Clean excess intersections extinguish point-local meridian detectors

FPRD Lab, 8 September 2026. Supporting ATTACK research note. AI-assisted and internally audited, without external specialist review or author endorsement. This does not reach the Publication threshold. Novelty and first priority are unassessed.

Outcome and claim boundary

The generic six-root filling construction is retained, but unrestricted nonvanishing is not proved. This continuation identifies a structural limit of the previous deformation method:

At a signed classical shape point where the tetrahedron and independent edge/filling equations have a smooth scheme-theoretic intersection of excess dimension, the mixed-sided formal-local quotient is annihilated by the deformation parameter. Consequently it is zero after that parameter is inverted.

The statement applies conditionally to the actual six-root relation presentation, not merely to an arbitrary collection of commuting operators. It rules out this particular point-local detector at every point satisfying its hypotheses. It does not imply that the original algebraic meridian cokernel vanishes. A completely explicit quantum-torus example has nonzero global quotient and zero generic formal-local quotients at all its classical intersection points.

A second obstruction is independent of completion. A nonzero cyclic peripheral quantum-torus module can have the correct nonempty classical filling support while its generic meridian operator is surjective. Its nonvanishing depends on an exact quantum resonance not visible in the classical fiber.

These results explain why neither classical shape existence, peripheral linearity, nor a naive continuation of the regular-intersection argument supplies the missing theorem. Panitch–Romo Conjecture 7.1 remains open in this work. The explicit algebraic examples below are not asserted to arise from manifold triangulations or to refute that conjecture.

1. The actual six-root target remains the object of interest

Use the conventions and named action of the preceding signed-six-root packet. Over

$$k=\mathbb Q(i,A^{1/2}),$$

the native GY target has all four central vertex values $iA$, the left-generated tetrahedron relations

$$f_T=z_T^{-2}+(z_T'')^2-1,$$

and right-generated actual edge relations $(H_e+A^2)S_6$. Its intrinsic filling target is isomorphic to

$$C_s^6=\operatorname{coker}(m+A^2:Q_G^6\longrightarrow Q_G^6). \tag{1}$$

The inherited isomorphism sends a filled class $[v]$ to $[(1-l^2)v]$, with $ml=A^2lm$. Thus a functional detecting the unit of (1) is not automatically a functional detecting the filled vacuum: the latter must detect $1-l^2$.

Central vertex elimination leaves one central involution $\sigma_T$ per tetrahedron, together with a rank-two shape quantum torus. Since 2 is invertible, the commuting idempotents

$$e_\varepsilon=2^{-N}\prod_T(1+\varepsilon_T\sigma_T), \qquad \varepsilon_T\in\{1,-1\}, \tag{2}$$

decompose the full ambient torus, its relation subspace, and its meridian cokernel into sign sectors. Some sectors may be zero. This is a finite central decomposition, not an identification of opposite roots and not a new choice of peripheral signs.

Fix one sector. Its ambient shape torus has $2N$ invertible variables and a nondegenerate integral skew form $\Omega$, one block

$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

for each ordered pair $(z_T,z_T'')$. All central and peripheral scalar powers must be retained when expressing the edge and filling monomials in this sector.

Choose $k_0=N-r+1$ independent classical moment rows: $N-r$ edge rows and the selected filling row. Write their actual quantum relations as $g_j$, including $m+A^2$. Their monomials commute because the edge/peripheral exponent span is isotropic. Each $g_j$ is a unit scalar times a Weyl monomial plus its prescribed scalar constant. The $f_i$’s commute because different tetrahedra commute. These facts, including the actual peripheral row, are inherited from GY, Neumann’s symplectic complex, and the signed-six-root packet.

The local obstruction below already holds with this subset of the relations. Imposing any remaining actual edge relations only takes a further quotient. No unproved independent-edge redundancy in a nontrivial sign sector is needed for the obstruction.

2. The deformation and its first syzygy obstruction

Let $p$ be a nonzero signed root point for the relations under consideration at $A=1$, and put

$$R=\widehat{\mathcal O}_{(\mathbb C^*)^{2N},p},\qquad W=R[[\hbar]],\qquad A=e^{\hbar}.$$

In centered logarithmic coordinates, use the constant Poisson star product

$$u\star v=uv+\frac{\hbar}{2}\{u,v\}+O(\hbar^2), \qquad \{u,v\}=\sum_{a,b}\Omega_{ab}\partial_a u\,\partial_b v. \tag{3}$$

The full exponential formula in the earlier regular-intersection proof defines the product at every order. Its coefficients are finite-order differential operators; it is $\hbar$-adically continuous. The substitution of the generic field into $\mathbb C((\hbar))$ is injective as before.

