\section{Cube squares, order independence, and finite matrices}
\label{trfb1odmscoc:sec-squares-order-matrices-scope}

The supplied geometric square is already an equality of Walker cobordisms.
Transporting it through the strict functor and the same comparison map at
every vertex gives an operator square.  Finite matrix identities then record
that square in chosen coordinates.

All notation and conventions are those of
Sections~\ref{trfb1odmscoc:sec-strict-operators-phases-traces} and the
standing assumptions.  In particular, every operator is formed with
the one map \(Q_{K,d}\), every source product and target product is written
right-factor first, and source topology always precedes its operator
image.

Here \(H_{a\mid A}=C(h_{a\mid A},0)\) is the marked source handleslide
cylinder, while \(T^K_{a\mid A}\) is its image on the fixed-level operator
fibre.  The set tag \(A\) belongs to both symbols: it records the source
vertex of the geometric edge before that edge is transported to an operator.

First transport each literal geometric square through the one fixed
comparison map and strict finite-level functor.  Then use the resulting
operator squares to compare adjacent orders, keeping the intermediate-set
tag attached to every factor.  The forward and reverse trace families are
telescoped only after that order-independence step.  Finally choose bases to
record the established operator identities as finite square and rectangular
matrices.  Thus the matrix formulas are coordinates on fixed-level
transport, not a new source of coherence or a comparison between levels.

\subsection{Literal operator squares}

\begin{proposition}[Every disjoint source face has a literal image]
\label{trfb1odmscoc:prop-operator-squares}
Let \(A\subseteq[s]\) and let \(a,b\notin A\) be distinct.  Then

\begin{equation}
 T^K_{b\mid A\cup\{a\}}T^K_{a\mid A}
 =T^K_{a\mid A\cup\{b\}}T^K_{b\mid A}
 \label{trfb1odmscoc:eq-operator-square}
\end{equation}

as a literal equality in
\(\operatorname{End}_{E_K}(\mathcal M_{K,d})\).
\end{proposition}

\begin{proof}
The geometric cube theorem supplies the Walker equality

\begin{equation}
 H_{b\mid A\cup\{a\}}H_{a\mid A}
 =H_{a\mid A\cup\{b\}}H_{b\mid A}.
 \label{trfb1odmscoc:eq-source-square}
\end{equation}

Every cylinder in \eqref{trfb1odmscoc:eq-source-square} has geometric
weight zero, and the source seam calculation shows that both composites
also have weight zero.  Apply
\eqref{trfb1odmscoc:eq-one-way-shadow} and the edge definition
\eqref{trfb1odmscoc:eq-edge-definition}; the result is exactly
\eqref{trfb1odmscoc:eq-operator-square}.  No matrix or operator equality
is used in the source direction.
\end{proof}

\paragraph{Smallest square.}
The first nontrivial instance has \(s=2\), \(A=\varnothing\), and
\(a=1,b=2\).  Its four vertices are
\(\varnothing,\{1\},\{2\},\{1,2\}\), and the proposition says that the two
routes from \(\varnothing\) to \(\{1,2\}\) give the same operator:
\(T^K_{2\mid\{1\}}T^K_{1\mid\varnothing}
=T^K_{1\mid\{2\}}T^K_{2\mid\varnothing}\).
This is the smallest picture in which the intermediate-set tag matters.  It
also shows what the fixed comparison map accomplishes: all four edge maps
act on the same finite fibre \(\mathcal M_{K,d}\), so the equality is obtained
before any basis is chosen.  The finite matrix statement below is only the
coordinate form of this already established operator square.

\subsection{Order independence and equality-only Coxeter closure}
\leavevmode\par\nobreak

For \(\pi=(\pi_1,\ldots,\pi_s)\in S_s\), set

\begin{equation}
 A_q^\pi:=\{\pi_1,\ldots,\pi_q\},
 \qquad A_0^\pi:=\varnothing,
 \label{trfb1odmscoc:eq-order-tags}
\end{equation}

and define

\begin{equation}
 T^K_\pi
 :=T^K_{\pi_s\mid A_{s-1}^\pi}\cdots
   T^K_{\pi_1\mid\varnothing}.
 \label{trfb1odmscoc:eq-order-product}
\end{equation}

The set tag records slides already applied to the right of the indicated
factor.

\begin{proposition}[Finite order closure]
\label{trfb1odmscoc:prop-order-closure}
For every \(\pi\in S_s\),

\begin{equation}
 T^K_\pi=T^K_{[s]}
 :=\rho^Q_{K,d}(H_{[s]}).
 \label{trfb1odmscoc:eq-common-order-operator}
\end{equation}

