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

\title{$S_3$-orbit decomposition of $S_1 \cap S_2$\\
       Do structurally different tires share the same canonical orbits?}
\author{}
\date{}

\newtheorem*{obs}{Observation}

\begin{document}
\maketitle

\section*{The question}

Step~2 (\texttt{tire\_fiber\_step2.tex}) reported $|S_1 \cap S_2|$ for
$23$ adjacent-tire pairs and found all $23$ compatible.  A follow-up
question is whether structurally different $(T_1, T_2)$ pairs that
share a cycle-length $|\gamma| = k$ have intersections with the
\emph{same orbit structure} --- i.e.\ whether a canonical pattern like
the rainbow $(a,b,c,b,c,a)$ persists when we vary $T_1$ and $T_2$, or
whether each pair gives its own pair-specific orbit.

Script: \texttt{experiments/orbit\_decomposition.py}; raw output:
\texttt{experiments/orbit\_decomposition\_data.txt}.

\section*{$S_3$-closure: a structural sanity check}

Permuting the three colors is a symmetry of proper edge $3$-coloring,
so for any fixed tire $T$, both $S_1$ and $S_2$ must be closed under
the diagonal $S_3$ action on $\{1,2,3\}^k$.  Hence so is $S_1 \cap
S_2$.

\begin{obs}[$S_3$-closure]
\label{obs:s3-closed}
In every one of the $23$ pairs, $S_1 \cap S_2$ is closed under the
$S_3$ color action.  This is structural rather than coincidental.
\end{obs}

\section*{Orbit size distribution}

For $\sigma \in \{1,2,3\}^k$ with $3$ distinct color-values, the $S_3$
orbit has size exactly $6$.  For $\sigma$ using $2$ distinct color values,
the orbit has size $6$ (since $S_3$ acts on the $\binom{3}{2}\cdot 2$ ways
of placing them).  For $\sigma$ constant (one color), the orbit has size $3$.

\begin{obs}[Uniform orbit sizes]
\label{obs:orbit-sizes}
Across all $23$ pairs, every $S_3$-orbit in $S_1 \cap S_2$ has size
exactly $6$, with one single exception: the constant orbit
$\{(1,\dots,1), (2,\dots,2), (3,\dots,3)\}$ of size $3$, which
appears only in the case $\gamma = 4$, $T_1 = T_2 = (4, -, \mathrm{SR})$.
\end{obs}

So intersection sizes are essentially $6 \cdot (\text{number of orbits})$
in all but one case.

\section*{The rainbow orbit reappears across structurally different pairs}

Combining $S_3$ color action with cyclic rotation of $\gamma$ gives a
larger symmetry group; the combined orbit of $(1,2,3,2,3,1)$ has size
$36$.  This single combined orbit appears in three different $(T_1, T_2)$
pairs at $\gamma = 6$:

\begin{center}
\begin{tabular}{l l l}
\toprule
$\gamma$ & $T_1$ & $T_2$ \\
\midrule
$6$ & $(6, (0,3), \mathrm{SP})$ & $(6, (0,3), \mathrm{SP})$ \\
$6$ & $(6, (0,3), \mathrm{SP})$ & $(6, (0,2)(3,5), \mathrm{SP})$ \\
$6$ & $(6, (0,3), \mathrm{SP})$ & $(3, -, \mathrm{SR})$ \\
\bottomrule
\end{tabular}
\end{center}

All three have $T_1 = (6, (0,3), \mathrm{SP})$ --- the side with the
\emph{antipodal} chord.  The three different $T_2$ structures range
from another antipodal-chord SP tire to a chord-less SP tire to a
chordless SR tire.  In every case the rainbow orbit is part of (or
all of) the intersection.

\begin{obs}[Rainbow orbit is $T_1$-driven at $\gamma=6$]
\label{obs:rainbow-source}
At $\gamma=6$, the rainbow combined orbit $(a,b,c,b,c,a)$ appears
\emph{iff} $T_1 = (6, (0,3), \mathrm{SP})$ --- the antipodal-chord
SP tire.  The two-chord SP tire $T_1 = (6, (0,2)(3,5), \mathrm{SP})$
yields different orbits, never the rainbow.  So the rainbow pattern
is associated with antipodal $O$-chord topology rather than with the
pair $(T_1, T_2)$ as a whole.
\end{obs}

