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Twisted Alexander polynomials and Reidemeister torsion

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Twisted chain complex and based Reidemeister torsion for link exteriors

Definition

Assume the Axiom of Choice as in the maximal-abelian-cover convention. Let L=K1∪⋯∪Kr⊂S3 be a nonempty oriented smooth link, with r≥1, compact exterior EL formed by deleting the interiors of pairwise-disjoint chosen closed normal-disk tubes, and basepoint b∈EL. For a knot, r=1 and EL=EK. The full meridian abelianization group is HL=H1(EL;Z)≅Zr, with the positive normal-disk meridians as its ordered basis. Choose a finite-rank free abelian group H of positive rank and an epimorphism χ:HL→H; write ψ=χ∘ab:G→H for the induced map from G=π1(EL,b). The identity choice H=HL is the full multivariable deck map; a coloring may identify its variables only through the explicitly stated map χ. Let F be a field, let d≥1, and let ρ:G→GLd(F) be a representation. Put Λ=F[H], the commutative Laurent-polynomial ring with monomial basis H, and Q=Frac⁡(Λ). On row vectors define the right F[G] action by v⋅g=vρ(g)ψ(g), where ψ(g) is central multiplication by its monomial in Λ. For a chosen finite based CW structure X on EL, orient and order its cells and choose one lift of each cell to the universal cover X~. The twisted cellular complex is

C∗ρ,ψ(X;Λd)=Λd⊗F[G]C∗(X~;F),

and its fraction-field complex is C∗ρ,ψ(X;Qd)=Q⊗ΛC∗ρ,ψ(X;Λd). The lifted cell bases and the standard coordinate basis of Fd give the based Q-vector-space bases. If this fraction-field complex is not acyclic, define its twisted torsion to be 0. If it is acyclic, choose in each degree a basis bi of Bi=im⁡∂i+1 and lifts b~i−1 of bi−1 under ∂i; with ci the based cellular basis, set

τ(C∗,c)=∏i[bi,b~i−1/ci](−1)i+1∈Q×,

where brackets denote the determinant of the change-of-basis matrix. The twisted Reidemeister torsion is this value modulo the subgroup Uρ,ψ={±det⁡(ρ(g))ψ(g)d:g∈G}⊂Q×. This exact determinant image is contained in the commonly used coarser unit subgroup {±ah:a∈det⁡(ρ(G)), h∈H}. The choice-free statement uses the full abelianization chain model first and specializes it only by the declared coefficient map χ.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The admissible Wada quotient of a deficiency-one exterior spine

Definition

Assume the Axiom of Choice convention of the twisted-chain definition. Let L be a nonempty oriented link, let EL be its chosen compact exterior, and let X be a finite based two-dimensional CW complex with one 0-cell, n≥2 oriented 1-cells x1,…,xn, and n−1 oriented 2-cells attached by based words r1,…,rn−1. The supplied data include a based cellular spine certificate: a finite sequence of elementary cellular expansions and collapses identifying the simple-homotopy type of X with the selected finite CW structure on EL. Thus an abstract group isomorphism π1(X)≅π1(EL) alone is not a spine certificate. Let G=π1(EL), let ρ:G→GLd(F) and ψ:G↠H be the representation and declared free-abelian deck map, and put Q=Frac⁡(F[H]). The associated deficiency-one presentation is ⟨x1,…,xn∣r1,…,rn−1⟩ through the certified spine map. For w∈Fn, define Fox derivatives recursively by ∂xk/∂xj=δkj, ∂(uv)/∂xj=∂u/∂xj+u ∂v/∂xj, and ∂xk−1/∂xj=−xk−1δkj. Extend them linearly to the group ring and let Φ:F[Fn]→Md(Q) be the ring homomorphism induced by xk↦ρ(xk)ψ(xk). Form the block Fox matrix J=(Φ(∂ri/∂xk))i,k, of size (n−1)d by nd. For a chosen column index j, let Jj^ be obtained by deleting its entire d-column block and let Dj=det⁡(ρ(xj)ψ(xj)−Id). The Wada quotient for this column is defined only when Dj≠0, and then is Wj=det⁡(Jj^)/Dj∈Q; a zero denominator makes that column inadmissible even when another column is valid. No abstract deficiency-one presentation is assigned exterior torsion unless its presentation complex is the stated based spine.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Torsion first homology forces twisted acyclicity for a deficiency-one exterior spine

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let X be a connected finite based two-dimensional CW complex with one 0-cell, n≥2 1-cells and n−1 2-cells, so χ(X)=0. Let G=π1(X), let ψ:G↠H≅Zm be an epimorphism with m≥1, let ρ:G→GLd(F) be a representation over a field F, set Λ=F[H] and Q=Frac⁡(Λ), and give Λd the right F[G]-action v⋅g=vρ(g)ψ(g), with ψ(g) acting by its central monomial. Form

C∗ρ,ψ(X;Λd)=Λd⊗F[G]C∗(X~;F),

where X~ is the universal cover, and set C∗ρ,ψ(X;Qd)=Q⊗ΛC∗ρ,ψ(X;Λd). If every element of H1(C∗ρ,ψ(X;Λd)) is annihilated by some nonzero element of Λ, then C∗ρ,ψ(X;Qd) is acyclic. No assumption that its degree-two homology vanishes before localization is required.

