A nonzero filling target with zero six-root vacuum

8 September 2026. Research note; internally checked, without external specialist review. The algebraic statements below retain the native convention and generic named-action/filling dependencies of the preceding notes. No unrestricted conjecture resolution or priority claim is made.

Result and its meaning

For the fixed two-tetrahedron figure-eight triangulation, the meridian vacuum vanishes before any specialization of opposite-root signs:

$$[1-l^2]=0\quad\text{in }C_\mu^6, \qquad [1]\ne0\quad\text{in }C_\mu^6.$$

Here $C_\mu^6$ is the native six-root meridian cokernel. The second assertion is retained from the previous convergent all-plus detector: a nonzero image detects a nonzero element upstairs. The first assertion is new here. It follows from a finite identity involving only squared moment relations. Their factors accommodate all central signs simultaneously; no sign components are enumerated or identified.

The retained filling isomorphism sends the empty skein to $[1-l^2]$, not to $[1]$. Thus its boundary-balanced filled target is nonzero but its descended trace is zero for meridian filling, which returns $S^3$. This does not contradict the literal nonzero-quotient requirement of Panitch–Romo, Conjecture 7.1. That statement does not explicitly require its descended map to be nonzero. Nor does the present calculation rule out a vacuum-preserving quotient that imposes fewer relations than the boundary-balanced construction.

1. Retained native presentation without sign specialization

Work over $k=\mathbb Q(i,A^{1/2})$, with generic $A$. The four native vertex monomials at each tetrahedron have value $B=iA$. Dividing the other three vertex relations by the base vertex gives

$$y=\sigma z,\quad y'=\sigma z',\quad y''=\sigma z'',\qquad \sigma^2=1.$$

This is elimination of relations already in the presentation, not the further quotient $\sigma=1$. The remaining local algebra is a rank-two quantum torus over the central involution ring. Global edge relations may impose additional central relations; none is needed for the zero-vacuum proof.

In grouped coordinates $u=(u_{z1},u_{z2};u_{w1},u_{w2})$, write

$$\Omega(u,v)=u_w\cdot v_z-u_z\cdot v_w, \qquad W(u)W(v)=A^{\Omega(u,v)/2}W(u+v), \qquad f_j=z_j^{-2}+w_j^2-1.$$

Tetrahedron relations are $Sf_j$, with multipliers on the left. Edge and filling relations have arbitrary multipliers on the right. This is a vector-space/module quotient, not a two-sided algebra quotient.

The actual collapsed raw rows, supplied by Dimofte–Garoufalidis, Sections 2.6 and 4.6, are

$$h_0=(2,2;1,1),\quad a=(1,1;1,0),\quad d=(-4,0;-2,0).$$

They represent one edge, the meridian, and the canonical longitude. Their native scalar factors are respectively $1,B^{-1},B^2$, multiplied by central products of the $\sigma_j$. These are the retained native half-root conventions, not a replacement by the source’s geometric eigenvalue signs. For meridian $a$, choose complementary longitude $-d$.

The preceding exact cyclic automorphism is

$$C(u_z,u_w)=(-u_w,u_z-u_w),\qquad \Psi(W(u))=B^{\sum u_w}W(Cu).$$

It fixes the central ring. It has order three, preserves Weyl multiplication, and satisfies $\Psi(f_j)=-w_j^{-2}f_j$, preserving the tetrahedron relation side. After this automorphism the selected moment and insertion data have the form

$$h=(-1,-1;1,1),\qquad s=(-1,0;0,1),\qquad b=(-2,0;2,0),$$

$$E=W(h)-\epsilon,\qquad M=W(s)+A^2\zeta,\qquad l'=\xi W(b), \qquad \epsilon^2=\zeta^2=\xi^2=1.$$

The three central units are the actual products of opposite-root signs. Their exponents need not be chosen: the following theorem holds for arbitrary such products. In particular $(l')^2=W(2b)$. The actual edge generator is a central unit multiple of $E$; the same applies to the normalized filling generator $M$.

2. A universal squared-moment certificate

Theorem. In the quantum torus over any commutative $k$-algebra, there are finite Laurent polynomials $U,V$, with coefficients already in $k$, such that

$$1-W(-4,0;4,0) =G+\bigl(W(2h)-1\bigr)U+\bigl(W(2s)-A^4\bigr)V, \tag{1}$$

where $G\in Sf_1+Sf_2$. Consequently this vacuum lies in

$$Sf_1+Sf_2+(W(h)-\epsilon)S+(W(s)+A^2\zeta)S \tag{2}$$

for any central involutions $\epsilon,\zeta$.

The full finite polynomials are in central-certificate.json. The construction below also defines them without relying on a computation.

