% Provenance: corrected synthesis of stage 23.
\section{Separating-Meridian Twists and Bridge Chambers}
\label{p4:retained:ch:separating-twists}

A separating compressing disk is dual to a bridge in a suitable trivalent
handlebody spine.  Its Dehn twist is therefore diagonal in the graph basis.
This section constructs that missing generic seed, distinguishes scalar
denominator-clearing factors from honest polarized endpoints, and adjoins all bridge powers to
the recoupling chamber groupoid.

Begin with the disk twist on one graph edge and use bridge parity to identify
the integral exponent available on a closed separating bridge.  The inverse
twist then presents three different tasks: clearing one scalar coefficient,
regularizing a bounded symmetric family of pairings, or transporting an
honest polarized endpoint lattice.  After those tasks are separated, the
bridge degrees can be added to the recoupling groupoid and completed with
restricted degree support.  The final compatibility discussion records the
local ribbon relation and its finite-level limitation, without extending the
verified generic bridge chamber to an unproved mapping-class presentation.

\subsection{Polynomial Gram entries}

Fix a closed genus-$g$ lollipop spine $G$ and its universal core
$L_\star=\bigoplus_\lambda\Lambda\mathbb U_\lambda$.  Write
\[
 \varrho_\lambda^\star=B_G(\mathbb U_\lambda,\mathbb U_\lambda).
\]

\begin{lemma}[Polynomial Gram diagonal]
\label{p4:retained:lem:c15-gram}
For every coloring $\lambda$, one has
$\varrho_\lambda^\star\in\Lambda$.
\end{lemma}

\begin{proof}
 The Gram diagonal is obtained by using one negative twist eigenvalue as a
test rather than dividing by it.  The contracted kernel is already
polynomial on the universal core, so multiplying back by the monomial
eigenvalue proves polynomiality in the coefficient ring without any
unstated cancellation.

Choose any standard handle whose loop color is $\ell$.  The diagonal
coefficient of the negative twist is $\mu_\ell^{-1}$.  The corresponding
contracted kernel on the universal core is polynomial, hence
$\mu_\ell^{-1}\varrho_\lambda^\star\in\Lambda$.  Multiplication by
$\mu_\ell\in\Lambda$ proves the claim; no division in $\Lambda$ is used.
\end{proof}

\subsection{The direct disk twist}

Let $e$ be an edge of $G$ with integer Jones--Wenzl color $j_e(\lambda)$,
and let $D_e$ be its transverse dual disk.  Denote the two disk twists by
$T_{e,+}$ and $T_{e,-}$.  Costantino's half twist on spin $s$ is
\cite{Costantino}
\[
 H_s=\mathrm i^{2s}x^{s(s+1)}\operatorname{id}_s.
\]
Use the formal identification $j=2s$ and $x^{1/2}=A$; this is a formal
square-root convention, not an analytic branch simplification.

\begin{theorem}[Source-matched disk-twist eigenvalue]
\label{p4:retained:thm:c15-eigenvalue}
On every fixed-color graph line,
\[
 T_{e,+}\mathbb U_\lambda=\mu_{j_e(\lambda)}\mathbb U_\lambda,
 \qquad
 T_{e,-}\mathbb U_\lambda=\mu_{j_e(\lambda)}^{-1}\mathbb U_\lambda,
\]
where
\[
 \mu_j=(-1)^jA^{j(j+2)}.
\]
\end{theorem}

\begin{proof}
 The disk twist is diagonal on the graph line because the transverse disk is
dual to one colored edge.  Squaring the half twist gives the full-twist
eigenvalue, and the bridge parity calculation then identifies the integral
exponent available on a separating bridge.  This supplies a local chamber
seed; it does not make a scalar denominator-clearing factor into an invertible polarized action.

