Polynomial bounds for chromatic number. I. Excluding a biclique and an induced tree

Let H be a tree. It was proved by Rodl that graphs that do not contain H as an induced subgraph, and do not contain the complete bipartite graph $K_{t,t}$ as a subgraph, have bounded chromatic number. Kierstead and Penrice strengthened this, showing that such graphs have bounded degeneracy. Here we give a further strengthening, proving that for every tree H, the degeneracy is at most polynomial in t. This answers a question of Bonamy, Pilipczuk, Rzazewski, Thomasse and Walczak.


Introduction
The Gyárfás-Sumner conjecture [4,13] asserts: 1.1 Conjecture: For every forest H, there is a function f such that χ(G) ≤ f (ω(G)) for every H-free graph G.
(We use χ(G) and ω(G) to denote the chromatic number and the clique number of a graph G, and a graph is H-free if it has no induced subgraph isomorphic to H.) One attractive feature of this conjecture is that it is best possible in a sense: for every graph H that is not a forest, there is no function f as in 1.1 (this is easily shown with a random graph).The conjecture has been proved for some special families of trees (see, for example, [2,5,6,7,9,10,11]) but remains open in general.
A class C of graphs is χ-bounded if there is a function f such that χ(G) ≤ f (ω(G)) for every graph G that is an induced subgraph of a member of C (see [12] for a survey).Thus the Gyárfás-Sumner conjecture asserts that, for every forest H, the class of all H-free graphs is χ-bounded.Esperet [3] asked whether, for every χ-bounded class, f can always be chosen to be a polynomial.Neither conjecture has been settled in general.
The complete bipartite graph with parts of cardinality s, t is denoted by K s,t .Let us define τ (G) to be the largest t such that G contains K t,t as a subgraph (not necessarily induced).It was proved by Rödl (mentioned in [8], and see also [6]) that the analogue of the Gyárfás-Sumner conjecture is true if we replace ω(G) by τ (G).That is: 1.2 For every forest H, there is a function f such that χ(G) ≤ f (τ (G)) for every H-free graph G.
This has the same attractive feature that the result is best possible (in the same sense).
This result was strengthened by Kierstead and Penrice.Let us say a graph G is d-degenerate (where d ≥ 0 is an integer) if every nonnull subgraph has a vertex of degree at most d; and the degeneracy ∂(G) of G is the smallest d such that G is d-degenerate.Then χ(G) ≤ ∂(G) + 1, and so the following result of Kierstead and Penrice [7] is a strengthening of 1.2: 1.3 For every forest H, there is a function f such that ∂(G) ≤ f (τ (G)) for every H-free graph G.
What about the analogue of Esperet's question: do 1.2 and 1.3 remain true if we require f to be a polynomial in τ (G)?This question was raised by Bonamy, Bousquet, Pilipczuk, Rzazewski, Thomassé and Walczak in [1], and they proved it when H is a path, that is: In this paper we answer the question completely.Our main result is: 1.5 For every forest H, there exists c > 0 such that ∂(G) ≤ τ (G) c for every H-free graph G.
We also look at a related question: what can we say about χ(G) and ∂(G) if G is H-free and does not contain K s,t as a subgraph?More exactly, if H, s are fixed, how do χ(G) and ∂(G) depend on t?We will show that the dependence is in fact linear in t: 1.6 For every forest H and every integer s > 0, there exists c > 0 such that for every graph G and every integer t > 0, if G is H-free and does not contain K s,t as a subgraph, then ∂(G) ≤ ct.
We also prove a weaker result, that under the same hypotheses, χ(G) ≤ ct, and for this the bound on c is a small function of s, H.
Finally, there is a second pretty theorem in the paper [1] of Bonamy, Pilipczuk, Rzazewski, Thomassé and Walczak: 1.7 Let ℓ be an integer; then there exists c > 0 such that ∂(G) ≤ τ (G) c for every graph G with no induced cycle of length at least ℓ.
We give a new proof of this, simpler than that in [1].
