% Provenance: corrected synthesis of stages 18--20; stage numbers occur only in this comment.
\section[Opposite Twists and Endpoint Chambers]
{Opposite Twists, Polynomial Endpoints, and Chamber Approximants}
\label{p4:retained:ch:opposite-endpoints}

The generic longitude twist has infinite columns in the Newton basis, so it
is not an endomorphism of the algebraic direct sum.  It nevertheless has
row-finite triangular coordinates, an exact generic inverse, polynomial
source and target endpoint bases, and integral Hopf-contracted kernels after
explicit separable clearing.  These facts produce an arbitrary-genus
chamber system for standard handles without pretending that one
non-Laurent primal lattice is invariant under both twist signs.
The one-holed-torus basis and twist conventions are matched to
\cite{GilmerMasbaum}, with the graphical normalization fixed by
\cite{Costantino}.

\subsection{Positive and negative longitude matrices}

Retain
\[
 \Lambda=\mathbb Z[A],\qquad q=A^2,\qquad \FA=\mathbb Q(A),
 \qquad J_h=q^h(q;q)_h.
\]
For boundary color $2c$, set
\[
 Q_n^{\prime(c)}=\frac{Q_n^{(c)}}{\{n\}!},\qquad
 \{j\}=(-A)^j-(-A)^{-j}.
\]
The positive and negative Kirby multipliers have coefficients
\[
 \gamma_k^+=
 \frac{(-A)^{(k^2+5k)/2}}
      {\prod_{j=1}^k(A^{2j+1}-1)},
 \qquad
 \gamma_k^-=A^{-3k}\gamma_k^+.
\]
With
\[
 C^l_{a,b}=(-1)^l
 \frac{\{a\}!\{b\}!\{a+b+1\}!}
 {\{a-l\}!\{b-l\}!\{a+b+1-l\}!\{l\}!},
\]
define, for $n\geq m$ and $k=l+n-m$,
\[
 b_{n,m,l}^{\pm(c)}=
 C^l_{k,m+c}\frac{\gamma_k^\pm}{\{k\}!}
 \frac{\{n\}!}{\{m\}!},
 \qquad
 b_{n,m}^{\pm(c)}=\sum_{l=0}^{m+c}b_{n,m,l}^{\pm(c)}.
\]
Then
\[
 t_\pm^\star Q_m^{\prime(c)}
 =\sum_{n=m}^{\infty}b_{n,m}^{\pm(c)}Q_n^{\prime(c)},
\]
and the diagonal entries are
\[
 b_{m,m}^{+(c)}=\mu_{c+m},\qquad
 b_{m,m}^{-(c)}=\mu_{c+m}^{-1},\qquad
 \mu_j=(-1)^jA^{j(j+2)}.
\]

Each coefficient $b_{n,m}^{\pm(c)}$ is a finite sum, but a fixed column
has infinitely many rows.  These formulas therefore define formal
lower-triangular matrices; the next proposition asserts an entrywise inverse
identity, not an endomorphism of the algebraic direct sum.

\begin{proposition}[Formal inverse identity]
\label{p4:retained:prop:c12-inverse}
The two infinite lower-triangular matrices $B^+$ and $B^-$ are formal
inverses: every matrix-product entry is a finite sum and equals
$\delta_{nm}$ in $\FA$.
\end{proposition}

\begin{proof}
 The inverse statement is checked entry by entry, rather than by multiplying
infinite operators on the direct sum.  For fixed source and target indices the
triangular shape leaves only finitely many intermediate colors.  Root-level
inverse twists identify that finite rational expression at infinitely many
admissible specializations, which forces the generic identity.

For fixed $(n,m)$ the product uses only $m\leq k\leq n$.  At every
sufficiently large admissible odd level, the chosen lifts are inverse
geometric twists, so this finite rational function of $A$ equals
$\delta_{nm}$.  Equality at infinitely many values of $A$ proves the
generic identity.
\end{proof}

\subsection{Endpoint-local products}

The following general lemma explains why row-finite triangular operators
act on product completions even when their columns are infinite.

\begin{theorem}[Two-sided triangular endpoint products]
\label{p4:retained:thm:c12-formal-products}
Let $V=\bigoplus_{h\geq0}V_h$, with every $V_h$ finite dimensional, and
let $T_{hk}=0$ for $h<k$ with $T_{hh}$ invertible.  There are finite free
lattices $L_h^-$ and $L_h^+=T_{hh}L_h^-$ for which the normalized matrix
$B$ of $T$ and its inverse $C=B^{-1}$ are lower unitriangular over
$\Lambda$.

