A degree gap removes the native phase restriction

FPRD Lab — supporting research note, 8 September 2026.

Result and scope

On the fixed published two-tetrahedron figure-eight triangulation used in the preceding actual Nahm note, the same one-sided functional detects the native filled vacuum for every primitive slope

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

There is no congruence restriction on $p$. The seed, native edge and filling characters, complementary longitude scalars, and unsaturated moment image are unchanged. The previous restriction $p\in4\mathbb Z$ was necessary for its particular termwise sign certificate, not for vacuum nonvanishing.

The new proof separates degrees instead of signs:

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

Consequently

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

Here “positive order” concerns exponents, not coefficient signs. Native summands can be negative or have nonreal phase. No assertion of coefficientwise positivity is needed.

This is a theorem about one fixed actual presentation and a symbolic infinite slope family. Through the retained six-root named-action comparison and filling reconstruction, it detects the actual named filled vacuum. Those earlier results remain dependencies, not independent external validation. This does not resolve Panitch–Romo Conjecture 7.1, establish triangulation independence, or construct a new closed-manifold invariant. This AI-assisted supporting note has not received external specialist or peer review; novelty and priority are unassessed.

1. Native presentation and actual rows

Let $k=\mathbb Q(i,A^{1/2})$, $K=\mathbb Q(i)((A^{1/2}))$, $B=iA$, and $\mathfrak q=A^4$. The slope coefficient $q$ is always an integer, distinct from $\mathfrak q$. Use grouped coordinates $(z_1,z_2;w_1,w_2)$ and

$$W(u)W(v)=A^{\Omega(u,v)/2}W(u+v),\qquad \Omega(u,v)=u_w\cdot v_z-u_z\cdot v_w.$$

The native all-plus central sector is the vector-space quotient

$$C_s=S\Big/\left(\sum_{j=1}^2 S f_j+ \sum_e(c_eW(h_e)+A^2)S+(c_sW(h_s)+A^2)S\right), \qquad f_j=z_j^{-2}+w_j^2-1. \tag{3}$$

Thus tetrahedron relations have arbitrary multipliers on the left, while edge and filling relations have arbitrary multipliers on the right. We do not replace (3) by a two-sided algebra quotient.

Dimofte and Garoufalidis’s Section 2.6 and Section 4.6 provide the labeled actual incidence and peripheral rows:

The native dictionary is $P(R)=(R_1-R_2;R_3-R_2)$, with scalar $B^{\sum R_2}$. Hence

$$h=P(E_1)=(2,2;1,1),\quad P(E_2)=-h, \quad u_\mu=(1,1;1,0),\quad u_\lambda=(-4,0;-2,0).$$

The source’s geometric longitude eigenvalue convention is not substituted for this native scalar dictionary. The latter and its named-action meaning are retained from the local collar comparison.

For the slope in (1), set

$$h_s=(p-4q,p;p-2q,0),\qquad c_s=B^{-p+2q}, \qquad \Lambda=\mathbb Zh+\mathbb Zh_s. \tag{4}$$

The exact moment character is

$$\chi(eh+nh_s)=(-A^2)^e\left(\frac{-A^2}{B^{-p+2q}}\right)^n. \tag{5}$$

Both actual edges are included: their characters $-A^2$ and $-A^{-2}$ are exactly inverse. No hidden consistency relation is dropped in (4).

Choose any integers $r,t$ with $qr-pt=1$. The complementary native longitude is

$$b=(\alpha,\beta)=(r-4t,r;r-2t,0),\quad l=c_lW(b),\quad c_l=B^\rho,\quad \rho=-r+2t. \tag{6}$$

Then $\Omega(h_s,b)=2$ and $m l=A^2l m$ for $m=c_sW(h_s)$. In particular, $l^2=c_l^2W(2b)$, with no additional self-Weyl factor. The retained filling reconstruction sends the filled vacuum to $[1-l^2]$, not $[1]$. Nonvanishing in the all-plus sector implies nonvanishing of its preimage upstairs; injectivity of the sector projection is neither asserted nor needed.

