FPRD research loop state

Latest completed mode

ATTACK — exact cyclic transport covers every nonmeridional filled vacuum; a finite certificate identifies the meridian all-plus boundary.

On the fixed published two-tetrahedron figure-eight datum, the native six-root meridian cokernel is nonzero for every primitive slope. The native filled vacuum is nonzero for every nonmeridional slope q!=0, using the retained exact named-action/filling reconstruction. One exact full-root cyclic map supplies the missing convergence cone; no signs or target relations are changed. For the meridian, the transformed all-plus average has nonzero unit, but an independent finite quantum-algebra certificate proves that the vacuum class 1-l² is zero in the all-plus sector. The full-six-root meridian vacuum remains unresolved: zero after central projection does not imply zero before projection. This is a supporting research note, not a Publication or an unrestricted conjecture resolution.

Current delivery state: the latest research is saved in the Site source and stable handoff. Public deployment is still pending the approval requested after the phase-free update; do not mistake saved versions for live releases. No author outreach is warranted. The earlier approved comparison publication is unchanged.

Preceding phase-free result, retained

On the fixed two-tetrahedron figure-eight datum published by Dimofte–Garoufalidis (arXiv:1202.6268v2, Sections 2.6 and 4.6), the unchanged one-sided seed now proves H(1-l²)=1+strictly positive A-order for every primitive slope pμ+qλ with q!=0 and p/q>-4. There is no congruence restriction on p. The exact unit and longitude degrees have the same strictly copositive quadratic part and interchanged nonnegative linear coefficients. The only possible zero-degree longitude endpoint has nonintegral second seed coordinate (1/(2q),-1/(2q)), so is unsupported. Coefficient positivity, a new local solution, and changes to the native character are unnecessary. Through the retained native named-action and filling reconstruction, this detects the actual named filled vacuum. The proof is symbolic in the slope parameters, not a census.

The preceding opposite-sign theorem remains valid for p divisible by four, but that phase certificate is no longer a necessary gate for this datum. The new odd-numerator control (p,q)=(1,1) has genuinely negative native unit summands, while its filled vacuum starts 1+mathfrak-q+2 mathfrak-q²+4 mathfrak-q³, mathfrak-q=A⁴. This is a supporting research note, not a Publication. No external specialist review or priority claim is made. The actual odd peripheral value and unrestricted conjecture remain open.

Retained +4 application beyond the earlier ordinary-index detector: -2h+h_s=(-4,0;0,-2) has character one and every positive ray multiple has ordinary seed constant coefficient one. That average fails on the unit, while the one-sided vacuum starts 1+mathfrak-q+3 mathfrak-q²+6 mathfrak-q³. No strict filling-angle certificate exists for this datum and slope. No exclusion from every earlier regular-intersection or clean-excess criterion is claimed.

Active program and retained publication

Program: https://fprd-lab.joshgay.chatgpt.site/research-programs/dehn-filled-quantum-trace

Retained Publication: https://fprd-lab.joshgay.chatgpt.site/publications/ck-degree-zero-descent

Panitch–Romo Conjecture 7.1 remains open in its unrestricted scope. No universal actual filled-vacuum nonvanishing or actual full-six-root counterexample has been proved. No new publication outreach is warranted by this note.

Chen and Kricker were separately contacted about the earlier comparison publication. No reply or endorsement is assumed. Keep the earlier publication and all established comparison hypotheses intact.

Current evidence packet

cyclic-polarization-attack/:

• FPRD-cyclic-polarization-and-meridian-boundary-2026-09-08.md: exact full-root cyclic map; all-input correct-sided transport; complementary degree gap; every nonmeridional vacuum; nonzero meridian unit; finite all-plus meridian vacuum certificate.
• verify-cyclic-polarization.py, exact-results.json: generic Laurent identities, an outside-cone original-native control, meridian series with certified tails, and full quantum-algebra verification of the certificate before and after inverse transport.
• meridian-certificate.json: eight tetrahedron left multipliers and explicit finite edge/filling right multipliers in the actual all-plus sector.
• internal-audit.md, source-ledger.md, README.md, build-release.py, manifests, archive: self-audit, source scope and influences, reproduction and integrity.