Facts & Assumptions

Given: AC; the finite based 2-complex and its cell counts; a positive-rank free abelian quotient ψ:G↠H; a field F; and the torsion hypothesis on H1 over Λ=F[H].

[A1]

AC is inherited from the maximal-abelian-cover convention used to define the coefficient complex. The finite-dimensional linear-algebra argument below itself uses no choice principle. The Axiom of Choice

[F1]

The twisted complex is formed from the chosen based universal-cover cellular chains using the right action v⋅g=vρ(g)ψ(g); the boundary of a lifted oriented 1-cell x is multiplication by ρ(x)ψ(x)−Id. Twisted chain complex and based Reidemeister torsion for link exteriors

Proof

1.1A1givenF2algebra

By [A1], use the prescribed AC-based twisted coefficient convention. Choose an ordered basis t1,…,tm of H. Then Λ is the Laurent-polynomial ring F[t1±1,…,tm±1]. Order exponent vectors lexicographically. For nonzero Laurent polynomials f,g, their finite supports have largest exponents a,b with nonzero coefficients fa,gb; the largest exponent of fg is a+b and its coefficient is fagb≠0. Thus Λ is an integral domain, and [F2] makes Q a field containing Λ.

2.1givenF1step 1.1algebra

Let z/s∈C1⊗ΛQ be a fraction-field cycle, with z∈C1 and 0≠s∈Λ. Since d1(z)/s=0 and each coordinate of C0 lies in the domain Λ, d1(z)=0. The torsion hypothesis gives 0≠f∈Λ and w∈C2 with fz=d2(w). Hence z/s=d2(w/(fs)), so every fraction-field 1-cycle is a boundary and H1(C∗⊗Q)=0.

2.2givenF1step 1.1algebra

Because ψ is onto a positive-rank free abelian group, some 1-cell generator x has ψ(x)=h≠1. Lift that oriented edge from the unique 0-cell to the universal-cover vertex; its cellular boundary is (x−1) times that vertex. Tensoring with the stated right action gives the corresponding d1 block A=ρ(x)h−Id. Its determinant is a polynomial in the distinct monomials 1,h,…,hd whose top coefficient is det⁡ρ(x)≠0, so det⁡A≠0 in Λ. Thus A is invertible over Q, this block makes d1:C1⊗Q→C0⊗Q surjective, and H0(C∗⊗Q)=0.

3.1F1step 2.1step 2.2algebra∎

The fraction-field chain dimensions are d(n−1),dn,d, whose alternating sum is d((n−1)−n+1)=0. Euler–Poincaré for this finite-dimensional complex gives dim⁡QH0−dim⁡QH1+dim⁡QH2=0. Steps 2.1 and 2.2 make the first two terms zero, hence H2=0; there are no chain groups above degree 2. The localized complex is acyclic.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Based twisted torsion changes only by determinant and deck units

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let X be a finite based CW model of a link exterior, let G=π1(X), let α(g)=ρ(g)ψ(g)∈GLd(Q) be its declared twisted coefficient representation, and suppose the based fraction-field complex is acyclic. Let U={±det⁡(α(g)):g∈G}. Changing the lift or orientation of any cell multiplies the based Reidemeister torsion by an element of U, so its class in Q×/U is unchanged. An elementary simple expansion or collapse, meaning addition or deletion of a pair of adjacent-dimensional cells whose lifted incidence block is ±g for some g∈G, has the same property.

Facts & Assumptions

Given: The finite based cellular model, its acyclic twisted complex, the fixed coefficient representation α, and the torsion convention of the preceding definition.

[A1]

AC is inherited from the maximal-abelian-cover convention; the finite basis-change argument uses only finite-dimensional linear algebra. The Axiom of Choice

[F1]

For an acyclic based complex with bases ci, torsion is the alternating product of determinants [bi,b~i−1/ci](−1)i+1; the bases are the chosen cellular lifts tensored with the fixed coefficient basis. Twisted chain complex and based Reidemeister torsion for link exteriors

[F2]

A lifted oriented cell is a basis vector in a free Q[G]-module; translating its lift by g applies α(g) to its coefficient block, and reversing its orientation multiplies that block by −Id. Twisted chain complex and based Reidemeister torsion for link exteriors

[F3]

The determinant map g↦det⁡(α(g)) is a group homomorphism to Q×, and U is its image together with the sign unit. Twisted chain complex and based Reidemeister torsion for link exteriors

Proof

1.1A1F1F2F3algebra

If a chosen lift of one q-cell is translated by g, [F2] changes exactly its d-vector block in cq by α(g). The only torsion bracket whose denominator basis changes is the degree-q bracket in [F1], so the torsion is multiplied by det⁡(α(g))(−1)q (or its inverse under the opposite bracket-order convention). This factor belongs to U by [F3]. Reversing the cell orientation gives the factor (−1)d(−1)q, also in U. Applying finitely many such changes multiplies by a product of elements of the subgroup U.

