% Provenance: corrected synthesis of the module CRT, restricted-completion, and common-approximant results developed in stages 15, 16, 24, and 25.
\section[Joint Completions and Finite-Stage Representatives]
{Simultaneous Disk-Adic and Cyclotomic Completions with Compatible Finite-Stage Representatives}
\label{p4:retained:ch:joint-completion-basics}

The chamber analysis dictates a non-Laurent coefficient lattice, but it
does not yet couple disk and cyclotomic convergence.  Start with a marked
boundary state and retain one finite representative for its later disk
reading and root-of-unity cap.  At every finite stage the two ideals are
comaximal, so the Chinese remainder theorem separates the two residues,
without freeness or flatness assumptions on the module.

The remaining issue is compatibility through the stages.  A single sequence
of finite polynomials must have increments in both ideals to represent both
limits.  This common-origin condition couples the representatives; it does
not identify the completed targets.  The section first constructs the finite
splitting and then states the exact approximation conditions that survive
passage to inverse limits.

\subsection{The coefficient ring and the joint regulator}

Use the standard polynomial, Laurent, and generic rings
\[
 \Lambda=\Z[A],\qquad R=\Z[A,A^{-1}],\qquad \FA=\Q(A),
\]
and put $q=A^2$.  For $N\geq0$, define
\[
 P_N(A)=(q;q)_N=\prod_{k=1}^{N}(1-A^{2k}),
 \qquad
 \mathfrak j_N=A^{2N}P_N(A).
\]
Thus $P_0=\mathfrak j_0=1$; this stage-zero value is used for height-zero
approximant terms.  Unless explicitly stated otherwise, the finite-stage CRT
statements below assume $N\geq1$.
We use the same symbols for the generated principal ideals when no
confusion is possible.  The filtrations are nested because
\[
 \mathfrak j_{N+1}=A^2(1-A^{2N+2})\mathfrak j_N
 \qquad(N\geq0).
\]
The factor $A^{2N}$ measures order at the disk origin; $P_N$ measures
cyclotomic divisibility.
The product ideals therefore define the simultaneous disk-adic and
cyclotomic filtration.  As in the local terminology introduced above,
``disk--cyclotomic'' abbreviates this product-ideal construction.

\begin{proposition}[No naive common skein base change]
\label{p4:retained:prop:c10-no-naive-base-change}
For $N\geq1$, the quotient $\Lambda/(\mathfrak j_N)$ is not a coefficient ring to which
the ordinary Kauffman two-handle theorem can be directly base changed:
the image of $A$ is not invertible.  If $A$ is localized first, then
\[
 \mathfrak j_NR=P_NR,
\]
so the disk factor disappears.  Joint finite stages must therefore be
understood as quotients of declared polynomial lattices inside a generic
$\FA$-skein space, not as Kauffman skein modules over
$\Lambda/(\mathfrak j_N)$.
\end{proposition}

\begin{proof}
 The base-change failure is detected before any skein calculation: the
polynomial quotient contains a nonzero class killed by the image of the
parameter, so that parameter is not invertible. Localizing first makes the
disk monomial a unit and removes the disk filtration. The joint stage must
therefore remain a polynomial-lattice quotient inside the generic skein
space.

The nonzero class of $A^{2N-1}P_N$ in
$\Lambda/(A^{2N}P_N)$ is killed by $A$, so $A$ is a zero divisor and cannot
be a unit.  In $R$, the factor $A^{2N}$ is a unit, giving the displayed
equality of ideals.
\end{proof}

This is the coefficient-ring distinction behind the raw and saturated
relations of Section~\ref{p4:retained:ch:raw-saturated}: first define
the genuine slide submodule over $R$, intersect its appropriate span with a
polynomial lattice, and only then reduce by $\mathfrak j_N$.

\subsection{Complementary CRT idempotents}

The two factors of $\mathfrak j_N$ are comaximal because $P_N(0)=1$.
The following explicit complementary idempotents implement the Chinese
remainder projection for every module.

