very high probability. In fact, we show something slightly stronger.
Lemma 2.10.3. There exists a universal constant 푐 > 0 such that the following holds. √︁ ln 푛 Let 푞 . Further, let 1 푛퐴 푛푞, 푛푞/3 푛퐵 2푛푞, and let 퐴 퐺 of size ≫ 푛 ≤ ≤ ≤ ≤ ⊆ 퐴 = 푛퐴. Let 퐵 be a set of size 푛퐵 chosen uniformly at random from 퐺 퐴. Then | | ∖ −100푛 with probability at least 1 2 퐵 we have that 푌퐴,퐵 min 푐 푛, 푐 푛퐴푛퐵 . − | | ≥ { · · } 59 Proof. Let 푐 > 0 be a constant to be chosen later. For disjoint sets 퐴, 퐵 퐺 let ⊆ 푈퐴,퐵 be the event that 푌퐴,퐵 < min 푐푛, 푐 퐴 퐵 . Let us choose 퐵 one element at | | { | || |} a time and analyse how the value of 푌퐴,퐵 changes. To be precise, for 1 푡 푛퐵 ≤ ≤ choose an element 푏푡 퐺 (퐴 퐵푡−1) uniformly at random, where 퐵0 := and ∈ ∖ ∪ ∅ . Let ⃒ ⃒ be the random variable that counts the 퐵푡 := 퐵푡−1 푏푡 푋푡 := ⃒푌퐴,퐵푡 푌퐴,퐵푡 1 ⃒ ∪ { } ∖ − “new” values −1 with , after step . If ⃒ ⃒ , we are done. 푎 푏 푎 퐴 푏 퐵 푡 ⃒푌퐴,퐵푡 1 ⃒ 푐푛 ∈ ∈ − ≥ So we may condition on the event that 푌퐴,퐵 < 푐푛 (and hence 푌퐴,퐵 < 푐푛 for any | | | 푡 | 푡 푛퐵) which we abbreviate with 푍. Then for a fixed element 푎 퐴 ≤ ∈
(︀ −1 ⃒ )︀ 푐푛 Pr 푎 푏푡 푌퐴,퐵푡 1 ⃒ 푍 < < 2푐 ∈ − 퐺 퐴 퐵푡−1 | − − |
for 푛 large enough since 퐴 + 퐵푡 3푛푞 = 표(푛). Hence, the expected size of | | | | ≤ 푋푡 conditioning on the event 푍 is at least (1 2푐) 퐴 and therefore, by Markov’s − | | Inequality,
(︂ )︂ (︂ )︂ 퐴 ⃒ 퐴 ⃒ Pr 푋푡 < | |⃒ 푍 = Pr 퐴 푋푡 > | |⃒ 푍 2 | | − 2
2 E( 퐴 푋푡 푍) · | | − | ≤ 퐴 | | 4푐. ≤
That is, the random variable 푋푡 takes value (at least) 퐴 /2 with probability (at | | least) 1 4푐, independent of the history 푋1, . . . , 푋푡−1. Let 푌1, . . . , 푌푛 be independent − 퐵 Bernoulli experiments that are one with probability 4푐 and zero otherwise. Then the sum ∑︀ 푌 is a lower bound on the number of steps that 푋 fails to have value 1≤푡≤푛퐵 푡 푡
60 at least 퐴 /2. Therefore, | |
(︀ ⃒ )︀ Pr(푈퐴,퐵) Pr 푌퐴,퐵 < 푐 퐴 퐵 ⃒ 푌퐴,퐵 < 푐푛 ≤ | | | || | | |
= Pr (푋푡 < 퐴 /2 at least (1 2푐)푛퐵 times) | | − (︃ )︃ ∑︁ Pr 푌푡 (1 2푐)푛퐵 ≤ ≥ − 1≤푡≤푛퐵 2푛퐵 (4푐)(1−2푐)푛퐵 ≤ · < 2−100푛퐵 for 푐 small enough.
Proof of Lemma 2.10.2. Let 푐 > 0 be the constant from Lemma 2.10.3 and let 푍 푞푛 denote the event that 푁 := 푅1 [ , 2푞푛] and that for every non-trivial partition | | ∈ 2 −1 퐴 ˙ 퐵 of 푅1 the set 푌퐴,퐵 = 푎 푏 : 푎 퐴, 푏 퐵 has size at least min 푐푛, 푐 퐴 퐵 . ∪ { ∈ ∈ } { | || |} We claim that the probability of 푍 is 1 표(1). Certainly, 푁 [ 푞푛 , 2푞푛] a.a.s. Let − ∈ 2 now 퐴 ˙ 퐵 = 푅 be a partition of 푅, let 푛퐴 := 퐴 , 푛퐵 := 퐵 and assume without ∪ | | | | loss of generality that 푛퐴 푛퐵. Note that by Lemma 2.10.3, the probability that ≤ −100푛 −20푛푞 푌퐴,퐵 < min 푐푛, 푐 퐴 퐵 is at most 2 퐵 2 . It follows that the probability | | { | || |} ≤ that 푍 fails to hold is at most
2푛푞 ∑︁ ∑︁ 2−20푛푞 + 표(1) 2푛푞2−19푛푞 + 표(1) 0 ≤ → 푁=푛푞/2 퐴⊆푅,|퐴|≤푁/2
since 푛푞 . We now condition on 푍. Fix now a non-trivial partition 푅1 = 퐴 ˙ 퐵, → ∞ ∪ say 퐴 = 푛퐴, 퐵 = 푛퐵, and assume without loss of generality that 푛퐴 푁/2. By | | | | ≤ assumption, there are at least min 푐푛, 푐 퐴 퐵 (nontrivial) solutions for 푎푦 = 푏 with { | || |} 푎 퐴 and 푏 퐵. Let 푋퐴,퐵 be the random variable that counts the number of ∈ ∈ elements in 푌퐴,퐵 that are chosen to be in 푅2 ... 푅퐶 . Since for a particular element ∪ ∪ 푦 푌퐴,퐵 the choices for 푅2, . . . , 푅퐶 are independent, and since choices for distinct ∈ elements in 푌퐴,퐵 are independent, it follows that
(퐶−1)|푌퐴,퐵 | Pr푅 ,...,푅 (푋퐴,퐵 = 0) = (1 푞) exp [ (퐶 1)푞 min 푐푛, 푐 퐴 퐵 ] . 2 퐶 − ≤ − − { | || |} 61 It follows that
Pr(푅1 is not connected) ⃒ = Pr(푅1 is not connected ⃒ 푍) + 표(1) ⃒ = Pr( a partition 퐴 퐵 = 푅1 such that 푋퐴,퐵 = 0 ⃒ 푍) + 표(1) ∃ ∪ 2푛푞 푁/2 ∑︁ ∑︁ ∑︁ ⃒ Pr푅 ,...,푅 (푋퐴,퐵 = 0 ⃒ 푍) + 표(1) ≤ 2 퐶 푁=푛푞/2 푛퐴=1 퐴∈ 푅1 (푛퐴) ⎛ ⎞ 2푛푞 (︂ )︂ (︂ )︂ ∑︁ ⎜ ∑︁ 푁 −(퐶−1)푞푐푛 (푁−푛 ) ∑︁ 푁 −(퐶−1)푞푐푛⎟ ⎜ 푒 퐴 퐴 + 푒 ⎟ + 표(1) ≤ ⎝ 푛퐴 푛퐴 ⎠ 푁=푛푞/2 1≤푛퐴≤푁/2 1≤푛퐴≤푁/2 푛퐴(푁−푛퐴)≤푛 푛퐴(푁−푛퐴)>푛 ⎛ ⎞ 2푛푞 ∑︁ ⎜ ∑︁ 푛 (ln 푁−Θ(푞푁)) 푁 −(퐶−1)푞푐푛⎟ ⎜ 푒 퐴 + 2 푒 ⎟ + 표(1) ≤ ⎝ ⎠ 푁=푛푞/2 1≤푛퐴≤푁/2 푛퐴(푁−푛퐴)≤푛
2 2(푛푞)2푒−Θ(푞 푛) + 2푛푞2푁 푒−(퐶−1)푞푐푛 + 표(1), ≤
2 since ln 푁 ln 푛 푞 푛/2 푞푁. Therefore, the probability that 푅1 is not connected ≤ ≪ ≤ is at most
exp [︀푂(ln(푞푛)) Θ(푞2푛)]︀ + exp [ln(2푛푞) + 2푛푞 (퐶 1)푐푛푞] + 표(1) = 표(1), − − −
again since ln(푛푞) 푞2푛 and if we choose 퐶 = 퐶(푐) 3/푐. ≪ ≥ We now prove condition (3) of Theorem 2.4.1. We need to analyse the family
(퐻) = 퐹 퐻[푁(푥)] : 푥 푉 (퐻), 퐹 is a connected component of 퐻[푁(푥)] ℱ { ⊆ ∈ }
and show that there is a 2-colouring of 퐾훼(퐻) that does not contain a monochromatic copy of 퐹 , for any 퐹 (퐻). Since 퐻 is vertex-transitive, the graph 퐻[푁(푥)] is the ∈ ℱ same graph for every 푥 퐺. Let us denote this graph by 퐻푁 . We show first that 퐻푁 ∈ is connected a.a.s. which implies that (퐻) consists of a single graph. ℱ
Lemma 2.10.4. Let 퐺, 푝, 푞, 퐻, 푅1, . . . , 푅퐶 be as above. Then, 퐻푁 is connected a.a.s.
62 Proof. Fix a vertex 푥 퐺. Note that the neighbourhood 푁(푥) = 푥.(푆 푆−1) ∈ ∪ is connected since 푥.푅1 is connected a.a.s. by Lemma 2.10.2, and since 푥.푅1 is a dominating set in 퐻.
Lemma 2.10.5. Let 퐺, 푝, 퐻 be as in Theorem 2.10.1. Then, 퐻푁 has at least 푚 = 푝3푛2(1 + 표(1)) edges a.a.s.
Proof. Fix 푥 퐺 and let 푆 be the set we pick by including each element of 퐺 ∈ with probability 푝, all choices being independent. We will count only the edges in 푥.푆 푁(푥). Note that an edge in 푥.푆 corresponds to a triple (푎, 푏, 푐) 푆3 such that ⊆ ∈ 푎 = 푏푐. We shall see that a.a.s the number of solutions to 푎 = 푏푐 with 푎, 푏, 푐 푆 ∈ is large. There are a total of 푛2 choices for 푏 and 푐, giving a total of 푛2 solutions for 푎 = 푏푐 in the group 퐺 of order n. For simplicity, let us only consider nontrivial triples, i.e. solutions for 푎 = 푏푐 where 푎, 푏, 푐 are distinct. There are still 푛2 푂(푛) − such solutions. For a nontrivial triple (푎, 푏, 푐) let 푋푎,푏,푐 be the indicator random variable which evaluates to one if and only if all of 푎, 푏, 푐 are chosen to be in 푆. ∑︀ Let 푋 = 푋푎,푏,푐 be the random variable counting the number of these solutions with 푎, 푏, 푐 chosen to be generators and 푎, 푏, 푐 distinct. The expected number of 푋 is 3 2 2 3 푝 푛 푂(푛푝 ) by linearity of expectation, as 푋푎,푏,푐 = 1 with probability 푝 for each − nontrivial triple. To see that 푋 is concentrated about its mean, we use the second moment method and bound the variance of 푋. For two (distinct) nontrivial triples ′ ′ ′ write (푎, 푏, 푐) (푎 , 푏 , 푐 ) if 푋푎,푏,푐 and 푋푎 ,푏 ,푐 are not independent. Let ∼ ′ ′ ′
∑︁ ∆ = Pr(푋푎,푏,푐 = 1 푋푎 ,푏 ,푐 = 1), ∧ ′ ′ ′ (푎,푏,푐)∼(푎′,푏′,푐′)
where the sum runs over all ordered pairs of non-trivial triples. We note that
Var(푋) E(푋) + ∆, see e.g. Chapter 4.3 in [3]. A pair (푋푎,푏,푐, 푋푎 ,푏 ,푐 ) is not ≤ ′ ′ ′ independent if and only if the sets 푎, 푏, 푐 and 푎′, 푏′, 푐′ have nonempty intersection. { } { } If two nontrivial triples 푎, 푏, 푐 and 푎′, 푏′, 푐′ intersect in one element then { } { } 5 3 Pr(푋푎,푏,푐 = 1 푋푎 ,푏 ,푐 = 1) = 푝 . The number of such (pairs of) triples is 푂(푛 ). If ∧ ′ ′ ′ two nontrivial triples 푎, 푏, 푐 and 푎′, 푏′, 푐′ intersect in two or three elements then { } { } 63 3 2 Pr(푋푎,푏,푐 = 1 푋푎 ,푏 ,푐 = 1) 푝 . The number of such (pairs of) triples is 푂(푛 ). ∧ ′ ′ ′ ≤ It follows that Var(푋) = 푂(푝3푛2) + 푂(푝5푛3) = 표(E(푋)2) since 푝 푛−2/3. There- ≫ fore, by Corollary 4.3.3 in [3], 푋 = E(푋)(1 + 표(1)) = 푝3푛2(1 + 표(1)) a.a.s.
We are ready to prove condition (3) and thus finish the proof of Theorem 2.10.1.
Let 퐻푁 be as before. By Lemma 2.10.4, 퐻푁 is a.a.s. connected, and by Lemma 3 2 2.10.5, 퐻푁 has a.a.s. at least 푝 푛 (1 + 표(1)) edges, and a.a.s. at most 2푛푝 vertices. The Ramsey number of a graph 퐺 with 푀 edges and 푁 vertices is at least 2(푀−1)/푁 (take a random colouring of the complete graph on 2(푀−1)/푁 vertices and see that the expected number of monochromatic copies of 퐺 is 표(1)).
Now, 훼(퐻) 푛 since 퐻 has 푛 vertices. Therefore, 훼(퐻) 푅(퐻푁 ) a.a.s., since √︁ ≤ ≪ 푝 ln 푛 , and it follows that Property (3) holds a.a.s. ≫ 푛
64 Chapter 3
The Erdős-Hajnal Conjecture for Rainbow Triangles
3.1 Introduction
A classical result of Erdős and Szekeres [36], which is a quantitative version of Ram-
sey’s theorem [79], implies that every graph on 푛 vertices contains a clique or an independent set of order at least 1 . In the other direction, Erdős [31] showed 2 log 푛 that a random graph on n vertices almost surely contains no clique or independent
set of order 2 log 푛. An induced subgraph of a graph is a subset of its vertices together with all edges with both endpoints in this subset. There are several results and conjectures indicat- ing that graphs which do not contain a fixed induced subgraph are highly structured. In particular, the most famous conjecture of this sort by Erdős and Hajnal [35] says that for each fixed graph 퐻 there is 휖 = 휖(퐻) > 0 such that every graph 퐺 on 푛 vertices which does not contain a fixed induced subgraph 퐻 has a clique or indepen- dent set of order 푛휖. This is in stark contrast to general graphs, where the order of the largest guaranteed clique or independent set is only logarithmic in the number of vertices. There are now several partial results on the Erdős-Hajnal conjecture. Erdős and
Hajnal [35] proved that for each fixed graph 퐻 there is 휖 = 휖(퐻) > 0 such that every
65 graph 퐺 on 푛 vertices which does not contain an induced copy of 퐻 has a clique √ or independent set of order 푒휖 log 푛. Fox and Sudakov [45], strengthening an earlier result of Erdős and Hajnal, proved that for each fixed graph 퐻 there is 휖 = 휖(퐻) > 0 such that every graph 퐺 on 푛 vertices which does not contain an induced copy of 퐻 has a balanced complete bipartite graph or an independent set of order 푛휖. All graphs on at most four vertices are known to satisfy the Erdős-Hajnal conjecture, and
Chudnovsky and Safra [18] proved it for the 5-vertex graph known as the bull. Alon,
Pach, and Solymosi [2] proved that if 퐻1 and 퐻2 satisfy the Erdős-Hajnal conjecture,
then for every 푣 of 퐻1, the graph formed from 퐻 by substituting 푣 by a copy of 퐻2 satisfies the Erdős-Hajnal conjecture. The recent survey [17] discusses many further related results on the Erdős-Hajnal conjecture. A natural restatement of the Erdős-Hajnal conjecture is that for every fixed red-
blue edge-coloring 휒 of a complete graph, there is an 휖 = 휖(휒) > 0 such that every red-blue edge-coloring of the complete graph on 푛 vertices without a copy of 휒 contains a monochromatic clique of order 푛휖. Indeed, for the graphs 퐻 and 퐺, we can color the edges red and the nonadjacent pairs blue. Erdős and Hajnal also proposed studying a multicolor generalization of their con-
jecture. It states that for every fixed 푘-coloring of the edges of 휒 of a complete graph, there is an 휖 = 휖(휒) > 0 such that every 푘-coloring of the edges of the complete graph on 푛 vertices without a copy of 휒 contains a clique of order 푛휖 which only uses 푘 1 √ − colors. They proved a weaker estimate, replacing 푛휖 by 푒휖 log 푛. Note that the case of two colors is what is typically referred to as the Erdős-Hajnal conjecture. Hajnal [59] conjectured that the following special case of the multicolor generaliza-
tion of the Erdős-Hajnal conjecture holds. There is 휖 > 0 such that every 3-coloring of the edges of the complete graph on 푛 vertices without a rainbow triangle (that is, a triangle with all its edges different colors), contains a set of order 푛휖 which uses at most two colors. We prove Hajnal’s conjecture, and further determine a tight bound on the order of the largest guaranteed 2-colored set in any such coloring. A Gallai coloring is a coloring of the edges of a complete graph without rainbow triangles, and a Gallai 푟-coloring is a Gallai coloring that uses 푟 colors.
66 Theorem 3.1.1. Every Gallai 3-coloring on 푛 vertices contains a set of order Ω(푛1/3 log2 푛) which uses at most two colors, and this bound is tight up to a constant factor.
To give an upper bound, we use lexicographic products. We will let [푚] = 1, . . . , 푚 denote the set consisting of the first 푚 positive integers. { } Definition 3.1.2. Given edge-colorings of and of using colors from 퐹1 퐾푚1 퐹2 퐾푚2
푅, the lexicographic product coloring 퐹1 퐹2 of 퐸(퐾푚 푚 ) is defined on any edge ⊗ 1 2 푒 = (푢1, 푣1), (푢2, 푣2) (where we take the vertex set of 퐾푚 푚 to be [푚1] [푚2]) to be { } 1 2 × 퐹1(푢1, 푢2) if 푢1 = 푢2, and otherwise 푣1 = 푣2 and it is defined to be 퐹2(푣1, 푣2). ̸ ̸
That is, there are 푚1 disjoint copies of 퐹2 and they are connected by edge colors defined by 퐹1. The upper bound in Theorem 3.1.1 is obtained by taking the lexicographic product of three 2-edge-colorings of the complete graph on 푛1/3 vertices, where each pair of colors is used in one of the colorings, and the largest monochromatic clique in each of the colorings is of order 푂(log 푛). A simple lemma in the next section shows that,
in a lexicographic product coloring 퐹 = 퐹1 퐹2, the largest set of vertices using only ⊗ colors red and blue (for example) in 퐹 has size equal to the product of the size of the
largest set of vertices using only colors red and blue in 퐹1 with the size of the largest
set of vertices using only colors red and blue in 퐹2. For any set 푆 of two of the three colors, the largest such set has order 푂(푛1/3)푂(log 푛)푂(log 푛) = 푂(푛1/3 log2 푛). In the other direction, we will utilize the following important structural result of Gallai [48] on edge-colorings of complete graphs without rainbow triangles.
Lemma 3.1.3. An edge-coloring 퐹 of a complete graph on a vertex set 푉 with 푉 2 | | ≥ is a Gallai coloring if and only if 푉 may be partitioned into nonempty sets 푉1, . . . , 푉푡
with 푡 > 1 so that each 푉푖 has no rainbow triangles under 퐹 , at most two colors are
used on the edges not internal to any 푉푖, and the edges between any fixed pair (푉푖, 푉푗) use only one color. Furthermore, any such substitution of Gallai colorings for vertices
of a 2-edge-coloring of a complete graph 퐾푡 yields a Gallai coloring.
Gallai colorings naturally arise in several areas including in information theory [70], in the study of partially ordered sets, as in Gallai’s original paper [48], and in
67 the study of perfect graphs [14]. There are now a variety of papers which consider Ramsey-type problems in Gallai colorings (see, e.g., [19, 47, 56, 58]). However, these works mainly focus on finding various monochromatic subgraphs in such colorings. Because it may be of independent interest to the reader, we first present a par- ticularly simple approach that will prove Hajnal’s conjecture, but will not give tight bounds. A graph is a cograph if it has at most one vertex, or if it or its complement is not connected, and all of its induced subgraphs have this property. In other words, the family of cographs consists of all those graphs that can be obtained from an isolated vertex by successively taking the disjoint union of two previously constructed
cographs, 퐺1 and 퐺2, or by the join of them that we get by adding all edges between
퐺1 and 퐺2. It was shown by Seinsche [87] that cographs are precisely those graphs which do not contain the path with three edges as an induced subgraph. It is easy to check by induction that every cograph is a perfect graph, that is, the chromatic number of every induced subgraph is equal to its clique number.
Proposition 3.1.4. In any Gallai 3-coloring of a complete graph, there is an edge- partition of the complete graph into three 2-colored subgraphs, each of which is a cograph.
Proof: This follows from Gallai’s structure theorem by induction on the number of vertices. The result is trivial for edge-colorings of complete graphs with fewer than two vertices, which serves as the base case. Using Lemma 3.1.3, we get a nontrivial vertex partition of the Gallai 3-coloring of the complete graph into parts 푉1, . . . , 푉푡 such that only two colors appear between the parts. By the induction hypothesis, we can partition the edge-set of the complete graph on 푉푖 into three cographs, each which is two-colored. For the two colors that go between the parts, we take the graph which is the join of the cographs in each 푉푖, that is, add all edges between the parts, and for each of the other two pairs of colors, we just take the disjoint union of the cographs of those two colors from each part. Since the join or disjoint union of cographs are cographs, this completes the proof by induction. 2
68 The following corollary verifies Hajnal’s conjecture and, apart from the two loga- rithmic factors, gives the lower bound in Theorem 3.1.1.
Corollary 3.1.5. Every Gallai 3-coloring of 퐸(퐾푛) contains a 2-colored clique with at least 푛1/3 vertices.
Proof: Indeed, applying Proposition 3.1.4, if the first cograph (which is 2-colored) contains a clique of order 푛1/3 then we are done; otherwise, it contains no clique of order 푛1/3 and, since cographs are perfect graphs, has chromatic number at most 푛1/3, in which case it contains an independent set of order 푛2/3. In this latter case, this independent set of order 푛2/3 in the first cograph contains in the second cograph a clique of order 푛1/3 or an independent set (which is a clique in the third cograph) of order 푛1/3. We thus get a clique of order 푛1/3 in one of the three cographs, which is a 2-colored set. 2
Improving the lower bound further to Theorem 3.1.1 appears to be considerably harder, and uses a different proof technique, relying on a weighted version ofRam- sey’s theorem and a carefully chosen induction argument. The weighted version of
Ramsey’s theorem shows that if each vertex of a complete graph on 푡 vertices is given a positive red weight and a positive blue weight whose product is one, then in any red-blue edge-coloring of 퐾푡, there is a red clique 푆 and a blue clique 푈 such that the product of the red weight of 푆 (the sum of the red weights of the vertices in 푆) and the blue weight of 푈 (the sum of the blue weights of the vertices in 푈) is Ω (︀log2 푡)︀. Note that this extends the quantitative version of Ramsey’s theorem as the case in which all the red and blue weights are one implies that there is a monochromatic clique of order Ω(log 푡).
We further consider a natural generalization of this problem to more colors, and give a tight bound in the next theorem. In order to state the result more succinctly,
69 we introduce some notation: for positive integers 푟 and 푠 with 푠 푟, let ≤ ⎧ ⎪ if or if and is even; ⎪1 1 = 푠 < 푟 푠 = 푟 1 푟 ⎪ − ⎪ ⎨⎪푠(푟 푠) if 1 < 푠 < 푟 1; 푐푟,푠 = − − ⎪ 3 if and is odd; ⎪1 + 푟 푠 = 푟 1 푟 ⎪ − ⎪ ⎩⎪0 if 푠 = 푟.
Theorem 3.1.6. Let 푟 and 푠 be fixed positive integers with 푠 푟. Every 푟-coloring ≤ of the edges of the complete graph on 푛 vertices without a rainbow triangle contains 푠 / 푟 푐 a set of order Ω(푛(2) (2) log 푟,푠 푛) which uses at most 푠 colors, and this bound is tight apart from the constant factor.
We next give a brief discussion of the proof of Theorem 3.1.6. The case 푠 = 푟 is trivial as the complete graph uses at most 푟 colors. The case 푠 = 1 is easy. Indeed, in this case, by the Erdős-Szekeres bound on Ramsey numbers for 푟 colors, there is a monochromatic set of order Ω(log 푛), where the implied positive constant factor depends on 푟. In the other direction, we give a construction which we conjecture is tight.
The Ramsey number 푟(푡) is the minimum 푛 such that every 2-coloring of the edges of the complete graph on 푛 vertices contains a monochromatic clique of order 푡. The bounds mentioned in the beginning of the introduction give 2푡/2 푟(푡) 22푡 for ≤ ≤ 푡 2. For 푟 even, consider a lexicographic product of 푟/2 colorings, each a 2-edge ≥ coloring of the complete graph on 푟(푡) 1 vertices with no monochromatic 퐾푡. This − gives a Gallai 푟-coloring of the edges of the complete graph on (푟(푡) 1)푟/2 vertices − with no monochromatic clique of order 푡. A similar construction for 푟 odd gives a Gallai 푟-coloring of the edges of the complete graph on (푡 1) (푟(푡) 1)(푟−1)/2 vertices − − with no monochromatic clique of order 푡. The following conjecture which states that these bounds are best possible seems quite plausible. It was verified by Chung and
Graham [19] in the case 푡 = 3.
Conjecture 3.1.7. Let 푁(푟, 푡) = (푟(푡) 1)푟/2 for 푟 even and − 70 푁(푟, 푡) = (푡 1) (푟(푡) 1)(푟−1)/2 for 푟 odd. For 푛 > 푁(푟, 푡), every 푟-coloring of the − − edges of the complete graph on 푛 vertices has a rainbow triangle or a monochromatic
퐾푡.
Having verified the easy cases 푠 = 1 and 푠 = 푟 of Theorem 3.1.6, for the rest of the chapter, we assume 1 < 푠 < 푟. A natural upper bound on the size of the largest set using at most 푠 colors comes from the following construction. We will let [푟] be the set of colors. Consider the complete graph on [푟], where each edge 푃 gets a positive integer weight 푛푃 such that the product of all 푛푃 is 푛. For each edge 푃 of this complete graph, we consider a 2-coloring 푐푃 of the edges of the complete graph on 푛푃 vertices using the colors in 푃 and whose largest monochromatic clique has order 푂(log 푛푃 ), which exists by Erdős lower bound [31] on Ramsey numbers. We then consider the Gallai 푟-coloring 푐 of the complete graph on 푛 vertices which is the lexicographic product of the (︀푟)︀ colorings of the form . For each set of colors, 2 푐푃 푆 the largest set of vertices in this edge-coloring of 퐾푛 using only colors in 푆 has order
∏︁ ∏︁ 푛푃 푂(log 푛푃 ). 푃 ∈푆 |푃 ∩푆|=1
The order of the largest set using at most 푠 colors in coloring 푐 is thus the maximum of the above expression over all subsets 푆 of colors of size 푠. Therefore, we want to choose the various 푛푃 to minimize this maximum. For 푠 < 푟 1, we give a second − moment argument which shows that the best choice is essentially that the 푛푃 are all 1/(푟) equal, i.e., 푛푃 = 푛 2 for all 푃 . In this case, the above expression, for each choice of 푆, matches the claimed upper bound in Theorem 3.1.6. The case 푠 = 푟 1 turns − 2/푟 out to be more delicate. For 푟 even, the optimal choice turns out to be 푛푃 = 푛 for 푃 an edge of a perfect matching of the complete graph with vertex set [푟], and otherwise 푛푃 = 1. For 푟 odd, we have three different edge weights. The graph on [푟] whose edges consist of those pairs with weight not equal to 1 consist of a disjoint union of a triangle and a matching with (푟 3)/2 edges. The edges of the triangle − each have weight 푛1/푟(log 푛)(푟−3)/2푟 and the edges of the matching each have weight 푛2/푟(log 푛)−3/푟. It is straightforward to check that these choices of weights give the
71 claimed upper bound in Theorem 3.1.6.
Similar to the case 푟 = 3 and 푠 = 2 mentioned above, using Gallai’s structure theorem, we observe that, in any 푟-coloring of the edges of the complete graph on 푛 vertices without a rainbow triangle, the complete graph can be edge-partitioned into (︀푟)︀ subgraphs, each of which is a -colored perfect graph. A simple argument then 2 2 푠 / 푟 shows that there is a vertex subset of at least 푛(2) (2) vertices which uses at most 푠 colors, which verifies the lower bound in Theorem 3.1.6 apart from the logarithmic factors. Improving the lower bound further to Theorem 3.1.6 is more involved, using a weighted version of Ramsey’s theorem and a carefully chosen induction argument to prove this.
