The minimal meridian quotient: a zero trace into a nonzero target

FPRD Lab · 8 September 2026 · ATTACK research note

Result

For the standard two-tetrahedron figure-eight triangulation, let $T:V=\operatorname{Sk}_k(S^3\setminus4_1)\to Q^6$ be the native six-root quantum trace, in the retained named-action conventions, and let $p:V\to\operatorname{Sk}_k(S^3)=k$ be meridian filling. Work at generic parameter over $k=\mathbb Q(i,A^{1/2})$.

We construct a skein $\kappa$ satisfying

$$\boxed{T(\kappa)=0,\qquad p(\kappa)=A^{-4}\ne0.} \tag{1}$$

Consequently every compatible meridian-filled trace obtained by a linear quotient of this particular unfilled target is zero. This includes the quotient imposing exactly the necessary relations:

$$\boxed{D_{\min}=Q^6/T(\ker p)=Q^6/\operatorname{im}T\ne0, \qquad \overline T:k\longrightarrow D_{\min}\text{ is zero}.} \tag{2}$$

The nonzero assertion in (2) has a separate proof: an explicit convergent index functional detects a root monomial outside the trace image. Thus replacing the earlier boundary-balanced target by the minimal compatible quotient does not restore the meridian vacuum.

The proof uses Panitch–Park’s published full six-root trace formulas, the Lab’s retained native peripheral action, and the tetrahedron-index recurrence and degree formula from the established literature. It does not lift zero backward through opposite-root identification. The calculation supplies one nontrivial compatible quotient in the literal sense of Panitch–Romo Conjecture 7.1, with a necessarily zero map. It neither resolves that conjecture for arbitrary manifolds and slopes nor supplies a nonzero invariant of the filled sphere. This is AI-assisted work with an internal self-audit; external review and novelty are unassessed.

1. The skeins and their framings

Let $\mu$ be the zero-framed meridian in the boundary collar. Let $b$ be the zero-framed oval around the lower clasp shown in Garoufalidis–Yu’s Figure 11, and let $b^{(2)}$ be its two parallel copies with the induced zero framing. Products involving $\mu$ mean boundary-collar action. They do not presume an algebra structure on a three-manifold skein module.

Define

$$\kappa=b^{(2)}+A^{-2}(\mu^2-1)b+A^{-4}\mu^2-1. \tag{3}$$

Panitch–Park use the same unframed oval with a ribbon framing one larger. Write their skeins as $B,B^{(2)}$. Their published p. 2310 explicitly records the framing change by $-1$ and its factor $-A^{-3}$. On two cables the correction is not the square of that factor. The annular skein identities are

$$b=-A^{-3}B,\qquad b^{(2)}=A^{-8}(B^{(2)}-1)+1. \tag{4}$$

For completeness, the two-strand Jones–Wenzl decomposition is $B^{(2)}=S_2(B)+1$. A positive full ribbon twist has eigenvalue $A^8$ on $S_2$ and eigenvalue one on the trivial summand. This proves (4). These follow as well from the two-strand Temperley–Lieb relation $e^2=de$, $d=-A^2-A^{-2}$, and the crossing $A1+A^{-1}e$; the verifier checks this identity. We use the published initial nine-state formula for $T(B^{(2)})$. An incorrect simplification at the end of the 2024 preprint’s cable calculation has been corrected in the journal edition and is not used here.

After restoring the removed figure-eight knot neighborhood, the oval and the meridian are split zero-framed unknots. This is visible after deleting the heavy knot from the cited planar figures. Therefore

$$p(\mu^2)=d^2,\quad p(b)=d,\quad p(\mu^2b)=d^3,\quad p(b^{(2)})=d^2.$$

Substitution in (3) gives

$$d^2+A^{-2}(d^2-1)d+A^{-4}d^2-1=A^{-4}. \tag{5}$$

In particular $\kappa$ is a nonzero skein, independently of the subsequent basis argument.

2. Coordinates and the forward map to the full target

Use grouped root coordinates $u=(z_0,z_1;w_0,w_1)$, where $w_j=z_j''$, with

$$W(u)W(v)=A^{\Omega(u,v)/2}W(u+v),\qquad \Omega(u,v)=\sum_{j=0}^1(u_{wj}v_{zj}-u_{zj}v_{wj}). \tag{6}$$

The tetrahedron relations are $f_j=z_j^{-2}+w_j^2-1$, with arbitrary left multipliers. Set

$$h=(1,1;-1,-1),\quad a=(1,0;0,-1),\quad m=W(a), \quad q=A^4,\quad S=m^2+m^{-2}. \tag{7}$$

The identified edge character is $W(h)=1$. For the proof in the full native target we will use only the squared relation $$E_2=W(2h)-1, \tag{8}$$ with arbitrary right multipliers.

