coloring_nested_tire_graphs: replace partial-dual Tait prop with complete-tire-dual variant; verify on octahedron

The previous Proposition (Tait correspondence on partial tire dual)
stated equality between non-equivalent 4-vertex-colorings of T and
non-equivalent 3-edge-colorings of D(T).  This is wrong as
empirically verified on the octahedron (n=m=3, O=C_3, spoke-only):
  - Octahedron: 96 4-vertex-colorings -> 4 classes mod S_4.
  - Partial tire dual C_6 ∘ K_1: 66 3-edge-colorings -> 11 classes
    mod S_3.

Replaces that proposition with a variant on the COMPLETE tire dual
D*(T) that incorporates non-annular constraints:

  Definition 1.13 (Complete tire dual):  Quotient D(T)'s leaves into
    non-annular-face vertices.  Outer leaves merge into a single
    outer-face vertex v_out of degree n; for each bounded face F of
    O interior to B_in, the corresponding inner leaves merge into
    v_F of degree |F|.  Equivalently, D*(T) is the planar dual of T.

  Proposition 1.14 (Tait correspondence on complete tire dual): the
    number of non-equivalent 4-vertex-colorings of T (mod S_4) equals
    the number of non-equivalent Tait colorings of D*(T) (mod S_3).
    A Tait coloring is an edge labelling by the three nonzero elements
    of Z_2 x Z_2 with XOR-to-0 at every vertex of D*(T).

  Remark 1.16 (octahedron verification): For octahedron tire,
    D*(T) is the cube Q_3.  Octahedron has 4 vertex-coloring classes;
    Q_3 has 24 proper 3-edge-colorings -> 4 Tait-coloring classes.
    Empirically verified via Sage:
      - chromatic_polynomial(octahedron)(4) = 96
      - chromatic_polynomial(L(Q_3))(3) = 24

The partial tire dual definition (Def 1.7) and its corona-graph
structure proposition (Prop 1.8) are unchanged.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/coloring_nested_tire_graphs/paper.aux

@@ -15,11 +15,10 @@
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 \citation{bauerfeld-pds}
 \citation{bauerfeld-pds}
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 \newlabel{rem:tire-component-degenerate}{{1.11}{7}}
 \newlabel{rem:tire-no-extra-hypotheses}{{1.12}{7}}
-\newlabel{prop:tait-tire}{{1.13}{7}}
-\newlabel{rem:tait}{{1.14}{7}}
+\newlabel{def:complete-tire-dual}{{1.13}{7}}
+\citation{Tait1880}
 \bibcite{Tait1880}{1}
 \bibcite{bauerfeld-pds}{2}
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@@ -27,5 +26,8 @@
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 \newlabel{tocindent2}{0pt}
 \newlabel{tocindent3}{0pt}
+\newlabel{prop:tait-tire-complete}{{1.14}{8}}
+\newlabel{rem:tait-construction}{{1.15}{8}}
+\newlabel{rem:tait-octahedron}{{1.16}{8}}
 \@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{8}{}\protected@file@percent }
 \gdef \@abspage@last{8}

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papers/coloring_nested_tire_graphs/paper.tex

@@ -487,25 +487,71 @@ boundary cycle (the link of $v_0$); the corresponding tire graph has
 degenerate outer boundary $\{v_0\}$.
 \end{remark}
 
