An actual one-sided Nahm detector for the filled vacuum

FPRD Lab — supporting research note, 8 September 2026.

Result and scope

The one-sided detector applies to genuine triangulation data. On the published two-tetrahedron figure-eight triangulation of Dimofte and Garoufalidis, it detects the native filled vacuum for every primitive slope

$$s=p\mu+q\lambda,\qquad q\ne0,\quad p\in4\mathbb Z,\quad p/q>-4. \tag{1}$$

Here $\mu,\lambda$ are the oriented meridian and canonical longitude specified by their raw rows below. No slope search is involved: the proof uses one fixed published datum and symbolic integer parameters. The key new sufficient condition is opposite coefficient signs for the unit and longitude-square sums. It does not require the earlier pure-$w$ longitude form.

For $(p,q)=(4,1)$, the resulting vacuum value is an explicit positive Nahm series. The ordinary tetrahedron-index average, with the same native moment character, fails even to converge on the unit because of an exact lattice ray. Thus this is an actual application beyond that earlier detector, not merely an abstract presentation satisfying a new criterion.

Through the retained native peripheral comparison and filling reconstruction, the detected class is the vacuum of the named descended trace. Those results remain dependencies, not new proofs here. The unrestricted Panitch–Romo Conjecture 7.1 remains open. We do not claim that this family lies outside every earlier regular-intersection result, that the formal series is triangulation-independent, or that this is a new closed-manifold invariant. This is an AI-assisted research note without external specialist review; novelty is unassessed.

1. Exact presentation and the retained seed

Work over $k=\mathbb Q(i,A^{1/2})$, with values in $K=\mathbb Q(i)((A^{1/2}))$. Write $B=iA$ and $\mathfrak q=A^4$, reserving $q$ for the slope coefficient. Coordinates are grouped as $(z_1,z_2;w_1,w_2)$. The quantum torus has Weyl basis

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

In the native all-plus central sector the target is the mixed-sided vector quotient

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

It is not a two-sided algebra quotient. Keep the actual image lattice $\Lambda=\sum_e\mathbb Z h_e+\mathbb Z h_s$, not its saturation, and the consistent exact character $\chi(h_r)=-A^2/c_r$. For $m=c_sW(h_s)$, a complementary native longitude $l=c_lW(b)$ is oriented so that $\Omega(h_s,b)=2$; then $ml=A^2lm$.

The retained filling reconstruction identifies the filled vacuum with $[1-l^2]$, not $[1]$. Detecting it in this sector detects its preimage in the original six-root target. We do not assert that projection onto the sector is injective.

The preceding one-sided seed note constructs

$$\ell_+(W(2a,2d))= \frac{(-1)^{\sum a_j} A^{2(\|a\|^2+\sum a_j-a\cdot d)}} {\prod_j(\mathfrak q)_{a_j}},\qquad a_j\ge0, \tag{3}$$

where $(\mathfrak q)_n=\prod_{j=1}^n(1-\mathfrak q^j)$. Set the seed to zero off the even lattice or when some $a_j<0$. These are classical Euler quantum-dilogarithm coefficients, not a new special function. The seed satisfies $\ell_+(a f_j)=0$ with exactly the side in (2).

The full-image average is

$$\mathcal H(a)=\sum_{\xi\in\Lambda}\chi(\xi)^{-1}\ell_+(W(\xi)a). \tag{4}$$

The retained convergence theorem states that (4) is unordered-summable on every algebraic input exactly when

$$Q(x,y)=\|x\|^2-x\cdot y>0 \quad\text{for }0\ne(x,y)\in(\Lambda\otimes\mathbb R)/2,\ x\ge0. \tag{5}$$

For sufficiency, strict positivity on the compact unit section of the cone supplies a quadratic lower bound on every affine input support; the remaining exponent is linear. Therefore only finitely many summands reach any prescribed degree. This justifies the reindexing $\mathcal H(W(\xi)a)=\chi(\xi)\mathcal H(a)$, while the seed recurrence gives $\mathcal H(a f_j)=0$. Thus (4) annihilates the original relation module, including all algebraic root inputs. The general necessity proof is in the retained note; the boundary and negative directions needed here are explicit below.

2. Opposite-phase lemma

Write $\chi(\xi)=\eta(\xi)A^{\nu(\xi)}$ and $c_l=\eta_l A^\rho$. Assume (5) and

$$\eta(\xi)=(-1)^{\sum x_j} \quad\text{whenever }\xi=2(x,y)\in\Lambda\cap2\mathbb Z^{2N}. \tag{6}$$

Then every supported unit summand has nonnegative real coefficients, and its leading coefficient is positive. The term $\xi=0$ is 1, so $\mathcal H(1)\ne0$.

