The longitude difference: exact coefficients and a strict-angle cancellation obstruction

FPRD Lab, 8 September 2026. Supporting ATTACK note. AI-assisted work, internally audited, without external specialist review or author endorsement. Novelty and first priority are unassessed. This is not a new Publication.

Outcome

The actual named-longitude gate is not closed. This note derives its exact shifted summand degree and coefficient, proves a character-twisted family of additional all-relations functionals, and locates a genuine limitation of the existing hypotheses.

There is a three-tetrahedron abstract mixed-sided presentation with all tetrahedron equations, exact six-root-form scalars, an unsaturated isotropic moment lattice, strictly positive filling angles, and the required edge/meridian/longitude commutation laws, for which

$$\mathcal F(1)=1+\text{positive }A\text{-order},\qquad [1-l^2]=0\quad\text{in the meridian cokernel}. \tag{1}$$

The second assertion is established by a finite correct-sided identity, not by a truncated series. Consequently every all-relations functional kills this witness’s filled-vacuum image. A second detector cannot repair an algebraically zero class.

This witness is not an actual triangulation. Its second designated edge row cannot have a nonnegative six-root incidence lift; the obstruction is proved explicitly below. Thus (1) refutes an inference from the abstract angle/index/peripheral data alone, not actual filling nonvanishing. The extra actual incidence structure, omitted from the previous detector’s proof, can matter decisively.

Panitch–Romo Conjecture 7.1 remains open in this work. The prior actual-cokernel detector and the earlier positive-geometric vacuum theorem retain their precise hypotheses and conclusions.

1. Exact retained actual setup

Use

$$k=\mathbb Q(i,A^{1/2}),\qquad K=\mathbb Q(i)((A^{1/2})),\qquad q=A^4.$$

In a fixed opposite-root sign sector of the native six-root target, the reduced root torus has lattice $L=\mathbb Z^{2N}$, coordinates $(a_j,b_j)$, and

$$W(u)W(v)=A^{\Omega_2(u,v)/2}W(u+v),\qquad \Omega_2(u,v)=\sum_j(u_{wj}v_{zj}-u_{zj}v_{wj}). \tag{2}$$

Thus $w_jz_j=Az_jw_j$, $W(a,b)=A^{ab/2}z^aw^b$, and the actual sector cokernel is

$$C=S\big/\left(\sum_j S f_j+\sum_r(c_rW(h_r)+A^2)S\right), \qquad f_j=z_j^{-2}+w_j^2-1. \tag{3}$$

This is a vector-space quotient by mixed-sided relation spaces, not a two-sided algebra quotient. All actual edge rows and the actual meridian row are retained. Their isotropic image lattice is $\Lambda$; it is not saturated. The exact assignments

$$\chi(h_r)=-A^2/c_r \tag{4}$$

must extend to a character on $\Lambda$, including phases. Let $\Lambda_0=\Lambda\cap2L$.

Retain the exact named longitude

$$l=c_lW(b),\qquad \Omega_2(h_e,b)=0,\qquad \Omega_2(h_s,b)=2. \tag{5}$$

The latter is $ml=A^2lm$. Neither $b$ nor $c_l$ is replaced to obtain a favorable answer. The intrinsic filling isomorphism sends the filled vacuum to $[1-l^2]$, not to $[1]$.

The established seed and average are

$$\ell_0(W(u))=0\ (u\notin2L),\qquad \ell_0(W(2a_1,2b_1,\ldots))=\prod_j I_\Delta(b_j,-a_j),$$ $$\mathcal F(W(u))=\sum_{\lambda\in\Lambda} \chi(\lambda)^{-1}A^{\Omega_2(\lambda,u)/2} \ell_0(W(u+\lambda)). \tag{6}$$

Here

$$I_\Delta(m,e)=\sum_{n\geq\max(0,-e)} \frac{(-1)^n q^{n(n+1)/2-(n+e/2)m}} {(q;q)_n(q;q)_{n+e}}. \tag{7}$$