\begin{lemma}[Integral B\'ezout identity and complementary CRT idempotents]
\label{p4:retained:lem:c10-bezout}
For $N\geq1$, let $b_N(A)$ be the truncation through degree $2N-1$ of the formal power
series $P_N(A)^{-1}$.  Then $b_N\in\Z[A]$ and there is an
$a_N\in\Z[A]$ such that
\[
 a_NA^{2N}+b_NP_N=1.
\]
The two elements
\[
 e_N^{\mathrm{disk}}=b_NP_N,
 \qquad
 e_N^{\mathrm{cyc}}=a_NA^{2N}=1-e_N^{\mathrm{disk}}
\]
are complementary orthogonal idempotents modulo $\mathfrak j_N$, with
\begin{align*}
 e_N^{\mathrm{disk}}&\equiv1\pmod {A^{2N}},&
 e_N^{\mathrm{disk}}&\equiv0\pmod {P_N},\\
 e_N^{\mathrm{cyc}}&\equiv0\pmod {A^{2N}},&
 e_N^{\mathrm{cyc}}&\equiv1\pmod {P_N}.
\end{align*}
\end{lemma}

\begin{proof}
 The constant term of \(P_N\) makes its truncated inverse integral, and the
truncation converts the formal inverse into a finite Bézout separator modulo
the disk power. The two complementary residues are then obtained by the
selector pair; no freeness of a later module is involved at this stage.

Recursive inversion of a polynomial with constant coefficient one has
integral coefficients.  Truncation gives $b_NP_N\equiv1\pmod {A^{2N}}$,
so $(1-b_NP_N)/A^{2N}$ is the required $a_N$.  The congruences are
immediate.  They imply idempotence and orthogonality modulo the product
ideal.
\end{proof}

\paragraph{The one-holed-torus model.}
For a marked one-holed torus, keep a boundary state as a vector in the
declared polynomial lattice before capping it with the complementary solid
torus. The cap supplies the root-of-unity reading, while the same finite
polynomial representative can be read at the disk origin. A single Laurent
skein module cannot carry both readings: localizing makes the disk factor a
unit, whereas retaining the polynomial parameter prevents a direct base
change of the ordinary skein relation. The module CRT below is the
finite-stage device that keeps these two readings attached to one
representative.  Figure~\ref{p4:fig:one-holed-torus-running} records this
model and the later root-fiber pairing and cap without adding an evaluation
map on the completed disk coordinate.

\subsection{The module Chinese remainder theorem}

\begin{theorem}[Module CRT]
\label{p4:retained:thm:c10-module-crt}
For $N\geq1$ and every $\Lambda$-module $M$, the diagonal map
\[
 \Theta_{M,N}:M/\mathfrak j_NM
 \longrightarrow
 M/A^{2N}M\times M/P_NM,
 \qquad [x]\longmapsto([x],[x]),
\]
is a natural isomorphism.  No flatness, torsion-freeness, finite generation,
or freeness hypothesis on $M$ is required.  Its inverse is
\[
 ([x],[y])\longmapsto
 [e_N^{\mathrm{disk}}x+e_N^{\mathrm{cyc}}y]_{\mathfrak j_N}.
\]
\end{theorem}

\begin{proof}
The two selectors show directly that the proposed inverse has the requested
disk and cyclotomic residues, proving surjectivity.  For injectivity it is
enough to prove
\[
 A^{2N}M\cap P_NM=A^{2N}P_NM.
\]
If $x=A^{2N}u=P_Nv$ lies in the intersection, the B\'ezout identity gives
\[
 x=(a_NA^{2N}+b_NP_N)x
   =a_NA^{2N}P_Nv+b_NP_NA^{2N}u,
\]
which lies in the product submodule.  The reverse inclusion is immediate.
Naturality follows because the selectors are central scalars.
\end{proof}

\begin{corollary}[CRT after relations]
\label{p4:retained:cor:c10-crt-after-relations}
For every submodule $X\subset M$,
\[
 \frac{M}{X+\mathfrak j_NM}
 \cong
 \frac{M}{X+A^{2N}M}\times
 \frac{M}{X+P_NM}.
\]
Equivalently,
\[
 (X+A^{2N}M)\cap(X+P_NM)=X+\mathfrak j_NM.
\]
\end{corollary}

\begin{proof}
 The relation submodule is absorbed first by passing from \(M\) to \(M/X\).
The module CRT on that quotient supplies both the product decomposition and
the intersection identity, so adding relations does not require a new
flatness or torsion argument.

Apply Theorem~\ref{p4:retained:thm:c10-module-crt} to the module $M/X$.
\end{proof}

In particular, the statement applies equally to raw or saturated full-slide
quotients.  It does not assert that either root fiber is the reduced RT
state space; that remains a separate non-flat specialization problem.

