Actual incidence, normal Euler degree, and a two-angle vacuum theorem

FPRD Lab, 8 September 2026. ATTACK research note. AI-assisted work, internally audited, without external specialist review or author endorsement. Novelty and first priority are unassessed.

Outcome

The named-longitude cancellation problem has a positive answer in a new, explicitly stated two-angle chamber. For an actual ideal triangulation, suppose there are:

1. a strictly positive angle structure with zero peripheral angular holonomy; and
2. a strictly positive angle structure with angular holonomy $2\pi$ on the chosen filling meridian.

Then the actual generic six-root filled vacuum, and hence its named descended quantum trace, are nonzero. No solution of the nonlinear shape equations or regular classical intersection is required in this criterion.

The structural step is an exact translation of every longitude-shifted summand into an integral $Q$-normal class $X_\lambda$:

$$\boxed{D_l(\lambda)=2\bigl(\delta(X_\lambda)-\chi_{\rm top}(X_\lambda)\bigr).} \tag{1}$$

The native central scalar, the filling eigenvalue, and the meridian–longitude commutator are all essential to $1$. The class has a fixed nonzero longitude component at the boundary. The first angle structure makes every degree in $1$ strictly positive; the second gives convergence and descent on every algebraic input. Therefore

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

A direct leading–trailing deformation supplies the second angle structure for all primitive slopes outside an explicit bounded region, once the first structure is given. This is a uniform proof, not a search over slopes or triangulations.

The unrestricted Panitch–Romo conjecture remains open. This is a conditional nonvanishing research note, not a new Publication, a new surgery invariant, or an identification with an existing closed-manifold index.

1. Retained target and what is being detected

Use $k=\mathbb Q(i,A^{1/2})$, $K=\mathbb Q(i)((A^{1/2}))$, and $B=iA$. In a central sign sector, the root torus has lattice $L=\mathbb Z^{2N}$ and

$$W(u)W(v)=A^{\Omega_2(u,v)/2}W(u+v),\qquad \Omega_2(u,v)=\sum_j(u_{wj}v_{zj}-u_{zj}v_{wj}). \tag{3}$$

Its meridian cokernel is the vector-space quotient

$$C_s=S\Big/\left(\sum_j S(z_j^{-2}+w_j^2-1) +\sum_e(c_eW(h_e)+A^2)S+(c_sW(h_s)+A^2)S\right). \tag{4}$$

The relation spaces have the displayed, different sides. No multiplication of quotient classes is used below. Retain the exact native operators

$$m=c_sW(h_s),\qquad l=c_lW(b),\qquad ml=A^2lm. \tag{5}$$

The earlier generic filling reconstruction, Section 8, sends the filled vacuum to $[1-l^2]$ in (4). Its cokernel unit $[1]$ is a different vector. The native six-root collar comparison supplies the actual named GY, CK, and original-PP action and is retained with its exact conventions and CK source-reading caveat.

Let $\Lambda=\sum_e\mathbb Zh_e+\mathbb Zh_s$, the actual isotropic image lattice, without saturation. The exact character is

$$\chi(h_e)=-A^2/c_e,\qquad\chi(h_s)=-A^2/c_s. \tag{6}$$

The established functional is

$$\mathcal F(W(u))=\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1} A^{\Omega_2(\lambda,u)/2}\ell_0(W(u+\lambda)), \tag{7}$$

where $\ell_0$ vanishes off $2L$ and

$$\ell_0(W(2a_1,2d_1,\ldots))=\prod_j I_\Delta(d_j,-a_j),\qquad q=A^4. \tag{8}$$

The global all-input theorem, Sections 4–7, proves that consistent character data and strict filling angles make (7) unordered-summable on every algebraic input, annihilate every relation in (4), and give $\mathcal F(1)=1+$ positive order. The seed, recurrence, and averaging method originate in the DGG and subsequent mathematical index literature; they are not new constructions here.