Prepared proof path: /evidence/dehn-filled-quantum-trace/cyclic-polarization-attack/FPRD-cyclic-polarization-and-meridian-boundary-2026-09-08.md. This and the preceding phase-free update await public deployment approval.

The preceding phase-free-nahm-attack/ retains:

• FPRD-phase-free-Nahm-vacuum-2026-09-08.md: unchanged native setup and seed, all-input descent, general complete-square degree identity, exact fractional longitude defect, and phase-free symbolic family proof.
• internal-audit.md, source-ledger.md: self-audit, retained primary-source input and access limits, formal dependencies, and intellectual influences.
• verify-phase-free-nahm.py, exact-results.json: exact generic Laurent polynomial identities; a single odd-numerator raw Weyl-sum control with Gaussian integer coefficients, original relation sides, odd roots, and rigorous tail bounds.
• README.md, build-release.py, manifests, and phase-free-nahm-attack.zip: reproducibility and integrity.

Proof: https://fprd-lab.joshgay.chatgpt.site/evidence/dehn-filled-quantum-trace/phase-free-nahm-attack/FPRD-phase-free-Nahm-vacuum-2026-09-08.md

The preceding actual-nahm-attack/ retains:

• FPRD-actual-Nahm-vacuum-2026-09-08.md: opposite-phase lemma, actual native datum, symbolic cone and phase proof, explicit +4 Nahm formula, and old-index failure ray.
• internal-audit.md, source-ledger.md: scope, source convention checks, dependencies and intellectual influences; self-audit only.
• verify-actual-nahm.py, exact-results.json: symbolic identities, original-sided native relations, supported odd roots and exact coefficients with quadratic tail bounds.
• README.md, build-release.py, manifests, and actual-nahm-attack.zip: reproducibility and integrity.

Proof: https://fprd-lab.joshgay.chatgpt.site/evidence/dehn-filled-quantum-trace/actual-nahm-attack/FPRD-actual-Nahm-vacuum-2026-09-08.md

The preceding one-sided-seed-attack/ retains:

• FPRD-one-sided-Nahm-detector-2026-09-08.md: complete seed, exact cone theorem, original-sided descent, graph/Nahm translation, conditional vacuum theorem, and abstract example.
• internal-audit.md, source-ledger.md: self-audit, dependencies, and intellectual influences; no external or independent specialist audit is asserted.
• verify-one-sided-seed.py, exact-results.json: exact local identities and global example checks with quadratic tail certificates; full-root and unsaturated-image checks.
• README.md, build-release.py, SHA256SUMS.txt, one-sided-seed-attack.zip, ARCHIVE-SHA256.txt, archive-validation.txt: reproducibility and integrity.

Proof: https://fprd-lab.joshgay.chatgpt.site/evidence/dehn-filled-quantum-trace/one-sided-seed-attack/FPRD-one-sided-Nahm-detector-2026-09-08.md

The preceding peripheral-projection-attack/ remains the integral classification of supported same-complete-character translates, exact peripheral scalar, and odd projection. Every ordinary remaining-cusp boundary-skein insertion has vacuum value zero; the actual odd part remains unresolved.

The preceding balanced-pairing-attack/ packet remains the integral normal-fiber and exact phase bridge, sourced balanced vanishing theorem, and one-cusp stopping result.

The preceding actual-incidence-attack/ packet remains the two-angle and uniform large-slope theorem. longitude-difference-attack/ retains the general phase formula, correct-sided shifted-character family, and incidence-infeasible abstract zero-vacuum witness. global-meridian-index-attack/ retains all-input convergence. six-root-peripheral-attack/ retains the native named-action comparison. Earlier positive-geometric and clean-excess results retain their scopes.

Retained exact presentation

Use $k=\mathbb Q(i,A^{1/2})$, $K=\mathbb Q(i)((A^{1/2}))$, $B=iA$, $q=A^4$. In a central sector,

$$C_s=S\Big/\left(\sum_jS(z_j^{-2}+w_j^2-1) +\sum_e(c_eW(h_e)+A^2)S+(c_sW(h_s)+A^2)S\right).$$

This is a mixed-sided vector-space quotient, not a two-sided algebra quotient. The four native vertex values are $iA$; opposite-root signs remain after central elimination. Retain the exact additive native peripheral action, generic parameter restriction, Eulerian normalization, and CK source-reading caveat. Do not reopen peripheral signs.