Squaring the half twist gives
\[
 H_s^2=\mathrm i^{4s}x^{2s(s+1)}
 =(-1)^jA^{j(j+2)}.
\]
This agrees with the standard Kauffman full-twist eigenvalue on a colored
edge.  The universal core vector is a scalar multiple of the same definite
edge-color graph vector, so it has the same eigenvalue.
\end{proof}

\begin{lemma}[Closed bridge parity]
\label{p4:retained:lem:c15-parity}
If $e$ is a bridge of a closed trivalent spine, then every admissible
coloring has $j_e=2a_e$ even, and
\[
 \mu_{j_e}=A^{4a_e(a_e+1)},\qquad
 \frac{j_e(j_e+2)}2=2a_e(a_e+1)\in\mathbb N.
\]
\end{lemma}

\begin{proof}
Remove the bridge and sum the vertex parity equations over one component.
Every internal edge is counted twice and the bridge once, forcing its color
to be even.  The formula follows from
Theorem~\ref{p4:retained:thm:c15-eigenvalue}.
\end{proof}

Conversely, a separating compressing disk can be made dual to a bridge:
cut along the disk, choose a spine on each side with one endpoint on the cut
disk, and join the two endpoints.  Thus every separating meridian has such
a seed in an appropriate marking.

If $w(j)=j(j+2)$, then for $n\in\mathbb Z$,
\[
 \operatorname{ord}_{A=0}(\mu_j^n)=nw(j),\qquad
 \operatorname{ord}_{A^{-1}=0}(\mu_j^n)=-nw(j),
\]
and every cyclotomic valuation of $\mu_j^n$ is zero.  Hence all nonmonomial
denominators in an $F$-conjugated disk twist come from the recoupling
coordinates, not from the twist eigenvalue.

\subsection{Three inequivalent responses to the inverse pole}

The raw inverse contracted coefficient is
\[
 B_G(T_{e,-}\mathbb U_\lambda,\mathbb U_\lambda)
 =(-1)^{j_e}A^{-w(j_e)}\varrho_\lambda^\star.
\]
If $v_\lambda=\operatorname{ord}_A\varrho_\lambda^\star$, the least
nonnegative one-sided scalar denominator-clearing factor is
\[
 r^-_\lambda=\max\{0,w(j_e)-v_\lambda\}.
\]
It regularizes one scalar coefficient but does not change the inverse
operator.

If both pairing arguments are multiplied by $A^r$, the least symmetric
exponent is
\[
 r_\lambda^{\mathrm{sym}}=
 \max\left\{0,
 \left\lceil\frac{w(j_e)-v_\lambda}{2}\right\rceil\right\}.
\]
For a closed bridge, $w(j_e)/2$ is a canonical norm-independent choice.

For the honest invertible action define polarized endpoint lattices
\[
 L_{e,k}=\bigoplus_\lambda
 \Lambda\mu_{j_e(\lambda)}^k\mathbb U_\lambda,
 \qquad k\in\mathbb Z.
\]
Then
\[
 T_{e,+}:L_{e,k}\xrightarrow{\sim}L_{e,k+1},
 \qquad
 T_{e,-}:L_{e,k+1}\xrightarrow{\sim}L_{e,k}.
\]

\begin{proposition}[Denominator-clearing factor versus endpoint]
\label{p4:retained:prop:c15-three}
For a nonvacuum color:
\begin{enumerate}
\item a one-sided denominator-clearing factor regularizes one coefficient but is not an inverse
twist;
\item a symmetric half-shift handles a bounded signed family of contracted
pairings but not arbitrary negative powers;
\item only the polarized endpoint sequence carries the exact inverse pair.
\end{enumerate}
No nonzero finite free $\Lambda$-lattice on the eigenline is preserved by
both signs.
\end{proposition}

\begin{proof}
 The three responses are separated by what they preserve.  A one-sided
denominator-clearing factor changes a coefficient, a symmetric shift handles only a bounded
family of pairings, and the polarized degree sequence is the object on
which the twist and its inverse act exactly.  The nonunit eigenvalue rules
out collapsing those degrees into one finite free lattice.

