Signed-angle convergence and detection away from the unit

FPRD Lab | 8 September 2026 | Supporting research note

AI-assisted derivation with internal self-audit. No independent-model, external specialist, or formal verification of the general theorem is claimed. Novelty is unassessed.

Result and scope

The strict-angle trace-cokernel argument separates into two obligations: convergence of a native index functional and nonvanishing of at least one of its values. Its value on the unit need not be nonzero.

We prove an exact convergence criterion allowing signed generalized angle certificates. Nonnegative angles together with this criterion retain the previous nonzero-unit argument and hence the lower bound

$$\dim_k(Q^6/\operatorname{im}\tau)\ge 2^r-1.$$

For signed certificates the same lower bound follows whenever the functional has any nonzero value. Convergence alone does not prove that nonvanishing. An exact one-tetrahedron algebraic control has F(1)=0 but F(w^{-2})=1, demonstrating why checking only the unit can reject a useful detector. The control is not asserted to come from a manifold triangulation.

This is a supporting extension of the preceding index and grading proofs, not a new index construction, a classification of triangulations, a nonzero filled invariant, or a resolution of Panitch–Romo Conjecture 7.1.

1. Retained presentation

Use the actual all-plus projection, field k=Q(i,A^{1/2}), lattice L=Z^{2N}, and Weyl rule

$$W(u)W(v)=A^{\Omega(u,v)/2}W(u+v),\qquad \Omega(u,v)=\sum_t(u_{wt}v_{zt}-u_{zt}v_{wt}).$$

The mixed-sided relations are Sf_t, with f_t=z_t^{-2}+w_t2-1, and (c_eW(h_e)+A^2)S. Keep the actual isotropic image lattice Lambda=sum Z h_e, not its saturation, and the exact character chi(h_e)=-A^2/c_e.

A rational generalized angle certificate theta has triple sums one and satisfies the actual edge equations, but its entries may be negative. Write vartheta(u)=sum_t(theta_{t1}u_{zt}+theta_{t3}u_{wt}). Then vartheta(lambda)=nu_A chi(lambda).

For actual torus-boundary triangulations, existence of generalized angles is established by GHHR Proposition A.2; that result even permits zero peripheral holonomy. No positivity is supplied by that theorem. Exact native phase consistency follows by the same argument as Section 7 of the face-defect packet: an integral edge dependency has equal local triple counts k_t; applying the generalized angle equations gives sum k_t=2 sum n_e, and consequently product chi(h_e)^{n_e}=1. This calculation uses the equations, not positivity.

Set T(a,b)=(-a,0,-b) blockwise, let Delta_t denote the all-one triple in block t, and define

$$H=T((\Lambda\cap2L)/2)+\sum_t\mathbb Z\Delta_t, \quad C=(H\otimes\mathbb R)\cap\mathbb R_{\ge0}^{3N}, \quad \delta(x)=\sum_t(x_{t1}x_{t2}+x_{t2}x_{t3}+x_{t3}x_{t1}).$$

The auxiliary H does not replace Lambda in the average. The restriction of theta to H is fixed by the native character and the triple sums; choosing another feasible generalized angle vector does not change it.

The established seed is ell(W(2a,2b))=product I_Delta(b_t,-a_t), zero on other parity classes, with q=A^4 and the tetrahedron index of the preceding packet. Consider the same unordered family

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

2. Exact signed-angle criterion

Theorem. With these data, (1) converges on every finite Laurent input if and only if

$$x\in C\setminus\{0\},\quad\delta(x)=0 \quad\Longrightarrow\quad\theta\cdot x>0. \tag{2}$$

Necessity already holds for convergence of the unordered family at the unit. Failure proves failure of this family, not zero target or impossibility of another detector.

