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

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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>
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didericis
2026-06-17 01:03:37 -04:00
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papers/heawood_restrictions_on_nested_tire_graph_duals/paper.aux
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\citation{bauerfeld-medial-tires}
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\citation{bauerfeld-nested-tires}
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\bibcite{bauerfeld-depth}{2}
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papers/heawood_restrictions_on_nested_tire_graph_duals/paper.log
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papers/heawood_restrictions_on_nested_tire_graph_duals/paper.tex
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@@ -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}