2.1F1F2F3step 1.1algebra∎

In an elementary expansion, order the new cell bases last and clear the old-cell entries of the new boundary column by adding multiples of old basis columns; these are triangular basis changes with determinant 1. Because the new q-cell is a free face, the lifted boundary matrices have the old blocks together with a single invertible diagonal block ±α(g) between the new (q+1)- and q-cell blocks; clearing the other entries in that new column by the described triangular basis changes splits off the two-term complex 0→Qd→±α(g)Qd→0. The torsion formula [F1] for this two-term summand is det⁡(α(g)) or its inverse according to the degree parity, hence lies in U by [F3]. The direct-sum torsion formula multiplies the old torsion by that factor, so the class is unchanged. A collapse is the inverse operation. Iterating proves invariance under every finite sequence of these elementary simple moves.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The evaluated Fox Jacobian is the twisted cellular boundary

Statement

Let X be the based presentation 2-complex with one vertex, oriented 1-cells x1,…,xn, and oriented 2-cells attached along words r1,…,rm in the free group Fn. Let G=⟨x1,…,xn∣r1,…,rm⟩ be its presented fundamental group, let ρ:G→GLd(F) be a representation, and let ψ:G→H be the declared free-abelian deck map. For Φ:F[Fn]→Md(Q) induced by xj↦ρ(xj)ψ(xj), the twisted differential ∂1 has blocks ρ(xj)ψ(xj)−Id, and the ith block row of ∂2 is

(Φ(∂ri∂x1),…,Φ(∂ri∂xn)),

where coefficient vectors are written as rows and differentials act on the right: the matrix of ∂1 is the stacked column of blocks ρ(xj)ψ(xj)−Id, while the matrix of ∂2 is the md-by-nd block matrix whose ith row is the displayed Fox row. The Fox derivatives use the product and inverse recursions in the Wada definition. Thus the evaluated Fox-Jacobian blocks are exactly the cellular boundary blocks of the based twisted complex of this presentation 2-complex.

Facts & Assumptions

Given: The finite based presentation 2-complex, its presentation generators and relators, ρ, ψ, and the twisted coefficient map Φ.

[A1]

AC is inherited from the twisted-chain convention; the finite word calculation uses no choice principle. The Axiom of Choice

[F1]

A presentation 2-complex has one vertex, one oriented 1-cell for each generator, and one oriented 2-cell attached by each relator word. Group presentation by generators and relations; Free group on a set of generators

[F2]

Fox derivatives satisfy ∂xk/∂xj=δkj, ∂(uv)/∂xj=∂u/∂xj+u∂v/∂xj, and ∂xk−1/∂xj=−xk−1δkj. The admissible Wada quotient of a deficiency-one exterior spine

[F3]

The group ring has basis indexed by group elements and multiplication [g][h]=[gh]; the coefficient map Φ is the ring homomorphism prescribed in the Wada definition. The group ring R[G] of finitely supported formal R-linear combinations of group elements; The group ring R[G] is a unital R-algebra with basis G, and each g∈G is a unit of R[G]; Twisted chain complex and based Reidemeister torsion for link exteriors

Proof

1.1F2F3giveninduction

For every word w∈Fn and each generator index j, the recursive Fox derivative satisfies ∑j=1n∂w∂xj(xj−1)=w−1. For w=xk this is xk−1, and for w=xk−1 it is −xk−1(xk−1)=xk−1−1. If it holds for u and v, then the product recursion in [F2] makes the sum for uv equal (u−1)+u(v−1)=uv−1. Induction on the length of a reduced word proves the identity.

1.2A1F1F3givenconstruct

Under the coefficient convention [A1], lift the vertex of X to v~. The lift of the oriented edge xj from v~ ends at xjv~, so its cellular boundary is (xj−1)v~. Tensoring with the right coefficient module of the twisted-chain definition sends this block to ρ(xj)ψ(xj)−Id. Thus the jth block of ∂1 is the stated matrix in the stacked-block convention.

2.1F1F2F3step 1.1construct

Write a relator ri as a finite word in letters xj±1. Along its lifted attaching path, a positive occurrence of xj after prefix u traverses the translate ue~j and contributes u to the xj coefficient; a negative occurrence xj−1 after prefix u traverses the reverse of uxj−1e~j and contributes −uxj−1. The product/inverse recursions [F2] give exactly these same contributions to ∂ri/∂xj term by term. Summing the finitely many occurrences shows ∂2(e~ri)=∑j(∂ri/∂xj)e~xj.