\subsection{Split inverse limits}

Define
\[
 \widehat M_{\mathrm{joint}}=
   \varprojlim_NM/\mathfrak j_NM,
 \qquad
 \widehat M_{\mathrm{disk}}=
   \varprojlim_NM/A^{2N}M,
 \qquad
 \widehat M_{\mathrm{cyc}}=
   \varprojlim_NM/P_NM.
\]

\begin{theorem}[Split joint completion]
\label{p4:retained:thm:c10-split-completion}
For every $\Lambda$-module $M$, there is a canonical natural isomorphism
\[
 \boxed{
 \widehat M_{\mathrm{joint}}
 \cong
 \widehat M_{\mathrm{disk}}\times\widehat M_{\mathrm{cyc}}.}
\]
For $M=\Lambda$, the disk factor is $\Z[[A]]$, while the cyclotomic factor
is
\[
 \varprojlim_N\Z[A]/((A^2;A^2)_N).
\]
\end{theorem}

\begin{proof}
For the disk factor we use the ordinary \(A\)-adic filtration.

The finite-stage CRT maps commute with every reduction map.  Take inverse
limits and use that finite products commute with inverse limits.  The ideals
$(A^{2N})$ are cofinal with the ordinary $A$-adic ideals, yielding
$\Z[[A]]$.
\end{proof}

The product theorem makes the independence of the two coordinates exact.
It also explains why a distinguished compatible disk--cyclotomic lift is extra data: the
cyclotomic coordinate alone does not choose the disk coordinate, as proved
in Theorem~\ref{p4:retained:thm:c07-no-continuous-map}.

\subsection{Strict common-origin approximants}

\begin{definition}[Finite-stage representative sequence]
A sequence $f_N\in M$ is a strict common-origin sequence when
\[
 f_{N+1}-f_N\in\mathfrak j_NM
\]
for every $N$.  The same finite representatives then define both the disk
and cyclotomic limits.
\end{definition}

\begin{theorem}[Criterion for a finite-stage representative sequence]
\label{p4:retained:thm:c10-common-criterion}
For a sequence $f_N\in M$, the following are equivalent:
\begin{enumerate}
\item $f_{N+1}-f_N\in\mathfrak j_NM$ for every $N$;
\item simultaneously,
\[
 f_{N+1}-f_N\in A^{2N}M,
 \qquad
 f_{N+1}-f_N\in P_NM
\]
for every $N$.
\end{enumerate}
The quotient version is obtained by adding one fixed relation submodule
$X$ to all three conditions.
\end{theorem}

\begin{proof}
 The common-origin condition is exactly the intersection of the disk and
cyclotomic increment conditions. The finite-stage CRT identifies that
intersection with the joint regulator ideal, while the quotient formulation
simply carries one fixed relation submodule through the same calculation.

Use the intersection identity in
Theorem~\ref{p4:retained:thm:c10-module-crt}, or its quotient form in
Corollary~\ref{p4:retained:cor:c10-crt-after-relations}.
\end{proof}

The word ``same'' is essential.  One disk-Cauchy sequence and an unrelated
cyclotomic-Cauchy sequence determine a pair in the split product, but they do
not exhibit a common kernel or a common family of finite polynomials.

\begin{theorem}[Constructive bicoercive cutoff]
\label{p4:retained:thm:c10-bicoercive-cutoff}
Let $T_x\in M$ for $x$ in a countable set $X$.  Suppose proper functions
$h,d:X\to\Z_{\geq0}$ satisfy
\[
 T_x\in P_{h(x)}M,
 \qquad
 T_x\in A^{2d(x)}M.
\]
Set
\[
 X_N=\{x:h(x)<N\ \text{or}\ d(x)<N\},
 \qquad
 f_N=\sum_{x\in X_N}T_x.
\]
Then $X_N$ is finite and $f_{N+1}-f_N\in\mathfrak j_NM$.
\end{theorem}

\begin{proof}
 The bicoercive cutoff uses two independent properness tests. Proper height
makes each finite lower set a finite marked-state sum, while the disk and
cyclotomic bounds force every new layer into both branch ideals. Their
intersection is the joint regulator ideal, so successive finite sums define
one common-origin limit.

Properness makes each of the two sublevel sets finite, hence also their
union.  Every term in the new layer has $h(x),d(x)\geq N$.  The layer lies
in both $P_NM$ and $A^{2N}M$, hence in their intersection
$\mathfrak j_NM$.
\end{proof}

A layerwise variant only asks that the sum of each new finite layer lie in
the two branch ideals.  It therefore permits exact equal-degree
cancellation without assuming termwise positivity.