The rest of the chapter is organized as follows. In the next section, we prove some basic properties of lexicographic product colorings. In Section 3.3, we give simple proofs of lower and upper bounds in the direction of Theorem 3.1.1 which match apart from two logarithmic factors. In order to close the gap and obtain Theorem 3.1.1, in Section 3.4 we prove a weighted extension of Ramsey’s theorem. We complete the proof of Theorem 3.1.1 in Section 3.5 by establishing a tight lower bound on the size of the largest 2-colored set of vertices in any Gallai 3-coloring of the complete graph on 푛 vertices. The remaining sections are devoted to the proof of Theorem 3.1.6. In Section 3.6, we prove the upper bound for Theorem 3.1.6. In Section 3.7, we give a simple proof of a lower bound which matches Theorem 3.1.6 apart from the logarithmic factors. In Section 3.8.1, using the second moment method, we establish an auxiliary lemma which gives a tight bound on the minimum possible number of nonzero weights in a graph with non-negative edge weights such that no set of 푠 vertices contains sufficiently more than the average weight of a subset of 푠 vertices. We give the lower bound for Theorem 3.1.6 in Section 3.8.2, which completes the proof of this theorem. The proofs of some of the auxiliary lemmas which involve lengthy calculations are deferred to the end. All logarithms in this chapter are base
2, unless otherwise indicated. All colorings are edge-colorings of complete graphs, unless otherwise indicated. For the sake of clarity of presentation, we systematically omit floor and ceiling signs whenever they are not crucial. We also do not makeany
72 serious attempt to optimize absolute constants in our statements and proofs.
3.2 Lexicographic product colorings
In this section, we will prove some simple results about lexicographic product colorings
(Definition 3.1.2). These will be useful in constructing examples of 푟-colorings that do not contain large vertex sets that use at most 푠 colors.
For such a lexicographic product coloring 퐹1 퐹2 with 퐹1 on 푚1 vertices and 퐹2 on ⊗ 푚2 vertices, we will view the vertex set interchangeably as [푚1 푚2] and [푚1] [푚2]. × × For the sake of brevity, we often refer to a lexicographic product coloring as simply a product coloring.
Definition 3.2.1. For 퐹 an edge-coloring of 퐾푛 and 푆 푅 a set of colors, we write ⊆ that a set 푍 of vertices is 푆-subchromatic in 퐹 if every edge internal to 푍 takes colors (under 퐹 ) only from 푆.
When 퐹 and 푆 are clear from context, we shall simply say that 푍 is subchromatic. We will write to be the size of the largest subchromatic set of vertices. 푔푆,퐹
If 퐹 is an edge-coloring constructed via a product of two other colorings 퐹1, 퐹2, then the next lemma allows us to determine 푔 in terms of 푔 and 푔 . 푆,퐹 푆,퐹1 푆,퐹2
Lemma 3.2.2. For any -colorings of respectively, and any 푟 퐹1, 퐹2 퐸(퐾푛1 ), 퐸(퐾푛2 ),
set 푆 푅 of colors, 푔 = 푔 푔 , where 퐹 = 퐹1 퐹2. ⊆ 푆,퐹 푆,퐹1 · 푆,퐹2 ⊗
Proof: Let 푍 a set of subchromatic vertices in 퐹 (so 푍 푉 (퐾푛 ×푛 )) be given. We ⊆ 1 2 will first show 푍 푔 푔 . | | ≤ 푆,퐹1 · 푆,퐹2 Take 푈 [푛1] to be the set of 푢 [푛1] such that there is some 푣 [푛2] with ⊆ ∈ ∈ (푢, 푣) 푍; that is, 푈 is the subset of [푛1] that is used in 푍. For any 푢 [푛1], take ∈ ∈ 푉푢 [푛2] to be the set of 푣 [푛2] such that (푢, 푣) 푍, that is, 푉푢 is the subset of ⊆ ∈ ∈ ⋃︀ [푛2] that is paired with 푢 in 푍. By construction, we have 푍 = 푢 푉푢. 푢∈푈 { } × Therefore, the set 푈 must be subchromatic in 퐹1, as given distinct 푢1, 푢2 푈 ∈ 73 there are 푣1, 푣2 so that (푢1, 푣1), (푢2, 푣2) 푍, and hence: ∈
퐹1(푢1, 푢2) = 퐹 ((푢1, 푣1), (푢2, 푣2)) 푆. ∈
Thus, 푈 푔 . | | ≤ 푆,퐹1 Furthermore, given 푢 푈 we must have that 푉푢 is subchromatic in 퐹2, as given ∈ distinct 푣1, 푣2 푉푢 we have that ∈
퐹2(푣1, 푣2) = 퐹 ((푢, 푣1), (푢, 푣2)) 푆. ∈
Therefore, 푉푢 푔 . | | ≤ 푆,퐹2 Hence,
⃒ ⃒ ⃒ ⋃︁ ⃒ ∑︁ ∑︁ 푍 = ⃒ 푢 푉푢⃒ = 푉푢 푔 = 푈 푔 푔 푔 . | | ⃒ { } × ⃒ | | ≤ 푆,퐹2 | | 푆,퐹2 ≤ 푆,퐹1 · 푆,퐹2 ⃒푢∈푈 ⃒ 푢∈푈 푢∈푈
Since 푍 was arbitrary, we get 푔 푔 푔 . 푆,퐹 ≤ 푆,퐹1 · 푆,퐹2 We now prove that 푔 푔 푔 , thus giving the desired result: take 푈 [푛1] a 푆,퐹 ≥ 푆,퐹1 푆,퐹2 ⊆ subchromatic set under 퐹1 and 푉 [푛2] a subchromatic set under 퐹2. We claim that ⊆ 푈 푉 is subchromatic under 퐹 . Consider any distinct pairs (푢1, 푣1), (푢2, 푣2) 푈 푉 . × ∈ × If 푢1 = 푢2 then ̸ 퐹 ((푢1, 푣1), (푢2, 푣2)) = 퐹1(푢1, 푢2) 푆, ∈ and if 푢1 = 푢2 then
퐹 ((푢1, 푣1), (푢2, 푣2)) = 퐹2(푣1, 푣2) 푆. ∈ If we choose 푈 to have size 푔 and 푉 to have size 푔 , we get 푔 푔 = 푆,퐹1 푆,퐹2 푆,퐹1 · 푆,퐹2 푈 푉 푔 . 2 | × | ≤ 푆,퐹 The next lemma states that the property of being a Gallai coloring is preserved under taking product colorings.
Lemma 3.2.3. If are Gallai -colorings of , respectively, then 퐹1, 퐹2 푟 퐸(퐾푛1 ), 퐸(퐾푛2 ) if 퐹 = 퐹1 퐹2 then 퐹 is a Gallai coloring. ⊗ 74 Proof: Let any three vertices 푢 = (푢1, 푢2), 푣 = (푣1, 푣2), 푤 = (푤1, 푤2) [푛1] ∈ × [푛2] be given. We will show that they do not form a rainbow triangle under 퐹 . If
푢1 = 푣1 = 푤1 then 퐹 (푢, 푣) = 퐹2(푢2, 푣2), 퐹 (푢, 푤) = 퐹2(푢2, 푤2), 퐹 (푣, 푤) = 퐹2(푣2, 푤2); therefore, 푢, 푣, 푤 do not form a rainbow triangle by the assumption that 퐹2 is a
Gallai coloring. If 푢1, 푣1, 푤1 are pairwise distinct then 퐹 (푢, 푣) = 퐹1(푢1, 푣1), 퐹 (푢, 푤) =
퐹1(푢1, 푤1), 퐹 (푣, 푤) = 퐹1(푣1, 푤1) and so 푢, 푣, 푤 do not form a rainbow triangle by the assumption that 퐹1 is a Gallai coloring. Otherwise, exactly one pair of 푢1, 푣1, 푤1 are equal. Assume without loss of generality that 푢1 = 푣1, 푢1 = 푤1, and 푣1 = 푤1. We ̸ ̸ have:
퐹 (푢, 푤) = 퐹1(푢1, 푤1) = 퐹1(푣1, 푤1) = 퐹 (푣, 푤), so again 푢, 푣, 푤 do not form a rainbow triangle. 2
The following corollary states that we may take a product of any number of 2- colorings and the result will be a Gallai coloring; since all 2-colorings are Gallai colorings, it follows by induction from the previous lemma.
Corollary 3.2.4. If 퐹1, . . . , 퐹푘 are 2-edge-colorings, then 퐹1 퐹푘 is a Gallai ⊗ · · · ⊗ coloring.
3.3 Simple bounds for three colors
In this section we will demonstrate simple upper and lower bounds in the case 푟 = 3 and 푠 = 2. We first apply the techniques of the previous section to demonstrate a Gallai 3-coloring with no large 2-colored vertex set; we will use 푅 for the set of colors.
Theorem 3.3.1. There is a Gallai 3-coloring on 푚 vertices so that, for every two colors 푆 (︀푅)︀, every vertex set 푍 using colors from 푆 satisfies 푍 (4/9 + ∈ 2 | | ≤ 표(1))푚1/3 log2 푚.
Proof: Take 푡 = 푚1/3 ; then 푡3 is at least 푚. For every pair of colors 푃 (︀푅)︀, take ⌈ ⌉ ∈ 2 퐹푃 to be a 2-coloring of 퐸(퐾푡) using colors from 푃 so that the largest monochromatic clique has size at most 2 log 푡. Such a coloring exists by the lower bound on Ramsey
75 numbers proved by Erdős and Szekeres in [36]. We define 퐹 a coloring on 푡3 vertices by taking 퐹 = 퐹{푅 ,푅 } 퐹{푅 ,푅 } 퐹{푅 ,푅 } where 푅1, 푅2, 푅3 are such that 푅 = 1 2 ⊗ 2 3 ⊗ 1 3 푅1, 푅2, 푅3 . This is a Gallai coloring by Corollary 3.2.4. Fixing any set 푆 of two { } colors, two of the above three colorings have 푆-subchromatic sets of size at most 2 log 푡, and the remaining one has size 푡, so the size of the largest 푆-subchromatic set in 퐹 is at most 푡(2 log 푡)2 by Lemma 3.2.2. Since 푆 is arbitrary, the size of the largest 푆-subchromatic set for any 푆 (︀푅)︀ is at most 푡(2 log 푡)2. ∈ 2 Restricting 퐹 to any 푚 vertices will be a 3-Gallai coloring with no subchromatic set of size larger than 푡(2 log 푡)2. Note that since 푡 = 푚1/3 , we have 푡 = (1 + 표(1))푚1/3, ⌈ ⌉ so
푡(2 log 푡)2 = (1 + 표(1))푚1/3(2 log(푚1/3))2 = (4/9 + 표(1))푚1/3 log2 푚.
2
We now proceed to prove that any Gallai 3-coloring on 푚 vertices contains a subchromatic set on two colors of size at least 푚1/3. Indeed, the next theorem is a strengthening of this statement, as it states that the geometric average over 푆 (︀푅)︀ ∈ 2 of must be at least 1/3. 푔푆,퐹 푚 We have three colors and we refer to them as red, blue, and yellow.
Theorem 3.3.2. For any Gallai -coloring on vertices, we have ∏︀ 3 퐹 푚 푆∈ 푅 푔푆,퐹 ( 2 ) ≥ 푚.
Proof: We proceed by induction on 푚 to prove the theorem. Define 푔 to be the size of the largest subchromatic set using only the colors blue and yellow, 표 to be the size of the largest subchromatic set using only the colors red and yellow, and 푝 to be the size of the largest subchromatic set using only the colors red and blue. (A note on nomenclature: 푔 stands for “green," as blue and yellow form green when mixed. Similarly, 표 stands for “orange" and 푝 for “purple.") We wish to show that 푔표푝 푚. ≥ If 푚 = 1, then 푔 = 표 = 푝 = 1 and 푔표푝 = 푚. Otherwise, 푚 > 1 and by the structure theorem for Gallai colorings there is a non- (︀푅)︀ trivial partition of the vertex set into parts 푉1, . . . , 푉푡 and a pair of colors 푄 ∈ 2 76 satisfying that for any distinct 푖, 푗 [푡] there is a 푞 푄 so that every edge between 푉푖 ∈ ∈ and 푉푗 has color 푞. Take 푚푖 to be the size of 푉푖. Take 푔푖 to be the size of the largest
set using only the colors blue and yellow from 푉푖, 표푖 to be the size of the largest set
using only the colors red and yellow from 푉푖, and 푝푖 to be the size of the largest set
using only the colors red and blue from 푉푖. Without loss of generality we assume that 푄 contains colors blue and yellow. We have ∑︀ . Indeed, we may combine all the largest sets using colors blue 푔 = 푖 푔푖 and yellow from each to obtain a set of size ∑︀ that only uses blue and yellow. 푉푖 푖 푔푖
Furthermore, 표 max푖 표푖 and 푝 max푖 푝푖. This gives: ≥ ≥
∑︁ ∑︁ ∑︁ 푔표푝 = 푔푖표푝 푔푖표푖푝푖 푚푖 = 푚, 푖 ≥ 푖 ≥ 푖 where the last inequality follows by the induction hypothesis applied to 퐹 restricted to 푉푖. 2
Note that we use 표 max푖 표푖 and 푝 max푖 푝푖. It is on these inequalities that ≥ ≥ we will, in the next sections, gain multiple factors of log 푚; if, for example, we find
some set 푈 [푡] satisfying that, for each distinct 푖, 푗 푈, the edges between 푉푖, 푉푗 ⊆ ∈ ∑︀ are all yellow, then 표 표푖. If it were the case that the 표푖, 푝푖 were all pairwise ≥ 푖∈푈 equal, then we would get by the Erdős-Szekeres bound for Ramsey numbers that
2 표푝 = Ω(log 푡 max푖 표푖푝푖); this motivates the approach in the next two sections, where we handle the general case in which it may not be true that the 표푖, 푝푖 are all pairwise equal.
3.4 A weighted Ramsey theorem
In this section we will prove a version of Ramsey’s theorem that will apply to graphs in which the weight of a vertex may depend on the color of the clique that contains the vertex. The next lemma is a convenient statement of a quantitative bound on the classical Ramsey Theorem.
77 Lemma 3.4.1. In every 2-coloring of the edges of 퐾푡, for some 푘 and ℓ there is a red clique of order 푘 and a blue clique of order ℓ with 푘ℓ 1 log2 푡. ≥ 4
Proof: Take 푘 to be the order of the largest red clique and ℓ to be the order of the largest blue clique. We must have
(︂푘 + ℓ)︂ 푡 < 푅(푘 + 1, ℓ + 1) . ≤ 푘
√ In Section 3.9 we indicate how to prove that (︀푘+ℓ)︀ 22 푘ℓ; combining this with the 푘 ≤ above inequality gives the desired result. 2
For the rest of this chapter, let 푀 := 216. The following lemma, which we call
the weighted Ramsey theorem, states that if vertex 푖 contributes weight 훼푖 to any
red clique in which it is contained and weight 훽푖 to any blue clique in which it is contained, then we may give a lower bound for the product of the sizes of the largest (weighted) red and blue cliques.
Lemma 3.4.2. Given a 2-coloring of the edges of a complete graph on 푡 vertices with
푡 푀 and vertex weights (훼푖, 훽푖), take 훾푖 = 훼푖훽푖 and 훾 = min푖 훾푖. There is a red ≥ clique 푆 and a blue clique 푈 with
(︃ )︃ (︃ )︃ ∑︁ ∑︁ 훾 2 훼푠 훽푢 log 푡. ≥ 32 푠∈푆 푢∈푈
Proof: We will dyadically partition the vertices based on their pair of weights (훼푖, 훽푖), and then apply the classical Erdős-Szekeres bound on Ramsey numbers in the form
of the previous lemma. That is, we will find a large set of vertices 퐴 so that any two vertices in 퐴 have similar values for 훼푖 and 훽푖. By applying Lemma 3.4.1 to this set we will obtain the desired result.
Take 훼 = max푖 훼푖 and 훽 = max푖 훽푖. 훾 2 If 훼훽 log 푡 we may take 푆 = 푖 with 훼푖 = 훼 and 푈 = 푗 with 훽푗 = 훽. ≥ 32 { } { } 1 2 Otherwise, 훼훽/훾 < log 푡. Observe that for each 푖 we have 훼푖 훼, 훽푖 훽, and 32 ≤ ≤ 훼푖훽푖 훾. ≥ 78 This gives 훾/훽 훼푖 훼 and 훾/훼 훽푖 훽. Note that we may partition [훾/훽, 훼] ≤ ≤ ≤ ≤ into 푚1 log(훼훽/훾)+1 intervals 퐼1, . . . , 퐼푚 such that, within any interval 퐼푖, we have ≤ 1 sup(퐼푖)/ inf(퐼푖) 2. Similarly, we may partition [훾/훼, 훽] into 푚2 log(훼훽/훾) + 1 ≤ ≤ intervals 퐼′ , . . . , 퐼′ with sup(퐼′)/ inf(퐼′) 2. By the pigeonhole principle, there 1 푚2 푖 푖 ≤ ′ ′ must be some pair (푗, 푗 ) such that, taking 퐴 := 푖 : 훼푖 퐼푗, 훽푖 퐼 , we have { ∈ ∈ 푗′ } 퐴 푡/(푚1푚2). | | ≥ Applying the previous lemma to 퐴, we get that there is a red clique 푆 of size 푘 1 2 and a blue clique 푈 of size ℓ with 푘ℓ log (푡/(푚1푚2)). ≥ 4 1 2 2 2 1 2 1/4 Note that, since 푡 푀, we get 푚1푚2 (log( log 푡)+1) = log ( log 푡) 푡 . ≥ ≤ 32 16 ≤ Therefore, we get
1 2 1 2 3/4 1 2 log (푡/(푚1푚2)) log (푡 ) log 푡. 4 ≥ 4 ≥ 8
Take 훼퐴 = min푖∈퐴 훼푖 and 훽퐴 = min푖∈퐴 훽푖. For any 푖 퐴, 훼푖 퐼푗 and hence ∈ ∈ 훼퐴 훼푖/2. Similarly, for any 푖 퐴 we have 훽퐴 훽푖/2. Therefore, fixing any 푖 퐴, ≥ ∈ ≥ ∈ 훼푖 훽푖 we get 훼퐴훽퐴 훾/4. Therefore, ≥ 2 2 ≥
(︃ )︃ (︃ )︃ (︃ )︃ (︃ )︃ ∑︁ ∑︁ ∑︁ ∑︁ 훾 2 훼푠 훽푢 훼퐴 훽퐴 = 푘훼퐴ℓ훽퐴 푘ℓ훾/4 log 푡. ≥ ≥ ≥ 32 푠∈푆 푢∈푈 푠∈푆 푢∈푈
2
Since, in the statement of the weighted Ramsey theorem, we take 훾 = min푖 훼푖훽푖, it provides good bounds when 훼푖훽푖 does not vary much between the vertices. Therefore, when we wish to use it in the upcoming sections, we will first dyadically partition the vertices based on 훼푖훽푖 and then apply the lemma to each partition.
Note that we chose 훾 = min푖 훼푖훽푖. We may hope to be able to use other functions of 훼푖, 훽푖 in this expression. However, it is not as robust as one may hope. In particular, we want to observe that the function 훼푖 + 훽푖 will not yield an analogous theorem, as if we have many vertices of weight (0, 1) and color all of the edges red, then the largest red clique has size 0 and the largest blue clique has size 1, but for each 푖 we
79 have 훼푖 + 훽푖 = 1. Fortunately, using 훼푖훽푖 will suffice for our purposes.
3.5 Tight lower bound for three colors
In this section we will show that any Gallai 3-coloring on 푚 vertices has a 2-colored set of size Ω(푚1/3 log2 푚). This matches the upper bound up to a constant factor. We will refer to the three edge colors as red, blue, and yellow.
For the rest of this section, fix an integer 푚 N. We remark that in this section ∈ there is an inductive argument for which it is important to note that 푚 remains fixed throughout. Let
⎧ 2 if 4/9 ⎪ 푐 log (퐶푛) 0 < 푛 푚 ⎨⎪ ≤ 푓(푛) := 푐2 log2(푚4/9) log2(퐶푛푚−4/9) if 푚4/9 < 푛 푚8/9 , ⎪ ≤ ⎩⎪ 푐3 log4(푚4/9) log2(퐶푛푚−8/9) if 푚8/9 < 푛 푚, ≤ where 퐷 = 22048, 퐶 = 2퐷8 , and 푐 = log−2(퐶2) = 퐷−16/4. We will have a further discussion about 푓 and its properties shortly. For now, simply note that 푓(푚) = Ω(log6 푚). We will prove the following theorem.
Theorem 3.5.1. For any 푛 [푚], a Gallai coloring 퐹 on 푛 vertices satisfies either ∈ 7/18 ∏︀ max푆 푔 푚 /8 or 푔 푛푓(푛). 푆,퐹 ≥ 푆 푆,퐹 ≥ Before we prove Theorem 3.5.1, we show how it implies the existence of a large subchromatic set.
Theorem 3.5.2. Every Gallai 3-coloring of 퐸(퐾푚) has a two colored set of size Ω(푚1/3 log2 푚).
7/18 1/3 2 Proof: By Theorem 3.5.1, we have that either max푆 푔 푚 /8 = Ω(푚 log 푚), 푆,퐹 ≥ or ∏︁ 푔 푚푓(푚) = 푐3푚 log4(푚4/9) log2(퐶푚1/9) 푆,퐹 ≥ 푆 80 푐3푚2−6(log4 푚)2−9(log2 푚) = 2−15푐3푚 log6 푚. ≥ As we have a lower bound on the product of three numbers, one of these numbers must
−5 1/3 2 1/3 2 be at least the cubed root. Hence, max푆 푔 2 푐푚 log 푚 = Ω(푚 log 푚), as 푆,퐹 ≥ desired. 2
We will now proceed with a further discussion about 푓. We call (0, 푚4/9], (푚4/9, 푚8/9], (푚8/9, 푚] the “intervals of 푓." Note that on each interval, 푓(푛) = 훾 log2(훿푛) for some constants 훾, 훿 (where 푚 is viewed as a constant). Intuitively, 퐶 is large so that we avoid the range of values in which log is poorly behaved, and 푐 is small both so that we may assume 푛 is large and to make the transitions between intervals easier. 푓 was chosen so that it satisfies certain properties, the more interesting of which we explic- itly enumerate below. All of these properties are formalizations of the statement “푓 does not grow too quickly."
Lemma 3.5.3. If 푚 퐶, then the following statements hold about 푓 for any integer ≥ 푛 with 1 < 푛 푚. ≤ 1. For any 훼 [︀ 1 , 1]︀ , 푓(훼푛) 훼푓(푛). ∈ 푛 ≥ [︀ 1 ]︀ ∑︀ 2. For any 훼1, 훼2, 훼3 , 1 such that 훼푖 = 1 we have, taking 푛푖 = 훼푖푛, ∈ 푛 푖
∑︁ 8 푛푓(푛) 푛 푓(푛 ) 푛푓(푛). 푖 푖 log 퐶 − 푖 ≤
7/18 푗 푖 2 푗 푖+ 8 푗 3. For 푖 0 and 푚 2 1 we have 푓(2 ) log (퐷2 ) 512푓(2 7 ). ≥ ≥ ≥ ≥ 4. For 1 휏 푛 퐷3휏, we have 푓(휏) 푓(푛)/2. ≤ ≤ ≤ ≥ 5. For any 훼 [︀ 1 , 1 ]︀ , 푓(훼푛) 16훼푓(푛). ∈ 푛 32 ≥ These properties are collectively referred to as “the facts about 푓" and are proved in Section 3.10. We now proceed with a proof of Theorem 3.5.1.
Proof of Theorem 3.5.1: We proceed by induction on 푛. Define 푔 to be the size of the largest set in 퐹 using only the colors blue and yellow, 표 to be the size of the
81 largest set in 퐹 using only the colors red and yellow, and 푝 to be the size of the largest set in 퐹 using only the colors red and blue. We wish to show that either 푔표푝 푛푓(푛) ≥ or max(푔, 표, 푝) 푚7/18/8. ≥ Our base cases are those 푛 for which 푓(푛) 1, as for these cases by Theorem ≤ 3.3.2 푔표푝 푛 푛푓(푛). Since 푐 = log−2(퐶2), any 푛 < 퐶 is a base case. ≥ ≥ If we are not in a base case, we have 푛 퐶 (and 푓(푛) 1). ≥ ≥
Since 퐹 is a Gallai coloring, there is a non-trivial partition 푉 (퐾푛) = 푉1 ... 푉푡 ∪ ∪ with 푉1 ... 푉푡 1 such that there is some 2-coloring 휒 of [푡] such that for | | ≥ ≥ | | ≥ every distinct 푖, 푗 [푡] and 푢 푉푖, 푣 푉푗, the color under 퐹 of 푢, 푣 is 휒(푖, 푗). ∈ ∈ ∈ { } Suppose, without loss of generality, that 휒 only uses the colors blue and yellow. The proof will split into three cases.
Cases 1 and 2, Preliminary discussion: These will be the cases in which 푉1 has a substantial portion of the vertices. Let 푈1 = 푉1, 푈2 denote the union of 푉푗 over
푗 = 1 such that 휒(1, 푗) is yellow, and 푈3 denote the union of 푉푗 over 푗 = 1 such that ̸ ̸ 휒(1, 푗) is blue. We have that 푈1, 푈2, 푈3 is a non-trivial partition of 푉 . Let 푛푖 = 푈푖 . | | Let 훼푖 = 푈푖 /푛 = 푛푖/푛 for 푖 = 1, 2, 3, so 훼1 + 훼2 + 훼3 = 1. | |
For 푖 = 1, 2, 3, let 퐹푖 be the coloring 퐹 restricted to 푈푖. Let 푔푖 be the size of the largest subchromatic set in 퐹푖 using only the colors blue and yellow, 표푖 be the size of the largest subchromatic set in 퐹푖 using only the colors red and yellow, and 푝푖 be the size of the largest subchromatic set in 퐹푖 using only the colors red and blue. Suppose without loss of generality 푛2 푛3, so 훼2 (1 훼1)/2 and max(훼1, 훼2) 1/3. By ≥ ≥ − ≥ the induction hypothesis, for 푖 = 1, 2, 3, we have that either one of 푔푖, 표푖, 푝푖 is at least 7/18 푚 /8, in which case we may use 푔 max푖 푔푖, 표 max푖 표푖, 푝 max푖 푝푖 to complete ≥ ≥ ≥ the induction, or
푔푖표푖푝푖 푛푖푓(푛푖). ≥
Assume we are in this latter case. Since the 푈푖 are connected only by yellow and blue edges, we may take the largest subchromatic set using only yellow and blue from each 푈푖, giving 푔 푔1 +푔2 +푔3 (in fact, equality holds). Since 푈1 and 푈2 are connected ≥ with yellow edges, we may take the largest subchromatic set using only red and yellow
82 from both 푈1 and 푈2, or we may simply take the largest such subchromatic set from
푈3, so we get 표 max(표1 + 표2, 표3). Similarly, 푝 max(푝1 + 푝3, 푝2). ≥ ≥ Note
푔표푝 푔1표푝 + 푔2표푝 + 푔3표푝 푔1(표1 + 표2)(푝1 + 푝3) + 푔2(표1 + 표2)푝2 + 푔3표3(푝1 + 푝3) ≥ ≥ 3 ∑︁ 푔1(표1 + 표2)푝1 + 푔2(표1 + 표2)푝2 + 푔3표3푝3 = 푔1표2푝1 + 푔2표1푝2 + 푔푖표푖푝푖. ≥ 푖=1
We thus have
3 ∑︁ √︀ √︀ 푔표푝 푔푖표푖푝푖 푔1표2푝1 + 푔2표1푝2 2 (푔1표2푝2)(푔2표1푝1) = 2 (푔1표1푝1)(푔2표2푝2) − 푖=1 ≥ ≥ √︀ √︀ 2 (푛1푓(푛1))(푛2푓(푛2)) 2√훼1훼2푛 푓(푛1)푓(푛2), ≥ ≥
where the second inequality is an instance of the arithmetic-geometric mean inequal- ity.
−1/4 Case 1: 훼1, 훼2 (log 퐶) . In this case, we have ≥
3 ∑︁ √︀ √︀ 푔표푝 푔푖표푖푝푖 2√훼1훼2푛 푓(푛1)푓(푛2) 2훼1훼2푛푓(푛) 2푛푓(푛)/ log 퐶 − 푖=1 ≥ ≥ ≥ 3 8 ∑︁ 푛푓(푛) 푛푓(푛) 푛 푓(푛 ), log 퐶 푖 푖 ≥ ≥ − 푖=1
where the second inequality is by the first fact about 푓, the third inequality is by
substituting lower bounds on 훼1 and 훼2, and the last inequality is by the second fact about 푓. Hence,
3 3 ∑︁ ∑︁ 푔표푝 푔푖표푖푝푖 + 푛푓(푛) 푛푖푓(푛푖) 푛푓(푛), ≥ 푖=1 − 푖=1 ≥
where the last inequality is by the induction on hypothesis applied to 푈푖 for 푖 = 1, 2, 3. This completes this case.
−1/4 Case 2: 훼1 (log 퐶) 훼2. Before we proceed with this case, we prove a simple ≥ ≥ 83 claim.
Claim 3.5.4. 푛푓(푛) + 푛1푓(푛1) 2푛1푓(푛) > 0. −
Proof: Note 푛푓(푛) 푛1푓(푛) = (1 훼1)푛푓(푛). Therefore, − −
2 푛1푓(푛1) 푛1푓(푛) 훼1푛1푓(푛) 푛1푓(푛) = 훼 푛푓(푛) 훼1푛푓(푛) = 훼1(1 훼1)푛푓(푛), − ≥ − 1 − − − where the first inequality follows from the first fact about 푓. From this we get
2 푛푓(푛) + 푛1푓(푛1) 2푛1푓(푛) (1 훼1)푛푓(푛) 훼1(1 훼1)푛푓(푛) = (1 훼1) 푛푓(푛) > 0. − ≥ − − − −
2
−1/4 In this case we have 훼1 1 (훼2 + 훼3) 1 2훼2 1 2(log 퐶) 1/2 and ≥ − ≥ − ≥ − ≥ hence
3 ∑︁ √︀ 푔표푝 푔푖표푖푝푖 2√훼1훼2푛 푓(푛1)푓(푛2) 8훼1훼2푛푓(푛) 4훼2푛푓(푛) − 푖=1 ≥ ≥ ≥
2(훼2 + 훼3)푛푓(푛) = 2(푛 푛1)푓(푛) ≥ −
2(푛 푛1)푓(푛) (푛푓(푛) + 푛1푓(푛1) 2푛1푓(푛)) ≥ − − − 3 ∑︁ = 푛푓(푛) 푛1푓(푛1) 푛푓(푛) 푛푖푓(푛푖), − ≥ − 푖=1 where the second inequality is by both the first fact about 푓 applied to 푓(푛1) and the fifth fact about 푓 applied to 푓(푛2), the third inequality is by 훼1 1/2, and the ≥ second-to-last one is by the claim. Hence, 3 3 ∑︁ ∑︁ 푔표푝 푔푖표푖푝푖 + 푛푓(푛) 푛푖푓(푛푖) 푛푓(푛), ≥ 푖=1 − 푖=1 ≥ where the last inequality is by the induction on hypothesis applied to 푈푖 for 푖 = 1, 2, 3. This completes this case.