Proof of sufficiency. Put l(x)=theta dot x. On the compact slice K={x in C: ||x||_1=1}, delta is nonnegative. If K is empty, C={0} and the following estimates are immediate. Otherwise, on its closed subset Z={delta=0}, condition (2) makes l strictly positive. If Z is nonempty, choose c>0 and a neighborhood U of Z in K with l>=c. On the compact complement K delta has a positive minimum eta. Also |l|<=M on K. Thus, for t>=1, on U we have

$$\delta(tx)+l(tx)\ge ct,$$

and on Ke have

$$\delta(tx)+l(tx)\ge\eta t^2-Mt\ge ct-B$$

for a finite B (for example (M+c)^2/(4 eta)). If Z is empty, use the latter bound on all of K with any c>0. Consequently

$$\delta(Y)+\theta\cdot Y\ge c\|Y\|_1-B \quad(Y\in H\cap\mathbb N^{3N}). \tag{3}$$

This is the step that replaces entrywise positivity of theta. Small nonzero vectors may have negative degree; proper growth, not positivity at every point, is proved.

For a supported input u, reduce T((u+lambda)/2) by its blockwise minima to x>=0. Its exact summand valuation is

$$d_u(x)=2\delta(x)+2\theta\cdot x+\vartheta(u)+\Omega_3(x,T(u)), \quad \Omega_3=\bigoplus_t\begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix}. \tag{4}$$

This is the sourced seed-degree formula with the exact character and Weyl factor included. All reduced x lie in one affine lattice T(v_0)+H. Dickson’s lemma supplies finitely many coordinatewise minima b. For x=b+Y, Y belongs to H and is nonnegative. Isotropy gives, exactly as in the preceding all-input proof,

$$d_u(b+Y)-d_u(b) =2\delta(Y)+2\theta\cdot Y +2 b^T(D_{\rm sym}-\Omega_3)Y \ge 2c\|Y\|_1-2B, \tag{5}$$

where D_sym has off-diagonal entries one and diagonal zero in each block. The mixed term is nonnegative entrywise, independently of theta’s signs. The map lambda to reduced x is injective. Thus only finitely many summands have valuation below any bound, on every input, including odd-root cosets. Reindexing is now justified, proving annihilation of both relation families on their original sides.

Necessity. If (2) fails, some nonzero x>=0 in C has delta(x)=0 and theta dot x<=0. Its support uses at most one coordinate per triple. Restricting to that support leaves rational linear equalities and inequalities. Choose a nonzero rational point and scale it into H, obtaining Y=T(mu/2)+sum d_t Delta_t with mu in Lambda intersect 2L. Because Y is reduced and nonzero, mu is nonzero. The unit summands at lambda=n mu have exact valuations

$$2n^2\delta(Y)+2n\theta\cdot Y=2n\theta\cdot Y\le0.$$

Each summand is nonzero. Infinitely many fail to tend to zero in valuation, so the unordered family cannot converge. Prescribed regrouping or regularization would be a different construction. This proves the theorem.

The argument is an extension of the previous Lab affine-input estimate, influenced by GHHR’s normal-cone convergence argument. It does not assert a new independent proof of the external tetrahedron-index identities.

3. What remains of the cokernel theorem

For theta>=0, condition (2) is exactly the earlier no-null-ray condition. Every nonzero supported unit term has 2delta(x)+2theta dot x>0, and therefore F(1)=1+positive A-order. The face-defect proof then works for this weaker certificate, without strict angles. No actual triangulation outside the strict-angle class is exhibited here.

More generally, retain the face-defect map d:L/2L -> D_Gamma, the integral normalizer P=Lambda^perp intersect L, dim d(P mod 2)>=r, and the theorem that the entire trace image has degree zero. Suppose (2) holds and F is not the zero functional. Since F has support in 2L+Lambda, select u_0 with d(u_0)=0 and F(W(u_0))!=0. For every nonzero defect f in d(P mod 2), choose an integral normalizing lift w_f and put

$$\Psi_f(v)=F(W(-w_f)v),\qquad x_f=W(w_f)W(u_0).$$

The same side-preserving normalizer argument proves descent. The evaluations are

$$\Psi_f(x_g)=\begin{cases}F(W(u_0)),&f=g,\\0,&f\ne g.\end{cases}$$

Every Psi_f annihilates the whole trace image. Hence dim(Q^6/im tau)>=2^r-1 and every minimal compatible filling quotient is nonzero. A nonzero value at the unit is sufficient but unnecessary. This is a conditional extension, not an assertion that F is nonzero for every signed-angle triangulation.