For clarity, the convergence hypotheses being retained are rational nonnegative triples $\theta_j=(\gamma_j,\beta_j,\alpha_j)$ with sum one and

$$\vartheta(u)=\sum_j(\gamma_ju_{zj}+\alpha_ju_{wj}),\qquad \vartheta|_\Lambda=\nu_A\chi. \tag{8}$$

Put $T(a,b)=(-a,0,-b)$, let $\Delta_j$ be the all-one triple in block $j$, and set

$$H=T(\Lambda_0/2)+\sum_j\mathbb Z\Delta_j,\qquad \mathcal C=(H\otimes\mathbb R)\cap\mathbb R_{\geq0}^{3N},$$ $$\delta(x)=\sum_j(x_{j1}x_{j2}+x_{j2}x_{j3}+x_{j3}x_{j1}).$$

Assume $x\in\mathcal C$, $\delta(x)=\theta\cdot x=0$ implies $x=0$. Strict angles suffice. The previous global detector theorem, Sections 4–7, proves order-independent summability for every algebraic input, all-relations descent, and $\mathcal F(1)=1+$ positive order. This note uses that proved theorem; it does not replace its convergence argument by finite boxes.

2. Exact longitude-shifted degree and leading coefficient

Write the native monomial scalars as

$$c_l=\eta_lA^\rho,\qquad \chi(\lambda)=\eta_\lambda A^{\vartheta(\lambda)}. \tag{9}$$

The phases $\eta_l,\eta_\lambda$ are retained exactly. Since $2b$ is even, precisely $\lambda\in\Lambda_0$ contribute to the longitude square. For such $\lambda$, define

$$v=b+\lambda/2,\quad m_j=\min T(v)_j,\quad x_\lambda=T(v)-\sum_jm_j\Delta_j,\quad M_\lambda=\sum_jm_j. \tag{10}$$

Each reduced triple has minimum zero. Use

$$J_\Delta(a,b,c)=(-q^{1/2})^{-b}I_\Delta(b-c,a-b).$$

Its leading $q^{1/2}$-degree is $\delta(x)-m$, and its leading coefficient is $(-1)^m$, for $m=\min(a,b,c)$ and $x=(a-m,b-m,c-m)$. See GHHR, arXiv:1604.02688v1, Section 2, equations (5)–(7). In our variables $I_\Delta(v_w,-v_z)=J_\Delta(-v_z,0,-v_w)$. Individual normalized series have increments in $A^4$.

Proposition 2.1 — the exact shifted summand

The $\lambda$-summand of $c_l^2\mathcal F(W(2b))$ has degree

$$\boxed{D_b(\lambda)=2\rho-\vartheta(\lambda)+\Omega_2(\lambda,b) +2\delta(x_\lambda)-2M_\lambda} \tag{11}$$

and leading coefficient

$$\boxed{s_b(\lambda)=\eta_l^2\eta_\lambda^{-1}(-1)^{M_\lambda}.} \tag{12}$$

Equivalently, with $\kappa=\rho+\vartheta(b)$ and

$$\Omega_3=\bigoplus_j \begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix},$$

the exact degree is

$$D_b(\lambda)=2\delta(x_\lambda)+2\theta\cdot x_\lambda +2\kappa+2\Omega_3(x_\lambda,T(b)). \tag{13}$$

Proof. The scalar and Weyl factors of (6) give the first three terms in (11). The product of the leading terms of $J_\Delta$ gives the remaining terms and (12). For (13), substitute $\lambda=2v-2b$, use $\theta\cdot T(v)=-\vartheta(v)$, $\theta\cdot\Delta_j=1$, $\Omega_3(Tv,Tb)=\Omega_2(v,b)$, and the fact that every $\Delta_j$ is in the radical of $\Omega_3$. These are exact identities, including the longitude’s quantum shift. ∎

The extra skew term in (13) cannot be discarded. Also, (11) is a summand valuation: the valuation of the whole sum need not be the minimum of these numbers.