\section*{Most-shared canonical orbits}

Other canonical combined orbits appear across many structurally
different pairs.  Top entries:

\begin{center}
\begin{tabular}{r l r l}
\toprule
$|\gamma|$ & canonical rep & \# distinct pairs & combined-orbit size \\
\midrule
$4$ & $(1,2,1,3)$         & $7$ & $12$ \\
$4$ & $(1,2,1,2)$         & $6$ & $6$  \\
$4$ & $(1,2,2,1)$         & $5$ & $12$ \\
$4$ & $(1,2,2,3)$         & $5$ & $24$ \\
$3$ & $(1,2,3)$           & $4$ & $6$  \\
$5$ & $(1,2,1,2,3)$       & $3$ & $30$ \\
$6$ & $(1,2,3,2,3,1)$ (rainbow) & $3$ & $36$ \\
\bottomrule
\end{tabular}
\end{center}

\begin{obs}[Universal orbits dominate]
\label{obs:universal-orbits}
At each cycle length $k$, the most-shared canonical orbit appears in
roughly $30$--$40\%$ of tested pairs at that $k$.  These ``universal''
orbits are typically of the form $(a, b, a, c)$ or $(a, b, a, b)$ at
$\gamma = 4$, $(a, b, c)$ at $\gamma = 3$, etc.  They survive across
chord placements and across SR/SP model choices.
\end{obs}

\section*{What this means for chain pigeonhole}

If $S_1 \cap S_2$ is always $S_3$-closed (Obs.~\ref{obs:s3-closed})
and always contains at least one full $S_3$-orbit (which, given
size $6$ for non-constant orbits, is essentially the same as saying
$|S_1 \cap S_2| > 0$), then \emph{chain pigeonhole at a single shared
cycle is not just a counting fact but a structural one}: the
intersection respects color symmetry, never collapses to a thin
exotic subset, and contains canonical orbits that recur across
structurally varied pairs.

This is a real upgrade on the step-$2$ data:
\begin{itemize}
  \item Step $2$ only reported $|S_1 \cap S_2| > 0$; it could in
        principle be a single weird configuration with no symmetry.
  \item The orbit decomposition shows the intersections always have
        the full $S_3$-symmetric shape, with at least one orbit of
        size $6$ (or $3$ in the trivial case).
  \item Several canonical orbits recur across structurally different
        pairs --- evidence that chain compatibility is detecting
        something structural about $\gamma$ and the tire-pair
        topology, not a coincidence of one specific configuration.
\end{itemize}

\section*{Conjecture suggested by the data}

\begin{obs}[Antipodal-chord rainbow conjecture --- \emph{refuted as
originally stated}; refined below]
\label{obs:antipodal-rainbow-conjecture}
\emph{Original (incorrect) statement.}
Let $T = (m, (0, m/2), \mathrm{SP})$ be a Steiner-poor tire whose
inner outerplanar graph $O$ is a cycle of length $m$ together with a
single antipodal chord (so $m$ is even).  Conjecture (originally,
without an outer-boundary precondition): the projection support
$\pi_D(\mathcal{C}(T))$ on the $|\gamma| = m$ inner-side spokes
always contains the combined orbit
\[
   \mathrm{Orbit}\bigl(\,(a, b, c, b, c, \dots, b, c, a)\,\bigr)
\]
under $S_3 \times C_m$, with $a$-positions at the two chord endpoints
and $b, c$ alternating elsewhere.  At $m = 6$ this is the rainbow
orbit of size $36$ that Obs.~\ref{obs:rainbow-source} witnessed.

