coloring_nested_tire_graphs: counterexamples to the loose conjecture as stated

Before attempting to prove the loose chain pigeonhole conjecture ("|π(T)| ≥ 6" for every non-trivial cut tire), looked for counterexamples and found TWO in the existing empirical data: (d, face) = (1, 1): 1 out spoke, |π(T)| = 3, orbit size [3]. (d, face) = (4, 0): 1 out spoke, |π(T)| = 3, orbit size [3]. Reason: a cut tire with exactly k = 1 in/out spoke has σ ∈ {1,2,3}. S_3 acts with stabilizer of size 2 on any single-color σ, so the orbit has size 6/2 = 3, never 6. The "|π(T)| ≥ 6" claim is automatically false for k = 1 tires. For k ≥ 2: σ can use 1 color (size-3 orbit) or ≥ 2 colors (size-6 orbit). |π(T)| ≥ 6 requires at least one multi-color σ to extend, which is not automatic but typically holds. Three refined conjectures proposed: 1. Restrict to k ≥ 2 spokes (avoids the trivial counterexample). 2. Weaken to "non-empty and S_3-closed" (very weak; needs the chain composition to preserve non-emptiness). 3. Just describe orbit sizes 3 or 6 (no useful claim). The two found counterexamples are at "side" faces in the chain; they don't break the bottom-line chain pigeonhole because the main chain runs through larger faces. To find harder counterexamples: look for k ≥ 2 cut tires whose face boundary forces all spoke colors equal (= |π(T)| = 3 with k ≥ 2). Such examples might exist but weren't found in the current data. Recommended next step: restrict the conjecture to k ≥ 2 and re-run the empirical sweep. Note: loose_conjecture_counterexamples.tex (3 pages). Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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+\documentclass[11pt]{article}
+\usepackage{amsmath,amssymb,amsthm}
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+
+\title{Counterexamples to the loose chain pigeonhole conjecture\\
+(and what they suggest about the right statement)}
+\author{}
+\date{}
+
+\newtheorem*{obs}{Observation}
+
+\begin{document}
+\maketitle
+
+\section*{The conjecture as stated}
+
+From \texttt{cut\_depth\_label.tex}, the loose chain pigeonhole
+conjecture says:
+
+\begin{quote}
+For every non-trivial cut tire $T$ (with $\ge 1$ in/out spoke), the
+joint projection $\pi(T)$ is non-empty, $S_3$-closed, and contains
+at least one full $S_3$-orbit (i.e.\ $|\pi(T)| \ge 6$).
+\end{quote}
+
+\section*{Counterexamples already in the data}
+
+Re-examining the cut tire tests on $G'_1$ of Holton-McKay \#0
+(\texttt{cut\_tire\_conjecture\_tests.tex}):
+
+\begin{center}
+\small
+\begin{tabular}{lr|rrr|r|cl}
+\toprule
+$d$ & face & out & in & total spokes & $|\pi(T)|$ & orbit sizes & \\
+\midrule
+$1$ & $0$ & $5$ & $0$ & $5$ & $93$ & $[3, 6^{15}]$ & ✓\\
+$1$ & $1$ & $1$ & $0$ & $1$ & $3$  & $[3]$ & \textbf{✗}\\
+$2$ & $0$ & $4$ & $3$ & $7$ & $126$ & $[6^{21}]$ & ✓\\
+$2$ & $1$ & $4$ & $3$ & $7$ & $126$ & $[6^{21}]$ & ✓\\
+$4$ & $0$ & $1$ & $0$ & $1$ & $3$ & $[3]$ & \textbf{✗}\\
+$4$ & $1$ & $2$ & $1$ & $3$ & $21$ & $[3, 6^3]$ & ✓\\
+$6$ & $0$ & $3$ & $2$ & $5$ & $93$ & $[3, 6^{15}]$ & ✓\\
+\bottomrule
+\end{tabular}
+\end{center}
+
+Two non-trivial cut tires have $|\pi(T)| = 3 < 6$:
+$(d, \text{face}) = (1, 1)$ and $(4, 0)$.  Both have exactly $1$
+spoke.  In each case, $\pi(T)$ is the $3$-element $S_3$-orbit of a
+single-color spoke configuration --- which has $S_3$-stabilizer of
+order $2$ (the two permutations fixing the chosen color and swapping
+the others), so the orbit has size $|S_3| / 2 = 3$.
+
+\section*{Why this happens (and is general)}
+
+For a cut tire with exactly $k$ in/out spokes, the projection
+$\pi(T) \subseteq \{1, 2, 3\}^k$.  $S_3$ acts by permuting colors.
+The $S_3$-orbit of $\sigma \in \{1,2,3\}^k$ has size:
+\begin{itemize}
+  \item $3$ if $\sigma$ uses exactly $1$ color (stabilizer of order $2$);
+  \item $6$ if $\sigma$ uses $\geq 2$ distinct colors (trivial stabilizer).