Proposition 2.2 — finite leading test and its exact stopping boundary

Let

$$d_* =\min_{\lambda\in\Lambda_0}D_b(\lambda),\qquad S_* =\sum_{D_b(\lambda)=d_*}s_b(\lambda). \tag{14}$$

These are well-defined and the indicated sum is finite, by all-input summability. If $S_*\ne0$, then the longitude value has degree $d_*$ and leading coefficient $S_*$. Consequently $\mathcal F(1-l^2)\ne0$ if any of the following holds:

• $d_*>0$, regardless of cancellation within the longitude series;
• $d_*<0$ and $S_*\ne0$;
• $d_*=0$ and $S_*\ne1$.

The cases $S_*=0$ at negative degree and $(d_*,S_*)=(0,1)$ require more information. The latter is not a merely technical ambiguity: Section 5 gives all-orders cancellation with exactly this leading data.

For an exact, non-asymptotic formulation, write each summand as

$$s_b(\lambda)A^{D_b(\lambda)}P_\lambda(A),\qquad P_\lambda\in1+A^4\mathbb Z[[A^4]], \tag{15}$$

where the notation means a unit series with exponents divisible by four. At each degree $t$, the longitude coefficient is the finite sum of $s_b(\lambda)$ times the coefficient of $A^{t-D_b(\lambda)}$ in that normalized series. Subtract it from the corresponding unit coefficient. This supplies an exact coefficient criterion; it is not a theorem that some coefficient must survive.

3. Difference equations, a second functional family, and failed shortcuts

The tetrahedron index satisfies both

$$q^{e/2}I(m+1,e)+q^{-m/2}I(m,e+1)=I(m,e),$$ $$q^{e/2}I(m-1,e)+q^{-m/2}I(m,e-1)=I(m,e). \tag{16}$$

The first is the right-tetrahedron annihilator used in (6). The second gives the additional seed identity

$$\ell_0((z_j^2+w_j^{-2}-1)s)=0. \tag{17}$$

These sourced recurrences, triality, and the two-dimensional space of solutions are in Garoufalidis, arXiv:1208.1663v2, Definition 3.1 and Theorem 3.2. The nonnegative-degree solution space is one-dimensional over its stated coefficient ring; choosing another uniformly nonnegative tetrahedron index is not automatically an independent route.

Equation (17) does not automatically hold with $\mathcal F$ in place of $\ell_0$. Moving $W(\lambda)$ across the two nonconstant terms multiplies them respectively by $A^{2\lambda_{wj}}$ and $A^{2\lambda_{zj}}$. Those factors depend on the summation variable. Treating (17) as a second unshifted global annihilator loses the moment character.

There is, however, an exact family that retains that character correctly.

Theorem 3.1 — left-seed translates with shifted convergence character

For $v\in L$, define

$$\chi_v(\lambda)=\chi(\lambda)A^{\Omega_2(\lambda,v)}. \tag{18}$$

Suppose $\chi_v$ admits a nonnegative-angle datum and the no-null-ray condition of Section 1, with $\nu_A\chi_v$ in place of $\nu_A\chi$. Then

$$\mathcal G_v(s)=\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1} \ell_0(W(v)W(\lambda)s) \tag{19}$$

converges on every algebraic input and annihilates every relation of the original presentation (3). In terms of the global construction made using $\chi_v$,

$$\mathcal G_v(s)=\mathcal F_{\chi_v}(W(v)s), \tag{20}$$

so its exact filled-vacuum test is

$$\mathcal G_v(1-l^2)=\mathcal F_{\chi_v}(W(v)) -c_l^2A^{\Omega_2(v,b)}\mathcal F_{\chi_v}(W(v+2b)). \tag{21}$$

