Polynomial Bounds for Chromatic Number. IV: A Near-polynomial Bound for Excluding the Five-vertex Path

A graph G is H-free if it has no induced subgraph isomorphic to H. We prove that a P5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_5$$\end{document}-free graph with clique number ω≥3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\omega \ge 3$$\end{document} has chromatic number at most ωlog2(ω)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\omega ^{\log _2(\omega )}$$\end{document}. The best previous result was an exponential upper bound (5/27)3ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(5/27)3^{\omega }$$\end{document}, due to Esperet, Lemoine, Maffray, and Morel. A polynomial bound would imply that the celebrated Erdős-Hajnal conjecture holds for P5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_5$$\end{document}, which is the smallest open case. Thus, there is great interest in whether there is a polynomial bound for P5\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_5$$\end{document}-free graphs, and our result is an attempt to approach that.


Introduction
If G, H are graphs, we say G is H-free if no induced subgraph of G is isomorphic to H; and for a graph G, we denote the number of vertices, the chromatic number, the size of the largest clique, and the size of the largest stable set by |G|, χ(G), ω(G), α(G) respectively.
The k-vertex path is denoted by P k , and P 4 -free graphs are well-understood; every P 4 -free graph G with more than one vertex is either disconnected or disconnected in the complement, which implies that χ(G) = ω(G).Here we study how χ(G) depends on ω(G) for P 5 -free graphs G.
The Gyárfás-Sumner conjecture [9,23] says: 1.1 Conjecture: For every forest H there is a function f such that χ(G) ≤ f (ω(G)) for every This is open in general, but has been proved [9] when H is a path, and for several other simple types of tree ([2, 10, 11, 12, 13, 16, 18]; see [17] for a survey).The result is also known if all induced subdivisions of a tree are excluded [16].
A class of graphs is hereditary if the class is closed under taking induced subgraphs and under isomorphism, and a hereditary class is said to be χ-bounded if there is a function f such that χ(G) ≤ f (ω(G)) for every graph G in the class (thus the Gyárfás-Sumner conjecture says that, for every forest H, the class of H-free graphs is χ-bounded).Louis Esperet [7] made the following conjecture: Esperet's conjecture was recently shown to be false by Briański, Davies and Walczak [1].However, this raises the further question: which χ-bounded classes are polynomially χ-bounded?In particular, the two conjectures 1.1 and 1.2 would together imply the following, which is still open: 1.3 Conjecture: For every forest H, there exists c > 0 such that χ(G) ≤ ω(G) c for every H-free graph G.This is a beautiful conjecture.In most cases where the Gyárfás-Sumner conjecture has been proved, the current bounds are very far from polynomial, and 1.3 has been only been proved for a much smaller collection of forests (see [14,19,21,22,20,4,15]).In [22] we proved it for any P 5 -free tree H, but it has not been settled for any tree H that contains P 5 .In this paper we focus on the case H = P 5 .
The best previously-known bound on the chromatic number of P 5 -free graphs in terms of their clique number, due to Esperet, Lemoine, Maffray, and Morel [8], was exponential: Here we make a significant improvement, showing a "near-polynomial" bound: (The cycle of length five shows that we need to assume ω(G) ≥ 3. Sumner [23] showed that χ(G) ≤ 3 when ω(G) = 2.) Conjecture 1.3 when H = P 5 is of great interest, because of a famous conjecture due to Erdős and Hajnal [5,6], that: This is open in general, despite a great deal of effort; and in view of [3], the smallest graph H for which 1.6 is undecided is the graph P 5 .Every forest H satisfying 1.3 also satisfies the Erdős-Hajnal conjecture, and so showing that H = P 5 satisfies 1.3 would be a significant result.
We use standard notation throughout.When X ⊆ V (G), G[X] denotes the subgraph induced on X.We write χ(X) for χ(G[X]) when there is no ambiguity.

