A one-sided Nahm functional for the filled quantum gluing module

FPRD Lab, 8 September 2026. Supporting research note. AI-assisted work; no external specialist review. Novelty is unassessed.

Result and limitation

The tetrahedron relation in the mixed-sided gluing module requires only one adjoint difference equation. Imposing both equations of the usual tetrahedron index unnecessarily restricts a search for a linear detector on this module. Euler’s quantum-dilogarithm coefficients give an explicit alternative seed satisfying the required equation, with support in a half-space.

This note proves an exact criterion for its lattice average to converge on every algebraic input. The criterion is strict positivity of one quadratic form on one rational polyhedral cone. The convergent average annihilates the original tetrahedron, edge, and filling relations, including the full root lattice and exact moment character. It is a congruence-restricted Nahm sum in a suitable lattice coordinate system.

A further phase and longitude criterion proves that this functional detects the filled vacuum, rather than only the cokernel unit. For actual native data satisfying these criteria and a complete strict angle certificate, no filling-angle certificate is needed. No actual triangulation outside the preceding two-angle chamber is exhibited or proved to satisfy the new criteria. The worked example is an abstract presentation, explicitly not triangulation data. Thus this is a constructed alternative detector and a conditional theorem, not an established enlargement of the topological nonvanishing region. Panitch–Romo Conjecture 7.1 and the preceding actual odd peripheral value remain open.

1. Original domain and conventions

Work over $k=\mathbb Q(i,A^{1/2})$, extending values to $K=\mathbb Q(i)((A^{1/2}))$, and put $q=A^4$. In one native central sector let

$$L=\mathbb Z^{2N},\qquad \Omega((u_z,u_w),(v_z,v_w))=u_w\cdot v_z-u_z\cdot v_w, \qquad W(u)W(v)=A^{\Omega(u,v)/2}W(u+v). \tag{1}$$

Coordinates are written as two $N$-vectors; this is just a reordering of the earlier interleaved convention. Let $\Lambda\subset L$ be the actual, unsaturated image of all edge and filling rows. It is isotropic. The exact consistent character is

$$\chi(h_r)=-A^2/c_r,\qquad \chi(\lambda)=\eta(\lambda)A^{\nu(\lambda)}, \tag{2}$$

where $\eta$ is the native root-of-unity phase and $\nu$ is additive. The target is

$$C_s=S/\mathcal R,\qquad \mathcal R=\sum_jS(z_j^{-2}+w_j^2-1) +\sum_r(c_rW(h_r)+A^2)S. \tag{3}$$

It is a vector-space quotient, not an algebra quotient. The first set of generators has multipliers on the left, the second on the right. The native named meridian and longitude satisfy

$$m=c_sW(h_s),\quad l=c_lW(b),\quad \Omega(h_s,b)=2,\quad ml=A^2lm. \tag{4}$$

The earlier filling reconstruction identifies the relevant vacuum with $[1-l^2]$. A nonzero functional on the unit alone does not suffice. A functional on this central-sector quotient pulls back to the original six-root target.

2. The half-space seed

Write $(q)_a=\prod_{j=1}^a(1-q^j)$, with $(q)_0=1$. Define

$$f_+(m,e)= \begin{cases} \displaystyle\frac{(-1)^e q^{e(e-1)/2+em/2}}{(q)_{-e}},&e\leq0,\\ 0,&e>0. \end{cases} \tag{5}$$

Lemma 1. This seed satisfies exactly the needed recurrence

$$q^{e/2}f_+(m+1,e)+q^{-m/2}f_+(m,e+1)-f_+(m,e)=0. \tag{6}$$

For $e<0$, division by the nonzero value in (5) makes the first two terms $q^e$ and $1-q^e$. At $e=0$, the values are $1,0,1$. For $e>0$ all terms vanish. This proves (6), including the support boundary.

In contrast, the other usual tetrahedron-index recurrence has, at $e=0$, the residual

$$q^{e/2}f_+(m-1,e)+q^{-m/2}f_+(m,e-1)-f_+(m,e) =-\frac{q^{1-m}}{1-q}\ne0. \tag{7}$$

Thus the rank-two space defined by both recurrences in Garoufalidis, Definition 3.1 and Theorem 3.2 does not contain this seed. That theorem does not prohibit its use for (3).

