Published proof of CK Conjecture 5.27

A degree-zero retraction closes the comparison kernel

Let $U$ be the full face-cone quantum torus and$S\subset U$ the six-root shape subtorus. Each shape root pairs two independent half-edge occurrences. Grading the ambient lattice by the difference in each pair gives $U_0=S$. Projection $P_0:U\to S$ is an $S$-bimodule map, so it preserves the mixed right-tetrahedron and left-edge sidedness.

The raw tetrahedron trinomial is not in$S$. With Chen–Kricker's variables, however, right multiplication by an explicit half-edge unit gives the exact identity

$$rh=-A^{-1/2}f+A^{-1/2}(v-v_0)\bigl(v^{-1}z^2-v_0^{-1}(z'')^{-2}\bigr).$$

Together with the vertex relation and tetrahedral symmetry, this identifies the complete ambient relation module with the extension of the six-root shape relations. Degree-zero projection therefore proves

$$\boxed{S\cap\mathcal R=J,\qquad Q_{CK}=S/J\xrightarrow{\;\cong\;}S/(S\cap\mathcal R)=Q_{PP}^{\mathrm{ext}}.}$$

This is the injectivity assertion of Chen–Kricker Conjecture 5.27 for the comparison map and target they define. No generic-parameter, positivity, regularity or nonvanishing assumption enters the argument. Chen and Kricker already prove surjectivity and the named-trace intertwining; the degree-zero proof supplies the missing reverse inclusion.

Signed six-root named-action bridge

The raw collar square commutes

Preserve each bare-edge label in the full Garoufalidis–Yu target rather than identifying opposite roots. Two named basis insertions give rows $a,b$with $B_6(a,b)=2$ and zero pairing with every edge row. If$L=W_G(a)$ and$M=W_G(b)$, then$LM=A^2ML$, and the torus skein relation extends the two basis formulas to every primitive slope:

$$E^G_{p\lambda+q\mu}=W_G(pa+qb)=A^{-pq}L^pM^q.$$

The local comparison uses Panitch–Park's unweighted marking reversal, with the exact ordinary-to-symmetric cup and cap factors $h^{-1}$ and$h$, where$h=iA^{3/2}$. Anchoring the same physical edges in the two opposite cyclic conventions gives$P:z\leftrightarrow z'',\ y\leftrightarrow y''$, with primed roots fixed. The split tensor, annular counit, corner relations and mixed-sided quotient all commute under this dictionary. Consequently,

$$\boxed{\overline D_P\tau_G^6=\tau_{GY}^{CK}=\tau_{CK}.}$$

After the existing Eulerian normalization, the original Panitch–Park trace therefore has the exact named action

$$\tau_{PP1}(K_\gamma\alpha)=\tau_{PP1}(\alpha)\bigl(E^{PP}_\gamma+(E^{PP}_\gamma)^{-1}\bigr),\quad E^{PP}_\gamma=i^{-\sum u_\gamma}\epsilon^{Pu_\gamma}W_x(Pu_\gamma).$$

A tempting weighted self-duality is not interchangeable with this map: using the same weighted flip on both sides of an unchanged cut leaves a factor$v^{-2s}$ and fails the standard annular counit. This is an exact failed route, not a defect in Panitch–Park's balanced-action theorem.

Positive geometric filling theorem

Definite transversality forces the filling quotient to survive

Let an ideal triangulation have $N$tetrahedra and $r\geq1$ torus ends. At the commutative fiber write $Z_j=z_j^2$for the ordinary shape squares. When every$\operatorname{Im} Z_j>0$, the tetrahedron tangent is the graph of the diagonal matrix

$$D_{jj}=\frac1{Z_j-1},\qquad \operatorname{Im}D_{jj}<0.$$

Neumann's symplectic complex makes the independent edge rows together with one primitive filling row a real isotropic space. A real isotropic space cannot meet this definite complex Lagrangian nontrivially. Equivalently, if$(P\mid Q)$ is the exponent matrix, then $P+QD$ has full row rank. Hence the joint Jacobian has rank

$$\boxed{\operatorname{rank}J=2N-r+1.}$$

Principal half-logarithms of the positive shapes give vertex product $i$ and edge value$-1$. The genuine surgery equation $H_s=\pm2\pi i$gives meridian value $-1$, while the complementary slope measures a loxodromic core and has square different from one. The earlier quantum-flatness theorem therefore yields

$$\boxed{Q_s\ne0\quad\text{and the descended GY trace is nonzero}.}$$

This covers strictly geometric positive-shape hyperbolic fillings, including sufficiently long surgeries when the parent is supplied with such a complete triangulation. It does not assert that every manifold or prescribed triangulation admits one. A mixed-orientation folded-shape family in the proof note has an actual excess-dimensional intersection, showing why isotropy without definite shape orientation is insufficient.

Global scaling compatibility

The two Panitch–Park branches are conjugate after root identification

Orient an Euler circuit in every component of the 4-regular dual multigraph. Assign scalar $i$to outgoing face occurrences and$-i$ to incoming ones. Each tetrahedron has two of each, so opposite bare edges acquire the same sign. Vertex triples change by −1, Lagrangian squares stay fixed, and paired-face factors cancel in every manifold-edge monomial.

This produces an isomorphism between the two identified Panitch–Park targets and conjugates their named traces. The proof includes self-glued faces and parallel dual edges. Together with the six-root collar square above, it now transports the native Garoufalidis–Yu named action to the original Panitch–Park branch.