2. The unchanged seed and all-input descent

For integer vectors $a,d$, define

$$\ell_+(W(2a,2d))= \frac{(-1)^{a_1+a_2}A^{2(\|a\|^2+a_1+a_2-a\cdot d)}} {(\mathfrak q)_{a_1}(\mathfrak q)_{a_2}},\qquad a\ge0, \tag{7}$$

where $(\mathfrak q)_n=\prod_{j=1}^n(1-\mathfrak q^j)$. Set the seed to zero off the even root lattice or when a coordinate of $a$ is negative. These are the classical Euler quantum-dilogarithm coefficients, as used in the preceding one-sided seed note.

For clarity, the required recurrence can be checked directly. On a supported monomial with $a_j>0$, the ratios of the two terms of $W(2a,2d)f_j$ to its unit term are

$$A^{-2d_j}\frac{\ell_+(W(2a-2e_j,2d))}{\ell_+(W(2a,2d))} =1-A^{-4a_j},\qquad A^{-2a_j}\frac{\ell_+(W(2a,2d+2e_j))}{\ell_+(W(2a,2d))} =A^{-4a_j}.$$

Their sum is one. At $a_j=0$ the first term is zero and the second ratio is one. Negative coordinates and unsupported parity give zero throughout. Thus $\ell_+(v f_j)=0$ for every algebraic input $v$.

Define the full, unsaturated image average

$$\mathcal H(v)=\sum_{\xi\in\Lambda}\chi(\xi)^{-1}\ell_+(W(\xi)v). \tag{8}$$

For $\xi=eh+nh_s=2(x,y)$, elimination gives

$$n=\frac{x_2-x_1}{2q},\quad e=x_2-\frac p2n, \quad y=Tx, \quad T=\begin{pmatrix}\frac12-\frac p{8q}&\frac p{8q}\\ \frac p{8q}&\frac12-\frac p{8q}\end{pmatrix}. \tag{9}$$

The quadratic part of the seed order is

$$Q(x)=\|x\|^2-x\cdot Tx =\frac{x_1^2+x_2^2}{2}+\frac p{8q}(x_1-x_2)^2 =x_1x_2+\frac{4+p/q}{8}(x_1-x_2)^2. \tag{10}$$

Under (1), this is strictly positive on the nonnegative orthant minus its origin. Compactness of the unit section gives $Q(a)\ge c\|a\|^2$ for $a\ge0$, with some $c>0$. For an arbitrary root input $W(u)$, put $\alpha_u=u_z/2$, $\beta_u=u_w/2$, and $a=x+\alpha_u$. Supported terms have $a\in\mathbb Z_{\ge0}^2$, with the exact image and second-coordinate parity also imposed. Their orders equal $2Q(a)$ plus a fixed linear function and constant. Thus only finitely many terms lie below any given order. Denominator inverses have constant one and only nonnegative powers of $A^4$. This proves unordered formal summability on every algebraic root input, not just on the vacuum.

Isotropy of $\Lambda$ now justifies reindexing to give

$$\mathcal H(W(\mu)v)=\chi(\mu)\mathcal H(v)\quad(\mu\in\Lambda).$$

Together with the seed recurrence, this annihilates every relation in (3) on its original side. The functional therefore descends to $C_s$ with values in $K$. A nonzero value proves the class is already nonzero over $k$.

The cone boundary is unchanged. At $p/q=-4$, the axes in (10) are null; below it they are negative. For $q=0$, the actual moment span has the vertical direction $2u_\mu-h=(0,0;1,-1)$. The retained exact cone theorem rules out all-input unordered convergence of this same seed there. These are limitations of this detector, not zero-vacuum conclusions.