Define mutually commuting operators on $W$:

$$D_i(w)=w\star f_i\quad(1\leq i\leq N),\qquad D_{N+j}(w)=g_j\star w\quad(1\leq j\leq k_0). \tag{4}$$

These sides are essential. The associated Koszul complex is a complex of $\mathbb C[[\hbar]]$-modules, not in general of $R$-modules. Let $s=(f_1^0,\ldots,f_N^0,g_1^0,\ldots,g_{k_0}^0)$, let $I=(s)\subset R$, and write

$$H=W\Big/\sum_aD_a(W). \tag{5}$$

For a classical syzygy $a=(a_a)$, $\sum_a a_as_a=0$, lift its coefficients to $W$. Its first obstruction is

$$\beta([a])= \left[\hbar^{-1}\sum_aD_a(\widetilde a_a)\bmod\hbar\right] \in R/I. \tag{6}$$

This is the first Koszul Bockstein, a $\mathbb C$-linear map

$$\beta:H_1(K_\bullet(s;R))\longrightarrow R/I.$$

Changing lifts adds a classical boundary to (6). Changing the syzygy by a Koszul boundary has no effect: the exactly commuting operators lift that boundary to an exact quantum cycle. This proves that (6) is well-defined. It is generally not $R/I$-linear; treating it as an ordinary multiplication map would erase the obstruction.

Lemma 2.1 — first-obstruction extinction

If $\beta$ is surjective, then

$$\hbar H=0,\qquad H[\hbar^{-1}]=0. \tag{7}$$

Proof. Put $d(v)=\sum_aD_a(v_a)$. Given $w\in W$, surjectivity of (6) supplies a lifted classical syzygy $v_0$ such that

$$dv_0=\hbar w\pmod{\hbar I+\hbar^2W}.$$

Subtracting $\hbar$ times suitable coefficient lifts removes the $\hbar I$ term. Hence there is $v\in W^{N+k_0}$ with

$$dv=\hbar w+\hbar^2w_1.$$

Apply the same construction to $-w_1$, multiply its coefficient vector by $\hbar$, and add it to $v$. Repeat. The corrections converge in the complete module $W^{N+k_0}$; continuity of $d$ gives a single coefficient vector $v_\infty$ with $dv_\infty=\hbar w$. Thus $\hbar W\subset\operatorname{im}d$. There are still only finitely many operators, each acting on one formal series. No closed-image assumption or convergence claim about an algebraic Laurent-polynomial space has been made. This proves (7). ∎

The opposite conclusion in the earlier regular-intersection theorem used $H_1(K(s;R))=0$ to prove $\hbar$-torsion-freeness. The present lemma isolates what can happen when actual non-Koszul syzygies appear.

3. Clean excess forces the obstruction to be onto

For a syzygy $a$, separate its tetrahedron and moment coefficients as $(a_i,b_j)$. Formula (3), applied to $u a$, gives

$$\beta(u[a])=u\beta([a])+V_a(u),\qquad V_a(u)=\sum_i a_i\{u,f_i^0\}\pmod I. \tag{8}$$

Indeed the derivative part before simplification is

$$\frac12\sum_i a_i\{u,f_i^0\} +\frac12\sum_j b_j\{g_j^0,u\}.$$

Differentiating $\sum_i a_if_i^0+\sum_jb_jg_j^0=0$ and restricting to $I=0$ makes the two sums equal. Every subprincipal scalar, including the derivative of $A^2$ and every retained central Weyl scalar, contributes only to $\beta([a])$, not to the vector field $V_a$.

The vector field is tangent to the intersection: it kills the tetrahedron functions because those Poisson-commute, and it kills the moment functions because on the intersection

$$\sum_i a_i\,df_i^0=-\sum_j b_j\,dg_j^0 \tag{9}$$

and the moments Poisson-commute. Thus (8) is well-defined on $R/I$. Equivalently, the ideal $I$ annihilates classical Koszul homology, by the usual wedge homotopy.

