Cyclic polarization and the meridian vacuum boundary

FPRD Lab — supporting research note, 8 September 2026.

Result and scope

One exact cyclic change of tetrahedron roots extends the preceding one-sided detector to the missing half of the slope line. Together with the phase-free degree-gap theorem, it proves the following for the fixed published two-tetrahedron figure-eight datum.

Theorem. For every primitive slope $p\mu+q\lambda$ with $q\ne0$, the native six-root filled vacuum is nonzero. The native meridian cokernel is nonzero for every primitive slope, including the meridian. The named filled-trace conclusion for $q\ne0$ retains the earlier exact peripheral comparison and filling reconstruction as dependencies.

The meridian exposes an essential distinction. In its native all-plus central sector, a finite algebraic certificate gives

$$[1-l^2]=0,\qquad [1]\ne0. \tag{1}$$

This is stronger than failure of one scalar detector. Every functional on that sector kills the meridian vacuum. However, the sector projection need not be injective: vanishing of the full six-root meridian vacuum is not established. Thus (1) is not a counterexample to the unrestricted nonzero-target conjecture, and it must not be lifted backward through the sector projection.

The construction changes neither the native signs nor the target. It provides an explicit isomorphism of the mixed-sided presentations, with the exact image lattice, coefficient character, peripheral scalar, and vacuum transported. It uses one fixed cyclic map and symbolic slope parameters, not a census of polarizations, slopes, or triangulations.

This is an AI-assisted supporting research note without external specialist or peer review. Novelty and priority are unassessed. Panitch–Romo Conjecture 7.1 remains open in its unrestricted scope; no triangulation independence or new closed-manifold invariant is claimed.

1. Retained native setup

Use $k=\mathbb Q(i,A^{1/2})$, $K=\mathbb Q(i)((A^{1/2}))$, $B=iA$, and $\mathfrak q=A^4$. Integer slope coefficient $q$ is distinct from $\mathfrak q$. In grouped coordinates,

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

The all-plus central sector is

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

It is a vector quotient: tetrahedron relations have arbitrary left multipliers; edge and filling relations have arbitrary right multipliers. The original root lattice is $\mathbb Z^4$. We never substitute a two-sided algebra quotient or saturate the actual moment image.

The source-fixed raw rows are those of Dimofte–Garoufalidis, Sections 2.6 and 4.6, as recorded with their native conventions in the preceding actual-data note:

Here $P(R)=(R_1-R_2;R_3-R_2)$ and the native scalar is $B^{\sum R_2}$. The geometric eigenvalue lift used in the source is not substituted for this combinatorial native dictionary.

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

$$h_s=(p-4q,p;p-2q,0),\quad c_s=B^{-p+2q}, \quad\Lambda=\mathbb Zh+\mathbb Zh_s,\quad\chi(h_r)=-A^2/c_r.$$

A complementary longitude satisfies $qr-pt=1$, and

$$b=(r-4t,r;r-2t,0),\quad c_l=B^{-r+2t},\quad l=c_lW(b). \tag{3}$$

Thus $\Omega(h_s,b)=2$ and $ml=A^2lm$. The retained native collar comparison and generic filling reconstruction identify the filled vacuum with $[1-l^2]$, not the cokernel unit. A nonzero image in (2) proves its preimage is nonzero upstairs. A zero image does not prove its preimage is zero.

2. An exact cyclic map on full roots

On each tetrahedron set

$$C(u_z,u_w)=(-u_w,u_z-u_w),\qquad \Psi(W(u))=B^{\sum_j u_{wj}}W(Cu). \tag{4}$$

The block matrix $C$ is integral, symplectic, and has order three. The additive scalar in (4) makes $\Psi$ a quantum-torus algebra automorphism, with $\Psi^3=1$. This holds on every root monomial, including odd roots. Locally,

$$\Psi(z)=w,\qquad \Psi(w)=B,W(-1,-1).$$

Most importantly, the exact generator identity is

$$\boxed{\Psi(z^{-2}+w^2-1)=-w^{-2}(z^{-2}+w^2-1).} \tag{5}$$

Indeed $\Psi(z^{-2})=w^{-2}$, $\Psi(w^2)=-A^2W(-2,-2)$, and $w^{-2}z^{-2}=A^2W(-2,-2)$. The factor in (5) is an invertible left factor, so $\Psi(Sf)=Sf$ with the required sidedness.