Explicit filling reduction

One meridian operator controls nonvanishing

Work over $k=\mathbb Q(i,A^{1/2})$with $A$ indeterminate. Suppose the boundary-linear trace has target$Q$ carrying invertible peripheral operators $m,l$with $ml=A^2lm$. Here $m$ is the filling meridian. In the opposite collar convention specified in the proof note, Frohman–Gelca Theorem 3gives the two solid-torus vacuum relations. They yield

$$\boxed{Q_s\cong\operatorname{coker}(m+A^2:Q\longrightarrow Q).}$$

The maps are explicit, with multiplication in the displayed order:

$$[q]\longmapsto[(1-l^2)q],\qquad [q]\longmapsto\left[\frac{m+A^{-2}}{A^{-2}-A^2}q\right].$$

Their representative independence and both inverse compositions are proved by exact identities in the noncommutative quantum torus. Thus a nonzero filled target is equivalent to failure of surjectivity of $m+A^2$. The regular-intersection theorem below gives a sufficient condition for that failure. A nonzero classical fiber by itself does not establish it.

Conditional nonvanishing theorem

A regular intersection survives quantization

For the identified GY target, take the tetrahedron equations, an independent set generating all edge relations, and the filling-meridian equation. Suppose their classical symbols at$A=1$ form a regular sequence at a common nonzero shape point. Full row rank of their joint Jacobian is a sufficient condition.

$$\boxed{\text{Regular classical intersection}\ \Longrightarrow\ Q_s\ne0\text{ at generic parameter}.}$$

The proof uses the actual sidedness of the GY quotient. Tetrahedron operators act on one side; edge and filling operators act on the other. These operators commute exactly. Classical regular-sequence syzygies then lift to prove that the formal local quotient has no$\hbar$-torsion under$A=e^{\hbar}$. An explicit map from the original generic meridian cokernel detects its nonzero unit in that formal quotient.

If the complementary longitude satisfies$l(p)^2\ne1$, the same map also detects the filled vacuum, so the descended GY trace is nonzero. Nonzero shape values alone do not imply the required regularity. The theorem does not claim that every filling in Panitch–Romo's intended class satisfies this hypothesis.

Clean-excess obstruction

Point-local deformation detectors can disappear

Fix a central sign sector of the actual six-root presentation and complete at a nonzero classical shape point. Let the tetrahedron relations act on the right and independent edge/filling relations act on the left. These operators commute exactly. Their first Koszul Bockstein is a differential operator on the classical intersection:

$$\beta(u[a])=u\beta([a])+V_a(u),\qquad V_a(u)=\sum_i a_i\{u,f_i^0\}.$$

If the scheme-theoretic intersection is smooth and has excess dimension, separate independence of the tetrahedron and moment blocks forces a syzygy whose tangent field$V_a$ is nonzero. On a smooth formal germ the resulting first-order operator is surjective. Successive coefficient corrections then give the all-orders conclusion

$$\boxed{\hbar H_p=0,\qquad H_p[\hbar^{-1}]=0.}$$

Thus the earlier regular-sequence detector cannot be continued through a clean excess locus. This is a conditional theorem for the actual relation system, not a vanishing theorem for its global Laurent-polynomial cokernel.

Two exact algebraic models delimit the inference. First,$T/(T(m-1)+(m-1)T)\cong k$globally, even though every corresponding generic formal-local detector vanishes. Second, on$k[z^{\pm1}]$ with$m(z^n)=-A^{c+2n}z^n$, the cokernel of $m+A^2$ is zero for $c=1$ and one-dimensional for $c=2$, although their classical filling fibers are identical. Exact quantum resonance, unavailable from classical support alone, decides survival. Neither model is claimed to arise from a manifold triangulation.

Global meridian-cokernel detector

A convergent orbit average survives beyond point-local geometry

Fix a central sign sector and retain every actual edge and filling row $(c_jW(h_j)+A^2)S$. Their exact values must define a character$\chi:\Lambda\to k^\times$ on the unsaturated image lattice. Extending the tetrahedron-index seed by zero across odd root cosets, define

$$\mathcal F(W(u))=\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1}A^{\Omega(\lambda,u)/2}\ell_0(W(u+\lambda)).$$

Suppose the exact character valuations admit nonnegative filling angles. On the associated homogeneous nonnegative cone, let$\delta$ be the double-arc quadratic degree. If no nonzero ray is invisible to both$\delta$ and the angle weight, then the sum converges coefficientwise on every algebraic root input. The tetrahedron recurrence and lattice reindexing give

$$\boxed{\mathcal F(sf_j)=0,\qquad \mathcal F((c_jW(h_j)+A^2)s)=0,\qquad \mathcal F(1)=1+\text{positive }A\text{-order}.}$$

Thus the actual native six-root meridian cokernel is nonzero under this criterion, without a classical shape point or regular intersection. Strict filling angles suffice but are not necessary. Conversely, a null ray produces infinitely many nonzero zero-valuation summands, so this particular unordered average cannot converge. That is an obstruction to the construction, not a vanishing theorem for the cokernel.

The theorem detects the cokernel unit, not the filled vacuum. For the exact named longitude $l=c_lW(b)$, the unresolved value is$\mathcal F(1-l^2)=\mathcal F(1)-c_l^2\mathcal F(W(2b))$. Nonvanishing of this difference is required before claiming a nonzero descended trace in the new chamber.

