Separating polynomial χ -boundedness from χ -boundedness

Extending the idea from the recent paper by Carbonero, Hompe, Moore, and Spirkl, for every function f : N → N ∪{∞} with f (1) = 1 and f ( n ) (cid:62) (cid:0) 3 n +13 (cid:1) , we construct a hereditary class of graphs G such that the maximum chromatic number of a graph in G with clique number n is equal to f ( n ) for every n ∈ N . In particular, we prove that there exist hereditary classes of graphs that are χ -bounded but not polynomially χ -bounded.


Introduction
Given a class of graphs C its χ-bounding function is the function χ C : N → N ∪ {∞} defined as where χ(G) and ω(G) denote, respectively, the chromatic number and the clique number of G.A class of graphs C is χ-bounded if there is a function f : N → N such that χ(G) f (ω(G)) for every graph G ∈ C, or equivalently if χ C (n) is finite for every n ∈ N. A class C is polynomially χ-bounded if such a function f can be chosen to be a polynomial.A class C is hereditary if it is closed under taking induced subgraphs.
A well-known and fundamental open problem, due to Esperet [6], has been to decide whether every hereditary χ-bounded class of graphs is polynomially χ-bounded.We provide a negative answer to this question.More generally, we prove that χ-bounding functions may be arbitrary, so long as they are bounded from below by a certain cubic function.
Theorem 1.Let f : N → N ∪ {∞} be such that f (1) = 1 and f (n) for every n 2. Then there exists a hereditary class of graphs G such that χ G (n) = f (n) for every n ∈ N.
On the other hand, χ-bounding functions are not entirely arbitrary.For instance, Scott and Seymour [11] proved that every hereditary class of graphs C with χ C (2) = 2 satisfies χ C (n) 2 2 n+1 .
The proof of Theorem 1 is heavily based on the idea used by Carbonero, Hompe, Moore, and Spirkl [2] in their very recent solution to another well-known problem attributed to Esperet [12].They proved that for every k ∈ N, there is a K 4 -free graph G with χ(G) k such that every triangle-free induced subgraph of G has chromatic number at most 4. Their proof, in turn, relies on an idea by Kierstead and Trotter [8], who proved in 1992 that the class of oriented graphs excluding an directed path on four vertices as an induced subgraph is not χ-bounded.We derive Theorem 1 from the following result, which also generalises the recent breakthrough of Carbonero, Hompe, Moore, and Spirkl [2].
Theorem 2. For every pair of integers n and k with k n 2, there exists a graph G with clique number n and chromatic number k such that every induced subgraph of G with clique number m < n has chromatic number at most 3m+1 3 .
In the case that n is a prime number, we obtain a better bound.Theorem 3.For every pair of integers p and k with p a prime and k p, there exists a graph G with clique number p and chromatic number k such that every induced subgraph of G with clique number m < p has chromatic number at most m+2 3 .
In the first version of this paper [1] we only proved a weaker version of Theorem 3 with m+2 3 replaced by m m 2 .Despite the worse bound obtained, this alternative proof may still be of independent interest.
Very recently, Girão, Illingworth, Powierski, Savery, Scott, Tamitegama, and Tan [7] independently proved another result generalising the fact that for every prime p, there are graphs with clique number p and arbitrarily large chromatic number whose induced subgraphs with clique number at most p − 1 have bounded chromatic number [1].They proved that for every graph F with at least one edge, there are graphs of arbitrarily large chromatic number and the same clique number as F in which every F -free induced subgraph has chromatic number at most some constant c F depending only on F .They also showed the analogous statement where clique number is replaced by odd girth.See [12] and [9] for recent surveys on χ-boundedness and polynomial χ-boundedness.