-\begin{proposition}[Tait correspondence on the partial tire dual]
-\label{prop:tait-tire}
-The number of non-equivalent proper $4$-vertex-colorings of a tire
-graph $T$ (modulo permutation of the four colors) equals the number
-of non-equivalent proper $3$-edge-colorings of its partial tire dual
-$D(T)$ (modulo permutation of the three colors).
+\begin{definition}[Complete tire dual]
+\label{def:complete-tire-dual}
+The \emph{complete tire dual} $D^{\ast}(T)$ of a tire graph $T$ is
+obtained from the partial tire dual $D(T)$ by quotienting its
+leaves into non-annular-face vertices:
+\begin{itemize}
+  \item Replace the $n$ outer leaves $\{\ell_e^{\mathrm{out}} :
+        e \in E(B_{\mathrm{out}})\}$ by a single \emph{outer-face
+        vertex} $v_{\mathrm{out}}$ of degree $n$.  Each former leaf
+        edge $\{d_f, \ell_e^{\mathrm{out}}\}$ becomes an edge
+        $\{d_f, v_{\mathrm{out}}\}$ of $D^{\ast}(T)$.
+  \item For each bounded face $F$ of $O$ interior to $B_{\mathrm{in}}$,
+        replace the $|F|$ inner leaves whose edges lie on $\partial F$
+        by a single \emph{inner-face vertex} $v_F$ of degree $|F|$,
+        in the same way.
+\end{itemize}
+Equivalently, $D^{\ast}(T)$ is the planar dual of $T$: one dual
+vertex per face of $T$ (annular triangle, outer face, or bounded
+interior face of $O$), one dual edge per edge of $T$.
+\end{definition}
+
+\begin{proposition}[Tait correspondence on the complete tire dual]
+\label{prop:tait-tire-complete}
+Let $T$ be a tire graph.  Then the number of non-equivalent proper
+$4$-vertex-colorings of $T$ (modulo permutation of the four colors)
+equals the number of non-equivalent \emph{Tait colorings} of
+$D^{\ast}(T)$ (modulo permutation of the three nonzero elements of
+$\mathbb{Z}_2 \times \mathbb{Z}_2$), where a Tait coloring is an
+edge-labelling by the three nonzero elements of
+$\mathbb{Z}_2 \times \mathbb{Z}_2$ such that at every vertex of
+$D^{\ast}(T)$ the XOR of incident labels vanishes.
 \end{proposition}
 
 \begin{remark}
-\label{rem:tait}
-Proposition~\ref{prop:tait-tire} is the tire-graph analogue of Tait's
-classical correspondence~\cite{Tait1880}: identifying the four colors
-with the elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$, the XOR of
-the two endpoint colors of an edge of $T$ lies in the three nonzero
-elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$ and assigns a proper
-$3$-edge-coloring to the corresponding edge of $D(T)$.  The annular
-triangles of $T$, encoded as the degree-$3$ vertices $d_f$ of $D(T)$,
-contribute the requirement that each $d_f$'s three incident edges
-carry three distinct colors.
+\label{rem:tait-construction}
+The correspondence is the classical Tait XOR map~\cite{Tait1880}:
+identifying the four colors with $\mathbb{Z}_2 \times \mathbb{Z}_2$,
+each edge of $T$ receives the XOR of its endpoint colors, which is
+nonzero and so a Tait color.  At any degree-$3$ vertex $d_f$ of
+$D^{\ast}(T)$ (an annular triangle of $T$), the XOR-to-$0$ condition
+forces the three incident edge colors to be distinct, recovering
+proper $3$-edge-coloring at $d_f$.  At the higher-degree dual
+vertices $v_{\mathrm{out}}$ (degree $n$) and $v_F$ (degree $|F|$),
+the XOR-to-$0$ condition is the non-trivial non-annular consistency
+constraint absent from the partial tire dual $D(T)$.
+\end{remark}
+
+\begin{remark}
+\label{rem:tait-octahedron}
+For the octahedron viewed as a tire graph with $n = m = 3$,
+$O = C_3$ (no chords), and the spoke-only annular triangulation,
+$D^{\ast}(T)$ is the cube $Q_3$: the six $d_f$ vertices form a
+$6$-cycle, $v_{\mathrm{out}}$ is connected to the three O-move
+$d_f$'s and $v_{\mathrm{in}}$ to the three I-move $d_f$'s, giving a
+$3$-regular bipartite graph on $8$ vertices.  The octahedron has
+$96$ proper $4$-vertex-colorings, hence $96 / 24 = 4$ equivalence
+classes modulo $S_4$; the cube has $24$ Tait colorings ($= 96 / 4$
+by Tait), hence $24 / 6 = 4$ equivalence classes modulo $S_3$.  The
+counts match, illustrating
+Proposition~\ref{prop:tait-tire-complete}.  In contrast, the partial
+tire dual $D(T) \cong C_6 \circ K_1$ has $2^6 + 2 = 66$ proper
+$3$-edge-colorings, i.e.\ $11$ classes modulo $S_3$, exceeding the
+four vertex-coloring classes of $T$; the excess is precisely the
+edge-colorings of $D(T)$ that violate the non-annular
+XOR-to-$0$ constraints captured only in $D^{\ast}(T)$.
 \end{remark}
 
 \begin{thebibliography}{9}