2. The actual incidence data and the angle polytope

Collapse each actual six-root raw row to its three opposite-pair totals. For a row $R\in\mathbb Z^{3N}$, write its block as $(r_j,s_j,t_j)$, and set

$$P(R)_j=(r_j-s_j,t_j-s_j),\qquad p(R)=\sum_js_j. \tag{9}$$

Let $E_e$ be the actual nonnegative edge-incidence rows, and let $R_\gamma$ be the additive native peripheral rows fixed by the retained collar dictionary. Then

$$h_e=P(E_e),\quad h_s=P(R_s),\quad b=P(R_l),$$ $$c_e=\epsilon_e B^{p(E_e)},\quad c_s=\epsilon_s B^{p(R_s)},\quad c_l=\epsilon_l B^{p(R_l)},\qquad\epsilon_\bullet\in\{1,-1\}. \tag{10}$$

The signs are exactly the central-sector signs. In the all-plus sector they are all $1$. No extra scalar is inserted into an edge or peripheral operator.

Set

$$J=\bigoplus_j\begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix}, \qquad\omega(U,V)=U^TJV. \tag{11}$$

Direct calculation gives $\Omega_2(PU,PV)=\omega(U,V)$. Every tetrahedron vector $\Delta_j=(1,1,1)_j$ is radical. The native incidence/peripheral structure gives

$$\omega(E_e,E_f)=\omega(E_e,R_\gamma)=0,\qquad \omega(R_s,R_l)=2. \tag{12}$$

The edge rows are not independent. For example, their sum is $2\sum_j\Delta_j$, because each local edge occurs once in the full edge list. Cusp dependencies give further relations. All of them remain in (4).

An angle vector $\theta\in\mathbb Q^{3N}$, normalized in units of $\pi$, satisfies

$$\theta\cdot\Delta_j=1,\quad\theta\cdot E_e=2,\quad\theta>0. \tag{13}$$

These conditions define the open angle polytope $\mathscr A$. Its peripheral holonomy map is

$$H_\theta(\gamma)=\theta\cdot R_\gamma. \tag{14}$$

A complete angular certificate is $\theta^0\in\mathscr A$ with $H_{\theta^0}=0$ on every boundary component. This is only an angular completeness condition, not a complete hyperbolic structure. A filling angular certificate is $\theta^s\in\mathscr A$ with $H_{\theta^s}(s)=2$. Other peripheral holonomies need not vanish for the theorem.

The translation into normal surfaces uses the established incidence dictionary: $E_e,R_\gamma,\Delta_j$ lie in $Q(\mathcal T;\mathbb R)$, with $\partial E_e=\partial\Delta_j=0$, $\partial R_\gamma=2\gamma$, and formal Euler values $-2,0,-1$, respectively. See GHHR, Sections 6–8 and Appendix A. This is substantive prior structure, not a consequence of a rowwise nonnegativity test. The proof of the degree identity below is supplied explicitly.

3. Global dependencies fix the all-plus character

Lemma 3.1

If a complete angular certificate exists, (6) is automatically consistent in the all-plus sector, on the full unsaturated image lattice.

Proof. Consider any integer dependency

$$\sum_en_eh_e+n_sh_s=0.$$

Pair it with $b$. Equations (12) give $2n_s=0$, so $n_s=0$. Put $R=\sum_en_eE_e$. Since $P(R)=0$, there are integers $d_j$ with $R=\sum_jd_j\Delta_j$. Taking the complete angular pairing gives

$$p(R)=\sum_jd_j=2\sum_en_e.$$

Since $-A^2=B^2$, the exact product of prescribed character values along the dependency is

$$B^{2\sum_en_e-p(R)}=1.$$

This includes its phase, not just its valuation. Hence the assignments descend from the free group on all rows to its image $\Lambda$. ∎

For another sign sector the remaining test is $\prod_e\epsilon_e^{n_e}=1$ for every edge dependency; it is not silently discarded. In particular, the sum-of-all-edges dependency imposes $\prod_j\sigma_j=1$, since each tetrahedron contributes three opposite bare roots. This is a necessary condition, not a complete classification of compatible sectors. The all-plus sector suffices for nonvanishing of the original full target.