The named operators satisfy

$$m=c_sW(h_s),\quad l=c_lW(b),\quad ml=A^2lm.$$

The intrinsic filling isomorphism sends the filled vacuum to $[1-l^2]$. It is not the cokernel unit.

Keep $L=\mathbb Z^{2N}$, $\Omega(u,v)=\sum_j(u_{wj}v_{zj}-u_{zj}v_{wj})$, and the unsaturated image $\Lambda=\sum_e\mathbb Zh_e+\mathbb Zh_s$. The exact character is $\chi(h_r)=-A^2/c_r$, including phase consistency. The retained ordinary-index seed vanishes off $2L$ and equals $\prod_jI_\Delta(d_j,-a_j)$ on $W(2a_1,2d_1,\ldots)$.

Current one-sided construction

Use grouped coordinates (z-vector,w-vector) for this section only. Put s(a)=sum a_j, q=A⁴. The new seed vanishes off 2L and when any a_j<0; otherwise

$$\ell_+(W(2a,2d))=\frac{(-1)^{s(a)}A^{2(\|a\|^2+s(a)-a\cdot d)}}{\prod_j(q)_{a_j}}.$$

It solves the plus recurrence required by Sf_j. Its minus-recurrence residual at e=0 is −q^(1−m)/(1−q), so the established two-recurrence rank-two classification does not restrict this seed. These local coefficients are classical Euler quantum-dilogarithm coefficients.

Define H(a)=sum_{lambda in Lambda} chi(lambda)^(-1) ell_+(W(lambda)a), keeping the actual unsaturated image. Let V=(Lambda tensor R)/2, C_+={(x,y) in V:x>=0}, Q=||x||²−x.y. All-input unordered convergence is equivalent to Q>0 on C_+ minus {0}. If the cone is zero, each input has finite support. A rational negative ray defeats the unit; a nonvertical rational null ray is defeated by a sufficiently shifted w-input, and a vertical null ray defeats one sign already at the unit. Rational null rays follow from the rational restricted Hessian kernel on the minimal face; no faces were enumerated.

For u=2(alpha,beta), lambda=2(x,y), supported only if lambda+u in 2L and x+alpha>=0, the exact summand order is

$$D_u=2Q+2s(x)-\nu_A\chi(\lambda)+4(\alpha-\beta)\cdot x +2(\|\alpha\|^2+s(\alpha)-\alpha\cdot\beta).$$

Strict cone positivity supplies quadratic coercivity on every affine support. Then H(a f_j)=0 and H(W(mu)a)=chi(mu)H(a) for every mu in Lambda, proving original-sided descent. No angle or shape point is required for this convergence theorem.

Projection to x is injective, so y=Tx on V_z; isotropy makes x.T x’ symmetric. The criterion is strict copositivity of I−T on V_z intersect the nonnegative orthant. In the full-rank one-cusp case V_z=R^N. This is not necessarily positive definiteness and not a claimed efficient decision procedure. Retain Gamma={x in Z^N:2(x,Tx) in Lambda}, not its saturation.

Sufficient vacuum certificate: the exact phase satisfies eta(2x,2Tx)=(-1)^s(x) on Gamma; the declared native longitude has b=(0,-t), nonzero integral t>=0, c_l=eta_l A^rho; rho>0. Unit summands then have positive coefficients, and the longitude-square/unit summand ratio is eta_l² A^(2rho)q(t.x). Hence H(1−l²) is nonzero. Under a native complete strict angular certificate and zero longitude holonomy, rho=p(R_l)=sum alpha_j t_j>0, so no separate filling angles are needed. The phase and pure-w form must be proved for the original data, not forced by redefinition.

For actual raw R=sum n_e E_e+n_s R_s with P(R) even, the exact phase condition is equivalently sum_j R_j1 congruent to 2(sum_e n_e+n_s) mod 4. This follows from eta(P(R))=(-1)^(sum n_e+n_s)i^(-p(R)) and 2s(x)=sum R_j1−p(R). Check it on the actual even image, not a saturation or a chosen new sign sector.