Exact longitude boundary

Strict angles alone do not prevent vacuum cancellation

For the exact named longitude $l=c_lW(b)$, write $c_l=\eta_lA^\rho$ and$\chi(\lambda)=\eta_\lambda A^{\vartheta(\lambda)}$. Reducing $T(b+\lambda/2)$ by its block minima gives an exact degree and leading coefficient for every longitude summand:

$$D_b(\lambda)=2\rho-\vartheta(\lambda)+\Omega_2(\lambda,b)+2\delta(x_\lambda)-2M_\lambda,\qquad s_b(\lambda)=\eta_l^2\eta_\lambda^{-1}(-1)^{M_\lambda}.$$

If $d_*$ is the minimum of these degrees and $S_*$ is the finite sum of leading coefficients there, then$d_*>0$, or$d_*<0$ with$S_*\ne0$, or$(d_*,S_*)=(0,S_*),\ S_*\ne1$proves $\mathcal F(1-l^2)\ne0$. The remaining case is genuine, not merely a weakness of the leading-term estimate.

A three-tetrahedron abstract mixed-sided presentation has exact native-form phases, an unsaturated isotropic moment lattice, strictly positive filling angles, the required longitude commutators, and $\mathcal F(1)\ne0$, yet a finite correct-sided identity proves

$$\boxed{1-l^2\in\sum_jSf_j+\sum_rE_rS.}$$

Hence every all-relations functional kills the filled-vacuum image in that sector. This is not a triangulation counterexample. A nonnegative six-root lift of its second edge would need at least eight prime incidences, while its exact scalar forces seven. More generally, an actual unscaled edge must satisfy

$$\nu_A\chi(h)+\sum_j\max(0,-h_{zj},-h_{wj})\le2.$$

The witness violates this necessary inequality, and no change of generators within its rank-two edge lattice repairs it. Thus the extracted angle, index and peripheral data do not by themselves prove vacuum survival; a positive theorem must use genuine global edge-incidence structure.

A further family of character-twisted index functionals also descends to the original target when its shifted character has its own nonnegative-angle and no-null-ray certificate. It does not evade an algebraically zero vacuum class.

Actual filled-vacuum theorem

Normal Euler degree prevents longitude cancellation

For genuine native incidence data, every contributing$\lambda\in\Lambda\cap2L$determines a representation-independent integral nonnegative$Q$-normal class$X_\lambda$. It has fixed nonzero longitude boundary component, and the exact native scalar, character valuation and meridian–longitude Weyl term cancel to give

$$\boxed{D_l(\lambda)=2\bigl(\delta(X_\lambda)-\chi_{\mathrm{top}}(X_\lambda)\bigr).}$$

Suppose the triangulation has a complete strict angular certificate $\theta^0$ with zero peripheral angular holonomy and a strict filling angular certificate $\theta^s$ with meridian holonomy two. The first turns the displayed degree into

$$D_l(\lambda)=2\delta(X_\lambda)+2\theta^0\!\cdot X_\lambda>0,$$

while the second supplies all-input convergence, relation descent and a unit term of coefficient one. Therefore

$$\boxed{\mathcal F(1-l^2)=1+\text{positive }A\text{-order}\ne0.}$$

This detects the actual full-six-root filled vacuum and its named descended traces without nonlinear shape equations or a regular classical intersection. A leading–trailing deformation gives the filling certificate uniformly for every primitive slope$s=pa+qd$ satisfying

$$p^2+q^2>C^2,\qquad C^2=\max_j\frac{(JR_a)_j^2+(JR_d)_j^2}{(\theta_j^0)^2}.$$

This is a sufficient chamber theorem, not a classification of exceptional slopes. Panitch–Romo's unrestricted conjecture remains open.

Balanced detector stopping theorem

The natural complete-angle pairing kills the vacuum exactly

With only a complete strict angular certificate, shifting the moment character by the negative longitude produces a convergent, correct-sided functional $\mathcal G_{-b}$on the original target. The actual unsaturated lattice is bijective with the complete integral normal boundary fibers, so its peripheral values have the exact expansion

$$F_0(l^r)=\sum_{n\in\mathbb Z}(-1)^nA^{rn}\mathcal Z(-rl-ns),$$

where $\mathcal Z(\gamma)=\sum_a I_{\mathcal T}^a(\gamma)$is the total extended normal index. Applying the published boundary-inversion identity only after this scalar- and phase-exact bridge gives $F_0(l^{-r})=F_0(l^r)$ and hence

$$\boxed{\mathcal G_{-b}(1-l^2)=0.}$$

This is not evidence that the vacuum or target vanishes. The same functional detects the longitude:$\mathcal G_{-b}(l)=c_l(1+\text{positive }A\text{-order})\ne0$. For a one-cusp triangulation, every monomial translate certified by the same complete angle vector likewise kills the vacuum: it is either a scalar multiple of this pairing or has empty parity support.

With another cusp, an unexhausted correct-sided test remains. For a native curve $t$ on that cusp it is

$$\mathcal G_{w-b}(1-l^2)=\frac{c_l}{c_t}\sum_n(-1)^nA^n\bigl[\mathcal Z(t-l-ns)-\mathcal Z(-t-l-ns)\bigr].$$

Simultaneous boundary inversion does not evaluate this separate-cusp difference. Its nonvanishing remains open.