Here is the exact even-coordinate dictionary from the published Panitch–Park shape variables to our GY-oriented coordinates. The map reverses multiplication, as required by the opposite module sides:

This is the retained collar/branch dictionary with the tetrahedra exchanged. Panitch–Park’s definition $\widehat X=-\widehat x^2$, together with $\widehat x\mapsto i\epsilon X$, removes the branch sign on every shape variable. The PP tetrahedron relation $\widehat Z''^{-1}+\widehat Z-1$ maps to $f_1$, and similarly for $Y$.

The actual PP edge uses two double-prime occurrences and one prime occurrence in each tetrahedron. Its squared shape monomial maps to $$[z_0^4(z'_0)^2z_1^4(z'_1)^2]=A^4W(2h).$$ Thus its edge value $A^4$ gives (8), with the displayed relation side. This is an actual incidence row, not a substituted abstract edge.

Panitch–Park Theorem 5.21 maps this even shape module forward into the full six-root module; injectivity of that map is unnecessary. Equivalently, after central vertex elimination an unsquared native edge has a central involution character $\epsilon$, and $$W(2h)-1=(W(h)-\epsilon)(W(h)+\epsilon). \tag{9}$$ The finite certificate below uses only even root monomials and (8), so it maps into the full target without setting any opposite-root involution to one.

3. Published trace representatives and the finite kernel identity

Write $Z'_j=(z'_j)^2=-A^2W(-2e_{zj}-2e_{wj})$. The published PP computations, transported by the preceding dictionary and corrected using (4), give

$$v_{1,n}=T(b)=-A^{-2}(z_0^{-2}w_1^2+w_0^2), \tag{10}$$

$$\begin{split} v_2=T(b^{(2)})={}&q^{-2}(Z'_0)^{-1}(Z'_1)^{-1} +q^{-1}\bigl(Z'_0(Z'_1)^{-1}+(Z'_0)^{-1}Z'_1\bigr)\\ &-(q^{-1}+q^{-2})\bigl((Z'_0)^{-1}+(Z'_1)^{-1}\bigr)+q^{-2}+1. \end{split} \tag{11}$$

The same formula (11) is GY’s equation (100), and its equation (99) gives the alternative representative $$v_1=-A^{-2}(z_0^{-2}w_1^2+w_0^2z_1^{-2}+w_0^2w_1^2).$$ The difference is the legitimate tetrahedron multiple $$v_{1,n}-v_1=A^{-2}w_0^2 f_1. \tag{12}$$

The retained native named-action theorem on every skein input gives $T(\mu^2x)=(S+2)T(x)$. Equality for the trace of the meridian alone would not justify this step. Its squared operator agrees with the published formula $T(\mu^2)=2+\widehat Y''\widehat Z^{-1}+\widehat Y''^{-1}\widehat Z$ under the dictionary. Since $\Omega(h,a)=0$, this operator preserves the edge relations.

Consequently a representative of $T(\kappa)$ is $$P_n=v_2+A^{-2}(S+1)v_{1,n}+A^{-4}(S+2)-1. \tag{13}$$

Here is a finite certificate. Set $$\begin{split} H_0&=A^{-4}W(0,0;2,4),\\ H_1&=A^{-4}W(0,0;2,2),\\ H_2&=A^{-8}W(2a)W(0,0;0,2),\\ H_3&=A^{-4}(S+1)W(0,0;2,0). \end{split}$$ There is a finite even Laurent polynomial $U$, explicitly recorded in kernel-certificate.json, such that $$\boxed{P_n=(H_0+H_2)f_0+(H_1+H_3)f_1+E_2U.} \tag{14}$$ All products in (14) have their original required sides. The companion verifier checks it first by Weyl multiplication and then by an independent ordered-monomial multiplication rule.