Normalization using only even moment powers

Put $\mathfrak q=A^4$. In the quotient by the right-multiplied squared moment relations, a Weyl monomial with even first two coordinates has

$$u=e h+n s+k,\quad e=-u_{z2},\quad n=-u_{z1}+u_{z2},\quad k=(0,0;u_{w1}+u_{z2},u_{w2}+u_{z1}).$$

Both $e,n$ are even. The two moments commute and hence

$$N(W(u))=A^{-\Omega(eh+ns,k)/2+2n}W(k). \tag{3}$$

This formula uses only $W(2h)=1$ and $W(2s)=A^4$, acting on the left of an arbitrary right multiplier. It makes no substitution inside an algebra quotient. The residual $W(u)-N(W(u))$ is explicitly in the right-multiplied relation span by the telescoping formula below.

Let $e_{a,b}$ denote the formal remaining monomial $W(0,0;a,b)$. Left multiplying the tetrahedron polynomials by $W(0,0;a,b)$ and applying (3) gives

$$e_{a+2,b}-e_{a,b}+A^{4-2a}e_{a,b-2}, \qquad e_{a,b+2}-e_{a,b}+A^{-4-2b}e_{a-2,b}. \tag{4}$$

Define $V_{j,k}=e_{2j+2,2k-2}$. The two normalized relations become

$$R_1(j,k)=V_{j+1,k}-V_{j,k}+\mathfrak q^{-j}V_{j,k-1},$$ $$R_2(j,k)=V_{j,k+1}-V_{j,k}+\mathfrak q^{-k}V_{j-1,k}.$$

Direct cancellation of these formal basis vectors gives

$$\begin{aligned} V_{-1,1}-V_{1,-1}={}&\mathfrak q\bigl(R_1(1,1)-R_2(2,0)-R_1(1,0)\\ &\quad+R_1(0,2)-R_2(1,1)+R_2(0,1)\bigr)\\ &-R_1(0,1)+R_2(1,0). \end{aligned} \tag{5}$$

For an algebraic derivation, subtract the two recurrences around a unit square. With $C_{j,k}=R_1(j,k+1)-R_2(j+1,k)+R_2(j,k)-R_1(j,k)$, the right side of (5) is $\mathfrak q(C_{1,0}+C_{0,1})+(\mathfrak q-1)(R_1(0,1)-R_2(1,0))$. Expanding cancels every basis vector except the two displayed on the left.

Replace each $R_j(r,t)$ in (5) by $W(0,0;2r+2,2t-2)f_j$, retaining its displayed coefficient. This defines the eight-term $G$. Formula (3) gives $N(G)=N(1-W(2b))=V_{-1,1}-V_{1,-1}$.

Explicit right multipliers

For a commuting monomial $H$, scalar $c$, and integer $r$, define

$$P_r(H,c)= \begin{cases} \sum_{j=0}^{r-1}H^jc^{r-1-j},&r\geq0,\\ -\sum_{j=r}^{-1}H^jc^{r-1-j},&r<0. \end{cases}$$

The empty sum is zero and $(H-c)P_r(H,c)=H^r-c^r$, also for negative $r$.

For each term $a_uW(u)$ of $1-W(2b)-G$, use the decomposition in (3) and add to $U,V$, respectively,

$$a_uA^{-\Omega(eh+ns,k)/2} P_{e/2}(W(2h),1)W(ns)W(k),$$ $$a_uA^{-\Omega(eh+ns,k)/2} P_{n/2}(W(2s),A^4)W(k).$$

Factoring $W(eh+ns)-A^{2n}$ in this order proves (1), since the summed normal forms vanish by (5). This is a finite polynomial identity; no convergence or completion is used.

Finally,

$$W(2h)-1=(W(h)-\epsilon)(W(h)+\epsilon),$$ $$W(2s)-A^4=(W(s)+A^2\zeta)(W(s)-A^2\zeta). \tag{6}$$

These factors place the original relation generators on the left of their right multipliers, as required. Equations (1) and (6) prove (2). This proof works over the universal involution ring itself, including any further quotient of that ring. It does not deduce zero upstairs from zero after a projection. ∎

3. Applying the identity to the full six-root target

Take the actual opposite-root involution ring and substitute its actual central products for $\epsilon,\zeta,\xi$. Equation (1) lies in the relation span of a single actual edge and the actual meridian, together with the two tetrahedron relations. Adding the remaining edge relations cannot invalidate membership. Inverse cyclic transport returns the identity to the original native presentation and the actual class $1-l^2$.

The reverse meridian orientation is covered without a slope search. Exchanging the two tetrahedra fixes $h$ and sends $s$ to $h-s$. Thus the exchanged squared meridian generator is

$$W(2h-2s)-A^4 =(W(-2s)-A^4)W(2h)+A^4(W(2h)-1).$$

The exchanged longitude-square exponent differs from the reversed exponent by $4h$, with zero symplectic pairing. The difference of their vacuum representatives is therefore a right multiple of $W(2h)-1$. This yields the same certificate for the reverse filling with arbitrary central characters. The verifier checks this reversed identity as well.