Integral classification and boundary-skein non-detection

Ordinary boundary curves cannot repair the balanced detector

With the same complete strict angle certificate, every monomial translate that has nonempty vacuum support reduces to a native insertion on the remaining cusps, up to an exact scalar. Integral normal homology proves this for the actual edge lattice; no saturation or additional half-integral classes are introduced.

$$w=\lambda_0+w_\gamma,\qquad \mathcal G_{w-b}=\chi(\lambda_0)A^{-n_0}c_\gamma^{-1}\mathcal G[E_\gamma].$$

Write $\mathscr B(P)=\mathcal G[P](1-l^2)$for the vacuum value after an insertion, and let$\sigma(E_\gamma)=E_{-\gamma}$reverse all remaining peripheral classes. The exact normal-index formula and GHHR's simultaneous boundary inversion imply

$$\boxed{\mathscr B\circ\sigma=-\mathscr B,\qquad \mathscr B(P)=\mathscr B\!\left(\frac{P-\sigma P}{2}\right).}$$

Frohman–Gelca's torus-skein embedding sends each boundary curve to $E_\gamma+E_{-\gamma}$. These sums are fixed by inversion. Inversion also preserves multiplication and Weyl factors, so every boundary-skein insertion has vacuum value zero, including arbitrary products of intersecting curves on a torus.

The oriented odd value is still open. Simultaneous inversion does not prove independent-cusp symmetry, and this detector failure does not prove that the filled vacuum vanishes. The two-angle nonvanishing theorem retains its scope.

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An alternative local seed

A one-sided Nahm functional with an exact convergence test

The tetrahedron relation needs only one adjoint recurrence. Coefficients of Euler's quantum-dilogarithm productsupply a half-space seed satisfying that recurrence. It fails the extra recurrence of the ordinary tetrahedron index, so the preceding balanced-index obstruction does not decide its value.

$$\ell_+(W(2a,2d))=\frac{(-1)^{\sum_j a_j}A^{2(\|a\|^2+\sum_j a_j-a\cdot d)}}{\prod_j(A^4;A^4)_{a_j}},\qquad a\geq0.$$

The seed is zero outside that support and outside the even root lattice. Average it over the full actual edge-and-meridian image, with the original native character. Write its real moment directions as $2(x,y)$. The unordered sum converges on every algebraic input exactly when

$$\boxed{\|x\|^2-x\cdot y>0\quad\text{for every nonzero moment direction with }x\geq0.}$$

This single cone condition gives quadratic growth of the summand degrees. A null ray is insufficient: a shifted input defeats it. When the condition holds, the functional annihilates every original tetrahedron, edge, and filling relation with the correct left and right multipliers. Projection to the first coordinates gives an exact congruence-restricted Nahm sum, preserving the unsaturated lattice.

A further native phase condition makes its unit summands positive. If the named longitude has exponent$b=(0,-t)$ with$t\geq0$, a complete strict angle certificate gives its scalar positive order. Every longitude-square summand then has greater order than its unit counterpart, proving$\mathcal H(1-l^2)\ne0$under these hypotheses, without a filling-angle assumption.

The original worked example is an abstract presentation. The actual-data result belowremoves the pure-longitude-coordinate restriction through an opposite-sign argument and certifies a genuine filling family. The unrestricted conjecture and the earlier odd peripheral value remain open.

Read the proof and scope, the internal audit, and the reproducibility packet. The source ledgerdistinguishes the classical local coefficients and Nahm-sum influences from the present global argument.

A genuine triangulation, not an abstract witness

The one-sided detector survives an actual boundary filling

On the fixed two-tetrahedron figure-eight triangulation published by Dimofte and Garoufalidis, the detector proves that the native filled vacuum survives for every primitive slope in the symbolic family

$$s=p\mu+q\lambda,\qquad \gcd(p,q)=1,\quad q\ne0,\quad p\in4\mathbb Z,\quad p/q>-4.$$

This opposite-sign result is retained with its original scope. The degree-gap extension removes the numerator congruence restriction using the same seed and native character.

The meridian and canonical longitude are fixed by the source's raw rows. Eliminating the actual edge-and-filling image gives the exact quadratic form

$$Q(x)=\frac{x_1^2+x_2^2}{2}+\frac{p}{8q}(x_1-x_2)^2.$$

It is strictly positive on the nonnegative cone precisely when$p/q>-4$. The native unit-phase condition holds when $p\in4\mathbb Z$. Every complementary longitude then makes its square summands negative, while the unit summands are positive. This proves$\mathcal H(1-l^2)\ne0$ without changing the native character or requiring a special longitude coordinate form. The original unsaturated image is retained.

For $4\mu+\lambda$, set$\mathfrak q=A^4$. The value is the sum of two positive Nahm series:

$$\mathcal H(1-l^2)=\sum_{u,v\ge0}\frac{\mathfrak q^{2u^2+2v^2-2uv+u}}{(\mathfrak q)_{2u}(\mathfrak q)_{2v}}+\sum_{u,v\ge0}\frac{\mathfrak q^{2u^2+2v^2-2uv+u+2v+1}}{(\mathfrak q)_{2u+1}(\mathfrak q)_{2v+1}}.$$

It starts $1+\mathfrak q+3\mathfrak q^2+6\mathfrak q^3+\cdots$. For the same native target, the ordinary-index average has an exact ray whose every positive multiple contributes constant coefficient one, so that average fails even on the unit. The one-sided seed excludes the ray. Strict filling angles are also impossible for this datum and slope.