\medskip
\noindent
\emph{The original statement is false:} the
counterexample-search log
(\texttt{experiments/counterexample\_search.log}) records the pair
\[
   T_1 = (m_1 = 3,\ (0, 3),\ \mathrm{SP}),
   \qquad
   T_2 = (k_2 = 3,\ -,\ \mathrm{SP}),
   \qquad
   |\gamma| = 6,
\]
where $T_1$ is the antipodal-chord SP tire at $\gamma = 6$ but with
\emph{small outer boundary} $m_1 = 3$.  Direct computation gives
$|\pi_D(\mathcal{C}(T_1))| = 18$.  The conjectured rainbow combined
orbit has size $36$, so the orbit cannot fit inside $\pi_D$; in fact
only $6$ of the $36$ rainbow-orbit elements lie in $\pi_D$, and the
literal generator pattern $(1, 2, 3, 2, 3, 1)$ is itself among the
$30$ \emph{missing} elements.  Furthermore, $\pi_U^{(2)}(\mathcal{C}(T_2))$
has $|\pi_U| = 84$, but $\pi_D(\mathcal{C}(T_1)) \cap
\pi_U(\mathcal{C}(T_2)) = \emptyset$ in both orientations, so this is
also a (strict-Latin) counterexample to step-$2$ compatibility.

\medskip
\noindent
\emph{Refined statement (now consistent with all data).}
The original conjecture omitted an outer-boundary precondition.  The
correct conjecture is:
\begin{quote}
\textbf{If additionally $m_1 = |B_{\mathrm{out}}^{(1)}| \geq m$
(i.e.\ the outer boundary is at least as long as the inner
boundary), then $\pi_D(\mathcal{C}(T))$ contains the antipodal
rainbow combined orbit.}
\end{quote}
The threshold $m_1 \geq m$ is the same saturation threshold from
step~$1$: when the dual cycle of $T$ has length $n_1 = m_1 + m
\geq 2m$, the spoke projection saturates the full Latin set on
$\gamma$.  Under this precondition all $23$ pairs of the original
step-$2$ tests and all $44$ strict-Latin pairs in the search log
with $m_1 \geq m$ \emph{and} $k_2 \geq m$ confirm the property.
\end{obs}

If the refined statement holds, this is a uniform structural property
of the antipodal-chord SP tire \emph{with sufficient outer cycle
length}.  The chain pigeonhole step at $|\gamma| = m$ on such a tire
reduces to ``$\pi_U$ of the other tire intersects this fixed orbit,''
a much smaller compatibility claim --- but only once both sides clear
the outer-boundary threshold.

\paragraph{Direct test.}  At $m = 4$ ($\theta(1, 2, 2) = K_4 - e$) the
antipodal-chord SP tire's $\pi_D$ support has size $36$ at $|\gamma| =
4$ (\texttt{tire\_fiber\_chords.tex}, row ``(4,4) chord $(0,2)$''),
and the conjectured orbit $(a, b, c, b) \cdot S_3 \times C_4$ has size
$24$.  Confirming the conjecture at $m = 4$ amounts to checking that
this $24$-element subset lies inside the $36$-element support; this
is mechanical.

\paragraph{Why antipodal?}  In the planar dual picture, the antipodal
chord of $O$ corresponds to the dual edge of a single
``maximally-separating'' chord in the tire's inner outerplanar graph:
it splits $\pi_1$ of the annulus most symmetrically.  Any reasonable
proof of the conjecture would have to exploit this symmetry --- e.g.\
via reflection invariance on the chord axis.

\section*{Caveats}

\begin{enumerate}
  \item Still empirical for $k \le 6$, chord counts $\le 2$, and the
        $23$ pairs from step~$2$.
  \item ``Same combined orbit'' is taken modulo $S_3 \times C_k$ (color
        permutation $\times$ cyclic rotation of $\gamma$).  Reflection
        of $\gamma$ is \emph{not} quotiented out --- some orbits would
        coincide if it were.
  \item The rainbow orbit's persistence is tied to the antipodal
        chord; a structural proof would need to show that the
        antipodal-chord $\mathrm{SP}$ tire's projection support
        $\pi_D$ always contains the rainbow orbit, independently of
        the outer tire.
\end{enumerate}

\end{document}