In this paper, all graphs are finite and have no loops or parallel edges.We denote by |H| the number of vertices of a graph H.
We use "G-adjacent" to mean adjacent in G, and "G-neighbour" to mean a neighbour in G, and so on.
2 Producing a path-induced rooted tree.
We will prove 1.5 in this section and the next.We need to show that if a graph G has degeneracy at least some very large polynomial in t (independent of G), and does not contain K t,t as a subgraph, then it contains any desired tree as an induced subgraph.We will show this in two stages: in this section we will show that G contains a large (with degrees a somewhat smaller polynomial in t) "path-induced" tree, and in the next section we will convert this to the desired induced tree."Pathinduced" means that each path of the tree starting at the root is an induced path of G; so we should be talking about rooted trees.Let us say this carefully.
A rooted tree (H, r) consists of a tree H and a vertex r of H called the root.A rooted subtree of (H, r) means a rooted tree (J, r) where J is a subtree of H and r ∈ V (J).The height of (H, r) is the length (number of edges) of the longest path of H with one end r.If u, v ∈ V (H) are adjacent and u lies on the path of H between v, r, we say v is a child of u and u is the parent of v.The spread of H is the maximum over all vertices u ∈ V (H) of the number of children of u. (Thus the spread is usually one less than the maximum degree.)Let H be a subgraph of G (not necessarily induced), where (H, r) is a rooted tree.We say that (H, r) is a path-induced rooted subgraph of G if every path of H with one end r is an induced subgraph of G.
Let ζ, η ≥ 1.The rooted tree (H, r) is (ζ, η)-uniform if • every vertex with a child has exactly ζ children; • every vertex with no child is joined to r by a path of H of length exactly η.
Let t, η ≥ 1 and ζ ≥ 2 be integers.Let (T, r) be a (ζ, η)-uniform rooted tree, where T is a subgraph of G.A vertex u ∈ V (G) \ V (T ) is t-bad for (T, r) if there is a vertex w ∈ V (T ), with ζ children in (T, r), such that u is G-adjacent to more than (t − 1)ζ/t of these children.We will often use the following: We omit the proof, which is clear.The second lemma is: Let G be a graph that does not contain K t,t as a subgraph, and let (T, r) be a (ζ, η)-uniform rooted tree, where T is a subgraph of G. Then at most Proof.For each w ∈ V (T ) that has ζ children, let C w be the set of its children in (T, r).Suppose that there are and so at most t(|C w |/t − 1) vertices in C w are G-nonadjacent to one of u 1 , . . ., u t .Consequently at least t vertices in C w are G-adjacent to all of u 1 , . . ., u t , contradicting that G does not contain K t,t as a subgraph.Thus there are at most t − 1 vertices in V (G) \ V (J) with more than |C w |(t − 1)/t G-neighbours in C w .So the number of vertices in V (G) \ V (T ) that are t-bad for (T, r) is at most (t − 1) times the number of vertices of T that have children, and so at most ζ η (t − 1) (since ζ ≥ 2).This proves 2.3.
The main result of this section is the following: 2.4 Let η > 0 be an integer and let c = (η + 1)!.Let ζ ≥ 2, and let (H, r) be a rooted tree of height at most η, and spread at most ζ.Let t ≥ 1 be an integer, and suppose that the graph G does not contain K t,t as a subgraph, and does not contain a rooted tree isomorphic to (H, r) as a path-induced rooted subgraph.Then ∂(G) ≤ (ζt) c .
Proof.We may assume that t ≥ 2. We proceed by induction on η.If η = 1, it follows that G has maximum degree at most ζ − 1, since it does not contain (H, r) as a path-induced rooted subgraph; and so ∂(G) ≤ ζ − 1 ≤ (ζt) c as required.So we may assume that η ≥ 2, and the result holds for all rooted trees with height less than η.Let c ′ = η! and ζ ′ = tζ η+1 .Let us say a limb is a (ζ ′ , η − 1)-uniform rooted tree that is a path-induced rooted subgraph of G.
