The transverse defect of a local 2-3 move

FPRD Lab | 9 September 2026 | Supporting research note

Result and scope

The local quantum-gluing map from two tetrahedra to three is not surjective. Its even cokernel has an exact, elementary description:

$$\boxed{\operatorname{coker}\phi^{\rm ev}_{2\to3} \cong \bigoplus_{\substack{\beta\in\mathbb Z^3\\\beta_1+\beta_2+\beta_3=-1}} k\,e_\beta.} \tag{1}$$

Here only the three local tetrahedron relations and the new interior edge relation are imposed on the three-tetrahedron side. Exterior edge relations are not imposed. This is a vector-space isomorphism, with an explicit action of the source quantum torus below, not an algebra structure on a mixed-sided quotient.

Consequently an index identity on the image of this move does not determine every target value. A finite rational functional detects the missing part and respects even the primitive root-edge relation. This identifies an actual local algebraic obstruction to reversing detection; it is not a counterexample to global native-index nonvanishing.

The general question from the preceding packet remains open: is the native functional nonzero for every 1-efficient triangulation of the solid torus or the product of a torus with an interval? We obtain a conditional directed-refinement extension, not an unrestricted answer. All results here remain research notes, with novelty unassessed and no independent full-proof or formal-language review.

1. Conventions and the move map

Work over $k=\mathbb Q(i,A^{1/2})$, with $q=A^4$ transcendental. On each of three tetrahedra put $Z_j=z_j^2$, $W_j=w_j^2$, so

$$Z_jW_j=q^{-1}W_jZ_j,\quad f_j=Z_j^{-1}+W_j-1,\quad E=Z_1Z_2Z_3.$$

Different blocks commute. Let $S_3$ be their Laurent quantum torus and set

$$R_3=S_3/\left(\sum_j S_3f_j+(E-q)S_3\right). \tag{2}$$

Tetrahedron relations have arbitrary multipliers on the left; the edge relation has arbitrary multipliers on the right. No two-sided-ideal argument is used.

Write $X(a,b)=\prod_{j=1}^3 Z_j^{a_j}W_j^{b_j}$, in this order. The source has two pairs $(U_1,V_1),(U_2,V_2)$, with the same commutation convention and relations $U_i^{-1}+V_i-1$. In these coordinates the map of Garoufalidis-Yu [1, Eq. (40)] becomes

$$\begin{array}{ll} \phi(U_1)=-qZ_1^{-1}W_1^{-1}W_3,& \phi(V_1)=-qW_2Z_3^{-1}W_3^{-1},\\ \phi(U_2)=-qW_1Z_3^{-1}W_3^{-1},& \phi(V_2)=-qZ_1^{-1}W_1^{-1}W_2. \end{array} \tag{3}$$

These four invertible monomials preserve the source commutation relations and commute with E. The phases in (3) include both squaring the root factor $iA$ and converting Weyl order to displayed order.

For completeness, descent of the first source tetrahedron relation has the finite correct-sided certificate

$$\begin{split} \phi(U_1^{-1}+V_1-1) ={}&-Z_1W_3^{-1}f_1-qZ_3^{-1}W_3^{-1}f_2-W_3^{-1}f_3\\ &-(E-q)Z_2^{-1}Z_3^{-1}W_3^{-1}. \end{split} \tag{4}$$

Swap blocks 1 and 3 for the second certificate. Since E commutes with the entire source image, multiplying (4) by any source image on the left preserves the required relation sides. Thus (3) induces the local map in (1).

2. The source image is exactly charge zero

Define the charge of $X(a,b)$ to be $c=b_1+b_2+b_3$. This grades the ambient torus, not the quotient (the tetrahedron relations mix charges).

The four exponent columns of (3), ordered by $(Z_1,W_1,Z_2,W_2,Z_3,W_3)$, are

$$C_1=(-1,-1,0,0,0,1),\quad C_2=(0,0,0,1,-1,-1),$$ $$C_3=(0,1,0,0,-1,-1),\quad C_4=(-1,-1,0,1,0,0).$$

Together with $h=(1,0,1,0,1,0)$, they are an integral basis of the charge-zero lattice. Indeed, for $v=(a_x,b_x,a_y,b_y,a_v,-b_x-b_y)$, its coefficients in this basis are

$$(a_y-b_y-b_x-a_v,\ a_x-b_x-a_v,\ a_y+b_x-a_x, \ b_y-a_x+b_x+a_v,\ a_y). \tag{5}$$

There is no saturation or finite-index omission. Factors corresponding to h are powers of E, which can be moved to the left and evaluated as q in (2). Therefore the image of the local source module is precisely the span in R3 of charge-zero monomials. Denote this image by V0.