Proof. The commutation identity $W(v)W(\lambda)=A^{\Omega_2(v,\lambda)}W(\lambda)W(v)$ proves (20) with exactly (18). The previous all-input theorem therefore proves summability. The functional $s\mapsto\ell_0(W(v)s)$ still kills every $Sf_j$, because all left multiples remain left multiples. Reindexing $\lambda\mapsto\lambda+h$, using isotropy, gives $\mathcal G_v(W(h)s)=\chi(h)\mathcal G_v(s)$. Thus all original moment relations are killed, not the shifted ones. Equation (21) follows from the Weyl product. ∎

For $v=b$, the meridian character valuation in the convergence problem changes by $+2$; for $v=-b$, it changes by $-2$. Accordingly the native affine meridian-angle right-hand side becomes four or zero, respectively. Original filling-angle feasibility does not imply either shifted feasibility. This is a genuine all-relations construction with an additional, explicitly stated convergence requirement, not an unconditional rescue theorem.

If $v\in\Lambda$, then $\chi_v=\chi$ and $\mathcal G_v=\chi(v)\mathcal F$, so these translates are not new detectors. Triality likewise changes tetrahedron coordinates and scalar data; it does not permit changing the named longitude or declaring the two values in (21) unequal.

Finally, an exact meridian shear of the longitude does not create a free normalization parameter. If $l_n=A^{-n}m^nl$, then $l_n^2=A^{-4n}m^{2n}l^2$. Every original meridian eigenfunctional satisfies

$$\mathcal F(l_n^2s)=\mathcal F(l^2s). \tag{22}$$

The factor $(-A^2)^{2n}$ cancels $A^{-4n}$. This checks the shear at the functional level; no target-wide naturality theorem is assumed or claimed.

4. A strict-angle algebraic witness with native-form phases

This section constructs one abstract algebraic witness to a logical implication. It does not search or enumerate manifolds, slopes, triangulations, or skein configurations.

Take three independent tetrahedron root blocks. Put

$$B=iA,\quad q=B^4=A^4,\quad x_j=z_j^2,\quad y_j=w_j^2, \qquad y_jx_j=q x_jy_j.$$

Use the three tetrahedron relations $f_j=x_j^{-1}+y_j-1$. Designate two moment rows as edges and one as meridian:

$$\begin{array}{c|c|c|c} \text{row}&W(h)&c&\chi(h)\\\hline e_1&y_1y_2^{-1}&-B^2=A^2&-1\\ e_2&y_1^{-2}y_3^{-2}&B^7&B^{-5}\\ s&y_3&B&B \end{array} \tag{23}$$

Each relation is precisely $(cW(h)+A^2)S$. Equivalently, up to its displayed nonzero scalar, the moment relations are

$$E_1=y_1y_2^{-1}+1,\qquad E_2=y_1^{-2}y_3^{-2}-B^{-5},\qquad E_3=y_3-B. \tag{24}$$

All their root rows have only $w$-coordinates, hence are isotropic. Their exact image lattice, in squared-$w$ coordinates, is

$$\Lambda=\{(0,2k_1,0,2k_2,0,2k_3):k_1+k_2\equiv0\pmod2\},$$ $$\chi(k)=(-1)^{k_2}B^{\frac32(k_1+k_2)+k_3}. \tag{25}$$

Indeed the three squared-coordinate generators are $(1,-1,0)$, $(-2,0,-2)$, $(0,0,1)$, of determinant $-2$. Any $k$ in (25) has coefficients $-k_2$, $-(k_1+k_2)/2$, $k_3-k_1-k_2$ in that basis. This proves the lattice and the exact character without adjoining an individual square root of $B^3$. It also makes the unsaturation explicit.

Choose strict angle triples

$$\theta_1=\theta_2=(1/8,1/8,3/4),\qquad \theta_3=(1/4,1/4,1/2). \tag{26}$$

Their restrictions to the rows in (23) are $0,-5,1$, exactly the character valuations. Every coordinate is positive, so the no-null-ray condition holds. The prior proof therefore supplies an all-input functional with $\mathcal F(1)=1+$ positive order for this presentation as well: that proof uses exactly the algebraic conditions just checked.

