\documentclass[11pt]{article}
\usepackage{amsmath,amssymb,amsthm}
\usepackage{graphicx}
\usepackage{geometry}
\usepackage{caption}
\usepackage{subcaption}
\geometry{margin=1in}

\title{Fiber decomposition of edge $3$-colorings\\
       of tire annular face connectors}
\author{}
\date{}

\newtheorem*{prop}{Proposition}
\newtheorem*{defn}{Definition}

\begin{document}
\maketitle

\section*{Why this note exists}

The menagerie note gives a closed form (Section~6) for $P_e(G, k)$ when
$G$ is a $2$-connected outerplanar graph with $\Delta \le 3$: a polygon
$C_n$ with a non-crossing chord matching $M$.  But the object we
actually want to count over, the \emph{tire annular face connector}
$T'_{f'}$ (Definition~1.16 of the main paper), is not that graph --- it
has extra pendant edges hanging off the cycle, one per non-chord
boundary vertex.  These pendants are the \emph{spokes}
(Definition~1.17).

This note explains the picture below: $T'_{f'}$ factors as a
\emph{core} (the polygon-with-matching $C_n + M = T'_{\mathrm{ann}}$
that the menagerie handles) plus a set of \emph{spoke pendants} reaching
into the neighbouring tires.  Each edge $3$-coloring of $T'_{f'}$ is
the data of (i) a core edge coloring and (ii) a spoke coloring, and the
spoke coloring is what compatibility-with-neighbours acts on.

\begin{figure}[h]
\centering
\includegraphics[width=0.92\textwidth]{fig_fiber_setup.png}
\caption{Anatomy of a tire annular face connector $T'_{f'}$ in the
spoke-only case.  \textbf{Left:} the core $T'_{\mathrm{ann}} = C_n + M$ (here
$\theta(1,3,3)$), the object the menagerie \S6 formula counts.
\textbf{Right:} $T'_{f'}$ adds one pendant spoke edge for each non-chord
boundary vertex; the spoke vertex is a dual vertex of a face of $G$
\emph{outside} the tire annulus.  In a nested-tire setting, that
outside face belongs to the next tire over.}
\label{fig:fiber-setup}
\end{figure}

\section*{Setup}

Let $T \subseteq G$ be a tire with annular faces $F_{\mathrm{ann}}$,
let $T'_{\mathrm{ann}} := G'[\{d_f : f \in F_{\mathrm{ann}}\}]$ be the
tire annular subgraph, fix a face $f'$ of $T'_{\mathrm{ann}}$ with
boundary walk $V(f')$, and let $T'_{f'}$ be the tire annular face
connector (Definition~1.16).  Write
\[
   S(f') \;:=\; V(T'_{f'}) \setminus V(f')
   \;=\; V_{\mathrm{out}}(T'_{f'}) \sqcup V_{\mathrm{in}}(T'_{f'})
\]
for the set of \emph{spokes} (outer and inner).  In the spoke-only
setting (Remark~1.16.1) each spoke $u \in S(f')$ contributes a single
pendant edge $vu$ to $T'_{f'}$ for a unique cycle vertex
$v = v(u) \in V(f')$.  The \emph{spoke edge set} is
\[
   E_S \;:=\; \{\, v(u)\,u \,:\, u \in S(f') \,\} \;\subseteq\; E(T'_{f'}),
\]
and $E(T'_{f'}) = E(T'_{\mathrm{ann}}) \sqcup E_S$.

\section*{Spoke configurations and fibers}

\begin{defn}[Spoke configuration]
A \emph{spoke configuration} on $T'_{f'}$ (with palette $[k]$) is a map
$\sigma : E_S \to [k]$.  Equivalently, $\sigma$ assigns a color to the
$G'$-edge from each spoke $u \in S(f')$ to its attachment vertex
$v(u) \in V(f')$.
\end{defn}

\begin{defn}[Fiber]
For $\sigma : E_S \to [k]$ the \emph{fiber} of $T'_{f'}$ over $\sigma$
is
\[
   N(T'_{f'};\,\sigma) \;:=\; \#\bigl\{\,\text{proper edge $k$-colorings of }
       T'_{f'} \text{ that restrict to } \sigma \text{ on } E_S\,\bigr\}.
\]
\end{defn}

Trivially
\[
   \boxed{\;P_e(T'_{f'},\,k) \;=\; \sum_{\sigma : E_S \to [k]}
                                    N(T'_{f'};\,\sigma)\;}
\]
and the support of the sum is the set of \emph{realisable} spoke
configurations.