Theorem 3.1 — actual-presentation clean-excess obstruction

Suppose at $p$:

1. The $N$ tetrahedron differentials are independent, and the selected $k_0$ moment differentials are independent.
2. The scheme-theoretic common zero locus is smooth of codimension $c<N+k_0$; equivalently, $R/I$ is a formal power-series ring and $I$ has $c$ generators with independent differentials.

Then the quotient (5) satisfies $\hbar H=0$. The same holds after adding the remaining actual edge relations. Consequently the preceding point-local deformation construction cannot detect a nonzero generic meridian class at $p$.

Proof. Smoothness lets one choose $c$ of the listed functions whose classes form a basis of $I/\mathfrak m I$. They generate $I$ by Nakayama’s lemma. Every remaining listed function is an $R$-linear combination of those $c$. Because there are more than $c$ functions, this supplies a syzygy $(a_i,b_j)$ nonzero at $p$.

Independence of each separate list implies

$$\theta=\sum_i a_i(p)df_i^0(p) =-\sum_jb_j(p)dg_j^0(p)\ne0.$$

Otherwise each block of coefficients would be zero. The ambient Poisson form is nondegenerate, so the tangent vector $V_a(p)$ corresponding to $\theta$ is nonzero.

On the smooth formal ring $R/I$, choose a coordinate $t$ in whose direction $V_a$ has a unit coefficient. Divide the operator on the right side of (8) by that unit. In the remaining formal coordinates $y$, its equation becomes

$$\partial_tu+\sum_j q_j(t,y)\partial_{y_j}u+q_0(t,y)u=v.$$

After prescribing $u(0,y)$, comparing coefficients of $t^n$ determines $(n+1)u_{n+1}(y)$ from the already known coefficients $u_0,\ldots,u_n$. Characteristic zero therefore gives a unique formal solution. Hence the operator, and thus $\beta$, is surjective. Apply Lemma 2.1. Further relations only take a quotient of an already $\hbar$-annihilated module. ∎

The proof uses only the two separately independent Poisson-commuting blocks and the stated scheme-smoothness condition. It therefore also applies to other opposite-side constraint pairs with those properties; the quantum-torus illustration in Section 5 uses this same elementary argument.

For the actual tetrahedron functions, the first independence assertion holds at every nonzero shape point: the $z_i''$ logarithmic derivative is $2(z_i'')^2\ne0$, with disjoint variable pairs. Independence of the selected moment rows gives the second. For one filled end among $r$ torus ends, the expected dimension is $r-1$, so the theorem concerns a smooth common scheme of dimension strictly greater than $r-1$.

This is stronger than observing a deficient Jacobian at one point, and narrower than covering all nonregular intersections. A singular complete intersection may still be a regular sequence and remains covered by the earlier nonvanishing theorem. A smooth reduced support with hidden nilpotents does not meet the scheme-smooth hypothesis above. Singular excess intersections need their own syzygy analysis.

The theorem is conditional on such a point in a chosen sector of an actual triangulation. This note neither asserts that every actual filling has such a point nor substitutes the algebraic example below for that existence obligation.

4. Exact folded-shape witness for the mechanism

The previous positive-geometric note already exhibited the following algebraic excess intersection. It is revisited as a deformation calculation, not as a manifold example. Use four invertible variables $(x,w,y,v)$, with Poisson brackets $\{x,w\}=-xw$, $\{y,v\}=-yv$, and all cross-tetrahedron brackets zero. Put

$$f_1=x^{-2}+w^2-1,\quad f_2=y^{-2}+v^2-1,\quad U=xy,\quad V=x^{-1}w^{-1}v,$$

$$g_1=U-1,\qquad g_2=V-i. \tag{10}$$

The moment rows $(1,0,1,0)$ and $(-1,-1,0,1)$ are independent and isotropic. The exact classical syzygy is

$$x^2f_1+f_2+x^2(U+1)U^{-2}g_1 -x^2w^2(V+i)g_2=0. \tag{11}$$

The moment equations give $y=x^{-1}$, $v=ixw$, so $f_2=-x^2f_1$. The common scheme is therefore the smooth curve

$$x^{-2}+w^2=1$$

in the invertible chart. For instance $(x,w,y,v)=(5/3,4/5,3/5,4i/3)$ is a nonzero root point; the full logarithmic Jacobian has rank 3, not 4.