84 −1/4 Case 3: 훼1 < (log 퐶) . This is the sparse case, when each part is at most a (log 퐶)−1/4 = 퐷−2 fraction of the total.
Take 푛푖 = 푉푖 . Take 퐹푖 to be the coloring 퐹 restricted to 푉푖. Take 푔푖 to be the | | size of the largest subchromatic set in 퐹푖 using only the colors blue and yellow, 표푖 to
be the size of the largest subchromatic set in 퐹푖 using only the colors red and yellow,
and 푝푖 to be the size of the largest subchromatic set in 퐹푖 using only the colors red and blue.
We reorder the 푉푖 so that if 푖 푗 then 표푖푝푖 표푗푝푗. ≤ ≤ −2 −1/4 −2 휏 −2 Take 휏 = log(2퐷 푛) , so max푖 푛푖 (log 퐶) 푛 퐷 푛 2 2퐷 푛. ⌊ ⌋ ≤ ≤ ≤ ≤ 푖 푖+1 Define, for 푖 휏, 퐼푖 := [2 , 2 ]. Take Φ(푖) = 푗 : 푛푗 퐼푖 . The Φ(푖) are dyadically ≤ { ∈ } partitioning the indices; we will eventually use these partitions to construct sets to which we will apply the weighted Ramsey theorem. Note that ∑︀ , so we have ∑︀ . 푔 = 푗 푔푗 푔표푝 = 푗 푔푗표푝 We now present the idea behind the argument for the rest of this case. Fix 푖 so 7 (휏−푖) −7/18 that Φ(푖) has at least 2퐷2 8 elements and 푖 log(푛푚 ) (we will show that ≥ most vertices 푣 are contained in 푉푗 as 푗 varies over the Φ(푖) that have this property). We will define a weighted graph whose vertices are the indices and whose coloring is
휒. Given an index 푗 its weight will be (표푗, 푝푗). If we find a yellow clique in 휒 then the sum of the 표푗 in the clique gives a lower bound on 표, and, similarly, if we find a blue clique in 휒 then the sum of the 푝푗 in the clique gives a lower bound on 푝. We will apply the weighted Ramsey theorem to half of the indices in Φ(푖) (to the indices that are larger than the median of Φ(푖), to be precise); from this, we will be able to ′ conclude that if 푗 is an index smaller than the median, then 표푝/(표푗푝푗) 퐷 푓(푛)/푓(푛푗) ≥ ′ ′ ′ for some large constant 퐷 and so 푔푗표푝 퐷 푔푗표푗푝푗푓(푛)/푓(푛푗) 퐷 푛푗푓(푛). We now ≥ ≥ proceed with the argument.
When we count, we wish to omit parts Φ(푖) that don’t satisfy desired properties; take
′ 7 (휏−푖) 퐵 := 푖 휏 : Φ(푖) 2퐷2 8 , { ≤ | | ≤ } 퐵′′ := 푖 log(푛푚−7/18) . { ≤ }
85 Take 퐵 = 퐵′ 퐵′′. We will show that a large fraction of the vertices are not contained ∪ in 푉푗 for 푗 Φ(푖) where 푖 ranges over 퐵. ∈
∑︁ ∑︁ ∑︁ 푖+1 7 (휏−푖) 7 휏 ∑︁ 푖 7 휏 1 (휏+1)/8 푛 2 (2퐷2 8 ) = 4퐷2 8 2 8 4퐷2 8 2 푗 21/8 1 ≤ 푖≤휏 푖≤휏 ≤ · 푖∈퐵′ 푗∈Φ(푖) − 1 256 8퐷 2휏 128퐷2휏 푛 푛/4, ≤ 21/8 1 ≤ ≤ 퐷 ≤ −
where the fourth inequality follows from 21/8 (1 + 1/16). ≥
∑︀ 7/18 Note, if 푔푖 푚 /8, then we may complete the induction; assume this is not 푖 ≥ 7/18 the case. In particular, we get 푡 푚 /8 (since 푔푖 1). Therefore, ≤ ≥
∑︁ ∑︁ ∑︁ ∑︁ −7/18 −7/18 푛푗 2푛푚 2푡푛푚 푛/4. ≤ ≤ ≤ 푖∈퐵′′ 푗∈Φ(푖) 푖∈퐵′′ 푗∈Φ(푖)
Hence, ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ 푛푗 푛푗 + 푛푗 푛/4 + 푛/4 푛/2. ≤ ≤ ≤ 푖∈퐵 푗∈Φ(푖) 푖∈퐵′ 푗∈Φ(푖) 푖∈퐵′′ 푗∈Φ(푖)
∑︀ ∑︀ As a corollary we get 푛푗 푛/2. 푖̸∈퐵 푗∈Φ(푖) ≥
For any fixed 푖 휏 such that 푖 퐵, take 훽푖 to be the median of Φ(푖) (if Φ(푖) has ≤ ̸∈ an even number of elements, take 훽푖 to be the larger of the two medians). Consider 7 (휏−푖) (표푗, 푝푗): 푗 Φ(푖), 푗 훽푖 . By 푖 퐵, this has at least 퐷2 8 푀 elements { ∈ ≥ } ̸∈ ≥ (recall from the weighted Ramsey theorem that 푀 = 216), so we get, by applying the 2 (︁ 7 (휏−푖))︁ weighted Ramsey theorem to this set, that 표푝 표훽 푝훽 log 퐷2 8 /32. Finally, ≥ 푖 푖 7/18 observe that either one of the 표푗, 푝푗, 푔푗 is at least 푚 /8 in which case we may conclude the induction, or by the induction hypothesis we may assume 표푗푝푗푔푗 ≥ 푛푗푓(푛푗). Therefore,
86 ∑︁ ∑︁ 2 (︁ 7 (휏−푖))︁ ∑︁ 2 (︁ 7 (휏−푖))︁ 푔푗표푝 푔푗표훽 푝훽 log 퐷2 8 /32 푔푗표훽 푝훽 log 퐷2 8 /32 ≥ 푖 푖 ≥ 푖 푖 푗∈Φ(푖) 푗∈Φ(푖) 푗∈Φ(푖):푗≤훽푖
∑︁ 2 (︁ 7 (휏−푖))︁ ∑︁ 2 (︁ 7 (휏−푖))︁ 푔푗표푗푝푗 log 퐷2 8 /32 푛푗푓(푛푗) log 퐷2 8 /32 ≥ ≥ 푗∈Φ(푖):푗≤훽푖 푗∈Φ(푖):푗≤훽푖
(︁ 7 )︁ ∑︁ (︀ log 푛푗 )︀ 2 (휏−log 푛푗 ) ∑︁ 휏 푛푗푓 2 log 퐷2 8 /32 16푛푗푓(2 ) ≥ ≥ 푗∈Φ(푖):푗≤훽푖 푗∈Φ(푖):푗≤훽푖 ∑︁ 8푛푗푓(푛), ≥ 푗∈Φ(푖):푗≤훽푖
′ where the third inequality is by 표푗푝푗 표푗 푝푗 for 푗 푗 , the fourth inequality is by ≤ ′ ′ ≤ the induction hypothesis applied to 푉푗, the sixth inequality is by the third fact about 푓, and the seventh inequality is by the fourth fact about 푓 and noting 2휏 퐷−3푛. ≥
We now consider for any set 퐽 Φ(푖): ⊆
∑︁ 푖 푛푗 2 퐽 . ≥ | | 푗∈퐽
∑︁ 푖+1 푛푗 2 Φ(푖) . ≤ | | 푗∈Φ(푖)
This gives: ∑︀ 푗∈퐽 푛푗 퐽 ∑︀ | | . 푛푗 ≥ 2 Φ(푖) 푗∈Φ(푖) | |
Noting that 푗 Φ(푖): 푗 훽푖 Φ(푖) /2: |{ ∈ ≤ }| ≥ | |
∑︁ 1 ∑︁ ∑︁ 8푛푗푓(푛) 8푛푗푓(푛) = 2푓(푛) 푛푗. ≥ 4 푗∈Φ(푖):푗≤훽푖 푗∈Φ(푖) 푗∈Φ(푖)
Therefore,
87 ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ ∑︁ 푔표푝 푔푗표푝 푔푗표푝 푔푗표푝 2푓(푛) 푛푗 ≥ ≥ ≥ ≥ 푗 푖≤휏 푗∈Φ(푖) 푖≤휏:푖̸∈퐵 푗∈Φ(푖) 푖≤휏:푖̸∈퐵 푗∈Φ(푖) ∑︁ ∑︁ 푛 = 2푓(푛) 푛푗 2푓(푛) = 푛푓(푛). ≥ 2 푖≤휏:푖̸∈퐵 푗∈Φ(푖)
We have thus concluded the induction. 2 We informally refer to 퐵′′ in the above proof as large if a large fraction of the ′′ vertices are contained in a 푉푗 for 푗 Φ(푖) where 푖 ranges over 퐵 . The case in which ∈ 퐵′′ was large easily implied the desired result. In extending this result in Section 3.8 to more colors, the primary difficulty is the following: when 푠 is not 2, it is not obvious that there is a large 푠-colored set as a result of 퐵′′ being large.
3.6 Upper bound for many colors
In this section we will give asymptotically tight upper bounds for how large of a
subchromatic set must exist in an edge coloring on 푚 vertices. We will first show how to construct such colorings from weighted graphs with vertex set 푅, and then we will choose such graphs to finish the construction. The next theorem states that
if we have a weighted graph on 푟 vertices with edge weights 푤푃 , then we can find a coloring so that is, up to logarithmic factors, ∏︀ . 퐹 푔푆,퐹 푃 ⊆푆 푤푃
Lemma 3.6.1. Given a weighted graph (푅, ) on 푟 vertices with integer edge weights 풫 ∏︀ 푤푃 푃 ∈풫 , taking 푚 := 푤푃 , there is a Gallai 푟-coloring on 푚 vertices so that { } 푃 ∈풫 for any 푆 푅, the size of the largest subchromatic set with colors in 푆 is at most ⊆ ∏︀ ∏︀ 푤푃 2 log 푤푃 . 푃 ∈풫:푃 ⊆푆 · 푃 ∈풫:|푃 ∩푆|=1
Proof: We may define a Gallai 푟-coloring on 푚 vertices as follows: take 푃1, . . . , 푃푘 an arbitrary enumeration of . For each edge 푃 , take 퐹푃 to be a 2-coloring of 퐸(퐾푤 ) 풫 푃 using colors from 푃 so that the largest monochromatic clique has order at most
2 log 푤푃 (such a coloring exists by the Erdős-Szekeres bound for Ramsey numbers
88 [36]). We define a coloring 퐹 on 푚 vertices by
퐹 = 퐹푃 퐹푃 퐹푃 . 1 ⊗ 2 ⊗ · · · ⊗ 푘
퐹 is a Gallai coloring by Corollary 3.2.4. Given any 푆 푅, note that 푔 = 푤푃 if ⊆ 푆,퐹푃 푃 푆, as 퐹푃 uses only colors from 푃 . If 푃 푆 = 1, then the largest subchromatic ⊆ | ∩ | set in 퐹푃 using colors from 푃 푆 is at most 2 log 푤푃 by choice of 퐹푃 , so 푔 ∩ 푆,퐹푃 ≤ 2 log 푤푃 . If 푃 푆 = 0, then 푔 = 1 as any two distinct vertices are connected by | ∩ | 푆,퐹푃 an edge the color of which is not in 푆. Therefore,
∏︁ ∏︁ ∏︁ 푔푆,퐹 = 푔푆,퐹 푤푃 2 log 푤푃 . 푃푖 ≤ · 푖 푃 ∈풫:푃 ⊆푆 푃 ∈풫:|푃 ∩푆|=1
2
The condition in the above lemma that the edge weights are integers is slightly cumbersome; we will now eliminate it.
Lemma 3.6.2. Let (푅, ) be a weighted graph on 푟 vertices (푟 3) and weights 풫 ≥ ∏︀ 푤푃 satisfying 푤푃 = 휔(1) for every 푃 . Letting 푚 := 푤푃 , if 푚 is an ∈ 풫 푃 ∈풫 integer and each 푤푃 satisfies 푤푃 휔(1), then there is a Gallai 푟-coloring on 푚 ≥ vertices such that, for any 푆 푅, the size of the largest subchromatic set is at most ⊆ ∏︀ ∏︀ (1 + 표(1)) 푤푃 2 log 푤푃 . 푃 ∈풫:푃 ⊆푆 · 푃 ∈풫:|푃 ∩푆|=1
′ ′ Proof: Take 푤 = 푤푃 . Since 푤푃 휔(1), we get 푤 (1 + 표(1))푤푃 . We may 푃 ⌈ ⌉ ≥ 푃 ≤ apply the previous lemma to the 푤′ to get an 푟-Gallai coloring on ∏︀ 푤′ 푚 푃 푃 푃 ≥ vertices so that for any 푆 푅 the size of the largest subchromatic set is at most ⊆
∏︁ ′ ∏︁ ′ ∏︁ ∏︁ 푤 2 log 푤 (1 + 표(1)) 푤푃 2 log 푤푃 . 푃 · 푃 ≤ · 푃 ∈풫:푃 ⊆푆 푃 ∈풫:|푃 ∩푆|=1 푃 ∈풫:푃 ⊆푆 푃 ∈풫:|푃 ∩푆|=1
Restrict this coloring to any 푚 vertices; it is still a Gallai 푟-coloring and for any ∏︀ 푆 푅 the size of the largest subchromatic set is at most (1 + 표(1)) 푤푃 ⊆ 푃 ∈풫:푃 ⊆푆 · ∏︀ 푃 ∈풫:|푃 ∩푆|=1 2 log 푤푃 . 2
89 Now, if we wish to obtain colorings without large subchromatic sets, we need only construct appropriate weighted graphs. Intuitively, we would like to minimize the number of edges in such a graph (while still being able to maintain that all the ∏︀ 푆 푅 have approximately the same value of 푤푃 ), as every edge creates extra ⊆ 푃 ⊆푆 log factors. This observation motivates the following bounds.
Theorem 3.6.3. There is a Gallai 푟-coloring on 푚 vertices such that for any 푆 (︀푅)︀ ∈ 푠 푠 / 푟 푐 the size of the largest subchromatic set is at most (1 + 표(1))푚(2) (2) log 푟,푠 푚, where
⎧ if ⎪ 푠(푟 푠) 푠 < 푟 1, ⎨⎪ − − 푐푟,푠 = 1 if 푠 = 푟 1 and r is even, ⎪ − ⎩⎪ (푟 + 3)/푟 if 푠 = 푟 1 and r is odd. − Proof: If 푠 < 푟 1, we may apply the previous lemma to a clique on 푟 vertices with − 1/ 푟 (︀푠)︀ edge weights 푚 (2). Any 푆 푅 of size 푠 has internal edges and 푠(푟 푠) edges ⊆ 2 − intersecting it in one vertex. By the previous lemma, we may find a Gallai 푟-coloring where the size of the largest subchromatic set is asymptotically at most:
푠(푟−푠) 푠 / 푟 (︁ (︁ 1/ 푟 )︁)︁ 푠 / 푟 푠(푟−푠) 푚(2) (2) 2 log 푚 (2) 푚(2) (2)(log 푚) . ≤
If 푠 = 푟 1 and 푟 is even, we may consider a perfect matching on 푟 vertices where − each edge has weight 푚2/푟; any subset of size 푟 1 contains 푟/2 1 edges and there is − − one edge with which it shares exactly one vertex. By the previous lemma, we may find a Gallai 푟-coloring where the size of the largest subchromatic set is asymptotically at most:
(푟/2−1)/(푟/2) 1/(푟/2) (푟/2−1)/(푟/2) 푠 / 푟 푚 2 log(푚 ) 푚 log 푚 = 푚(2) (2) log 푚. ≤
If 푠 = 푟 1 and 푟 is odd, we may consider a graph formed by taking the disjoint − union of a triangle on 3 vertices and a matching with (푟 3)/2 edges. The edges − 1/푟 (푟−3)/2푟 of the triangle will each have weight 푤1 := 푚 (log 푚) and the edges of the 2/푟 −3/푟 matching will each have weight 푤2 := 푚 (log 푚) . Note that the product of the weights is 푤3푤(푟−3)/2 = 푚. Let 푆 푅 of size 푠 = 푟 1 be given. 1 2 ⊆ − 90 If the vertex not contained in 푆 is part of the triangle, then 푆 contains (푟 3)/2 − edges of weight 푤2 and 1 edge of weight 푤1. Furthermore, there are two edges each of weight 푤1 that 푆 intersects in one vertex. In the graph obtained from the previous lemma the size of the largest subchromatic set taking colors from 푆 is asymptotically at most:
(푟−3)/2 2 (푟−2)/푟 −(푟−3)/푟 1/푟 (푟−3)/2푟 2 푤1푤2 (2 log 푤1) = 푚 (log 푚) (2 log(푚 (log 푚) )) 푚(푟−2)/푟(log 푚)−(푟−3)/푟(log 푚)2 ≤ 푠 / 푟 (푟+3)/푟 = 푚(2) (2)(log 푚) .
If the vertex not contained in 푆 is part of the matching then 푆 contains (푟 5)/2 − edges of weight 푤2 and 3 edges of weight 푤1. Furthermore, there is one edge of weight
푤2 that intersects 푆 in one vertex. In the graph obtained from the previous lemma the size of the largest subchromatic set taking colors from 푆 is asymptotically at most:
3 (푟−5)/2 (푟−2)/푟 3/푟 2/푟 −3/푟 푤1푤2 (2 log 푤2) = 푚 (log 푚) (2 log(푚 (log 푚) )) 푚(푟−2)/푟(log 푚)3/푟(log 푚) ≤ 푠 / 푟 (푟+3)/푟 = 푚(2) (2)(log 푚) .
2
3.7 Weak lower bound for many colors
We now provide a simple lower bound for the largest size of a subchromatic set in any
푟-coloring of 퐸(퐾푚) that shows our upper bounds are tight up to polylogarithmic factors; we show that any Gallai 푟-coloring on 푚 vertices contains a subchromatic 푠 / 푟 set of size at least 푚(2) (2). The following is a common generalization of Hölder’s inequality that we will find useful.
91 Lemma 3.7.1. If is a finite set of indices and, for each 푆 , 푔푆 is a function 풮 ∈ 풮 mapping [푡] to the non-negative reals, then
(︃ )︃|풮| ∏︁ ∑︁ ∑︁ ∏︁ 1/|풮| 푔푆(푖) 푔푆(푖) ≥ 푆∈풮 푖 푖 푆∈풮
Using the above lemma, we will prove a lower bound on the product of the 푔푆,퐹 for 퐹 a Gallai 푟-coloring. This will easily imply the desired lower bound.
Theorem 3.7.2. For any Gallai 푟-coloring 퐹 on 푚 vertices,
∏︁ 푟 2 푔 푚(푠−2). 푆,퐹 ≥ − 푆∈ 푅 ( 푠 )
Proof: Take . We proceed by induction on . If , then each is 푔푆 = 푔푆,퐹 푚 푚 = 1 푔푆 1 푟 2 as is their product, while (푠−2) is also . If , we may find some pair of colors 푚 − 1 푚 > 1
푄 and some non-trivial partition of the vertices 푉1, . . . , 푉푡 such that for each pair of distinct 푖, 푗 in [푡], there is a 푞 푄 so that all of the edges between 푉푖 and 푉푗 have ∈ color 푞.
Define, for 푖 [푡], 퐹푖 to be the restriction of 퐹 to 푉푖. Take 푔 := 푔 . By 푆,푖 푆,퐹푖 ∈ 푟 2 ∏︀ (푠−2) induction, for each 푖 we have 푔 푚 − , where 푚푖 = 푉푖 . 푆 푆,푖 ≥ 푖 | | Note that if 푄 푆 then 푔 ∑︀ 푔 , since we may combine the largest subchro- ⊆ 푆 ≥ 푖 푆,푖 matic sets from each 퐹푖. For every 푆 we have 푔 max푖 푔 , so 푆 ≥ 푆,푖
푟 2 (︃ )︃ (︃ )︃(푠−2) 1/ 푟 2 − ∏︁ ∏︁ ∑︁ ∏︁ ∑︁ ∏︁ (푠−2) ∏︁ 푔 푔 푔 푔 − 푔 푆 ≥ 푆,푖 푆 ≥ 푆,푖 푆 푆 푆:푄⊆푆 푖 푆:푄̸⊆푆 푖 푆:푄⊆푆 푆:푄̸⊆푆 푟 2 푟 2 (︃ )︃(푠−2) (︃ )︃(푠−2) 1/ 푟 2 1/ 푟 2 − 1/ 푟 2 − ∑︁ ∏︁ (푠−2) ∏︁ (푠−2) ∑︁ ∏︁ (푠−2) = 푔 − 푔 − 푔 − 푆,푖 푆 ≥ 푆,푖 푖 푆:푄⊆푆 푆:푄̸⊆푆 푖 푆 푟 2 (︃ )︃(푠−2) ∑︁ − 푟 2 (푠−2) 푚푖 = 푚 − , ≥ 푖 where the first inequality follows by 푔 ∑︀ 푔 if 푄 푆, the second inequality 푆 ≥ 푖 푆,푖 ⊆ 92 follows by the preceding lemma and noting 푆 = (︀푟−2)︀, the third inequality follows | | 푠−2 by 푔 푔 , and the fourth inequality follows by the induction hypothesis. 2 푆 ≥ 푆,푖 Note that, in proving this bound, if 푆 푄 = 1, we simply use 푔 푔 . As in | ∩ | 푆 ≥ 푆,퐹푖 the case, if we can find a set of indices so that between any 푟 = 3, 푠 = 2 푉푖1 , . . . , 푉푖푘 two of them the edges use the color contained in 푆 푄, we may obtain a stronger ∩ lower bound on . 푔푆 We now conclude the argument.
Theorem 3.7.3. In any Gallai 푟-coloring 퐹 on 푚 vertices, there is some 푆 (︀푅)︀ ∈ 푠 푠 / 푟 with 푔 푚(2) (2) 푆,퐹 ≥
푟 2 Proof: By the previous theorem, ∏︀ (푠−2). As this is a product over 푆∈ 푅 푔푆,퐹 푚 − ( 푠 ) ≥ (︀푟)︀ numbers, there must be some with 푠 푆
푟 2 / 푟 푠 / 푟 푔 푚(푠−2) (푠) = 푚(2) (2). 푆,퐹 ≥ −
2
3.8 Lower bound for many colors
In this section we show that our upper bounds on sizes of subchromatic sets in Gallai
colorings are tight up to constant factors (where we view 푟 and 푠 as constant).
3.8.1 Discrepancy lemma in edge-weighted graphs
The lemma in this subsection has the following form: either a given weighted graph
has many edges of non-zero weight or it has some set 푆 of size 푠 whose weight is significantly larger than average. In the next subsection we will show howtoreduce
the problem of lower bounding the size of the largest subchromatic set in a Gallai 푟- coloring to a problem regarding the number of non-zero edges in a graph that doesn’t
contain vertex subsets 푆 whose weight is significantly larger than average, so this lemma will be useful.
93 (︀푅)︀ ∑︀ Lemma 3.8.1. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . Take ∈ 2 ≥ 푃 (︀푟)︀ 푎0 = if 푠 < 푟 1, 푎0 = 푟/2 if 푠 = 푟 1 and 푟 is even, and 푎0 = (푟 + 3)/2 if 2 − − 푠 = 푟 1 and 푟 is odd. Either there are at least 푎0 pairs 푃 with 푤푃 > 0 or there is − some 푆 푅 of size 푠 satisfying ⊆ ⎛ ⎞ (︃ (︂ )︂2)︃−1 (︀푠)︀ ∑︁ 푟 2 푤푃 ⎝1 + 4푟 ⎠ 푤. ≥ 2 (︀푟)︀ 푃 ⊆푆 2
The proof of the above lemma uses elementary techniques along with the second moment method and is deferred to Section 3.11.
3.8.2 Proof of lower bound for many colors
Let
(︀푟−2)︀ 푟 푠 푑 = 푠−1 = − , (︀푟−2)︀ 푠 1 푠−2 − (︂푟)︂3 퐶 = 32푟 푑, 2 (︂ (︂푟 2)︂ )︂−1 훿 = 4 − 퐶 , 푠 2 (︂− )︂(︂ )︂−1 (︂(︂ )︂ )︂−1 (︂(︂ )︂ )︂−1 −1 푟 2 푟 2 푟 −1 푟 훿0 = 퐶 − − + 1 = 퐶 푑 + 1 , 푠 1 푠 2 2 2 − − (︂ )︂−1 푟 − 푟 −1 푟 2 −( )−2 (︀ −1 )︀ (2) 훿1 = 2 2 훿 + 1 − , 0 푠 1 − 2 1/푑 푐 = (훿/4) 훿1 .
푑 is an appropriately chosen scaling factor; why it is appropriate will become evident
later. 퐶 should be thought of as a large constant, and 훿, 훿0, 훿1, and 푐 should be thought of as small constants. We provide some bounds on the above; although we will not explicitly reference these, they are useful for verifying various inequalities:
94 푟−1 푑 푟, ≤ ≤ 퐶 4푟8, ≤ 훿 2−8푟, ≥ −11 훿0 푟 /4, ≥ −11푟3 훿1 2 , ≥ 4 푐 2−12푟 . ≥
When we constructed the upper bound via product colorings, there was a weighted graph (namely the one used to construct the coloring) so that for any 푆 푅 we could ⊆ approximate the size of the largest clique using colors from 푆 by the product of the weights of edges contained in 푆. It is tempting to believe that the structure of any Gallai coloring 퐹 can be approximated this way. Though this is not true in general, the next theorem states that if it is not true then ∏︀ must be large. Take for 푆 푔푆,퐹 the rest of this chapter 2 28푟 22 푚0 := 2 . (3.1)
Theorem 3.8.2. If 푚 푚0, then for any Gallai coloring 퐹 on 푛 푚 vertices, there ≥ ≤ (︀푅)︀ are 푓 1, 휖 0, , and, for 푃 , weights 푤푃 [1, ) satisfying: ≥ ≥ 풫 ⊆ 2 ∈ 풫 ∈ ∞ 1. For every (︀푅)︀, ∏︀ 푆 푠 푔푆,퐹 푃 ∈ 푆 ∩풫 푤푃 . ∈ ≥ (2)
∏︀ −휖 2. 푤푃 푚 푛. 푃 ∈풫 ≥
푟 2 3. ∏︀ (푠−2) 푆∈ 푅 푔푆,퐹 (푛푓) − . ( 푠 ) ≥ 4. 푓 (log 푚)퐶휖. ≥ 푎푑 5. Taking 푎 to be the size of , 푓 (︀푐 log2 푚)︀ . 풫 ≥ From the above theorem we will quickly be able to conclude Theorem 3.1.1. Note that, if is large enough, then by condition (3) we conclude that ∏︀ is large 푓 푆∈ 푅 푔푆,퐹 ( 푠 ) and so some is large. Otherwise, by condition (4) we have an upper bound on the 푔푆,퐹
95 size of 휖, so by conditions (1) and (2) the structure of the coloring is well-approximated
by the 푤푃 . This latter case will allow us to apply our work on weighted graphs from the previous subsection to get a lower bound on 푎, and then we will apply condition (5) to conclude that some is large. 푔푆,퐹 Proof: We will write for . We will take for any in (︀푅)︀ but not in 푔푆 푔푆,퐹 푤푃 = 1 푃 2 (︀푅)︀ ∏︀ ∏︀ ; this way, for any 푇 , we have 푤푃 = 푤푃 . 풫 ⊆ 2 푃 ∈푇 ∩풫 푃 ∈푇 We proceed by induction on 푛. Base Case: If 푛 = 1, then we may take 푓 = 1, 휖 = 0, and = . Letting 푎 = 푃 = 0, 풫 ∅ | | 1. For every (︀푅)︀, ∏︀ . 푆 푠 푔푆 = 1 = 푃 ∈ 푆 푤푃 ∈ (2) 2. ∏︀ −휖 . 푃 ∈ 푅 푤푃 = 1 = 푚 푛 ( 2 )
푟 2 3. ∏︀ (푠−2). 푆∈ 푅 푔푆 = 1 = (푛푓) − ( 푠 )
4. 푓 = 1 = (log 푚)퐶휖.
푎푑 5. 푓 = 1 = (︀푐 log2 푚)︀ .
Preliminary discussion: If 푛 > 1, there is some pair of colors 푄 = 푄1, 푄2 and { } there is a non-trivial partition 푉 (퐾푛) = 푉1 ... 푉푡 with 푉1 ... 푉푡 such that ∪ ∪ | | ≥ ≥ | | (︀푡)︀ there is some 2-coloring 휒 : 푄 such that for every distinct 푖, 푗 [푡] and 푢 푉푖, 2 → ∈ ∈ 푣 푉푗, the color under 퐹 of 푢, 푣 is 휒(푖, 푗) (which is in 푄). ∈ { } 휖 퐶 log(훼푚 ) Given 휖 > 0, define 푓휖(ℓ) := (log 푚) log 푚 , where 훼 = ℓ/푛. Note that we may 퐶휖+퐶 log 훼 rewrite 푓휖(ℓ) = (log 푚) log 푚 ; we will move between the two expressions freely.