(1) For each vertex u, there are at most ζ − 1 G-neighbours v of u with the property that there is a limb (J, v) of G such that u / ∈ V (J) and u is not t-bad for (J, v).
Suppose there are ζ such vertices v 1 , . . ., v ζ , and let the corresponding limbs be But then adding u to the union of these trees gives a (ζ, η)-uniform rooted tree, and it is path-induced in G, and contains a rooted induced subgraph isomorphic to (H, r), a contradiction.This proves (2).
Let P be the set of vertices v of G such that there is a limb with root v, and let Q = V (G) \ P .For each v ∈ P , there is at least one limb with root v; select one, and call it (J v , v).For each edge e with at least one end in P , select one such end, and call it the head of e.
• Let A be the set of all edges with both ends in Q; • Let B be the set of all edges uv of G with head v, such that u / ∈ V (J v ), and u is not t-bad for (J v , v); • Let C be the set of all edges uv of G with head v, such that u / ∈ V (J v ), and u is t-bad for (J v , v); • Let D be the set of all edges uv of G with head v, such that u ∈ V (J v ).For each v ∈ P , there are at most ζ η−1 (t − 1) edges uv ∈ C with head v by 2.3, and so

Thus every edge of G belongs to exactly one of
For each v ∈ P , since (J v , v) is path-induced, there are at most ζ ′ edges uv ∈ D with head v, and so Summing, we deduce that and so some vertex of G has degree at most 2 ( Since this also holds for every non-null induced subgraph of G, we deduce that We recall that ζ ′ = tζ η+1 ; and so Of the four terms on the right side, the sum of the second and third is at most the fourth, so and since c ′ ≥ 2, the second term here is at most the first, so (since we may assume that t ≥ 2, and so and 2c ′ + 2 ≤ c, and so ∂(G) ≤ (ζt) c .This proves 2.4.
We remark that 2.4 implies 1.4, and a strengthening:

Growing a tree
If (T, r) is a rooted tree and v ∈ V (T ), the height of v in (T, r) is the number of edges in the path between v, r; and so the height of (T, r) is the largest of the heights of its vertices.Let (T, r) be a rooted tree, and let (S, r) be a rooted subtree.The graph obtained from T by deleting all the edges of S is disconnected, and each of its components contains a unique vertex of S; for each v ∈ V (S), let T v be the component that contains v ∈ V (S).We call the rooted tree (T v , v) the decoration of S at v in T .
Let G be a graph, let (S, r) be a rooted tree, and let ζ ≥ 2 and η ≥ 1.We say that (S, r) is (ζ, η)-decorated in G if S is an induced subgraph of G with height at most η − 1, and there is a rooted tree (T, r) with the following properties: • (S, r) is a rooted subtree of (T, r), and (T, r) is a path-induced rooted subgraph of G; • for each u ∈ V (S) and v ∈ V (T ) \ V (S), if u, v are G-adjacent then they are T -adjacent; Thus, informally, T is obtained from S by attaching to S uniform trees rooted at each vertex of S. Note that T is only required to be path-induced: the various uniform trees that are attached to S might have edges between them.In view of 2.4, if we have a graph G with huge degeneracy that does not contain K t,t , then it contains a (ζ, η)-uniform rooted tree (T, r) as a path-induced rooted subgraph; and consequently there is a one-vertex rooted tree (S, r) that is (ζ, η)-decorated in G.The next result shows that if we start with ζ large enough, then by reducing ζ we can grow S into any larger tree that we wish, and that will prove 1.5.
3.1 Let η, t ≥ 1 and ζ ≥ 2 be integers, let G be a graph that does not contain K t,t as a subgraph, and let (S ′ , r) be a (ζ ′ , η)-decorated rooted tree in G, where ζ ′ ≥ ζ η |S ′ |t η+1 .Let p ∈ V (S ′ ) with height in (S ′ , r) less than η.Then there is a G-neighbour q of p, with q ∈ V (G) \ V (S ′ ), and with no other G-neighbour in V (S ′ ), such that, if S denotes the tree obtained from S ′ by adding q and the edge pq, then (S, r) is a (ζ, η)-decorated rooted tree in G.