\subsection{Restricted completions and infinite matrices}

Let $X$ be countable and let
\[
 L(X)=\Lambda^{(X)}
\]
be the free direct-sum lattice.  Its joint completion is naturally the
restricted product
\[
 \widehat L_{\mathrm{joint}}(X)=
 \left\{(a_x)\in\prod_{x\in X}\widehat\Lambda_{\mathrm{joint}}:
 \begin{array}{l}
 \text{for every }N,\text{ all but finitely many }a_x\\[-1mm]
 \text{vanish modulo }\mathfrak j_N
 \end{array}
 \right\}.
\]

\begin{theorem}[Restricted-column criterion]
\label{p4:retained:thm:c10-restricted-columns}
A matrix $U=(u_{yx})$ over $\widehat\Lambda_{\mathrm{joint}}$ defines a
continuous map
\[
 U:\widehat L_{\mathrm{joint}}(X)\longrightarrow
 \widehat L_{\mathrm{joint}}(Y)
\]
if and only if every column is restricted:
\[
 \text{for all }x,N,\qquad
 u_{yx}\equiv0\pmod{\mathfrak j_N}
 \quad\text{for all but finitely many }y.
\]
Restricted matrices are closed under composition.
\end{theorem}

\begin{proof}
 At a fixed stage, a restricted-product input has finite support, so
continuity is exactly the requirement that each surviving column have finite
output support modulo \(\mathfrak j_N\).  The same finite-support observation
applied to a composite proves closure under composition.  This is a
completion-theoretic support test, not a basis-change identity.

Modulo $\mathfrak j_N$, an input has finite support.  Its image has finite
support precisely when each of the finitely many columns it uses has finite
support at that stage.  Conversely, the image of a basis vector is one
column, so continuity forces that column to be restricted.  For a composite
and one fixed input column, only finitely many intermediate indices survive
modulo $\mathfrak j_N$; the finite union of the corresponding restricted
output supports is finite.
\end{proof}

This criterion detects a basic failure of the ordinary color-basis Hopf
kernel.  After removing a Laurent monomial, its entry is
\[
 G_{mn}(q)=1+q+\cdots+q^{(m+1)(n+1)-1}.
\]
For a fixed $n$,
\[
 G_{mn}(1)=(m+1)(n+1)\ne0
\]
for infinitely many $m$.  Thus the column is not restricted even modulo
$1-q$, and multiplying rows or columns by monomials cannot change the value
at $q=1$.  A non-diagonal cyclotomic basis is necessary.

\subsection{A diagonal annular seed}

In the representation ring, define
\[
 P_n^{\mathrm H}=\prod_{i=0}^{n-1}
 \left(V_1-v^{2i+1}-v^{-2i-1}\right),
\]
and in its even subalgebra define
\[
 S_n^{\mathrm H}=\prod_{i=1}^{n}
 \left(V_1^2-(v^i+v^{-i})^2\right).
\]
These are bases of two different modules.  Their Hopf pairing is diagonal:
\[
 \langle P_m^{\mathrm H},S_n^{\mathrm H}\rangle
   =\delta_{mn}\{2m+1\}_{2m}.
\]
These two integral bases and their diagonal pairing are Habiro's
construction \cite{Habiro}.
With $q=v^2$, the polynomial part of the diagonal norm, up to sign and a
Laurent monomial, is
\[
 D_n(q)=q^n\prod_{j=2}^{2n+1}(1-q^j).
\]

\begin{theorem}[Annular joint tail]
\label{p4:retained:thm:c10-annular-tail}
For every $n\geq N\geq0$,
\[
 D_n(q)\in q^N(q;q)_N\Z[q].
\]
\end{theorem}

\begin{proof}
 The annular tail is certified coefficientwise by two independent factors: the
leading monomial supplies the disk order, and counting each cyclotomic factor
in the product supplies the required \(P_N\)-divisibility. This is a pairing
tail estimate, not an assertion that the two annular bases define one
invertible operator.

The monomial supplies $q^N$.  For $d\geq2$, the exponent of the cyclotomic
polynomial $\Phi_d(q)$ in the product is
\[
 \left\lfloor\frac{2n+1}{d}\right\rfloor
 \geq\left\lfloor\frac Nd\right\rfloor,
\]
its exponent in $(q;q)_N$.  For $d=1$, the product has $2n$ factors and
again supplies at least the required $N$.
\end{proof}

This is a pairing-tail theorem, not an invertible annular $S$ operator: the
two displayed families are not bases of the same module.  A relative
operator still needs source and target lattices, boundary colors, inverse
basis shifts, and the restricted-column property.