On
\[
 \mathcal P_\pm=\prod_{h\geq0}
  (\widehat\Lambda_J\otimes_\Lambda L_h^\pm)
\]
the maps $T$ and $T^{-1}$ are continuous inverse homeomorphisms.  For
normalized dual-coordinate products the covector push-forward is
\[
 (T_\#y)_k=\sum_{h\geq k}\frac{J_h}{J_k}C_{hk}^*y_h,
\]
and
\[
 \langle x,y\rangle_\pm=\sum_hJ_hy_h(x_h)
\]
is convergent and invariant under $(T,T_\#)$.
\end{theorem}

\begin{proof}
 The endpoint lattices absorb the diagonal denominators first, and the
successive denom\-inator-clearing factors absorb the finitely many lower-triangular blocks at each
height.  The remaining inverse is integral by finite triangular recursion.
The $J$-filtration then suppresses the distant tail in the dual action, so
finite-support pairing invariance extends to the product completion.

Choose reference lattices $M_h$.  Clearing the finitely many blocks
$T_{hh}^{-1}T_{hk}$ with $k<h$ by factors $\Delta_h$, and putting
$D_h=\prod_{j\leq h}\Delta_j$, gives
\[
 L_h^-=D_h^{-1}M_h,\qquad L_h^+=T_{hh}L_h^-,\qquad
 B_{hk}=\frac{D_h}{D_k}T_{hh}^{-1}T_{hk}\in\operatorname{Mat}(\Lambda).
\]
Finite triangular recursion makes $C$ integral and row finite.  For the
dual tail use the exact identity
\[
 \frac{J_h}{J_k}
 =J_{h-k}\genfrac{[}{]}{0pt}{}{h}{k}_q\in(J_{h-k}).
\]
Hence the terms with $h\geq k+N$ vanish modulo $J_N$.  Ordinary finite
matrix algebra proves inversion and pairing invariance on finite-support
sequences, and continuity and separatedness extend the identities.
\end{proof}

This theorem is formal: arbitrary denominator-clearing factors $D_h$ can have unbounded disk
or cyclotomic divisors.  The explicit endpoint theorem below removes this
defect for a standard longitude coordinate matrix.

\subsection{Closed matrices and the universal polynomial endpoint pair}

Put
\[
 d=n-m,\qquad M=m+c,\qquad
 \tau_{n,m}=\frac{d(n+m+1)}2,
\]
\[
 \Xi(c,n,m)=\frac{d^2+3d+M^2+2Md+M}{2},\qquad
 D_{M,d}(A)=\prod_{j=1}^{d}(1-A^{2(M+j)+1}).
\]

\begin{theorem}[Closed raw matrices]
\label{p4:retained:thm:c12-raw}
The divided-Newton matrices satisfy
\[
 b_{n,m}^{+(c)}=
 (-1)^d(-A)^{M-\tau_{n,m}}q^{\Xi(c,n,m)}
 \frac{\genfrac{[}{]}{0pt}{}{n}{m}_q}{D_{M,d}(A)},
\]
\[
 b_{n,m}^{-(c)}(A)=(-1)^d b_{n,m}^{+(c)}(A^{-1}).
\]
Their exact $A$-orders are
\[
 v_+=M(M+2)+d(m+2c)+\frac{d(d+5)}2,\qquad
 v_-=-M(M+2)-dm+\frac{d(d-1)}2.
\]
\end{theorem}

\begin{proof}
 The raw coefficient calculation separates three tasks: convert the brace
quotients to a Gaussian polynomial, collect the Laurent monomial, and leave
the finite denominator $D_{M,d}$ visible.  Because the two non-monomial
factors have constant coefficient one, the resulting orders are read from
the collected monomial without an unproved cancellation claim.