Note also that 푓휖(ℓ) is an increasing function of ℓ. We will need some lemmas about
푓휖, all of which are formalizations of the statement “푓휖 does not grow too quickly".
Lemma 3.8.3. The following statements hold about 푓휖 for every choice of 휖 0, ≥ 푚 푚0, and 1 < 푛 푚. ≥ ≤ 1. For any 1 , 훼 [ 푛 , 1] ∈ 푟 2 1/(2( − )) 푓휖(훼푛) 훼 푠 2 푓휖(푛). ≥ −
In particular, 푓휖(훼푛) 훼푓휖(푛). ≥ 96 1 2. For any 훼1, 훼2, 훼3 [ , 1] with 훼1 + 훼2 + 훼3 = 1, taking 푛푖 = 훼푖푛, ∈ 푛
∑︁ −3/4 푛푓휖(푛) 푛푖푓휖(푛푖) + 3(log 푚)푛푓휖(푛). ≤ 푖
푟 2 훿 푗 푖 2/( − ) 1/4 푗 (︀푟)︀ 푖+2푗 3. For 푖 0 and 푚 2 1 we have 푓휖(2 ) log 푠 2 ((log 푚)2 ) 256 푓휖(2 ). ≥ ≥ ≥ − ≥ 2
−1 4. For any 훼 log 푚, 푓휖(훼푛) 푓휖(푛)/2. ≥ ≥
We will refer to the above collectively as the facts about 푓휖; we prove them in Section 3.12. The proof will split into four cases. Cases 1 and 2, Preliminary discussion: For these cases, a simple numerical claim will be useful.
Claim 3.8.4. For positive reals 푎, 푏 with 푎 1, we have ≤
(1 + 푎)푏 1 + 푎푏/2. ≥
Proof: Since 0 푎 1 we have 1 + 푎 푒푎/2. Then ≤ ≤ ≥
(1 + 푎)푏 푒푎푏/2 1 + 푎푏/2. ≥ ≥
2
Cases 1 and 2 will be those cases in which 푉1 is large.
Take 푈1 = 푉1. Take 푈2 to be the union of the 푉푗 such that the edges between 푉1
and 푉푗 are of color 푄1. Take 푈3 to be the union of the 푉푗 such that the edges between
푉1 and 푉푗 are of color 푄2. We may assume without loss of generality that 푈2 푈3 . | | ≥ | | Then 푈1, 푈2, 푈3 is a partition of 푉 . { } Define, for 푖 = 1, 2, 3, 퐹푖 to be the restriction of 퐹 to 푈푖. Take 푔 := 푔 . 푆,푖 푆,퐹푖
Define 푛푖 = 푈푖 and 훼푖 = 푛푖/푛. By the induction hypothesis, for each 퐹푖 there are | | appropriate choices of 푓푖, 휖푖, 푖, and 푤푃,푖. Take 푎푖 = 푖 . 풫 |풫 | 97 The general approach for these cases as well as for Case 3 will be to choose some
index 푖 and simply use the same graph to approximate our coloring. That is, we will take = 푖 and 푤푃 = 푤푃,푖, and then we will show that 휖 and 푓 may be chosen 풫 풫 appropriately.
We now proceed: if, for some index 푖, we take = 푖 and 푤푃 = 푤푃,푖, since 푔 풫 풫 푆 ≥ (︀푅)︀ ∏︀ ∏︀ 푔 , we will have property (1): for every 푆 , 푔 푔 푤푃,푖 = 푤푃 . 푆,푖 ∈ 푠 푆 ≥ 푆,푖 ≥ 푃 ⊆푆 푃 ⊆푆 Furthermore,
∏︁ ∏︁ −휖푖 −휖푖 −휖푖−log(1/훼푖)/ log 푚 푤푃 = 푤푃,푖 푚 푛푖 = 푚 훼푖푛 = 푚 푛. ≥ 푃 ∈ 푅 푃 ∈ 푅 ( 2 ) ( 2 )
If we take log(1/훼푖) , then the above shows that property (2) will hold. 휖 = 휖푖 + log 푚 Define (︂ (︁ )︁ )︂ 퐶 휖 + log(1/훼푖) 푖 log 푚 2 푎푖푑 푥푖 := max (log 푚) , (푐 log 푚) .
If 푖 is the index minimizing 푥푖, then we will take:
log(1/훼 ) 휖 = 휖 + 푖 , 푖 log 푚
= 푖, 풫 풫
푤푃 = 푤푃,푖,
and 푓 = 푥푖; we will show that this satisfies properties (4) and (5), so choosing 푖 to minimize 푥푖 minimizes our 푓. Take 푎 = . We have already observed that properties |풫| (1) and (2) will hold.
(︁ log((푐 log2 푚)푎푑) )︁ Take ′ . Note that 퐶휖′ . In this 휖 = max 휖, 퐶 log log 푚 푓 = 푥푖 = (log 푚) = 푓휖′ (푛) case properties (4) and (5) hold by the choice of 푓:
푓 (log 푚)퐶휖′ (log 푚)퐶휖. ≥ ≥ 2 푎푑 퐶휖 퐶 log((푐 log 푚) ) log((푐 log2 푚)푎푑) 2 푎푑 푓 (log 푚) ′ (log 푚) 퐶 log log 푚 = 2 = (푐 log 푚) . ≥ ≥
We have only to show that, with this choice of 푓, property (3) holds. We claim
98 that each 푓푖 satisfies 푓푖 푓휖 (푛푖). ≥ ′
′ log 훼푖 2 푎 푑 If for some 푖 we have 휖푖 < 휖 + , then we must have (푐 log 푚) 푖 푓 = log 푚 ≥ 퐶휖 (log 푚) ′ , for otherwise we would have 푥푖 < 푓, contradicting our choice of 휖. There- fore, for such an index 푖,
2 푎푖푑 퐶휖′ 푓푖 (푐 log 푚) (log 푚) = 푓휖 (푛) 푓휖 (푛푖). ≥ ≥ ′ ≥ ′
′ log 훼푖 Otherwise, 휖푖 휖 + so ≥ log 푚
log 훼푖 퐶휖푖 퐶(휖′+ ) 푓푖 (log 푚) (log 푚) log 푚 = 푓휖 (푛푖). ≥ ≥ ′
We have, for each 푆 satisfying 푄 푆, that 푔 푔 + 푔 + 푔 . For each 푆 ⊆ 푆 ≥ 푆,1 푆,2 푆,3 satisfying 푄1 푆, we have ∈
푔 max(푔 + 푔 , 푔 ) 푔 + 푔 . 푆 ≥ 푆,1 푆,2 푆,3 ≥ 푆,1 푆,2
Similarly, if 푄2 푆 then 푔 푔 + 푔 . Finally, for all 푆 we have 푔 max푖 푔 . ∈ 푆 ≥ 푆,1 푆,3 푆 ≥ 푆,푖 We have by the generalization of Hölder’s inequality (Lemma 3.7.1):
푟 2 ⎛ 1/ 푟 2 ⎞(푠−2) (︃ )︃ (푠−2) − ∏︁ ∏︁ ∑︁ ∏︁ ∑︁ ∏︁ ∏︁ − 푔푆 푔푆,푖 푔푆 ⎝ 푔푆,푖 푔푆 ⎠ . ≥ ≥ 푆 푆:푄⊆푆 푖 푆:푄̸⊆푆 푖 푆:푄⊆푆 푆:푄̸⊆푆
Therefore, we need only check that
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − 푔 푔 푛푓. 푆,푖 푆 ≥ 푖 푆:푄⊆푆 푆:푄̸⊆푆
(︀푅)︀ Fix 푇 so that 푇 푄 = 푄1 . We get 푔 푔 + 푔 . ∈ 푠 ∩ { } 푇 ≥ 푇,1 푇,2 −1/2 Case 1: 훼1, 훼2 log 푚. The argument from the weak lower bound case applied ≥ 99 here only gives:
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − ∑︁ 푔 푔 푛푖푓휖 (푛푖). 푆,푖 푆 ≥ ′ 푖 푆:푄⊆푆 푆:푄̸⊆푆 푖
The main idea behind solving this case is to observe that it is sufficient to gain a constant factor on either the largest or second largest term of the above sum. Consider:
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − 푔푆,푖 푔푆 푖 푆:푄⊆푆 푆:푄̸⊆푆 1/ 푟 2 1/ 푟 2 1/ 푟 2 (︃ )︃ (푠−2) (︃ )︃ (푠−2) (︃ )︃ (푠−2) ∏︁ − ∏︁ − ∏︁ − (푔 + 푔 ) 푔 + (푔 + 푔 ) 푔 + 푔 . ≥ 푇,1 푇,2 푆,1 푇,1 푇,2 푆,2 푆,3 푆̸=푇 푆̸=푇 푆
We will handle the case 푔 푔 ; the case where 푔 푔 has a symmetric 푇,1 ≤ 푇,2 푇,1 ≥ 푇,2 argument. Then, since 푔 + 푔 2푔 , the previous is at least: 푇,1 푇,2 ≥ 푇,1
푟 2 푟 2 푟 2 1/ 푟 2 ∏︁ 1/( − ) ∏︁ 1/( − ) ∏︁ 1/( − ) (푠−2) 푠 2 푠 2 푠 2 2 − 푔푆,1 − + 푔푆,2 − + 푔푆,3 − 푆 푆 푆 1/ 푟 2 (︁ 푟 2 )︁ 1/ 푟 2 ∑︁ ∏︁ (푠−2) 1/( − ) ∏︁ (푠−2) = 푔 − + 2 푠 2 1 푔 − 푆,푖 − − 푆,1 푖 푆 푆 ∑︁ 1/ 푟 2 (푠−2) 푛푖푓푖 + ((1 + 1) − 1)푛1푓1 ≥ 푖 − −1 ∑︁ (︂ (︂푟 2)︂)︂ 푛푖푓휖 (푛푖) + 2 − 푛1푓휖 (푛1) ≥ ′ 푠 2 ′ 푖 − (︂ (︂ )︂)︂−1 ∑︁ 푟 2 −1/2 −1/2 푛푖푓휖 (푛푖) + 2 − (log 푚)푛푓휖 ((log 푚)푛) ≥ ′ 푠 2 ′ 푖 − (︂ (︂ )︂)︂−1 ∑︁ 푟 2 −1/2 푛푖푓휖 (푛푖) + 4 − (log 푚)푛푓휖 (푛) ≥ ′ 푠 2 ′ 푖 − ∑︁ −3/4 푛푖푓휖′ (푛푖) + 3(log 푚)푛푓휖′ (푛) 푛푓휖′ (푛) = 푛푓, ≥ 푖 ≥
where the first follows from the induction hypothesis, the second follows from Claim
3.8.4, the third follows from the lower bound on 훼1, the fourth follows from the fourth
100 fact about 푓휖 , the fifth follows from 푚 푚0, and the sixth follows from the second ′ ≥ fact about . 푓휖′ −1/2 Case 2: log 푚 훼2. In this case we have 훼1 = 1 (훼2 + 훼3) 1 2훼2 ≥ − ≥ − ≥ 1 2 log−1/2 푚 3/4. − ≥ Again, the argument from the weak lower bound only gives:
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − ∑︁ 푔 푔 푛푖푓휖 (푛푖). 푆,푖 푆 ≥ ′ 푖 푆:푄⊆푆 푆:푄̸⊆푆 푖
The main idea behind this case is to observe that it is sufficient to gain either a factor
of (1 + 8훼2) on the first term (which is much larger than the others) or a factor of −1/ 2 푟 2 ( (푠−2)) − on the second term. We will do the former if is large enough, 4훼2 푔푇,2 /푔푇,1 and otherwise we may accomplish the latter.
(︀푟−2)︀ If 푔 16 훼2푔 , we have 푇,2 ≥ 푠−2 푇,1
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − 푔푆,푖 푔푆 푖 푆:푄⊆푆 푆:푄̸⊆푆 1/ 푟 2 (︃ )︃ (푠−2) ∏︁ ∏︁ − 푔 푔 ≥ 푆,1 푆 푆:푄⊆푆 푆:푄̸⊆푆 1/ 푟 2 푟 2 (︂ (︂ )︂ )︂ (푠−2) 푟 2 1/ 푟 2 ∏︁ 1/( − ) 푟 2 − ∏︁ 1/( − ) (푠−2) 푠 2 푠 2 (푔 + 푔 ) 푔푆,1 − 1 + 16 − 훼2 푔푆,1 − ≥ 푇,1 푇,2 − ≥ 푠 2 푆̸=푇 − 푆 (︃ (︀푟−2)︀ )︃ 16 푠−2 훼2 1 + (︀푟−2)︀ 푛1푓1 = 푛1푓1 + 8훼2푛1푓1, ≥ 2 푠−2
where the last inequality is by Claim 3.8.4.
We know 푓푖 푓휖 (푛푖), so the above is at least: ≥ ′
2 2 푛1푓휖 (푛1) + 8훼2푛1푓휖 (푛1) 훼 푛푓휖 (푛) + 8훼2훼 푛푓휖 (푛) ′ ′ ≥ 1 ′ 1 ′ 2 (1 2훼2) 푛푓휖 (푛) + 4훼2푛푓휖 (푛) ≥ − ′ ′
> 푛푓휖′ (푛) = 푛푓,
101 where the first inequality follows from the first fact about and the second inequality 푓휖′ from substituting lower bounds on 훼1. (︀푟−2)︀ Otherwise, we have 푔 16 훼2푔 , so 푇,2 ≤ 푠−2 푇,1
1/ 푟 2 1/ 푟 2 (︃ )︃ (푠−2) (︃ )︃ (푠−2) − 1/ 푟 2 − ∑︁ ∏︁ ∏︁ ∏︁ (푠−2) ∏︁ 푔 푔 푔 − + (푔 + 푔 ) 푔 . 푆,푖 푆 ≥ 푆,1 푇,1 푇,2 푆,2 푖 푆:푄⊆푆 푆:푄̸⊆푆 푆 푆̸=푇
Then the latter term is at least:
1/ 푟 2 1/ 푟 2 (︃ )︃ (푠−2) (︃ )︃ (푠−2) ∏︁ − 1 ∏︁ − 푔 푔 푔 푇,1 푆,2 ≥ (︀푟−2)︀ 푆,2 푆̸=푇 16 푠−2 훼2 푆 1/ 푟 2 (︃ )︃ (푠−2) − −1/ 2 푟 2 1 ( (푠−2)) (︀푟−2)︀ 푛2푓2 4훼2 − 푛2푓2 ≥ 16 푠−2 훼2 ≥ −1/ 2 푟 2 −1/ 2 푟 2 (︂ 1/ 2 푟 2 )︂ ( (푠−2)) ( (푠−2)) ( (푠−2)) 4훼 − 푛2푓휖 (푛2) 4훼 − 푛2 훼 − 푓휖 (푛) ≥ 2 ′ ≥ 2 2 ′
= 4푛2푓휖′ (푛),
where the third inequality follows from the upper bound on 훼2 and from 푚 푚0 and ≥ the fifth inequality from the first fact about . 푓휖′ Therefore,
1/ 푟 2 (︃ )︃ (푠−2) − 1/ 푟 2 ∑︁ ∏︁ ∏︁ ∏︁ (푠−2) 푔 푔 푔 − + 4푛2푓휖 (푛) 푛1푓휖 (푛1) + 4푛2푓휖 (푛) 푆,푖 푆 ≥ 푆,1 ′ ≥ ′ ′ 푖 푆:푄⊆푆 푆:푄̸⊆푆 푆
2 훼 푛푓휖 (푛) + 4훼2푛푓휖 (푛) ≥ 1 ′ ′ 2 (1 2훼2) 푛푓휖 (푛) + 4훼2푛푓휖 (푛) ≥ − ′ ′
푛푓휖 (푛) = 푛푓, ≥ ′
where the third inequality follows from the first fact about . 푓휖′ Cases 3 and 4, Preliminary discussion: These will be the cases in which none of
the 푉푖 are large. For these cases, we will take 푛푖 = 푉푖 and 훼푖 = 푛푖/푛. We will take | | 102 퐹푖 to be 퐹 restricted to 푉푖 and take 푔 = 푔 . By induction, for each 퐹푖 there are 푆,푖 푆,퐹푖 appropriate choices of 푓푖, 휖푖, 푖, and 푤푃,푖. Take 푎푖 = 푖 . 풫 |풫 | Since in these cases we have many indices, we will be able to apply the weighted Ramsey theorem to appropriately selected subsets of them. The rest of the prelimi- nary discussion for Cases 3 and 4 is based on doing so.
푖 푖+1 For each non-negative integer 푖 let 퐼푖 := [2 , 2 ]. Take Φ(푖) = 푗 : 푛푗 퐼푖 . The { ∈ } Φ(푖) form a dyadic partition of the indices which will eventually determine how the indices are clustered when we apply the weighted Ramsey theorem. Take 퐵′′ := 푖 { ≤ log(푛푚−훿) . } −1/2 ′ −1/2 휏 −1/2 Take 휏 = log(2(log 푚)푛) , so max푖 푛 (log 푚)푛 2 2(log 푚)푛 ⌊ ⌋ 푖 ≤ ≤ ≤ and Φ(푖) is empty for 푖 > 휏.
′ ′ For any pair 푇, 푇 satisfying 푄 푇 = 푄1 and 푄 푇 = 푄2 , take 푔푖 = ∩ { } ∩ { } ∏︀ and ∏︀ . Take to 푆̸∈{푇,푇 } 푔푆,푖 , 표푖 = 푔푇,푖 , 푝푖 = 푔 푔 = 푆̸∈{푇,푇 } 푔푆 , 표 = 푔푇 , 푝 = 푔 퐺푇,푇 ′ 푇 ′,푖 ′ 푇 ′ ′ 2 1/4 (휏−푖)/2 be the set of indices 푗 with 표푝 표푗푝푗 log ((log 푚)2 )/32, where 푖 is such that ≥ 푗 Φ(푖). ∈ is the collection of indices for which is substantially larger than 퐺푇,푇 푗 푔푇 푔 ′ 푇 ′ , where the meaning of “substantially larger" depends on the size of . The 푔푇,푗 푔 푛푗 푇 ′,푗 following lemma states that almost all vertices are contained in some 푉푗 as 푗 varies over the indices of . 퐺푇,푇 ′
Claim 3.8.5. ∑︀ . 푗∈퐺 푛푗 (1 훿1)푛 푇,푇 ′ ≥ −
Proof: Take ′ to be a reordering of so that if then 푉푖 푖≤푡 푉푖 푖≤푡 푖 푗 푔푇,푉 푔푇 ,푉 { } { } ≤ 푖′ ′ 푖′ ≤ . That is, the ′ are in increasing order based on the value of . 푔푇,푉 푔푇 ,푉 푉푖 푔푇,푉 푔푇 ,푉 푗′ ′ 푗′ 푖′ ′ 푖′ Let ′ ′ ′ ′ ′ ′ be defined as before (so, for example, ′ ). 푔푗, 표푗, 푝푗, Φ(푖) , 푛푗, 퐺푇,푇 표푖 = 푔푇,푉 ′ 푖′ When we count, we wish to omit certain intervals that do not satisfy desired properties. Let
퐵 := 푖 휏 : Φ(푖)′ 4훿−1(log1/4 푚)2(휏−푖)/2 . { ≤ | | ≤ 1 }
⋃︀ ′ We will show that a large fraction of the vertices are not contained in 푗∈Φ(푖) 푉 ; 푖∈퐵 ∪ ′ 푗 103 that is, most vertices are not contained in ′ where is an index in ′ for some 푉푗 푗 휑(푖) 푖 퐵. ∈
∑︁ ∑︁ ∑︁ ∑︁ 푛′ 2푖+1(4훿−1(log1/4 푚)2(휏−푖)/2) = 8훿−1(log1/4 푚)2휏/2 2푖/2 푗 ≤ 1 1 푖∈퐵 푗∈Φ(푖)′ 푖≤휏 푖≤휏 8훿−1(log1/4 푚)2휏/24 2휏/2 = 32훿−1(log1/4 푚)2휏 64훿−1(log−1/4 푚)푛 ≤ 1 · 1 ≤ 1
훿1푛/2. ≤
Thus,
∑︁ ∑︁ ′ 푛 훿1푛/2. (3.2) 푗 ≤ 푖∈퐵 푗∈Φ(푖)′
′ For any fixed 푖 휏 such that 푖 퐵, enumerate Φ(푖) as 휑푖,1, 휑푖,2, . . . , 휑푖,|Φ(푖) | ≤ ̸∈ ′ with ′ ′ ′ ′ if . That is, this enumeration is so that the 표휑 푝휑 표휑 푝휑 푗1 푗2 푖,푗1 푖,푗1 ≤ 푖,푗2 푖,푗2 ≤ ′ are listed in increasing order with respect to their ′ ′ values. Take to be 푉휑푖,푗 표푗푝푗 훽푖 ′ ′ ′ 휑푖,(1−훿 /4)|Φ(푖) |. Consider (표 , 푝 ): 푗 Φ(푖) , 푗 훽푖 . By 푖 퐵, this has at least 1 ′ { 푗 푗 ∈ ≥ } ̸∈ (log1/4 푚)2(휏−푖)/2 푀 elements. We get by applying the weighted Ramsey theorem ≥ to this set (with the coloring given by 휒) that:
(︁ )︁ log2 (log1/4 푚)2(휏−푖)/2 표푝 표′ 푝′ . ≥ 훽푖 훽푖 32
For any ′, we have that if then ′ ′ ′ ′ , so the above is at least 푗 Φ(푖) 푗 훽푖 표푗푝푗 표훽푖 푝훽푖 2 ∈1/4 (휏 푖)/2 ≤ ≤ ′ ′ log ((log 푚)2 − ) so ′ . 표푗푝푗 , 푗 퐺 32 ∈ 푇,푇 ′ If 푖 퐵 we obtain ̸∈
∑︁ ∑︁ 훿1 푛′ 푛′ Φ(푖)′ 2푖+1 푗 ≤ 푗 ≤ 4 | | 푗∈Φ(푖)′:푗̸∈퐺′ 푗∈Φ(푖)′:푗>훽푖 푇,푇 ′ ′ 푖 훿1 ∑︁ = Φ(푖) 2 훿1/2 푛푗 | | ≤ 2 푗∈Φ(푖)′
104 Therefore,
∑︁ ′ ∑︁ ′ 푛 (1 훿1/2) 푛 . 푗 ≥ − 푗 푗∈Φ(푖)′∩퐺′ 푗∈Φ(푖)′ 푇,푇 ′ Thus,
∑︁ ′ ∑︁ ∑︁ ′ ∑︁ ∑︁ ′ 푛 푛 (1 훿1/2) 푛 푗 ≥ 푗 ≥ − 푗 푗∈퐺′ 푖̸∈퐵 푗∈Φ(푖)′∩퐺′ 푖̸∈퐵 푗∈Φ(푖)′ 푇,푇 ′ 푇,푇 ′ 2 (1 훿1/2) 푛 (1 훿1)푛, ≥ − ≥ −
where the third inequality follows from (3.2).
As ∑︀ ′ ∑︀ , this completes the proof of the claim. 푗∈퐺 푛푗 = 푗∈퐺 푛푗 푇,푇 ′ 푇,푇 ′ 2
−1/2 ∑︀ ∑︀ ′ (︀푅)︀ Case 3: 훼1 log 푚 and 푛푗 푛/2. Fix any pair 푇, 푇 with ≤ 푖∈퐵′′ 푗∈Φ(푖) ≤ ∈ 푠 ′ 푇 푄 = 푄1 and 푇 푄 = 푄2. The idea behind this case will be to choose some subset ∩ ∩ 퐺푎 of the indices so that, as 푗 varies over 퐺푎, the value of 푎푗 does not change, to intersect this set with , and to use this to show that ∏︀ is large. In this case, 퐺푇,푇 ′ 푆 푔푆
as in Cases 1 and 2, we will simply choose some 푗 appropriately and take = 푗 and 풫 풫 푤푃 = 푤푃,푗. [︀(︀푟)︀]︀ Note that 푠푖 , so by the pigeonhole principle there must be some value 푠 ∈ 2 such that −1 −1 ∑︁ ∑︁ (︂푟)︂ ∑︁ ∑︁ (︂푟)︂ 푛푗 푛푗 푛/2, ≥ 2 ≥ 2 푖̸∈퐵′′ 푗∈Φ(푖):푠푖=푠 푖̸∈퐵′′ 푗∈Φ(푖)
where the last inequality follows by the assumptions for Case 3. Take 퐺푎 to be the
set of indices 푗 with 푎푗 = 푎 and, taking 푖 to be the index with 푗 Φ(푖), 푖 is not in ∈ −1 ′′ ∑︀ (︀푟)︀ log(1/훼푖) 퐵 ; by the above, 푛푗 푛/2. Then take 휖 = min푖∈퐺 휖푖 + . Note 푗∈퐺푎 ≥ 2 푎 log 푚 that this is the same as taking log(1/훼푖) where is an index in minimizing 휖 = 휖푖 + log 푚 푖 퐺푎
(︂ (︁ )︁ )︂ 퐶 휖 + log(1/훼푖) 푖 log 푚 2 푎푖푑 푥푖 := max (log 푚) , (푐 log 푚) ,
as all the 푎푖 are equal to 푎.
Take, with 푖 as above, = 푖 and 푤푃 = 푤푃,푖. Then, as in Cases 1 and 2, 풫 풫 105 properties (1) and (2) hold. Furthermore, as in Cases 1 and 2, taking
(︃ )︃ log (︀(푐 log2 푚)푎푑)︀ 휖′ = max 휖, , 퐶 log log 푚
we have for 푖 퐺푎 that 푓푖 푓휖 (푛푖), and taking 푓 = 푓휖 (푛), properties (4) and (5) ∈ ≥ ′ ′ hold. We need only check that property (3) holds.
Then take 퐺 = 퐺푎 퐺푇,푇 . ∩ ′ We have
(︃ −1 )︃ (︃ −1 )︃ ∑︁ ∑︁ ∑︁ (︂푟)︂ (︂푟)︂ 푛푗 푛 푛푗 푛푗 /2 훿1 푛 /4 푛. (3.3) ≥ − − ≥ 2 − ≥ 2 푗∈퐺 푗̸∈퐺푎 푗̸∈퐺 푇,푇 ′
Take ∏︀ and ∏︀ . 푔푖 = 푆̸∈{푇,푇 } 푔푆,푖 , 표푖 = 푔푇,푖 , 푝푖 = 푔 푔 = 푆̸∈{푇,푇 } 푔푆 , 표 = 푔푇 , 푝 = 푔 ′ 푇 ′,푖 ′ 푇 ′ We have:
푟 2 푟 2 ∑︁ 1/( − ) ∑︁ ∑︁ 1/( − ) (푔푗표푝) 푠 2 (푔푗표푝) 푠 2 − ≥ − 푗 푖 푗∈퐺∩Φ(푖) 1/ 푟 2 (︃ 2 1/4 )︃ (푠−2) ∑︁ ∑︁ log ((log 푚)2(휏−푖)/2) − 푔푗표푗푝푗 ≥ 32 푖 푗∈퐺∩Φ(푖)
푟 2 푟 2 ∑︁ ∑︁ 2/( − ) 1/4 (휏−푖)/2 −1/( − ) 푛푗푓푗 log 푠 2 ((log 푚)2 )32 푠 2 ≥ − − 푖 푗∈퐺∩Φ(푖)
푟 2 푟 2 ∑︁ ∑︁ 2/( − ) 1/4 (휏−푖)/2 −1/( − ) 푛푗푓휖 (푛푗) log 푠 2 ((log 푚)2 )32 푠 2 ≥ ′ − − 푖 푗∈퐺∩Φ(푖)
푟 2 ∑︁ ∑︁ 푖 2/( − ) 1/4 (휏−푖)/2 −1 푛푗푓휖 (2 ) log 푠 2 ((log 푚)2 )32 ≥ ′ − 푖 푗∈퐺∩Φ(푖) (︂ )︂ ∑︁ ∑︁ 푟 휏 8 푛푗푓휖 (2 ) ≥ 2 ′ 푖 푗∈퐺∩Φ(푖) (︂ )︂ ∑︁ ∑︁ 푟 −1/2 8 푛푗푓휖 (푛 log 푚) ≥ 2 ′ 푖 푗∈퐺∩Φ(푖) ∑︁ ∑︁ (︂푟)︂ 4 푛푗푓휖 (푛) 푛푓휖 (푛) = 푛푓, ≥ 2 ′ ≥ ′ 푖 푗∈퐺∩Φ(푖)
where the second inequality follows from 퐺 퐺푇,푇 , the sixth inequality follows by ⊆ ′ 106 the third fact about , the eighth inequality follows by the fourth fact about , and 푓휖′ 푓휖′ the ninth inequality follows from (3.3).
Note that here we used only one pair 푇, 푇 ′; we can afford to do this because we gain a large amount due to 퐵′′ being small. In the next case, we will use all of the relevant pairs.