The coefficients in (5) are classical. Euler’s second sum gives

$$(qt;q)_\infty=\sum_{a\geq0} \frac{(-1)^a q^{a(a+1)/2}t^a}{(q)_a}. \tag{8}$$

See NIST DLMF 17.5.1. The new issue here is the exact global average and its domain, not this local series.

Define the linear seed on root monomials by

$$\ell_+(W(u))=0\quad(u\notin2L),\qquad \ell_+(W(2a,2d))= \begin{cases} \displaystyle\frac{(-1)^{s(a)}A^{2(\|a\|^2+s(a)-a\cdot d)}}{\prod_j(q)_{a_j}},&a\geq0,\\ 0,&\text{otherwise}, \end{cases} \tag{9}$$

where $s(a)=\sum_j a_j$. Equation (6) and the Weyl rule show

$$\ell_+\bigl(S(z_j^{-2}+w_j^2-1)\bigr)=0. \tag{10}$$

Indeed $W(u)f_j=A^{-u_{wj}}W(u-2e_{zj})+A^{-u_{zj}}W(u+2e_{wj})-W(u)$; for even $u$ this is (6), while all terms for odd parity vanish.

The usual tetrahedron index can also be recovered as a sum of translated copies of (5):

$$I_\Delta(m,e)=\sum_{n\geq\max(e,0)} \frac{q^{-nm/2}}{(q)_n}\,f_+(m,e-n). \tag{11}$$

Substituting (5) and setting $r=n-e$ gives the defining tetrahedron-index sum directly. This is an identity for a fixed local input; no interchange with a global moment sum is asserted. The omitted summation in (11) is precisely why the new seed need not inherit the two-recurrence inversion arguments.

3. Exact all-input convergence criterion

Let

$$V=\{(x,y)\in\mathbb R^{2N}:2(x,y)\in\Lambda\otimes\mathbb R\}, \qquad\mathcal C_+=\{(x,y)\in V:x\geq0\}, \qquad Q(x,y)=\|x\|^2-x\cdot y. \tag{12}$$

The factor 2 is retained to match the even seed. It does not replace $\Lambda$ by its saturation. Define

$$\mathcal H(a)=\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1} \ell_+(W(\lambda)a). \tag{13}$$

Theorem 2. The unordered sum (13) exists in $K$ for every $a\in S$ if and only if

$$Q(v)>0\qquad(0\ne v\in\mathcal C_+). \tag{14}$$

If $\mathcal C_+=\{0\}$, this condition is vacuous and every input has finite support. Whenever (14) holds, $\mathcal H$ descends to (3).

Here unordered convergence means that for every degree bound only finitely many nonzero summands have order below it. Cancellation by a chosen order of summation is not permitted.

Proof: exact degree

For an input $u=2(\alpha,\beta)$, with half-integral $\alpha,\beta$, and a supported summand $\lambda=2(x,y)$, the conditions are $u+\lambda\in2L$ and $x+\alpha\geq0$. Denominators in (9) have order zero. Expansion of the seed exponent and the Weyl factor gives the exact order

$$D_u(\lambda)=2Q(x,y)+2s(x)-\nu(\lambda) +4(\alpha-\beta)\cdot x +2\bigl(\|\alpha\|^2+s(\alpha)-\alpha\cdot\beta\bigr). \tag{15}$$

The $y\cdot\alpha$ terms cancel. This cancellation matters for both directions of the proof.

Proof: sufficiency

Supported points lie in an affine lattice inside $x\geq-\alpha$; its recession cone is (12). If an unbounded sequence of such points is divided by its norm, every limit lies in $\mathcal C_+$ on the unit sphere. Strict positivity and compactness make $Q$ bounded below there by a positive constant. Consequently, for all sufficiently large supported points, $Q\geq c\|(x,y)\|^2$ for some $c>0$. The other terms in (15) are affine. Therefore $D_u\to+\infty$. Lattice discreteness gives local finiteness in every degree. If the recession cone is zero, the support polyhedron is bounded instead. Finite linear combinations of monomials now give all of $S$.