Every edge or filling generator transforms to

$$\Psi(c_rW(h_r)+A^2)=c'_rW(h'_r)+A^2, \quad h'_r=Ch_r,\quad c'_r=c_rB^{\sum h_{r,wj}}. \tag{6}$$

Thus $\Lambda'=C\Lambda$ is the exact unsaturated image and

$$\chi'(C\xi)=B^{-\sum\xi_{wj}}\chi(\xi). \tag{7}$$

All right multipliers remain right multipliers. Equations (5)–(7) give an actual isomorphism of the mixed-sided quotients. There is no change of quantum parameter. On raw triples the operation is the cyclic permutation $(R_1,R_2,R_3)\mapsto(R_2,R_3,R_1)$, and (6) changes the prime-sum scalar to the new prime sum. This is why its phases are the original native phases in new coordinates, rather than adjustable signs.

For an average $\mathcal H'$ constructed on the transformed presentation, define

$$\widetilde{\mathcal H}(v)=\mathcal H'(\Psi(v)). \tag{8}$$

Then $\widetilde{\mathcal H}$ descends to the original target whenever $\mathcal H'$ does, and it evaluates the original vacuum through $\Psi(1-l^2)=1-(l')^2$.

3. Transformed actual data and convergence

Put $d=2q-p$. The transformed rows and scalars simplify to

$$h'=(-1,-1;1,1),\qquad c'_1=c'_2=-A^2,$$

$$h'_s=(d,0;-2q,p),\quad c'_s=1,\quad b'=(-r+2t,0;-2t,r),\quad c'_l=1. \tag{9}$$

Both edge characters are one, and the filling character is $-A^2$. Therefore

$$\Lambda'=\mathbb Zh'+\mathbb Zh'_s, \qquad\chi'(eh'+nh'_s)=(-A^2)^n. \tag{10}$$

The simplification $c'_l=1$ is the output of the exact map, not a normalization imposed on the target.

Use the unchanged one-sided seed

$$\ell_+(W(2a,2b))= \frac{(-1)^{a_1+a_2}A^{2(\|a\|^2+a_1+a_2-a\cdot b)}} {(\mathfrak q)_{a_1}(\mathfrak q)_{a_2}} \quad(a\in\mathbb Z_{\ge0}^2,\ b\in\mathbb Z^2), \tag{11}$$

zero otherwise, and average over (10) with the exact character. The one-sided seed theorem supplies its local recurrence and exact all-input convergence criterion.

For $d\ne0$, write $eh'+nh'_s=2(x,y)$. Direct elimination gives

$$e=-2x_2,\quad n=\frac{2(x_1-x_2)}d, \quad y=T'x,\qquad T'=\frac1d \begin{pmatrix}-2q&p\\p&-2q\end{pmatrix}.$$

The exact quadratic form is

$$\boxed{Q'(x)=(x_1+x_2)^2+\frac{2q}{d}(x_1-x_2)^2.} \tag{12}$$

On $x\ge0$, it is strictly positive away from zero precisely when

$$1+\frac{2q}{d}>0. \tag{13}$$

If $q/d\ge0$, positivity is immediate. Otherwise use $|x_1-x_2|\le x_1+x_2$; an axis proves necessity. At $d=0$, the filling row is a nonzero vertical vector, so the all-input criterion fails. These statements concern the transformed seed, not target vanishing.

Whenever (13) holds, strict positivity on the compact cone section supplies a quadratic lower bound on every shifted support. Each monomial input contributes only finitely many terms below any degree, with all exact root parity and image conditions imposed. The seed recurrence and justified isotropic image reindexing then annihilate every tetrahedron, edge and filling relation on its original side. Applying (8) gives all-input descent to the original presentation.

4. Vacuum degree gap and coverage of all nonmeridional slopes

The transformed character valuation is

$$\nu_A\chi'(2x,2T'x)=\frac4d(x_1-x_2).$$

Write $b'=(\alpha,\beta)$, with $\alpha=(-r+2t,0)$. Bezout’s identity gives

$$\beta-T'\alpha=\left(-\frac2d,\frac2d\right),\qquad 2\alpha\cdot(\beta-T'\alpha)+\frac4d(\alpha_1-\alpha_2)=0. \tag{14}$$

The exact complete-square identity from the preceding phase-free proof now yields

$$D'_0(x)=2Q'(x)+\left(2-\frac4d\right)x_1+ \left(2+\frac4d\right)x_2,$$

$$D'_l(a)=2Q'(a)+\left(2+\frac4d\right)a_1+ \left(2-\frac4d\right)a_2,\qquad a=x+\alpha. \tag{15}$$

The constant in the longitude formula is exactly zero with the scalar in (9). If (13) and $|d|>2$ hold, all four linear coefficients are positive. Every nonzero supported unit or longitude term has positive order. The unit endpoint contributes one. The only possible zero-coordinate longitude term would require second seed coordinate $(-2/d,2/d)$, which is nonintegral, so it is unsupported. Therefore

$$\boxed{\widetilde{\mathcal H}(1-l^2)=1+\text{positive }A\text{-order}\ne0.} \tag{16}$$

Positive order refers to exponents, not coefficient signs. Formula (16) is a sufficient transported vacuum theorem; we do not assert necessity of $|d|>2$ for a nonzero vacuum.