Historical applicability boundary: the original pure-w certificate had only a three-block abstract example, h=(2,2,0;0,0,0), chi(h)=A⁴, b=(0,0,0;−1,0,−2), c_l=iA, Q=2k². That example remains abstract. The new opposite-phase lemma below removes the pure-w requirement and supplies actual data; it does not retroactively make the old example a triangulation.

Retained opposite-phase actual Nahm theorem

Under all-input convergence and the native unit phase eta(2x,2y)=(-1)^sum(x), every unit summand has nonnegative coefficients. For any native longitude b, every longitude-square summand has constant phase epsilon_l=eta_l²(-1)^sum(b_z). If epsilon_l=-1, H(1-l²) is nonzero regardless of the difference of supports or scalar valuations. For a raw native longitude R_l, epsilon_l=(-1)^sum(R_l,unprimed). No signs are redefined.

Use grouped coordinates. Actual raw rows: E1=((2,0,1),(2,0,1)), E2=((0,2,1),(0,2,1)), R_mu=((0,-1,0),(1,0,0)), R_lambda=((-2,2,0),(0,0,0)). Collapsed rows are h=(2,2;1,1), -h, a=(1,1;1,0), d=(-4,0;-2,0), with scalar exponents 0,4,-1,2 respectively. Omega(a,d)=-2. These are the declared Dimofte–Garoufalidis labels and canonical longitude, not a differently twisted GY example. Do not replace the native B^(prime sum) scalar with the source’s SL2 eigenvalue lift.

For s=p mu+q lambda, h_s=(p-4q,p;p-2q,0), c_s=B^(-p+2q), Lambda=Z h+Z h_s. The two actual edge characters are exactly inverse. A complement r mu+t lambda satisfies qr-pt=1, b=(r-4t,r;r-2t,0), c_l=B^(-r+2t), Omega(h_s,b)=2.

For q!=0, y=T x with diagonal T_jj=1/2-p/(8q), off-diagonal p/(8q). The exact cone form is Q=(x1²+x2²)/2+p(x1-x2)²/(8q), strictly positive on the nonnegative cone iff p/q>-4. At equality the axes are null; below it they are negative. At q=0 a vertical ray prevents all-input convergence. These failures do not imply vacuum vanishing.

Even moment parity is e even and np even. The native phase criterion is 2e+n(p-2q-2)=0 mod4 on that full image, equivalent for primitive slopes to p in 4Z. Then q is odd, Gamma={x in 2Z²:x2-x1 in 2q Z}, and every complement has r odd, hence epsilon_l=-1. Therefore the actual filled vacuum survives for gcd(p,q)=1, q!=0, p in 4Z and p/q>-4. Sector detection detects the vacuum upstairs; no injectivity of sector projection is asserted.

For (p,q)=(4,1), choose b=a, c_l=-iA^-1. Writing x=(2u,2v), the positive vacuum formula is

H(1-l²)=sum_{u,v>=0} q^(2u²+2v²-2uv+u)/((q)(2u)(q)(2v)) + sum_{u,v>=0} q^(2u²+2v²-2uv+u+2v+1)/((q)(2u+1)(q)(2v+1)), q=A⁴.

It starts 1+q+3q²+6q³+10q⁴+17q⁵. The all-orders positivity proof, not finite coefficients, proves nonvanishing. The original-character ordinary-index ray -2h+h_s has seed I_Delta(0,2k)I_Delta(-k,0)=1+positive order for all k>0, so its unordered unit sum fails. The new seed excludes the ray. Strict filling angles would imply 2gamma1+alpha2=0. No triangulation independence, first-priority claim, or unrestricted conjecture resolution follows.

Phase-free degree-gap theorem

Keep every actual row, scalar and unsaturated image from the preceding section. Write v=(1-1/q,1+1/q); then nu_A chi(2x,2Tx)=v.x. For a general full-graph native insertion b=(alpha,beta), c_l=eta_l A^rho, put Delta_b=beta-T alpha and kappa_b=2 alpha.Delta_b+v.alpha+2rho. With h0=2*1-v and a=x+alpha, exact Weyl expansion gives

$$D_0(x)=2Q(x)+h_0\cdot x,\qquad D_l(a)=2Q(a)+(h_0-4\Delta_b)\cdot a+\kappa_b.$$

If Q is strictly copositive, both linear vectors are componentwise nonnegative, kappa_b=0, and a=0 is outside the longitude support, the unit has constant one while every longitude term has positive order. Delta_b nonintegral is sufficient to exclude that endpoint when b is integral. This is a polarization-relative algebraic criterion, not a claimed coordinate-independent topological object.

