didericis
0303225f39
- Name the edges of the reduced-dual construction (merged, spike, sides)
via a new definition; use these names in lem:chord-apex.
- Add lem:pentagonal-externals with full exhaustive proof: any proper
3-edge-colouring near a pentagonal face of a cubic plane graph has its
five external edges forming, up to cyclic rotation, the pattern
(a, b, c, c, c) with {a, b, c} = {1, 2, 3} (iff).
- Cite the new lemma in the chord-apex proof scaffold as the lifting step.
- Remove the icosahedron experimental remark.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
301 lines
13 KiB
TeX
301 lines
13 KiB
TeX
%% filename: amsart-template.tex
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%% American Mathematical Society
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%% AMS-LaTeX v.2 template for use with amsart
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%% ====================================================================
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\documentclass{amsart}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{proposition}[theorem]{Proposition}
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\theoremstyle{definition}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{example}[theorem]{Example}
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\theoremstyle{remark}
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\newtheorem{remark}[theorem]{Remark}
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\numberwithin{equation}{section}
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\begin{document}
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\title{Dual Decomposition of Minimal Counterexamples}
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% author one information
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\author{Eric Bauerfeld}
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\address{}
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\curraddr{}
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\email{}
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\thanks{}
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\subjclass[2010]{Primary }
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\keywords{four colour theorem, plane triangulation, dual graph, cubic planar
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graph, edge connectivity, cyclic edge cut, Tait colouring, $3$-edge-colouring}
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\date{}
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\dedicatory{}
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\begin{abstract}
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% TODO: abstract.
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\end{abstract}
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\maketitle
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\section{The minimal counterexample}
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Throughout, a \emph{triangulation} is a simple plane graph, with a fixed
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embedding, in which every face --- including the outer face --- is bounded by a
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triangle. We first reduce to triangulations, then record the degree properties a
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smallest counterexample must have.
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\begin{lemma}[Reduction to triangulations]
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\label{lem:triangulate}
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If every triangulation is properly $4$-vertex-colourable, then so is every plane
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graph.
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\end{lemma}
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\begin{proof}
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Let $H$ be a plane graph. Add edges to $H$, maintaining planarity, until no
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further edge can be added; the result is a triangulation $H^{+}$ on the same
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vertex set with $E(H) \subseteq E(H^{+})$. A proper $4$-colouring of $H^{+}$
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restricts to a proper $4$-colouring of $H$, since every edge of $H$ is an edge of
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$H^{+}$.
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\end{proof}
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By Lemma~\ref{lem:triangulate}, if the Four Colour Theorem fails then it fails for
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some triangulation. We may therefore make the following assumption.
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\begin{definition}[Minimal counterexample]
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\label{def:minimal}
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Let $G$ be a triangulation on the fewest vertices that admits no proper
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$4$-vertex-colouring. We call $G$ a \emph{minimal counterexample}. By minimality,
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every triangulation on fewer than $|V(G)|$ vertices is properly
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$4$-colourable.
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\end{definition}
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\begin{remark}
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Since every triangulation on at most four vertices is properly $4$-colourable
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(the largest being $K_4$), a minimal counterexample has $|V(G)| \ge 5$; the degree
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bound below sharpens this to $|V(G)| \ge 12$.
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\end{remark}
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\begin{lemma}[Minimum degree]
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\label{lem:mindeg}
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A minimal counterexample $G$ has minimum degree $\delta(G) \ge 5$.
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\end{lemma}
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\begin{proof}
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Suppose some vertex $v$ has $\deg(v) = d \le 4$.
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If $d \le 3$, let $G' = G - v$. Then $G'$ is a plane graph on fewer vertices, so
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by Definition~\ref{def:minimal} and Lemma~\ref{lem:triangulate} it has a proper
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$4$-colouring. The at most three neighbours of $v$ use at most three colours, so
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a fourth colour is free for $v$, extending the colouring to $G$ --- a
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contradiction.
