% Provenance: corrected synthesis of stage 21, with generic recoupling deferred to the corrected convention of the next section. \section[Endpoint-Valuation Geometry and the Universal Core] {Two-Sector Endpoint-Valuation Geometry and the Universal Contracted Core} \label{p4:retained:ch:stokes-core} The polynomial endpoints of Section~\ref{p4:retained:ch:opposite-endpoints} are not two bases of one disk-regular module. They are opposite endpoint-valuation sectors. This section computes their exact inside and outside valuations, places one integral scalar core inside every endpoint chamber, proves that the raw twist cannot preserve that core, and constructs the regulated pairing which survives this no-go. The local recoupling and norm inputs use the integral graphical conventions of \cite{Costantino,GilmerMasbaum}; all new endpoint-valuation statements are proved below. The section follows the obstruction that forces the different constructions. We first compare the inside- and outside-disk valuations, which identifies the two endpoint sectors. We then take common denominator-clearing factors and form one universal contracted core. The next step shows that this core lies in every endpoint lattice but is not preserved by the raw twists; that failure is the scalar-primal obstruction. Finally, an output regulator and finite-lower-ideal approximation produce the rigged pairing. Thus the common core is a shared contracted object, while the endpoint lattices and their twist actions remain distinct. \subsection{Exact inside and outside valuations} For $c,n\geq0$ define \[ \mathscr U_{c,n}(X,z)= \frac{\prod_{j=1}^{n} (1+X^{2(c+j)}z+X^{4(c+j)})} {\prod_{j=1}^{n}(1-X^{2j}) \prod_{j=1}^{n}(1-X^{2(c+j)+1})}. \] It belongs to $\mathbb Z[[X]][z]$. Let $\operatorname{ord}_0$ be $A$-order, and let $\operatorname{ord}_\infty$ be the $B$-order after $A=B^{-1}$. \begin{lemma}[Coefficient valuation] \label{p4:retained:lem:c13-Uorder} For $0\leq k\leq n$, \[ \operatorname{ord}_X[z^k]\mathscr U_{c,n}(X,z)=2ck+k(k+1). \] \end{lemma} \begin{proof} Every denominator is a unit at \(X=0\), so the lowest \(X\)-order in the \(z^k\) coefficient comes from selecting the \(z\)-term in the first \(k\) numerator factors and the constant term in the rest. Summing those \(k\) exponents gives the displayed valuation. Every denominator factor has constant coefficient one. The unique lowest term in the $z^k$ coefficient selects the $z$ term from the first $k$ numerator factors and $1$ from the rest, with order $\sum_{j=1}^k2(c+j)=2ck+k(k+1)$. \end{proof} Recall \[ \mathsf E_n^-=\frac{A^{-\binom n2}}{(Aq^{c+1};q)_n}Q_n^{\prime(c)}, \qquad \mathsf E_n^+=\mu_{c+n}\mathsf E_n^-. \] \begin{theorem}[Exact two-sector polarization] \label{p4:retained:thm:c13-polarization} With $N_n=n(n+1)/2$, \begin{align*} \mathsf E_n^-(A) &=(-1)^{n+N_n}A^{-n(n+2c)}\mathscr U_{c,n}(A,z),\\ \mathsf E_n^+(A) &=(-1)^{c+N_n}A^{c(c+2)+2n}\mathscr U_{c,n}(A,z),\\ \mathsf E_n^-(B^{-1}) &=(-1)^{n+N_n}B^{n(n+1)}\mathscr U_{c,n}(B,z). \end{align*} Consequently \begin{align*} \operatorname{ord}_0[z^k]\mathsf E_n^- &=-n(n+2c)+2ck+k(k+1),\\ \operatorname{ord}_0[z^k]\mathsf E_n^+ &=c(c+2)+2n+2ck+k(k+1),\\ \operatorname{ord}_\infty[z^k]\mathsf E_n^- &=n(n+1)+2ck+k(k+1),\\ \operatorname{ord}_\infty[z^k]\mathsf E_n^+ &=n(n+1)+2ck+k(k+1)-(c+n)(c+n+2). \end{align*} Thus $\mathsf E^+$ is the inside-disk endpoint and $\mathsf E^-$ is the outside-disk endpoint; the opposite endpoint has an unbounded pole. \end{theorem} \begin{proof} The endpoint formulas come from one common Newton product, but the monomial factored from that product is different in the inside and outside chambers. The coefficient valuation records the remaining regular factor, and the endpoint exponents then expose the unbounded pole at the opposite end. The same calculation tensorizes handle by handle; it does not identify the two analytic sectors. Factoring the lowest monomial from each Newton factor gives \[ Q_n^{(c)}=A^{-2cn-n(n+1)} \prod_{j=1}^n(1+A^{2(c+j)}z+A^{4(c+j)}). \] Use \[ \{n\}!