−1/2 ∑︀ ∑︀ Case 4: 훼1 log 푚 and 푛푗 푛/2. In this case there are many ≤ 푖∈퐵′′ 푗∈Φ(푖) ≥ vertices contained in very small parts; this is the case where we will not simply take
to be some 푖. 풫 풫
The idea behind this case is to choose a set 퐺푎 of many indices 푗 with similar ∑︀ values of 푗, 푤푃,푗, and 휖푗. We will be able to take 푤푄 = 푤푄,푗, which is a 풫 푗∈퐺푎 significant improvement over simply taking for some 푗 each 푤푃 = 푤푃,푗. We will
intersect 퐺푎 with some collection of 퐺푇,푇 where each 푇 with 푇 푄 = 푄1 and each ′ ∩ { } ′ ′ ′ 푇 with 푇 푄 = 푄2 will occur exactly once (so we pair up the sets 푇, 푇 with ∩ { } ′ 푇 푄 = 푄1 and 푇 푄 = 푄2 ; if an index is in 퐺푇,푇 , then we have gained a large ∩ { } ∩ { } ′ factor on that index). This allows us to lower bound ∏︀ . 푆 푔푆 We may partition into at most −1 intervals of length at most . Fur- [0, 1] 훿0 + 1 퐽푖 훿0 thermore, we may partition into at most −1 intervals with [1, 푚] 훿0 +1 퐻푖 sup(퐻푖)/ inf(퐻푖) 푚훿0 . ≤ We partition the indices in ⋃︀ Φ(푖) by saying two indices 푗, 푗′ are in the same 푖∈퐵′′ part if and only if 휖푗, 휖푗 are in the same interval 퐽푖, 푗 = 푗 , and for each 푃 푗 = ′ 풫 풫 ′ ∈ 풫 푗 , 푤푃,푗 and 푤푃,푗 are in the same interval 퐻푖. 풫 ′ ′ 푟 −1 푟 +1 Then the total number of possible partitions is at most (2) (2) . There- 2 (훿0 + 1) ⋃︀ fore, there is some part 퐺푎 Φ(푖) where ⊆ 푖∈퐵′′ (︂ )︂ ∑︁ −(푟) −1 −(푟)−1 ∑︁ ∑︁ −(푟)−1 −1 −(푟)−1 푟 2 푛푗 2 2 (훿 +1) 2 푛푗 2 2 (훿 +1) 2 푛 = 2 − 훿1푛. ≥ 0 ≥ 0 푠 1 푗∈퐺 푎 푖∈퐵′′ 푗∈Φ(푖) −
∑︀ Then take 휖0 = max푖∈퐺 휖푖. Take 푤푄 = 푤푄,푖 and for 푃 = 푄 take 푤푃 = 푎 푖∈퐺푎 ̸ min푖∈퐺 푤푃,푖. Take = 푖 푄 for any 푖 퐺푎. Take 푎 = and note 푎 푎푖 + 1 푎 풫 풫 ∪ { } ∈ |풫| ≤ for any 푖 퐺푎. We check that property (1) holds. For each 푆 with 푄 푆, ∈ ⊆ 107 ∑︁ ∑︁ ∏︁ ∑︁ ∏︁ ∏︁ ∏︁ 푔푆 푔푆,푖 푤푃,푖 푤푄,푖 min 푤푃,푖 = 푤푄 푤푃 = 푤푃 . ≥ ≥ ≥ 푖∈퐺푎 푖 푖 푃 ⊆푆 푖∈퐺푎 푃 ⊆푆,푃 ̸=푄 푃 ⊆푆,푃 ̸=푄 푃 ⊆푆
For each 푆 with 푄 푆, fixing any 푖 퐺푎, ̸⊆ ∈
∏︁ ∏︁ ∏︁ 푔 푔 푤푃,푖 min 푤푃,푖 = 푤푃 . 푆 ≥ 푆,푖 ≥ ≥ 푖∈퐺 푃 ⊆푆 푃 ⊆푆 푃 ⊆푆
Now, we choose (︀푟)︀ and check that property (2) holds: 휖 = 2 훿0 + 휖0
∏︁ ∑︁ ∏︁ ∑︁ ∏︁ −훿0 푤푃 = 푤푄,푖 푤푃 푤푄,푖 (푚 푤푃,푖) ≥ 푃 푖∈퐺푎 푃 ̸=푄 푖∈퐺푎 푃 ̸=푄
푟 푟 −(( )−1)훿0 ∑︁ ∏︁ −(( )−1)훿0−휖0 ∑︁ = 푚 2 푤푃,푖 푚 2 푛푖 ≥ 푖∈퐺푎 푃 푖∈퐺푎 (︂ )︂ 푟 푟 2 푟 −(( )−1)훿0−휖0 −( )훿0−휖0 푚 2 2 − 훿1푛 푚 2 푛, ≥ 푠 1 ≥ −
′ 훿0 where the first inequality is valid since for 푗, 푗 퐺푠 we have 푤푃,푗/푤푃,푗 푚 , the ∈ ′ ≤ second inequality follows by choice of , and the last inequality uses 휖0 = max푗∈퐺푎 휖푗
훿0 (︀푟−2)︀ 푚 2 훿1 which follows from 푚 푚0 (푚0 is defined in Equation 3.1) and the ≥ 푠−1 ≥ choices of 훿0 and 훿1.
(︀푅)︀ (︀푅)︀ Fix a bijection 휋 between 푆 : 푆 푄 = 푄1 and 푆 : 푆 푄 = 푄2 { ∈ 푠 ∩ { }} { ∈ 푠 ∩ { }} (one such bijection takes any 푆 in the first set and removes 푄1 and adds 푄2.) Take (︀푅)︀ 퐺 to be the intersection of 퐺푎 and all sets of the form 퐺푆,휋(푆) where 푆 satisfies ∈ 푠 (︀푟−2)︀ 푆 푄 = 푄1 . There are pairs 푆, 휋(푆), so by Claim 3.2 we have ∩ { } 푠−1
∑︁ ∑︁ ∑︁ ∑︁ (︂ (︂푟 2)︂ (︂푟 2)︂ )︂ 푛푗 푛푗 푛푗 2 − 훿1 − 훿1 푛 ≥ − ≥ 푠 1 − 푠 1 푗∈퐺 푗∈퐺푎 푆:푆∩푄={푄1} 푗̸∈퐺푆,휋(푆) − −
(︂푟 2)︂ − 훿1푛 훿1푛. ≥ 푠 1 ≥ −
Note that, if 푗 퐺, then for any 푆 with 푆 푄 = 푄1 , since 퐺 퐺푆,휋(푆), we ∈ ∩ { } ⊆ 108 have 푔 푔 푔 푔 log2((log1/4 푚)2(휏−푖)/2) 푔 푔 log2(2(휏−푖)/2) where 푖 is 푆 휋(푆) ≥ 푆,푗 휋(푆),푗 ≥ 푆,푗 휋(푆),푗 such that 푗 Φ(푖). However, if 푗 퐺 then 푖 퐵′′, so ∈ ∈ ∈
(︂2휏 )︂1/2 2(휏−푖)/2 = 2푖 (︃ )︃1/2 푛 log−1/2 푚 푚훿/4. ≥ 푛푚−훿 ≥
Therefore, log(2(휏−푖)/2) 훿(log 푚)/4. ≥ This gives:
1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − 푔푆,푗 푔푆 푗 푆:푄⊆푆 푆:푄̸⊆푆 1/ 푟 2 (︃ )︃ (푠−2) ∑︁ ∏︁ ∏︁ − 푔 푔 ≥ 푆,푗 푆 푗∈퐺 푆:푄⊆푆 푆:푄̸⊆푆 1/ 푟 2 ⎛ ⎞ (푠−2) ∑︁ ∏︁ ∏︁ − ⎝ 푔푆,푗 푔푆 ⎠ ≥ 푗∈퐺 푆:|푄∩푆|̸=1 푆:|푄∩푆|=1 1/ 푟 2 ⎛ ⎞ (푠−2) ∑︁ ∏︁ ∏︁ − = ⎝ 푔푆,푗 푔푆 푔휋(푆) ⎠
푗∈퐺 푆:|푄∩푆|̸=1 푆:푄∩푆={푄1} 1/ 푟 2 ⎛ ⎞ (푠−2) 2 − ∑︁ ∏︁ ∏︁ 훿 2 ⎝ 푔푆,푗 (log 푚)푔푆,푗 푔 ⎠ ≥ 16 휋(푆),푗 푗∈퐺 푆:|푄∩푆|̸=1 푆:푄∩푆={푄1} 1/ 푟 2 (︃ )︃ (푠−2) ∑︁ 2 푟 2 ∏︁ − (푠−1) = (훿(log 푚)/16) − 푔푆,푗 푗∈퐺 푆
∑︁ 2 푟 2 / 푟 2 (푠−1) (푠−2) 푛푗푓푗 (훿(log 푚)/4) − − ≥ 푗∈퐺
∑︁ 2푑 = 푛푗푓푗 (훿(log 푚)/4) . 푗∈퐺
Take 푓 ′ = (log 푚)퐶휖, 푓 ′′ = (푐 log2 푚)푎푑 and 푓 = max(푓 ′, 푓 ′′). Note that 푓 푓 ′ ≥ guarantees that property (4) holds and 푓 푓 ′′ guarantees that property (5) holds, so ≥ we need only check that property (3) holds. There will be two cases, that in which
109 푓 = 푓 ′ and that in which 푓 = 푓 ′′.
′ 퐶(휖0−훿0) If 푓 = 푓 , for each 푗 퐺푎 we have 휖푗 휖0 훿0, so 푓푗 (log 푚) . Then we ∈ ≥ − ≥ get:
∑︁ 2푑 ∑︁ 퐶(휖0−훿0) 2푑 푛푗푓푗 (훿(log 푚)/4) 푛푗(log 푚) (훿(log 푚)/4) ≥ 푗∈퐺 푗∈퐺
푟 (︁ )︁2푑 ∑︁ 퐶(휖0+( )훿0) 1/2 푛푗(log 푚) 2 훿(log 푚)/4 ≥ 푗∈퐺
푟 (︁ )︁2푑 퐶(휖0+( )훿0) 1/2 훿1푛(log 푚) 2 훿(log 푚)/4 ≥ 퐶 휖 + 푟 훿 퐶휖 ′ 푛(log 푚) ( 0 (2) 0) = 푛(log 푚) = 푛푓 = 푛푓. ≥
′′ 2 (푎−1)푑 Otherwise, 푓 = 푓 . For each 푗 퐺푎 we have 푓푗 (푐 log 푚) , as 푎 푎푗 + 1. ∈ ≥ ≤ This gives:
∑︁ 2푑 ∑︁ 2 (푎−1)푑 2푑 푛푗푓푗 (훿(log 푚)/4) 푛푗(푐 log 푚) (훿(log 푚)/4) ≥ 푗∈퐺 푗∈퐺
2푑 −푑 2 푎푑 ∑︁ 2푑 −푑 2 푎푑 = (훿/4) 푐 (푐 log 푚) 푛푗 (훿/4) 푐 (푐 log 푚) 훿1푛 ≥ 푗∈퐺 푛(푐 log2 푚)푎푑 = 푛푓 ′′. ≥
2
(︀푟)︀ Take 푎0 to be if 푠 < 푟 1, 푟/2 if 푠 = 푟 1 and 푟 is even, and (푟 + 3)/2 if 2 − − 2 푎0푑 푠 = 푟 1 and 푟 is odd. Take 푓0 = (푐 log 푚) . The following theorem states that − either some is large or their product is large. 푔푆
푟 2 ∏︀ ( − ) Theorem 3.8.6. If 푚 푚0, either 푔 (푚푓0) 푠 2 or there is some 푆 푅 of ≥ 푆 푆 ≥ − ⊆ (푠)/(푟) size 푠 with 푔 (푚푓0) 2 2 . 푆 ≥
Proof: Choose 푓, 휖, , 푤푃 as given by the previous theorem, then we need only show 풫 −1 (︁ (︀푟)︀2)︁ (︀푟)︀ 퐶휖 푓 푓0. If 휖 16푟 then we have 퐶휖 2 푑 2푎0푑 so 푓 (log 푚) ≥ ≥ 2 ≥ 2 ≥ ≥ ≥ 2푎0푑 (log 푚) 푓0. ≥ 110 (︁ )︁−1 Otherwise, (︀푟)︀2 . Define a weighted graph on vertex set where an 휖 < 16푟 2 푅 (︀푅)︀ edge 푒 has weight log 푤푒. Note that this graph has non-negative edge weights ∈ 2 and if an edge is not in , then it has weight 0. 풫 By Lemma 3.8.1, either this graph has at least 푎0 edges or there is some set 푆 on 푠 vertices with ⎛ ⎞ (︃ (︂ )︂2)︃−1 (︀푠)︀ ∑︁ 푟 2 ∑︁ log 푤푃 ⎝1 + 4푟 ⎠ log 푤푃 . ≥ 2 (︀푟)︀ 푃 ⊆푆 2 푃
2 푎푝0푑 If the graph has at least 푎0 edges, then 푎0 so 푓 (푐 log 푚) , as desired. |풫| ≥ ≥ Otherwise, there is some 푆 so that
⎛ ⎞ (︃ (︂ )︂2)︃−1 (︀푠)︀ ∑︁ 푟 2 ∑︁ log 푤푃 ⎝1 + 4푟 ⎠ log 푤푃 . ≥ 2 (︀푟)︀ 푃 ⊆푆 2 푃
Then we have
푠 (︂ (︁ 2)︁ 1)︂ (︂ )︂ 푠 푟 − (2) (︁ 2)︁ 1 (2) 1+ 4푟( ) (1−휖) 1+ 4푟 푟 − 2 푟 (2) 푟 ∏︁ ∏︁ (2) (2) 푤푃 푤 푚 ≥ 푃 ≥ 푃 ⊆푆 푃 (︂ )︂(︂ )︂ 푠 (︂ )︂ 푠 (︁ 2)︁ 1 (︁ 2)︁ 1 (2) (︁ 2)︁ 1 (2) 1− 16푟 푟 − 1+ 4푟 푟 − 1+ 8푟 푟 − (2) (2) 푟 (2) 푟 푠 푟 (2) (2) ( )/( ) 푚 푚 (푚푓0) 2 2 , ≥ ≥ ≥
where the second to last inequality follows from (1 + 푏)(1 푏/4) 1 + 푏/2 for any − ≥ 푏 [0, 1]. 2 ∈ The previous theorem easily implies a general lower bound for the largest value of . 푔푆
(푠)/(푟) Theorem 3.8.7. If 푚 푚0, there is some 푆 푅 of size 푠 with 푔 (푚푓0) 2 2 . ≥ ⊆ 푆 ≥
푟 2 ∏︀ ( − ) Proof: By the previous theorem, either such an 푆 exists or 푔 (푚푓0) 푠 2 . 푆⊆푅 푆 ≥ − In this latter case, since this is the product of (︀푟)︀ numbers, there must be some 푠 푆 푟 2 푟 푠 푟 ( − )/( ) ( )/( ) with 푔 (푚푓0) 푠 2 푠 = (푚푓0) 2 2 . 2 푆 ≥ − 111 Before we proceed, note that:
(︂푟 2)︂ (︂푟 2)︂ 푟 푠 푑 = − / − = − . 푠 1 푠 2 푠 1 − − − We now simply rewrite the statement of the previous theorem in more familiar notation.
Theorem 3.8.8. Every Gallai coloring of a complete graph on 푚 vertices has a vertex (︁ 푠 / 푟 푐 )︁ subset using at most 푠 colors of order Ω 푚(2) (2) log 푟,푠 푚 , where
⎧ if ⎪ 푠(푟 푠) 푠 < 푟 1, ⎨⎪ − − 푐푟,푠 = 1 if 푠 = 푟 1 and r is even, ⎪ − ⎩⎪ (푟 + 3)/푟 if 푠 = 푟 1 and r is odd. −
Proof: If 푠 < 푟 1 and 푚 푚0, by Theorem 3.8.7 the coloring has a subchromatic − ≥ 푠 푠 / 푟 (︀ 2 )︀( )푑 (︀푠)︀ 푟−푠 set of order at least 푚(2) (2) 푐 log 푚 2 . As 2 푑 = 푠(푠 1) = 푠(푟 푠), this 2 − 푠−1 − gives the desired bound in this case.
If 푠 = 푟 1, 푟 is even, and 푚 푚0, by Theorem 3.8.7, the coloring has a − ≥ 푠 / 푟 (︀ 2 )︀(푟/2)푑 subchromatic set of order at least 푚(2) (2) 푐 log 푚 . As
(︂푠)︂(︂푟)︂−1 푠(푠 1) 푟 푠 (푟 1)(푟 2) 1 2(푟/2) 푑 = 푟 − − = 푟 − − = 1, 2 2 푟(푟 1) 푠 1 푟(푟 1) 푟 2 − − − − this gives the desired bound in this case.
If 푠 = 푟 1, 푟 is odd, and 푚 푚0, by Theorem 3.8.7, the coloring has a − ≥ 푠 / 푟 (︀ 2 )︀((푟+3)/2)푑 subchromatic set of order at least 푚(2) (2) 푐 log 푚 . As
(︂푠)︂(︂푟)︂−1 푠(푠 1) 푟 푠 (푟 1)(푟 2) 1 2((푟+3)/2) 푑 = (푟+3) − − = (푟+3) − − = (푟+3)/푟, 2 2 푟(푟 1) 푠 1 푟(푟 1) 푟 2 − − − − this gives the desired bound in this case. 2
3.9 Proof of Lemma 3.4.1
√ Lemma 3.9.1. For all 푘, ℓ 1 we have (︀푘+ℓ)︀ 22 푘ℓ ≥ ℓ ≤ 112 푘+ℓ 푘 ℓ 푘+ℓ Proof: We first observe that (︀푘+ℓ)︀ (푘+ℓ) , or equivalently that 푘 ℓ (푘+ℓ) . ℓ ≤ 푘푘ℓℓ 푘! ℓ! ≤ (푘+ℓ)! To see this, note that the function 푓(푛) = 푛 log 푛 log(푛!) is super-additive (over the − positive integers), that is it satisfies 푓(푛1)+푓(푛2) 푓(푛1+푛2). This follows from using ≤ Stirling’s formula, which is equivalent to 푓(푛) = 푛 log 푒 1 log 푛 1 log 2휋 +표(1), and − 2 − 2 known estimates on the 표(1) term as well as the sub-additive nature of the logarithm
function. Taking 푛1 = 푘, 푛2 = ℓ gives
(푘 log 푘 log(푘!)) + (ℓ log ℓ log(ℓ!)) (푘 + ℓ) log(푘 + ℓ) log((푘 + ℓ)!). − − ≤ −
Exponentiating both sides gives 푘푘 ℓℓ (푘+ℓ)푘+ℓ , as desired. 푘! ℓ! ≤ (푘+ℓ)!
푘+ℓ √ Therefore, it now suffices to show that (푘+ℓ) 22 푘ℓ. This is equivalent to 푘푘ℓℓ ≤ showing that, for all 푘, ℓ 1, ≥ √ √ (︂푘 + ℓ)︂푘/ 푘ℓ (︂푘 + ℓ)︂ℓ/ 푘ℓ 4. 푘 ℓ ≤
Taking 푥 = 푘/ℓ, note that 푥 > 0 and that we may rewrite the left hand side of the above as √ √ (︀ −1)︀ 푥 푥 1 푔(푥) := 1 + 푥 (1 + 푥) − .
Note that 푔(푥) = 푔(1/푥), so it now suffices to show that, for all 푥 1, we have ≥ 푔(푥) 4. Note that 푔′(1) = 0. We claim that, for 푥 [1, 1.5] we have 푔′′(푥) < 0 and ≤ ∈ for 푥 [1.5, 5] we have 푔′(푥) < 0. These claims may be verified numerically, as the ∈ inequalities are strict and the relevant functions are uniformly continuous over their
respective compact sets. This gives that, in the interval [1, 5], 푔(푥) is maximized at √ √ 1 1 푔(1) = 4. Then, for 푥 > 5, since 1+푥−1 푒푥− we have (1+푥−1) 푥 푒 푥− < 1.6 and, √ ≤ √ ≤ 1 in this range, 1+푥 2.5 푥 so we have (1+푥) 푥− 2.5, giving that 푔(푥) 2.5 1.6 = 4 ≤ ≤ ≤ · for 푥 > 5. 2
113 3.10 Proof of Lemma 3.5.3
For convenience, we restate both the definition of 푓 and the statement of the lemma here:
⎧ 2 if 4/9 ⎪ 푐 log (퐶푛) 0 < 푛 푚 ⎨⎪ ≤ 푓(푛) := 푐2 log2(푚4/9) log2(퐶푛푚−4/9) if 푚4/9 < 푛 푚8/9 , ⎪ ≤ ⎩⎪ 푐3 log4(푚4/9) log2(퐶푛푚−8/9) if 푚8/9 < 푛 푚, ≤ where 퐷 = 22048, 퐶 = 2퐷8 , and 푐 = log−2(퐶2) = 퐷−16/4.
Lemma 3.10.1. If 푚 퐶, then the following statements hold about 푓 for any integer ≥ 푛 with 1 < 푛 푚. ≤
1. For any 훼 [︀ 1 , 1]︀, we have 푓(훼푛) 훼푓(푛). ∈ 푛 ≥
[︀ 1 ]︀ ∑︀ 2. For any 훼1, 훼2, 훼3 , 1 such that 훼푖 = 1 we have, taking 푛푖 = 훼푖푛, ∈ 푛 푖
∑︁ 8 푛푓(푛) 푛 푓(푛 ) 푛푓(푛). 푖 푖 log 퐶 − 푖 ≤
7/18 푗 푖 2 푗 푖+ 8 푗 3. For 푖 0 and 푚 2 1, we have 푓(2 ) log (퐷2 ) 512푓(2 7 ). ≥ ≥ ≥ ≥
4. For 1 휏 푛 퐷3휏, we have 푓(휏) 푓(푛)/2. ≤ ≤ ≤ ≥
5. For any 훼 [︀ 1 , 1 ]︀, we have 푓(훼푛) 16훼푓(푛). ∈ 푛 32 ≥
4/9 8/9 Proof: Observe that 푓(푛) has two points of discontinuity: 푝0 = 푚 and 푝1 = 푚 .
Recall that the three intervals of 푓 are (0, 푝0], (푝0, 푝1], (푝1, 푚]; name these 퐼0, 퐼1, 퐼2, respectively.
If 푡 is either 푝0 or 푝1, then we have 푓+(푡) := lim + 푓(푛) lim 푓(푛) =: 푓−(푡). 푛→푡 ≤ 푛→푡− Observe further that, if 푛 is in some interval 퐼 of 푓, then for any 푡 퐼 we have ∈ 푓(푡) = 훾 log2(훿푡) for constants 훾, 훿 with 훿푡 퐶. ≥ Proof of Fact 1: We first argue that it is sufficient to showFact 1 in the case that both 푛 and 훼푛 are in the same interval of 푓. Intuitively, the points of discontinuity
114 only help us. If 푛 is in 퐼1, 훼푛 is in 퐼0, and we have shown that Fact 1 holds when 푛 and 훼푛 are in the same interval, then
훼푛 훼푛 훼푛 푝0 푓(훼푛) 푓+(푝0) 푓−(푝0) 푓(푛) = 훼푓(푛). ≥ 푝0 ≥ 푝0 ≥ 푝0 푛
The case where 푛 is in 퐼2 and 훼푛 is in 퐼1 and the case where 푛 is in 퐼2 and 훼푛 is in
퐼0 hold by essentially the same argument.
We next show Fact 1 in the case that both 푛 and 훼푛 are in the same interval 퐼 of 푓. We have, choosing 훾 and 훿 to be such that 푓(푡) = 훾 log2(훿푡) on 퐼, that
푓(훼푛) = 훾 log2(훼훿푛),
훼푓(푛) = 훾훼 log2(훿푛) = 훾(√훼 log(훿푛))2.
Thus, it is sufficient to show that log(훼훿푛) √훼 log(훿푛) 0. Note that equality − ≥ holds if 훼 = 1. We consider the first derivative with respect to 훼; we will show that it is negative for 훼 4 . The first derivative is: ≥ ln2(훿푛)
1 1 2 √훼 ln(훿푛) log(훿푛) = − . 훼 ln(2) − 2√훼 훼 ln(4)
Note that the above is negative if 2 √훼 ln(훿푛) 0, which is equivalent to − ≤ 훼 4 . ≥ ln2(훿푛) [︁ ]︁ Therefore, for 훼 4 , 1 , assuming 훼푛 퐼, we have 푓(훼푛) 훼푓(푛). If ∈ ln2(훿푛) ∈ ≥ 훼 < 4 with 훼푛 퐼 then, ln2(훿푛) ∈
4 4 훼푓(푛) < 훾 log2(훿푛) = 훾 log2 퐶훾 훾 log2(훿훼푛) = 푓(훼푛), ln2(훿푛) ln2(2) ≤ ≤ where the first inequality follows by the assumed upper bound on 훼 and the last one by 훼푛 in 퐼 (and so 훿훼푛 퐶). ≥
Proof of Fact 2: Let 훾, 훿 be such that for 푡 in the interval 퐼푗 containing 푛 we have 2 −1 푓(푡) = 훾 log (훿푡). We define a new function 푓2 whose domain is [푛 log 퐶, 푛]. For any
푡 in the domain of 푓2 that is in 퐼푗, we define 푓2(푡) = 푓(푡), and for any 푡 in the domain of
115 2 −1 푓2 that is not in 퐼푗, we define 푓2(푡) = 훾 log 퐶. If there is some point 푡 in [푛 log 퐶, 푛] −1 −1 that is not in 퐼푗, then 푡 must be in 퐼푗−1, as 푡 푛 log 퐶 푝푗−1 log 퐶 > 푝푗−2. Then ≥ ≥ note that we have chosen, for 푡 not in 퐼푗, 푓2(푡) = 푓+(푝푗−1). Therefore, 푓2 is continuous.
Also, 푡푓2(푡) is convex (this is easy to see by looking at the first derivative). The main
idea behind the proof will be to replace 푓 by 푓2 and then apply convexity to get the bounds.
We claim 푓(푡) 푓2(푡) for all 푡 in the domain of 푓2. If 푡 is in 퐼푗, then 푓2(푡) = 푓(푡). ≥ −1 −1 Otherwise, 푡 is in 퐼푗−1. For any 푡 [푛 log 퐶, 푛], note that 훿푡 훿푛 log 퐶 ∈ ≥ ≥ 퐶 log−1 퐶. Therefore,
훾 훾 푓(푡) = log2(훿푡푚4/9) log2 (︀푚4/9퐶 log−1 퐶)︀ 푐 log2(푚4/9) ≥ 푐 log2(푚4/9)
훾 2 훾 log 퐶 = 푓2(푡), ≥ 푐 ≥
−1 where the first equality follows by 푡 퐼푗−1 and the first inequality by 훿푡 log 퐶. ∈ ≥ −1 Take 푆 = 푖 : 훼푖 log 퐶 . For 푖 푆 we have that 훼푖푛 is in the domain of 푓2. { ≥ } ∈ ∑︀ ∑︀ ∑︀ −1 Take 휅 such that 훼푖 = 휅. Since 훼푖 = 1, 휅 = 1 훼푖 1 3 log 퐶. 푖∈푆 푖 − 푖̸∈푆 ≥ − Hence,
∑︁ ∑︁ ∑︁ ∑︁ 휅 (︂ 휅 )︂ 푛푖푓(푛푖) 푛푖푓(푛푖) 푛푖푓2(푛푖) 푛푓2 푛 ≥ ≥ ≥ 푆 푆 푖 푖∈푆 푖∈푆 푖∈푆 | | | | (︂ )︂ (︂ −1 )︂ 휅 1 3 log 퐶 2 = 휅푛푓2 푛 휅푛푓2 − 푛 휅푛푓2(푛/4) 휅훾푛 log (훿푛/4), 푆 ≥ 3 ≥ ≥ | | where the third inequality follows Jensen’s inequality applied to the convex function
푡푓2(푡) and the fourth inequality holds since 푓2 is an increasing function.
This gives
∑︁ 2 2 2 2 휅푛푓(푛) 푛푖푓(푛푖) 휅훾푛 log (훿푛) 휅훾푛 log (훿푛/4) = 휅훾푛(log (훿푛) log (훿푛/4)). − ≤ − − 푖∈푆
116 We now consider
log2(훿푛) log2(훿푛/4) = (log(훿푛) + log(훿푛/4))(log(훿푛) log(훿푛/4)) − −
2 log(훿푛) log 4 = 4 log(훿푛). ≤
Noting that log(훿푛) log 퐶, we get ≥ 4 4 휅훾푛(4 log(훿푛)) 훾푛 log2(훿푛) = 훾푛푓(푛). ≤ log 퐶 log 퐶
Thus,
∑︁ ∑︁ 푛푓(푛) 푛푖푓(푛푖) (1 휅)푛푓(푛) + 휅푛푓(푛) 푛푖푓(푛푖) − ≤ − − 푖 푖∈푆 1 4 8 푛푓(푛) + 푛푓(푛) 푛푓(푛). ≤ 2 log 퐶 log 퐶 ≤ log 퐶
Proof of Fact 3: Take 훾, 훿 such that for 푡 in the interval of 푓 containing 2푖 we have
2 . Take ′ 8 . If 푖 and 푖+푗′ are in the same intervals of , then we 푓(푡) = 훾 log (훿푡) 푗 = 7 푗 2 2 푓 get
푓(2푖) log2(퐷2푗) = 훾 log2(훿2푖) log2(퐷2푗) = 훾 log2(2푖+log 훿) log2(2푗+log 퐷)
2 (푗+log 퐷)(푖+log 훿) 2 푖+푗 푖+ 8 푗 = 훾 log (2 ) 512훾 log (훿2 ′ ) = 512푓(2 7 ), ≥ where the last inequality follows from 푖+log 훿 log 퐶 log 퐷 and 푗 +log 퐷 log 퐷. ≥ ≥ ≥ Therefore,
log 퐷 log 퐷 (푗 +log 퐷)(푖+log 훿) (푗 +푖+log 훿) (2푗 +푖+log 훿) 512(푗′ +푖+log 훿). ≥ 2 ≥ 4 ≥
If 2푖 and 2푖+푗′ are in different intervals of 푓, then they are in adjacent intervals
117 since 2푗′ 푚4/9. Therefore, ≤
푖+ 8 푗 푖+푗 2 4/9 2 −4/9 푖+푗 2 푖 2 푗 푓(2 7 ) = 푓(2 ′ ) = 푐훾 log (푚 ) log (훿푚 2 ′ ) 푐훾 log (훿2 ) log (2 ′ ) ≤ 1 1 2푐 (︀훾 log2(훿2푖))︀ log2(2푗) 푓(2푖) log2(2푗) 푓(2푖) log2(퐷2푗), ≤ ≤ 512 ≤ 512 where the first inequality follows by the fact thatif 푎0 푎1 푏1 푏0 2 and ≥ ≥ ≥ ≥ if 푎0푏0 = 푎1푏1 then (log 푎0)(log 푏0) (log 푎1)(log 푏1). To see this last fact about ≤ logarithms, one may take the logarithm of both sides and apply the concavity of the logarithm function.