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If $d = 4$, again $4$-colour $G - v$. If the four neighbours of $v$ use at most
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three colours we extend as before, so assume they receive all four colours; let
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$v_1, v_2, v_3, v_4$ be the neighbours in cyclic order around $v$, coloured
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$1,2,3,4$. Consider the subgraph induced by the colour classes $1$ and $3$, and
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let $K$ be its connected component containing $v_1$. If $v_3 \notin K$, swap
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colours $1$ and $3$ on $K$; now no neighbour of $v$ is coloured $1$, freeing it
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for $v$. If $v_3 \in K$, then a $1$--$3$ Kempe chain joins $v_1$ to $v_3$, and
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this chain together with $v$ encloses exactly one of $v_2, v_4$; hence the
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$2$--$4$ component containing $v_2$ cannot also reach $v_4$, and swapping colours
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$2$ and $4$ on it frees colour $2$ for $v$. Either way the colouring extends to
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$G$, a contradiction.
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Hence $\delta(G) \ge 5$.
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\end{proof}
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\section{The reduced dual}
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Write $G'$ for the dual of $G$: since $G$ is a triangulation, $G'$ is a cubic
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plane graph in which each vertex of $G$ corresponds to a face of $G'$, each face
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of $G$ to a vertex of $G'$, and each edge to a dual edge. A vertex of $G$ of
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degree $k$ corresponds to a $k$-gonal face of $G'$.
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By Lemma~\ref{lem:mindeg}, $\delta(G) \ge 5$, and Euler's formula gives
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$\sum_{u \in V(G)}(6 - \deg u) = 12$, so $G$ has a vertex of degree exactly $5$
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(indeed at least twelve). Fix such a vertex $v$. Its dual face $F_v$ is a
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pentagon, bounded by the five dual vertices corresponding to the five faces of
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$G$ incident to $v$.
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\begin{definition}[Reduced dual]
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\label{def:reduced-dual}
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Let $v$ be a degree-$5$ vertex of $G$ with pentagonal dual face $F_v$, and fix an
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index $i \in \{0,1,2,3,4\}$. The \emph{reduced dual} $\widehat{G}'_{v,i}$ is the
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plane graph obtained from $G'$ as follows.
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\begin{enumerate}
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\item Delete the five dual vertices on the boundary of $F_v$, together with all
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edges incident to them. Each deleted vertex is cubic, with two edges on
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$\partial F_v$ and one edge leaving $F_v$; deleting the five boundary
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vertices therefore removes the five external edges as well, dropping their
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five outer endpoints from degree $3$ to degree $2$. These five degree-$2$
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vertices lie on the boundary of a single face $F$ of the resulting graph.
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\item List the five degree-$2$ vertices in clockwise order around $F$ as
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$A = (A_0, A_1, A_2, A_3, A_4)$.
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\item Add a new vertex $v_n$ and join it to $A_i$, $A_{i+1}$, and $A_{i+2}$
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(indices mod $5$) by three new edges.
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\item Add a new edge between $A_{i+3}$ and $A_{i+4}$ (indices mod $5$).
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\end{enumerate}
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\end{definition}
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\begin{remark}
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Steps (3) and (4) restore cubicity: $A_i, A_{i+1}, A_{i+2}$ each gain one edge to
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$v_n$ and $A_{i+3}, A_{i+4}$ each gain the new edge, so all five return to degree
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$3$, and $v_n$ has degree $3$. Since $A_i,\dots,A_{i+2}$ and $A_{i+3}, A_{i+4}$
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are each consecutive along $\partial F$, the new vertex and edge can be drawn
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inside $F$ without crossings, so $\widehat{G}'_{v,i}$ is again a cubic plane
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graph. The construction depends on the choice of $i$ up to the rotational
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symmetry of $A$.
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\end{remark}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step1.png}\hfill
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\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step2.png}\\[0.5em]
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\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step3.png}\hfill
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\includegraphics[width=0.48\textwidth]{fig_reduced_dual_step4.png}
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\caption{The four steps of Definition~\ref{def:reduced-dual}, illustrated on
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$G' = $ the dodecahedron (dual of the icosahedron) with $F_v$ the inner
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pentagon and $i = 0$. Top left: delete the five boundary vertices of $F_v$,
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leaving five degree-$2$ vertices on a new face $F$. Top right: order them
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clockwise as $A_0,\dots,A_4$. Bottom left: add $v_n$ joined to $A_0, A_1, A_2$.