For $l^2=c_l^2W(2b)$, a supported summand again has $\xi$ even. Its seed sign differs from the unit sign by $(-1)^{\sum b_{zj}}$, independently of $\xi$. Weyl factors are powers of $A$, so the total coefficient phase is

$$\varepsilon_l=\eta_l^2(-1)^{\sum b_{zj}}. \tag{7}$$

If $\varepsilon_l=-1$, all supported longitude-square summands have negative real coefficients. Hence

$$\boxed{\mathcal H(1-l^2)\ne0.} \tag{8}$$

The two supports need not be equal. There is no cancellation because one subtracts a series with nonpositive coefficients from a nonzero series with nonnegative coefficients. All series are bounded below and coefficientwise finite by (5). No sign-sector redefinition, positivity of $\rho$, or pure-$w$ assumption is used.

For a native raw row $R_l=((r_j,s_j,t_j))$, the retained dictionary is

$$P(R_l)=(r-s;t-s)=b,\qquad c_l=B^{\sum s_j}.$$

Consequently (7) reduces to the intrinsic parity of this declared row:

$$\varepsilon_l=(-1)^{\sum_j r_j}. \tag{9}$$

This is a sufficient sign certificate, not a universal parity assertion for arbitrary triangulations.

3. Published actual incidence and native conventions

Use the cyclic labeling in Dimofte–Garoufalidis, arXiv:1202.6268v2, Section 2.6, dated 25 October 2012. Its two edge rows and meridian row, together with the longitude from Section 4.6, are

They collapse to

$$h=(2,2;1,1),\quad P(E_2)=-h,\quad a=(1,1;1,0),\quad d=(-4,0;-2,0). \tag{10}$$

Direct calculation gives $\Omega(h,a)=\Omega(h,d)=0$, $\Omega(a,d)=-2$. Both actual edges are retained: their scalars are $1,A^4$, and their characters are $-A^2,-A^{-2}$, exactly inverse. Thus removing the redundant edge does not lose a central consistency condition.

For $s=p\mu+q\lambda$,

$$h_s=(p-4q,p;p-2q,0),\qquad c_s=B^{-p+2q},\qquad \Lambda=\mathbb Zh+\mathbb Z h_s. \tag{11}$$

A complementary curve $r\mu+t\lambda$ is specified by $qr-pt=1$. Its native operator has

$$b=(r-4t,r;r-2t,0),\qquad c_l=B^{-r+2t},\qquad \Omega(h_s,b)=2. \tag{12}$$

This sign convention is deliberate. The source’s geometric $\mathrm{SL}_2$ longitude $\ell=-e^v$ is not simply our native operator: we use its raw combinatorial row and the scalar $B^{\text{prime sum}}$. The squared native roots correspond to the source’s ordinary shape variables. We also do not identify these labels with a differently relabeled or meridionally twisted longitude in another example. The passage from these raw rows to the native named action is the retained local collar comparison, not an inference from scalar volume or classical holonomy agreement.

4. Exact cone and phase calculations for the whole family

Let $q\ne0$. If $e h+n h_s=2(x,y)$, elimination yields

$$n=\frac{x_2-x_1}{2q},\qquad e=x_2-\frac p2n,\qquad y=T x,\quad T=\begin{pmatrix}\frac12-\frac p{8q}&\frac p{8q}\\ \frac p{8q}&\frac12-\frac p{8q}\end{pmatrix}. \tag{13}$$

Thus

$$Q(x,Tx)=\frac12(x_1^2+x_2^2)+\frac p{8q}(x_1-x_2)^2. \tag{14}$$

On $x\ge0$, (14) is strictly positive away from zero exactly when $p/q>-4$. For a negative ratio use $(x_1-x_2)^2\le x_1^2+x_2^2$; for a nonnegative ratio positivity is immediate. At ratio −4 the form is $x_1x_2$ and has null axes; below −4 an axis is negative. These are rational directions in the actual moment space. When $q=0$, the primitive meridian case instead contains the vertical direction $2a-h=(0,0;1,-1)$, so it does not satisfy all-input convergence. These are detector limitations, not vacuum-vanishing conclusions.

The actual parity condition $e h+n h_s\in2\mathbb Z^4$ is exactly

$$e\equiv0\pmod2,\qquad np\equiv0\pmod2. \tag{15}$$

Using the exact native character, the unit phase condition (6) becomes

$$2e+n(p-2q-2)\equiv0\pmod4 \quad\text{for every pair satisfying (15).} \tag{16}$$

For primitive $(p,q)$, (16) holds exactly when $p\in4\mathbb Z$. Indeed, then $q$ is odd and $p-2q-2$ is divisible by four. If $p$ is odd, take $e=0,n=2$, giving residue two. If $p\equiv2\pmod4$, $q$ is odd and $e=0,n=1$ gives residue two.

In the favorable family the even image is the proper lattice

$$\Gamma=\{x\in2\mathbb Z^2:x_2-x_1\in2q\mathbb Z\}. \tag{17}$$

This follows directly from (13) and is not replaced by $\mathbb Z^2$. Native phase positivity also follows directly: the even moment has $e$ even, the filling character has phase +1, and both $x_j$ are even.