Proof.For each v ∈ V (S ′ ), let h(v) denote the height of v in (S ′ , r).Since (S ′ , r) is (ζ ′ , η)-decorated in G, it follows that S ′ is an induced subgraph of G, and there is a rooted tree (T ′ , r) such that • (S ′ , r) is a rooted subtree of (T ′ , r), and (T ′ , r) is a path-induced rooted subgraph of G; For each v ∈ V (S ′ ), let (T v , v) be the decoration of S ′ at v in T ′ .Since T p is (ζ ′ , η −h(p))-uniform, and h(p) < η, it follows that p has ζ ′ children in (T p , p).We need to select one of these children, say q, to add to S ′ , forming S. Any one of them would make a larger induced tree when added to S ′ , since (S ′ , r) is a (ζ, η)-decorated.But in order to make the new rooted tree (S, r) (ζ, η)-decorated, we will delete from T ′ all vertices of T ′ that are G-adjacent and not T ′ -adjacent to q; and doing so must not destroy too much of T ′ .
For each v ∈ V (S ′ ), let (S v , v) be a (tζ, η − h(v))-uniform rooted subtree of (T v , v).By 2.3, there are at most (tζ) η−h(v) (t − 1) < t η+1 ζ η vertices not in V (S v ) that are t-bad for (S v , v), and so there fewer than ζ η |S ′ |t η+1 ≤ ζ ′ children of p in (T p , p) that are t-bad for one of the rooted trees (S v , v) (v ∈ V (S ′ )).Hence there is at least one child q of p in (T p , p) that is t-bad for none of the trees (S v , v) (v ∈ V (S ′ )).Moreover we claim that we can choose q such that q / ∈ V (S p ).This is automatic if (S p , p) has height at least two, since then every child of p in (S p , p) is bad for (S p , p), so we may assume that (S p , p) has height one, that is, h(p) = η − 1. Consequently no child of p in (T p , p) is t-bad for (S p , p), and so the number that are t-bad for one of the rooted trees (S v , v) This proves that we can choose q such that q / ∈ V (S p ).Let Q be the component containing q of the graph obtained from T ′ by deleting V (S); thus (Q, q) is (ζ ′ , η − h(p) − 1)-uniform, and so we may choose a (ζ, η − h(p) − 1)-uniform rooted subtree (R q , q) of (Q, q).Note that q has no neighbours in V (Q) except its neighbours in T ′ , since (T ′ , r) is path-induced.Since q is not t-bad for any of the rooted trees (S v , v) (v ∈ V (S ′ )), it follows by 2.2 that for each v there is a Let S be the tree induced on V (S ′ ) ∪ {q}, and let T be the union of T ′ , the trees R v (v ∈ V (S ′ ) ∪ {q}) and the edge pq.Then S satisfies the theorem, because the tree T exists.This proves 3.1.
We deduce 1.5, which we restate in a strengthened form: For every rooted tree (H, r) with height at most η and spread at most ζ, let c = (η + 3)!|H|; then ∂(G) ≤ (|H|ζt) c for every H-free graph G that does not contain K t,t as a subgraph.
Proof.It suffices to prove the statement for all trees H, and it is helpful to assign a root s to H, so (H, s) is a rooted tree.Choose η ≥ 1 and ζ ≥ 2 such that (H, s) has height at most η and spread at most ζ.Let H have k vertices.Define ζ k = ζ, and for i = k − 1, k − 2, . . ., 1 let ζ i = iζ η i+1 t η+1 .Let G be an H-free graph that does not contain K t,t as a subgraph.Suppose that G contains a one-vertex rooted tree that is (ζ 1 , η)-decorated in G. Choose a maximal rooted subtree (F, s) of (H, s) such that there is a rooted subtree (S, r) of G, isomorphic to (F, s), such that (S, r) is (ζ i , η)decorated in G, where i = |F |.By 3.1, i = k; and so G contains an induced subgraph isomorphic to H, a contradiction.