4. An intrinsic normal class for each longitude summand

Only $\lambda\in\Lambda_0=\Lambda\cap2L$ contribute to $c_l^2\mathcal F(W(2b))$. Choose any integer expression

$$\lambda=\sum_en_eh_e+n_sh_s, \quad R=\sum_en_eE_e+n_sR_s, \quad U=R_l+\tfrac12R. \tag{15}$$

For each block put $t_j=\max(U_{j1},U_{j2},U_{j3})$, and define

$$\boxed{X_\lambda=\sum_jt_j\Delta_j-U.} \tag{16}$$

Lemma 4.1 — well-defined integral structural translation

The vector (16) is independent of the chosen expression (15). It is an integral nonnegative $Q$-normal class, with minimum coordinate zero in every tetrahedron. The map $\lambda\mapsto X_\lambda$ is injective, and

$$\partial X_\lambda=-2l-n_ss,\qquad \chi_{\rm top}(X_\lambda)=\sum_en_e-\sum_jt_j. \tag{17}$$

Here $s,l$ in the boundary formula denote the underlying oriented homology classes, not their quantum operators. In particular $X_\lambda\ne0$.

Proof. Although $U$ may have half-integral entries, its coordinate differences are integral: $P(U)=b+\lambda/2\in L$. Thus subtracting each block from its maximum gives integral entries. The actual matching equations hold because $U$ and every $\Delta_j$ are in the real matching space.

Changing the expression for $\lambda$ changes $R$ by a tetrahedron vector. Indeed its $P$-image is zero; pairing with $b$ first shows that its meridian coefficient is unchanged. Adding a constant to all entries in a block of $U$ changes its maximum by the same constant, leaving (16) unchanged.

Moreover

$$P(X_\lambda)=-b-\lambda/2, \tag{18}$$

which proves injectivity. The boundary and Euler formulas follow by applying their linear maps to (16). The coefficient of $l$ in the boundary is $-2$, so the class is nonzero. Independently, $X_\lambda=0$ would give $b+\lambda/2=0$; pairing with $h_s$ would give $2=0$, by isotropy. ∎

This map uses the actual image lattice. It does not assert that every integral $Q$-normal class of the displayed boundary type is in its image; homological and parity restrictions are preserved.

Theorem 4.2 — exact native Euler-degree cancellation

For every compatible sign sector and every $\lambda\in\Lambda_0$, the exact $A$-degree of its longitude summand is given by equation (1).

Proof. Write $E=\sum_en_e$, $S_R=p(R)$, $T=\sum_jt_j$, and $\rho=p(R_l)$. The native character valuation is

$$\nu_A\chi(\lambda)=2E+2n_s-S_R. \tag{19}$$

Set $v=b+\lambda/2=P(U)$. The seed uses the triple $(-v_z,0,-v_w)$. In a block this is $-U+U_{j2}\Delta_j$; its minimum is $U_{j2}-t_j$. Consequently the sum $M$ of these minima is

$$M=\rho+\tfrac12S_R-T. \tag{20}$$

The prior exact longitude summand formula, Proposition 2.1, gives

$$D_l(\lambda)=2\rho-\nu_A\chi(\lambda) +\Omega_2(\lambda,b)+2\delta(X_\lambda)-2M. \tag{21}$$

Here $\Omega_2(\lambda,b)=2n_s$, exactly. Substitution yields

$$\begin{aligned} D_l(\lambda) &=2\rho-(2E+2n_s-S_R)+2n_s+2\delta(X_\lambda) -2(\rho+S_R/2-T)\\ &=2\bigl(\delta(X_\lambda)-E+T\bigr) =2\bigl(\delta(X_\lambda)-\chi_{\rm top}(X_\lambda)\bigr). \end{aligned}$$

No phase was adjusted. The leading coefficient remains exactly

$$\eta_l^2\eta_\lambda^{-1}(-1)^M, \quad c_l=\eta_lA^\rho,\quad \chi(\lambda)=\eta_\lambda A^{\nu_A\chi(\lambda)}. \tag{22}$$

There is no need to assume that these coefficients are positive. ∎

For comparison, the same calculation for the unit gives

$$D_1(\lambda)=2\bigl(\delta(X^0_\lambda) -\chi_{\rm top}(X^0_\lambda)-n_s\bigr), \tag{23}$$

where $X^0_\lambda$ is formed from $R/2$, without $R_l$. The term $-2n_s$ does not disappear for the unit. This is why the complete certificate alone is not the convergence certificate for the original filling average.

5. The two-angle nonvanishing theorem

Theorem 5.1

Let an actual oriented ideal triangulation with torus boundary carry a complete angular certificate $\theta^0$ and a filling angular certificate $\theta^s$ for one primitive filling slope $s$. Then the generic actual six-root filled vacuum is nonzero. Under the retained native comparison and filling isomorphism, the named descended GY, CK, and original-PP traces are nonzero.