3. A completion-of-the-square identity for native insertions

The following calculation isolates what replaces the sign certificate. Suppose a full graph moment presentation has symmetric $T$, and

$$\nu_A\chi(2x,2Tx)=v\cdot x,\qquad h_0=2\mathbf1-v.$$

For a native insertion $l=c_lW(\alpha,\beta)$, write $c_l=\eta_l A^\rho$ and set

$$\Delta_b=\beta-T\alpha,\qquad \kappa_b=2\alpha\cdot\Delta_b+v\cdot\alpha+2\rho.$$

Let $a=x+\alpha$ denote its supported seed coordinate. The exact orders of a unit summand and an $l^2$ summand are

$$D_0(x)=2Q(x)+h_0\cdot x,\qquad D_l(a)=2Q(a)+(h_0-4\Delta_b)\cdot a+\kappa_b. \tag{11}$$

To verify the second formula without suppressing a Weyl factor, expand the raw exponent before translating:

$$2Q(x)+2\mathbf1\cdot x-v\cdot x +4(\alpha-\beta)\cdot x +2(\|\alpha\|^2+\mathbf1\cdot\alpha-\alpha\cdot\beta)+2\rho. \tag{12}$$

The Weyl contribution is $\Omega(2(x,Tx),2(\alpha,\beta))/2 =2(Tx\cdot\alpha-x\cdot\beta)$; it cancels the other mixed $Tx$-term in the seed expansion. Substituting $x=a-\alpha$ yields (11).

This gives a useful sufficient criterion. If $Q$ is strictly copositive, $h_0$ and $h_0-4\Delta_b$ are componentwise nonnegative, $\kappa_b=0$, and $a=0$ is not in the longitude support, then (2) follows independently of all coefficient phases. For an integral native $b$, the condition $\Delta_b\notin\mathbb Z^N$ suffices to exclude $a=0$, because its second seed coordinate would be $\Delta_b$.

This criterion is relative to the declared polarization and native scalar data; no coordinate-independent topological interpretation is claimed.

4. Exact degree separation for all primitive numerators

From (5) and (9),

$$\nu_A\chi(\xi)=2e+(2+p-2q)n =\left(1-\frac1q\right)x_1+\left(1+\frac1q\right)x_2. \tag{13}$$

Therefore the unit order is exactly

$$\boxed{D_0(x)=2Q(x)+\left(1+\frac1q\right)x_1 +\left(1-\frac1q\right)x_2.} \tag{14}$$

Because $q$ is a nonzero integer, $|q|\ge1$; both linear coefficients are nonnegative, including negative $q$. Every supported unit term has integral $x\ge0$. Equation (10) makes (14) strictly positive unless $x=0$. The graph projection is injective, so $x=0$ is exactly $\xi=0$. Its coefficient is one. No native sign has entered this argument.

For every complement in (6), Bezout’s identity gives the exact defect

$$\boxed{\Delta_b=\beta-T\alpha =\left(\frac1{2q},-\frac1{2q}\right).} \tag{15}$$

Also

$$\kappa_b=\frac{\alpha_1-\alpha_2}{q} +\left(1-\frac1q\right)\alpha_1 +\left(1+\frac1q\right)\alpha_2+2(-r+2t)=0.$$

The native scalar in (6) is essential to this cancellation. Omitting it would leave constant $-2\rho$. Substitution in (11) now yields

$$\boxed{D_l(a)=2Q(a)+\left(1-\frac1q\right)a_1 +\left(1+\frac1q\right)a_2.} \tag{16}$$

This is the same quadratic form with the two nonnegative linear coefficients interchanged. Every supported nonzero $a\ge0$ consequently has strictly positive order.

There is one remaining case, and it is an integral support question rather than a sign question. If $a=0$, then $x=-\alpha$. Its second seed coordinate would be

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

which is not integral for any integer $q\ne0$. The seed is therefore zero. This excludes the only possible degree-zero longitude contribution. It is not enough merely to say that $2b\notin\Lambda$: the seed need not vanish on a monomial $W(0,2d)$ with nonzero integral $d$. The fractional second coordinate is the needed stronger statement.