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Bottom right: add the chord $A_3 A_4$, giving the cubic plane graph
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$\widehat{G}'_{v,0}$.}
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\label{fig:reduced-dual-steps}
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\end{figure}
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\begin{definition}[Edges of the reduced dual]
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\label{def:edge-names}
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The four edges added in steps (3) and (4) of Definition~\ref{def:reduced-dual}
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are named as follows. The chord $A_{i+3}A_{i+4}$ is the \emph{merged edge}; the
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edge $A_{i+1}v_n$ is the \emph{spike edge}; and the edges $A_iv_n$ and
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$A_{i+2}v_n$ are the \emph{side edges}. In the $i = 0$ case of
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Figure~\ref{fig:reduced-dual-steps} these are $\{A_3, A_4\}$, $\{A_1, v_n\}$,
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and $\{A_0, v_n\}, \{A_2, v_n\}$ respectively.
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\end{definition}
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We will use the following structural fact about proper $3$-edge-colourings near
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a pentagonal face of a cubic plane graph; it is stated for a generic such graph
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$H$, not specifically for the reduced dual.
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\begin{lemma}[Pentagonal externals]
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\label{lem:pentagonal-externals}
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Let $H$ be a cubic plane graph and $F$ a pentagonal face of $H$, with
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$\partial F$ traversed clockwise as $u_0, u_1, u_2, u_3, u_4$. For each $i$
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let $f_i$ be the unique edge of $H$ incident to $u_i$ that does not lie on
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$\partial F$. An assignment $\varphi$ of colours from $\{1, 2, 3\}$ to the ten
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edges incident to $\{u_0, \dots, u_4\}$ is proper at every $u_i$ if and only if
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there is some index $j$ such that
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\[
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\varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2})
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\quad\text{and}\quad
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\{\varphi(f_{j+3}), \varphi(f_{j+4})\}
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= \{1, 2, 3\} \setminus \{\varphi(f_j)\},
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\]
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indices mod $5$.
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\end{lemma}
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\begin{proof}
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Write $e_i = u_i u_{i+1}$ for the boundary edges of $\partial F$ (indices mod
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$5$). A colouring $\varphi$ is proper at every $u_i$ if and only if at each
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$u_i$ the three incident edges $e_{i-1}, e_i, f_i$ receive three distinct
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colours; whenever this holds, $\varphi(f_i)$ is forced to be the unique colour
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in $\{1, 2, 3\} \setminus \{\varphi(e_{i-1}), \varphi(e_i)\}$, and $\varphi$
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restricts to a proper $3$-edge-colouring of the cycle $\partial F$.
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\textbf{($\Rightarrow$)} The line graph of $\partial F$ is $C_5$, whose
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maximum independent set has size $2$, so no colour appears more than twice on
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$\partial F$; and since $\partial F$ is an odd cycle, all three colours appear.
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The colour multiset on $(\varphi(e_0), \dots, \varphi(e_4))$ is therefore
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$(2, 2, 1)$, with the singleton at a unique position. Cyclically shifting
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indices we may place this position at $0$; let $c$ be the singleton colour.
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The remaining four edges form the path $e_1 e_2 e_3 e_4$, which by propriety
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alternates between the other two colours, so for some labelling
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$\{a, b, c\} = \{1, 2, 3\}$,
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\[
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(\varphi(e_0), \varphi(e_1), \varphi(e_2), \varphi(e_3), \varphi(e_4))
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= (c, a, b, a, b).