The brace quotient is
\[
 \frac{\{n\}!}{\{m\}!(q;q)_d}
 =(-1)^d(-A)^{-\tau_{n,m}}
 \genfrac{[}{]}{0pt}{}{n}{m}_q.
\]
Substitution into the terminating contracted formula proved below gives the
positive identity; mirror substitution gives the negative one.  Both the
Gaussian polynomial and $D_{M,d}$ have constant coefficient one, so the
remaining monomial powers give the exact orders.
\end{proof}

Define
\[
 f(n)=\binom n2,\qquad
 F_n^{(c)}=(Aq^{c+1};q)_n,
\]
and the endpoint vectors
\[
 \mathsf E_n^-=\frac{A^{-f(n)}}{F_n^{(c)}}Q_n^{\prime(c)},
 \qquad
 \mathsf E_n^+=\mu_{c+n}\mathsf E_n^-.
\]

\begin{theorem}[Polynomial endpoint pair]
\label{p4:retained:thm:c12-endpoints}
In the source and target endpoint bases, the two twist matrices are
\[
 \widehat\beta^+_{n,m}
 =(-1)^{\tau_{n,m}}\genfrac{[}{]}{0pt}{}{n}{m}_q,
\]
\[
 \widehat\beta^-_{n,m}
 =(-1)^{d+\tau_{n,m}}A^{d(d-1)}
 \genfrac{[}{]}{0pt}{}{n}{m}_q.
\]
They lie in $\mathbb Z[A]$, are independent of $c$, are lower
unitriangular, and are formal inverses.
\end{theorem}

\begin{proof}
 The endpoint basis is chosen to contain exactly the denominator string that
appears in the raw coefficient.  Conjugating by those basis factors therefore
turns the two twist matrices into signed Gaussian matrices.  Finite
$q$-binomial inversion supplies the opposite direction and explains why the
result is polynomial and independent of the boundary color.

Because $D_{M,d}=F_n^{(c)}/F_m^{(c)}$, endpoint conjugation cancels the
entire denominator.  If
$G_{n,m}=\genfrac{[}{]}{0pt}{}{n}{m}_q$ and
$R_{nn}=(-1)^{n(n+1)/2}$, then
$\widehat\beta^+=RGR^{-1}$.  The finite $q$-binomial inversion formula
\[
 (G^{-1})_{n,m}=(-1)^{n-m}q^{\binom{n-m}{2}}
 \genfrac{[}{]}{0pt}{}{n}{m}_q
\]
gives $\widehat\beta^-=RG^{-1}R^{-1}$.
\end{proof}

The factor $F_n^{(c)}$ is an $A$-adic unit.  At the admissible odd-prime
branch $A=-q^{(p+1)/2}$ it is also a unit on every admissible color block.
This is a cutoff-dependent finite-level statement, not an untruncated
branch-independent Habiro operator.

There is a sharp obstruction to using one bounded-below diagonal chart.

\begin{proposition}[Diagonal two-sided no-go]
\label{p4:retained:prop:c12-diagonal-nogo}
If source and target weights $s_m,t_m\in \FA^\times$ make both raw twist
directions entrywise $A$-regular, and
$a_m=\operatorname{ord}_A(s_m)$, $b_m=\operatorname{ord}_A(t_m)$, then
\[
 b_m-a_m=(m+c)(m+c+2),\qquad
 a_m-a_{m+d}\geq dm+\frac{d(d-1)}2.
\]
In particular $a_n\leq a_0-n(n-1)/2$.  The exponent in
Theorem~\ref{p4:retained:thm:c12-endpoints} is optimal among diagonal coordinate
weights.
\end{proposition}

\begin{proof}
 The no-go calculation reads the diagonal eigenvalue relation first, then
uses the lowest off-diagonal order of the positive twist to force a
monotone loss in the source weights. Iterating that inequality gives the
quadratic bound, and the explicit endpoint chart attains it; the obstruction
is therefore a coordinate-weight obstruction, not a failure of the formal
inverse identity.