Through the retained native comparison and filling reconstruction, this detects the named descended-trace vacuum. It does not prove the unrestricted Panitch–Romo conjecture, triangulation independence, or a new closed-manifold invariant. We do not claim exclusion from every earlier shape-based criterion. This remains an AI-assisted supporting note without external specialist review; novelty is unassessed.

Read the proof and exact conventions, the internal audit, the source ledger, and the reproducibility packet.

An exact degree gap, with native phases unchanged

The numerator restriction is unnecessary

The same one-sided detector now proves native filled-vacuum nonvanishing on the fixed published figure-eight triangulation for every primitive slope

$$s=p\mu+q\lambda,\qquad \gcd(p,q)=1,\quad q\ne0,\quad p/q>-4.$$

No new seed or change of sign sector is needed. The earlier divisibility condition was sufficient for termwise sign separation; it was not necessary for the vacuum to survive. With the native longitude scalar and Weyl factor included, the exact unit and longitude-square summand degrees are

$$D_0(x)=2Q(x)+(1+1/q)x_1+(1-1/q)x_2,$$$$D_l(a)=2Q(a)+(1-1/q)a_1+(1+1/q)a_2.$$

The quadratic form is strictly positive on the nonnegative orthant away from zero. Both linear coefficients are nonnegative because $q$ is a nonzero integer. Every nonzero supported term therefore has positive degree. The unit's zero term contributes one. A hypothetical zero-coordinate longitude term would require the second seed coordinate

$$\beta-T\alpha=\left(\frac1{2q},-\frac1{2q}\right),$$

which is never integral and so is outside the seed support. Thus

$$\mathcal H(1-l^2)=1+\text{strictly positive }A\text{-order}\ne0.$$

Positive order means positive exponents, not positive coefficients. Native signs may vary. The proof retains the unsaturated image and establishes convergence on every algebraic root input before using the original mixed-sided relations.

As an odd-numerator control, the slope$\mu+\lambda$ has a negative native unit summand, but its filled vacuum begins$1+\mathfrak q+2\mathfrak q^2+4\mathfrak q^3+6\mathfrak q^4+\cdots$, where $\mathfrak q=A^4$. The all-orders degree proof, not these initial coefficients, establishes nonvanishing.

This covers the entire strict convergence cone of this seed on the declared datum. It does not resolve Panitch–Romo's unrestricted conjecture, prove triangulation independence, or identify a new closed-manifold invariant. The named-trace conclusion retains the earlier exact comparison and filling reconstruction as dependencies. This is an AI-assisted supporting note without external specialist review; novelty is unassessed.

Read the phase-free proof, the internal audit, the source and dependency ledger, and the reproducibility packet.

Exact root transport and an actual sector boundary

Every nonmeridional vacuum survives

On the fixed published two-tetrahedron figure-eight datum, the native meridian cokernel is now proved nonzero for every primitive filling slope. The filled vacuum is nonzero for every nonmeridional slope$p\mu+q\lambda$,$q\ne0$. This follows from one exact cyclic change of tetrahedron roots:

$$\Psi(W(u_z,u_w))=(iA)^{\sum_j u_{wj}}W(-u_w,u_z-u_w).$$

The map has order three on full roots. It preserves each tetrahedron relation with the required left multiplier,

$$\Psi(f_j)=-w_j^{-2}f_j,\qquad f_j=z_j^{-2}+w_j^2-1.$$

Every native edge, filling and longitude scalar is transported exactly. With $d=2q-p$, the new quadratic form is

$$Q'(x)=(x_1+x_2)^2+\frac{2q}{d}(x_1-x_2)^2.$$

For negative $p/q$, this gives a convergent average and a strict vacuum degree gap. Combined with the preceding detector for nonnegative$p/q$, it covers the entire nonmeridional slope line without a slope or polarization census.

The meridian has a different outcome. Its transformed average still has nonzero unit value, beginning$-\mathfrak q^2$, where$\mathfrak q=A^4$. An independent finite certificate, made from eight tetrahedron relations and explicit edge and filling multiples, proves

$$[1-l^2]=0,\qquad [1]\ne0\quad\text{in the all-plus meridian sector}.$$

This is actual sector vanishing, stronger than a failed scalar detector. The subsequent squared-moment certificate lifts it to the full six-root module without reversing the central projection. The native target itself is nonzero.

The named-trace conclusion retains the exact earlier comparison and filling reconstruction as dependencies. This is a supporting result for one fixed actual triangulation, not a resolution or counterexample to the unrestricted Panitch–Romo conjecture. This AI-assisted note has no external specialist review; novelty is unassessed.

Read the cyclic transport and meridian proof, the finite quantum-algebra certificate, the internal audit, the source ledger, and the reproducibility packet.