Proof of Fact 4: Choose 훾, 훿 such that for 푡 in the same interval 퐼푗 as 휏 we have 2 푓(푡) = 훾 log (훿푡). If 푛 and 휏 are in different intervals of 푓, then 푛 must be in 퐼푗+1 as 푛/휏 퐷3 < 푚4/9. Furthermore, for 푛 and 휏 to be in different intervals, we must ≤ have 퐷3훿휏 퐶푚4/9 and so 훿휏 푚4/9. This gives: ≥ ≥
푓(휏) = 훾 log2(훿휏) 훾 log2(푚4/9) 푐 log2(푚4/9)훾 log2(퐶퐷3) ≥ ≥ 푐 log2(푚4/9)훾 log2(훿푚−4/9푛) = 푓(푛), ≥ where the second inequality follows by 퐶퐷3 퐶2 and 푐 = log−2(퐶2). ≤ Otherwise, 휏 and 푛 are in the same interval. Then
푓(푛) = 훾 log2(훿푛) 훾 log2(훿퐷3휏) 훾 log2((훿휏)4/3) 2훾 log2(훿휏) = 2푓(휏), ≤ ≤ ≤ where the second inequality follows by 훿휏 퐶 and 퐶1/3 퐷. ≥ ≥ Proof of Fact 5: Let 훼 1 be given. Consider: ≤ 32 1 1 16훼푓(푛) = (32훼푓(푛)) 푓(32훼푛) 푓(훼푛), 2 ≤ 2 ≤ where the first inequality is by the first fact about 푓 and the second inequality is by the fourth fact about 푓.
2
118 3.11 Proof of Lemma 3.8.1
We argue that if a weighted graph on 푟 vertices deviates in structure from the complete graph with edges of equal weight and if 푠 < 푟 1, then there is some set of vertices − 푆 of size 푠 so that the sum of the weights of the edges contained in 푆 is substantially larger than average.
(︀푅)︀ ∑︀ Lemma 3.11.1. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . ∈ 2 ≥ 푃 (︀푟)︀−1 Then if 푠 < 푟 1, if some 푤푃 differs from 푤 by at least 퐹 , then there is some − 2 푆 푅 of size 푠 satisfying ⊆ (︀푠)︀ (︀푠)︀ ∑︁ 2 2 퐹 푤푃 푤 + . ≥ (︀푟)︀ (︀푟)︀2 푃 ⊆푆 2 푟 2
Proof: We will directly handle the case 푠 = 푟 2, from which the other cases will − follow. We are interested in finding an 푆 푅 of size 푟 2 with a large value for the ⊆ − total weight of edges in . For each we give this value a name: ∑︀ . 푆 푆 푍푆 = 푃 ⊆푆 푤푃 (︀푅)︀ Note that 푍푆 is closely related the following: for 푄 , define 푌푄 to be the weight ∈ 2 of edges incident to at least one vertex of : ∑︀ . Then, if we 푄 푌푄 = 푃 ∈ 푅 :푃 ∩푄̸=∅ 푤푃 ( 2 )
take 푆 to be 푅 푄, we have 푌푄 +푍푆 is the total weight of all the edges. Thus, to show ∖ that there is a large 푍푆, it is sufficient to show that there is asmall 푌푄. Towards this (︀푅)︀ end, choose 푄 uniformly at random; we will now compute the variance of 푌푄. ∈ 2 (︀푅)︀ Take, for 푃 , 푋푃 to be 푤푃 if 푃 푄 = and 0 otherwise. Then, taking ∈ 2 ∩ ̸ ∅ ∑︀ , we get 푤 = 푃 푤푃
2푟 3 E [푋푃 ] = Pr[푃 푄 = ]푤푃 = (︀푟−)︀ 푤푃 . ∩ ̸ ∅ 2
By linearity of expectation, we have
∑︁ 2푟 3 ∑︁ 2푟 3 E [푌푄] = E [푋푃 ] = (︀푟−)︀ 푤푃 = (︀푟−)︀ 푤, 푃 ∈ 푅 2 푃 ∈ 푅 2 ( 2 ) ( 2 )
119 and
[︀ 2]︀ ∑︁ ∑︁ E 푌푄 = E [푋푃 푋푃 ′ ] 푃 ∈ 푅 푃 ∈ 푅 ( 2 ) ′ ( 2 ) ∑︁ [︀ 2 ]︀ ∑︁ ∑︁ = E 푋푃 + E [푋푃 푋푃 ′ ] 푅 2 푃 ∈ 푣∈푅 (푃,푃 )∈ 푅 :푣∈푃,푣∈푃 ,푃 ̸=푃 ( 2 ) ′ ( 2 ) ′ ′ ∑︁ + E [푋푃 푋푃 ′ ] , 2 (푃,푃 )∈ 푅 :푃 ∩푃 =∅ ′ ( 2 ) ′
where the last equality follows by partitioning the pairs 푃, 푃 ′ into those which are equal, those which are distinct but intersect in some vertex 푣, and those which are disjoint.
We now look at these terms individually.
[︀ 2 ]︀ 2 2푟 3 2 E 푋푃 = Pr[푃 푄 = ]푤푃 = (︀푟−)︀ 푤푃 . ∩ ̸ ∅ 2
For 푃 = 푣, 푢 , 푃 ′ = 푣, 푢′ distinct and intersecting, the event 푃 푄 = and { } { } ∩ ̸ ∅ 푃 ′ 푄 = can occur if either 푣 푄 or 푄 = 푢, 푢′ ; the first of these has probability ∩ ̸ ∅ ∈ { } 푟−1 and the second has probability 1 , and they are disjoint events. So, if and ′ 푟 푟 푃 푃 (2) (2) intersect in a vertex we get:
푟 E [푋푃 푋푃 ′ ] = (︀푟)︀푤푃 푤푃 ′ . 2
If ′ are disjoint, then for to be non-zero we must have that has an 푃, 푃 푋푃 푋푃 ′ 푄 element from and an element from ′, which occurs with probability 4 , so in this 푃 푃 푟 (2) case: 4 E [푋푃 푋푃 ′ ] = (︀푟)︀푤푃 푤푃 ′ . 2
Therefore, taking for the (weighted) degree to be ∑︀ , 푣 푅 푑(푣) 푃 ∈ 푅 :푣∈푃 푤푃 ∈ ( 2 ) [︀ 2]︀ is equal: E 푌푄
120 ∑︁ 2푟 3 2 ∑︁ ∑︁ 푟 ∑︁ 4 (︀푟−)︀ 푤푃 + (︀푟)︀푤푃 푤푃 ′ + (︀푟)︀푤푃 푤푃 ′ 푅 2 2 2 푅 2 푃 ∈ 푣∈푅 (푃,푃 )∈ 푅 :푣∈푃,푣∈푃 ,푃 ̸=푃 (푃,푃 )∈ :푃 ∩푃 =∅ ( 2 ) ′ ( 2 ) ′ ′ ′ ( 2 ) ′ ⎛ ⎞ 2푟 3 ∑︁ 푟 ∑︁ ∑︁ = − 푤2 + ⎜푑(푣)2 푤2 ⎟ (︀푟)︀ 푃 (︀푟)︀ ⎝ − 푃 ⎠ 2 푃 ∈ 푅 2 푣∈푅 푃 ∈ 푅 :푣∈푃 ( 2 ) ( 2 ) ⎛⎛ ⎞2 ⎞
4 ∑︁ ∑︁ 2 ∑︁ 2 + ⎜⎜ 푤푃 ⎟ 푑(푣) + 푤 ⎟ (︀푟)︀ ⎝⎝ ⎠ − 푃 ⎠ 2 푃 ∈ 푅 푣∈푅 푃 ∈ 푅 ( 2 ) ( 2 ) 2푟 3 ∑︁ 푟 ∑︁ 푟 ∑︁ 4 4 ∑︁ 4 ∑︁ = − 푤2 + 푑(푣)2 2 푤2 + 푤2 푑(푣)2 + 푤2 (︀푟)︀ 푃 (︀푟)︀ − (︀푟)︀ 푃 (︀푟)︀ − (︀푟)︀ (︀푟)︀ 푃 2 푃 ∈ 푅 2 푣∈푅 2 푃 ∈ 푅 2 2 푣∈푅 2 푃 ∈ 푅 ( 2 ) ( 2 ) ( 2 )
4 2 1 ∑︁ 2 푟 4 ∑︁ 2 =(︀푟)︀푤 + (︀푟)︀ 푤푃 + (︀−푟)︀ 푑(푣) 2 2 푃 ∈ 푅 2 푣∈푅 ( 2 )
Note that ∑︀ 2 is minimized subject to the constraint ∑︀ 푣∈푅 푑(푣) 푣∈푅 푑(푣) = 2푤 when the 푑(푣) are pairwise equal by the Cauchy-Schwarz inequality, so ∑︀ 푑(푣)2 푣∈푅 ≥ ∑︀ (︀ 2푤 )︀2, so the above is at least 푣∈푅 푟
2 4 1 ∑︁ 푟 4 ∑︁ (︂2푤)︂ 푤2 + 푤2 + − (︀푟)︀ (︀푟)︀ 푃 (︀푟)︀ 푟 2 2 푃 ∈ 푅 2 푣∈푅 ( 2 ) 2 4 1 ∑︁ 푟 4 (︂2푤)︂ = 푤2 + 푤2 + 푟 − (︀푟)︀ (︀푟)︀ 푃 (︀푟)︀ 푟 2 2 푃 ∈ 푅 2 ( 2 ) 8푟 16 1 ∑︁ = − 푤2 + 푤2 . 푟(︀푟)︀ (︀푟)︀ 푃 2 2 푃 ∈ 푅 ( 2 )
The variance of 푌푄 satisfies:
(︃ )︃2 Var [︀ 2]︀ 2 8푟 16 2 1 ∑︁ 2 2푟 3 (푌푄) = E 푌푄 E [푌푄] − 푤 + 푤푃 − 푤 − ≥ 푟(︀푟)︀ (︀푟)︀ − (︀푟)︀ 2 2 푃 ∈ 푅 2 ( 2 )
121 1 ∑︁ 1 = 푤2 푤2. (︀푟)︀ 푃 − (︀푟)︀2 2 푃 ∈ 푅 2 ( 2 )
Note we may rewrite the variance as:
1 ∑︁ 2 1 2 Var (푌푄) 푤 푤 = Var (푤푄) . ≥ (︀푟)︀ 푃 − (︀푟)︀2 2 푃 ∈ 푅 2 ( 2 )
If some is far from (︀푟)︀ , then the variance will be large. As- 푤푃 푤/ 2 = E [푤푄] sume that for some ′ there is a non-zero real so that (︀푟)︀ . Note 푃 퐹 푤푃 ′ = 푤/ 2 + 퐹 Var Var [︀ (︀푟)︀]︀ and that (︀푟)︀ 2 is a non-negative (푌푄) (푤푄) = E (푤푄 푤/ (푤푄 푤/ ) ≥ − 2 − 2 random variable. If ′ (which occurs with probability (︀푟)︀−1), then this random 푄 = 푃 2 variable has value 2, so its expectation is at least (︀푟)︀−1 2. That is, the variance of 퐹 2 퐹 is at least (︀푟)︀−1 2. 푤푄 2 퐹
Thus, there must be some 푄 so that
⃒ ⃒ ⃒ ⃒ (︂ )︂−1/2 ⃒ 2푟 3 ⃒ 푟 ⃒푌푄 (︀푟−)︀ 푤⃒ 퐹 퐹/푟. ⃒ − 2 ⃒ ≥ 2 ≥
If (2푟−3)푤 , since there are (︀푟)︀ different and the average is (2푟−3)푤 , there 푌푄 푟 퐹/푟 2 푌푄 푟 − (2) ≥ (2) must be some 푄′ so that
2푟 3 퐹 (︂ (︂푟)︂)︂−1 푌푄 − 푤 − 푟 퐹. ′ − (︀푟)︀ ≤ 푟 (︀(︀푟)︀ 1)︀ ≤ − 2 2 2 − The other case is that
2푟 3 (︂ (︂푟)︂)︂−1 푌푄 (︀푟−)︀ 푤 퐹/푟 푟 퐹. − 2 ≤ − ≤ − 2
Therefore, there is some with (2푟−3)푤−퐹/푟 . 푄 푌푄 푟 ≤ (2)
Define 푆 = 푅 푄. We get 푍푆 + 푌푄 = 푤 so 푍푆 = 푤 푌푄. By the above, there is ∖ − some 푆 with (︀푟−2)︀ (2푟 3)푤 퐹/푟 2 푤 + 퐹/푟 푍푆 푤 − (︀푟)︀ − = (︀푟)︀ . ≥ − 2 2 122 푠 푟 2 푤+퐹/푟 Taking as above, choosing a random ′ (︀푆)︀, we get that (2) ( −2 ) . 푆 푆 푠 E [푍푆′ ] 푟 2 푟 ∈ ≥ ( −2 ) (2) Therefore, there must be some 푆′ (︀푆)︀ with ∈ 푠
(︀푠)︀ (︀푠)︀ (︀푠)︀ (︀푠)︀ 2 2 퐹 2 2 퐹 푍푆 푤 + 푤 + . ′ ≥ (︀푟)︀ 푟(︀푟)︀(︀푟−2)︀ ≥ (︀푟)︀ (︀푟)︀2 2 2 2 2 푟 2
2
The case where 푠 < 푟 1 in Lemma 3.8.1 is an immediate corollary. − (︀푅)︀ ∑︀ Lemma 3.11.2. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . If ∈ 2 ≥ 푃 (︀푟)︀ 푠 < 푟 1, then either there are at least pairs 푃 (i.e. all of them) with 푤푃 > 0 or − 2 there is some 푆 푅 of size 푠 satisfying ⊆ ⎛ ⎞ (︃ (︂ )︂2)︃−1 (︀푠)︀ ∑︁ 푟 2 푤푃 ⎝1 + 4푟 ⎠ 푤. ≥ 2 (︀푟)︀ 푃 ⊆푆 2
Proof: Assume there is some ′ with . We may apply the previous lemma 푃 푤푃 ′ = 0 with (︀푟)︀, since differs from (︀푟)︀ by . This gives that there is some set 퐹 = 푤/ 2 푤푃 ′ 푤/ 2 퐹 푆 푅 of size 푠 satisfying: ⊆ ⎛ ⎞ (︃(︀푠)︀ (︀푠)︀ )︃ (︃ (︂ )︂2)︃−1 (︀푠)︀ ∑︁ 2 2 푟 2 푤푃 + 푤 ⎝1 + 4푟 ⎠ 푤. ≥ (︀푟)︀ (︀푟)︀3 ≥ 2 (︀푟)︀ 푃 ⊆푆 2 푟 2 2 2
The following lemma states that if in a weighted graph there is a vertex whose
degree deviates from the average, then there is a set 푆 푅 of size 푟 1 so that the ⊆ − sum of the weights of the edges contained in 푆 is substantially larger than average.
(︀푅)︀ ∑︀ Lemma 3.11.3. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . For ∈ 2 ≥ 푃 ∑︀ 푣 푅, define 푑(푣) := 푤푃 . If there is some 푣 푅 for which 푑(푣) differs from ∈ 푃 :푣∈푃 ∈ 2푤/푟 by at least 퐹 , then there is some 푆 푅 of size 푟 1 with ⊆ − (︀푟−1)︀ ∑︁ 2 푤푃 푤 + 퐹/푟. ≥ (︀푟)︀ 푃 ⊆푆 2
123 Proof: Choose a vertex 푣 for which 푑(푣) 2푤/푟 퐹 . If 푑(푣) 2푤/푟 퐹 , then | − | ≥ ≤ − we may take 푆 = 푉 푣 . This gives: ∖ { } (︀푟−1)︀ (︀푟−1)︀ ∑︁ 2 2 푤푃 = 푤 푑(푣) 푤 (2푤/푟 퐹 ) = 푤 + 퐹 푤 + 퐹/푟. − ≥ − − (︀푟)︀ ≥ (︀푟)︀ 푃 ⊆푆 2 2
Otherwise, we have 푑(푣) 2푤/푟 + 퐹 . Since ∑︀ 푑(푢) = 2푤, ≥ 푢
∑︁ 푑(푢) 2푤 (2푤/푟 + 퐹 ) = ((푟 1)/푟)2푤 퐹. ≤ − − − 푢̸=푣
Since the average is 2푤/푟, there is some 푢 with
(︂푟 1 )︂ 푑(푢) − 2푤 퐹 /(푟 1) = 2푤/푟 퐹/(푟 1) 2푤/푟 퐹/푟. ≤ 푟 − − − − ≤ −
We may take 푆 = 푉 푢 . We get: ∖ { } (︀푟−1)︀ ∑︁ 2 푤푃 = 푤 푑(푢) 푤 (2푤/푟 퐹/푟) = 푤 + 퐹/푟. − ≥ − − (︀푟)︀ 푃 ⊆푆 2
2
A strengthening of the case 푠 = 푟 1 and 푟 is even in Lemma 3.8.1 is a corollary. −
(︀푅)︀ ∑︀ Lemma 3.11.4. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . ∈ 2 ≥ 푃 2 Either there are at least 푟/2 pairs 푃 for which 푤푃 푤/푟 or there is some 푆 푅 of ≥ ⊆ size 푟 1 with ⎛ ⎞ − (︃ (︂ )︂2)︃−1 (︀푟−1)︀ ∑︁ 푟 2 푤푃 ⎝1 + 4푟 ⎠ 푤. ≥ 2 (︀푟)︀ 푃 ⊆푆 2
2 Proof: If there are fewer than 푟/2 pairs 푃 for which 푤푃 푤/푟 , then we must ≥ have that there is some vertex 푣 not adjacent to any such pair. For this 푣, 푑(푣) ≤ (푟 1)푤/푟2 푤/푟. The previous lemma gives that there is some set 푆 of size 푟 1 − ≤ − with: ⎛ ⎞ (︃(︀푟−1)︀ )︃ (︃ (︂ )︂2)︃−1 (︀푟−1)︀ ∑︁ 2 1 푟 2 푤푃 + 푤 ⎝1 + 4푟 ⎠ 푤. ≥ (︀푟)︀ 푟2 ≥ 2 (︀푟)︀ 푃 ⊆푆 2 2
124 2
Finally, we prove a strengthening of the case 푠 = 푟 1 and 푟 is odd in Lemma − 3.8.1:
(︀푅)︀ ∑︀ Lemma 3.11.5. Given weights 푤푃 for 푃 with 푤푃 0, take 푤 = 푤푃 . If 푟 ∈ 2 ≥ 푃 is odd either there are at least (푟 + 3)/2 pairs 푃 for which 푤푃 > 0 or there is some 푆 푅 of size 푟 1 with ⊆ − ⎛ ⎞ (︃ (︂ )︂2)︃−1 (︀푟−1)︀ ∑︁ 푟 2 푤푃 ⎝1 + 4푟 ⎠ 푤. ≥ 2 (︀푟)︀ 푃 ⊆푆 2
(︂ −1)︂ 푟 1 (︁ 2)︁ − Proof: Assume there is no of size with ∑︀ (︀푟)︀ ( 2 ) 푆 푅 푟 1 푃 ⊆푆 푤푃 1 + 4푟 2 푟 푤. ⊆ − ≥ (2) 푟 1 In this case there is no of size with ∑︀ ( −2 ) 3 , as 푆 푅 푟 1 푃 ⊆푆 푤푃 푟 푤 + 푤/(4푟 ) ⊆ − ≥ (2) (︂ −1)︂ 푟 1 (︁ 2)︁ − this latter term is larger than (︀푟)︀ ( 2 ) . 1 + 4푟 2 푟 푤 (2) We define an unweighted graph 퐺 = (푉, 퐸) by taking 푉 = 푅 and a possible edge (︀푅)︀ 3 푒 is in 퐸 if and only if 푤푒 푤/(4푟 ). By the previous lemma, 퐺 has at least ∈ 2 ≥ 2 푟/2 edges 푒 satisfying 푤푒 푤/푟 , and, indeed, the proof of the previous lemma shows ≥ that every vertex must have degree at least 1. Since 푟 is odd, 퐺 must have at least 2 (푟 + 1)/2 edges 푒 with 푤푒 푤/푟 , and so it must have some vertex 푣 incident to two ≥ such edges. Fix two neighbors of so that and both have weight at least 푣1, 푣2 푣 푤{푣,푣1} 푤{푣,푣2} 2 푤/푟 . We claim that both 푣1 and 푣2 have degree at least 2 in 퐺. Assume at least one
of them, without loss of generality 푣1, has degree one. We must have 푑(푣) 2푤/푟 + ≤ 푤/(2푟2), for otherwise we have a contradiction by Lemma 3.11.3. However, this gives 2 2 that, since 푤{푣,푣 } 푤/푟 , we must have 푤{푣,푣 } 푑(푣) 푤{푣,푣 } 2푤/푟 푤/(2푟 ). 2 ≥ 1 ≤ − 2 ≤ − 3 Then all other 푃 incident to 푣1 have weight at most 푤/(4푟 ), so
3 2 3 2 푤{푣,푣 } 푤{푣,푣 } + (푟 2)푤/(4푟 ) 2푤/푟 푤/(2푟 ) + 푟푤/(4푟 ) = 2푤/푟 푤/(4푟 ). 1 ≤ 1 − ≤ − −
Then by Lemma 3.11.3 we have reached a contradiction.
125 Therefore, we must have that there are at least 3 vertices of degree 2 in 퐺 and that every vertex has degree at least 1. Then the sum of the degrees is at least 6 + (푟 3) = 푟 + 3 and so the number of edges of 퐺 must be at least (푟 + 3)/2, as − desired. 2
3.12 Proof of Lemma 3.8.3
For convenience, we restate both the lemma and definition of 푓휖 here: (︁ 휖 )︁ 퐶 log(훼푚 ) log 푚 푓휖(ℓ) = (log 푚) , where 훼 = ℓ/푛. Recall also 푚0 from Equation 3.1.
Lemma 3.12.1. The following statements hold about 푓휖 for every choice of 휖 0, ≥ 푛 > 1, and 푚 푚0. ≥
1. For any 훼 [ 1 , 1], we have ∈ 푛
푟 2 1/(2( − )) 푓휖(훼푛) 훼 푠 2 푓휖(푛). ≥ −
In particular, 푓휖(훼푛) 훼푓휖(푛). ≥
1 2. For any 훼1, 훼2, 훼3 [ , 1] with 훼1 + 훼2 + 훼3 = 1, taking 푛푖 = 훼푖푛 we have ∈ 푛
∑︁ −3/4 푛푓휖(푛) 푛푖푓휖(푛푖) + 3(log 푚)푛푓휖(푛). ≤ 푖
푟 2 훿 푗 푖 2/( − ) 1/4 푗 (︀푟)︀ 푖+2푗 3. For 푖 0 and 푚 2 1, we have 푓휖(2 ) log 푠 2 ((log 푚)2 ) 256 푓휖(2 ). ≥ ≥ ≥ − ≥ 2
−1 4. For any 훼 log 푚, we have 푓휖(훼푛) 푓휖(푛)/2. ≥ ≥ Proof of 1: Note
퐶(휖+ log 훼 ) 퐶 log 훼 퐶휖 퐶 log log 푚 푓휖(훼푛) = (log 푚) log 푚 = (log 푚) log 푚 (log 푚) = 훼 log 푚 푓휖(푛).
−1 Since 0 < 훼 1, it is sufficient to show that 퐶(log log 푚)/ log 푚 (︀2(︀푟−2)︀)︀ . This ≤ ≤ 푠−2 (︀푟−2)︀ 1/3 holds because 퐶, log log 푚 and 2 are at most log 푚, since 푚 푚0. 푠−2 ≥ 126 Proof of 2: Note that the function ℓ푓휖(ℓ) is a convex function of ℓ. Indeed,
퐶휖 (log ℓ)/ log 푚 퐶휖 (log log 푚)/ log 푚 푓휖(ℓ) = (log 푚) (log 푚) = (log 푚) ℓ .
That is, 푓휖(ℓ) is a polynomial of degree greater than 0 in ℓ, so ℓ푓휖(ℓ) is a polynomial of degree greater than 1 in ℓ and so is convex. −3/4 ∑︀ Take 푆 = 푖 : 훼푖 log 푚 . Take 휅 such that 훼푖 = 휅. Note 휅 { ≥ } 푖∈푆 ≥ 1 2 log−3/4 푚. −
∑︁ ∑︁ ∑︁ 휅 (︂ 휅 )︂ 푛푖푓휖(푛푖) 푛푖푓휖(푛푖) 푛푓휖 푛 ≥ ≥ 푆 푆 푖 푖∈푆 푖∈푆 | | | | (︂ 휅 )︂ (︂1 2 log−1 푚 )︂ = 휅푛푓휖 푛 휅푛푓휖 − 푛 휅푛푓휖(푛/4), 푆 ≥ 3 ≥ | | where the second inequality follows by Jensen’s inequality applied to the convex function ℓ푓휖(ℓ). This gives
∑︁ 휅푛푓휖(푛) 푛푖푓휖(푛푖) 휅푛푓휖(푛) 휅푛푓휖(푛/4) = 휅푛(푓휖(푛) 푓휖(푛/4)). − ≤ − − 푖∈푆
We now consider
퐶휖 퐶(휖+ log(1/4) ) 퐶휖 (︀ −2퐶/ log 푚)︀ 푓휖(푛) 푓휖(푛/4) = (log 푚) (log 푚) log 푚 = (log 푚) 1 (log 푚) . − − −
The second factor satisfies:
1 (log 푚)−2퐶/ log 푚 = 1 2−2퐶(log log 푚)/ log 푚 − − 1 (1 2퐶(log log 푚)/ log 푚) = 2퐶(log log 푚)/ log 푚 log−3/4 푚, ≤ − − ≤
where the first inequality follows by 2푥 1 + 푥 for 푥 0. ≥ ≤ Thus,
127 ∑︁ ∑︁ 푛푓휖(푛) 푛푖푓휖(푛푖) (1 휅)푛푓휖(푛) + 휅푛푓휖(푛) 푛푖푓(푛푖) − ≤ − − 푖 푖∈푆 −3/4 −3/4 −3/4 2(log 푚)푛푓휖(푛) + (log 푚)푛푓휖(푛) 3(log 푚)푛푓휖(푛). ≤ ≤
Proof of 3: We prove a slightly stronger statement. Take ′ 1 so 푗 = 푗 + 4 (log log 푚) that 2푗′ = (log 푚)1/42푗. We will show that
푟 2 푖 2/( − ) 푗′ 푖+2푗′ 푓휖(2 ) log 푠 2 (2 ) 푓휖(2 ). − ≥
This is indeed stronger than the original statement as 푓휖(ℓ) is an increasing function of ℓ.
Consider
(︂ )︂ log(2푖/푛) 퐶 휖+ 푖 log 푛 푖 log 푚 퐶휖 퐶 −퐶 푓휖(2 ) = (log 푚) = (log 푚) (log 푚) log 푚 (log 푚) log 푚 .
Similarly,
푖+2푗′ log 푛 푖+2푗′ 퐶휖 퐶 −퐶 푓휖(2 ) = (log 푚) (log 푚) log 푚 (log 푚) log 푚 .