For the actual figure-eight data and every complement qr-pt=1,

$$\Delta_b=\left(\frac1{2q},-\frac1{2q}\right),\quad \kappa_b=0,$$

$$D_0(x)=2Q(x)+\left(1+\frac1q\right)x_1+\left(1-\frac1q\right)x_2, \quad D_l(a)=2Q(a)+\left(1-\frac1q\right)a_1+\left(1+\frac1q\right)a_2.$$

All nonzero supported x,a lie in the nonnegative integral orthant. For any nonzero integer q the linear coefficients are nonnegative, and Q>0 there when p/q>-4. The unit endpoint x=0 is exactly the image zero and contributes one. A longitude endpoint a=0 would require second seed coordinate Delta_b, which is not integral, and is therefore absent. Merely showing 2b not in Lambda is insufficient; nonzero pure-w seeds are allowed. Including 2rho from the native scalar is essential: deleting it leaves constant -2rho.

Thus H(1-l²)=1+strictly positive A-order for all gcd(p,q)=1, q!=0, p/q>-4, independently of phases. Positive order does not mean positive coefficients. The original opposite-sign certificate still fails outside p in 4Z; the new proof makes it unnecessary. No seed, sector, moment character or multiplier side is changed. All-input convergence comes before reindexing; this proof is not restricted to squared-root inputs.

At the single odd-numerator control (p,q)=(1,1), hs=(-3,1;-1,0), cs=iA, chi(hs)=iA, b=(1,1;1,0), cl=-iA^-1. The moment 4h+2hs=(2,10;2,4) has negative unit summand, despite positive low-order aggregated coefficients. The exact filled vacuum starts 1+mathfrak-q+2 mathfrak-q²+4 mathfrak-q³+6 mathfrak-q⁴+9 mathfrak-q⁵+14 mathfrak-q⁶. The computation uses one actual control and certified tails, not a slope census. No all-orders coefficient positivity is inferred.

Exact cyclic transport and the meridian boundary

Use the single blockwise map C(z,w)=(-w,z-w), Psi(W(u))=B^(sum u_w) W(Cu). It is an integral symplectic full-root automorphism of order three, and Psi(f_j)=-w_j^-2 f_j. The left unit is on the correct side. Raw triples cycle (R1,R2,R3)->(R2,R3,R1), so every edge/peripheral scalar transforms exactly with its new prime sum. Lambda’=C Lambda is unsaturated, and chi’(Cxi)=B^(-sum xi_w)chi(xi). Pull back the convergent transformed average by Psi to act on the original native target.

Put d=2q-p. The transformed data are h’=(-1,-1;1,1), both edge scalars -A², hs’=(d,0;-2q,p), cs’=1, b’=(-r+2t,0;-2t,r), cl’=1. Both edge characters are one; chi’(e h’+n hs’)=(-A²)^n. These scalar simplifications are outputs of the exact map, not choices of new signs.

For d!=0, e=-2x2, n=2(x1-x2)/d, and y=T’x with T’=[[-2q,p],[p,-2q]]/d. The cone form is

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

Its exact strict copositivity condition is 1+2q/d>0. At d=0 the actual moment span has a vertical ray. Under strict copositivity, all-input quadratic coercivity proves descent before reindexing. The exact native longitude defect is (-2/d,2/d), with zero complete-square constant. The unit and longitude degrees are

$$D'_0=2Q'+(2-4/d)x_1+(2+4/d)x_2,\quad D'_l=2Q'+(2+4/d)a_1+(2-4/d)a_2.$$

If |d|>2, the coefficients are positive and the only zero-coordinate longitude endpoint has nonintegral second seed coordinate. Hence the pulled-back original vacuum value is 1+positive A-order. For q!=0, use the preceding original polarization when p/q>=0; when p/q<0, q/d>0 and |d|=2|q|+|p|>=3, so the cyclic detector applies. This proves every nonmeridional native filled vacuum is nonzero on this fixed datum, without enumerating slopes or polarizations.