To explain the short identity, formally normalize the commuting edge and meridian monomials, retaining $m$ as an operator variable. In the even block let $V_{r,s}=W(0,0;2r,2s)$. The normalized tetrahedron relations are $$\begin{split} R_0(r,s)&=V_{r+1,s}-V_{r,s}+m^{-2}q^{-r}V_{r,s-1},\\ R_1(r,s)&=V_{r,s+1}-V_{r,s}+m^2q^{-s}V_{r-1,s}. \end{split}$$ The elementary cancellation $$V_{2,2}-(1-q^{-1}S)V_{1,1}+q^{-1} =R_0(1,2)+R_1(1,1)+q^{-1}m^2R_0(0,1) \tag{15}$$ produces the first three multiples. Equation (12) produces $H_3f_1$. Normalization is used to discover (14), not to assume multiplication of quotient classes.

For an explicit finite definition of $U$, put $P=P_n-H_3f_1$, $G=H_0f_0+H_1f_1+H_2f_0$, and expand $P-G=\sum c_{u,t}A^tW(u)$. Each $u$ is even. For $e=u_{z1}$, $k_u=u-eh$, define $$U=\sum c_{u,t}A^{t-\Omega(eh,k_u)/2} \frac{W(2h)^{e/2}-1}{W(2h)-1}\,W(k_u). \tag{16}$$ The fraction denotes the finite Laurent geometric sum, also for negative integers; no edge relation is inverted. The edge-normalized residual is exactly zero. Telescoping therefore proves (14) in the unquotiented quantum torus. This establishes (1) in the full target by the forward map of Section 2.

4. The minimal quotient and every compatible map

For any linear maps $T:V\to Q$, $p:V\twoheadrightarrow W$, the minimal required relation space is $T(\ker p)$. The induced map to $D_{\min}=Q/T(\ker p)$ has kernel $p(\ker T)$, with exact sequence $$0\longrightarrow W/p(\ker T)\longrightarrow D_{\min} \longrightarrow Q/\operatorname{im}T\longrightarrow0. \tag{17}$$ Indeed, $T(v)\in T(\ker p)$ precisely when $v-v'\in\ker T$ for some $v'\in\ker p$; applying $p$ proves the kernel statement. The cokernel is the quotient by the whole trace image.

Here $W=k$, and (1) gives $p(\ker T)=k$. More concretely, $$r=1-A^4\kappa\in\ker p,\qquad T(r)=T(1)=1. \tag{18}$$ For arbitrary $v\in V$, $v-A^4p(v)\kappa\in\ker p$ and has trace $T(v)$. Thus $T(\ker p)=\operatorname{im}T$, proving the equality in (2). If a quotient map $q:Q^6\to D$ satisfies $qT=\widetilde T p$, then $A^{-4}\widetilde T(1)=\widetilde T(p\kappa)=qT\kappa=0$. Hence $\widetilde T=0$ for every such quotient, not only the previously selected balanced one.

5. A nonzero quotient detected outside the trace image

It remains to prove $Q^6/\operatorname{im}T\ne0$. For this nonzero claim it is legitimate to project forward to the all-plus identified sector $$Q_+=S/(Sf_0+Sf_1+(W(h)-1)S). \tag{19}$$ The second edge is redundant in this actual identified presentation, as in GY-index Section 3.3.

Use the established tetrahedron index $$I_\Delta(m,e)=\sum_{j\ge\max(-e,0)} \frac{(-1)^j q^{j(j+1)/2-(j+e/2)m}}{(q;q)_j(q;q)_{j+e}}.$$ Define $\ell_0(W(2x_0,2x_1;2y_0,2y_1))= I_\Delta(y_0,-x_0)I_\Delta(y_1,-x_1)$, and set it to zero off $2\mathbb Z^4$. Its published recurrence gives $\ell_0(Sf_j)=0$. Let $$F(v)=\sum_{n\in\mathbb Z}\ell_0(W(nh)v). \tag{20}$$

This converges on every algebraic input. To check convergence explicitly, each supported translate of a fixed monomial has index triples $(-k+O(1),0,k+O(1))$ in both tetrahedra, where $n=2k+\varepsilon$ with fixed $\varepsilon$. The published minimum-degree formula for $J_\Delta$ gives total $A$-degree $8k^2+O(|k|)$; the Weyl prefactor adds only a linear term. Thus every coefficient has finitely many contributions, and the degrees are bounded below. This also justifies reindexing for the edge relation. The seed recurrence handles arbitrary left tetrahedron multiples, so $F$ is a well-defined functional on (19).

On the unit, the formula is particularly simple: $$F(1)=\sum_{k\in\mathbb Z}I_\Delta(-k,-k)^2 =1+\text{positive }A\text{-order}\ne0. \tag{21}$$ For $k\ne0$, the exact degree of the squared summand is $8k^2+4|k|>0$; for $k=0$, $I_\Delta(0,0)$ has constant coefficient one.