A finite identity before central specialization

A nonzero filling target with zero six-root vacuum

For the fixed two-tetrahedron figure-eight triangulation, the meridian vacuum vanishes in the full native six-root module, not merely its all-plus sector:

$$[1-l^2]=0\text{ in }C_\mu^6,\qquad [1]\ne0\text{ in }C_\mu^6.$$

After the exact cyclic transport, the eight tetrahedron terms combine using only even powers of the edge and meridian operators. Finite Laurent polynomials give the universal identity

$$1-W(2b)=G+(W(2h)-1)U+(W(2s)-A^4)V,\qquad G\in Sf_1+Sf_2.$$

For arbitrary central involutions, the squared moment factors are

$$W(2h)-1=(W(h)-\epsilon)(W(h)+\epsilon),\quad W(2s)-A^4=(W(s)+A^2\zeta)(W(s)-A^2\zeta).$$

These factors retain the required right multipliers for edge and filling relations. The identity therefore holds over the central sign algebra itself. No sign assignments are enumerated. Nonzero unit follows separately by pulling back the earlier convergent all-plus detector.

The retained filling isomorphism sends the empty skein to$[1-l^2]$, not the module unit. Meridian filling returns the three-sphere, so this particular boundary-balanced descended trace is zero, despite its nonzero target. Panitch–Romo Conjecture 7.1asks for a nontrivial quotient and a commuting map; it does not explicitly require that map to be nonzero. This is not a counterexample to the conjecture.

A smaller compatible quotient may behave differently. For the unfilled trace and filling map$T:V\to Q,\ p:V\twoheadrightarrow W$, the minimal required linear relation space is$T(\ker p)$. Its quotient satisfies

$$0\longrightarrow W/p(\ker T)\longrightarrow Q/T(\ker p)\longrightarrow Q/\operatorname{im}T\longrightarrow0.$$

Thus meridian vacuum survival in that minimal quotient is exactly the question whether filled bracket evaluation vanishes on the actual trace kernel. The kernel witness belownow settles that question. The result retains the named-action and generic filling dependencies and is an AI-assisted research note without external specialist review; novelty is unassessed.

Read the proof, the finite central certificate, the internal audit, the source ledger, and the reproducibility packet.

An actual trace-kernel witness

The minimal quotient survives, but every compatible meridian trace is zero

For the standard figure-eight triangulation, let $b$be the zero-framed clasp oval, $b^{(2)}$ its parallel cable, and $\mu$ the boundary meridian. The explicit skein

$$\kappa=b^{(2)}+A^{-2}(\mu^2-1)b+A^{-4}\mu^2-1$$

lies in the actual full six-root trace kernel, but its meridian-filled bracket is nonzero:

$$T(\kappa)=0,\qquad p(\kappa)=A^{-4}.$$

The proof starts with Panitch–Park’s published full-root trace formulas, corrects the two-cable ribbon framing, and gives a finite identity using tetrahedron left multiples and a squared edge relation with its right multiplier. It retains the named meridian action on arbitrary inputs. Opposite-root signs are preserved.

Thus every compatible meridian-filled trace from the sphere’s skein module is zero, including the trace into the quotient imposing only the necessary filling relations. A separate convergent index functional detects a root-parity class outside the entire actual trace image, proving that this minimal compatible target remains nonzero:

$$D_{\min}=Q^6/T(\ker p)=Q^6/\operatorname{im}T\ne0,\qquad\overline T=0.$$

This supplies a nontrivial quotient in the literal sense of Panitch–Romo Conjecture 7.1 for this meridian instance. The unrestricted conjecture remains open. The cohomological grading below now extends the nonzero-cokernel argument to the strict-angle class. This is an AI-assisted research note with no external specialist review; novelty is unassessed.

Read the proof, the finite native certificate, the internal audit, the source ledger, and the reproducibility packet.

Cohomological degrees outside the image

Strict angles give a nonzero trace cokernel for every filling

Let an actual connected ideal triangulation have $r\geq1$torus cusps and a strict angle structure. At generic parameter, its full six-root quantum-trace target contains at least$2^r-1$ independent classes outside the entire trace image:

$$\boxed{\dim_k\bigl(Q^6_{\mathcal T}/\operatorname{im}\tau\bigr)\geq2^r-1>0.}$$

Half-face parities define a grading on the dual graph. Closed traced skeins have degree zero, by Panitch–Park’s published face-even theorem. The integral lattice of exponents commuting with every edge has at least $r$ independent degrees outside that zero part. These degrees extend to classes of$H^1(M;\mathbb F_2)$.

Strict edge angles prove the exact native character is consistent, including its fourth-root phase, and make the index functional converge on every algebraic input. Translating it by an integral edge-normalizing monomial detects a nonzero degree and annihilates the whole trace image. No filling angles or slope bounds enter.

$$D_p=Q^6/\tau(\ker p)\twoheadrightarrow Q^6/\operatorname{im}\tau\ne0.$$

Thus the minimal compatible quotient is nonzero for every Dehn filling. The common quotient on the right supports a compatible zero filled trace. This verifies the literal nonzero-quotient condition in this generic strict-angle class; it supplies no nonzero filled invariant and leaves the unrestricted conjecture open. The result is a research note, with no external specialist review or priority claim.

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Universal endpoint construction

The generic torus theorem restores algebraic gluing

Gunningham–Jordan–Vazirani Theorem 6.1makes the empty object a generator of the even endpoint block of the generic SL₂ torus skein category. The relative skein functor of an unmarked three-manifold vanishes in the odd endpoint block. This does not restrict the homology classes of its closed links.