Preservation by both signs would make $\mu_j$ and $\mu_j^{-1}$
endomorphism scalars and hence units, impossible for $j>0$.  For a fixed
symmetric denominator-clearing factor the $k$-fold negative coefficient has order
$2r+v_\lambda-kw(j)\to-\infty$.  The endpoint identities are immediate
from their definition.
\end{proof}

The common intersection of the three degrees $-1,0,1$ is the fully shifted
line $L_{e,1}$, not the half-shift used for a bilinear contraction.

\subsection{Simultaneous bridge scalarization}

Let $E$ be a finite set of bridge edges and define
\[
 w_e(\lambda)=j_e(\lambda)(j_e(\lambda)+2),\qquad
 W_E(\lambda)=\sum_{e\in E}w_e(\lambda).
\]
Every term is even.  Put
\[
 \mathbb S_\lambda^E=A^{W_E(\lambda)/2}\mathbb U_\lambda,
\]
and, for $\boldsymbol\delta\in\{-1,0,1\}^E$,
\[
 T_E^{\boldsymbol\delta}=
 \prod_{e\in E}T_{e,+}^{\delta_e}.
\]
The factors commute because their dual disks are disjoint.

\begin{theorem}[Separating-edge contracted core]
\label{p4:retained:thm:c15-separating-core}
For every square-free signed separating multitwist,
\[
 B_G(T_E^{\boldsymbol\delta}\mathbb S_\mu^E,
                 \mathbb S_\lambda^E)
 =\delta_{\lambda\mu}
 A^{W_E(\lambda)+\sum_e\delta_ew_e(\lambda)}
 \varrho_\lambda^\star\in\Lambda.
\]
The all-negative corner has zero additional exponent, so the half-shift is
sharp without further Gram divisibility.
\end{theorem}

\begin{proof}
 The separating core calculation is diagonal after bridge parity.  Orthogonality
reduces the contracted pairing to one coloring, the half-shift contributes
one copy of the bridge weight, and the signed power contributes the second
copy or removes it at the negative corner.  The remaining Gram entry is
polynomial by the earlier lemma, so every square-free corner is integral.

Bridge parity removes all signs from the eigenvalues.  Orthogonality makes
the kernel diagonal, and the exponent is
$\sum_e(1+\delta_e)w_e(\lambda)\geq0$.  The remaining Gram factor is
polynomial by Lemma~\ref{p4:retained:lem:c15-gram}.
\end{proof}

\paragraph{Worked checkpoint.}
For one closed bridge of color $j=2$,
\[
 w(2)=8,\qquad \mu_2=A^8,\qquad \frac{w(2)}2=4.
\]
After extracting the polynomial Gram entry
$\varrho_\lambda^\star$, the canonically shifted contracted factors at
$\delta=-1,0,1$ are respectively
\[
 1,\qquad A^8,\qquad A^{16}.
\]
The all-negative corner has exponent zero exactly, while the other two
corners have positive disk order.

A standard-handle triangular twist changes only its active loop excess and
fixes all bridge colors.  Hence its nonzero contracted entries have the
same $W_E$ on source and output.

\begin{corollary}[Combined standard and separating family]
\label{p4:retained:cor:c15-combined}
For every square-free signed standard multitwist $T_{S,\varepsilon}$,
\[
 B_G(T_E^{\boldsymbol\delta}T_{S,\varepsilon}\mathbb S_\mu^E,
                     \mathbb S_\lambda^E)
 =A^{W_E(\lambda)+\sum_e\delta_ew_e(\lambda)}
 \kappa^{\star,S,\varepsilon}_{\lambda\mu}\in\Lambda
\]
on the inherited triangular support, and it vanishes elsewhere.
\end{corollary}

\begin{proof}
A standard-handle triangular twist changes only its active loop excess and
fixes all bridge colors.  The preceding separating-edge theorem therefore
applies without changing the bridge weight, and the separating twists
contribute exactly \(\sum_e\delta_e w_e(\lambda)\).  This gives the
displayed formula on the inherited triangular support and zero elsewhere.
\end{proof}

For a relative graph, bridge parity can be odd because external colors
contribute.  Replacing $W_E/2$ by $\lceil W_E/2\rceil$ leaves residual
order zero or one.  The closed Heegaard sector needs no ceiling.

