Bipartite graphs whose squares are not chromatic-choosable

TL;DR

This paper constructs bipartite graphs whose squares are not chromatic-choosable, providing counterexamples to the conjecture that all such squares are chromatic-choosable, and shows the discrepancy can be arbitrarily large.

Contribution

It presents the first known bipartite graphs with non-chromatic-choosable squares, disproving a conjecture and demonstrating the potential for large differences between chromatic and list chromatic numbers.

Findings

Constructed bipartite graphs with non-chromatic-choosable squares

Showed the difference between chromatic and list chromatic numbers can be arbitrarily large

Counterexamples to the conjecture that all bipartite graph squares are chromatic-choosable

Abstract

The square $G^2$ of a graph $G$ is the graph defined on $V(G)$ such that two vertices $u$ and $v$ are adjacent in $G^2$ if the distance between $u$ and $v$ in $G$ is at most 2. Let $\chi(H)$ and $\chi_{\ell}(H)$ be the chromatic number and the list chromatic number of $H$, respectively. A graph $H$ is called {\em chromatic-choosable} if $\chi_{\ell} (H) = \chi(H)$. It is an interesting problem to find graphs that are chromatic-choosable. Motivated by the List Total Coloring Conjecture, Kostochka and Woodall (2001) proposed the List Square Coloring Conjecture which states that $G^2$ is chromatic-choosable for every graph $G$. Recently, Kim and Park showed that the List Square Coloring Conjecture does not hold in general by finding a family of graphs whose squares are complete multipartite graphs with partite sets of unbounded size. It is a well-known fact that the List Total Coloring Conjecture is true if the List Square Coloring Conjecture holds for special class of bipartite graphs. On the other hand, the counterexamples to the List Square Coloring Conjecture are not bipartite graphs. Hence a natural question is whether $G^2$ is chromatic-choosable or not for every bipartite graph $G$. In this paper, we give a bipartite graph $G$ such that $\chi_{\ell} (G^2) \neq \chi(G^2)$. Moreover, we show that the value $\chi_{\ell}(G^2) - \chi(G^2)$ can be arbitrarily large.