At the meridian (p,q)=(1,0), choose r=0,t=-1. Then T’x=(-x2,-x1), Q’=(x1+x2)^2, hs’=(-1,0;0,1), b’=(-2,0;2,0), cl’=1. The full average still converges. Its unit series is

$$H'(1)=\sum_{a,b\ge0}\frac{(-1)^{a+b}\mathfrak q^{(a+b)(a+b-1)/2+2a}}{(\mathfrak q)_a(\mathfrak q)_b}.$$

The longitude-square series interchanges a,b, so H’(1-l’^2)=0. Euler summation gives

$$H'(1)=(\mathfrak q)_\infty\sum_{a\ge1}\frac{(-1)^a\mathfrak q^{a(a-1)/2+2a}}{(\mathfrak q)_a(\mathfrak q)_{a-1}}=-\mathfrak q^2+\text{higher order}\ne0.$$

The reversed meridian orientation interchanges the degree coordinates and gives the same nonzero unit series. Thus every primitive slope has a nonzero native cokernel, even though meridian vacuum survival requires a separate argument.

Finite meridian sector certificate, stronger than detector failure: after quotienting by the left-acting moment characters W(h’)=1 and W(hs’)=-A², put e_(a,b)=[W(0,0;a,b)]. The integral full-root normal form gives relations e_(a+2,b)-e_(a,b)+A^(4-2a)e_(a,b-2)=0 and e_(a,b+2)-e_(a,b)+A^(-4-2b)e_(a-2,b)=0. In the even-even block V_(j,k)=e_(2j+2,2k-2), define

$$R_1(j,k)=V_{j+1,k}-V_{j,k}+\mathfrak q^{-j}V_{j,k-1},\quad R_2(j,k)=V_{j,k+1}-V_{j,k}+\mathfrak q^{-k}V_{j-1,k}.$$

The meridian vacuum is v=V_(-1,1)-V_(1,-1), and

$$v=\mathfrak q[R_1(1,1)-R_2(2,0)-R_1(1,0)+R_1(0,2)-R_2(1,1)+R_2(0,1)]-R_1(0,1)+R_2(1,0).$$

This has been lifted to 1-W(-4,0;4,0)=G+(W(h’)-1)U+(W(hs’)+A²)V in the full quantum Laurent algebra, with explicit finite U,V in meridian-certificate.json. The exact Weyl identity and its inverse cyclic image both check. Thus [1-l²]=0 in the actual all-plus meridian sector while [1]!=0. Exchanging tetrahedra gives the reversed-orientation statement modulo the edge relation.

Do not lift sector vanishing backward. The full-six-root meridian vacuum is not proved zero, and the native target is proved nonzero. This is not a counterexample to the unrestricted nonzero-target conjecture. The next mathematical object is the central sign algebra before specialization, not another all-plus scalar functional or a sign-sector census.

Retained two-angle vacuum theorem

For an actual triangulation, raw rows satisfy $P(R)_j=(R_{j1}-R_{j2},R_{j3}-R_{j2})$, $p(R)=\sum_jR_{j2}$. All-plus native scalars are $B^{p(R)}$. Edge/peripheral/tetrahedron normal classes have Euler values $-2,0,-1$, boundary values $0,2\gamma,0$, and zero mod-2 homology.

A complete strict certificate $\theta^0$ has tetrahedron sums one, edge pairings two, and all peripheral holonomies zero. A filling certificate $\theta^s$ additionally has meridian pairing two. The earlier theorem proves

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

when both certificates exist. Its normal longitude degree is $2(\delta(X)-\chi_{\rm top}(X))>0$; the filling certificate supplies original-character all-input convergence. A nonnegative filling certificate with the exact prior no-null-ray condition can replace strict filling positivity.