\subsection{Polarized bridge degrees and literal approximants}

For $\mathbf k\in\mathbb Z^E$, define
\[
 m_{\mathbf k}(\lambda)=\prod_{e\in E}
 \mu_{j_e(\lambda)}^{k_e},
\]
\[
 \mathbb U_{\mathbf k,\lambda}=m_{\mathbf k}(\lambda)\mathbb U_\lambda,
 \qquad
 \mathbb U_{\mathbf k}^{\lambda}=
 m_{\mathbf k}(\lambda)^{-1}\mathbb U^\lambda.
\]
Then
\[
 \mathbb U_{\mathbf k+\boldsymbol\delta}^{\lambda}
 \bigl(T_E^{\boldsymbol\delta}
 \mathbb U_{\mathbf k,\mu}\bigr)=\delta_{\lambda\mu}.
\]
For a finite lower ideal $C$, set
\[
 X_{\mathbf k,C}(x)=\sum_{\lambda\in C}
 x_\lambda\mathbb U_{\mathbf k,\lambda},
\]
\[
 F_{\mathbf k,C}(y)=\sum_{\lambda\in C}
 J_{H_G(\lambda)}y_\lambda\mathbb U_{\mathbf k}^{\lambda}.
\]

The index $\mathbf k$ records the current bridge-twist degree.  The factor
$m_{\mathbf k}$ changes the primal frame and its inverse changes the cotest,
so the pairing remains Kronecker; $X_{\mathbf k,C}$ and
$F_{\mathbf k,C}$ are the corresponding finite primal and regulated cotest
sums.

\begin{theorem}[Bridge-degree finite representatives]
\label{p4:retained:thm:c15-bridge-common}
For every integer power vector $\boldsymbol\delta$,
\[
 F_{\mathbf k+\boldsymbol\delta,C}(y)
 \bigl(T_E^{\boldsymbol\delta}X_{\mathbf k,C}(x)\bigr)
 =\sum_{\lambda\in C}J_{H_G(\lambda)}x_\lambda y_\lambda.
\]
The polynomial is independent of the bridge degree.  If $C\subset C'$, its
increment is in $J_{\nu(C)}\Lambda$ and hence is simultaneously small in
the disk and cyclotomic topologies.
\end{theorem}

\begin{proof}
 The bridge-degree identity is labelwise contragredient cancellation.  The
finite lower-ideal increment is controlled by the height regulator, so the
same diagonal polynomial survives both disk and cyclotomic projection.
The actual root-specialized kernel still retains its operator coefficient; only the
transported coordinate pairing is degree-independent.

The target cotest cancels the source multiplier label by label.  Every new
label has height at least $\nu(C)$, so its regulator is a multiple of
$J_{\nu(C)}$.
\end{proof}

The root-specialized contracted net retains the actual operator coefficient:
\[
 \sum_{\lambda\in C}\sum_{\mu\preceq_S\lambda}
 x_\lambda y_\mu J_{H_G(\lambda)}
 A^{W_E(\lambda)+\sum_e\delta_ew_e(\lambda)}
 \kappa^{\star,S,\varepsilon}_{\lambda\mu}.
\]
It is polynomial and has the same tail estimate, but unlike the coordinate
pairing it is not degree-independent.