Proof. Work first in the all-plus sector. Lemma 3.1 proves exact character consistency. The equations for $\theta^s$ give

$$\sum_j\bigl(\theta^s_{j1}(r_j-s_j) +\theta^s_{j3}(t_j-s_j)\bigr)=2-p(R_\bullet)$$

on each actual edge and meridian row. These are precisely the character-valuation equations for the earlier all-input theorem. Strict positivity also gives its no-null-ray condition. Thus (7) descends on the correct sides and $\mathcal F(1)=1+$ positive order.

For the longitude, the complete certificate gives directly from (16)

$$\theta^0\cdot X_\lambda =T-\theta^0\cdot R_l-\tfrac12\theta^0\cdot R =T-E=-\chi_{\rm top}(X_\lambda). \tag{24}$$

Therefore

$$D_l(\lambda)=2\delta(X_\lambda)+2\theta^0\cdot X_\lambda>0. \tag{25}$$

The strict inequality follows from $X_\lambda\in\mathbb Z_{\ge0}^{3N}\setminus\{0\}$. More explicitly, if $\varepsilon=\min\theta^0>0$, every summand has degree at least $2\varepsilon$. All-input summability makes the longitude sum well-defined; cancellation among its summands cannot create a term of lower degree. Hence its constant coefficient is zero, while the unit’s is one. This proves (2).

The functional pulls back from the central all-plus quotient to the actual full six-root cokernel and still detects $1-l^2$. The retained filling isomorphism then detects the filled vacuum. Its named quantum trace is consequently nonzero; the retained collar comparison transfers that statement with its stated conventions. ∎

The proof also works in any other compatible sector with the same angular certificates: only the coefficient phases change, and (25) is unaffected. It does not claim all central sectors are compatible.

In fact, the longitude sum alone is unordered-summable from $\theta^0$ without $\theta^s$: (25) gives $D_l(\lambda)\ge2\varepsilon\|X_\lambda\|_1$, and (18) is injective. A bounded degree therefore permits only finitely many nonnegative integral $X_\lambda$. This does not define the original functional on all inputs or normalize its unit. For the latter purposes, the strict filling certificate can be replaced by the prior theorem’s nonnegative filling certificate together with its no-null-ray condition.

The nonlinear shape equations, a hyperbolic filling solution, a regular sequence at a shape point, and a classical value $l^2\ne1$ are absent from these hypotheses. No actual example separating this chamber from the earlier positive-geometric chamber is asserted.

6. All sufficiently large primitive slopes, without enumeration

The relevant intrinsic object is the peripheral holonomy image of the angle polytope,

$$\mathscr H=\{H_\theta:\theta\in\mathscr A\}. \tag{26}$$

For a fixed slope $s$, the filling certificate asks whether the linear hyperplane $H(s)=2$ meets this open convex set. The complete certificate places $0$ in its interior in peripheral directions. This produces a uniform result.

Fix native integral basis rows $R_a,R_d$ on the chosen cusp with $\omega(R_a,R_d)=2$. For coprime integers $p,q$, put

$$R_s=pR_a+qR_d,\quad r^2=p^2+q^2,\quad u=JR_a,\quad v=JR_d,$$ $$\boxed{\theta^{p,q}=\theta^0+\frac{pv-qu}{r^2}.} \tag{27}$$

The perturbation preserves every tetrahedron sum and every edge angle equation, by (11)–(12). It has

$$\theta^{p,q}\cdot R_s =\frac{\omega(pR_a+qR_d,-qR_a+pR_d)}{r^2}=2. \tag{28}$$

It also preserves zero holonomy on other cusps. It is thus a filling certificate whenever all coordinates in (27) are positive. An exact rational test is

$$r^2\theta^0_j+p v_j-q u_j>0\qquad(1\le j\le3N). \tag{29}$$

Corollary 6.1 — uniform bounded exceptional region

Define the explicit rational number

$$C^2=\max_j\frac{u_j^2+v_j^2}{(\theta^0_j)^2}. \tag{30}$$

For every primitive slope with $p^2+q^2>C^2$, (27) is strictly positive, and the actual named generic filled vacuum is nonzero.