The unit has constant coefficient one; the longitude square has none. All sums are coefficientwise finite by Section 2. Equation (2) follows for the entire family (1), completing the proof. The argument applies to every complementary pair $r,t$ with $qr-pt=1$, and preserves the original image rather than enlarging its parity support.

5. An odd-numerator control

Take the single control $(p,q)=(1,1)$, which the previous sign certificate does not cover, and use $(r,t)=(1,0)$. Then

$$h_s=(-3,1;-1,0),\quad c_s=iA,\quad \chi(h_s)=iA, \quad b=u_\mu,\quad c_l=-iA^{-1}.$$

The moment $4h+2h_s=(2,10;2,4)$ has a negative unit summand: its seed sign is positive and its character phase is negative. Thus the old termwise positivity hypothesis really fails. This does not imply that any coefficient of the aggregated unit series is negative.

Direct evaluation of (7)–(8), with the raw lattice and phases retained, gives

$$\begin{aligned} \mathcal H(1)&=1+\mathfrak q^3+2\mathfrak q^4+5\mathfrak q^5+8\mathfrak q^6+\cdots,\\ \mathcal H(l^2)&=-\mathfrak q-2\mathfrak q^2-3\mathfrak q^3-4\mathfrak q^4-4\mathfrak q^5-6\mathfrak q^6+\cdots,\\ \mathcal H(1-l^2)&=1+\mathfrak q+2\mathfrak q^2+4\mathfrak q^3+6\mathfrak q^4+9\mathfrak q^5+14\mathfrak q^6+\cdots. \end{aligned}$$

These are reproducibility controls, not a finite substitute for the proof in Section 4. No all-orders coefficient-positivity assertion is inferred from them.

The verifier uses exact Laurent polynomial coefficient dictionaries to prove (14)–(16) for arbitrary parameters, substituting $r=(1+pt)/q$, not sampling slopes. A separate raw Weyl-sum evaluator uses Gaussian integer coefficients and exact rational exponents for the control. It verifies both actual edge relations, the filling relation, original-sided tetrahedron relations, and supported odd-root inputs. For each input its complete-square degree is $2Q(a)+L\cdot a+c$, with

$$2Q(a)\ge\|a\|^2,\qquad D(a)\ge\tfrac12\|a\|^2-\tfrac12\|L\|^2+c$$

at this control. This proves the finite tail cutoffs used for every displayed coefficient and relation check. Computation does not decide a topology census.

6. Dependencies, significance, and next question

The substantive extension is removal of the numerator congruence restriction throughout the exact convergence cone of this fixed seed. The previous opposite-sign proof remains valid within its stated subfamily; its sufficient phase condition is not falsified. The new degree/support proof makes that condition unnecessary for this actual datum.

Dimofte–Garoufalidis supply the labeled triangulation and peripheral rows. Euler’s quantum-dilogarithm coefficients supply the local seed’s formal and intellectual origin. The prior FPRD one-sided construction supplies the orbit-averaging approach; the preceding actual-data note supplies the convention-fixed application. Its failed phase certificate motivates the present degree separation. The native six-root collar comparison and generic filling reconstruction are explicit retained dependencies for the named trace conclusion. Underlying Panitch–Park, Garoufalidis–Yu and Chen–Kricker trace results remain credited through that comparison. No GHHR boundary inversion, separate-cusp symmetry, or additional angular certificate is used in this proof. Details and source-access limits are recorded in the source ledger.

The next decisive question is an exact transport to a different local polarization or solution outside this seed’s convergence cone, preserving the raw edge/filling character, named peripheral operators, relation sides, and convergence on every root input. A formal cyclic relabeling or reversal of $A$ does not establish that transport. It must be proved, or its failure preserved. Repeating numerator congruence cases, sampling slopes, or silently changing native phases cannot advance that question. The earlier actual inversion-odd full-index value and the unrestricted Panitch–Romo conjecture remain open.