To cover every primitive slope with $q\ne0$, no search is needed. If $p/q\ge0$, the preceding original-polarization theorem applies because $p/q>-4$. If $p/q<0$, then

$$q/d>0,\qquad |d|=2|q|+|p|\ge3.$$

Thus (13) and (16) apply. This proves the nonmeridional part of the theorem on the fixed actual triangulation. It also extends the original detector across its null boundary $p/q=-4$.

A single control outside the old cone is $(p,q)=(-5,1)$. The verifier evaluates the original native Weyl sum through (8), including $c_s=B^7$ and supported odd-root phases. Its vacuum starts

$$1+\mathfrak q+\mathfrak q^2+2\mathfrak q^3+3\mathfrak q^4 +4\mathfrak q^5+6\mathfrak q^6+\cdots.$$

The old quadratic form is negative on an axis for this datum and slope. This control verifies actual transport across the previous convergence boundary; the symbolic proof above, not a list of sampled fillings, proves coverage.

5. The meridian average is nonzero but kills its vacuum

For $(p,q)=(1,0)$, take $(r,t)=(0,-1)$. Then

$$d=-1,\quad h'_s=(-1,0;0,1),\quad b'=(-2,0;2,0),\quad c'_l=1, \quad T'\begin{pmatrix}x_1\\x_2\end{pmatrix} =\begin{pmatrix}-x_2\\-x_1\end{pmatrix}. \tag{17}$$

Here $Q'(x)=(x_1+x_2)^2$, which is strictly positive on the nonnegative orthant minus zero. Thus the average converges on every input. The even unit support is all $x\in\mathbb Z_{\ge0}^2$: $e=-2x_2$ and $n=-2(x_1-x_2)$ are integral and even. Its phase is $(-1)^{x_1+x_2}$. Setting $N=a+b$,

$$\mathcal H'(1)=\sum_{a,b\ge0} \frac{(-1)^{a+b}\mathfrak q^{N(N-1)/2+2a}} {(\mathfrak q)_a(\mathfrak q)_b}. \tag{18}$$

The longitude-square support is likewise all nonnegative integral $a,b$, with the same phase and with exponent $N(N-1)/2+2b$. Interchanging $a,b$ therefore proves the exact identity

$$\mathcal H'(1-(l')^2)=0. \tag{19}$$

This functional is not zero. Euler’s identity, applied after grouping the already unordered-summable terms by $a$, gives

$$\mathcal H'(1)=(\mathfrak q)_\infty \sum_{a\ge1} \frac{(-1)^a\mathfrak q^{a(a-1)/2+2a}} {(\mathfrak q)_a(\mathfrak q)_{a-1}} =-\mathfrak q^2+\text{higher order}\ne0. \tag{20}$$

Indeed the inner sum in $b$ is $(\mathfrak q^a;\mathfrak q)_\infty$. Its $a=0$ term is zero; the $a=1$ term has unique lowest order two. This use of Euler’s product is legitimate in the formal topology. Reversing the meridian orientation, $(p,q)=(-1,0)$, interchanges the two degree coordinates and gives the same nonzero unit series. Hence the native cokernel is nonzero for the meridian as well as all nonmeridional slopes.

6. A finite actual-sector certificate for meridian vacuum vanishing

The stronger conclusion (1) does not rely on (19). We now give a finite identity in the relation module.

Let $H=W(h')$, $S_0=W(h'_s)$ at (17). The normalized moment generators are $H-1$ and $S_0+A^2$, both with arbitrary right multipliers. The actual first edge generator is $-A^2(H-1)$, an invertible scalar multiple. The second edge is redundant with the same exact character.

First quotient only by these moment relations. Write $e_{a,b}=[W(0,0;a,b)]$. Every full-root monomial has the exact normal form

$$[W(z_1,z_2;w_1,w_2)] =(-A^2)^{-z_1+z_2} A^{\{z_1(w_1+z_2)+z_2(w_2+z_1)\}/2} e_{w_1+z_2,w_2+z_1}. \tag{21}$$

To derive it, decompose the exponent into $-z_2h'+(-z_1+z_2)h'_s+(0,0;w_1+z_2,w_2+z_1)$, and use the Weyl factor and the moment characters. The moment rows together with the two $w$-rows form an integral basis; no lattice saturation is involved.

