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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}