Quantize $U,V$ as Weyl monomials, use right multiplication by $f_1,f_2$, and left multiplication by $U-1,V-i$. For the syzygy $a$ displayed in (11), direct Poisson differentiation gives

$$\beta([a])=-2x^2w^2, \qquad \beta(u[a])=-2(x^2-1)\left(x\frac{du}{dx}+u\right) \tag{12}$$

on a local branch of the curve with coordinate $x$. In particular, the derivative coefficient is a unit at every point of this invertible chart. At $x_0=5/3$, the formal function

$$u(x)=-\frac1{4x}\log\!\left(\frac{4(x-1)}{x+1}\right) \tag{13}$$

is defined by the formal logarithm of a unit with constant term 1, and (12) gives $\beta(u[a])=1$. Theorem 3.1 gives the exact all-orders conclusion $\hbar H=0$, not only the first-order relation.

Changing the quantized moment constants to $U-e^{\alpha\hbar}$ and $V-i e^{\beta_0\hbar}$ changes the scalar part to

$$-2(1+\beta_0)x^2w^2-2\alpha x^2, \tag{14}$$

while leaving the nonzero derivative coefficient unchanged. Thus the local obstruction is not cured by adjusting the quantum scalar branch. Actual incidence-derived constants cannot be adjusted freely, and this observation does not propose doing so.

The supplied exact verifier checks (11), isotropy, the scheme reduction, the Jacobian rank, the Poisson formula (12), the two subprincipal coefficients in (14), and a rational formal-series solution of the unit equation. No row in (10) is asserted to be an actual edge or filling row of a manifold.

5. Local extinction does not imply algebraic extinction

Let

$$T=k\langle m^{\pm1},l^{\pm1}\mid ml=A^2lm\rangle, \qquad C=T\Big/(T(m-1)+(m-1)T). \tag{15}$$

As a vector space, $C\cong k$, and $[1]\ne0$. To prove this, order monomials as $m^a l^b$. Left relations set $m^{a+1}l^b\equiv m^al^b$; right relations set $A^{-2b}m^{a+1}l^b\equiv m^al^b$. Hence every $b\ne0$ class vanishes and every $b=0$ class equals $[1]$. The functional

$$F\!\left(\sum_{a,b}c_{a,b}m^al^b\right)=\sum_a c_{a,0} \tag{16}$$

annihilates both relation spaces and sends 1 to 1, proving the claim.

Nevertheless every formal-local detector at $m=1$, $l=l_0\ne0$ vanishes after $A=e^{\hbar}$, $\hbar^{-1}$ is adjoined. The intersection is the smooth one-dimensional scheme $m=1$, and the two opposite-side copies are a clean excess pair. One can see the torsion directly. With $v=\log(l/l_0)$,

$$(m-1)\star v-v\star(m-1)=2\hbar m.$$

Subtracting $2\hbar(m-1)$ exhibits $2\hbar$ in the relation image. Theorem 3.1 supplies annihilation of the entire local quotient. The global functional (16) cannot extend to these formal germs: they contain logarithmic primitives unavailable among finite Laurent polynomials.

Thus even extinction at all classical points in this example does not prove extinction of the original mixed-sided algebraic quotient. A supposed faithful-localization step would be false. The example is a warning against making that inference for $C_s^6$.

6. A cyclic quantum-resonance obstruction to a universal detector

There is also a completely algebraic obstruction to inferring nonvanishing from boundary action and classical support alone. On $Q_c=k[z^{\pm1}]$, for an integer $c$, define

$$l(z^n)=z^{n+1},\qquad m(z^n)=-A^{c+2n}z^n. \tag{17}$$

Then $ml=A^2lm$, both generators are invertible, and $Q_c$ is cyclic as a left $T$-module, generated by 1. The filling operator is diagonal:

$$(m+A^2)z^n=(A^2-A^{c+2n})z^n. \tag{18}$$

Since $A$ is indeterminate,

$$\operatorname{coker}(m+A^2;Q_c)= \begin{cases} k\cdot[z^{(2-c)/2}],&c\equiv2\pmod2,\\ 0,&c\not\equiv2\pmod2. \end{cases} \tag{19}$$

The first line is the one-dimensional span of the indicated class. For nonresonant $c$, the inverse of (18) divides each coefficient by its displayed nonzero scalar; a finite Laurent polynomial remains finite. For resonant $c$, coefficient extraction at the unique zero eigenvalue detects the cokernel.