The diagonal eigenvalues force the first equality.  Substitution into the
positive off-diagonal order gives the second inequality; setting $m=0$
gives the bound.  The displayed endpoint pair attains equality.
\end{proof}

\subsection{Hopf-contracted kernels}

Let $\eta_{c,n}$ be the relative norm from
Theorem~\ref{p4:retained:thm:c11-norm}.  Define
\[
 \mathcal K_{n,m,l}^{\pm(c)}=
 \eta_{c,n}C^l_{l+n-m,m+c}
 \frac{\gamma_{l+n-m}^\pm}{\{l+n-m\}!},
 \qquad
 \mathcal K_{n,m}^{\pm(c)}=\sum_{l=0}^{m+c}
 \mathcal K_{n,m,l}^{\pm(c)}.
\]
Thus
$\mathcal K_{n,m}^{\pm(c)}=B_c(t_\pm^\star Q_m^{(c)},Q_n^{(c)})$.

\begin{theorem}[Terminating product and Laurent cancellation]
\label{p4:retained:thm:c12-kernel}
For $n\geq m$,
\[
 \mathcal K_{n,m}^{+(c)}=
 \eta_{c,n}\frac{(-A)^Mq^{\Xi(c,n,m)}}
 {(q;q)_d(Aq^{M+1};q)_d},
\]
and
\[
 \mathcal K_{n,m}^{-(c)}(A)=
 (-1)^cA^{-2c(c+1)}\mathcal K_{n,m}^{+(c)}(A^{-1}).
\]
Both kernels lie in $\mathbb Z[A,A^{-1}]$ for all indices.
\end{theorem}

\begin{proof}
 The Hopf pairing converts the infinite twist column into a finite sum indexed
by the contracted color.  A terminating $q$-identity evaluates that sum, and
the relative norm supplies precisely the factors needed to cancel the
remaining Pochhammer denominator.  The positive calculation and its mirrored
negative version are performed before the common separable clearing is
introduced.

After converting braces to $q$-products, the positive finite sum is a
terminating $q$-Chu--Vandermonde sum:
\[
 \sum_{l=0}^{M}
 \frac{(q^{-M};q)_l(q^{n+c+2};q)_l}
 {(Aq^{d+1};q)_l(q;q)_l}q^l.
\]
Its finite product evaluation, followed by cancellation of the two
Pochhammer strings, gives the first formula.  Mirroring each summand gives
the second.  For Laurent polynomiality, use
\[
 \frac{\eta_{c,n}}{\eta_{c,m}}
 =\frac{\{n\}!}{\{m\}!}
 \prod_{r=1}^d\{2M+2r+1\}
 \prod_{r=1}^d\{2c+m+1+r\}^{+}.
\]
The first quotient cancels $(q;q)_d$ up to a Gaussian polynomial, and
\[
 \frac{\{j\}}{1-A^j}=-(-A)^{-j}(1+A^j)
\]
cancels each factor of $(Aq^{M+1};q)_d$.  The mirror identity handles the
negative kernel.
\end{proof}

The vacuum column is especially simple:
\[
 \mathcal K_{n,0}^{+(0)}=
 A^{-n(n+1)}\prod_{j=3}^{2n+2}(1+A^j),
 \qquad
 \mathcal K_{n,0}^{-(0)}=
 A^{-n(n+4)}\prod_{j=3}^{2n+2}(1+A^j).
\]
\paragraph{Worked checkpoint.}
For $c=m=0$ and $n=1$ these formulas give
\[
 \mathcal K_{1,0}^{+(0)}=A^{-2}(1+A^3)(1+A^4),
 \qquad
 \mathcal K_{1,0}^{-(0)}=A^{-5}(1+A^3)(1+A^4).
\]
Since $R_0(1)+R_0(0)=8+2=10$, the common clearing produces
\[
 A^{10}\mathcal K_{1,0}^{+(0)}=A^8(1+A^3)(1+A^4),
 \qquad
 A^{10}\mathcal K_{1,0}^{-(0)}=A^5(1+A^3)(1+A^4),
\]
both in $\mathbb Z[A]$.

One separable two-sided polynomial shift is
\[
 R_c(s)=2(s+c+1)^2.
\]

\begin{corollary}[Separable kernel clearing]
\label{p4:retained:cor:c12-separable}
For both signs and every $n\geq m$,
\[
 A^{R_c(n)+R_c(m)}\mathcal K_{n,m}^{\pm(c)}\in\mathbb Z[A].
\]
\end{corollary}

\begin{proof}
 The uniform clearing is obtained by bounding the valuation of each summand,
then writing \(n=m+d\) so the proposed separable exponent dominates both
sign-specific losses coefficientwise. Laurent polynomiality finishes the
integrality claim; the sharper cancellation in the positive case is not needed
for this uniform statement.