The main proof
We denote the set of nonnegative real numbers by R + , and the set of nonnegative integers by Z + .Let f : Z + → R + be a function.We say In this section we show that if a function f satisfies a certain inequality, then it is a binding function for all P 5 -free graphs.Then at the end we will give a function that satisfies the inequality, and deduce 1.5.
is not in A and has a neighbour and a non-neighbour in A. It is complete to A if it is adjacent to every vertex of A. We begin with the following: 2.1 Let G be P 5 -free, and let f be a near-binding function for G. Let G be connected, and let X be a cutset of G. Then Proof.We may assume (by replacing X by a subset if necessary) that X is a minimal cutset of G; and so G \ X has at least two components, and every vertex in X has a neighbour in V (B), for every component B of G \ X.Let B be one such component; we will prove that ), from which the result follows.
Choose v ∈ X (this is possible since G is connected), and let N be the set of vertices in B adjacent to v. Let the components of B \ N be R 1 , . . ., R k , S 1 , . . ., S ℓ , where R 1 , . . ., R k each have chromatic number more than f (⌊ω(G)/2⌋), and S 1 , . . ., S ℓ each have chromatic number at most f (⌊ω(G)/2⌋).Let S be the union of the graphs S 1 , . . ., S ℓ ; thus χ(S) ≤ f (⌊ω(G)/2⌋).For 1 ≤ i ≤ k, let Y i be the set of vertices in N with a neighbour in V (R i ), and let Let y ∈ Y i .Thus y has a neighbour in V (R i ); suppose that y is mixed on R i .Since R i is connected, there is an edge ab of R i such that y is adjacent to a and not to b.Now v has a neighbour in each component of G \ X, and since there are at least two such components, there is a vertex u ∈ V (G) \ (X ∪ V (B)) adjacent to v.But then u-v-y-a-b is an induced copy of P 5 , a contradiction.This proves (1).
From the minimality of I, for each i ∈ I there exists y i ∈ Y i such that for each j ∈ I \ {i} we have that y i / ∈ Y j ; and so the vertices y i (i ∈ I) are all distinct.For each i ∈ I choose r i ∈ V (R i ).For all distinct i, j ∈ I, if y i , y j are nonadjacent, then r i -y i -v-y j -r j is isomorphic to P 5 , a contradiction.Hence the vertices y i (i ∈ I) are all pairwise adjacent, and adjacent to v; and so All the vertices in N \ Y are adjacent to v, and so Since there are no edges between any two of the graphs G[N \ Y ], R 1 , . . ., R k , their union (Z say) has clique number at most ω(G) − 1 and so has chromatic number at most f (ω(G) − 1).But V (B) is the union of Y, V (S) and V (Z); and so This proves 2.1.

2.2
Let Ω ≥ 1, and let f : Z + → R + be non-decreasing, satisfying the following: • f is a binding function for every P 5 -free graph H with ω(H) ≤ Ω; and Then f is a binding function for every P 5 -free graph G.
Proof.We prove by induction on |G| that if G is P 5 -free then f is a binding function for G. Thus, we may assume that G is P 5 -free and f is near-binding for G.If G is not connected, or ω(G) ≤ Ω, it follows that f is binding for G, so we assume that G is connected and ω(G) > Ω.Let us write w = ω(G) and m = ⌊w/2⌋.If χ(G) ≤ f (w) then f is a binding function for G, so we assume, for a contradiction, that: (1) χ(G) > f (w − 1) + (w + 2)f (m).
We claim that f for each integer x > 4. If that is true, then by 2.2 with Ω = 4, we deduce that χ(G) ≤ f (ω(G)) for every P 5 -free graph G, and so 1.5 holds.Thus, it remains to show that f (x − 1) + (x + 2)f (⌊x/2⌋) ≤ f (x) for each integer x > 4.This can be verified by direct calculation when x = 5, so we may assume that x ≥ 6.