=(-1)^{n(n+3)/2}A^{-n(n+1)/2}(q;q)_n \] and the endpoint definitions. The closed forms follow, and Lemma~\ref{p4:retained:lem:c13-Uorder} gives the coefficient orders. Their minima are at $k=0$. \end{proof} The handlewise tensor product gives the same polarization in every finite genus. A sector-constant graph normalization does not alter it; uniform valuation under a change of marking is a separate recoupling problem. \subsection{The all-subset denominator-clearing data} Fix a genus-$g$ lollipop marking $G$. Let $\mathcal C_G$ be its finite lower-ideal cutoff lattice, and let $\mathcal D_{G,S}$ be the denomi\-nator-clearing factor lattice of Section~\ref{p4:retained:ch:opposite-endpoints} for every nonempty $S\subset[g]$. Choose a positive-leading denominator-clearing factor $D^S_{\theta_S(\lambda)}$ in every complementary sector and set \[ D_\star(\lambda)= \operatorname{lcm}_{\varnothing\neq S\subset[g]} D^S_{\theta_S(\lambda)}. \] The sign cube $\mathbb B_g=\{-,+\}^g$, ordered by $-<+$, and the product \[ \mathfrak A_G= \mathcal C_G\times\mathbb B_g\times \prod_{\varnothing\neq S\subset[g]}\mathcal D_{G,S} \] form a directed distributive lattice under coordinatewise intersection or union, minimum or maximum, and gcd or lcm. This statement is for each fixed finite genus; it contains no stabilization assertion. Define \[ R_a(b)=2(a+b+1)^2,\qquad p_b(A)=A^{b(b+1)/2}\{b\}! =(-1)^{b(b+3)/2}(q;q)_b. \] For the generic divided orthogonal lollipop vector $\widetilde b_\lambda$, put \begin{equation} \mathbb U_\lambda= D_\star(\lambda)A^{\sum_{i=1}^gR_{a_i}(b_i)} \left(\prod_{i=1}^gp_{b_i}(A)\right)\widetilde b_\lambda, \qquad L_\star=\bigoplus_{\lambda}\Lambda\mathbb U_\lambda. \label{p4:retained:eq:c13-core} \end{equation} For a nonempty active set $S$, the subset-specific scalar vector is \[ \widehat V^S_\lambda= D^S_{\theta_S(\lambda)} A^{\sum_{i\in S}R_{a_i}(b_i)} \left(\prod_{i\in S}\{b_i\}!\right) \widetilde b_\lambda. \] \begin{theorem}[Universal square-free contracted core] \label{p4:retained:thm:c13-universal-core} For every nonempty $S\subset[g]$, \[ \mathbb U_\lambda=E_S(\lambda)\widehat V^S_\lambda, \] where \[ E_S(\lambda)= \frac{D_\star(\lambda)}{D^S_{\theta_S(\lambda)}} A^{\sum_{j\notin S}R_{a_j}(b_j) +\sum_{i\in S}b_i(b_i+1)/2} \prod_{j\notin S}p_{b_j}(A)\in\Lambda. \] If $T_{S,\delta}$ is any square-free signed standard multitwist, then \[ \kappa^{\star,S,\delta}_{\lambda\mu} :=B_G(T_{S,\delta}\mathbb U_\mu,\mathbb U_\lambda) \in\Lambda \] on the triangular range and is zero otherwise. One lattice therefore supports all $3^g-1$ nontrivial square-free signed contracted kernels. \end{theorem} \begin{proof} The universal core is assembled by taking the lcm of the finite-sector denomi\-nator-clearing factors and adjoining the fixed factorial factors from the complementary handles. Dividing by a selected subset vector leaves the displayed polynomial multiplier; the previously established subset kernel then gives integrality by bilinearity. The construction is therefore a common contracted core, not an invariant primal module. The selected subset denominator-clearing factor divides $D_\star$. Dividing \eqref{p4:retained:eq:c13-core} by $\widehat V^S_\lambda$ gives the displayed polynomial $E_S$. Bilinearity expresses the new kernel as $E_S(\lambda)E_S(\mu)$ times the integral subset kernel of Theorem~\ref{p4:retained:thm:c12-scalar}. \end{proof} The theorem concerns contracted pairings; it does not claim $T_{S,\delta}L_\star\subset L_\star$. \subsection{The common core inside every endpoint} Put $F_b^{(a)}=(Aq^{a+1};q)_b$ and define \begin{align*} m^-_{a,b}&=(-1)^{b(b+3)/2} A^{R_a(b)+\binom b2}F_b^{(a)}(q;q)_b,\\ m^+_{a,b}&=(-1)^{b(b+3)/2+a+b} A^{R_a(b)+\binom b2-(a+b)(a+b+2)}F_b^{(a)}(q;q)_b. \end{align*} The positive exponent simplifies to \[ (a+b+1)^2+1+\frac{b(b-1)}2\geq2b. \] \begin{theorem}[Loop-coercive endpoint inclusions] \label{p4:retained:thm:c13-endpoint-inclusion} For every sign vertex $\varepsilon\in\mathbb B_g$, \[ \mathbb U_\lambda=M_\lambda^\varepsilon \mathsf e_\lambda^\varepsilon, \qquad M_\lambda^\varepsilon= D_\star(\lambda)\prod_{i=1}^g m_{a_i,b_i}^{\varepsilon_i}\in\Lambda, \] and \[ M_\lambda^\varepsilon\in \left(\prod_{i=1}^gJ_{b_i}\right)\Lambda. \] Thus $L_\star$ lies integrally in every endpoint lattice. \end{theorem} \begin{proof} The endpoint inclusion is checked one handle at a time. Dividing the core by an endpoint vector produces the explicit multiplier \(m^\pm\); its factorial string and the stated exponent bound give \(J_b\) divisibility. Multiplying over handles and by the global denominator-clearing factor places the same core in each endpoint chamber, while the nonunit multipliers explain why this inclusion is not equality. Divide the core vector by the endpoint scalar handle by handle. The displayed $m^\pm$ result. Each contains $(q;q)_b$ and at least $A^{2b}$, hence a factor $J_b$. \end{proof} \paragraph{Worked checkpoint.} For one handle and unit denominator-clearing factor, the vacuum multipliers are \[ m^-_{0,0}=m^+_{0,0}=A^2. \] For the first loop excitation they are \[ m^-_{0,1}=A^8(1-A^2)(1-A^3), \qquad m^+_{0,1}=-A^5(1-A^2)(1-A^3). \] Their exact $A$-orders are $8$ and $5$, while both contain $J_1=A^2(1-A^2)$. The example displays both the opposite-sector asymmetry and the common joint divisibility. The inclusion is one way: these multipliers are nonunits. They are coercive when loop colors grow, but not when only stick or tree labels grow; the output factor $J_{H_G(\lambda)}$ below supplies uniform decay. \subsection{Closed endpoint-valuation images} \begin{lemma}[Cofinal colon estimate] \label{p4:retained:lem:c13-colon} For every nonzero $z\in\Lambda$ and every $k$, there is $N=N(z,k)$ such that \[ (J_N):z\subset(J_k). \] Multiplication by $z$ on $\widehat\Lambda_J$ is therefore injective, a homeomorphism onto its image, and has closed image. \end{lemma} \begin{proof} The closed-image statement is a valuation-growth argument in the coefficient UFD. A fixed multiplier can remove only finitely much of the irreducible content of \(J_N\), so sufficiently deep regulators force the residual quotient into any prescribed \(J_k\). This gives injectivity and continuity of multiplication and, with completeness, closedness; coordinatewise application then handles the product sector. In the UFD $\Lambda$, $(J_N):z$ is generated by $J_N/\gcd(J_N,z)$. The fixed polynomial $z$ removes only finitely many powers of finitely many irreducibles, whereas their multiplicities in $J_N$ grow without bound. Choose $N$ so that the residual divisor contains $J_k$. This also makes division continuous on the image; completeness and separatedness make that image closed. \end{proof} Let \[ \mathcal S_\star=\prod_\lambda\widehat\Lambda_J\mathbb U_\lambda, \qquad \mathcal P_\varepsilon= \prod_\lambda\widehat\Lambda_J\mathsf e_\lambda^\varepsilon. \] The ambient identity induces \[ \iota_\varepsilon(x_\lambda)_\lambda =(M_\lambda^\varepsilon x_\lambda)_\lambda. \] \begin{corollary}[Closed endpoint-valuation sector] Each $\iota_\varepsilon$ is a homeomorphism onto the closed product submodule \[ \prod_\lambda M_\lambda^\varepsilon\widehat\Lambda_J \mathsf e_\lambda^\varepsilon. \] It is not onto the full endpoint product. \end{corollary} \begin{proof} Apply Lemma~\ref{p4:retained:lem:c13-colon} coordinatewise. Product topology requires no uniform bound. Non-surjectivity follows because nonvacuum multipliers are divisible by $A$. \end{proof} \subsection{Why the scalar core is not a primal twist lattice} Fix a handle with boundary half-color $c$ and write \[ \widehat V_n=DA^{R_c(n)}\{n\}!