Direct summand valuation gives sufficient exponents
\[
 E_{\Sigma,+}=n^2+(4c+2m+1)n+3c^2+2cm+5c+2m,
\]
and $E_{\Sigma,-}=E_{\Sigma,+}+3(n+c)$.  Subtracting these from
$R_c(n)+R_c(m)$ gives polynomials with nonnegative coefficients after
writing $d=n-m$.  Laurent polynomiality then proves the claim.  The exact
full positive order contains additional cancellation, but is not needed for
this uniform separable bound.
\end{proof}

\subsection{Arbitrary-genus cutoff and endpoint chambers}

For a fixed lollipop marking $G$, encode each dynamic coloring injectively
by a multidegree $d_G(\lambda)\in\mathbb N^r$ and order the admissible set
componentwise.  Let $\mathcal C_G$ be its finite lower ideals.  Intersection
and union make $\mathcal C_G$ a directed distributive lattice; every
principal lower ideal is finite.  The height
\[
 H_G(\lambda)=\max_jd_G(\lambda)_j
\]
is proper and monotone, and $C_N=\{H_G<N\}$ is a cofinal sequence.

For a sign vector $\varepsilon\in\{-,+\}^g$, define
\[
 \mathsf e_\lambda^\varepsilon=
 \left(\prod_{i=1}^g\alpha_{a_i,b_i}^{\varepsilon_i}\right)
 \widetilde b_\lambda,
\]
where
\[
 \alpha^-_{a,n}=\frac{A^{-\binom n2}}{F_n^{(a)}},
 \qquad \alpha^+_{a,n}=\mu_{a+n}\alpha^-_{a,n}.
\]
For a finite lower ideal $C$, let
$L_\varepsilon(C)=\bigoplus_{\lambda\in C}\Lambda
\mathsf e_\lambda^\varepsilon$ and take the inverse limit under coordinate
projection.

\begin{theorem}[Endpoint chamber cube]
\label{p4:retained:thm:c12-cube}
If $\varepsilon_s=-$, the positive standard twist on handle $s$ is a
continuous isomorphism from the $\varepsilon$ product to the product with
the $s$th sign changed to $+$; the negative twist is its inverse.  The
local matrices are those of Theorem~\ref{p4:retained:thm:c12-endpoints}.  Twists on
distinct handles tensor and commute, producing a $g$-dimensional Boolean
endpoint cube.
\end{theorem}

\begin{proof}
 The endpoint cube is assembled at finite lower ideals. A fixed output color
uses only finitely many lower source colors, so each truncation sees the
finite Gaussian inverse pair; coordinate projections commute with those
finite matrices. The tensor factors belong to distinct handles, which is
what makes the endpoint changes commute without turning the cube into an
integral action by repeated same-sign powers.

For any fixed output color only lower source colors occur.  A lower ideal
therefore contains every source coordinate needed to compute each output.
Every finite truncation is an exact Gaussian inverse pair, and coordinate
projection is compatible with the matrices.  Distinct handles act on
distinct tensor factors.
\end{proof}

The cube is square-free, not an integral $\mathbb Z^g$ action: repeated
same-sign twists introduce a negative off-diagonal disk order.  Arbitrary
powers require retaining an orbit degree.

\subsection{Spectator denominator-clearing factors and scalar chambers}

First single out one handle $s$.  Let its stick half-color be
$a_s=\kappa$, let
$\sigma=(a,b_{\widehat s},c_{\rm tree})$ collect the spectator labels, and
write
\[
 r_\sigma=\left\lfloor
 \frac{-e+\sum_i a_i}{2}\right\rfloor
\]
when the trunk color is $2e$.  Unprime only the active factor:
\[
 V_{\sigma,n}=\{n\}!\,\widetilde b_{(a,b_{\widehat s},n,c_{\rm tree})}.
\]
Higher-genus orthogonality factorizes as
\[
 B_G(V_{\sigma,n},V_{\tau,m})
 =\delta_{\sigma\tau}\delta_{nm}\rho_\sigma(A)\eta_{\kappa,n},
\]
where
\[
 \rho_\sigma=(1-q)^{-2r_\sigma}
 \langle DG(a,c_{\rm tree})\rangle
 \frac{\{1\}_q}{\{2\kappa+1\}_q}
 \prod_{i\ne s}
 \left(
  \frac{\eta_{a_i,b_i}}{\{b_i\}!^2}
  \frac{\{1\}_q}{\{2a_i+1\}_q}
 \right).
\]
This factor is independent of $n$ and $m$.