Nonzero target is a separate retained fact. The all-plus meridian functional from the preceding cyclic note has $H(1)=-\mathfrak q^2+\text{higher order}\ne0$, with a proved all-input quadratic convergence bound. Pullback through the central projection proves $[1]\ne0$ in the full cokernel. No converse to projection is used.

Under the retained generic filling isomorphism

$$Q_\mu\longrightarrow C_\mu^6,\qquad [x]\longmapsto[(1-l^2)x],$$

the image of the empty skein is zero, while $Q_\mu\ne0$. Meridian filling restores the original $S^3$; its ordinary skein module over $k$ is generated by the empty skein, as follows from crossing resolution and the loop relation. Hence this particular descended trace is the zero map. This consequence retains all hypotheses of the native comparison and boundary-balanced filling construction; the finite identity alone is not a new proof of those geometric inputs.

4. The exact next distinction: minimal compatible versus boundary-balanced

There is a simple structural translation that prevents overstating this failure. Let $T:V\to Q$ be the unfilled trace and $p:V\twoheadrightarrow W$ the topological filling map, all over $k$. Among linear quotients of $Q$ that admit a commuting descended map, the universal one is

$$D=Q/T(\ker p). \tag{7}$$

Its induced trace satisfies

$$\ker\overline T=p(\ker T),\qquad 0\longrightarrow W/p(\ker T)\longrightarrow D \longrightarrow Q/\operatorname{im}T\longrightarrow0. \tag{8}$$

Indeed, $T(v)\in T(\ker p)$ precisely when $v$ differs from an element of $\ker T$ by an element of $\ker p$. This proves the kernel identity; the final quotient is then immediate. No claim about a stronger quotient category or integral coefficient ring follows merely from (7).

For meridian filling, $W=k$ and $p$ is Kauffman-bracket evaluation after filling. Thus the minimal compatible trace is nonzero exactly when

$$\ker T\subseteq\ker p. \tag{9}$$

The boundary-balanced construction kills $XQ+YQ$, not just $T(\ker p)$. The inclusion $T(\ker p)\subseteq XQ+YQ$ follows from its retained commuting square. Equality has not been proved. Consequently the zero vacuum calculated here does not settle (9).

This is a real logical distinction. For $V=k^2$, $Q=k^3$, $T(a,b)=(a,b,0)$, and $p(a,b)=a$, the minimal quotient kills only the second coordinate, and the descended trace is nonzero. The larger quotient killing the first two coordinates is also nonzero, but its descended trace is zero.

For the actual meridian construction, its nonzero target and zero descended trace imply $Q/\operatorname{im}T\ne0$. As a linear quotient this alone admits a zero compatible map for any filling of that same triangulated complement. It illustrates why a nonzero quotient is weaker than a useful nonzero trace, not an unrestricted solution or refutation of the source conjecture. The next decisive question is (9) for the actual named trace: construct a trace-kernel skein with nonzero filled bracket, or prove bracket evaluation factors through the trace image.

Dependencies, verification, and scope

• The fixed datum is classical: Dimofte–Garoufalidis, The quantum content of the gluing equations, arXiv:1202.6268v2 (25 October 2012), Sections 2.6 and 4.6. Those sections were checked again on 8 September 2026. The native square-root dictionary is retained from the Lab’s preceding notes, not attributed to that paper.
• Central-vertex elimination and the named six-root comparison are retained from the local collar note. Its GY, CK, and PP formal dependencies and convention qualifications remain in force.
• The cyclic map, eight-relation starting combination, and nonzero all-plus meridian unit are retained from the cyclic-polarization note. The new contribution is the squared-moment lift, not a new local seed or a new triangulation.
• The filling isomorphism is Theorem 8.1 of the generic boundary reconstruction, with its Frohman–Gelca and excision inputs.
• Panitch–Romo, On the length conjecture for twist knots, arXiv:2609.02706v1 (2 September 2026), Section 7, was checked again on 8 September 2026. Its literal nontrivial-quotient condition must not be silently strengthened to nonzero trace.

verify-central-meridian.py constructs the full finite certificate over formal central involutions and verifies it both in Weyl multiplication and ordered-monomial multiplication. These are two arithmetic checks in one internally authored verifier, not independent specialist review. It also checks inverse cyclic transport and reversed orientation. The proof is the finite identity and its derivation, not a count of tests.

No sign-sector enumeration, manifold census, numerical truncation, alteration of native signs, root-of-unity claim, triangulation-invariance claim, or new Publication is involved. The unrestricted Panitch–Romo problem remains open. The superseded question of the full meridian vacuum in this particular boundary-balanced model is answered: it is zero.