Put $\mathcal C=\operatorname{SkCat}_k(T^2)$,$e=\varnothing$ and$B=\operatorname{End}_{\mathcal C}(e)$. A $B$-linear trace to$Q$ has the full endpoint target

$$\widetilde Q(X)=\operatorname{Hom}_{\mathcal C}(e,X)\otimes_B Q,\qquad \widetilde Q(f)(h\otimes q)=fh\otimes q.$$

Finite boundary-tangle frames turn every endpoint operation into a finite matrix of closed torus skeins. Cups, caps, endpoint transport and exchange act by composition, with their order and skein relations retained. The trace extends by an explicit frame-independent formula.

Combining this reconstruction with Gunningham–Jordan–Safronov excision gives, for arbitrary unmarked pieces glued along a torus,

$$\operatorname{Sk}_k(M_1)\otimes_B\operatorname{Sk}_k(M_2)\cong\operatorname{Sk}_k(M_1\cup_{T^2}M_2).$$

The conclusion holds at generic parameter. The earlier obstruction at $A=-1$remains valid and prevents an unrestricted integral version. No fixed integral torsion annihilator is claimed.

Uniform classical obstruction

Empty-sector gluing is not injective

Let $M$ be the exterior of any hyperbolic knot in$S^3$, and fill its meridian with a solid torus $V_\mu$. At the classical Kauffman-bracket parameter$A=-1$, the natural map

$$\boxed{\ \operatorname{Sk}_{-1}(V_\mu)\underset{\operatorname{Sk}_{-1}(T^2)}{\otimes}\operatorname{Sk}_{-1}(M)\longrightarrow\operatorname{Sk}_{-1}(S^3)\ \text{is not injective}.\ }$$

Choose complete holonomy with$U=\rho(\mu)=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$. Because the image is non-elementary, some$g$ has$G=\rho(g)=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$with $c\ne0$. For$\gamma=\mu g\mu g^{-1}$,

$$\operatorname{tr}(UGUG^{-1})=2-c^2.$$

The cusp holonomy and a solid-torus representation nevertheless have the same character on every boundary word. Their evaluations therefore define a functional on the balanced tensor product. It sends$x=1\otimes([K_\gamma]+2[\varnothing])$to $c^2\ne0$, while filling sends $x$ to zero. The proof uses the Przytycki–Sikora trace evaluation and the complete-holonomy normalization discussed by Dunfield–Friedl–Jackson.

What fails

Closed peripheral traces forget a matrix equation

Closed curves on the boundary see the semisimplified character. They cannot distinguish$\rho(\mu)=I+N$, with$N\ne0$ and$N^2=0$, from$\rho(\mu)=I$. Dehn filling requires the latter matrix equation, not merely equality of its peripheral traces. A nonperipheral group element detects the discarded nilpotent part.

At this classical specialization, adding more closed slopes cannot repair the construction. Generic parameter behaves differently: the finite torus-category factorizations recover all contributing endpoint relations from the empty sector.

The theorem rejects a universal Laurent-ring isomorphism that specializes compatibly at $A=-1$. The generic localization is now separately proved above; the classical witness was never shown to survive localization. The obstruction also does not refute Panitch–Romo Conjecture 7.1.

Categorical and algebraic descriptions agree

The full filling coend has an explicit generic model

Categorical excision retains the full boundary skein category. The generic torus theorem supplies the generation hypothesis needed to identify that coend with ordinary torus-algebra base change. For a boundary-linear target,

$$Q_s=\operatorname{Sk}_k(V_s)\otimes_B Q\cong Q/I_sQ.$$

If the full peripheral quantum torus acts, the explicit maps above reduce this quotient further to the meridian cokernel. Costantino–Lê's marked-surface splitting theorem is not used as a closed-torus excision theorem.

Signed angles and a detector beyond the unit

Convergence and nonvanishing are separate questions

The native index average converges on every finite Laurent input precisely when its signed generalized angle certificate is positive on every nonzero zero-quadratic ray of the actual moment cone:

$$x\in C\setminus\{0\},\quad\delta(x)=0 \quad\Longrightarrow\quad\theta\cdot x>0.$$

The proof retains the unsaturated edge lattice, native signs and both relation sides. With nonnegative angles, this criterion also gives a nonzero unit value and extends the preceding cokernel theorem. With signed angles, nonvanishing needs a separate argument. No actual triangulation outside the strict-angle class is exhibited here.

The grading proof only requires a nonzero functional, not a nonzero value at the unit. An exact abstract one-block control has $F(1)=0$ but $F(w^{-2})=1$. This is an algebraic control, not a manifold counterexample. It shows why a failed unit test need not exhaust the detector.

This supporting note extends the earlier affine-input estimate and draws on GHHR's normal-surface convergence framework. General native nonvanishing under the signed criterion was left open in that note; the normal-sector extension below settles the nonexceptional class. Read the proof, internal audit, exact reproducibility packet, and provisional confirmation cards.

Closed normal sectors and actual nonperipheral detectors

Beyond the strict-angle hypothesis

The actual even edge-image lattice is exactly the lattice of all closed integral normal classes modulo tetrahedral solutions. The native phase agrees as well, so the unit average is the sum of every closed normal-homology sector, not just the trivial sector. This identifies the signed convergence criterion with 1-efficiency.