Proof
First, we show that Theorem 2 implies Theorem 1.
Proof of Theorem 1 assuming Theorem 2. Fix a function f : for every n 2. By Theorem 2, for every pair of integers n and k with k n 2, there exists a graph H n,k with clique number n and chromatic number k such that every induced subgraph of . Combining these inequalities, we conclude that and the theorem follows.
The rest of the paper is devoted to proving Theorem 2. We begin with the following lemma.
Lemma 4. For every positive integer k, there is a graph G k and an acyclic orientation of its edges with the following properties: (2) for every pair of vertices u and v, there is at most one directed path from for any two distinct vertices u and v such that there is a directed path from u to v in G k .
Various well-known constructions of triangle-free graphs with arbitrarily high chromatic number, such as Zykov's [13] and Tutte's [3,4], satisfy the condition of Lemma 4 once the edges are oriented in a way that follows naturally from the construction.See [2] and [8] for an explicit construction of the graphs G k with the appropriate acyclic orientations, based on Zykov's construction.It is only implicit that the acyclic orientations of the graphs in [2] and [8] satisfy all of the properties in the conclusion of Lemma 4, so for the sake completeness we provide a proof based on Tutte's construction.
Proof of Lemma 4. We proceed by induction on k.The base case k = 1 follows by taking a single-vertex graph as G 1 .For the induction step, assume G k−1 is an acyclically oriented graph satisfying conditions (1)-( 4) for k − 1.To construct G k , begin with a stable set S with |S| = (k − 1)(|V (G k−1 )| − 1) + 1, and for every subset X of S with |X| = |V (G k−1 )|, add an isomorphic copy G X of G k−1 (with the same orientation as in G k−1 ) and an arbitrary perfect matching between the vertices in X and the vertices of G X , oriented from X to G X .This clearly preserves acyclicity of the orientation.Since every vertex in S has at most one edge to each copy G X of G k−1 , condition (2) is preserved.Any directed path on k − 1 vertices in G X extends to a directed path on k vertices in G k by adding a vertex from S, so (3) holds.Any colouring of the copies G X of G k−1 with a common palette of k − 1 colours extends to a k-colouring of G k by using a single new colour on S, which shows that χ(G k ) χ(G k−1 ) + 1 and condition (4) is preserved.Finally, suppose there exists a The fact that X and G X are connected by a perfect matching implies that at most k − 2 colours are used on G X , which contradicts the fact that For the rest of the argument, we fix an arbitrary sequence (G k ) k∈N of graphs given by Lemma 4. Now, for every pair of positive integers k and p, where p is a prime number, we construct a graph G k,p by adding edges to G k as follows.
Let be the directed reachability order of the vertices of G k , that is, u v if and only if there is a (unique) directed path from u to v in G k .Since the orientation of G k given by Lemma 4 is acyclic, is indeed a partial order.For every pair of vertices u and v in G k such that u v, let d(u, v) be the length of the unique directed path from u to v in G k (i.e., the number of edges in that path).The graph G k,p has the same vertex set as G k and has the set {uv : u < v and d(u, v) ≡ 0 (mod p)} as the edge set.We consider each such edge uv as oriented from u to v. Since the original (oriented) edges uv of G k satisfy u < v and d(u, v) = 1, the graph G k,p contains G k as a subgraph.Furthermore, every edge of G k,p connects vertices with different colours in a k-colouring φ of G k claimed in Lemma 4. Therefore, Next, we examine cliques in G k,p (and its induced subgraphs).Since G k,p is acyclic, every clique of G k,p induces a transitive tournament.Given a clique C of an acyclic oriented graph, we let t(C) be the first vertex of the transitive tournament induced by C. We call t(C) the tail of C. Given a clique C of G k,p , we let r(C) be the subset of Z p such that r(C) ≡ {d(t(C), v) : v ∈ C} (mod p).We call r(C) the residue of the clique C. Note that 0 is always contained in A rotation of a subset X of Z p is a subset of Z p of the form X + a = {x + a : x ∈ X} for any a ∈ Z p .A subset of Z p is rooted if it contains 0. The rotation X + a of a rooted subset X of Z p is rooted if and only if −a ∈ X.Let ∼ p be the equivalence relation on the rooted subsets of Z p such that X ∼ p Y whenever Y is a rotation of X.Let [X] p denote the equivalence class of X in ∼ p .For every proper rooted subset X of Z p (such that X = Z p ), since p is a prime, all rotations X + a of X with a ∈ Z p are distinct, and in particular |[X] p | = |X|.Order every equivalence class arbitrarily, and for every proper rooted subset X of Z p , let c(X) ∈ {1, . . ., |X|} denote the position of X in this ordering.Lemma 6.For every positive integer k, every prime p, and every induced subgraph G of G k,p with clique number m < p, we have χ(G) m+2 3 .
Proof.We will colour the vertices of G by triples of integers (a, b, c) with m a b c 1. Since there are m+2 3 choices for such a triple, this will be a m+2

3
-colouring of G.For each vertex v of G, let a(v) be the maximum size of a clique in G with tail v. Thus m a(v) 1.Let B(v) be the intersection of the residues of all cliques of size a(v) with tail v in G. Since 0 belongs to the residue of every clique, we have 0 We have m a(v) b(v) c(v) 1 for every v, so it remains to show that ψ is a proper colouring of G.
Suppose for the sake of contradiction that some two vertices u and v of G with ψ(u) = ψ(v) are connected by an edge of G oriented from u to v. Let d ∈ Z p be such that d(u, v) ≡ d (mod p).Since u and v are adjacent in G, we have d = 0. Observe that if X is the residue of a clique with tail v, then (X + d) ∪ {0} is the residue of a clique with tail u.Therefore, since a(u) = a(v), the residue of every clique of size a(v) with tail v must contain −d.Thus −d ∈ B(v) and if X is the residue of a clique of size a(v) with tail v, then X + d is the residue of a clique of the same size with tail u.By combining Lemma 5 and Lemma 6, we have so far proven Theorem 3. Next we extend the construction to non-primes in order to prove Theorem 2.
For every triple of positive integers k, n, p with p prime and p > 2n, we construct a graph G k,n,p by removing the edges uv of G k,p with d(u, v) ≡ ±1, . . ., ±(n − 1) (mod p).We will now determine the clique number and the chromatic number of G k,n,p . .
Theorem 2 now follows from Lemma 7, Lemma 9, and the following theorem of Schur [10] on the gaps between prime numbers.Theorem 10.For every integer n 2, there is a prime p such that 2n < p < 3n.