3. Only one additional charge slice is needed

Use brackets for classes in (2). From the tetrahedron relations,

$$[vW_t]=[v]-[vZ_t^{-1}]. \tag{6}$$

Every positive-charge class is consequently generated by charge-zero classes, by repeatedly writing a monomial as $vW_1$.

For a charge-c monomial v, telescoping the three factors $Z_j^{-1}$ gives

$$(1-q^{-c-1})[v] =\sum_{t=1}^3[vZ_1^{-1}\cdots Z_{t-1}^{-1}W_t]. \tag{7}$$

The scalar is essential: $vE^{-1}=q^{-c}E^{-1}v$, and left evaluation of $E^{-1}$ contributes another $q^{-1}$. Every term on the right has charge c+1. For c less than -1, the scalar is nonzero, so induction reduces all negative charges to -1. At c=-1 the scalar vanishes. Thus R3 is spanned by V0 and the charge-minus-one slice.

Modulo V0, apply (6) to charge-minus-one monomials. Since $X(a,b)Z_t^{-1}=q^{-b_t}X(a-e_t,b)$, one gets

$$[X(a,b)]=q^{-a\cdot b}[W^b]\pmod{V_0},\qquad \sum b_j=-1. \tag{8}$$

This proves that the classes in (1) span. A generic-parameter assumption is needed in (7); there is no root-of-unity claim.

4. Exact dual functionals prove independence

For each integer triple s with $s_1+s_2+s_3=1$, define

$$H_s(X(a,b))= \begin{cases} \displaystyle\frac{q^{s\cdot a}}{\prod_j(q;q)_{-s_j-b_j}},&b_j\le -s_j\text{ for every }j,\\ 0,&\text{otherwise}. \end{cases} \tag{9}$$

Here $(q;q)_n=\prod_{j=1}^n(1-q^j)$. Formula (9) is a rational function for each finite Laurent input, not a regularized infinite average.

Its one-block factor obeys

$$h_s(a,b+1)=(1-q^{-s-b})h_s(a,b),\qquad h_s(a+1,b)=q^s h_s(a,b).$$

These identities, including the zero at b=-s, directly prove $H_s(vf_t)=0$ and $H_s((E-q)v)=0$ for all monomials v. The support condition forces c to be at most -1, so Hs kills V0. At charge -1,

$$\boxed{H_s(W^\beta)=\delta_{s,-\beta}.} \tag{10}$$

Equations (8)-(10) prove (1), including linear independence. In particular $H_{(1,0,0)}(W_1^{-1})=1$, whereas this functional is zero on the entire source image.

The proof also supplies a finite normal form. If c is nonnegative, the class modulo V0 is zero. Otherwise

$$[X(a,b)]\bmod V_0 =\sum_{\substack{d\in\mathbb N^3\\|d|=-1-c}} \frac{q^{-(b+d)\cdot a}}{\prod_j(q;q)_{d_j}}e_{b+d}. \tag{11}$$

Only finitely many d occur. This formula is obtained by applying the separating family (10) to the already-proved spanning expansion; it is not an interchange of infinite sums.

The primitive edge sign is retained

In the root torus let $P=z_1z_2z_3$, so the native new-edge relation is $(P+A^2)S$ and $P^2=E$. Extend Hs to the parity sectors $2\mathbb Z^6+\mathbb Zh$ by

$$\widetilde H_s(P^\epsilon X(a,b))=(-A^2)^\epsilon H_s(X(a,b)), \quad\epsilon\in\{0,1\}, \tag{12}$$

and set it to zero in all other parity sectors. Ordered monomials have a unique such expression when supported. Since $H_s(Ex)=qH_s(x)$ and $(-A^2)^2=q$, (12) annihilates $(P+A^2)r$ on the right side. It annihilates every rf_t because f_t has even parity. This proves that the witness was not created by incorrectly lifting a zero through squaring the edge relation. Opposite-edge sign sectors are not asserted to have a move isomorphism; the local all-plus projection already suffices for this obstruction.