\subsection{Literal Poincar\'e finite-stage representatives}

The modified Poincar\'e series has the exact summands
\[
 T_n(q)=q^n(q^n;q)_n.
\]
This is the distinguished joint formula in Habiro's source
\cite{Habiro}.
For $n\geq1$,
\[
 (q^n;q)_n
 =\frac{(q;q)_{2n-1}}{(q;q)_{n-1}}
 =(q;q)_n\genfrac{[}{]}{0pt}{}{2n-1}{n}_q,
\]
and therefore
\[
 T_n(q)=q^n(q;q)_n\genfrac{[}{]}{0pt}{}{2n-1}{n}_q
 \in q^n(q;q)_n\Z[q].
\]

\begin{theorem}[Poincar\'e finite-stage representatives]
\label{p4:retained:thm:c10-poincare-common}
The literal partial sums
\[
 B_N(q)=\sum_{n=0}^{N-1}q^n(q^n;q)_n
\]
satisfy
\[
 B_{N+1}-B_N\in q^N(q;q)_N\Z[q].
\]
They therefore define one joint element whose disk and cyclotomic
projections come from the same finite polynomials.
\end{theorem}

\begin{proof}
 The displayed factorization puts the Poincaré summand at stage \(n\)
in both required ideals: its leading power supplies the disk order, and its
cyclotomic product supplies the divisibility.  The literal partial sums
therefore represent both limits through the same finite polynomials.

The increment is $T_N$, and the preceding factorization gives the required
divisibility.
\end{proof}

This explicit compatible disk--cyclotomic lift is special data supplied by the formula.  It is not
obtained from the closed cyclotomic coordinate by a continuous map.

% Keep the first contents page clear of the title-page metadata footnotes.
\addtocontents{toc}{\protect\pagebreak}
\subsection{Proper-height approximants after the slide quotient}

\leavevmode\newline
Let $L=\bigoplus_{\lambda\in X}\Lambda u_\lambda$ be a declared polynomial
lattice, let $H:X\to\Z_{\geq0}$ be proper, and let
$\mathscr X\subset L$ be either a raw or saturated full-slide relation
lattice.  Put
\[
 C_N=\{\lambda:H(\lambda)<N\}.
\]
For a polynomial coefficient family $x=(x_\lambda)$, define
\[
 v_N(x)=
 \left[
 \sum_{\lambda\in C_N}
 \mathfrak j_{H(\lambda)}x_\lambda u_\lambda
 \right]
 \in L/(\mathscr X+\mathfrak j_NL).
\]

\begin{theorem}[Quotient-valued finite-stage representatives]
\label{p4:retained:thm:c10-quotient-approximants}
The classes $v_N(x)$ are compatible under the stage transition maps and
define an element
\[
 \varprojlim_NL/(\mathscr X+\mathfrak j_NL).
\]
More generally, if $C\subset C'$ are finite lower sets and
\[
 \nu(C)=\min_{\lambda\notin C}H(\lambda),
\]
then the corresponding finite sums differ by an element of
$\mathfrak j_{\nu(C)}L$.
\end{theorem}

\begin{proof}
 Proper height makes each finite stage a finite marked-state sum, while the
regulator \(\mathfrak j_N\) makes the new height layer invisible at the
preceding stage.  This gives a common joint limit in the quotient; an exact
groupoid arrow transports it only when it transports both the lattice and
the declared relation submodule.

The new terms from stage $N$ to stage $N+1$ have height $N$ and hence carry
the factor $\mathfrak j_N$.  They vanish at stage $N$.  The same argument
with $\nu(C)$ proves the lower-set statement.
\end{proof}

The finite-stage module itself need not have finite rank.  Finiteness of the
approximant comes from properness of $H$.  If an exact $\Lambda$-linear
groupoid arrow transports both $L$ and $\mathscr X$, it transports every
$v_N$ and its joint limit exactly.  This comparison is based at one declared
source chamber; it does not compare independently chosen height sections or
cotests in unrelated chambers.

Finally, a scalar sequence appears only after choosing compatible cotests
\[
 \epsilon_N:L\longrightarrow\Lambda/(\mathfrak j_N)
\]
which annihilate $\mathscr X$.  The universal property of the quotient says
that this condition is both necessary and sufficient for descent.  The CRT,
restricted completion, and proper-height regulator construct the domain of
such a cap; they do not manufacture the nonzero root-specialized cap, its one-sided
handlebody covariance, or its stabilization normalization.