Lemma 5 .
r(C) since t(C) ∈ C. Furthermore |C| = |r(C)|, otherwise there would exist two distinct vertices u, v ∈ C such that d(t(C), u) ≡ d(t(C), v) (mod p), and so d(u, v) ≡ 0 (mod p), which would contradict the fact that u and v are adjacent.This observation allows us to determine the clique number of G k,p .For every positive integer k and every prime p k, the graph G k,p has clique number p. Proof.Since G k contains a directed path on k vertices and p k, the graph G k,p contains a clique of size p.Conversely, if C is a clique in G k,p , then |C| = |r(C)| |Z p | = p.
Hence B(u) ⊆ B(v) + d, and since b(u) = b(v), we further conclude that B(u) = B(v) + d.Since 0 belongs to the residue of every clique, both B(u) and B(v) are rooted and B(u) ∼ p B(v).Thus B(u) = B(v), as c(u) = c(v).However, since b(u) = b(v) m < p and d = 0, we have B(u) = B(v) + d = B(v), which is a constradiction.This shows that ψ is a proper colouring of G, as desired.

Lemma 7 .Lemma 8 .Lemma 9 . 3 .
Let k, n, and p be positive integers with p prime, p > 2n, and k n.Then G k,n,p has clique number n and chromatic number k.Proof.We have χ(G k ) = χ(G k,p ) = k.Since G k is a subgraph of G k,n,p and G k,n,p is a subgraph of G k,p , it follows that χ(G k,n,p ) = k.Next, we determine the clique number of G k,n,p .LetI = {n, n + 1, . . ., p − n} ⊂ Z p .Notethat uv is an edge of G k,n,p if and only if u < v and d(u, v) / ∈ {0} ∪ I (mod p).Since G k contains a directed path on k vertices and n k, the graph G k,n,p has clique number at least n.It remains to show that G k,n,p has clique number at most n.Let C be a clique in G k,n,p , and let v = t(C).Consider a vertex x ∈ C. Since −I = I (with arithmetic modulo p), we can observe that r(C) must be disjoint fromI + d(v, x).Indeed, if there is a vertex y ∈ C with d(v, y) ∈ I + d(v, x), then either y < x and d(y, x) = d(v, x) − d(v, y) = −(d(v, y) − d(v, x)) ∈ −I = I (mod p), or x < y and d(x, y) = d(v, y) − d(v, x) ∈ I (mod p),and in either case xy is not an edge of G k,m,p , contradicting the fact that C is a clique.Thus the set r(C) is disjoint from x∈C (I + d(v, x)), which implies |r(C)| + | x∈C (I + d(v, x))| p.For every y ∈ r(C) ∩ {1, . . ., n − 1}, we have p − n + y ∈ x∈C (I +d(v, x)), and similarly, for every y ∈ r(C)∩{−1, . . ., −n+1}, we have n+y ∈ x∈C (I +d(v, x)).Therefore, since d(v, x) ≡ d(v, y) (mod p) for any distinct x, y ∈ C, we have x∈C (I + d(v, x)) |I| + |r(C) \ {0}| = |r(C)| + p − 2n.Finally using the fact that |r(C)| = |C|, we conclude that |C| n, as desired.Next we examine the maximum size of a clique in an induced subgraph of G k,n,p that is induced by the vertices of a clique in G k,p .This will allow us to compare the chromatic number of induced subgraphs of G k,p and G k,n,p that have the same vertex set.Let k, n, and p be positive integers with p prime, p > 2n, and k n.Then for every clique C of G k,p , the induced subgraph G k,n,p [C] of G k,n,p contains a clique of size at least n p |C|.Proof.Let C be a clique in G k,p .For each i ∈ Z p , let J i = {i, i + 1, . . ., i + n − 1} ⊂ Z p .Since each i ∈ Z p is contained in exactly n of the p sets J 0 , . . ., J p−1 , by the pigeon-hole principle, there existsi ∈ Z p such that |r(C) ∩ J i | n p |r(C)|.Let C i = {v ∈ C : d(t(C), v) ∈ J i (mod p)}.It follows that |C i | = |r(C) ∩ J i | n p |r(C)| = n p |C|.It remains to show that C i is a clique in G k,n,p .Let x and y be distinct vertices in C i .Since x, y ∈ C, they are adjacent in G k,p , so d(t(C), x) ≡ d(t(C), y) (mod p), and we can assume without loss of generality that x < y.It follows that d(x, y) = d(t(C), y) − d(t(C), x) ∈ {±1, . . ., ±(n − 1)} (mod p), as d(t(C), x), d(t(C), y) ∈ J i (mod p).Hence x and y are adjacent in G k,n,p .We conclude that C i is indeed a clique in G k,n,p .Let k, n, and p be positive integers with p prime, p > 2n, and k n, and let G be an induced subgraph of G k,n,p with m = ω(G) < n.Then χ(G) mp/n +2 Proof.Let G = G k,p [V (G)].Lemma 8 yields ω(G ) mp/n .The fact that G is a subgraph of G and Lemma 6 yield χ(G) χ(G ) mp/n +2 3