3.1F1F3step 1.2step 2.1algebra∎

Tensor the displayed cellular boundary with the coefficient module and extend scalars to Q. Since tensoring sends each group-ring coefficient to its image under the ring homomorphism Φ, the resulting ith block row is exactly (Φ(∂ri/∂x1),…,Φ(∂ri/∂xn)). This is the asserted twisted cellular boundary matrix.

LemmaStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Torsion first homology makes the localized exterior spine complex acyclic

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let L be a nonempty oriented link and let X be a certified finite based two-dimensional spine of its compact exterior with one 0-cell, n≥2 1-cells and n−1 2-cells. Let ψ:π1(X)↠H≅Zm be an epimorphism with m≥1, let ρ:π1(X)→GLd(F) be a representation over a field, and form the twisted chain complex over Λ=F[H] and its localization over Q=Frac⁡(Λ). If its degree-one homology over Λ is a torsion module, then its fraction-field complex is acyclic.

Facts & Assumptions

Given: AC; a based two-dimensional simple spine of a nonempty link exterior with one 0-cell, n 1-cells and n−1 2-cells; a positive-rank free abelian deck quotient; a field-valued representation; and the torsion hypothesis on degree-one homology.

[A1]

AC is inherited from the full abelianization convention. The Axiom of Choice

[F1]

The based twisted complex is built from the exterior spine and its prescribed coefficient action; a certified spine has the same based simple-homotopy type as the selected exterior cell structure. Twisted chain complex and based Reidemeister torsion for link exteriors; The admissible Wada quotient of a deficiency-one exterior spine

[F2]

For a finite based two-dimensional complex with one 0-cell, n 1-cells and n−1 2-cells, positive-rank free-abelian deck map and torsion H1 over the Laurent coefficient ring, localization yields an acyclic complex. Torsion first homology forces twisted acyclicity for a deficiency-one exterior spine

Proof

1.1A1F1F2givenalgebra

The stated cell counts give χ(X)=1−n+(n−1)=0, and the two-dimensional spine has no cellular chain groups above degree 2. The declared deck group has positive rank, the coefficient field and representation are as in [F1], and the degree-one torsion hypothesis is exactly the remaining module assumption of [F2].

2.1F1F2step 1.1∎

Apply [F2] to this X, ρ, and ψ. It gives Hi(C∗ρ,ψ(X;Qd))=0 in every degree; hence the fraction-field complex is acyclic.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The admissible Wada deleted minor equals based twisted torsion

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let X be a certified based deficiency-one exterior spine with one 0-cell, n≥2 1-cells and n−1 2-cells, let ψ:π1(X)↠H≅Zm have m≥1, let ρ:π1(X)→GLd(F), and let H1(C∗ρ,ψ(X;F[H]d)) be torsion over F[H]. Form the Fox block matrix J and a chosen Wada column quotient Wj=det⁡(Jj^)/Dj with Dj=det⁡(ρ(xj)ψ(xj)−Id)≠0. Then the fraction-field complex is acyclic, det⁡(Jj^)≠0, and Wj equals its based twisted Reidemeister torsion in Q×/Uρ,ψ. Every admissible deleted column gives the same class. This conclusion applies only to the specified cellular spine; an abstract presentation of the same group is insufficient.

Facts & Assumptions

Given: AC; the certified based deficiency-one exterior spine; a positive-rank free abelian deck epimorphism; a field-valued representation; torsion degree-one homology over the Laurent ring; and an index j with Dj≠0.

[A1]

AC is inherited from the maximal-abelian-cover convention. The Axiom of Choice

[F1]

J has blocks Φ(∂ri/∂xk), Dj=det⁡(ρ(xj)ψ(xj)−Id), and Wj=det⁡(Jj^)/Dj only for Dj≠0. The admissible Wada quotient of a deficiency-one exterior spine

[F2]

With row coefficient vectors, ∂1 sends y=(y1,…,yn) to ∑kykAk for Ak=ρ(xk)ψ(xk)−Id, and ∂2 acts on the right by the Fox block matrix J. The evaluated Fox Jacobian is the twisted cellular boundary

[F3]

Under the positive-rank deck and cell-count hypotheses, torsion H1 implies the fraction-field complex is acyclic. Torsion first homology forces twisted acyclicity for a deficiency-one exterior spine

[F4]

Changing cell lifts, orientations or certified simple-spine structure changes the torsion only by Uρ,ψ. Based twisted torsion changes only by determinant and deck units

[F5]

For an acyclic based complex, torsion is the alternating product of determinants of boundary bases and their lifts in the chosen cellular bases. Twisted chain complex and based Reidemeister torsion for link exteriors

Proof

1.1A1givenF3

By [A1] and [F3], the fraction-field complex 0→Q(n−1)d→∂2Qnd→∂1Qd→0 is acyclic. Therefore ∂2 is injective and its image is ker⁡∂1, both of dimension (n−1)d.