4. Exact control: zero unit, nonzero functional

Take one algebraic block, Lambda=Z(2,0), chi(2,0)=A^4, theta=(2,-1,0). The relations are Sf and (z^2-A4)S. These are abstract moment data, not an actual edge-incidence claim.

Here C={x>=0:x_2=x_3}; its only nonzero delta-zero directions are on the first coordinate, where theta dot x=2x_1>0. Thus the signed criterion proves all-input convergence, despite the negative angle.

There is also an exact closed evaluation:

$$F(W(2a,2b))= \begin{cases} 0,&b\ge0,\\ q^{a(1+b/2)}/(q;q)_{-b-1},&b\le-1, \end{cases} \tag{6}$$

and F vanishes on other parity classes. In particular F(1)=0 whereas F(w^{-2})=1.

To derive (6), set e=-a-n in the average over lambda=(2n,0). Its weight is q^{-n(1+b/2)}. Expanding the seed and putting k=j+e gives two nonnegative summation indices. After including that weight, their exponents are j(j-1)/2-bj+k, up to the fixed factor q^{a(1+b/2)}. This is a summable double family for every fixed b. Euler’s product expansions therefore give

$$F(W(2a,2b))=q^{a(1+b/2)} \frac{(q^{-b};q)_\infty}{(q;q)_\infty}.$$

For b>=0 the numerator has a zero factor; for b<=-1 cancel the tails to obtain (6). The Euler expansions themselves follow by matching coefficients in the product recurrences for (z;q)_infinity and its reciprocal; no analytic continuation or divergent regrouping is used. Formula (6) also directly satisfies the seed-adjoint and moment recurrences, so the correct-sided descent can be checked without the product calculation.

The accompanying exact rational verifier checks those recurrences and independently computes finite coefficient windows of the two decisive double sums, with the quadratic/linear tail cutoff proved in the source. This is two computational descriptions within one internal audit, not two independent reviewers.

5. Failure boundary and next decisive question

The previous proposed gate “find a functional with nonzero unit” was stronger than the grading argument needs. The right gate is find one nonzero value of a correctly descending functional. A failed unit test must not be reported as a zero-functional theorem.

The signed convergence criterion is exact for this unordered average, but its nonvanishing on every actual triangulation satisfying (2) is unresolved. The next ATTACK should establish an actual nonzero monomial value from native incidence or determine an exact obstruction. GHHR’s 1-efficient convergence theorems are important comparison inputs, but identifying their normal lattices with every affine root-input lattice and proving a nonzero native value must not be assumed merely from convergence of a scalar index. No manifold, slope, or collision-stratum census is called for.

Sources and intellectual credit

• Prior face-defect packet, Sections 1-9: exact native signs, whole-image grading and integral rank bound. Formal dependencies.
• Prior global-index packet, Sections 3-7: seed recurrence, exact valuation and affine-input mixed-term identity. Formal dependencies and direct influence on this proof.
• GHHR, The 3D-index and normal surfaces, arXiv:1604.02688v1, Section 8 and Appendix A: normal-surface convergence framework and generalized-angle existence, freshly inspected 8 September 2026. Their convergence framework influenced the question; no priority is claimed for coercivity or the index seed.
• Garoufalidis, The 3D index of an ideal triangulation and angle structures: established tetrahedron-index framework, retained through the preceding source-audited packet.

See maturity-pilot.md for the separate provisional confirmation assessment. Neither band assignments nor computational check counts measure the significance of this note.