For $a,b\geq0$, put
\[
 E(a,b)=a^2+2ab+a+\frac{b^2+3b}{2}
\]
and
\[
 S_{a,b}(q)=
 \frac{(q;q^2)_{a+b+1}(-q;q)_{2a+b+1}}
 {(q;q)_b\genfrac{[}{]}{0pt}{}{2a}{a}_{q^2}
  (1-q^{4a+2})}.
\]

\begin{theorem}[Polynomial divided-spectator factor]
\label{p4:retained:thm:c12-spectator}
One has
\[
 \frac{\eta_{a,b}}{\{b\}!^2}
 \frac{\{1\}_q}{\{2a+1\}_q}
 =(-1)^aq^{-E(a,b)}S_{a,b}(q),
\]
with
\[
 S_{a,b}(q)\in\mathbb Z[q],\qquad
 S_{a,b}(0)=1,\qquad
 \operatorname{ord}_{1-q}S_{a,b}=a.
\]
\end{theorem}

\begin{proof}
 The factorization is an inventory of cyclotomic exponents after the exact norm
substitution. Separating odd and even indices reduces each exponent to a
residue calculation, while the \(q=1\) order is read from the remaining
linear factors. This proves polynomiality and the stated initial/order data
without promoting the factor to a new endpoint operator.

Substitute the exact norm factorization of
Theorem~\ref{p4:retained:thm:c11-norm}.  For a cyclotomic polynomial $\Phi_j(q)$,
the exponent in the resulting quotient is
\[
 \sigma_j=f_j(a,b)-2\left\lfloor\frac bj\right\rfloor
 +\mathbf1_{j\mid2}-\mathbf1_{j\mid(4a+2)},
\]
where $f_j(a,b)$ is the numerator exponent from the norm
factorization.  If $j>1$ is odd, write $a=kj+r$ and $b=tj+s$; the
remaining residue contribution is
\[
 k+\left\lfloor\frac{r+s+(j+1)/2}{j}\right\rfloor
 -\left\lfloor\frac{2r}{j}\right\rfloor
 -\mathbf1_{r=(j-1)/2}\geq0.
\]
If $j=2h>2$, write $a=kh+r$ and $b=2ht+s$; the residue contribution is
\[
 k+\left\lfloor\frac{2r+s+1+h}{2h}\right\rfloor
 -\left\lfloor\frac{2r}{h}\right\rfloor
 -\mathbf1_{h\ {\rm odd},\,r=(h-1)/2}\geq0.
\]
In each case the numerator floor cancels the only possible subtraction
when the residue crosses the midpoint; the remaining $k$ is nonnegative.
For $j=1,2$ the exponent is exactly $a$.  Thus the quotient is a
polynomial, its constant coefficient is one, and its $(1-q)$-order is
$a$.
\end{proof}

The theorem cancels every divided local spectator factorial.  The remaining
possible denominators lie in the trunk factor, the active quantum-dimension
factor, and the doubled-tree graph gauge; they are finite in each fixed
spectator sector.

Let $S\subset\{1,\dots,g\}$ be nonempty, let $\theta$ collect all
complementary labels, and unprime the active Newton factors:
\[
 V^S_{\theta,\mathbf n}=
 \left(\prod_{i\in S}\{n_i\}!\right)\widetilde b_\lambda.
\]
For a signed square-free product
$T_{S,\delta}=\prod_{i\in S}t_{\delta_i,i}^\star$, higher-genus
orthogonality gives
\[
 B_G(T_{S,\delta}V^S_{\theta,\mathbf m},
             V^S_{\vartheta,\mathbf n})
 =\delta_{\theta\vartheta}\rho_\theta^S
  \prod_{i\in S}\mathcal K_{n_i,m_i}^{\delta_i,(a_i)}.
\]
Here $\rho_\theta^S\in \FA^\times$ is independent of the active indices.