Thus G contains no one-vertex rooted tree that is (ζ 1 , η)-decorated in G. Hence G contains no (ζ 1 , η)-uniform rooted tree as a path-induced rooted subgraph, and so by 2.4 (applied with (H, r) , and so Consequently In this section we prove 1.6, and before that we prove a weaker statement, with ∂(G) replaced by χ(G).For the latter we need the following lemma: 4.1 Let J be a digraph such that every vertex has outdegree at most k.Then the undirected graph underlying J has chromatic number at most 2k + 1.
Proof.Let G be the undirected graph underlying J. Since every subgraph of G has the property that its edges can be directed so that it has outdegree at most k, it follows that every such subgraph H has at most k|H| edges; and therefore (if it is non-null) has a vertex of degree at most 2k.Consequently G is 2k-degenerate, and so is (2k + 1)-colourable.This proves 4.1.
We use 4.1 to prove the following (which we include here because the proof gives a relatively small constant c, although the fact that some c exists follows from 1.6): 4.2 Let H be a tree and s ≥ 1 an integer, and let c = (2s|H|) s+|H| .Then for every H-free graph G and every integer t ≥ 1, if G does not contain K s,t as a subgraph then χ(G) ≤ ct.
Proof.We will prove this by induction on |H| (for the same value of s).Let H be a tree and s ≥ 0 an integer, and suppose the theorem holds for all smaller trees and the same value of s.We may assume that |H| ≥ 3, since the theorem is true for trees with at most two vertices; let p ∈ V (H) have degree one, and let q be its H-neighbour.Let H ′ be obtained by deleting p from H. Let Let t ≥ 1 be an integer, and let G be an H-free graph not containing K s,t as a subgraph.We will show that χ(G) ≤ ct.Suppose that this is false, and choose a minimal induced subgraph G ′ of G with χ(G ′ ) > ct.It follows that every vertex of G ′ has degree at least ct (since c is an integer).
Let v ∈ V (G ′ ).We say a subset X ⊆ V (G ′ ) \ {v} is a v-bag if there is an isomorphism from H ′ to G[X ∪ {v}] that maps q to v. (Thus each v-bag has cardinality |H| − 2.) Let v ∈ V (G ′ ), and suppose that there are s − 1 pairwise disjoint v-bags, say X 1 , . . ., X s−1 .Since G is H-free, every G-neighbour u of v either belongs to X i or has a G-neighbour in X i , for 1 ≤ i ≤ s − 1.In particular, every G-neighbour u of v not in X 1 ∪ • • • ∪ X s−1 has a G-neighbour in each of X 1 , . . ., X s−1 .But for each choice of x i ∈ X i (1 ≤ i ≤ s − 1) there are at most t − 1 G-neighbours of v G-adjacent to each of x 1 , . . ., x s−1 (since they are also all adjacent to v, and G has no K s,t subgraph).Consequently there are at most (t − 1) Since ct = c + c(t − 1), and (s − 1)(|H| − 2) < c, and (t − 1)(|H| − 2) s−1 ≤ c(t − 1), this contradicts (1); so there is no such choice of X 1 , . . ., X s−1 .
Choose an integer r maximum such that there are r pairwise disjoint v-bags, say X 1 , . . ., X r .
Let J be the digraph with vertex set V (G ′ ) in which every vertex in Y v is J-adjacent from v, for each v ∈ V (G ′ ).Thus J has maximum outdegree at most (s − 2)(|H| − 2), and so by 4.1, the undirected graph J ′ underlying J has chromatic number at most 2(s − 2)(|H| − 2) + 1; and so V (G ′ ) = V (J ′ ) can be partitioned into 2(s − 2)(|H| − 2) + 1 sets each of which is a stable set of J ′ .Let Z be one of these sets.Then G[Z] is H ′ -free (because otherwise there would be a vertex v ∈ Z, and a subset X ⊆ Z \ {v}, and an isomorphism from H ′ to G[X ∪ {v}] mapping q to v, and hence with X ∩ Y v = ∅; but no vertex of Y v belongs to Z, since Z is stable in J ′ ).From the inductive hypothesis, χ(Z) ≤ c ′ t, and hence (1).This proves 4.2.
