English

Dominating $K_t$-Models

Combinatorics 2024-05-24 v1

Abstract

A \textit{dominating K_t-model} in a graph GG is a sequence (T1,,Tt)(T_1,\dots,T_t) of pairwise disjoint non-empty connected subgraphs of GG, such that for 1i<jt1 \leqslant i<j \leqslant t every vertex in TjT_j has a neighbour in TiT_i. Replacing "every vertex in TjT_j" by "some vertex in TjT_j" retrieves the standard definition of KtK_t-model, which is equivalent to KtK_t being a minor of GG. We explore in what sense dominating KtK_t-models behave like (non-dominating) KtK_t-models. The two notions are equivalent for t3t \leqslant 3, but are already very different for t=4t = 4, since the 1-subdivision of any graph has no dominating K4K_4-model. Nevertheless, we show that every graph with no dominating K4K_4-model is 2-degenerate and 3-colourable. More generally, we prove that every graph with no dominating KtK_t-model is 2t22^{t-2}-colourable. Motivated by the connection to chromatic number, we study the maximum average degree of graphs with no dominating KtK_t-model. We give an upper bound of 2t22^{t-2}, and show that random graphs provide a lower bound of (1o(1))tlogt(1-o(1))t\log t, which we conjecture is asymptotically tight. This result is in contrast to the KtK_t-minor-free setting, where the maximum average degree is Θ(tlogt)\Theta(t\sqrt{\log t}). The natural strengthening of Hadwiger's Conjecture arises: is every graph with no dominating KtK_t-model (t1)(t-1)-colourable? We provide two pieces of evidence for this: (1) It is true for almost every graph, (2) Every graph GG with no dominating KtK_t-model has a (t1)(t-1)-colourable induced subgraph on at least half the vertices, which implies there is an independent set of size at least V(G)2t2\frac{\lvert V(G) \rvert}{2t-2}.

Keywords

Cite

@article{arxiv.2405.14299,
  title  = {Dominating $K_t$-Models},
  author = {Freddie Illingworth and David R. Wood},
  journal= {arXiv preprint arXiv:2405.14299},
  year   = {2024}
}
R2 v1 2026-06-28T16:36:49.247Z