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face_monochromatic_pairs: reframe Lemma 5.2 as a non-existence result

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The previous statement "Heawood is constant on K through merged" was
strictly stronger than what the proof actually established without
Conjecture 5.3. Restate the lemma in the contrapositive direction:

  If h_phi is constant on V(K), then no edge e in E(K) admits a face
  F of G'^hat and edges e_1, e_2 on dF realising the clause-(3) arc
  of Conjecture 5.1 at the endpoints of e.

Proof structure is mostly preserved (same F_R/F_L geometry, same case
split on phi(e) in {a, b}, same reading-off of cyclic colour orders).
The hypothesis "h_phi(v_0) != h_phi(v_1)" becomes "h_phi(v_0) =
h_phi(v_1)", which flips the conclusion: the same-coloured non-e
edges at v_0, v_1 land on opposite faces of e instead of the same
face. No dependency on Conjecture 5.3 or Theorem 4.X.

Redraw the figure to match the new lemma: both vertices labelled
h_phi = +1, both showing CW order (a, b, c), and the same-colour pair
(b-edges in Case A, a-edges in Case B) drawn on opposite sides of e.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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didericis
2026-05-24 22:31:10 -04:00
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commit 037d987c7d
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papers/face_monochromatic_pairs/paper.tex
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@@ -662,67 +662,64 @@ merged edge, such that:
\end{enumerate}
\end{conjecture}
\begin{lemma}[Heawood number is constant on the Kempe cycles through the merged edge]
\begin{lemma}[A Heawood-constant Kempe cycle does not admit the clause-(3) arc]
\label{lem:kempe-heawood-constant}
Let $G$ be a minimal counterexample to the Four Colour Theorem, fix a
reduced dual $\widehat{G}'_{v,i}$ of $G' = \mathrm{dual}(G)$, and let
$\varphi$ be a proper $3$-edge-colouring of $\widehat{G}'_{v,i}$. Set
$a := \varphi(\mathrm{merged})$. Then for each
$b \in \{1, 2, 3\} \setminus \{a\}$, every vertex of the
$\{a, b\}$-Kempe cycle of $\varphi$ through the merged edge has the same
Heawood number $h_\varphi$.
$a := \varphi(\mathrm{merged})$ and let $K$ be the $\{a, b\}$-Kempe
cycle of $\varphi$ through the merged edge for some
$b \in \{1, 2, 3\} \setminus \{a\}$. If $h_\varphi$ is constant on
$V(K)$, then no edge $e \in E(K)$ admits a face $F$ of
$\widehat{G}'_{v,i}$ and two non-incident edges
$e_1, e_2 \in \partial F$ such that
$\varphi(e_1) = \varphi(e_2)$ and $e$ is the unique edge of
$\partial F$ between $e_1$ and $e_2$ along one of the two arcs of
$\partial F$ --- that is, no edge of $K$ admits the clause-(3) arc of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
with $e_1, e_2$ at its two endpoints.
\end{lemma}
\begin{proof}
Fix $b \in \{1, 2, 3\} \setminus \{a\}$, let $K$ be the $\{a, b\}$-Kempe
cycle of $\varphi$ through the merged edge, and let $c$ be the third
colour. Suppose for contradiction that $h_\varphi$ is not constant on
$V(K)$. Since $K$ is a closed cycle, there exist consecutive vertices
$v_0, v_1 \in V(K)$, joined by an edge $e \in E(K)$, with
$h_\varphi(v_0) \neq h_\varphi(v_1)$. After possibly swapping
$v_0, v_1$, take $h_\varphi(v_0) = +1$ and $h_\varphi(v_1) = -1$. By
Definition~\ref{def:heawood-number}, the clockwise cyclic colour order
at $v_0$ is $(a, b, c)$ (an even cyclic permutation), and at $v_1$ it is
$(a, c, b)$ (an odd one).
Let $c$ be the third colour. Fix any edge $e \in E(K)$ joining
$v_0, v_1 \in V(K)$. By hypothesis $h_\varphi(v_0) = h_\varphi(v_1)$;
after possibly relabelling we may take
$h_\varphi(v_0) = h_\varphi(v_1) = +1$, so by
Definition~\ref{def:heawood-number} the clockwise cyclic colour order
at $v_0$ and at $v_1$ is the same even cyclic class $(a, b, c)$.
Let $F_R, F_L$ be the two faces of $\widehat{G}'_{v,i}$ on the two sides
of $e$, with $F_R$ on the right side as one walks from $v_0$ to $v_1$.
For a vertex $v \in \{v_0, v_1\}$, the non-$e$ edge of $\partial F_R$ at
$v$ is the next-clockwise edge from $e$ around $v_0$ (since at $v_0$ the
right side coincides with the clockwise next edge from $e$) and the
next-counter-clockwise edge from $e$ around $v_1$ (since at $v_1$ the
orientation of $e$ is reversed, so the right side coincides with the
counter-clockwise next edge from $e$).
Let $F_R, F_L$ be the two faces of $\widehat{G}'_{v,i}$ on the two
sides of $e$, with $F_R$ on the right side as one walks from $v_0$ to
$v_1$. For a vertex $v \in \{v_0, v_1\}$, the non-$e$ edge of
$\partial F_R$ at $v$ is the next-clockwise edge from $e$ around $v_0$
(since at $v_0$ the right side coincides with the clockwise next edge