Therefore, it is sufficient to show that
(︂ )︂ 퐶 푖 2/ 푟 2 푗 푟 퐶 푖+2푗′ (log 푚) log 푚 log (푠−2)(2 ′ ) 256 (log 푚) log 푚 , − ≥ 2
or equivalently that
(︂ )︂ 2/ 푟 2 푗 푟 퐶 2푗′ log (푠−2)(2 ′ ) 256 (log 푚) log 푚 . − ≥ 2
Taking logarithms of both sides, we see that it is sufficient to have
(︂푟 2)︂ (︂ (︂푟)︂)︂ 2(log 푗′)/ − 2퐶푗′(log log 푚)/(log 푚) + log 256 , 푠 2 ≥ 2 − 128 or equivalently
(︂푟 2)︂ (︂ (︂푟)︂)︂ 2(log 푗′)/ − 2퐶푗′(log log 푚)/(log 푚) log 256 0. 푠 2 − − 2 ≥ −
We consider the first derivative of this with respect to 푗′: it is 2(ln(2)푗′)−1/(︀푟−2)︀ 푠−2 − 2퐶(log log 푚)/(log 푚). Note that this derivative is monotone decreasing for 푗′ ∈ [1, ), so the minimum of 2(log 푗′)/(︀푟−2)︀ 2퐶푗′(log log 푚)/(log 푚) log (︀256(︀푟)︀)︀ ∞ 푠−2 − − 2 must be achieved at either the largest or smallest possible value of 푗′. We have 1/4 1/4 assumed 푚훿 log 푚 2푗′ log 푚, so 2훿 log 푚 훿(log 푚) + (log log 푚)/4 푗′ ≥ ≥ ≥ ≥ ≥ (log log 푚)/4. We consider the two extrema. If 푗′ = (log log 푚)/4, we have
(︂푟 2)︂ 퐶 (︂ (︂푟)︂)︂ 2 log((log log 푚)/4)/ − log2(log 푚)/(log 푚) log 256 푠 2 − 2 − 2 − (︂푟 2)︂ (︂ (︂푟)︂)︂ 2 log((log log 푚)/4)/ − 1 log 256 0, ≥ 푠 2 − − 2 ≥ − where the last inequality follows from 푚 푚0. ≥ If 푗′ = 2훿 log 푚:
(︂푟 2)︂ (︂ (︂푟)︂)︂ 2 log(2훿 log 푚)/ − 4훿퐶(log log 푚) log 256 푠 2 − − 2 − (︂푟 2)︂ (︂푟 2)︂ (︂ (︂푟)︂)︂ =2 log(2훿 log 푚)/ − (log log 푚)/ − log 256 0, 푠 2 − 푠 2 − 2 ≥ − − where the last inequality follows from 푚 푚0. ≥ −1 Proof of 4: Since 푓휖 is increasing, it is sufficient to show this for 훼 = log 푚. Then
−1 퐶(휖−(log log 푚)/(log 푚)) 푓휖(log 푚) = (log 푚)
2 = (log 푚)퐶휖2−(log log 푚) / log 푚
퐶휖 −1 (log 푚) 2 = 푓휖(푛)/2. ≥
129 130 Chapter 4
Packing Vertex-Disjoint Monochromatic Copies of Sparse Graphs
4.1 Introduction
Let 퐾푛 be a complete graph on 푛 vertices whose edges are colored with 푟 colors (푟 1). How many monochromatic cycles (single vertices and edges are considered ≥ to be cycles) are needed to partition the vertex set of 퐾푛? This question received much attention in the last few years. Let 푝(푟) denote the minimum number of monochro- matic cycles needed to partition the vertex set of any 푟-colored 퐾푛. It is not obvious that 푝(푟) is a well-defined function. That is, it is not obvious that there always isa partition whose cardinality is independent of 푛. However, in [33] Erdős, Gyárfás and Pyber proved that there exists a constant 퐶 such that 푝(푟) 퐶푟2 log 푟 (throughout ≤ this chapter log denotes the natural logarithm). Furthermore, in [33] (see also [53]) the authors conjectured that 푝(푟) = 푟.
The special case 푟 = 2 of this conjecture was asked earlier by Lehel and, for 푛 푛0, ≥ was first proved by Łuczak, Rödl, and Szemerédi [73]. Allen improved on the valueof
푛0 [1] and recently Bessy and Thomassé [6] proved the original conjecture for all values
131 of 푛 with 푟 = 2. For general 푟 the current best bound is due to Gyárfás, Ruszinkó,
Sárközy, and Szemerédi [54] who proved that, for 푛 푛0(푟), we have 푝(푟) 100푟 log 푟. ≥ ≤ For 푟 = 3, in [55] it was proved that all but 표(푛) of the vertices may be covered by 3 monochromatic cycles. Surprisingly, Pokrovskiy [77] found a counterexample to the conjecture for all 푟 3. However, in the counterexample, all but one vertex can be ≥ covered by 푟 vertex-disjoint monochromatic cycles. Thus, a slightly weaker version of the conjecture still can be true, say that, apart from a constant number of vertices, the vertex set can be covered by 푟 vertex-disjoint monochromatic cycles. Let us also note that the above problem was generalized in various directions; for hypergraphs (see [57] and [84]), for complete bipartite graphs (see [33] and [61]), for graphs which are not necessarily complete (see [4] and [83]), and for vertex partitions by monochromatic connected 푘-regular subgraphs (see [85] and [86]). Another area that attracted much interest is the study of Ramsey numbers for bounded degree graphs. For a graph 퐺, the Ramsey number 푅(퐺) is the smallest positive integer 푁 such that, if the edges of a complete graph 퐾푁 are partitioned into two color classes, then one color class has a subgraph isomorphic to 퐺. The existence of such a positive integer is guaranteed by Ramsey’s classical result [79]. Determining
푅(퐺) even for very special graphs is notoriously hard (see e.g. [50] or [78]). In 1975, Burr and Erdős [11] raised the problem that every graph 퐺 with 푛 vertices and maximum degree ∆ has a linear Ramsey number, so 푅(퐺) 퐶(∆)푛, for some ≤ constant 퐶(∆) depending only on ∆. This was proved by Chvátal, Rödl, Szemerédi and Trotter [20] in one of the earliest applications of Szemerédi’s celebrated Regularity
Lemma [89]. Because the proof uses the Regularity Lemma, the bound on 퐶(∆) is quite weak; it is of tower type in ∆. This was improved by Eaton [30] who proved, using a variant of the Regularity Lemma, that the function 퐶(∆) can be taken to be of the form 22푂(Δ) . Soon after, Graham, Rödl, and Ruciński [49] improved this further to 퐶(∆) ≤ 푂(Δ log2 Δ) 푂(Δ log Δ) 2 and for bipartite graphs 퐶퐵(∆) 2 . They also proved that there ≤ are bipartite graphs with 푛 vertices and maximum degree ∆ for which the Ramsey number is at least 2Ω(Δ)푛. Recently, Conlon [22] and, independently, Fox and Su-
132 dakov [45] have shown how to remove the log ∆ factor in the exponent, achieving Θ(Δ) an essentially best possible bound of 퐶퐵(∆) = 2 in the bipartite case. For the non-bipartite graph case, the current best bound is due to Conlon, Fox, and Sudakov
[25] 퐶(∆) 2푂(Δ log Δ). Similar results have been proven for hypergraphs: [26, 27, 75] ≤ use the hypergraph regularity lemma and [24] improves the bounds by avoiding the regularity lemma.
Similar results also hold for 푎-arrangeable graphs. An 푎-arrangeable graph is one in which the vertices may be ordered as 푣1, . . . , 푣푛 such that, for any index 푖, if we consider those neighbors of 푣푖 in the set 푣푖+1, . . . , 푣푛 , they have at most 푎 neighbors { } in the set 푣1, . . . , 푣푖 . Chen and Schelp [16] proved that, for every 푎, there is some { } constant 퐶(푎) so that the Ramsey number of any 푎-arrangeable graph on 푛 vertices is at most 퐶(푎)푛. The best bound that is known for 퐶(푎), again due to Graham, Rödl, and Ruciński [49], is 퐶(푎) 2퐶푎 log2 푎. ≤ It is a natural question (initiated by András Gyárfás) to combine the studies of packing monochromatic cycles and of computing Ramsey numbers of sparse graphs and ask how many monochromatic members from a bounded-degree graph family are needed to partition the vertex set of a 2-edge-colored 퐾푁 . In this chapter we study this problem and related questions. Given = 퐹1, 퐹2,... a sequence of graphs, ℱ { } we say it is a proper graph sequence if 퐹푛 is a graph on 푛 vertices. We say it has some graph property if every graph of has that property (e.g. is bipartite if 퐹푛 ℱ ℱ is bipartite for every 푛). We derive the following lower bound from the result that, for 푛 sufficiently large, there are bipartite graphs on 푛 vertices of maximum degree at most ∆ with Ramsey number 2Ω(Δ)푛 [49]. We prove this bound in Section 4.6.
Theorem 4.1.1. There exists an absolute constant 푐 such that, for every ∆, there is a bipartite proper graph sequence with maximum degree at most ∆ and, for every ℱ 푛 sufficiently large, there isa 2-edge-coloring of 퐾푛 so that covering the vertices of 푐Δ 퐾푛 using monochromatic copies of graphs from requires at least 2 such copies. ℱ This matches the upper bound we find.
133 Theorem 4.1.2. There exists an absolute constant 퐶 such that, for every ∆ and ev- ery bipartite proper graph sequence with maximum degree at most ∆, every 2-edge- ℱ colored complete graph can be partitioned into at most 2퐶Δ vertex-disjoint monochro- matic copies of graphs from . ℱ We can actually prove that if has chromatic number at most 푘 then the bound ℱ above may be taken to be 2퐶푘Δ. If 푘 log ∆, then this is the best bound we know ≤ how to get. However, for larger values of 푘, we can get a better bound that depends only on ∆.
Theorem 4.1.3. There exists an absolute constant 퐶 such that, for every ∆ and every proper graph sequence with maximum degree at most ∆, every 2-edge-colored ℱ complete graph can be partitioned into at most 2퐶Δ log Δ vertex-disjoint monochromatic copies of graphs from . ℱ If we are interested in graphs with bounded arrangeability rather than bounded degree, we get a slightly weaker bound, assuming the graphs satisfy an additional degree-bound, which is required for one of the tools we use.
Theorem 4.1.4. There exists an absolute constant 퐶 such that, for every 푎 there and every 푎-arrangeable proper graph sequence = 퐹1, 퐹2,... satisfying that the ℱ { } maximum degree of 퐹푛 is at most √푛/ log 푛, every 2-edge-colored complete graph on at least 푁 vertices can be partition into at most 2퐶푎6 vertex-disjoint monochromatic copies of graphs from . ℱ Theorems 4.1.2 and 4.1.3 give, perhaps surprisingly, that we have the same phe- nomenon for these classes of graphs as for cycles; we can partition into monochromatic graphs from such that the average size of the parts (which are each monochromatic ℱ copies of graphs from ) is roughly as large as the size of the single largest monochro- ℱ matic graph we know how to find (i.e. the one given by our best bounds for Ramsey numbers). It is particularly interesting to note that the conditions of Theorem 4.1.4 are satisfied with high probability by 퐹푛 = 퐺(푛, 푑/푛) for 푑 a constant, as Fox and Sudakov [46] showed that, with high probability, the arrangeability of 퐺(푛, 푑/푛) is Θ(푑2).
134 It would be desirable to close the gap between the upper bound given by Theorem 4.1.3 and the lower bound in Theorem 4.1.1, though doing so may require improved bounds for the Ramsey numbers of bounded degree graphs.
Finally, let us mention one interesting special case of Theorem 4.1.3. The 푘푡ℎ power of a cycle 퐶 is the graph obtained from 퐶 by joining every pair of vertices with distance at most 푘 in 퐶. Density questions for powers of cycles have generated a lot of interest; in particular the famous Pósa-Seymour conjecture (see e.g. [15, 37, 38, 39, 40, 65, 68, 69, 71]). Theorem 4.1.3 implies the following result on the partition number by monochromatic powers of cycles.
Corollary 4.1.5. There exists an absolute constant 퐶 so that, for every 푘, ev- ery 2-colored complete graph can be partitioned into at most 2퐶푘 log 푘 vertex-disjoint monochromatic 푘th powers of cycles.
4.2 Notation
푉 (퐺) and 퐸(퐺) denote the vertex-set and the edge-set of the graph 퐺. (퐴, 퐵, 퐸) denotes a bipartite graph 퐺 = (푉, 퐸), where 푉 = 퐴 퐵 and 퐸 퐴 퐵. For a graph ∪ ⊂ × 퐺 and a subset 푈 of its vertices, 퐺 푈 is the restriction to 푈 of 퐺. 푁(푣) is the set of | neighbors of 푣 푉 . Hence, 푁(푣) = 푑푒푔(푣) = 푑푒푔퐺(푣), the degree of 푣. 훿(퐺) stands ∈ | | for the minimum and ∆(퐺) for the maximum degree in 퐺. When 퐴, 퐵 are subsets of 푉 (퐺), we denote by 푒(퐴, 퐵) the number of edges of 퐺 with one endpoint in 퐴 and the other in 퐵. We write 푑푒푔(푣, 푈) = 푒( 푣 , 푈) for the number of edges from 푣 to 푈. { } For non-empty 퐴 and 퐵, 푒(퐴, 퐵) 푑(퐴, 퐵) = 퐴 퐵 | || | is the density of the graph between 퐴 and 퐵.
Definition 4.2.1. The bipartite graph 퐺 = (퐴, 퐵, 퐸) is 휀-regular if
푋 퐴, 푌 퐵, 푋 > 휀 퐴 , 푌 > 휀 퐵 imply 푑(푋, 푌 ) 푑(퐴, 퐵) < 휀. ⊂ ⊂ | | | | | | | | | − | 135 We will often say simply that “the pair (퐴, 퐵) is 휀-regular” with the graph 퐺 implicit.
Definition 4.2.2. (퐴, 퐵) is (휀, 푑, 훿)-super-regular if it is 휀-regular, satisfies 푑(퐴, 퐵) ≥ 푑, and 푑푒푔(푎) > 훿 퐵 푎 퐴, 푑푒푔(푏) > 훿 퐴 푏 퐵. | | ∀ ∈ | | ∀ ∈
Definition 4.2.3. Given a 푘-partite graph 퐺 = (푉, 퐸) with 푘-partition 푉 = 푉1 ∪ (︀푘)︀ ... 푉푘, the 푘-cylinder 푉1 ... 푉푘 is 휀-regular ((휀, 푑, 훿)-super-regular) if all the ∪ × × 2 pairs of subsets (푉푖, 푉푗), 1 푖 < 푗 푘, are 휀-regular ((휀, 푑, 훿)-super-regular). If we ≤ ≤ wish to say a cylinder is (휀, 훿, 훿)-super-regular, we simply say it is (휀, 훿)-super-regular, and in this case it is not necessary to check the density condition. Given 훼 0, the ≥ 푘-cylinder 푉1 ... 푉푘 is 훼-balanced if, for every 푖 < 푗, 푉푖 푉푗 훼 min( 푉푖 , 푉푗 ). × × || |−| || ≤ | | | |
We say a graph 퐺 on 푛 vertices is 푎-nicely-arrangeable if it is 푎-arrangeable and satisfies that ∆(퐺) √푛/ log 푛. We say it is 휒-chromatically equitable if there is a ≤ proper 휒-coloring of the vertices of 퐺 (one where no two adjacent vertices have the same color) in which the size of any two color classes differs by at most 1. We say it is 푅-linearly-Ramsey if 푅(퐺) 푅푛. ≤
4.3 Regularity and blow-up lemmas
Some of our main tools for finding monochromatic copies of graphs are regularity and blowup lemmas. Regularity lemmas allow us to find monochromatic regular cylinders, and blowup lemmas allow us to cover the vertices of such a cylinder with a sparse graph. Instead of the Regularity Lemma of Szemerédi [89], we will use the following lemmas which Conlon and Fox [23] argued as consequences of the Duke, Lefmann, and Rödl weak Regularity Lemma [29].
Lemma 4.3.1 ([29] and Lemma 5.3 in [23]). For each 0 < 휀 < 1/2, given any graph
퐺 = (푉, 퐸) on 푛 푘 vertices we may find disjoint sets of vertices 푉1, . . . , 푉푘 so that ≥ 136 the induced 푘-partite cylinder is 0-balanced and 휀-regular; the size of each part is at
2 5 least 1 푘 휀− 2푘 휀 푛.
We will use the following corollary of this lemma.
Lemma 4.3.2 (Lemma 5.4 in [23]). For each 0 < 휀 < 1/2, given any 2-colored 2푘 complete graph on 푛 2 vertices we may find vertex-disjoints sets 푉1, . . . , 푉푘 so ≥ that the induced multipartite graph is, in one of the colors (say in red), an (휀, 1/2, 0)- super-regular 0-balanced cylinder (i.e. one with no minimum degree constraint and
4푘 5 parts of equal size), where the size of each part is at least 1 2 휀− . 2(22푘) 휀 푛
Indeed, to get this one applies Lemma 4.3.1 for the red subgraph with 22푘 in place of 푘 to get an 휀-regular 22푘-cylinder. Then we may consider the complete graph whose vertices 푖 correspond to the parts of the cylinder 푉푖 and we color the edge (푖, 푗) by 2푘 the majority color in the pair (푉푖, 푉푗). We then apply 푅(퐾푘) 2 and use the fact ≤ that, if (푉푖, 푉푗) is regular in one color, then it is also regular in the other color. Our main tool for dealing with bounded degree graphs is a quantitative version of the Blow-up Lemma (see [66, 67, 82]).
Lemma 4.3.3 (Quantitative Blow-up Lemma). For every constant 훼 there exists a constant 퐶 = 퐶(훼) such that, given a graph 푅 of order 푟 2 and positive parameters ≥ , , and , for any (︀ 훿 Δ)︀퐶 the following holds. Let us replace the vertices 훿 푑 ∆ 0 < 휀 < 푟Δ 푑 of 푅 with pairwise disjoint 훼-balanced sets 푉1, 푉2, . . . , 푉푟 (blowing up). We construct ⋃︀ ′ two graphs on the same vertex set 푉 = 푉푖. The graph 푅 is obtained by replacing all edges of 푅 with copies of the complete bipartite graph, and a sparser graph 퐺 is constructed by replacing the edges of 푅 with some (휀, 푑, 훿)-super-regular pairs. If a graph 퐻 with ∆(퐻) ∆ is embeddable into 푅′, then it is embeddable into 퐺. ≤
To deal with 푎-nicely-arrangeable graphs, we use a version of the blowup lemma due to Böttcher, Kohayakawa, Taraz, and Würfl [9]. It is worth noting that, in their paper, the authors of [9] allow more parameters and compute more explicit bounds than what we state below.
137 Theorem 4.3.4 (Blow-up Lemma for Arrangeable Graphs [9]). For every constant 훼 there exists a constant 퐶 = 퐶(훼) such that, given a graph 푅 of order 푟 2 and positive ≥ parameters 훿 and 푎, if we take ∆푅 to be a bound on the maximum degree of 푅, then, 4 −퐶푎2Δ −퐶푎4Δ2 2퐶푎 푟퐶 −퐶푎2 −퐶 for any 0 < 휀 < 훿 푅 2 푅 , taking 푛0 := 2 훿 휖 , the following holds. · Let us replace the vertices of 푅 with pairwise-disjoint 훼-balanced sets 푉1, 푉2, . . . , 푉푟
(blowing up) each of size at least 푛0. We construct two graphs on the same vertex ⋃︀ ′ set 푉 = 푉푖. The graph 푅 is obtained by replacing all edges of 푅 with copies of the complete bipartite graph, and a sparser graph 퐺 is constructed by replacing the edges of 푅 with some (휀, 훿)-super-regular pairs. If an 푎-nicely-arrangeable graph 퐻 is embeddable into 푅′, then it is embeddable into 퐺.
4.4 Proofs of Theorems 4.1.2, 4.1.3, and 4.1.4
We basically follow the greedy-absorbing proof technique that originated in [33] and is used in many papers in this area (e.g. [54], [61], [86]).
Given an 푅-linearly-Ramsey proper graph sequence , we will show how to de- ℱ compose any 2-edge-coloring of a 퐾푛 into useful structures, in pursuit of the goal of eventually decomposing it into monochromatic copies of graphs from . ℱ The structures we will use are monochromatic copies from , monochromatic ℱ super-regular cylinders, and something resembling a union of almost-complete mul- tipartite graphs. Such a decomposition is useful as we will use a blow-up lemma to cover the super-regular cylinders and it is easy to greedily embed graphs into almost- complete multipartite graphs. For technical reasons, we don’t actually use complete multipartite graphs, but a structure that will serve a similar purpose that we call a branching degree cylinder.
Definition 4.4.1. Given positive integers 푘1, . . . , 푘ℓ, the (푘1, . . . , 푘ℓ)-branching tree is a rooted tree defined recursively as follows: the ()-branching tree is a single vertex, the
root, and the (푘1, . . . , 푘ℓ+1)-branching tree is the graph obtained from the (푘1, . . . , 푘ℓ)-
branching tree by adding 푘ℓ+1 neighbors to each leaf (and the root vertex remains the
same). If 푘 := 푘1 = 푘2 = = 푘ℓ, we simply say it is a 푘-branching tree with ℓ levels. ··· 138 In other words, the branching tree is a full and complete rooted tree, and the
parameters (푘1, . . . , 푘ℓ) determine the branching factor at each level.
Definition 4.4.2. Given a 2-edge-coloring of a 퐾푛 and positive integers 푘1, . . . , 푘ℓ, a
(푘1, . . . , 푘ℓ)-branching (1 훾) degree cylinder with scaling 휀 assigns to every vertex 푣 − of the (푘1, . . . , 푘ℓ)-branching tree a collection of vertices from the 퐾푛; if 푣 and 푤 are
not ancestors of each other in the branching tree, then 푉푣 and 푉푤 are disjoint, and
otherwise, if 푤 is an ancestor of 푣, then 푉푣 푉푤. Furthermore, for every pair 푤 and ⊆ 푣 with 푤 the parent of 푣, there is some color and some set 푆푣 푉푤 푉푣 so that every ⊆ ∖ vertex in 푉푣 has degree at least (1 훾) 푆푣 to 푆푣 in that color; we call 푆푣 the parental − | | neighborhood of 푉푣. Finally, 푉푣 휀 푆푣 . If 푘 := 푘1 = 푘2 = = 푘ℓ, then we say it | | ≤ | | ··· is a 푘-branching cylinder with ℓ levels.
We are now ready to state the way in which we will decompose the vertices of a
2-edge-colored 퐾푛.
Lemma 4.4.3. There exists an absolute constant 퐶 such that, given any 2-edge- coloring of a complete graph, any 푘, ℓ > 0, any 0 < 휀 < 1/2 and any 0 < 훿 < 1/2 휀, − and any 푅-linearly-Ramsey proper graph sequence , we may find ℱ ∑︀ℓ−1 푘푖푅2퐶푘휀−퐶 vertex-disjoint monochromatic copies of graphs from ∙ 푖=0 ℱ A 푘-branching (1 훿 푘휀) degree cylinder with scaling 푘휀 and with ℓ levels in ∙ − − which, for 푣 a vertex of the underlying branching tree, 푉푣 is the corresponding
set of vertices with parental neighborhood 푆푣, and
Inside of each 푉푤 with 푤 not a leaf of the branching tree, a (2푘휀, 1/2 2푘휀, 훿 푘휀) ∙ − − super-regular cylinder 퐶푤 with 푘 parts which are 푘휀-balanced and which satisfies
that, for any 푣 a child of 푤, 푆푣 퐶푤. ⊆ The above structures satisfy two additional properties. The first is that every vertex of the complete graph is contained in one of the monochromatic copies of a graph from
, one of the super-regular cylinders, or in 푉푣 for 푣 a leaf in the branching tree. The ℱ second is that, for every 푉푤 with 푤 not a leaf of the branching tree, the associated
139 super-regular cylinder 퐶푤 does not intersect 푉푣 for any 푣 a descendant of 푤 and 퐶푤 does not intersect any of the monochromatic copies of graphs from . ℱ We first prove the above decomposition lemma, and then we show how toapply it to prove Theorems 4.1.2, 4.1.3, and 4.1.4.
4.4.1 Proof of Lemma 4.4.3
To prove the above theorem, we will proceed by induction on ℓ. One essential part of the inductive step is the following lemma.
Lemma 4.4.4. There exists an absolute constant 퐶 such that, given any 푅-linearly- Ramsey proper graph sequence , any 휀 > 0, any positive integer 푘, and any 2-edge- ℱ 퐶푘 −퐶 coloring of a 퐾푛, we may partition the vertices into at most 푅2 휀 monochromatic copies of graphs from and sets 푉1, . . . , 푉푘 so that the multipartite graph induced by ℱ one of the colors is a 0-balanced, (휀, 1/2 휀, 0)-super-regular cylinder. − To prove Lemma 4.4.4, we first find a monochromatic 휀/2-regular cylinder so that the density between any two pairs is at least 1/2 by Lemma 4.3.2; then, we will cover most of the remaining vertices with monochromatic copies of graphs from . We ℱ simply add the uncovered vertices to the cylinder; they do not significantly harm the regularity. To this end, we prove the following observation about 푅-linearly-Ramsey proper graph sequences.
Lemma 4.4.5. For any 푅-Ramsey proper graph sequence, given a 2-edge-colored 퐾푛 and an 훼 > 0, we may cover all but an 훼 fraction of the vertices of the 퐾푛 using at most 2푅 log(푒/훼) vertex-disjoint monochromatic copies of graphs from . ℱ
Proof. Given any 2-coloring of a 퐾푛 for 푛 2푅 we may find a monochromatic ≥ copy of 퐹⌊푛/푅⌋. Removing the vertices of this copy leaves a 2-edge-coloring of a clique on (1 1/푅)푛 (1 1/(2푅))푛 vertices; iterating this procedure 푎 := 2푅 log(푒/훼) ⌈ − ⌉ ≤ − times gives a 2-edge coloring on at most (1 1/(2푅))푎푛 푒푎/(2푅)푛 훼푛 vertices, − ≤ ≤ assuming that at each iteration the graph had at least 2푅 vertices. If at any point the graph failed to have at least 2푅 vertices, then we may remove each of the remaining
140 vertices as monochromatic copies of 퐹1; in either case, the total number of monochro- matic copies of graphs from used was at most 푎 + 2푅 = 2푅 log(1/훼) + 2푅 = ℱ 2푅 log(푒/훼). 2 Proof of Lemma 4.4.4.
By Theorem 4.3.2, either 푛 < 22푡 or we may find a monochromatic (휀/2, 1/2, 0)-
4푘 5 super-regular, -balanced cylinder where each part is at least an 1 2 (휀/2)− 0 훼 := 2(22푘) (휀/2) fraction of the vertices. In the former case, we may simply take each vertex to
be a monochromatic copy of 퐹1 and take the cylinder to be empty. In the lat-
ter case, we find a 0-balanced cylinder on vertex set 푉1, . . . , 푉푘 in which each part has size at least 훼푛. Take 휀′ := 휀2훼/4. Then, by Lemma 4.4.5, using at most 2푅 log(푒/휀′) = 푅2푂(푘)휀−푂(1) copies of graphs from , we may cover all but an 휀′ frac- ℱ ′ tion of those vertices of the 퐾푛 that are outside of 푉1, . . . , 푉푘. There are 푛 vertices ′ remaining outside of 푉1, . . . , 푉푘; we wish to assume 푛 is divisible by 푘, which we ′ may do by covering up to 푘 vertices by copies of 퐹1. Then, we simply add 푛 /푘 of 2 the remaining vertices to each of 푉1, . . . , 푉푘; since we are adding fewer than 휀 훼푛/4 vertices to each part (each part has size 훼푛), the new cylinder is (휀, 1/2 휀, 0)-super- − regular. 2
Proof of Lemma 4.4.3. We proceed by induction on ℓ. If ℓ = 0, then we may take
the whole vertex set of the 퐾푛 to be a 푘-branching cylinder with 0 levels. Otherwise, if , by induction we may decompose the into at most ∑︀ℓ−2 푖 퐶푘 −퐶 ℓ > 0 퐾푛 푖=0 푘 푅2 휀 vertex-disjoint monochromatic copies from where 퐶 is the constant from Lemma ℱ 4.4.4, into a 푘-branching 푘휀-scaling (1 훿 푘휀) degree cylinder with ℓ 1 levels − − − with vertex sets 푉푣 where 푣 is a vertex of the corresponding branching tree and their corresponding parental neighborhoods 푆푣, and in each 푉푤 for 푤 not a leaf of the branching tree there is a (2푘휀, 1/2 2푘휀, 훿 푘휀)-regular cylinder 퐶푤 satisfying that − − it does not intersect 푉푣 for any descendant 푣 of 푤, that it does not intersect any of the monochromatic copies of graphs from , and that it contains 푆푣 for any 푣 a child ℱ of 푤. In order to prove Lemma 4.4.3 with ℓ levels rather than with ℓ 1 levels, we wish − to extend the branching degree cylinder. To do so, for each of the 푘ℓ−1 leaves 푤 of the
141 푘-branching tree with ℓ-1 levels we apply Lemma 4.4.4 to decompose 푉푤 into at most 푅2퐶푘휀−퐶 monochromatic copies of graphs from and into vertex-disjoint 0-balanced ℱ sets 퐶′ , . . . , 퐶′ so that the induced cylinder 퐶′ in one of the colors is (휀, 1/2 휀, 0)- 1 푘 − super-regular. We now extract low-degree vertices from this cylinder. We define
′ ′ sets 퐷푖 recursively; 퐷푖 is the set of vertices in 퐶 퐶 that have degree at most ∖ 푖 ′ ′ (훿 휀) 퐶 to 퐶 , minus those vertices that appear in 퐷1, . . . , 퐷푖−1. We then take the − | 푖| 푖 sets corresponding to the children of 푤 in the ℓ-level 푘-branching tree to be 퐷푖 with ′ parental neighborhoods 퐶푖 := 퐶 퐷푖 and take 퐶푤 := 퐶1 퐶푘 to be the cylinder 푖 ∖ × · · · × corresponding to 푉푤. Note that, for each 푗 = 푖, by 휀-regularity at most an 휀-fraction ̸ ′ ′ ′ of the vertices of 퐶 fail to have degree at least (훿 휀) 퐶 to 퐶 . Therefore, each 퐷푖 푗 − | 푖| 푖 ′ has size at most (푘 1)휀 퐶 . This implies the cylinder 퐶푤 is (푘 1)휀 < 푘휀-balanced. − | 푖| − Furthermore, since each pair 퐶′, 퐶′ was (휀, 1/2 휀, 0)-super-regular and we removed 푖 푗 − all vertices that had degree to 퐶′ less than (훿 휀) 퐶′ , in the process we removed at 푖 − | 푖| ′ ′ most (푘 1)휀 퐶 vertices from 퐶 and so 퐶푣 is (2푘휀, 1/2 2푘휀, 훿 푘휀)-super-regular, − | 푗| 푗 − − as desired. 2
4.4.2 Applying Lemma 4.4.3
We use the three structures found in Lemma 4.4.3, namely the monochromatic copies of graphs from , the super-regular cylinders, and the branching degree cylinder, to ℱ cover all of the vertices with monochromatic graphs from . We now explain how to ℱ use the latter two structures to do this. To cover vertices in a super-regular cylinder, we will apply one of the blow-up lemmas. To that end, we wish to show how to embed few copies of graphs from ℱ into a balanced complete multipartite graph. It is easier to embed graphs that are, in some sense, balanced.
Definition 4.4.6. A graph 퐺 is 푘-chromatically equitable if it has a proper 푘-vertex- coloring (i.e. one in which no two adjacent vertices receive the same color) in which every color class has the same size.