2.1givenF2step 1.1algebra

In the ordered 1-cell basis, [F2] gives the stacked d1 blocks A1,…,An; a row vector y=(y1,…,yn) lies in its kernel exactly when ∑kykAk=0. Since Aj is invertible, projection pj^:ker⁡d1→Q(n−1)d onto all non-j blocks is an isomorphism: for arbitrary non-j row coordinates yk, the unique missing coordinate is yj=−(∑k≠jykAk)Aj−1. Thus pj^∂2 is an isomorphism, and its row-coefficient matrix is Jj^ by [F2]. In particular det⁡(Jj^)≠0.

3.1F1F2F4F5step 2.1algebra

Let c0,c1,c2 be the standard cellular bases. In degree zero choose b0=c0. In degree one choose b1=∂2c2, and lift b0 to degree one using the j-block Aj−1 and zero in the other blocks. After moving the j-block to the final position, the matrix of [b1,b~0] in c1 is block triangular with diagonal blocks Jj^ and Aj−1; its determinant is ±det⁡(Jj^)/Dj. The degree-zero and degree-two torsion brackets equal 1, so [F5] gives τ=±Wj. The sign lies in Uρ,ψ by [F4], proving the equality of classes.

4.1F1F4step 3.1algebra∎

The same construction applies to every index k with Dk≠0. Each quotient class equals the single torsion class of the same based complex by step 3.1; hence all admissible columns give the same class in Q×/Uρ,ψ.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Compute the twisted Burau determinant for a colored braid closure

Definition

Inputs. An integer n≥2; a finite Artin word β∈Bn in the positive-half-twist convention where σi sends the ith meridian xi to xixi+1xi−1 and xi+1 to xi; a surjective color map c from the cycles of the endpoint permutation to {1,…,μ}, written on strands as c1,…,cn and constant around each cycle; a field F with encoded exact arithmetic and equality; a positive integer d and invertible matrices ρ(xi)∈GLd(F); and compatible finite certificates of based deficiency-one 2-spines for the closure exterior X and the closure-with-axis exterior Y. The X spine has generators gi=x1⋯xi and relators ri=gi−1β∗(gi) for 1≤i<n. The Y spine has generators g1,…,gn,x and relators si=β∗(gi)(x−1gix)−1 for 1≤i≤n; x is the meridian of the axis. The certificates include a based cell structure on T with one vertex, 1-cells x,gn, and the commutator 2-cell; the one-vertex, one-core-edge structure on V=D2×S1; and compatible cell bases for X=Y∪TV. These spine and gluing certificates are geometric input and are not recognized by the procedure. Put H=⟨t1,…,tμ⟩≅Zμ, ψ(xi)=tci, gn=x1⋯xn, Λ=F[H], and Q=Frac⁡(Λ).

Instructions.

  1. Read the supplied X and Y spine/presentation correspondences and compatible gluing as certified geometric input; check their finite cell counts and relator lists, but do not attempt to recognize whether the data describe exterior spines. Return not-applicable if either certificate is absent. Compute the induced free-group automorphism β∗ by substituting the displayed Artin rules in the declared word order. For the mapping-torus checks, set the axis meridian's representation matrix to ρ(x)=Id. Return not-applicable if the coloring is not preserved on endpoints, if the representation fails any closure relator, or if it fails any mapping-torus relator si=β∗(gi)(x−1gix)−1. Equivalently the braid invariance checks ρ(β∗(xi))=ρ(xi) must hold for all meridians.

  2. Set ψ(x)=1 on the axis meridian. Verify the deck map on the relators ri and the axis relators si, and verify the colors are surjective, so ψ descends to both spine groups and is an epimorphism. Compute the positive-meridian full abelianization basis first, then use the explicit color map HL→H; no unannounced equal-variable or total-linking specialization is performed.

  3. Rewrite β∗(gi) in the free basis g1,…,gn, using xi=gi−1−1gi and g0=1. Compute the block Fox matrix Bβ with (i,j) block Φ(∂β∗(gi)/∂gj), using the product and inverse recursions of the Wada definition and Φ(g)=ρ(g)ψ(g). The braid fixes the boundary word gn, so its last block row is (0,…,0,Id). Let B‾ρ(β) be the upper-left (n−1)d-square block.

  4. Compute D=det⁡(ρ(gn)ψ(gn)−Id). If D=0, return not-applicable: this designated deleted column is inadmissible, even if another column of the Fox matrix could be used.

  5. Form the twisted cellular boundary matrices d1,d2 of the closure-exterior spine X from the Fox lemma. Use exact Gaussian elimination over Q to check the chain identity d1∘d2=0 (the row-matrix product JAstack=0). If this identity fails, return not-applicable because the encoded data do not define a chain complex. If it holds but the ranks are not rank⁡(d1)=d and rank⁡(d2)=(n−1)d, return 0 under the declared torsion convention. Otherwise compute N=det⁡(B‾ρ(β)−I(n−1)d) and return the rational function N/D as the twisted Burau determinant value, modulo the determinant/deck units of the based torsion convention.