Write $\rho_\theta^S=N_\theta^S/Q_\theta^S$ in lowest terms.  A
\emph{spectator denominator-clearing factor} is a nonzero
$D_\theta^S\in\Lambda$ such that
$D_\theta^S\rho_\theta^S\in\Lambda$.

\begin{proposition}[Lattice of denominator-clearing factors]
Every denominator-clearing factor is a multiple of $Q_\theta^S$, and every nonzero multiple is
a denominator-clearing factor.  Modulo signs, gcd and lcm give a directed distributive lattice
with minimum $Q_\theta^S$.  Taking the coordinatewise product over sectors
preserves these properties.
\end{proposition}

\begin{proof}
 The denominator-clearing factor statement is a divisibility calculation in the coefficient UFD.
Writing the spectator factor in lowest terms isolates the exact denominator;
coprimality forces every denominator-clearing factor to contain it, and multiplication by any
extra factor preserves clearing. The gcd/lcm operations and the
coordinatewise sector product then give the stated directed lattice.

In the UFD $\mathbb Z[A]$, $Q\mid DN$ and $(N,Q)=1$ imply $Q\mid D$.
The converse is immediate, and the gcd/lcm claims follow in a UFD.
\end{proof}

Set
\[
 \widehat V^{S,D}_{\theta,\mathbf n}
 =D_\theta^SA^{\sum_{i\in S}R_{a_i}(n_i)}
  V^S_{\theta,\mathbf n}.
\]

\begin{theorem}[Integral square-free scalar chamber]
\label{p4:retained:thm:c12-scalar}
For every sign choice and $\mathbf m\leq\mathbf n$,
\[
 B_G(T_{S,\delta}\widehat V^{S,D}_{\theta,\mathbf m},
             \widehat V^{S,D}_{\theta,\mathbf n})
 =D_\theta^S(D_\theta^S\rho_\theta^S)
 \prod_{i\in S}
 \left[A^{R_{a_i}(n_i)+R_{a_i}(m_i)}
 \mathcal K_{n_i,m_i}^{\delta_i,(a_i)}\right]
 \in\Lambda.
\]
\end{theorem}

\begin{proof}
 The scalar chamber proof separates spectator denominators from active twist
orders. The sector denominator-clearing factor first makes the spectator factor integral, while
the uniform local shift makes each active Hopf-contracted kernel polynomial;
the product formula then assembles those independent factors over the
selected handles. This is a square-free chamber construction, not a claim
that one diagonal chart is stable under arbitrary twist powers.

The first two factors are polynomial by the definition of a denominator-clearing factor, and
each bracketed local factor is polynomial by
Corollary~\ref{p4:retained:cor:c12-separable}.
\end{proof}

For polynomial coefficient families define, on a finite lower ideal $C$,
\[
 \mathsf G_C=
 \sum_{(\theta,\mathbf n)\in C}
 \sum_{\mathbf m\leq\mathbf n}
 x_{\theta,\mathbf n}y_{\theta,\mathbf m}
 J_{H_G(\theta,\mathbf n)}
 B_G(T_{S,\delta}\widehat V_{\theta,\mathbf m},
                    \widehat V_{\theta,\mathbf n}).
\]

\begin{theorem}[Arbitrary-genus common scalar approximants]
\label{p4:retained:thm:c12-common-net}
If $C\subset C'$ and
$\nu(C)=\min_{\lambda\notin C}H_G(\lambda)$, then
\[
 \mathsf G_{C'}-\mathsf G_C\in J_{\nu(C)}\Lambda.
\]
In particular, the height sequence has increments in $J_N\Lambda$, so the
same finite polynomials converge in the disk and cyclotomic branches.
\end{theorem}

\begin{proof}
 The common-net estimate is finite before it is completed. A larger lower
ideal adds only finitely many outputs and finitely many lower sources, and
the height of every new output supplies the corresponding $J$-factor.
Consequently the same scalar finite sums are Cauchy in the disk and
cyclotomic branches; the conclusion belongs to the common chamber because
the construction retains both filtrations.

There are finitely many outputs in $C'$, and each has finitely many lower
sources.  Every newly added output has height at least $\nu(C)$; the exact
quotient $J_h/J_{\nu(C)}\in\Lambda$ supplies the divisibility.
\end{proof}

This scalar chamber and the polynomial endpoint cube are distinct charts.
The additional scalar shift satisfies
\[
 R_c(m)-R_c(n)<0\qquad(n>m),
\]
and destroys the sharp two-sided endpoint inequality.\newline An endpoint-valuation bridge,
not a diagonal identification, is needed to combine them.