Proof: necessity, including rational null directions

If $Q<0$ somewhere on the rational cone, it is negative on a rational point of a face containing that point. Scale it into $\Lambda\cap2L$. The unit input along positive multiples of this lattice vector has degree tending to $-\infty$.

Suppose instead $Q\geq0$ on the cone but $Q(v)=0$ for a nonzero real point. A nonzero rational null point exists. To see this without enumerating faces, take the minimal face containing $v$. In its rational linear span, $v$ is a relative-interior local minimum of the restricted quadratic form. Its restricted gradient vanishes. The kernel of that rational linear system contains rational points arbitrarily near $v$; such points remain in the relative interior, and their quadratic value is zero. Scale one into the even image lattice, writing it as $2(x,y)$ with integral $x\geq0$.

If $x\ne0$, choose the single input $u=2(0,K\mathbf1)$, with an integer $K$ so large that

$$2s(x)-\nu(2x,2y)-4Ks(x)<0. \tag{16}$$

All positive multiples of the ray remain supported, and (15) tends linearly to $-\infty$. If $x=0$, both signs of the vertical ray belong to the cone. On the unit input their degrees are $-n\nu(2x,2y)$; one sign has nonpositive degree for every $n$. Neither case is unordered summable. A null ray might permit selected inputs, but cannot permit all inputs. This establishes the exact necessity in (14).

Proof: original sidedness

For every tetrahedron generator, $\mathcal H(af_j)=0$ term by term using (10). For $\mu\in\Lambda$, isotropy and the now-justified change of summation variable give

$$\mathcal H(W(\mu)a) =\sum_{\lambda\in\Lambda}\chi(\lambda)^{-1} \ell_+(W(\lambda+\mu)a) =\chi(\mu)\mathcal H(a). \tag{17}$$

Equations (2) and (17) annihilate every $(c_rW(h_r)+A^2)a$. Odd root shifts are included in this full-lattice reindexing. This proves descent to precisely (3). No angle or classical shape point was used. ∎

4. Structural translation to a Nahm sum

Under (14), projection $(x,y)\mapsto x$ is injective on $V$: a nonzero vertical vector would violate (14). Write $y=Tx$ on the rational subspace $V_z\subset\mathbb R^N$. Isotropy says that $x\cdot Tx'$ is symmetric in $x,x'$. Thus

$$Q(x,Tx)=x\cdot(I-T)x>0 \quad(0\ne x\in V_z\cap\mathbb R_{\geq0}^N). \tag{18}$$

For a one-cusp full-rank moment space, $V_z=\mathbb R^N$, and (18) is strict copositivity of the symmetric matrix $I-T$. This means positivity on the nonnegative orthant, not positive definiteness on all of $\mathbb R^N$. It is an exact characterization; no claim of a cheap algorithm for deciding copositivity is made.

The actual even image gives the possibly proper sublattice

$$\Gamma=\{x\in\mathbb Z^N:2(x,Tx)\in\Lambda\}. \tag{19}$$

Then

$$\mathcal H(1)=\sum_{x\in\Gamma\cap\mathbb Z_{\geq0}^N} \frac{(-1)^{s(x)}\eta(2x,2Tx)^{-1} A^{2x\cdot(I-T)x+2s(x)-\nu(2x,2Tx)}}{\prod_j(q)_{x_j}}. \tag{20}$$

This is a congruence-restricted Nahm sum. Odd inputs give the corresponding affine parity fibers, not a saturated replacement. Quadratic exponents over products of $(q)_n$ are established Nahm-sum objects; see Garoufalidis–Zagier, Asymptotics of Nahm sums at roots of unity. Their asymptotic results are not needed for Theorem 2. Their Neumann–Zagier/Nahm viewpoint is an influence on this translation.