Proof. Cauchy–Schwarz gives

$$\frac{|p v_j-q u_j|}{r^2} \le\frac{\sqrt{u_j^2+v_j^2}}r<\theta^0_j.$$

Apply Theorem 5.1. ∎

This is a basis-dependent sufficient bound, not an optimal geometric slope length or a classification of exceptional fillings. Condition (29), or the exact angle-polytope feasibility problem, can be sharper. Neither requires generating a list of slopes.

The real vector $(-qR_a+pR_d)/r^2$ in (27) is used only to deform angles. It does not replace the actual integral complementary longitude in (5), change its scalar, or select a different physical operator.

Leading–trailing angle deformations and their twice-intersection holonomy law are established prior tools; see Futer–Guéritaud, Lemmas 4.4–4.5 and Section 6. Equations (27)–(30) give the direct normalized calculation needed here. No novelty claim is made for the deformation method or for large-surgery phenomena.

7. What the attack does not establish

The two certificates have different jobs and cannot be merged by notation. $\theta^0$ gives positive normal Euler degree for the longitude; $\theta^s$ gives all-input convergence and the unit normalization. Complete angles alone do not satisfy the native meridian equation. Filling angles alone are insufficient, as the preceding strict-angle algebraic cancellation witness demonstrates.

The actual matching/Euler dictionary is more than nonnegative rows or the inequality $\nu_A\chi(h)+\sum_j\max(0,-h_{zj},-h_{wj})\le2$. The preceding witness fails that necessary inequality; it is still not an actual triangulation counterexample. The present theorem is not obtained merely by excluding that witness. It gives a structural positive pairing on every actual input satisfying the two certificates.

The attempted broader inference from ordinary 1-efficiency to positivity of every shifted degree is not used. Efficiency controls certain normal surfaces and convergence; the displayed positive linear Euler certificate is the precise extra input in (25). Generalized normal surface replacement alone does not justify dropping Euler or double-arc corrections.

There is a further available pairing even when the second certificate is missing. In the prior character-twisted family, $v=-b$ changes the meridian valuation by $-2$ and leaves all edge valuations unchanged. Therefore $\theta^0$ is a strict convergence certificate for

$$\mathcal G_{-b}(s')=\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1} \ell_0(W(-b)W(\lambda)s'). \tag{31}$$

It descends to the original relations by the previous family theorem. This observation does not prove that (31) detects the vacuum. Its exact remaining test is

$$\mathcal G_{-b}(1-l^2) =\mathcal F_{\chi_{-b}}(W(-b)) -c_l^2\mathcal F_{\chi_{-b}}(W(b)). \tag{32}$$

No unproved boundary-inversion symmetry or asymmetry is invoked to evaluate (32). A zero value would not prove the class zero.

Nor is (7) identified here with the surgery invariant in Celoria–Hodgson–Rubinstein, Theorem 6.2, which has its own standard-cusp and efficiency assumptions. An identity of that kind must preserve the exact insertion, phases, image lattice, and domain.

8. Evidence, significance, and next question

The accompanying code verifies the universal polynomial cancellations, the collapse/skew dictionary, fractional-lift integrality, and the explicit angle-deformation bound using exact rational arithmetic. An optional certificate checker validates supplied raw-row and angle data; it explicitly does not certify that arbitrary matrices are realized by a topological triangulation. No manifold, slope, triangulation, normal-surface support, or skein-configuration search is performed.

The mathematical advance is the exact native normal-class translation and the resulting conditional filled-vacuum theorem. The earlier detector only established a nonzero cokernel unit. This note now separates that unit from the actual longitude square throughout the two-angle chamber. The sourced degree/Euler machinery, index seed and averaging, leading–trailing deformations, and native trace comparisons are substantial dependencies and influences. This synthesis has not undergone a comprehensive priority or external specialist audit.

Next mode: ATTACK. Preserve the exact normal-class map (16) and the two-angle theorem. The decisive remaining problem is vacuum detection outside the filling-angle slice or without a complete positive angular certificate. In the complete-angle case, determine the exact value or structural vanishing mechanism of (32), or construct an alternative all-relations pairing that detects $1-l^2$. Do not treat the bounded region excluded by the sufficient estimate as a new census problem. Panitch–Romo Conjecture 7.1 remains open in the unrestricted scope stated in their current paper, Section 7.