Note that, if a graph 퐺 is 푘-colorable, then the graph obtained by taking 푘 disjoint
142 copies of 퐺, which we denote by 푘퐺, is 푘-chromatically equitable. To see this, we may take any 푘-coloring of 퐺 and cyclically permute the colors throughout the 푘 copies to get a chromatically equitable coloring.
Lemma 4.4.7. Given a proper graph sequence = 퐹1, 퐹2,... with chromatic ℱ { } number at most 푘 and given a complete 1/(푘 + 1)-balanced 푘 + 1-partite graph, we can cover the vertices using at most (푘 + 1)2 vertex-disjoint copies of graphs from . ℱ
Proof. Let 퐶1, . . . , 퐶푘+1 be the parts of the complete graph. For each index
푖 [푘 + 1], define 푣푖 = 퐶푖 and take 푣 = max푖(푣푖). Then, for each set 푆 [푘 + 1] of ∈ | | ⊆ size 푘, define 푤푆 := 푣 푣푖 where 푖 is the unique index not contained in 푆. For each − such set , we take a copy of (disjoint from all the copies chosen so far) that 푆 푘퐹푤푆
uses 푤푆 vertices from each 푉푖 with 푖 푆. We use 푘 + 1 such graphs, one for each 푆. ∈ Note that, because the structure is 1/(푘 + 1)-balanced, for every step corresponding
to some 푆 with 푖 푆 we remove at most a 1/(푘 + 1) fraction of 퐶푖, so there are ∈ always enough vertices in 퐶푖 to choose the next graph to be vertex-disjoint from the previous ones, and so the procedure does not fail.
After this procedure, the number of uncovered vertices in 퐶푖 is
∑︁ ∑︁ ∑︁ 푣푖 푤푆 = 푣푖 푤푆 + 푤[푘+1]∖{푖} = 푣 푤푆. − − − 푆:푖∈푆 푆 푆
That is, after this procedure, each 퐶푖 has the same number of uncovered vertices, say 2 푛. We cover these with a copy of (푘 + 1)퐹푛. We use a total of (푘 + 1) copies of graphs from . 2 ℱ Finally, we wish to show how, given a branching degree cylinder, to cover the
vertices of the 푉푣 with 푣 a leaf in the corresponding branching tree.
Lemma 4.4.8. Given a 2-edge-coloring of a 퐾푛, a 푑-degenerate 푘-chromatically equi- table proper graph sequence , and a 푘-branching (1/(32푘2푑2))-scaling (1 1/(32푘2푑2)) ℱ − degree cylinder with 2푘 3 levels which assigns the set 푉푣 to a vertex 푣 of the corre- − sponding branching tree with 푆푣 the parental neighborhood, we may cover all the 푉푣 with 푣 a leaf of the branching tree using at most 푘2푘 monochromatic copies of graphs
143 from the vertices of which are contained entirely in the various 푆푣 or in some 푉푣 ℱ with 푣 a leaf, and, in doing so, for any 푣 with parent 푤 we use either at most a 휀 fraction of the vertices of 푆푣 or we use all of 푉푤.
Proof. For every path 푃 = (푣0, . . . , 푣2푘−3) from the root to a leaf in the branching tree, we define, for every vertex 푣푖 with 0 푖 < 2푘 3, a corresponding color: this ≤ − 2 2 color is red if the vertices of 푉푣 all have red-degree at least (1 1/(32푘 푑 )) 푉푣 to 푖+1 − | 푖 | 2 2 푉푣 , and otherwise they all have blue-degree at least (1 1/(32푘 푑 )) 푆푣 to 푆푣 and 푖 − | 푖 | 푖 the color is blue. This gives a sequence of 2푘 3 colors; exactly one of the two colors − must occur at least 푘 1 times in this sequence. Then take, for 0 푗 < 푘 1, 푣푃 to − ≤ − 푗 be the vertex of the path 푃 corresponding to the 푗th occurrence of the more common color in the color sequence. Take to be . Take . 푣푃푘 1 푣2푘−3 푉푃푗 := 푉푣푃 − 푗
Now, we claim that, given any vertex sets 푉0 푉1 푉푘−1 with 푉푖 4 푉푖+1 ⊇ ⊇ · · · ⊇ | | ≥ | | and satisfying that, for every 푖 > 0, we have that the vertices of 푉푖 have degree at 2 least (1 1/(16푘푑 )) 푉푖−1 to 푉푖−1, there is a 푘-partite subgraph with vertex-disjoint − | | ′ ′ parts 푉 푉푖 with 푉 = 푉푘−1 where each part has size at least 푉푘−1 in which, for 푖 ⊆ 푘−1 | | any pair 푖 = 푗, we have that the vertices of 푉푖 have degree larger than (1 1/(2푑)) 푉푗 ̸ − | | to 푉푗. To see this, note that, for any 푖 and for every 푗 > 푖, the average degree of a 2 vertex in 푉푖 to 푉푗 is at least (1 1/(16푘푑 )) 푉푗 . By Markov’s inequality, at most a − | | 1/(4푘푑) fraction of the vertices of 푉푖 fail to have degree at least (1 1/(4푑)) 푉푗 to − | | 푉푗. We now proceed as follows: for each 푖, starting with 푖 = 0 and proceeding in
increasing order, we may remove at most a 1/(4푑) fraction of the vertices of 푉푖 to
ensure that every vertex of 푉푖 has degree more than (1 1/(4푑)) 푉푗 to 푉푗 for every − | | 푗 > 푖. For every pair of distinct 푖, 푗, at some point in this process 푉푖 has satisfied that
every vertex has degree more than (1 1/(4푑)) 푉푗 to 푉푗. After this point, at most a − | | 1/(4푑) fraction of the vertices of 푉푗 were removed by the process, so at the end of the
process we must have that the vertices of 푉푖 have degree more than (1 1/(2푑)) 푉푗 − | | to (what remains of) 푉푗. Given this structure, we may simply greedily embed a copy of into this structure, using all of and at most vertices of 푘퐹|푉푘 1| 푉푘−1 (푘 푖) 푉푘−1 − − | | each 푉푖. We now proceed as follows. Iterate over the various distinct paths 푃 in an arbitrary
144 order. We wish to cover the vertices of by few graphs from . By the previous 푣푃푘 1 − ℱ argument, we may do so if, for each , we have that each vertex of has degree 푗 > 푖 푉푃푗 2 at least (1 1/(16푘푑 )) 푉푃 to 푉푃 ; this was true before we started iterating. If this − | 푖 | 푖 is true, we will then use the previous argument to find a graph that covers and 푉푃푘 1 − uses at most vertices from . Throughout this process, we remove at (푘 푖) 푉푃푘 1 푉푃푖 − | − | most a (︃푘−푗−1 )︃ ∑︁ 휀푘−푗−1 푘푖 2(휀푘)푘−푗−1 2휀푘 < 1/(16푘푑2) 푖=0 ≤ ≤
2 fraction of any 푉푃 for 푗 < 푘 1. Therefore, we will remove at most a 1/(16푘푑 ) 푗 − fraction of the vertices from each throughout the process, and so the necessary 푉푃푖 condition will hold and we will cover each . 푉푃푘 1 2 −
We now prove Theorem 4.1.2 and Theorem 4.1.3.
Proof. Let a proper graph sequence with maximum degree at most ∆ and chro- ℱ matic number at most 푘 be given. We know that is 푅-Ramsey linear for some 푅, ℱ and we know by [45] that 푅 is at most 2퐶1푘Δ and by [25] that 푅 is at most 2퐶2Δ log Δ.
Take 푅 to be the minimum of 2퐶1푘Δ and 2퐶2Δ log Δ. Take, for some sufficiently large constant 퐶, 휀 = 푅−퐶 . Take ℓ = 2푘 3. Take 훿 = 1/(64푑2푘2). Apply Lemma 4.4.3 to − obtain 푘푂(푘)푅2푂(푘)휀−푂(1) = 휀−푂(1) copies of graphs from , a 푘-branching 푘휀-scaling ℱ (1 훿 푘휀) (1 2훿) cylinder with ℓ levels, and, for each 푤 in the underlying branch- − − ≥ − ing tree, a (2푘휀, 1/2 2푘휀, 훿 푘휀) super-regular 0-balanced cylinder 퐶푤, where these − − structures have the properties described in Lemma 4.4.3. We now apply Lemma 4.4.8 to the branching cylinder; this allows us to cover all of the vertices of the 푉푣 where 푣 is leaf of the branching tree and 푉푣 is the corresponding vertex set of the branching cylinder using at most 푘2푘 monochromatic copies of graphs from . Furthermore, for ℱ any 푤 in the branching tree, either 푉푤 is entirely covered, or the only vertices from
푉푤 that are used are from 퐶푤 and we use at most a 푘휀 fraction of 퐶푤. At this point, the only vertices that we have not covered with monochromatic copies of graphs from
are those vertices in the various 퐶푤 from which we’ve used at most a 푘휀 fraction of ℱ the vertices. Notice that, because each 퐶푤 is a (2푘휀, 1/2 2푘휀, 훿 푘휀) super-regular − − 0-balanced 푘 + 1-partite cylinder, after removing at most a 푘휀 fraction of 퐶푤, the
145 result is at least a (1/(푘 + 1))-balanced (푅−Ω(퐶), 1/4, 훿/2)-cylinder, so if 퐶 is large enough we may apply Lemma 4.3.3 and Lemma 4.4.7 to cover the remainder of the cylinder using at most (푘 + 1)2 monochromatic copies of graphs from . We used a ℱ total of 푅푂(퐶) monochromatic copies of graphs from when applying Lemma 4.4.3, ℱ we used at most 푘푂(푘) 푅푂(1) copies of graphs from when applying Lemma 4.4.8, ≤ ℱ and we used at most (푘 + 1)2 monochromatic copies of graphs from from each of ℱ the 푘푂(푘) vertices of the branching tree when applying Lemma 4.4.7, so we used a total of 푅푂(퐶) monochromatic copies of graphs from . If 푘 log ∆, then this value ℱ ≤ is 2푂(퐶푘Δ); in particular, when 푘 = 2 as in the case of Theorem 4.1.2, we use 2푂(퐶Δ) monochromatic copies from . For all values of 푘, we use 2푂(퐶Δ log Δ) monochromatic ℱ copies of graphs from , giving the bound in Theorem 4.1.3. 2 ℱ The proof of Theorem 4.1.4 is very similar, though it has some additional compli- cations.
Proof. Let a proper graph sequence which is 푎-nicely-arrangeable with chromatic ℱ number at most 푘 be given. We know that is 푅-Ramsey linear for some 푅 = ℱ 2푂(푎 log2 푎). Take, for some sufficiently large constant 퐶, 휀 = 2−퐶푎4푘2 . Take ℓ = 2푘 3. − Take 훿 = 1/(64푑2푘2). Apply Lemma 4.4.3 to obtain 푘푂(푘)푅2푂(푘)휀−푂(1) = 휀−푂(1) copies of graphs from , a 푘-branching 푘휀-scaling (1 훿 푘휀) (1 2훿) cylinder with ℓ ℱ − − ≥ − levels, and, for each 푤 in the underlying branching tree, a (2푘휀, 1/2 2푘휀, 훿 푘휀) − − super-regular 0-balanced cylinder 퐶푤, where these structures have the properties de- scribed in Lemma 4.4.3. We now apply Lemma 4.4.8 to the branching cylinder; this
allows us to cover all of the vertices of the 푉푣 where 푣 is leaf of the branching tree 2푘 and 푉푣 is the corresponding vertex set of the branching cylinder using at most 푘 monochromatic copies of graphs from . Furthermore, for any 푤 in the branching ℱ tree, either 푉푤 is entirely covered, or the only vertices from 푉푤 that are used are
from 퐶푤 and we use at most a 푘휀 fraction of 퐶푤. At this point, the only vertices that we have not covered with monochromatic copies of graphs from are those ℱ vertices in the various 퐶푤 from which we’ve used at most a 푘휀 fraction of the vertices.
Notice that, because each 퐶푤 is a (2푘휀, 1/2 2푘휀, 훿 푘휀) super-regular 0-balanced − − 푘 + 1-partite cylinder, after removing at most a 푘휀 fraction of 퐶푤, the result is at
146 least a (1/(푘 +1))-balanced (2−Ω(퐶푎4푘2), 1/4, 훿/2)-cylinder, so if 퐶 is large enough and
퐶푤 has enough vertices, we may apply Lemma 4.3.4 and Lemma 4.4.7 to cover the remainder of the cylinder using at most (푘 +1)2 monochromatic copies of graphs from 4 2푂(푎 ) . However, if 퐶푤 doesn’t have enough vertices, that is it has 2 vertices, then ℱ we may cover it using 2푂(푎4) monochromatic copies of graphs from using Lemma ℱ 4.4.5. We used a total of 2푂(퐶푘2푎4) = 2푂(퐶푎6) monochromatic copies of graphs from when applying Lemma 4.4.3, we used at most 푘푂(푘) copies of graphs from when ℱ ℱ applying Lemma 4.4.8, and we used at most 2푂(푎4) monochromatic copies of graphs from from each of the 푘푂(푘) vertices of the branching tree when applying Lemma ℱ 4.4.7, so we used a total of 2푂(퐶푎6) monochromatic copies of graphs from , giving ℱ the bound in Theorem 4.1.4.
4.5 Concluding Remarks
There are various interesting potential generalizations of Theorem 4.1.3. One may
ask if the theorem holds for 푟 colors for any positive integer 푟.
Conjecture 4.5.1. For every positive integer 푟 there exists a constant 퐶푟 (depending on 푟) such that, for every ∆-bounded sequence , every 푟-edge-colored complete graph ℱ can be partitioned into at most 2Δ퐶푟 vertex-disjoint monochromatic graphs from . ℱ
Since bounds on Ramsey numbers were key in proving the theorem for 푟 = 2, it is worth noting that Conlon, Fox, and Sudakov [25] proved that, for any fixed number
of colors 푟, for any graph 퐺 on 푛 vertices of maximum degree ∆ the Ramsey number
2 퐶푟Δ on 푟 colors 푅푟(퐺) is at most 2 푛. The primary difficulty in adapting our proof to this setting is in constructing a (1 훾) branching cylinder when there are more than − 2-colors. Finally, let us mention that since by now both the Regularity Lemma and the Blow-up Lemma have been generalized to hypergraphs (see [81] and [62], respec- tively), perhaps we can generalize our result to hypergraphs as well.
147 4.6 Lower Bound
We wish to show that there exists a ∆-bounded sequence = 퐹1, 퐹2,... and, for 푛 ℱ { } sufficiently large, a two-edge-coloring of 퐾푛 that cannot be partitioned into fewer than Ω(Δ) 2 monochromatic copies of graphs from . To see this, for every 푛 take 퐺푛 to be ℱ a graph on 푛 vertices of degree at most ∆ and, for 푛 sufficiently large, with Ramsey number at least 2Ω(Δ)푛, as given by the result of Graham, Rödl and Ruciński [49]. We define 푖 recursively; take 0 . Then define 푖 to be the disjoint union of 푖 1 퐹2 퐹2 = 퐺1 퐹2 퐹2 − 푖 푖 with 푖 1 . For integers of the form with , define 푖 to be the disjoint 퐺2 − 2 + 푗 푗 < 2 퐹2 +푗 union of 퐹2푖 with an independent set on 푗 vertices. Under this definition, each 퐹푛 is a graph on 푛 vertices with maximum degree at most ∆. Furthermore, for 푛0 < 푛1, 푖 퐹푛 is a subgraph of 퐹푛 . Finally, taking 푖 to be the largest integer with 2 푛, 퐹푛 0 1 ≤ Ω(Δ) 푖−1 Ω(Δ) contains a copy of 푖 1 and so has Ramsey number at least (for 퐺2 − 2 2 = 2 푛
푛 sufficiently large). Take = 퐹1, 퐹2,... . Now, for 푁 sufficiently large, take a 2- ℱ { } edge-coloring of a complete graph on 2Ω(Δ)푁 vertices without a monochromatic copy of 퐹푁 (this is possible by the condition on the Ramsey number). Since the sequence of graphs is increasing, this coloring also does not contain a monochromatic copy of any 퐹푛 for 푛 > 푁. Therefore, any partition of the vertex set into monochromatic copies of graphs from must use at least 2Ω(Δ) such copies. ℱ
148 Bibliography
[1] P. Allen, Covering two-edge-coloured complete graphs with two disjoint monochromatic cycles, Combinatorics, Probability and Computing 17 (2008), 471–486.
[2] N. Alon, J. Pach, and J. Solymosi, Ramsey-type theorems with forbidden sub- graphs, Combinatorica 21 (2001), 155–170.
[3] N. Alon and J. Spencer, The probabilistic method, John Wiley & Sons (2000).
[4] J. Balog, J. Barát, D. Gerbner, A. Gyárfás, and G.N. Sárközy, Partitioning edge- 2-colored graphs by monochromatic paths and cycles, Combinatorica.
[5] J. Beck, On size Ramsey number of paths, trees and circuits, Algorithms Combin. 5 (1990), 34–45.
[6] S. Bessy and S. Thomassé, Partitioning a graph into a cycle and an anticycle, a proof of Lehel’s conjecture, J. Combin. Theory Ser. B 100 (2010), 176–180.
[7] B. Bollobás, Extremal Graph Theory, Academic Press, London (1978).
[8] B. Bollobás, Random graphs, 2nd edition, Cambridge University Press (2001).
[9] J. Böttcher, Y. Kohayakawa, A. Taraz, and A. Würfl, An Extension of the Blow- up Lemma to arrangeable graphs, arXiv:1305.2059.
[10] S. Burr, Some Ramsey-infinite graphs in the diagonal case, Ars Combinatoria 21 (1986), 173–178.
149 [11] S. Burr and P. Erdős, On the magnitude of generalized Ramsey numbers for graphs, Coll. Math. Soc. Janós Bolyai 10, Infinite and Finite Sets (1975), 214– 240.
[12] S. Burr, P. Erdős, and L. Lovász, On graphs of Ramsey type, Ars Combinatoria 1 (1976), 167–190.
[13] S. Burr, J. Nešetřil, and V. Rödl, On the use of senders in generalized Ramsey theory for graphs, Discrete Math. 54 (1985), 1–13
[14] K. Cameron, J. Edmonds, and L. Lovász, A note on perfect graphs, Period. Math. Hungar. 17 (1986), 173–175.
[15] P. Chau, L. DeBiasio, and H. Kierstead, Pósa’s conjecture for graphs of order at
least 2 108, Random Structures Algorithms 39 (2011), 507–525. × [16] G. Chen and R. Schelp, Graphs with linearly bounded Ramsey numbers, J. Combin. Theory Ser. B 57 (1993), 138–149.
[17] M. Chudnovsky, The Erdős-Hajnal conjecture - a survey, J. Graph Theory 75 (2014), 178–190.
[18] M. Chudnovsky and S. Safra, The Erdős-Hajnal conjecture for bull-free graphs, J. Combin. Theory, Ser. B 98 (2008), 1301–1310.
[19] F. Chung and R. L. Graham, Edge-colored complete graphs with precisely colored subgraphs, Combinatorica 3 (1983), 315–324.
[20] V. Chvátal, V. Rödl, E. Szemerédi and W. Trotter, The Ramsey number of a graph with bounded maximum degree, J. Combin. Theory Ser. B 34 (1983), 239–243.
[21] D. Conlon, A new upper bound for diagonal Ramsey numbers, Annals of Math- ematics 170 (2009), 941–960
[22] D. Conlon, Hypergraph packing and sparse bipartite Ramsey numbers, Combin. Probab. Comput. 18 (2009), 913–923.
150 [23] D. Conlon and J. Fox, Bounds for graph regularity and removal lemmas, Geo- metric and Functional Analysis 22 (2012), 1191–1256.
[24] D. Conlon, J. Fox, and B. Sudakov, Ramsey numbers of sparse hypergraphs, Random Structures Algorithms 35 (2009), 1–14.
[25] D. Conlon, J. Fox, and B. Sudakov, On two problems in graph Ramsey theory, Combinatorica 32 (2012), 513–535.
[26] O. Cooley, N. Fountoulakis, D. Kühn and D. Osthus, 3-uniform hypergraphs of bounded degree have linear Ramsey numbers, J. Combin. Theory Ser. B 98 (2008), 484–505.
[27] O. Cooley, N. Fountoulakis, D. Kühn and D. Osthus, Embeddings and Ramsey numbers of sparse k-uniform hypergraphs, Combinatorica 28 (2009), 263–297.
[28] A. Dudek and V. Rödl, On 퐾푠-free subgraphs in 퐾푠+푘-free graphs and vertex Folkman numbers, Combinatorica 31 (2011), 39–53.
[29] R. A. Duke, H. Lefmann, and V. Rödl, A fast approximation algorithm for computing the frequencies of subgraphs in a given graph. SIAM J. Comput. 24 (1995), 598–620.
[30] N. Eaton, Ramsey numbers for sparse graphs, Discrete Math. 185 (1998), 63–75.
[31] P. Erdős, Some remarks on the theory of graphs, Bull. Amer. Math. Soc. 53 (1947), 292–294.
[32] P. Erdős, R. Faudree, C. Rousseau, and R. Schelp, The size Ramsey number, Period. Math. Hungar. 9 (1947), 145–161
[33] P. Erdős, A. Gyárfás, and L. Pyber, Vertex coverings by monochromatic cycles and trees, J. Combin. Theory Ser. B 51 (1991), 90–95.
[34] P. Erdős and A. Hajnal, On chromatic number of graphs and set-systems, Acta Mathematica Hungarica 17, 61–99.
151 [35] P. Erdős and A. Hajnal, Ramsey-type theorems, Discrete Appl. Math. 25 (1989), 37–52.
[36] P. Erdős and G. Szekeres, A combinatorial problem in geometry, Compositio Math. 2 (1935), 463–470.
[37] G. Fan and R. Häggkvist, The square of a hamiltonian cycle, SIAM J. Disc. Math. (1994), 203–212.
[38] G. Fan and H. A. Kierstead, The square of paths and cycles, J. Combin. Theory Ser. B 63 (1995), 55–64.
[39] G. Fan and H. A. Kierstead, Hamiltonian square-paths, J. Combin. Theory Ser. B 67 (1996), 167–182.
[40] R. J. Faudree, R. J. Gould, M. S. Jacobson and R. H. Schelp, On a problem of Paul Seymour, Recent Advances in Graph Theory (1991), 197–215.
[41] J. Fox, A. Grinshpun, A. Liebenau, Y. Person, T. Retter, and T. Szabó, Graphs with a low minimum degree minimal Ramsey graph.
[42] J. Fox, A. Grinshpun, A. Liebenau, Y. Person, and T. Szabó, What is Ramsey- equivalent to a clique? J. Combin. Theory Ser. B (2014).
[43] J. Fox and K. Lin, The minimum degree of Ramsey-minimal graphs, J. Graph Theory 54 (2006), 167–177.
[44] J. Fox, A. Grinshpun, and J. Pach, The Erdős-Hajnal conjecture for rainbow triangles, J. Combin. Theory Ser. B (2014).
[45] J. Fox and B. Sudakov, Density theorems for bipartite graphs and related Ramsey-type results, Combinatorica 29 (2009), 153–196.
[46] J. Fox and B. Sudakov, Two remarks on the Burr-Erdős conjecture.
[47] S. Fujita, C. Magnant, and K. Ozeki, Rainbow generalizations of Ramsey theory: a survey, Graphs Combin. 26 (2010), 1–30.
152 [48] T. Gallai, Transitiv orientierbare Graphen, Acta Math. Acad. Sci. Hungar 18 (1967), 25–66. English translation in [74].
[49] R.L. Graham, V. Rödl and A. Ruciński, On graphs with linear Ramsey numbers, J. Graph Theory 35 (2000), pp. 176–192.
[50] R.L. Graham, B.L. Rothschild, and J.H. Spencer, Ramsey Theory, John Wiley & Sons, New York, (1990).
[51] A. Grinshpun, R. Raina, and R. Sengupta, Minimum degress of minimal Ramsey graphs for almost-cliques.
[52] A. Grinshpun and G. Sárközy, Monochromatic bounded degree subgraph parti- tions.
[53] A. Gyárfás, Covering complete graphs by monochromatic paths, Irregularities of Partitions 8 (1989), 89–91.
[54] A. Gyárfás, M. Ruszinkó, G.N. Sárközy and E. Szemerédi, An improved bound for the monochromatic cycle partition number, J. Combin. Theory Ser. B 96 (2006), 855–873.
[55] A. Gyárfás, M. Ruszinkó, G.N. Sárközy and E. Szemerédi, Partitioning 3-colored complete graphs into three monochromatic cycles, Electronic J. of Combinatorics 18 (2011), 53.
[56] A. Gyárfás and G. Sárközy, Gallai colorings of non-complete graphs, Discrete Math. 310 (2010), 977–980.
[57] A. Gyárfás, G.N. Sárközy, Monochromatic path and cycle partitions in hyper- graphs, Electronic Journal of Combinatorics 20 (2013), 18.
[58] A. Gyárfás, G. Sárközy, A. Sebő, and S. Selkow, Ramsey-type results for Gallai colorings. J. Graph Theory 64 (2010), 233–243.
[59] A. Hajnal, Rainbow Ramsey theorems for colorings establishing negative parti- tion relations, Fund. Math. 198 (2008), 255–262.
153 [60] A. Hajnal and E. Szemerédi, Proof of a conjecture of P. Erdős, Combinatorial theory and its applications (1969), 601–623.
[61] P. Haxell, Partitioning complete bipartite graphs by monochromatic cycles, J. Combin. Theory Ser. B 69 (1997), 210–218.
[62] P. Keevash, A hypergraph blow-up lemma, Random Structures and Algorithms 39 (2011), 275–376.
[63] H. A. Kierstead and A. V. Kostochka, A short proof of the Hajnal-Szemerédi theorem on equitable colourings, Combinatorics, Probability and Computing 17 (2008), 265–270.
[64] Y. Kohayakawa, V. Rödl, M. Schacht, and E. Szemerédi, Sparse partition uni- versal graphs for graphs of bounded degree, Adv. Math. 226 (2011), 5041–5065.
[65] J. Komlós, G. N. Sárközy, and E. Szemerédi, On the square of a Hamiltonian cycle in dense graphs, Random Structures and Algorithms, 9 (1996), 193–211.
[66] J. Komlós, G. N. Sárközy, and E. Szemerédi, Blow-up Lemma, Combinatorica 17 (1997), 109–123.
[67] J. Komlós, G. N. Sárközy, and E. Szemerédi, An algorithmic version of the Blow- up Lemma, Random Structures and Algorithms 12 (1998), 297–312.
[68] J. Komlós, G. N. Sárközy, and E. Szemerédi, On the Pósa-Seymour conjecture, Journal of Graph Theory 29 (1998), 167–176.
[69] J. Komlós, G. N. Sárközy, and E. Szemerédi, Proof of the Seymour conjecture for large graphs. Annals of Combinatorics 1 (1998), 43–60.
[70] J. Körner and G. Simonyi, Graph pairs and their entropies: modularity problems, Combinatorica 20 (2000), 227–240.
[71] I. Levitt, G.N. Sárközy, and E. Szemerédi, How to avoid using the Regularity Lemma; Pósa’s Conjecture revisited, Discrete Mathematics, 310 (2010), 630–641.
154 [72] L. Lovász, On chromatic number of finite set-systems, Acta Mathematica Hun- garica 19 (1968), 59–67.
[73] T. Łuczak, V. Rödl, and E. Szemerédi, Partitioning two-colored complete graphs into two monochromatic cycles, Combinatorics, Probability and Computing 7 (1998), 423–436.
[74] F. Maffray and M. Preissmann, A translation of Gallai’s paper: ‘Transitiv Ori- entierbare Graphen’, Perfect Graphs, Wiley, New York (2001), 25–66.
[75] B. Nagle, S. Olsen, V. Rödl and M. Schacht, On the Ramsey number of sparse 3-graphs, Graphs Combin. 27 (2008), 205–228.
[76] J. Nešetřil V. Rödl, The Ramsey property for graphs with forbidden complete subgraphs, J. Combinatorial Theory Ser. B 20 (1976), 243–249.
[77] A. Pokrovskiy, Partitioning edge-coloured complete graphs into monochromatic cycles and paths, ArXiv:1205.5492v1
[78] S.P. Radziszowski, Small Ramsey numbers, Electronic Journal of Combinatorics (2002).
[79] F. P. Ramsey, On a problem of formal logic, Proc. London Math. Soc. 30 (1930), 264–286.
[80] V. Rödl and E. Szemerédi, On size Ramsey numbers of graphs with bounded degree, Combinatorica 20 (2000), 257–262.
[81] V. Rödl and M. Schacht, Regular partitions of hypergraphs: Regularity Lemmas, Combinatorics, Probability and Computing 16 (2007), 833–885.
[82] G. Sárközy, A quantitative version of the blow-up lemma.
[83] G. Sárközy, Monochromatic cycle partitions of edge-colored graphs, Journal of graph theory 66 (2011), 57–64.
[84] G. Sárközy, Improved monochromatic loose cycle partitions in hypergraphs.
155 [85] G. Sárközy and S. Selkow, Vertex partitions by connected monochromatic 푘- regular graphs, J. Combin. Theory Ser. B 78 (2000), 115–122.
[86] G.N. Sárközy, S. Selkow, and F. Song, An improved bound for vertex parti- tions by connected monochromatic k-regular graphs, Journal of Graph Theory 73 (2013), 127–145.
[87] D. Seinsche, On a property of the class of 푛-colorable graphs, J. Combinatorial Theory Ser. B 16 (1974), 191–193.
[88] T. Szabó, P. Zumstein, and S. Zürcher, On the minimum degree of minimal Ramsey graphs, J. Graph Theory 64 (2010), 150–164
[89] E. Szemerédi, Regular partitions of graphs, Colloques Internationaux 260 (1976), 399–401.
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