\paragraph{What the fiber count really is.}
Fixing $\sigma$ pins one extra color at each cycle vertex $v(u)$: the
spoke edge $v(u)u$ contributes a forbidden color
$\sigma(v(u)u)$ at $v(u)$.  After this, the remaining problem is to
proper-edge-color the core $T'_{\mathrm{ann}} = C_n + M$ subject to
vertex-level forbidden colors --- precisely the constrained transfer
matrix problem in the ``General method'' paragraph of menagerie \S6,
with the chord-color part of the forbidden-set assignment replaced by
the spoke-color part.

If we write $C^{(v)}_\sigma \subseteq [k]$ for the set of forbidden
colors at vertex $v$ coming from spokes (so $|C^{(v)}_\sigma|$ is the
number of spokes attached at $v$; in the spoke-only case it is $0$ or
$1$), then
\[
   N(T'_{f'};\,\sigma) \;=\; \sum_{(c_1,\dots,c_r)\in[k]^r}
       N\!\bigl(C_n;\,\mathrm{forbidden}_{M}(c_1,\dots,c_r)
                       \cup C^{(\cdot)}_\sigma,\,k\bigr),
\]
i.e.\ the menagerie sum but with the extra spoke-forbidden colors
appended to the forbidden sets at each cycle vertex.

\paragraph{Special case: $E_S = \emptyset$.}
If there are no spokes (e.g.\ the spoke-free setting), $\sigma$ is the
empty map and the unique fiber is
$N(T'_{f'};\,\sigma_\emptyset) = P_e(T'_{\mathrm{ann}}, k)$, which is
exactly the menagerie \S6 formula.

\section*{Small worked example: $C_4$ with four spokes at $k=3$}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_fiber_c4_example.png}
\caption{Three spoke configurations $\sigma$ on $C_4$ with four
pendant spokes, with cycle vertices labelled $v_0, v_1, v_2, v_3$ and
spoke vertices $u_0, u_1, u_2, u_3$.  \textbf{Left:} the constant
configuration $\sigma = (1,1,1,1)$ (all spokes red) is realisable with
fiber count $N(\sigma) = 2$ --- the cycle alternates between the two
remaining colors in two ways.  \textbf{Middle:} the alternating
configuration $\sigma = (1,2,1,2)$ is \emph{infeasible} ($N(\sigma) =
0$); the constraints at $v_0$ and $v_1$ force $c(e_0) = 3$, but the
constraint at $v_2$ then demands $c(e_1) \in \{2, 3\} \setminus
\{c(e_0)\}$ contradicting the $v_1$ constraint.  \textbf{Right:} the
non-uniform configuration $\sigma = (1,1,2,2)$ has fiber count
$N(\sigma) = 1$ --- a single cycle coloring extends it.}
\label{fig:fiber-c4}
\end{figure}

Let $T'_{f'}$ be the $4$-cycle $C_4 = v_0 v_1 v_2 v_3$ with one pendant
spoke $v_i u_i$ at each cycle vertex.  Each $v_i$ has degree $3$, so
proper edge $3$-coloring forces $\{c(s_i),\,c(e_{i-1}),\,c(e_i)\} =
\{1,2,3\}$ at every $v_i$, where $s_i := v_i u_i$ and $e_i$ is the
cycle edge from $v_i$ to $v_{i+1}$ (indices mod $4$).

Fix a spoke configuration $\sigma = (\sigma_0,\sigma_1,\sigma_2,\sigma_3)$.
At each $v_i$, the two cycle edges $e_{i-1}, e_i$ are the two colors of
$[3] \setminus \{\sigma_i\}$ in some order.  This gives the constraints
\[
   c(e_i) \in [3] \setminus \{\sigma_i, \sigma_{i+1}\},
   \qquad i = 0,1,2,3,
\]
together with $c(e_{i-1}) \ne c(e_i)$ at each $v_i$.