To prove 1.6, we will need the following strengthening of 1.3, also proved in [7]: 4.3 For every forest H, and every integer s > 0, there is a tree S such that for every H-free graph G, if G contains S as a subgraph, then G contains K s,s as a subgraph.Now we prove 1.6, which we restate: 4.4 For every forest H and every integer s > 0, there exists c > 0 such that for every graph G and every integer t > 0, if G is H-free and does not contain K s,t as a subgraph, then ∂(G) < ct.
Proof.Let S be as in 4.3, and let c = |S| s ; we will show that c satisfies the theorem.Let t > 0 be an integer, and let G be an H-free graph that does not contain K s,t as a subgraph.Suppose that ∂(G) ≥ ct, and choose G minimal with these properties: then every vertex of G has degree at least ct.
(φ 2 , Q 2 ) is a shift of (φ 1 , Q 1 ).Similarly we can choose an infinite sequence (φ i , Q i ) (i = 1, 2, 3 . ..) such that each φ i ∈ A and each (φ, Q i ) is a shift of its predecessor.Let v i be the root of φ i for each i.Then v i , v i+1 , . . ., v i+η are the vertices in order of Q i for each i; and so form an induced path of G. Since G is finite, there exists j > 0 such that v j is adjacent to one of v 1 , . . ., v j−2 ; choose a minimum such value of j, and choose i ≤ j − 2 maximum such that v i , v j are adjacent.Then {v i , . . ., v j } induces a cycle of G of length more than η, a contradiction.
So the second case holds, that is, A i is empty for some i.Choose k + 1 minimum such that A k+1 = ∅.For 1 ≤ i ≤ k let X i be the set of all vertices v such that v is the root of a member of A i and not the root of any member of A i+1 .Thus the sets X 1 , . . ., X k are pairwise disjoint.Let X 0 be the set of vertices that are not the root of any member of A 1 ; so the sets X 0 , . . ., X k form a partition of V (G).For each edge e of G with an end in one of X 1 , . . ., X k , choose i maximum such that e has an end in X i , let v be an end of e in X i , and call v the head of e.For each v ∈ X i , choose φ v ∈ A i with root v. (Thus φ v / ∈ A i+1 from the definition of X i .)Define • A is the set of all edges of G with both ends in X 0 ; • B is the set of all edges uv with head v such that u / ∈ V (φ v ) and u is not bad for φ v ; • C is the set of all edges uv with head v such that u / ∈ V (φ v ) and u is bad for φ v ; • D is the set of all edges uv with head v such that u ∈ V (φ v ).
For each u ∈ V (G), with u ∈ X i say, there do not exist t η neighbours v of u such that uv has head v and belongs to B, since there is no (t, η)-infusion of (H, r) with root u that is derived from members of A i .Hence |B| ≤ t η |G|.
For each v ∈ V (G), there are at most t η η neighbours u of v such that the edge uv has head v and belongs to C, by 5.2; so |C| ≤ t η η |G|.
Finally, for each v ∈ V (G), there are at most t η neighbours u of v such that the edge uv has head v and belongs to D; so |D| ≤ t η |G|.

2. 5
If H is a path, and t ≥ 1 is an integer, and G is H-free and does not contain K t,t as a subgraph, then∂(G) ≤ (2t) |H|! .Proof.Let ζ = 2, and η = |E(H)|.Let r be one end of H. Then G does not contain (H, r) as a path-induced rooted subgraph, and so ∂(G) ≤ (2t) |H|! by 2.4.This proves 2.5.