The braid convention is shown for two adjacent downward-oriented strands. The left strand crosses over the right strand for σi; the gap records the underpass. The word, color map, and algebraic action above, rather than the diagram, determine the computation.

xixi+1xi+1xi

The displayed generator interchanges endpoint positions, with the original left strand passing over; the colors must follow the same permutation along each link component.

TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

The braid-axis torsion product gives the twisted Burau determinant

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let β∈Bn be a geometric braid with n≥2 strands, let X be the compact exterior of its oriented closure, and let Y be the compact exterior of the closure together with the braid axis. Suppose finite based deficiency-one two-dimensional spine certificates are supplied for X and Y, with compatible cell bases for the filling decomposition X=Y∪TV, where T is the boundary torus and V≅D2×S1 is the filled solid torus. The certificate for X has generators g1,…,gn, where gi=x1⋯xi are cumulative positive meridians, and relators ri=gi−1β∗(gi) for 1≤i<n; the certificate for Y has generators g1,…,gn,x, where x is the axis meridian, and relators si=β∗(gi)(x−1gix)−1 for 1≤i≤n. The gluing certificate gives T its one-vertex cell structure with 1-cells x,gn and one commutator 2-cell, and gives V its one-vertex and one-core-edge structure, with compatible based cell bases. Assume these are actual compatible based spine and gluing certificates; an abstract group isomorphism or matching relator list alone is insufficient.

Let HL=H1(X;Z) have its ordered positive-meridian basis, let χ:HL↠H≅Zm be the declared color map with m≥1, and write ψ for the induced deck map. Let F be a field, let ρ:π1(X)→GLd(F) be a representation, and use the coefficient systems on Y,T,V pulled back or extended across the filling; thus x acts as the identity and the core gn of V acts by Mn=ρ(gn)ψ(gn). Put D=det⁡(Mn−Id) and assume D≠0. Compute the reduced colored Burau block B‾ρ(β) in the cumulative-meridian basis as in the finite procedure, and put N=det⁡(B‾ρ(β)−I(n−1)d). For the declared convention that torsion is zero on a nonacyclic fraction-field complex, if the twisted complex of X is acyclic then its torsion is represented by N/D modulo the exact determinant and deck-unit subgroup of the coefficient system. If the complex is nonacyclic then N=0 and τ(X)=N/D=0 in Q. The acyclic identity follows from the braid-axis filling product.

The diagram records only the supplied gluing certificate; it is schematic and is not a substitute for the cell or spine data.

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Facts & Assumptions

Given: AC; the actual compatible based spine and filling certificates; the full positive-meridian abelianization followed by the explicit color map; a field-valued representation of π1(X); and D≠0.

[A1]

The maximal-abelian-cover and twisted-chain conventions explicitly assume AC. The Axiom of Choice; Twisted chain complex and based Reidemeister torsion for link exteriors

[F1]

For a based acyclic twisted complex, torsion is the alternating determinant product in the cellular bases; nonacyclic torsion is defined to be zero. Twisted chain complex and based Reidemeister torsion for link exteriors

[F2]

For a certified deficiency-one exterior spine with torsion degree-one homology over the Laurent ring and an admissible deleted column, the Fox minor divided by its boundary determinant equals the based torsion class. The admissible Wada deleted minor equals based twisted torsion

[F3]

The finite braid procedure fixes the positive Artin convention, the cumulative meridians gi, the closure relators, the reduced Fox block B‾ρ(β), and the numerator and denominator N,D. Compute the twisted Burau determinant for a colored braid closure

[F4]

The closure and axis exteriors, their meridian generators, and the Artin action use the stated geometric braid and punctured-disk conventions; the assignment of generator automorphisms extends to a well-defined Artin representation and agrees with the geometric action on meridians; the standard positive puncture meridians form a free basis. The closure of a geometric braid; Artin automorphisms of the free group; The Artin representation on a free group; The geometric action on meridians is the Artin representation; Standard meridians of a punctured disk; The punctured-disk fundamental group is free on the standard meridians

[F5]

The based Fox block matrix is the cellular boundary matrix of the based presentation complex under the row-coefficient convention. The evaluated Fox Jacobian is the twisted cellular boundary

[F6]

The colored-braid algorithm forms the full positive-meridian basis before applying the declared map χ. Compute the twisted Burau determinant for a colored braid closure

[F7]

Filling the axis torus kills its meridian in the fundamental group, and the gluing fundamental group is determined by the inclusions of the common boundary torus. Seifert–van Kampen identifies the fundamental group with a group pushout; Knot exterior

[F8]

The Laurent coefficient ring is an integral domain embedded in its fraction field, so clearing a nonzero denominator in a free module is valid. The field of fractions Frac⁡(D)=(D∖{0})−1D of an integral domain; Frac⁡(D) is a field and d↦d/1 embeds the integral domain D

[F9]