Arbitrary tetrahedron left multiples reduce to the relations

$$e_{a+2,b}-e_{a,b}+A^{4-2a}e_{a,b-2}=0,$$

$$e_{a,b+2}-e_{a,b}+A^{-4-2b}e_{a-2,b}=0. \tag{22}$$

These are a complete relation presentation after moment reduction. To see completeness, an arbitrary monomial left multiplier is a scalar times a moment monomial times a $w$-monomial. The moment monomial can be replaced by its character on the left of the entire product with $f_j$. Thus no additional type of tetrahedron multiple is omitted.

In the even-even parity block define

$$U_{n,m}=e_{2n,2m},\qquad V_{j,k}=U_{j+1,k-1}.$$

The two relations become the symmetric recurrence vectors

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

$$R_2(j,k)=V_{j,k+1}-V_{j,k}+\mathfrak q^{-k}V_{j-1,k}. \tag{23}$$

By (21), the specified meridian vacuum is

$$v=e_{0,0}-e_{4,-4}=V_{-1,1}-V_{1,-1}.$$

It has the following finite certificate:

$$\begin{aligned} v={}&\mathfrak q\,[R_1(1,1)-R_2(2,0)-R_1(1,0) +R_1(0,2)-R_2(1,1)+R_2(0,1)]\\ &-R_1(0,1)+R_2(1,0). \end{aligned} \tag{24}$$

This identity is verified by substituting (23); every other basis coefficient cancels. For a conceptual derivation, set

$$C(j,k)=R_1(j,k+1)-R_2(j+1,k)+R_2(j,k)-R_1(j,k).$$

Then $V_{1,0}-V_{0,1}=R_1(0,1)-R_2(1,0)$, while $v=\mathfrak q[C(1,0)+C(0,1)]+(\mathfrak q-1)(V_{1,0}-V_{0,1})$, giving (24).

The supplied meridian-certificate.json also lifts (24) to a checked identity in the full quantum Laurent algebra:

$$1-W(-4,0;4,0)=G+(W(h')-1)U+(W(h'_s)+A^2)V. \tag{25}$$

Here $G$ is exactly the eight left tetrahedron multiples in (24): $R_j(r,s)$ lifts to $W(0,0;2r+2,2s-2)f_j$. The explicit finite polynomials $U,V$ are in the certificate file. For reproducibility, they are constructed by applying (21) to each residual monomial and telescoping $X^n-c^n$ as a multiple of $X-c$, including negative integers $n$. The verifier checks (25) coefficient by coefficient with exact Weyl multiplication and then checks it after the inverse cyclic map. Thus the certificate is not merely a relation in a scalar approximation.

Equation (25), together with the nonzero unit value (20), proves (1) for the all-plus central sector. For reversed meridian orientation, exchange the tetrahedra. The filling row becomes $h'-h'_s$, equivalent modulo the edge relation to $-h'_s$; the exchanged squared longitude differs from the reversed complementary squared longitude by $4h'$, with zero Weyl pairing. Hence the same sector-vanishing conclusion holds for that orientation.

The symmetric recurrence presentation is a useful structural reduction. The current certificate needs only the specified odd vector in (24); we do not need a classification of all recurrence solutions or all odd classes.

7. What is proved, and the next question

For this fixed actual triangulation:

The table distinguishes target nonvanishing from survival of the named vacuum. The meridian example shows that nonzero target and nonzero unit do not force a nonzero vacuum, even for genuine triangulation incidence. It does not contradict the earlier angle-chamber theorems, whose hypotheses are not supplied for this case, or the earlier source-reading limitations.

The new mathematical ingredients are the exact full-root cyclic presentation map, its native character simplification, the complementary degree-gap theorem, and the finite meridian sector certificate. Euler coefficients, the source triangulation, and the underlying named-trace framework are credited in the source ledger. The computations use generic Laurent identities, one outside-cone transport control, and the meridian boundary control with rigorous quadratic tails. They do not replace the symbolic proofs or establish novelty.

The next ATTACK must address the full six-root meridian vacuum through its central sign algebra. The all-plus certificate prevents every further all-plus functional from detecting it. Determine whether the original vacuum is supported in complementary central sectors, or whether the finite certificate lifts before central specialization. Seek an intrinsic central-algebra argument rather than enumerating sign assignments, triangulations, or fillings. The earlier inversion-odd full-index question and unrestricted Panitch–Romo conjecture remain unresolved.