from $e$) and the next-counter-clockwise edge from $e$ around $v_1$
(since at $v_1$ the orientation of $e$ is reversed, so the right side
coincides with the counter-clockwise next edge from $e$).
\emph{Case~A: $\varphi(e) = a$.} The CW order $(a, b, c)$ at $v_0$ makes
the next-CW edge from $e$ the colour-$b$ edge at $v_0$; the CW order
$(a, c, b)$ at $v_1$ makes the next-CCW edge from $e$ the colour-$b$
edge at $v_1$. Let $e_1, e_2$ be these colour-$b$ edges at $v_0$ and
$v_1$ respectively. Then $e_1, e_2 \in \partial F_R$, they are
non-incident (their endpoints other than $v_0, v_1$ are distinct
vertices), and $e$ is the unique edge of $\partial F_R$ lying between
$e_1$ and $e_2$ along one of the two arcs of $\partial F_R$. Both
$e_1, e_2$ lie on $K$ (the colour-$b$ edge at any $K$-vertex is a
$K$-edge), so $e_1, e_2$, and the merged edge are on a common
$\{a, b\}$-Kempe cycle, and $\varphi(e_1) = \varphi(e_2) = b \neq a$
means neither equals the merged edge.
\emph{Case~A: $\varphi(e) = a$.} In the CW order $(a, b, c)$ at $v_0$
the next-CW edge from $e$ has colour $b$; in the same CW order
$(a, b, c)$ at $v_1$ the next-CCW edge from $e$ has colour $c$ (since
CCW-next from $a$ in cyclic order $(a, b, c)$ is $c$). Hence the
non-$e$ edge of $\partial F_R$ at $v_0$ has colour $b$, while the
non-$e$ edge of $\partial F_R$ at $v_1$ has colour $c$ --- these
differ. Symmetrically, the non-$e$ edges of $\partial F_L$ at $v_0$
and $v_1$ have colours $c$ and $b$ respectively, again different.
Hence the colour-$b$ edges at $v_0$ and $v_1$ lie on opposite faces
of $e$, and the same for the colour-$c$ edges; no face of
$\widehat{G}'_{v,i}$ contains two same-coloured non-$e$ edges at
$\{v_0, v_1\}$.
\emph{Case~B: $\varphi(e) = b$.} By the symmetric reasoning applied to
$F_L$, the colour-$a$ edges at $v_0$ and $v_1$ both lie on
$\partial F_L$, with $e$ as the unique edge of $\partial F_L$ between
them on one arc; both lie on $K$, and if neither $v_0$ nor $v_1$ is an
endpoint of the merged edge (which can be arranged by choosing the
differing-Heawood pair $(v_0, v_1)$ appropriately on $K$), neither
colour-$a$ edge equals the merged edge.
\emph{Case~B: $\varphi(e) = b$.} By the analogous reasoning, the
non-$e$ edges of $\partial F_R$ at $v_0$ and $v_1$ have colours $c$
and $a$ respectively, and those of $\partial F_L$ have colours $a$
and $c$. The colour-$a$ edges at $v_0, v_1$ thus lie on opposite
faces of $e$, and so do the colour-$c$ edges.
Either way, the cyclic colour orders at $v_0, v_1$ force a face $F$ of
$\widehat{G}'_{v,i}$ and two non-incident edges $e_1, e_2 \in \partial
F$, both of the same colour and both on $K$ together with the merged
edge, such that $e_1$ and $e_2$ lie on a common arc of $\partial F$ with
exactly one edge of $\partial F$ between them.
The triple $(F, e_1, e_2)$ then satisfies clauses~(1)--(3) of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.
In either case, no face $F$ of $\widehat{G}'_{v,i}$ has two same-coloured
non-$e$ edges at $\{v_0, v_1\}$ on $\partial F$, so the clause-(3) arc
(with $e$ as the unique $\partial F$-edge between $e_1$ and $e_2$ at the
endpoints of $e$) cannot be realised.
\end{proof}
\begin{figure}[h]
@@ -731,15 +728,17 @@ Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.
\caption{The two cases in the proof of
Lemma~\ref{lem:kempe-heawood-constant}. Vertices $v_0, v_1$ are
consecutive on the $\{a, b\}$-Kempe cycle $K$, joined by an edge $e$,
with $h_\varphi(v_0) = +1$ (clockwise colour order $(a, b, c)$) and
$h_\varphi(v_1) = -1$ (clockwise order $(a, c, b)$). \emph{Left
(Case~A):} when $\varphi(e) = a$, the two $b$-edges at $v_0, v_1$ lie on
the same face $F$, with $e$ as the unique $\partial F$-edge between
them. \emph{Right (Case~B):} when $\varphi(e) = b$, the two $a$-edges at
$v_0, v_1$ lie on the opposite face $F$ instead, again with $e$ between
them on one arc. In either case $(F, e_1, e_2)$ witnesses clauses
(1)--(3) of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}.}
with the lemma's hypothesis $h_\varphi(v_0) = h_\varphi(v_1) = +1$ ---
so both vertices share the clockwise colour order $(a, b, c)$.
\emph{Left (Case~A):} when $\varphi(e) = a$, the colour-$b$ edge at
$v_0$ lies south of $e$ (on $\partial F_R$) and the colour-$b$ edge at
$v_1$ lies north of $e$ (on $\partial F_L$); the two would-be witness
edges are on opposite faces, so no face of $\widehat{G}'_{v,i}$ contains
both. \emph{Right (Case~B):} when $\varphi(e) = b$, the colour-$a$
edges at $v_0, v_1$ are likewise on opposite sides of $e$. In either
case the clause-$(3)$ arc of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
cannot be realised at $e$.}
\label{fig:lemma-kempe-heawood}
\end{figure}