Adjacent applications of
\eqref{trfb1odmscoc:eq-operator-square} give equality replacements
\(\tau_t\), and these replacements close around

\begin{equation}
 \tau_t^2=1,
 \qquad
 \tau_t\tau_u=\tau_u\tau_t\quad(|t-u|>1),
 \qquad
 \tau_t\tau_{t+1}\tau_t
 =\tau_{t+1}\tau_t\tau_{t+1}.
 \label{trfb1odmscoc:eq-equality-coxeter}
\end{equation}

The first two families use \(1\leq t,u\leq s-1\); the third uses
\(1\leq t\leq s-2\).  These are equalities between replacement paths,
not operators and not an action of \(S_s\).
\end{proposition}

\begin{proof}
The geometric cube theorem proves, for each \(\pi\), the literal source identity

\[
 H_{\pi_s\mid A_{s-1}^\pi}\cdots
 H_{\pi_1\mid\varnothing}=H_{[s]}.
\]

Apply the strict representation and use
\eqref{trfb1odmscoc:eq-order-product}.  This proves
\eqref{trfb1odmscoc:eq-common-order-operator} directly, including the
fact that every product has total Walker weight zero.

An adjacent transposition changes precisely the two factors spanning one
square at the correct intermediate set tag; Proposition
\ref{trfb1odmscoc:prop-operator-squares} replaces those two factors by
the other side.  Repeating the same equality replacement returns to the
original word, which gives the first family in
\eqref{trfb1odmscoc:eq-equality-coxeter}.  Replacements in disjoint word
positions commute, giving the second family.  The two length-three
replacement paths in the last family have the same final order word;
each intermediate word equals \(T^K_{[s]}\) by
\eqref{trfb1odmscoc:eq-common-order-operator}, so the two paths close.
This proves equality-only closure without assigning a linear map to any
\(\tau_t\).  When \(s=1\), no index range is nonempty.  When \(s=2\),
only the involution range is nonempty.
\end{proof}

\subsection{Forward and reverse telescoping}

\begin{proposition}[All-phase and all-weight telescoping]
\label{trfb1odmscoc:prop-telescoping}
For every phase assignment, every integer base-weight vector, and every
integer phase-weight vector,

\begin{align}
 T^K_{[s]}\,
 \mathsf J^K_\varnothing(\epsilon;\mathbf c,\mathbf u^+)
 &=\mathsf J^K_{[s]}(\epsilon;\mathbf c,\mathbf u^+),
 \label{trfb1odmscoc:eq-forward-telescope}\\
 \mathsf R^K_{[s]}(\eta;\mathbf r,\mathbf u^-)
 T^K_{[s]}
 &=\mathsf R^K_\varnothing(\eta;\mathbf r,\mathbf u^-).
 \label{trfb1odmscoc:eq-reverse-telescope}
\end{align}
\end{proposition}

\begin{proof}
Fix an order \(\pi\).  Iterating
\eqref{trfb1odmscoc:eq-J-covariance} in chronological order gives, for
every \(1\leq q\leq s\),

\[
 T^K_{\pi_q\mid A_{q-1}^\pi}\cdots
 T^K_{\pi_1\mid\varnothing}
 \mathsf J^K_\varnothing(\epsilon;\mathbf c,\mathbf u^+)
 =\mathsf J^K_{A_q^\pi}(\epsilon;\mathbf c,\mathbf u^+).
\]

At \(q=s\), the operator product is \(T^K_\pi=T^K_{[s]}\) and
\(A_s^\pi=[s]\), proving
\eqref{trfb1odmscoc:eq-forward-telescope}.  Starting at the terminal
vertex and iterating \eqref{trfb1odmscoc:eq-R-covariance} from the left
instead gives

\[
 \mathsf R^K_{A_s^\pi}(\eta;\mathbf r,\mathbf u^-)
 T^K_{\pi_s\mid A_{s-1}^\pi}\cdots
 T^K_{\pi_1\mid\varnothing}
 =\mathsf R^K_\varnothing(\eta;\mathbf r,\mathbf u^-),
\]

which is \eqref{trfb1odmscoc:eq-reverse-telescope}.  The argument never
changes a phase assignment or any integer weight, so it includes empty
phase words, zero weights, and arbitrary negative weights.  Their scalar
content remains exactly \eqref{trfb1odmscoc:eq-J-weight} and
\eqref{trfb1odmscoc:eq-R-weight}.
\end{proof}

\subsection{Fibrewise finite matrices}

At this point the operator identities have already been proved on the fixed
finite-level fibres.  Choosing the old-spine basis now produces square arrays
for endomorphisms and rectangular arrays for the forward and reverse maps
between the low and high fibres.  These matrices record the transported
relations in coordinates; they do not define the comparison map, remove its
choice dependence, or introduce a comparison between different levels.