Define the exact named operators

$$m=By_3,\qquad l=B^{-1}W(-1,0,-1,0,1,1). \tag{27}$$

Writing the longitude row as $b$, one checks

$$\Omega_2(h_{e_1},b)=\Omega_2(h_{e_2},b)=0, \qquad\Omega_2(h_s,b)=2. \tag{28}$$

Thus the longitude commutes with both edge operators and $ml=A^2lm$, with the original meridian character $-A^2$.

The scalars are not tuned rational functions of an index value. They are literal six-root central-elimination scalars. In the sector $(\sigma_1,\sigma_2,\sigma_3)=(-1,1,1)$, let each local raw row be ordered $(z,z',z'',y,y',y'')$. The following lifts reduce to (23) and (27):

Here a displayed $0$ means the zero six-tuple. The local elimination rule is $W_6(u)=\sigma^{u_y+u_{y'}+u_{y''}}B^s W_2(r-s,t-s)$, with $r=u_z+u_y$, $s=u_{z'}+u_{y'}$, $t=u_{z''}+u_{y''}$. The verifier checks every displayed lift. The signed second row is deliberately not represented as an actual edge-incidence row; its failure is intrinsic, as Section 6 proves.

5. An all-orders, correct-sided vacuum cancellation certificate

Put $a=y_1$, $b_0=y_2$, $X=x_3$, $Y=y_3$. The longitude in (27) has

$$l^2=-x_1^{-1}x_2^{-1}XY. \tag{29}$$

The minus sign is forced: $c_l^2=B^{-2}=-A^{-2}$ while $W(2b)=A^2x_1^{-1}x_2^{-1}XY$.

Theorem 5.1 — finite mixed-sided certificate

In the quantum torus itself,

$$\begin{aligned} 1-l^2={}&XYx_2^{-1}f_1+XY(1-a)f_2+Xf_3\\ &+E_1\,b_0(a-1)XY+E_2\,B^5a^2XY\\ &+E_3\bigl(B^3Y^{-2}(Y+B)XY-B^{-1}X\bigr). \end{aligned} \tag{30}$$

Consequently $[1-l^2]=0$ in (3), while $[1]\ne0$.

Proof. The tetrahedron elimination identity is

$$x_1^{-1}x_2^{-1}-(1-a)(1-b_0) =x_2^{-1}f_1+(1-a)f_2.$$

Next,

$$(1-a)(1-b_0)-(1-a^2)=E_1b_0(a-1),$$ $$a^2-B^3=E_2(-B^5a^2)+E_3(-B^3Y^{-2}(Y+B)).$$

The remaining expression is settled by the exact local difference identity

$$1+(1-B^3)XY=Xf_3-E_3B^{-1}X. \tag{31}$$

To verify (31), use $XY=B^{-4}YX$; its right-hand side is $1+XY-B^{-1}YX$. Combining these identities gives (30). Every $f_j$ has its multiplier on the left, and every $E_r$ its multiplier on the right. Since (24) differs from (23) only by invertible scalars, (30) is a certificate in exactly the required relation space. The nonzero unit follows from strict-angle index detection, already verified in Section 4. ∎

Thus not only $\mathcal F$, but every linear functional annihilating all relations, including any convergent $\mathcal G_v$, satisfies $F(1-l^2)=0$ in this witness. This is stronger than finding a zero value for one chosen series. The assertion is for the specified sector of (3). It does not claim that the vacuum vanishes in every sign sector of the associated full six-root Laurent lift, or rule out another compatible sector as a detector for an actual input.