\subsection{The bridge-extended chamber theorem}

Adjoin at every marked spine the bridge generators and their inverses to
the algebraic recoupling groupoid.  Quotient by exact pure recoupling
relations, literal twist inverse relations, commutation for disjoint disks,
and coordinate naturality
\[
 T_e^{(G')}=U_{G'G}T_e^{(G)}U_{GG'}
\]
whenever both markings describe the same disk twist.  Retain all nonunit
central degrees.  Let $D_G^{\rm br}$ be the resulting degrees from the base
marking to $G$ and set
\[
 \mathfrak A_G^{\mathrm{orb,br}}=
 \bigoplus_{\gamma\in D_G^{\rm br}}^{\mathrm{lat}}
 (\mathfrak A_0\sqcup\{\bot\}).
\]

\begin{theorem}[Arbitrary-genus bridge chamber]
\label{p4:retained:thm:c15-bridge-chamber}
For every fixed finite genus:
\begin{enumerate}
\item the bridge orbit chamber is a directed distributive lattice;
\item every verified recoupling or bridge arrow acts by an exact lattice
isomorphism and polarized $\Lambda$-linear transport;
\item the $J$-completion of the external orbit--Rees sum is the restricted
product with finite degree support modulo every $J_N$;
\item the transported finite polynomial of
Theorem~\ref{p4:retained:thm:c15-bridge-common} is independent of degree; and
\item every separating compressing-disk twist and all its powers occur
after choosing a bridge marking in the verified recoupling component.
\end{enumerate}
\end{theorem}

\begin{proof}
 The bridge chamber is built by adjoining exact degree shifts to the
recoupling orbit.  The finite-support and restricted-product arguments
transport unchanged, while the bridge marking realizes every separating
compressing disk in the verified component.  Keeping the nonunit degree
visible is what lets all bridge powers coexist with the common completion.

The bridge seed is diagonal and algebraic in both directions over $\FA$ and
acts by an endpoint degree shift.  The finite-support lattice and restricted
completion arguments of Section~\ref{p4:retained:ch:recoupling-groupoid} apply
unchanged.  Contragredient cancellation proves the common polynomial.  The
bridge realization following Lemma~\ref{p4:retained:lem:c15-parity} gives the final
claim.
\end{proof}

This covers separating meridians, not separating curves which do not bound
disks in the chosen handlebody.

\subsection{Ribbon consistency and the chain limitation}

The phased braiding satisfies the balancing identity
\[
 \theta_c=\theta_a\theta_b(R_c^{ab})^2,
 \qquad
 \theta_s=\mathrm i^{4s}x^{2s(s+1)}.
\]
With new channels as rows and old channels as columns, the correct
conjugations are
\[
 D_{\rm old}^{(\rm new)}=FD_{\rm old}F^{-1},
 \qquad
 D_{\rm new}^{(\rm old)}=F^{-1}D_{\rm new}F.
\]
Unequal-boundary blocks distinguish this orientation from the reverse one.

For three equal inner colors $a$ and total color $d$, let
$B_1=R$, $B_2=F^{-1}RF$, and define
\[
 A_{12}=B_1^2,\qquad A_{13}=B_2B_1^2B_2^{-1},\qquad A_{23}=B_2^2.
\]
The braid and ribbon identities imply
\[
 A_{12}A_{13}A_{23}=(B_1B_2)^3
 =\frac{\theta_d}{\theta_a^3}\operatorname{id}.
\]
Restoring the two inner-boundary twists in each pair curve gives the local
ribbon-lantern identity
\[
 T_{12}T_{13}T_{23}=\theta_a^3\theta_d\operatorname{id}.
\]
This is a verified local compatibility relation, not a complete
arbitrary-label map\-ping-class presentation.  The braid and pair-curve arrows
used in this certificate are not adjoined to the bridge-only groupoid above;
the identity is an external compatibility certificate, not an additional
quotient relation in that groupoid.

At prime level $p\geq5$, for an admissible integer
$0\leq c\leq(p-3)/2$, the one-holed-torus chain relation
\cite{GilmerMasbaum} gives
\[
 (tt^*)^6=q^{-6}T_\delta=A^{-12}T_\delta.
\]
This root-level central character does not select a canonical generic disk
degree: adding multiples of the root order changes the formal $A$-valuation
without changing the root scalar.  The untruncated opposite-torus/Hopf
operator also has infinite columns, so no generic chain relation is inferred
from this finite-level identity.