\paragraph{An exact \(K=1\) matrix.}
The central weight-one cylinder gives a small-level check with no unspecified
entries.  At \(K=1\), the normalization in
\cref{trfb1odmscoc:prop-root} has
\[
 \kappa_1=\exp(\pi i/4)=\zeta_8.
\]
Hence, in any ordered old-spine basis \(\mathcal B_{1,d}\),
\[
 \bigl[\rho^Q_{1,d}(W_Y)\bigr]_{\mathcal B_{1,d}}
 =\zeta_8 I_{r_{1,d}},
 \qquad r_{1,d}=\dim_{E_1}\mathcal M_{1,d}.
\]
This exact scalar matrix is already enough to expose the coefficient-field
issue: \(\kappa_1\notin F_1\), so the matrix belongs naturally over \(E_1\),
not uniformly over the raw field \(F_1\).  It is a normalization example, not
a computation of a handleslide-edge matrix.

\begin{proposition}[Fixed-level matrix realization]
\label{trfb1odmscoc:prop-matrices}
Order the finite old-spine basis \(\mathcal B_{K,d}\).  For
\(\epsilon,\eta\in\{0,1\}^k\) and
\(\mathbf u^+,\mathbf u^-\in\mathbb Z^k\), every edge, phase, and
total-order operator has a finite matrix over \(E_K\):

\begin{equation}
 [T^K_{a\mid A}]_{\mathcal B_{K,d}},
 \qquad
 [P^K_A(\epsilon,\mathbf u^+)]_{\mathcal B_{K,d}},
 \qquad
 [P^K_A(\eta,\mathbf u^-)]_{\mathcal B_{K,d}},
 \qquad
 [T^K_{[s]}]_{\mathcal B_{K,d}}.
 \label{trfb1odmscoc:eq-high-matrices}
\end{equation}

After Proposition~\ref{trfb1odmscoc:prop-low-fin} has established
\(\dim_{E_K}\mathcal V^-_{K,g}<\infty\), choose any basis of the low
module.  The maps \(\mathsf J^K_A\) and \(\mathsf R^K_A\) then have finite
rectangular matrices.  In these bases, every equality proved above remains
a literal matrix equality.
\end{proposition}

\begin{proof}
The reduced spine basis is finite by
Proposition~\ref{trfb1odmscoc:prop-cap-red}, and every high operator is an
\(E_K\)-linear endomorphism of its span.  This proves
\eqref{trfb1odmscoc:eq-high-matrices}.  The separate quotient-first
argument in Proposition~\ref{trfb1odmscoc:prop-low-fin} proves low
finiteness before a low basis is chosen, so the forward and reverse maps
have finite rectangular coordinate arrays afterward.  Passing an already
proved equality to coordinates preserves equality.  A change of the high
basis simultaneously conjugates all high endomorphisms and gives the
corresponding one-sided base change for the rectangular maps.  Nothing in
this argument relates different values of \(K\).
\end{proof}

\subsection{Main theorem}

The theorem below is the algebraic payoff of keeping one comparison
isomorphism in view.
The geometric cube has already supplied literal Walker equalities; the
remaining task is to show that the same normalized map carries every vertex
to the same finite-level fibre.  Once that is done, squares, route products,
and telescoping identities survive as literal operator equalities rather than
only as projective or basis-dependent statements.

Reusing the same comparison map lets its inverse cancel between successive
edge images.  Strict functoriality therefore carries each geometric
two-face to an operator square, and adjacent replacements give finite order
independence.  The endpoint covariance identities then make the phase-weight
traces telescope.  Bases are chosen afterward to express these operator
identities as finite square and rectangular matrix identities at the fixed
level.

\begin{theorem}[Fixed-level disjoint-cube operator transport]
\label{trfb1odmscoc:thm-T0}
Uniformly for every datum quantified in the standing assumptions, the following
nine statements hold.