Take $$w=(0,0;1,-1),\qquad \eta(u)=u_{z1}+u_{w0}\pmod2.$$ Then $\Omega(w,h)=0$, $\eta(h)=\eta(a)=0$, and $\eta(w)=1$. The tetrahedron relations are even, so $\eta$ grades (19). The functional $$\Psi(v)=F(W(-w)v) \tag{22}$$ descends on the original relation sides: $W(-w)$ commutes with the edge monomial, and preserves the left tetrahedron relation space. It annihilates the entire $\eta=0$ subspace by parity support, while $$\Psi(W(w))=F(1)\ne0. \tag{23}$$

Bullock–LoFaro’s twist-knot basis specializes to the spanning family $\mu^n b^{(j)}$, $n\ge0$, $0\le j\le2$, for the figure-eight exterior. Its images have $\eta=0$: (10)–(11) are even, $T(1)=1$, and the named meridian operator $m+m^{-1}$ has $\eta(a)=0$. Thus $\Psi$ kills the whole actual trace image. Pulling $\Psi$ back from (19) gives a nonzero functional on $Q^6/\operatorname{im}T$. This proves the final assertion of (2).

This use of a projected nonzero functional is logically different from using a projected zero to infer a native kernel relation. The former is sufficient for Section 5; the latter was avoided in Sections 2–3 by the explicit squared-edge certificate.

6. What remains

The minimal meridian quotient for this fixed triangulation is now settled: it is nonzero, and its compatible filled trace is zero. No further detector can restore that trace while retaining the same unfilled named trace and commuting quotient requirement.

The next structural question is whether the unused parity character in Section 5 has an intrinsic description for general triangulations. A general nonzero quotient annihilating the full trace image would automatically give a compatible zero filled trace for every filling. Establishing the precise scope of that phenomenon would distinguish the literal nonzero-quotient conjecture from the stronger demand for a nonzero filled invariant. This requires a general image-grading theorem and a convergent detector; a census of triangulations or slopes would not answer it.

The 2024 PP preprint conjectured unrestricted injectivity of its trace. The published 2026 version replaces that with Question 7.3, asking which triangulations have an injective trace, and gives a separate degenerate splitting example. We therefore do not advertise (1) as a disproof of a current universal published conjecture. It gives a specific noninjectivity certificate for the fixed figure-eight trace, under the retained native-action input.

Sources and reproducibility

Formal source dependencies:

1. Panitch–Park, 3d quantum trace map, AGT 26 (2026), 2255–2314, Definition 5.16, Lemma 5.17, Definition 5.18, Theorem 5.21, Sections 6.2.1–6.2.3 and Figures 12–18. The current journal PDF, including printed pp. 2307 and 2310, was read and visually checked on 8 September 2026. It is the authority for the full-root traces and framing statement.
2. Garoufalidis–Yu, The 3d-index of the 3d-skein module via the quantum trace map, arXiv:2406.04918v1, Section 3.3 and equations (98)–(100), Figure 11; and A quantum trace map for 3-manifolds, arXiv:2403.12424v1, equation (108). These supplied the identified formula used to discover the finite relation and its independent convention check.
3. Bullock–LoFaro, The Kauffman bracket skein module of a twist knot exterior, AGT 5 (2005), 107–118, Theorem 2.1 and Figure 1: the actual skein-module spanning family used to kill the full trace image.
4. Garoufalidis, The 3D index of an ideal triangulation and angle structures, Definition 3.1 and Section 4.5, for the DGG tetrahedron-index recurrence; Garoufalidis–Hodgson–Hoffman–Rubinstein, The 3D-index and normal surfaces, Section 2, for its exact degree formula. The tetrahedron index and lattice-averaging approach are theirs, not new constructions of this note.
5. The retained native collar and peripheral-action proof, especially Sections 5–7: the full named meridian action on arbitrary input and branch/side transport. The preceding central-meridian note supplied the minimal-quotient question and exact sequence, not the actual trace-kernel witness.
6. Panitch–Romo, On the length conjecture for twist knots, arXiv:2609.02706v1, Conjecture 7.1, for the external problem’s literal scope.

The rank-three twist-knot basis and the familiar loss of reducible character information motivated seeking a short trace relation. The proof does not infer any quantum kernel from a classical character variety or an A-polynomial. Run python3 verify-minimal-quotient.py; the finite coefficient certificate, exact results, self-audit, and source ledger are included in the accompanying archive.