\widetilde b_n =Z_n^-\mathsf E_n^-=Z_n^+\mathsf E_n^+. \] Direct substitution gives, for $n=m+d$, \begin{align*} \mathsf M^+_{n,m} &=(-1)^{m+c} \frac{A^{(m+c)(m+c+2)-d(2(m+c)+d+1)}} {(q;q)_d(Aq^{m+c+1};q)_d},\\ \mathsf M^-_{n,m} &=(-1)^{m+c} \frac{A^{-(m+c)(m+c+2)-d^2-4d(m+c+1)}} {(q;q)_d(Aq^{m+c+1};q)_d}. \end{align*} \begin{proposition}[Scalar-primal no-go] \label{p4:retained:prop:c13-primal-nogo} Neither signed standard twist preserves the scalar product lattice. In the vacuum column $c=m=0$, the off-diagonal $A$-orders are respectively $-d(d+1)$ and $-d^2-4d$. No finite free non-Laurent lattice can carry both a finite twist truncation and its inverse unless the determinant is a unit. \end{proposition} \begin{proof} The scalar-primal obstruction is already visible in the vacuum column: constant-term-one denominators leave the displayed negative monomial orders unchanged. The finite-height determinant packages the same defect invariantly; if a matrix and its inverse preserved one finite free lattice, their determinants would be mutual units, contradicting the nonunit twist eigenvalue. The denominator products have constant coefficient one, so the displayed monomial exponents are exact. They are negative for every $d>0$. More invariantly, the determinant through height $N$ is \[ \prod_{n=0}^{N}\mu_{c+n}=\pm A^{\Delta_{c,N}}, \] with $\Delta_{c,N}>0$ outside the vacuum. If both a matrix and its inverse were endomorphisms of one finite free $\Lambda$-lattice, their determinants would be inverse units, a contradiction. \end{proof} The obstruction is the algebraic shadow of the two topological fillings: positive and negative longitude twists require opposite endpoint chambers. The scalar core can therefore be paired with both fillings after the stated regulator is inserted, but it cannot itself be the common primal lattice on which both twist directions act integrally. \subsection{The rigged endpoint-valuation pairing and finite-stage representatives} Define the weighted cotest product \[ \mathcal S_\star^\dagger= \prod_{\lambda}J_{H_G(\lambda)} \widehat\Lambda_J\mathbb U_\lambda. \] For a nonempty $S$ and sign choice $\delta$, set \[ (\mathsf K_{S,\delta}^Jy)_\lambda =J_{H_G(\lambda)} \sum_{\mu\preceq_S\lambda} \kappa^{\star,S,\delta}_{\lambda\mu}y_\mu. \] \begin{theorem}[Rigged endpoint-valuation bridge] \label{p4:retained:thm:c13-rigged} The displayed formula defines a continuous map $\mathcal S_\star\to\mathcal S_\star^\dagger$. For arbitrary coefficient families, the finite-lower-ideal net \[ \mathfrak B_C^{S,\delta}(x,y)= \sum_{\lambda\in C} \sum_{\mu\preceq_S\lambda} x_\lambda y_\mu J_{H_G(\lambda)} \kappa^{\star,S,\delta}_{\lambda\mu} \] converges in $\widehat\Lambda_J$ and defines a continuous bilinear pairing. On finite-support vectors it is precisely the root-specialized Hopf-contracted kernel with the regulator inserted once on the output cotest. \end{theorem} \begin{proof} The rigged bridge uses the universal core kernel only after the output regulator has been inserted. Proper height makes the surviving output rows finite modulo each \(J_N\), and row-finiteness makes each such output depend on finitely many inputs. This stagewise finiteness proves convergence and continuity on the product, while the finite-support calculation identifies the resulting pairing with the root-specialized contracted kernel. All kernel entries are integral by Theorem~\ref{p4:retained:thm:c13-universal-core}, and every output row is finite. Modulo $J_N$, only the finitely many outputs with $H_G