In particular $Q_1$ has zero generic cokernel, while $Q_2$ has one-dimensional generic cokernel detected by its constant coefficient. Both have an integral Laurent lattice and the same classical specialization at $A=1$: $m=-1$ on the entire nonzero space $\mathbb C[z^{\pm1}]$. Their classical filling quotients are consequently nonzero and identical. Exact quantum shift data, not just classical support, decide the answer.

This rules out any theorem that would produce a nonzero meridian eigenfunctional for every nonzero cyclic peripheral module from only its quantum-torus law and its nonempty classical filling fiber. It does not rule out a detector using the additional actual tetrahedron and incidence relations. Neither $Q_1$ nor $Q_2$ is asserted to be such a target.

Even in the resonant case the filled vacuum test remains separate: coefficient extraction at $n=(2-c)/2$ detects $1-l^2=1-z^2$ only when $n=0$ or $n=2$.

7. Useful failures and source positioning

The following shortcuts fail for explicit reasons.

• Continue regular flatness using only exact commutation. Non-Koszul syzygies acquire the differential (8); at clean excess it is onto, giving torsion, not flatness.
• Detect with evaluation at any nondegenerate classical point. The folded model has nonzero invertible shapes and a smooth nonempty intersection, but all its formal point-local generic detectors vanish.
• Conclude the actual generic target is zero from those vanishing detectors. The global quotient (15) is a counterexample to that inference.
• Use a universal peripheral eigenfunctional. The cyclic module $Q_1$ has the correct classical filling eigenvalue but no nonzero generic eigenfunctional.
• Replace ordinary filling by a derived tensor product and declare success. Derived homology can retain different degrees. A construction in a different degree or completion is not the prescribed algebraic cokernel without an explicit comparison and a nonzero image in the required degree.
• Declare a conjecture counterexample from degenerate-chart remarks. GY’s published Section 1.4 warns that its target can degenerate, including in triangulations with univalent edges. That warning was checked, but it is not used here as an actual six-root/PP filled counterexample or as a resolution of the intended domain of PR 7.1.

The broad relationship between deformation-quantization modules, derived Lagrangian intersections and de Rham-type phenomena is established literature, not a new discovery here. In particular, Gunningham–Safronov, arXiv:2312.07595v3, Theorem 7.1 and Section 7.5 treat derived tensor products and clean intersections in a substantially richer setting. We do not invoke that theorem to identify the present algebraic quotient with their sheaf-theoretic object. The elementary proofs above instead track its particular mixed sides, first obstruction and formal completion. Classical Koszul homotopies are also standard; see Stacks Project, Section 15.29, especially Lemmas 15.29.5–15.29.6.

8. Exact frontier and recommended next mode

The new theorem is a conditional obstruction to point-local deformation detection for the actual relation system. It is not a universal obstruction to the actual generic cokernel itself. No new class of unrestricted filled six-root nonvanishing has been proved. The earlier regular-intersection and strictly positive geometric results, named peripheral action, CK comparison, and their scope remain unchanged.

Recommended next mode: ATTACK — global algebraic detection for the actual six-root meridian cokernel. Seek a Laurent/residue, global difference-equation, or suitably justified cycle-valued detector that survives the actual quantum moment shifts and annihilates every actual relation. A successful candidate must map the original algebraic quotient to a nonzero target, and must separately test $1-l^2$ if nonzero descended trace is claimed. Merely producing a nonzero derived stalk, a formal classical solution, or an abstract peripheral module is insufficient.

A tightly bounded alternative is to analyze a genuinely singular excess locus by its explicit first syzygy obstruction; Theorem 3.1 does not cover those loci. Do not restart the generic third-sign problem, enumerate manifolds/slopes/triangulations/skein configurations, infer nonvanishing from target comparison, or assume author confirmation. No contact with authors was made in this continuation.

The source ledger and hostile audit record the proof dependencies, finite-verification boundary and unassessed publication significance. Reproduce the exact algebra checks with python3 verify-clean-excess.py.