External relative-cellular gluing input. For the actual braid-axis filling X=Y∪T(D2×S1) with agreeing local coefficients, use finite CW representatives relative to the common boundary torus, with the based simple comparisons to the supplied exterior spines. The cellular inclusion and difference maps give the degreewise split exact gluing sequence. The based comparison is with those relative models, not with a core circle as a boundary-containing subcomplex. This is the geometric gluing/excision operation used in Conway, Burau Maps and Twisted Alexander Polynomials, arXiv:1510.06678v2, proof of Theorem 3.15, printed/PDF pp. 16–17; its based-chain multiplicativity is Proposition 2.4, printed/PDF p. 7. This finite relative-cellular comparison is an explicit external input, not a locally reconstructed relative-CW theorem. In the present proof the solid-torus comparison is additionally computed in step 3.1. The cellular operation applies after the declared color map as well: its degreewise splittings use individual lifted cells, so applying the coefficient representation preserves exactness; the only inverse needed in the solid-torus computation is the assumed invertible Mn−Id. No acyclicity is assumed before that specialization. Conway, Proposition 2.4 and proof of Theorem 3.15

Proof

1.1A1givenF3F4F6F7

Let α(g)=ρ(g)ψ(g) denote the coefficient matrix. Under [A1], the supplied filling certificate and [F7] identify π1(X) as the quotient of π1(Y) that kills the axis meridian x. Pulling the coefficient system back from X therefore gives α(x)=Id and ψ(x)=1; the core of V is represented on T by gn, with matrix Mn. The algorithm forms the full positive-meridian basis on X before applying χ, while the compatible filling certificate specifies the resulting pulled-back or extended coefficient systems on Y,T,V and makes them agree on overlaps. The map π1(Y)→π1(X) is surjective, so the pulled-back deck map on Y is still onto and the coefficient-unit images for the two exteriors agree. Since D≠0, Mn−Id is invertible.

2.1F1F5step 1.1algebra

First compute the core-circle spine Vcore with one 0-cell and one 1-cell for gn. Its twisted boundary is Mn−Id, so this core complex is acyclic and its torsion is D−1. This core is not used as a CW pair containing T; a separate relative model and its torsion comparison are established before applying the gluing sequence. Give T the standard one-vertex CW structure with 1-cells x,gn and one commutator 2-cell xgnx−1gn−1. The Fox boundary blocks are d1=(0,Mn−Id)T and d2=(Id−Mn,0): hence d1 is onto, d2 is injective onto ker⁡d1, and T is acyclic. In the torsion determinant product, the factors det⁡(Id−Mn) and det⁡((Mn−Id)−1) cancel up to (−1)d, so τ(T)=(−1)d.

2.2F3F5step 1.1algebra

For X, Fox differentiation of ri=gi−1β∗(gi) gives block row Mi−1(B~i,∗−ei), where ei is the ith identity block row and Mi=ρ(gi)ψ(gi). Thus, after deleting the gn column, the Fox block is diag⁡(M1−1,…,Mn−1−1)(B‾ρ(β)−I), whose determinant is (∏i=1n−1det⁡(Mi−1))N. The prefactor is a nonzero coefficient unit. The d1 block for gn is Mn−Id, which is invertible, so d1 is onto and projection from ker⁡d1 to the first n−1 coordinate blocks is an isomorphism. By [F5], the projected d2 is the displayed deleted Fox block. Therefore N≠0 implies d2 injective and, by dimensions and d1d2=0, the complex is acyclic; if the complex is acyclic, the same projection makes d2 injective and forces N≠0. Hence nonacyclicity is equivalent to N=0.

3.1F1F5givenstep 1.1step 2.1constructalgebra

Give D2 its boundary vertex, boundary edge x, and meridian disk cell a, and give S1 its vertex and core edge gn. Their product is a finite CW model Vrel of the actual solid torus containing exactly the specified torus CW structure as its boundary subcomplex. Besides the cells of T, it has the 2-cell a×{∗}, still denoted a, and the 3-cell b=a×gn. Put A=Mn−Id and let q denote the torus 2-cell, with the orientation in step 2.1. For a coefficient row v, the product boundaries are ∂a(v)=x(v),∂q(v)=x(−vA),∂b(v)=q(v)+a(vA). The first formula is the meridian disk boundary, the second is the torus Fox boundary from step 2.1, and the third is the oriented boundary of disk times core edge; its two terms have cancelling images under ∂. Replace the degree-two block q(v) by q′(v)=q(v)+a(vA). This block-triangular basis change has determinant one, ∂q′=0, and ∂b(v)=q′(v). Split off the unit two-term pair (b,q′), then the unit pair (a,x). The remaining based complex is exactly the core complex with ∂gn(v)=vA. Each split pair has identity incidence and torsion one up to the sign unit, and the triangular change contributes determinant one. Hence Vrel has the same homology and torsion class as Vcore, so it is acyclic and its torsion is D−1 up to the declared units. This calculation supplies the missing based comparison while retaining T in the actual gluing model.