5. A sufficient certificate for the filled vacuum

Convergence does not alone imply a nonzero vacuum. The following conditions retain the actual native phase and longitude; they do not authorize changing either.

Assume (14), and suppose

$$\eta(2x,2Tx)=(-1)^{s(x)}\quad(x\in\Gamma). \tag{21}$$

It suffices to check this equality of characters on a basis of the actual even image. In particular all summands of (20) have nonnegative integer coefficients and positive leading coefficient. The set of degrees has a minimum $d_*$, so $\mathcal H(1)\ne0$, with positive leading coefficient.

There is also an exact raw-incidence form of (21). For $R=\sum_en_eE_e+n_sR_s$, put $e(R)=\sum_en_e$, $r(R)=\sum_jR_{j1}$, and $p(R)=\sum_jR_{j2}$. On the all-plus native sector, $\eta(P(R))=(-1)^{e(R)+n_s}i^{-p(R)}$. If $P(R)\in2L$, then $2s(x)=r(R)-p(R)$, so (21) is equivalent to

$$r(R)\equiv2\bigl(e(R)+n_s\bigr)\pmod4 \quad\text{for every }P(R)\in\Lambda\cap2L. \tag{21a}$$

This is a congruence on the actual even image, not an imposed sign choice. Exact character consistency makes it independent of the chosen expression for the row. It supplies a direct incidence test for the remaining applicability question.

Suppose the named longitude in the declared polarization has

$$b=(0,-t),\quad 0\ne t\in\mathbb Z_{\geq0}^N, \qquad c_l=\eta_l A^\rho. \tag{22}$$

No lattice change is made to achieve (22); it is a hypothesis on the native data. The support of $\mathcal H(l^2)$ is identical to that of (20), and the exact ratio of its summand to the unit summand is

$$c_l^2q^{t\cdot x}=\eta_l^2A^{2\rho}q^{t\cdot x}. \tag{23}$$

This includes both the Weyl factor and the changed seed exponent.

Theorem 3. Under (14), (21), and (22), if $\rho>0$, then

$$\nu_A\mathcal H(l^2)\geq d_*+2\rho>d_*, \qquad \mathcal H(1-l^2)\ne0. \tag{24}$$

Hence the actual filled vacuum and named descended trace are nonzero whenever actual native data satisfy these hypotheses. The inequality follows from $t\cdot x\geq0$, while positivity in (20) prevents cancellation of its lowest coefficient. The phase $\eta_l$ is unrestricted. If the shifted sum were zero, the conclusion would still follow.

For completeness, when $\rho=0$ and $\eta_l^2=-1$, coefficient positivity also proves nonvanishing. When $\rho=0$ and $\eta_l^2=1$, it proves nonvanishing provided a supported $x$ has $t\cdot x>0$. Indeed

$$1-\prod_jq^{t_jx_j} =\sum_j\left(\prod_{i<j}q^{t_ix_i}\right) (1-q^{x_j})\sum_{h=0}^{t_j-1}q^{hx_j}. \tag{25}$$

A nonzero term has $x_j>0$, so $1-q^{x_j}$ cancels a denominator factor. The resulting coefficients are nonnegative and at least one term is nonzero. Empty sums for $t_j=0$ contribute zero.

Complete angles supply the positive longitude scalar

For native raw peripheral rows, put $B=iA$, $R_l=(r_j,s_j,t'_j)_j$, and $p(R_l)=\sum_js_j$. The earlier dictionary gives $b=P(R_l)$ and $c_l=B^{p(R_l)}$. A complete strict certificate $(\gamma_j,\beta_j,\alpha_j)$ has zero longitude angular holonomy. Under (22), $r_j=s_j$, $t'_j-s_j=-t_j$, so

$$0=\theta^0\cdot R_l=p(R_l)-\sum_j\alpha_jt_j, \qquad \rho=p(R_l)=\sum_j\alpha_jt_j>0. \tag{26}$$

Therefore a complete strict certificate together with (14), (21), and the pure-$w$ longitude condition (22) implies (24). There is no filling-angle assumption in this implication. Its realizability for additional actual fillings is unresolved; a formal conditional implication must not be reported as such an example.