The leading terms cancel in the most favorable possible way

The exact summand degrees in (25) are separable. For the unit they are

$$d_0(k)=U(k_1)+U(k_2)+V(k_3),$$ $$U(k)=\begin{cases}2k^2+k/2&k\geq0,\\-3k/2&k<0,\end{cases} \qquad V(k)=\begin{cases}2k^2+k&k\geq0,\\-k&k<0.\end{cases} \tag{32}$$

For the exact longitude square they are

$$D_b(k)=P(k_1)+P(k_2)+R(k_3),$$ $$P(k)=\begin{cases}2k^2+k/2&k\geq0,\\-11k/2&k<0,\end{cases} \qquad R(k)=\begin{cases}2k^2+5k&k\geq0,\\-k&k<0.\end{cases} \tag{33}$$

To derive these formulas, apply (11) to the three triples $(1,0,-k_1)$, $(1,0,-k_2)$, $(-1,0,-k_3-1)$. The raw scalar-plus-Weyl contribution is $-2-7(k_1+k_2)/2+k_3$. Reducing the first two triples splits at $k=0$; reducing the third splits at $k_3=0$ versus $k_3\leq-1$. This gives (33). The unit uses triples $(0,0,-k_j)$ and character contribution $-3(k_1+k_2)/2-k_3$, giving (32).

All four functions are nonnegative and vanish only at zero. Hence each global value has exactly one minimum-degree lattice summand, at $k=0$. For the longitude that summand is

$$c_l^2 I_\Delta(0,1)^2 I_\Delta(1,-1) =I_\Delta(0,1)^3, \tag{34}$$

where triality gives $I_\Delta(1,-1)=-A^2I_\Delta(0,1)$. Its leading coefficient is one, exactly as for the unit summand $I_\Delta(0,0)^3$. Thus $(d_*,S_*)=(0,1)$, without even a tie among longitude minimizers. All subsequent difference coefficients cancel by (30). A unique leading longitude summand and strict filling angles therefore do not suffice to separate the two series in an abstract presentation.

The companion script independently evaluates both values exactly through $A^{16}$, using the coercive piecewise formulas (32)–(33) to bound every contributing lattice point and a quadratic-tail bound for (7). The agreement is a normalization check. Equation (30), not those coefficients, proves all-orders equality.

6. Why this is not an actual triangulation counterexample

Actual native edge rows have nonnegative local six-root incidence counts. For any such local lift with reduced coordinates $(a,b)=(r-s,t-s)$, nonnegativity of $r,s,t$ forces

$$s\geq\max(0,-a,-b). \tag{35}$$

Our second designated edge has reduced $w$-coordinates $-4$ in blocks one and three, and zero $z$-coordinates. Any nonnegative six-root lift would therefore require total prime count $\sum s_j\geq4+4=8$. Its exact scalar is $B^7$, requiring $\sum s_j=7$. Opposite-root signs can change a phase but not this exponent. This is impossible.

The displayed signed lift is consequently not a poor choice that can be replaced by a nonnegative lift with the same exact operator. There is no such lift. The witness deliberately does not satisfy this necessary actual-incidence condition, and no actual manifold, slope, or triangulation is claimed.

More generally, an actual edge with no additional source scalar must satisfy the necessary inequality

$$\nu_A\chi(h)+\sum_j\max(0,-h_{zj},-h_{wj})\leq2. \tag{36}$$

It follows by adding (35) to $\nu_A\chi(h)=2-\sum s_j$. In the witness the left-hand side for $e_2$ is $-5+8=3>2$. The basic global detector proof does not use (36). Full actual incidence also imposes stronger global constraints; (36) is necessary, not a sufficient realization test.

Even changing the generators within this specified edge lattice cannot repair its incidence failure. A general edge-lattice row has squared-$w$ coordinates

$$k=(a-2d,-a,-2d),\qquad \nu_A\chi(k)=-5d,\qquad a,d\in\mathbb Z. \tag{37}$$

For $d>0$, the negative parts of the first two coordinates sum to at least $2d$, and the third contributes another $2d$. The left side of (36) is therefore at least $-5d+8d=3d>2$. For $d<0$, the valuation term alone is $5|d|>2$, and the negative-part contribution is nonnegative. Thus an edge-lattice row obeying (36) must have $d=0$; in fact then (36) also forces $|a|\leq1$. Such rows span at most the rank-one subgroup generated by $e_1$, not the specified rank-two edge lattice. There is no alternative nonnegative-edge generating set for this same edge-lattice/character datum. This is not a classification of all presentations whose abstract quotients might be isomorphic.