Finally $qr-pt=1$, with $q$ odd and $p$ divisible by four, forces $r$ odd. The sum of the unprimed coordinates of $rR_\mu+tR_\lambda$ is $r-2t$, also odd. Therefore $\varepsilon_l=-1$ for every complementary curve (12). The opposite-phase lemma proves (8) throughout (1). This establishes the claimed actual vacuum result, with the retained named-action and filling-reconstruction dependencies.

5. An explicit boundary filling: $4\mu+\lambda$

Set $(p,q)=(4,1)$, and choose the complement $\mu$. Then

$$h_s=(0,4;2,0),\quad c_s=-A^{-2},\quad\chi(h_s)=A^4, \quad b=a,\quad c_l=-iA^{-1}.$$

Here $T=\left(\begin{smallmatrix}0&1/2\\1/2&0\end{smallmatrix}\right)$, $\Gamma=2\mathbb Z^2$. Write $x=(2u,2v)$, $y=(v,u)$. Substitution in (3)–(4), including the exact Weyl factor and $c_l^2=-A^{-2}$, gives

$$\mathcal H(1)=\sum_{u,v\ge0} \frac{\mathfrak q^{2u^2+2v^2-2uv+u}} {(\mathfrak q)_{2u}(\mathfrak q)_{2v}}, \tag{18}$$

$$\mathcal H(l^2)=-\sum_{u,v\ge0} \frac{\mathfrak q^{2u^2+2v^2-2uv+u+2v+1}} {(\mathfrak q)_{2u+1}(\mathfrak q)_{2v+1}}. \tag{19}$$

Both exponents are at least $u^2+v^2$, proving formal convergence directly. Hence

$$\mathcal H(1-l^2)=1+\mathfrak q+3\mathfrak q^2+6\mathfrak q^3 +10\mathfrak q^4+17\mathfrak q^5+\cdots\ne0. \tag{20}$$

The nonvanishing proof is the all-orders sign argument, not this finite expansion.

For comparison, the ordinary-index functional uses $\ell_\Delta(W(2a,2d))=\prod_j I_\Delta(d_j,-a_j)$. The actual image contains

$$v_0=-2h+h_s=(-4,0;0,-2),\qquad\chi(v_0)=1. \tag{21}$$

For every positive integer $k$, its unit summand is

$$\ell_\Delta(W(kv_0))=I_\Delta(0,2k)I_\Delta(-k,0) =1+\text{positive }A\text{-order}. \tag{22}$$

In the defining index series the summation index zero gives coefficient one in each factor; every positive summation index has strictly positive degree. Thus infinitely many distinct terms have order zero. The ordinary unordered average cannot converge on 1. The one-sided seed excludes this ray because its first $a$-coordinate is $-2k$. This pinpoints the actual gain: changing the required local solution removes a genuine global nonconvergence ray without changing any relation or moment character.

There is also no strict filling-angle certificate for this slope in the declared triangulation. Write angles $(\alpha_j,\beta_j,\gamma_j)$, with each triple summing to one. The edge equation and the filling equation imply $2\gamma_1+\alpha_2=0$, incompatible with strict positivity. The earlier strict two-angle theorem therefore does not cover this case. We make no exclusion claim about earlier shape-based or clean-intersection criteria.

6. Verification, dependencies, and next question

verify-actual-nahm.py uses exact rational and integer arithmetic. It verifies the raw rows, symbolic graph and quadratic identities, the $4$-filling series independently from the raw Weyl sum, both actual edge relations, the filling relation, the correct tetrahedron multiplier side, and a genuinely supported odd-root input. Every series cutoff uses a quadratic tail bound; no sampled slopes stand in for the symbolic proof.

Formal dependencies are the preceding one-sided seed/cone theorem, the native six-root collar comparison, and the filling reconstruction of the vacuum. Dimofte–Garoufalidis provide the actual labeled incidence and peripheral rows. Euler’s quantum-dilogarithm coefficients provide the seed’s intellectual and formal origin. Panitch–Park, Garoufalidis–Yu, and Chen–Kricker provide the underlying named quantum-trace framework; their results and convention caveats are retained through the cited bridge. The normal-index work of Garoufalidis–Hodgson–Hoffman–Rubinstein influenced the earlier detector comparison, but its boundary-inversion theorem is not needed for (8).

The next decisive question is whether a one-sided solution or a finite combination of such solutions can remove the native unit-phase obstruction when $p\notin4\mathbb Z$, while retaining the exact original character and all-input convergence. Merely changing the phase by hand changes the target and is not allowed. A broader alternative is to express (6) and (9) intrinsically in the actual raw lattice and peripheral parity data, without a census of triangulations. The unrestricted conjecture and the earlier inversion-odd full-index value remain unresolved.