\paragraph{Total count.}
For $C_4$ alone, $P_e(C_4, 3) = P_v(C_4, 3) = 2^4 + 2 = 18$.  At $k=3$,
each spoke is forced (the unique color avoiding its endpoint's two
cycle-edge colors), so $P_e(T'_{f'}, 3) = 18$ as well: the pendants
contribute a factor of~$1^4$.

\paragraph{Distribution over $\sigma$.}
The map ``edge $3$-coloring of $T'_{f'}$'' $\mapsto$ ``induced
$\sigma$'' is surjective onto its image, and
\[
   \sum_{\sigma \in [3]^4} N(T'_{f'};\,\sigma) \;=\; 18.
\]
The fiber sizes for the three configurations in
Figure~\ref{fig:fiber-c4} can be checked directly from the constraints
above; doing the same for every $\sigma \in [3]^4$ would partition the
$18$ colorings according to the spoke pattern they induce.

\section*{Why this matters for nested tires}

\begin{figure}[h]
\centering
\includegraphics[width=0.95\textwidth]{fig_nesting_compatibility.png}
\caption{Two adjacent tires $T^{\mathrm{outer}}, T^{\mathrm{inner}}$ in
a common maximal planar $G$, sharing the cycle $\gamma$.  The cycle
vertex $v \in V(\gamma)$ is a dual vertex for an annular face in
\emph{both} tires.  The $G'$-edges at $v$ split into: two cycle edges
inside $T'^{\mathrm{outer}}_{\mathrm{ann}}$, two cycle edges inside
$T'^{\mathrm{inner}}_{\mathrm{ann}}$, and (in the spoke-only case)
\emph{one shared spoke edge} $vu$ pointing across the shared boundary.
That single spoke edge is simultaneously a spoke of $T'^{\mathrm{outer}}_{f'}$
and a spoke of $T'^{\mathrm{inner}}_{f'}$, so any global edge coloring
of $G'$ assigns it a single color from both sides at once.}
\label{fig:nesting-compatibility}
\end{figure}

The whole point of the fiber decomposition is that, in a nested-tire
configuration, the spoke edge set of the inner tire is (a subset of) the
spoke edge set of the outer tire \emph{regarded from the other side}.
Specifically, if two tires $T^{\mathrm{outer}}, T^{\mathrm{inner}}$ share a
cycle $\gamma$ in $G$, then for $v \in V(\gamma)$ the unique
$G'$-edge at $v$ that points out of $\gamma$'s annular faces serves as
an outer-spoke edge of one tire's face connector and an inner-spoke
edge of the other's (Figure~\ref{fig:nesting-compatibility}).

So an edge $3$-coloring of the whole region decomposes as: a coloring
of each tire's face connector, with the spoke configurations matched
on the shared cycle.  Writing
$\sigma_{\mathrm{shared}}$ for the common spoke configuration on the
shared boundary, this is
\[
   \#\{\text{joint colorings on } T^{\mathrm{outer}} \cup
       T^{\mathrm{inner}}\}
   \;=\; \sum_{\sigma_{\mathrm{shared}}}
         N(T'^{\mathrm{outer}}_{f'};\, \sigma_{\mathrm{shared}},\, *)
         \cdot
         N(T'^{\mathrm{inner}}_{f'};\, \sigma_{\mathrm{shared}},\, *)
\]
(with $*$ standing for the other-side spoke configuration, which is
free).  This is the conductivity step $\phi_B$ of the chain pigeonhole
sketch from the conversation that prompted this note: the inner tire's
profile, projected to its outer-side spoke configurations, must
intersect the outer tire's profile, projected to its inner-side spoke
configurations.

\paragraph{Where this leads.}
Two natural quantitative questions:
\begin{enumerate}
  \item For a fixed face connector $T'_{f'}$ and a fixed boundary side,
        what is the distribution of $|N(T'_{f'};\,\sigma)|$ over
        $\sigma$?  How concentrated is it, and how large is the
        realisable support?
  \item For two nested tires to admit a joint edge $3$-coloring it
        suffices that the supports of $\sigma$ overlap.  When can we
        guarantee this from the cycle length alone (which controls
        $|[k]^{|E_S|}|$ on the shared side)?
\end{enumerate}
The fiber form makes both questions concrete: the first is a fiber
distribution to compute, the second is a covering / pigeonhole on $\sigma$.

\end{document}