This sharpens the logical boundary:

• The algebraic global-detector hypotheses plus exact native-form phases and the peripheral commutation laws do not force $[1-l^2]\ne0$.
• The same assertion for the actual named GY/PP presentation has not been disproved or proved here.
• Any positive actual theorem must exploit further actual structure, or an additional verified hypothesis that excludes certificates such as (30). It cannot be a consequence of only the previously extracted abstract data.

7. Useful failed approaches and literature limits

The following attempts were resolved rather than left as implicit assumptions.

Drop the longitude skew term. This changes (11)–(13) and loses the actual $ml=A^2lm$ shift. The witness’s equality depends on that shift and the exact minus sign in (29).

Use strict angles and unique minimizers to detect the vacuum. Equations (26), (32)–(34), and the all-orders certificate refute that abstract implication. Strict angles still detect the cokernel unit.

Invoke the second tetrahedron recurrence after averaging without changes. Conjugation introduces lattice-dependent factors. The corrected construction is (18)–(21), with additional convergence conditions.

Change the longitude by a meridian shear. The exact quantum normalization makes its squared value unchanged under every original meridian eigenfunctional, as (22) shows.

Find a second all-relations functional to distinguish an algebraically zero witness. Impossible by (30). For actual inputs on which only the first detector vanishes, a second functional remains a legitimate possibility; the witness does not rule that out.

Identify the difference with a known surgery index and import nonvanishing. No such comparison is proved here. Celoria–Hodgson–Rubinstein, arXiv:2509.09886v2, Theorem 6.2 proves a surgery formula with specified standard-cusp and efficiency hypotheses. Its Section 15 still distinguishes unresolved closed-manifold questions. These results supply important prior art, not a universal nonzero-value theorem for the exact named functional (6). No computational census from that paper is used.

Promote the witness to a conjecture counterexample. Section 6 proves a necessary actual-incidence condition fails. Such promotion would be false. Nor does zero vacuum in this witness imply that its filled target is zero: its meridian cokernel unit is nonzero, so the retained abstract filling isomorphism gives a nonzero filled target with zero specified vacuum image.

8. Scope, evidence, and next mode

The new results are the exact actual longitude coefficient formulas and finite leading test; the shifted-character all-relations family; shear invariance at the functional level; and the strict-angle algebraic cancellation witness with a finite correct-sided certificate and a proved actual-incidence exclusion. These are supporting results about the active method’s reach and failure boundary, not a recognized open-conjecture resolution or a claimed major new invariant.

The archive includes a standard-library noncommutative verifier, exact results, proof-boundary audit, pinned-source ledger, checksums, and a proposed handoff. The verifier’s all-orders certificate is independent of the finite series calculation. No topological enumeration was performed. No new author contact, author endorsement, or external review is claimed.

Next mode: ATTACK — actual-incidence anti-cancellation. Retain the named $l=c_lW(b)$. The most focused remaining task is to use genuine nonnegative edge incidence and its global dependencies, together with the sourced normal-surface/index structure, to prove that the actual difference’s signed coefficient sequence cannot cancel identically, or to identify an actual all-relations obstruction. Equations (11)–(15) are the exact coefficient gate; (36) identifies one actual input absent from the failed abstract implication. A proved comparison with an efficient-filling index could help only if it preserves this exact insertion, phases, image lattice, and all-input descent and supplies a genuinely applicable nonvanishing theorem.

The character-twisted family (19) is available where its extra feasibility condition holds. It does not bypass algebraic vacuum membership. At null rays, the previous unordered average still diverges; no regularization or residue detector is established in this run. Do not infer actual target vanishing from a divergent or zero pairing, or actual filled-vacuum survival from $\mathcal F(1)\ne0$.