\begin{enumerate}
\item The root, coefficient-ring, field, rank, anomaly, involution, and
  weight-zero \(S^3\) identities hold, including the exact positive
  square-root branch and the \(K=1\) obstruction to the raw field.
\item The weight-zero old handlebody, its complementary pairing witness,
  the finite reduced old fibre, its admissible spine basis, and the old
  disk kernel are as constructed in
  Section~\ref{trfb1odmscoc:sec-root-reduced-crosswalk}; the low module is
  finite by Proposition~\ref{trfb1odmscoc:prop-low-fin}.
\item The universal map factors through the reduced quotient, its pairing
  is exactly \(\mathcal D_K^{-1}\) times the reduced complementary pairing,
  both kernel inclusions hold, and the result is the single isomorphism
  \(Q_{K,d}\) used at every vertex.
\item Equations \eqref{trfb1odmscoc:eq-rhoQ-product}--
  \eqref{trfb1odmscoc:eq-one-way-shadow} and
  \eqref{trfb1odmscoc:eq-word-correction}--
  \eqref{trfb1odmscoc:eq-weight-zero-word} give the strict corrected
  extended functor with its central character.
\item Equations \eqref{trfb1odmscoc:eq-edge-definition}--
  \eqref{trfb1odmscoc:eq-edge-word} and
  \eqref{trfb1odmscoc:eq-common-phase}--
  \eqref{trfb1odmscoc:eq-phase-conjugacy} give every edge, its exact
  inverse, all completed phases, their labelled orders, and their weights.
\item Equations \eqref{trfb1odmscoc:eq-J-definition}--
  \eqref{trfb1odmscoc:eq-R-covariance} give both independently typed
  forward and reverse families, phase factorizations, total-weight
  scalars, and covariances.
\item Every supplied disjoint source face has the literal operator square
  \eqref{trfb1odmscoc:eq-operator-square}, only in the source-to-target
  direction.
\item Equations \eqref{trfb1odmscoc:eq-common-order-operator}--
  \eqref{trfb1odmscoc:eq-reverse-telescope} give finite order
  independence, equality-only Coxeter closure with the exact vacuities,
  and forward/reverse telescoping for all phases and weights.
\item Proposition~\ref{trfb1odmscoc:prop-matrices} gives only the claimed
  fixed-\(K\), \(E_K\)-linear finite matrices and no cross-level, refined,
  topological, or closed-manifold promotion.
\end{enumerate}
\end{theorem}

\begin{proof}
Item one is Proposition~\ref{trfb1odmscoc:prop-root}.
Item two combines Propositions~\ref{trfb1odmscoc:prop-cap-red} and
\ref{trfb1odmscoc:prop-low-fin}; item three combines
Propositions~\ref{trfb1odmscoc:prop-q-pair} and
\ref{trfb1odmscoc:prop-q-iso}.  Items four through six are
Propositions~\ref{trfb1odmscoc:prop-strict-representation},
\ref{trfb1odmscoc:prop-word-correction},
\ref{trfb1odmscoc:prop-edge-operator},
\ref{trfb1odmscoc:prop-phases}, and
\ref{trfb1odmscoc:prop-trace-maps}.  The last three items are
Propositions~\ref{trfb1odmscoc:prop-operator-squares},
\ref{trfb1odmscoc:prop-order-closure},
\ref{trfb1odmscoc:prop-telescoping}, and
\ref{trfb1odmscoc:prop-matrices}.  Every use of the geometric cube theorem
in those proofs applies a supplied source equality; every target
conclusion is then obtained by the strict fixed-level functor.
\end{proof}

\subsection{Boundary of the cube-branch theorem}

Theorem~\ref{trfb1odmscoc:thm-T0} includes the degenerate edge cases
needed for a uniform statement: \(K=1\), genus \(g=1\), one or two active
slides, spectator handles, empty phase words, and arbitrary integral
Walker weights.  The theorem is nevertheless deliberately fixed-level
and choice-dependent.  More precisely:

\begin{itemize}
\item it is proved at one level \(K\) and supplies no comparison between
  different levels;
\item every operator uses one fixed comparison map \(Q_{K,d}\), with no claim
  that another admissible comparison map gives the identical matrix;
\item the source relations are the supplied relations among disjoint
  slide directions; source equality implies operator equality, but no
  converse is asserted;
\item the five overlapping directions of the pentagon with complete markings
  in Section~\ref{sec:complete-pentagon} are not instances of these
  disjoint directions; and
\item no Cerf-completeness, stabilization, closed-manifold invariance,
  refined theory, WRT--Habiro comparison, or \(\widehat Z\) statement is
  deduced here.
\end{itemize}