Combining this bridge with Garoufalidis–Hodgson–Hoffman–Rubinstein's published normal-surface positivity theorem proves, for a connected oriented manifold with $r\ge1$ torus boundary components and a 1-efficient ideal triangulation, excluding the solid torus and $T^2\times I$:

$$F(1)=1+\text{positive }A\text{-order},\qquad \dim_k(Q^6/\operatorname{im}\tau)\ge 2^r-1.$$

The generic native comparison and image-grading hypotheses are retained. Every minimal compatible filling quotient is nonzero; this does not supply a nonzero filled invariant. The normal-surface positivity theorem is an external proof dependency, not a new result here.

For the two published exceptional triangulations, the unit value is zero but an explicit nonperipheral root input survives:

$$F_{\mathrm{solid}}(w_1^2)=-I_\Delta(0,0)\ne0, \qquad F_{\mathrm{product}}(w_1^{-2})=I_\Delta(0,0)^2\ne0.$$

These are actual incidence data with degree-one edges, which rule out nonnegative angles. The resulting cokernel bounds are one and three, respectively. No assertion for all triangulations of the exceptional manifolds follows from those two calculations. Proof and sources, reproducibility packet, and provisional confirmation cards.

An exact local obstruction to reverse transport

What a two-to-three move does not carry

The local quantum-gluing map does not reach every target class. With only the new interior edge and the local tetrahedron relations imposed, its even cokernel has an explicit basis:

$$\operatorname{coker}\phi^{\mathrm{ev}}_{2\to3} \cong\bigoplus_{\beta\in\mathbb Z^3,\;\beta_1+\beta_2+\beta_3=-1} k e_\beta.$$

A finite normal form and explicit quantum-difference operators describe this missing space. Rational functionals separate its basis and respect the primitive edge sign. Thus equality of index values on the transported image cannot by itself transport an arbitrary detector backwards.

This is a local result, not a global vanishing example: exterior edge relations have not yet been imposed. The known exceptional detectors do propagate along directed 2-3 and 0-2 refinements that remain 1-efficient, using Garoufalidis–Yu's preprint index compatibility theorem. No path to every triangulation is assumed. Proof and sources, reproducibility packet, and provisional confirmation cards.

Next mode: structural attack

Which local detecting classes survive the exterior relations?

The remaining nonvanishing question within the 1-efficient class concerns arbitrary triangulations of the solid torus and the product of a torus with an interval. Their scalar boundary indices can be zero, and the local move map misses an explicit transverse space. The next question is how actual exterior edge relations act on that space, and whether the native index detects a surviving class when its restriction to the transported image vanishes. An intrinsic compressing-disk or product construction remains an alternative. No local-to-global survival or connectivity is assumed.

A nonzero filled trace is a separate obligation. For the fixed figure-eight meridian, the actual kernel witness already forces every compatible trace to be zero. The investigation continues through structural constructions, without a census of manifolds or slopes.

Dependencies and guardrails

What survives the audit

• Garoufalidis–Yu Theorem 5.7 supplies the named boundary-linear trace after opposite-root identification. The native six-root extension retains the source's prequotient standard-arc spanning, off-diagonal annihilation and Type II annular factor as explicit dependencies.
• The ambient normalizer lemma proves comparison-kernel invariance for right multiplication by shape elements that commute with the right gluing relations. However, Chen–Kricker's comparisonuses an extended GY target before opposite-root identification. The published degree-zero proof makes their comparison isomorphic. The signed additive lift and collar dictionary now connect this target to the native named GY action at generic parameter.
• Panitch–Park's UV–IR paper uses a modified target and records two scaling branches. Chen and Kricker's theorem supplies the named extended-trace intertwining in its normalized branch; the new injectivity argument makes that comparison invertible. The Eulerian normalization and six-root collar square transfer the named action to the original Panitch–Park branch.
• Garoufalidis–Yu already note that degenerate triangulations can make their quantum gluing module vanish. This is a scope qualification, not a new counterexample to the intended conjecture.
• The clean-excess theorem concerns formal-local completions. Their vanishing is not a faithful test for the global Laurent-polynomial quotient; the explicit Fourier-functional model proves that inference false.
• The global detector uses the established DGG tetrahedron index, the exact degree formula of Garoufalidis–Hodgson–Hoffman–Rubinstein, and the lattice-averaging strategy of Garoufalidis–Yu. The new contribution is the all-root actual-character descent, the all-input weak-angle estimate and its exact null-ray boundary; it is not claimed as a new surgery invariant.
• The longitude analysis retains the exact Weyl phase, the Garoufalidis two-recurrence system, and the Garoufalidis–Hodgson–Hoffman–Rubinstein degree and sign. Its cancellation witness is deliberately excluded from the topological class by a proved incidence obstruction.
• The two-angle theorem uses the standard actual$Q$-normal matching, boundary and formal-Euler dictionary. Complete angles control the longitude degree; filling angles separately control the original unit functional. Neither certificate is silently replaced by a nonlinear shape solution.
• The balanced stopping theorem uses GHHR's integral normal-homology theorem to cover every boundary fiber without saturating the moment lattice, and GHHR's published boundary inversion only after that bridge. Root inversion is false termwise for the tetrahedron seed.
• The multiple-cusp classification also uses GHHR's integral image and kernel theorem. Boundary-skein non-detection combines the resulting odd vacuum functional with Frohman–Gelca's inversion-fixed torus-skein algebra. Independent-cusp symmetry of the actual index remains unproved here.
• Panitch–Park and Panitch–Romo remain the formal sources for the square-root target and the Dehn-filling conjecture.