Multi-colouring the Mycielskian of Graphs Wensong Lin ∗ Daphne Der-Fen Liu † Xuding Zhu ‡ December 1, 2016 Abstract A k-fold colouring of a graph is a function that assigns to each vertex a set of k colours, so that the colour sets assigned to adjacent vertices are disjoint. The k-th chromatic number of a graph G, de- noted by χk(G), is the minimum total number of colours used in a k-fold colouring of G. Let µ(G) denote the Mycielskian of G. For any positive integer k, it holds that χ (G)+1 χ (µ(G)) χ (G) + k k ≤ k ≤ k [5]. Although both bounds are attainable, it was proved in [7] that if k 2 and χ (G) 3k 2, then the upper bound can be reduced by 1, ≥ k ≤ − i.e., χ (µ(G)) χ (G)+k 1. We conjecture that for any n 3k 1, k ≤ k − ≥ − there is a graph G with χk(G) = n and χk(µ(G)) = n + k. This is equivalent to conjecturing that the equality χk(µ(K(n, k))) = n + k holds for Kneser graphs K(n, k) with n 3k 1. We confirm this ≥ − conjecture for k =2, 3, or when n is a multiple of k or n 3k2/ ln k. ≥ Moreover, we determine the values of χ (µ(C )) for 1 k q. k 2q+1 ≤ ≤ ∗Department of Mathematics, Southeast University, Nanjing 210096, P.R. China. Sup- ported in part by NSFC under grant 10671033. Email: [email protected]. †Corresponding Author. Department of Mathematics, California State University, Los Angeles, CA 90032, USA. Supported in part by the National Science Foundation under grant DMS 0302456. Email: [email protected]. ‡Department of Applied Mathematics, National Sun Yat-sen University, Kaohsi- ung, and National Center for Theoretical Sciences, Taiwan. Supported in part by the National Science Council under grant NSC95-2115-M-110-013-MY3. Email: [email protected]. 1 1 Introduction In search of graphs with large chromatic number but small clique size, My- cielski [6]introduced the following construction: Let G be a graph with vertex set V and edge set E. Let V be a copy of V , V = x : x V , and let u { ∈ } be a new vertex. The Mycielskian of G, denoted by µ(G), is the graph with vertex set V V u and edge set E0 = E xy : xy E ux : x V . ∪ ∪{ } ∪{ ∈ }∪{ ∈ } The vertex u is called the root of µ(G); and for any x V , x is called the ∈ twin of x. For a graph G, denote χ(G) and ω(G), respectively, the chromatic number and the clique size of G. It is straightforward to verify that for any graph G with ω(G) 2, we have ω(µ(G)) = ω(G) and χ(µ(G)) = χ(G)+1. ≥ Hence, one can obtain triangle free graphs with arbitrarily large chromatic number, by repeatedly applying the Mycielski construction to K2. Multiple-colouring of graphs was introduced by Stahl [10], and has been studied extensively in the literature. For any positive integers n and k, we denote by [n] the set 0, 1,...,n 1 and [n] the set of all k-subsets of [n]. { − } k A k-fold n-colouring of a graph G is a mapping, f : V [n] , such that for → k any edge xy of G, f(x) f(y)= . In other words, a k-fold colouring assigns ∩ ∅ to each vertex a set of k colours, where no colour is assigned to any adjacent vertices. Moreover, if all the colours assigned are from a set of n colours, then it is a k-fold n-colouring. The k-th chromatic number of G is defined as χ (G) = min n : G admits a k-fold n-colouring . k { } The k-fold colouring is an extension of conventional vertex colouring. A 1-fold n-colouring of G is simply a proper n-colouring of G, so χ1(G)= χ(G). 