5. A concrete representation, not a list of residual cases

The source torus acts on the cokernel by left multiplication: it centralizes E and preserves both the relation space and V0. In the basis of (1), (3) gives

$$\begin{array}{ll} U_1e_\beta=-q^{\beta_1}e_{\beta+(-1,0,1)},& V_1e_\beta=-q^{\beta_3}e_{\beta+(0,1,-1)},\\ U_2e_\beta=-q^{\beta_3}e_{\beta+(1,0,-1)},& V_2e_\beta=-q^{\beta_1}e_{\beta+(-1,1,0)}. \end{array} \tag{13}$$

Thus the missing space is an explicit quantum-difference representation on an affine copy of $\mathbb Z^2$. This is an action of the source torus, not of the source mixed-sided module as an algebra. Exterior gluing relations can impose additional relations on this space and may annihilate its classes. No global descent of Hs is claimed without those checks.

6. What can be transported for actual triangulations

Garoufalidis-Yu [1, Proposition 5.5] prove the all-input index identities

$$I_2=I_3\phi_{2\to3},\qquad I_0=I_2\phi_{0\to2}, \tag{14}$$

when the stated homology condition holds and the triangulations are 1-efficient. These are stronger than scalar boundary-index invariance, but their direction matters. Their move maps and index compatibility are external proof inputs to this paragraph; the obstruction (1) was proved above.

For the solid torus and the torus product, boundary homology surjects onto manifold homology modulo two, so that homology condition holds. The native F on even inputs agrees with I in (14). To see the normalization, write $u=2P(S_0)$, with zero middle coordinate in each block, and reindex $\lambda=-2P(S)$ over the actual integer edge lattice. Primitivity in [1, Lemma 3.3] justifies this reindexing for these manifolds. If $S=\sum k_eR_e$, $K=\sum k_e$, and p is its total middle coordinate, then

$$\chi(\lambda)^{-1}=(-1)^pA^{4K-2p},\qquad J(T((u+\lambda)/2))=(-A^2)^pJ(-S_0+S).$$

Their product is $A^{4K}J(-S_0+S)$; the Weyl factor is $q^{\omega(S_0,S)/2}$. This is exactly [1, Eq. (67)]. The earlier all-input convergence theorem justifies the reindexing without regularization. Neither a scalar match nor an assumed saturation is being substituted for this calculation.

Conditional refinement corollary. Start from either of the two actual exceptional controls in the preceding packet [2]. Along any finite sequence of 2-3 and 0-2 moves whose endpoints at every step are 1-efficient, the image of its known detecting insertion still has the same nonzero value by (14). The generic native cokernel bound therefore persists on this directed refinement class, using the prior grading theorem. All images are finite Laurent inputs, so no new completion is required.

This does not cover arbitrary undirected paths, or assert that every exceptional triangulation lies in this directed class. Even assuming ordinary connectivity would not reverse (14): (1) gives a local nonzero quotient invisible on the transported image. The Hs are not identified with the native index; the example proves failure of that inference, not native-index failure.

Next decisive question

Determine how the actual exterior edge relations act on (13), and whether the native index has a nonzero value on the surviving charge-minus-one part whenever its charge-zero restriction vanishes. An intrinsic compressing-disk/product construction may identify the necessary class. This replaces an unjustified reverse-move inference with an explicit algebraic problem; no triangulation enumeration or unrestricted connectivity premise is introduced.

Sources and evidence

1. S. Garoufalidis and T. Yu, The 3d-index of the 3d-skein module via the quantum trace map, arXiv:2406.04918; author-hosted revision dated 6 August 2025, Sections 3.3-3.5, 4.1-4.2, 5.3-5.4. Inspected 9 September 2026. The author’s publication list lists this as a preprint; do not conflate it with their published A quantum trace map for 3-manifolds. The move formulas influenced this attack; their full index compatibility and homology lemma are proof dependencies of the directed corollary, not independent checks of our new local calculation.
2. Closed normal sectors and nonzero quantum-trace cokernels, including its native comparison, grading and convergence dependencies. Supplies the two actual starting detectors and the retained global hypotheses.
3. S. Garoufalidis, C. D. Hodgson, N. R. Hoffman and J. H. Rubinstein, The 3D-index and normal surfaces, arXiv text, Sections 5 and 9. This source distinguishes scalar invariance from connectivity. No fresh theorem resolving connectivity is used here.

The accompanying verifier checks exact relation sides, the integer lattice inverse, the resonant normal form and symbolic commutation exponents. Rational evaluation controls are not a formal proof of the generic formulas. The proof above supplies those formulas; an independent reconstruction remains a maturity gate.