For a fixed cusp basis $R_a,R_d$ with pairing two, the deformation

$$\theta^{p,q}=\theta^0+\frac{pJR_d-qJR_a}{p^2+q^2}$$

is a filling certificate whenever it is positive. A uniform sufficient condition is

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

Thus the actual named filled vacuum is nonzero for all sufficiently large primitive slopes on a triangulation carrying a complete strict certificate. This is a sufficient chamber, not an exceptional-slope classification. Do not enumerate its bounded complement.

Retained exact normal-index bridge

Let $n(\lambda)=\Omega(\lambda,b)/2$. The balanced convergence character is

$$\chi_0(\lambda)=\chi(\lambda)A^{-2n(\lambda)} =(-1)^nB^{2E-p(R)}, \quad R=\sum_en_eE_e+nR_s,\quad E=\sum_en_e.$$

Do not discard the residual phase $(-1)^n$. Write $F_0$ for the full unsaturated average with $\chi_0$. The complete strict certificate supplies all-input convergence, and

$$\mathcal G_{-b}(a)=F_0(W(-b)a)$$

descends to the original target.

For every integer $r,n$, the map

$$\lambda\longmapsto X=\max\bigl((R+rR_l)/2\bigr)\Delta-(R+rR_l)/2$$

is a bijection from $\{\lambda\in\Lambda:n(\lambda)=n,\ \lambda+rb\in2L\}$ to the minimum-zero integral normal classes with $\partial X=-rl-ns$. The maxima are blockwise. Surjectivity follows from the integer identity

$$2X+nR_s+rR_l\in\mathbb E+\mathbb T,$$

by GHHR Theorem 7.1; it does not saturate $\Lambda$. All mod-2 homology classes in that boundary fiber are included.

Let $\mathcal Z(\gamma)=\sum_aI_{\mathcal T}^a(\gamma)$, the total GHHR boundary index. Then

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

Every summand has exact degree $2(\delta(X)-\chi_{\rm top}(X))+rn$, and leading coefficient $(-1)^{n-\chi_{\rm top}(X)}$. The prior all-input theorem justifies grouping the individual unordered-summable terms into these fibers.

Retained balanced-pairing failure theorem

GHHR’s published Remark 8.3, printed p. 312, states $I^a(-\gamma)=I^a(\gamma)$. The published source and arXiv version were freshly checked. This is a formal external dependency, not a new independent inversion proof.

After the exact bridge, legitimate reindexing $n\mapsto-n$ yields $F_0(l^{-r})=F_0(l^r)$. Therefore

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

This is not termwise root inversion: the seed itself is not inversion-invariant. It is not zero target evidence. In fact

$$\mathcal G_{-b}(l)=c_lF_0(1) =c_l(1+\text{positive }A\text{-order})\ne0.$$

Thus the actual meridian cokernel is nonzero from the complete certificate alone; the actual vacuum remains a separate question.

For one torus boundary, every monomial left translate certified by the same complete vector also kills the vacuum. Writing $w=v+b$, orthogonality forces $w\in\Lambda\otimes\mathbb R$. If $w\in\Lambda$, the functional is $\chi(w)A^{-\Omega(-b,w)/2}\mathcal G_{-b}$. If $w\notin\Lambda$, integral normal homology forces both vacuum parity supports to be empty. The same dichotomy holds in multiple cusps when $w\in\Lambda\otimes\mathbb R$, but does not exhaust additional other-cusp directions.

The all-plus vanishing theorem has not been extended to every central sign sector. No all-purpose pairing has been excluded.

Preserved failures and limits

The preceding abstract strict-filling-angle example has a nonzero cokernel unit and an algebraically zero specified vacuum, but violates actual nonnegative edge incidence and lacks a complete nonnegative certificate. It remains an abstract obstruction only.

No target-wide nonvanishing or zero result follows from a scalar index alone. No new surgery invariant or identification with the Celoria–Hodgson–Rubinstein Dehn-filling index is asserted. Duan–Garoufalidis–Gu’s general factorization-based symmetry route is conjectural and was not used as proof. No researcher contact is authorized by this ordinary research note.