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\]
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Reading off the forced values of $\varphi(f_i)$,
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\[
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\varphi(f_0) = a, \quad
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\varphi(f_1) = b, \quad
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\varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c,
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\]
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which is the lemma's pattern at $j = 2$ (the cyclic shift maps this back to
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the corresponding $j$ in the original indexing). This case is the unique
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proper $3$-edge-colouring of $\partial F$ up to cyclic shift and permutation of
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$\{1, 2, 3\}$ (since $5 \cdot 3! = 30 = P(C_5, 3)$, the chromatic polynomial of
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$C_5$ at $3$), so it exhausts every proper $\varphi$.
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\textbf{($\Leftarrow$)} The lemma's hypothesis is invariant under cyclic
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shifts of indices and under permutations of $\{1, 2, 3\}$, so we may assume
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$j = 2$, $\varphi(f_2) = \varphi(f_3) = \varphi(f_4) = c$, $\varphi(f_0) = a$,
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and $\varphi(f_1) = b$, with $\{a, b, c\} = \{1, 2, 3\}$. Propriety at $u_i$
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and $u_{i+1}$ requires $\varphi(e_i) \notin \{\varphi(f_i), \varphi(f_{i+1})\}$,
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which gives
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\[
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\varphi(e_0) = c, \quad
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\varphi(e_1) = a, \quad
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\varphi(e_2) \in \{a, b\}, \quad
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\varphi(e_3) \in \{a, b\}, \quad
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\varphi(e_4) = b.
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\]
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The remaining propriety condition $\varphi(e_{i-1}) \neq \varphi(e_i)$ holds
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automatically at $u_0, u_1, u_4$, forces $\varphi(e_2) = b$ at $u_2$, and then
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forces $\varphi(e_3) = a$ at $u_3$. The resulting triples
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$(\varphi(e_{i-1}), \varphi(e_i), \varphi(f_i))$ at $u_0, u_1, u_2, u_3, u_4$
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are
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\[
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(b, c, a), \quad (c, a, b), \quad (a, b, c), \quad (b, a, c), \quad (a, b, c),
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\]
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each a permutation of $\{1, 2, 3\}$, so $\varphi$ is proper at every $u_i$.
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\end{proof}
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\begin{remark}
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The two-element condition $\{\varphi(f_{j+3}), \varphi(f_{j+4})\}
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= \{1,2,3\}\setminus\{\varphi(f_j)\}$ cannot be dropped: a 3-colouring
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satisfying $\varphi(f_j) = \varphi(f_{j+1}) = \varphi(f_{j+2})$ alone need not
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extend, e.g.\ $(1,1,1,1,2)$.
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\end{remark}
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Since $\widehat{G}'_{v,i}$ is the dual of a triangulation on fewer vertices than
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$G$, it is $3$-edge-colourable by the minimality of $G$. The following lemma
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constrains every such colouring.
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\begin{lemma}
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\label{lem:chord-apex}
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Let $G$ be a minimal counterexample, and let $\widehat{G}'_{v,i}$ be a reduced
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dual of its dual $G'$. Then in every proper $3$-edge-colouring of
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$\widehat{G}'_{v,i}$, the merged edge and the spike edge receive the same
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colour.
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\end{lemma}
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\begin{proof}
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% TODO. Intended argument: given a proper 3-edge-colouring of
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% \widehat{G}'_{v,i}, the colour at each A_k of the new edge there (one of the
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% three v_n-edges for k in {i, i+1, i+2}, or the chord for k in {i+3, i+4})
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% determines the colour of f_k in any lift to G' --- it is the unique colour
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% not used by the two original edges at A_k. Because the chord has a single
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% colour, f_{i+3} and f_{i+4} agree. If the merged edge and the spike edge
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% receive distinct colours, the three v_n-edges contribute three distinct
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% colours to f_i, f_{i+1}, f_{i+2}, and the resulting f-vector has shape
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% (X, Y, Z, W, W) with X, Y, Z distinct and W \neq Y. By
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% Lemma~\ref{lem:pentagonal-externals} applied to G' at the face F_v, this
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% extends to a proper 3-edge-colouring of G' --- contradicting that G' (the
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% dual of the minimal counterexample G) is not 3-edge-colourable. Hence the
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% merged edge and the spike edge must share a colour.
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\end{proof}
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\end{document}
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