Add connected tire clusters with two-cluster-per-vertex proposition

Define a connected tire cluster (union of same-depth tires joined by
shared vertices, transitive closure), prove same-depth tires meet only
in vertices, and prove every vertex lies in at most two clusters (one at
each of two consecutive depths) -- the bounded coarsening of the
unbounded per-vertex tire count.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>

papers/heawood_restrictions_on_nested_tire_graph_duals/paper.tex

@@ -98,6 +98,107 @@ along the boundary cycles of a nested tire graph, and to formulate a
 chain-pigeonhole programme in this Heawood labelling parallel to the
 medial programme of \cite{bauerfeld-medial-tires}.
 
+\section{Connected tire clusters}
+\label{sec:tire-clusters}
+
+The tire treads at a fixed depth partition the depth-$d$ faces of $G$
+\cite{bauerfeld-nested-tires}, but distinct depth-$d$ tires need not be
+vertex-disjoint: a single vertex of $G$ may lie on the source-side
+boundary of several depth-$d$ tires at once (this occurs exactly when
+the depth-$d$ faces around that vertex are split into more than one arc
+by depth-$(d{-}1)$ faces).  We organise the depth-$d$ tires by this
+sharing.
+
+\begin{lemma}[Same-depth tires meet only in vertices]
+\label{lem:same-depth-vertex-meet}
+Let $T \neq T'$ be two distinct tire treads at the same depth $d$ in
+$\mathcal{T}(G, S)$, arising from connected components $C', C''$ of the
+depth-$d$ dual subgraph $G'_d$.  Then $T$ and $T'$ share no edge of $G$;
+any intersection $V(T) \cap V(T')$ consists of isolated vertices.
+\end{lemma}
+
+\begin{proof}
+An edge $e$ of $G$ shared by two depth-$d$ annular faces $f_1, f_2$ is,
+by definition of the inner dual \cite[Definition~1.3]{bauerfeld-nested-tires},
+a dual edge of $G'$ joining $d_{f_1}$ and $d_{f_2}$; since $\delta(d_{f_1})
+= \delta(d_{f_2}) = d$, this edge lies in $G'_d$, so $d_{f_1}$ and
+$d_{f_2}$ belong to the same component of $G'_d$.  Hence no edge of $G$
+is shared by annular faces of two \emph{different} components, and
+distinct depth-$d$ tires share no edge.  Their intersection is therefore
+a set of isolated vertices.
+\end{proof}
+
+\begin{definition}[Connected tire cluster]
+\label{def:connected-tire-cluster}
+Fix a nested tire decomposition $\mathcal{T}(G, S)$ and a depth $d$.  On
+the set of depth-$d$ tire treads define the relation
+\[
+   T \sim T' \quad\Longleftrightarrow\quad V(T) \cap V(T') \neq \varnothing .
+\]
+A \emph{connected tire cluster} at depth $d$ is the subgraph of $G$
+\[
+   \mathsf{K} \;=\; \bigcup_{i} T_i \;\subseteq\; G
+\]
+obtained as the union (of underlying plane graphs) of the tires in a
+single connected component $\{T_i\}$ of the transitive closure of
+$\sim$.  A cluster consisting of a single tire is \emph{trivial}; the
+connected tire clusters at depth $d$ partition the depth-$d$ tires.
+\end{definition}
+
+\begin{remark}
+\label{rem:cluster-cut-vertices}
+By Lemma~\ref{lem:same-depth-vertex-meet} the constituent tires of a
+connected tire cluster are joined only at shared vertices, each of which
+is a cut vertex of $\mathsf{K}$; a connected tire cluster is thus a
+``cactus of tires'' and is in general \emph{not} itself a tire graph,
+since the annulus structure of \cite[Definition~1.5]{bauerfeld-nested-tires}
+fails at each such pinch.  The shared (cut) vertices are precisely the
+vertices that belong to more than one depth-$d$ tire.
+\end{remark}
+
+A single vertex may belong to several tires at one depth --- the
+high-degree case where its depth-$d$ faces split into many arcs --- so
+the number of \emph{tires} through a vertex is unbounded.  Clustering
+collapses exactly this multiplicity: all tires through a vertex at a
+fixed depth share that vertex, hence lie in one cluster.  The cluster
+count is therefore controlled.
+
+\begin{proposition}[A vertex meets at most two clusters]
+\label{prop:two-clusters-per-vertex}
+Every vertex $v \in V(G)$ belongs to at most two connected tire
+clusters, namely at most one at each of the two consecutive depths
+$\ell_G(v) - 1$ and $\ell_G(v)$.  In particular a source vertex
+($\ell_G(v) = 0$) belongs to a single cluster.
+\end{proposition}
+
+\begin{proof}
+Write $\ell = \ell_G(v)$.
+
+\emph{Step 1: every bounded face incident to $v$ has dual depth
+$\ell - 1$ or $\ell$.}  Let $f$ be a bounded triangular face with
+$v \in V(f)$.  Then
+$\delta_G(d_f) = \min_{u \in V(f)} \ell_G(u) \le \ell_G(v) = \ell$.
+The other two vertices of $f$ are adjacent to $v$ in $G$, and the level
+function $\ell_G(\cdot) = \mathrm{dist}_G(\cdot, S)$ is $1$-Lipschitz
+along edges, so each has level at least $\ell - 1$; hence
+$\delta_G(d_f) \ge \ell - 1$.  Thus $\delta_G(d_f) \in \{\ell-1, \ell\}$
+(only $\delta_G(d_f) = 0$ when $\ell = 0$), so $v$ bounds faces of, and
+therefore belongs to tires of, no depth other than $\ell - 1$ or
+$\ell$.
+
+\emph{Step 2: at each depth, all tires through $v$ lie in one cluster.}
+Fix $d \in \{\ell-1, \ell\}$ and let $T, T'$ be depth-$d$ tires with
+$v \in V(T) \cap V(T')$.  Then $V(T) \cap V(T') \ne \varnothing$, so
+$T \sim T'$ in the sense of
+Definition~\ref{def:connected-tire-cluster}, and all depth-$d$ tires
+containing $v$ lie in a single connected component of $\sim$ --- one
+connected tire cluster $\mathsf{K}_d$.
+
+Combining the two steps, $v$ belongs to at most the clusters
+$\mathsf{K}_{\ell-1}$ and $\mathsf{K}_{\ell}$, i.e.\ to at most two
+connected tire clusters; when $\ell = 0$ only $\mathsf{K}_0$ occurs.
+\end{proof}
+
 \section{Heawood restrictions on the tire dual}
 \label{sec:heawood-restrictions}