6. An exact abstract example and its boundary

Take three tetrahedron blocks, no edge rows, and one moment row

$$h=(2,2,0;0,0,0),\qquad c_s=B^{-2}=-A^{-2}, \qquad\chi(h)=A^4. \tag{27}$$

Let $b=(0,0,0;-1,0,-2)$, $c_l=B$. Then $\Omega(h,b)=2$. The image lattice is exactly $\mathbb Zh$, with $x=(k,k,0)$, $y=0$. Its supported cone is $k\geq0$ and $Q=2k^2$, while the phase condition is $1=(-1)^{2k}$. Consequently

$$\mathcal H(1)=\sum_{k\geq0}\frac{q^{k^2}}{(q)_k^2},\qquad \mathcal H(1-l^2)=\sum_{k\geq0} \frac{q^{k^2}(1+A^2q^k)}{(q)_k^2} =1+A^2+\text{terms of order at least }A^4. \tag{28}$$

Raw rows realizing the declared scalars are

$$R_s=((1,-1,-1),(1,-1,-1),(0,0,0)),\quad R_l=((1,1,0),(0,0,0),(0,0,-2)). \tag{29}$$

They have zero pairing with the strict angle triples

$$\theta^0=((1/2,1/6,1/3),(1/2,1/4,1/4),(1/3,1/3,1/3)). \tag{30}$$

But a filling angular certificate would require $\gamma_1+\gamma_2=2$, impossible for strict angle triples. This shows that the alternative seed and its certificates are algebraically consistent beyond the strict filling-angle condition. It does not give a manifold: there are no actual edge-incidence or normal-homology data in this example. In particular it is neither a triangulation counterexample nor evidence that (26) is realized in an additional topological chamber.

The verifier computes (28) through a declared order with a proved tail cutoff $4k^2$. It also checks a nontrivial tetrahedron recurrence after global summation, the original moment eigenvalue, a parity-empty odd input, and a proper even-image lattice. These checks supplement the proofs; they do not establish topological applicability.

7. Audit and next decisive question

The construction removes the extra second recurrence at the local level; it does not remove any relation in (3). The proof covers the support boundary, native central phases, odd inputs, unsaturated image, all-input convergence, rational null rays, and the original left/right multipliers. It distinguishes strict positivity on a cone from positivity on an entire vector space. The new convergence condition cannot be replaced by mere nonnegativity: (16) supplies an input that defeats every nonvertical null ray.

The remaining gate is concrete: establish or refute the joint realizability of (14), (21), and (22) for actual native triangulation data with a complete strict certificate beyond the known two-angle chamber. A successful case must retain all raw incidence, peripheral, and phase information; arbitrary symplectic matrices or the abstract example above do not answer it. If these sufficient certificates are incompatible with genuine incidence, retain Theorem 2 and use the exact signed Nahm sum (20) to investigate cancellation without assuming (21) or (22). The earlier inversion-odd index value is still an alternative unresolved route.

Sources and intellectual credit

• Euler’s second sum, as recorded in NIST DLMF 17.5.1, supplies the local quantum-dilogarithm coefficients. Lemma 1 is checked directly here.
• Stavros Garoufalidis, The 3D index of an ideal triangulation and angle structures, arXiv:1208.1663v2, 7 October 2015, supplies the two-recurrence comparison and tetrahedron-index definition used in (11). Its positivity/rank theorem is background, not a restriction on the one-recurrence module dual.
• Stavros Garoufalidis and Don Zagier, Asymptotics of Nahm sums at roots of unity, arXiv:1812.07690, provide the established Nahm-sum setting and influence on the graph-coordinate translation. No asymptotic or block-factorization theorem from that paper is imported.
• The FPRD native peripheral bridge, global averaging packet, and actual-incidence packet supply the retained target, filling-vacuum identification, and raw scalar dictionary, with their original external dependencies. The peripheral-projection packet motivates leaving the balanced family.

Primary sources were checked on 8 September 2026. No external novelty, priority, endorsement, or unrestricted conjecture-resolution claim is made.