0 It is known [8] and easy to see that for any k, k 1, χ 0 (G) χ (G)+ ≥ k+k ≤ k χ (G) χ 0 (G). This implies k χ(G). The fractional chromatic number of G is k k ≤ 2 defined by χ (G) χ (G) = inf k : k = 1, 2,... f { k } Thus χ (G) χ(G) (cf. [8]). f ≤ For a graph G, it is natural to ask the following two questions: 1. What is the relation between the fractional chromatic number of G and the fractional chromatic number of the Mycielskian of G? 2. What is the relation between the k-th chromatic number of G and the k-th chromatic number of the Mycielskian of G? The first question was answered by Larsen, Propp and Ullman [4]. It turned out that the fractional chromatic number of µ(G) is determined by the frac- tional chromatic number of G: For any graph G, 1 χf (µ(G)) = χf (G)+ . χf (G) The second question is largely open. Contrary to the answer of the first question in the above equality, the k-th chromatic number of µ(G) is not 0 0 determined by χk(G). There are graphs G and G with χk(G)= χk(G ) but χ (µ(G)) = χ (µ(G0)). So it is impossible to express χ (µ(G)) in terms of k 6 k k χk(G). Hence, we aim at establishing sharp bounds for χk(µ(G)) in terms of χ (G). Obviously, for any graph G and any positive integer k, χ (µ(G)) k k ≤ χk(G)+ k. Combining this with a lower bound established in [5] we have: χ (G) + 1 χ (µ(G)) χ (G)+ k. (1) k ≤ k ≤ k Moreover, it is proved in [5] that for any k both the upper and the lower bounds in (1) can be attained. On the other hand, it is proved in [7] that if χk(G) is relatively small compared to k, then the upper bound can be reduced. 3 Theorem 1 [7] If k 2 and χ (G)= n 3k 2, then χ (µ(G)) n+k 1. ≥ k ≤ − k ≤ − In this article, we prove that for graphs G with χk(G) relatively large com- pared to k, then the upper bound in (1) cannot be improved. We conjecture that the condition n 3k 2 in Theorem 1 is sharp. ≤ − Conjecture 1 If n 3k 1, then there is a graph G with χ (G) = n and ≥ − k χk(µ(G)) = n + k. A homomorphism from a graph G to a graph G0 is a mapping f : V (G) → V (G0) such that f(x)f(y) E(G0) whenever xy E(G). If f is a homomor- ∈ ∈ phism from G to G0 and c0 is a k-fold n-colouring for G0, then the mapping defined as c(x)= c0(f(x)) is a k-fold n-colouring of G. Thus χ (G) χ (G0). k ≤ k For positive integers n 2k, the Kneser graph K(n, k) has vertex set ≥ [n] in which x y if x y = . It follows from the definition that a graph k ∼ ∩ ∅ G has a k-fold n-colouring if and only if there is a homomorphism from G to K(n, k). In particular, if k0 = qk for some integer q, then it is easy to 0 show that χk0 (K(n, k)) = qn. If k is not a multiple of k, then determining χk0 (K(n, k)) is usually a difficult problem. The well-known Kneser-Lov´asz Theorem [3] gives the answer to the case for k0 = 1: χ(K(n, k)) = n 2k +2. − 0 For k 2, the values of χ 0 (K(n, k)) are still widely open. ≥ k Notice that, a homomorphism from G to G0 induces a homomorphism from µ(G) to µ(G0). Hence, we have max χ (µ(G)) : χ (G)= n = χ (µ(K(n, k))). { k k } k Therefore, Conjecture 1 is equivalent to Conjecture 2 If n 3k 1, then χ (µ(K(n, k))) = n + k. ≥ − k In this paper, we confirm Conjecture 2 for the following cases: 4 n is a multiple of k (Section 2), • n 3k2/ ln k (Section 2), • ≥ k 3 (Section 3). • ≤ It was proved in [5] that the lower bound in (1) is sharp for complete graphs K with k n. That is, if k n, then χ (µ(K )) = χ (K )+1= n ≤ ≤ k n k n kn + 1. In Section 4, we generalize this result to circular complete graphs Kp/q (Corollary 10). Also included in Section 4 are complete solutions of the k-th chromatic number for the Mycielskian of odd cycles C with k q. 2q+1 ≤ 2 Kneser graphs with large order In this section, we prove for any k, if n = qk for some integer q 3 or ≥ n 3k2/ ln k, then χ (µ(K(n, k))) = n + k. ≥ k In the following, the vertex set of K(n, k) is denoted by V . The Myciel- skian µ(K(n, k)) has the vertex set V V u . For two integers a b, let ∪ ∪{ } ≤ [a,b] denote the set of integers i with a i b. ≤ ≤ Lemma 2 For any positive integer k, χk(µ(K(3k, k))) = 4k. Proof. Suppose to the contrary, χ (µ(K(3k, k))) 4k 1. Let c be a k-fold k ≤ − colouring of µ(K(3k, k)) using colours from the set [0, 4k 2].