New multiple-cusp classification and odd projection

Let $\Gamma_o=\bigoplus_{T\ne T_0}H_1(T;\mathbb Z)$. For $w=v+b\in L\cap\Lambda^\perp$, nonempty vacuum seed support is equivalent to

$$w=\lambda_0+w_\gamma,\quad\lambda_0\in\Lambda,\quad \gamma\in\Gamma_{\rm adm}:= \{\gamma:\iota_*(\gamma-l)\in\operatorname{span}_{\mathbb F_2}\{\iota_*s\}\}.$$

The representation is unique for fixed additive native rows. The integral proof chooses an integral normal lift V of w and forms the minimum-zero class X from (V−R_l+R)/2. Integrality of its boundary forces the other-cusp and meridian coefficients of ∂V to be even; the doubled normal identity forces [V]₂=0. GHHR’s integer kernel theorem then gives the decomposition. No saturation or assumed half-integral peripheral lift is used. Unsupported translates vanish on the vacuum only, not necessarily on all other inputs.

For $n_0=n(\lambda_0)$, the exact all-input scalar is

$$\mathcal G_{w-b} =\chi(\lambda_0)A^{-n_0}c_\gamma^{-1}\mathcal G[E_\gamma], \quad\mathcal G[P](a)=F_0(W(-b)Pa),\quad E_\gamma=c_\gamma W(w_\gamma).$$

On the other-cusp quantum torus define $\sigma(E_\gamma)=E_{-\gamma}$, fixing A and coefficients. This is an algebra automorphism: it preserves product order and Weyl factors. For the original-target vacuum functional

$$\mathscr B(P)=\mathcal G[P](1-l^2),$$

the exact retained normal bridge gives

$$\mathscr B(E_\gamma)=c_l\sum_n(-1)^nA^n [\mathcal Z(\gamma-l-ns)-\mathcal Z(-\gamma-l-ns)], \qquad \mathscr B\circ\sigma=-\mathscr B.$$

Consequently $\mathscr B(P)=\mathscr B((P-\sigma P)/2)$. Frohman–Gelca’s torus-skein image is generated by $E_\gamma+E_{-\gamma}$ on each cusp, hence every ordinary remaining-cusp boundary-skein insertion has vacuum value zero. This covers arbitrary noncommuting products without assuming independent-cusp symmetry of Z. It does not extend σ to the whole mixed-sided target or assert zero value on all target inputs.

The formal boundary witnesses Z₀,Z₁ have common unit normalization, simultaneous inversion, admissible Lagrangian mod-2 support, finite summability and positive nonunit order, but the oriented test is respectively 0 and A⁴. They do not satisfy any claimed actual tetrahedron presentation and are not triangulation counterexamples. They prove only that the listed boundary constraints cannot decide the odd value.

Next mode and decisive question

ATTACK — determine the full-six-root meridian vacuum through its central sign algebra before specialization.

The exact cyclic-polarization gate is now closed: one full-root isomorphism with the exact scalar character supplies the missing half of the slope line. Together with the phase-free theorem, every q!=0 vacuum is detected on the fixed actual datum. The meridian cokernel is also nonzero, but its all-plus vacuum is algebraically zero by a finite certificate. Do not seek another all-plus detector for this vacuum; the certificate excludes every such functional. The next decisive gate is whether the original meridian vacuum survives in complementary central components, or whether the finite certificate lifts before central specialization. Seek a single intrinsic central-algebra argument, preserving the native moment action and relation sides, rather than enumerating sign assignments, slopes, or triangulations. A zero specialized image is not a zero full-six-root class.

The alternative actual odd peripheral value B restricted to T_o^- remains unresolved. Do not repeat inversion-even boundary insertions, same-complete-character translate classification, saturation constructions, or conjectural separate-cusp Weyl symmetry as if they solve it. A differentiated character has an inhomogeneous moment relation. The new half-space seed is an explicit different one-recurrence solution; do not incorrectly force it into the two-recurrence rank-two space. Reassembling the full index by the extra local sum requires a separate global interchange proof.

Retain all earlier actual two-angle and uniform large-slope nonvanishing results. Do not enumerate manifolds, slopes, triangulations, normal supports, or residual configurations. Panitch–Romo Conjecture 7.1 remains open; no new Publication or author outreach is warranted.

Other programs retain their latest paused or withdrawn status. Selfridge remains paused; the semicomplete-digraph target remains withdrawn; the JT two-boundary work is not restarted by this continuation.