+\end{itemize}
+
+For $k = 1$: every $\sigma \in \{1, 2, 3\}$ uses exactly $1$ color,
+so every orbit has size $3$, and $|\pi(T)| \in \{0, 3\}$.  The
+conjecture's ``$|\pi(T)| \ge 6$'' is automatically false for any
+non-empty $\pi(T)$ on a $1$-spoke tire.
+
+For $k \ge 2$, $\sigma$ can use $1$ or $\geq 2$ colors:
+\begin{itemize}
+  \item Single-color $\sigma$ contributes a size-$3$ orbit.
+  \item Multi-color $\sigma$ contributes size-$6$ orbits.
+\end{itemize}
+$|\pi(T)| \ge 6$ requires at least one multi-color $\sigma$, which
+is not automatic but typically holds when $k$ is large or the tire
+allows enough flexibility.
+
+\section*{Why the conjecture's chain half can still fail}
+
+Even if we restrict to $k \ge 2$ cut tires (avoiding the trivial
+$k = 1$ counterexample), the chain pigeonhole conclusion is not
+immediate:
+
+\begin{itemize}
+  \item Chain composition involves \emph{compatibility} between
+        adjacent cut tires.  A full $S_3$-orbit at one layer projects
+        through the in/out-spoke ↔ face-boundary bijection to some
+        subset at the adjacent layer; this projection might land in
+        a size-$3$ orbit even if the source was size-$6$.
+  \item The chain at $d_{\max}$ terminates (no in spokes); the
+        innermost cut tire's structure constrains everything outward.
+  \item Sufficiency for $\mathcal{R}_0 \cap \mathcal{R}_1 \neq
+        \emptyset$ requires the chain to preserve the $S_3$-orbit
+        structure all the way to the cut, on both sides, and have a
+        common orbit there.
+\end{itemize}
+
+The counterexamples at $1$-spoke tires don't break the chain at
+those depths because they're at non-essential ``side'' faces
+(face $1$ at $d = 1$, face $0$ at $d = 4$; the main chain runs
+through the larger faces).  But they show the per-tire claim
+is not universal.
+
+\section*{Refined conjectures}
+
+In view of these counterexamples, three natural refinements:
+
+\begin{enumerate}
+\item \textbf{Restrict to $\ge 2$ spokes.}  ``For every cut tire
+      $T$ with $\geq 2$ in/out spokes total, $|\pi(T)| \ge 6$ and
+      $\pi(T)$ contains a full $S_3$-orbit.''  This avoids the
+      $k = 1$ trivial case.  Still empirically open whether
+      $k = 2$ cut tires always have $\ge 6$ projections.
+
+\item \textbf{Weaken to non-emptiness.}  ``For every non-trivial
+      cut tire, $\pi(T)$ is non-empty and $S_3$-closed.''  This
+      is much weaker but might still suffice for chain pigeonhole
+      \emph{provided} chain composition preserves non-emptiness ---
+      which is itself the hard step.
+
+\item \textbf{Allow size-$3$ orbits.}  ``For every non-trivial cut
+      tire, $\pi(T)$ is non-empty and $S_3$-closed; the orbits in
+      $\pi(T)$ have size $3$ or $6$.''  This is just a description
+      of the empirical findings, not a useful conjecture --- chain
+      pigeonhole at a size-$3$-orbit layer is more restrictive than
+      at a size-$6$ one.
+\end{enumerate}
+
+\section*{Where to look for harder counterexamples}
+
+The two counterexamples found are at low-spoke tires.  More
+substantive counterexamples would look like:
+
+\begin{itemize}
+  \item A $k \ge 2$ spoke cut tire whose face boundary is structured
+        such that only single-color $\sigma$'s extend, leaving
+        $|\pi(T)| = 3$.  This would require the face boundary to
+        force all spoke colors equal --- an unusual constraint.
+  \item A chain in which adjacent compatibility kills all size-$6$
+        orbits, leaving only size-$3$ at the cut layer.  This would
+        be a chain-level counterexample to the bottom-line
+        pigeonhole.
+\end{itemize}
+
+Neither is observed in the present data.  But the data is from a
+single $6$-cut on a single Holton-McKay graph.  A broader empirical
+sweep (multiple cuts, multiple Holton-McKay graphs, larger $G'$)
+might surface harder counterexamples.
+
+\paragraph{Suggested next step.}  Restrict the conjecture to
+$k \ge 2$ tires and re-run the empirical test.  If still empirically
+true after broader sweeps, attempt to prove the restricted form via
+the existing Prop 1.13 lower bound (which gives $\ge 2^n + 2(-1)^n$
+colorings for spoke-only cut tires of cycle length $n$) plus a
+combinatorial argument that the multi-color $\sigma$'s are non-empty.
+
+\end{document}