4.1F9givenstep 2.1step 3.1algebra

Use the relative-cellular gluing operation [F9] for the actual decomposition in the supplied certificate. For V use Vrel from step 3.1, not its core spine. For the exterior pieces the operation uses the compatible relative CW representatives and the supplied based simple-spine comparisons. Equivalently, a boundary map into a spine is represented by its finite cellular mapping cylinder: each added prism over a torus cell pairs with its source face, with lifted incidence a unit; these pairs are removed in decreasing cell dimension. Their torsion changes lie in the declared unit subgroup. Thus the based torsion classes used below are those of the supplied spines, without identifying the boundary torus with a subcomplex of a core circle or of an abstract presentation. In the common relative models each cell of T occurs in both pieces, and each other cell occurs in exactly one. Inclusion into the first piece and its negative into the second, followed by the difference quotient to the union, therefore give the degreewise split short exact sequence 0⟶C∗(T)⟶C∗(Yrel)⊕C∗(Vrel)⟶C∗(Xrel)⟶0. Choose the shared lifts compatibly; any permitted adjustment changes only the declared determinant/deck units. Since T and Vrel are acyclic by steps 2.1 and 3.1, the long exact homology sequence identifies H∗(Y;Qd) with H∗(X;Qd). Thus Y is acyclic exactly when X is. Subsequent references to the gluing sequence mean these relative models and their certified based comparisons.

5.1F1step 2.1step 4.1algebra

In the acyclic case, choose boundary bases in the short exact sequence compatibly: lift a basis of each quotient boundary space, append a basis of the subcomplex boundary space, and lift the resulting boundary bases to chain groups. The change-of-basis matrices are block triangular with diagonal blocks the matrices for the subcomplex and quotient, so their determinants multiply degree by degree; the alternating torsion product therefore gives τ(Y)τ(V)=±τ(T)τ(X). As τ(T)=±1 and both signs lie in the declared unit subgroup, step 2.1 yields τ(Y)≐τ(X)D.

5.2givenF1F8step 4.1algebra

Suppose X and hence Y are acyclic over Q. The finite based chain modules over Λ=F[H] are free. If z∈C1(Y;Λd) is a 1-cycle, acyclicity over Q gives z/1=d2(y)/s for some y∈C2(Y;Λd) and nonzero s∈Λ. By [F8], equality in the fraction field means that some nonzero u∈Λ annihilates sz−d2(y); since C1(Y;Λd) is free over the domain Λ, multiplication by u is injective, so sz=d2(y). Every degree-one homology class is therefore killed by a nonzero element of Λ.

6.1F2F3F5step 5.2algebra

Apply [F2] to the certified spine of Y and delete the gn column, whose denominator is D≠0. For si=β∗(gi)(x−1gix)−1, the g1,…,gn Fox blocks evaluate to B~−I, where B~ is the full cumulative-meridian Fox matrix, and the x-column blocks are Id−Mi with Mi=ρ(gi)ψ(gi). Because β∗(gn)=gn, the last row of B~−I is zero; after deleting its gn column, the remaining matrix has upper-left block B‾ρ(β)−I, arbitrary upper-right block, zero lower-left block, and lower-right block Id−Mn. It is block upper triangular, so its determinant is (−1)dND. Dividing by D gives (−1)dN, which represents N modulo the sign unit. Consequently τ(Y)≐N.

7.1F1step 5.1step 6.1step 2.2algebra∎

If the complex is acyclic, combine steps 5.1 and 6.1 to obtain τ(X)D≐N, hence τ(X)≐N/D. If it is nonacyclic, step 2.2 gives N=0, and the declared convention sets τ(X)=0=N/D. This proves the statement in both branches.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A certified knot-spine Wada quotient equals twisted Reidemeister torsion

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let K⊂S3 be an oriented knot, let EK be its compact exterior, let X be a finite based deficiency-one two-dimensional spine with one vertex, n≥2 oriented 1-cells and n−1 oriented 2-cells, certified by a based cellular simple-homotopy equivalence with EK, and let ψ:π1(EK)↠H≅Zm have m≥1. Let ρ:π1(EK)→GLd(F) be a representation over a field, and assume the degree-one twisted homology over Λ=F[H] is a torsion module. For any admissible generator column of the Fox block matrix of the certified spine, its deleted-minor Wada quotient represents the based twisted Reidemeister torsion of EK in Q×/Uρ,ψ, where Q=Frac⁡(Λ) and Uρ,ψ={±det⁡(ρ(g))ψ(g)d:g∈π1(EK)}. All admissible columns give the same class. If two such deficiency-one spines are related by a supplied finite based simple cellular equivalence carrying the same representation and deck map, their Wada quotient classes also agree. A group-presentation isomorphism alone is not such an equivalence.

Facts & Assumptions

Given: AC; the finite based knot-exterior spine certificate; a positive-rank free-abelian deck epimorphism; a field-valued representation; degree-one Laurent-ring homology torsion; and an admissible deleted column.

[A1]

AC is inherited from the maximal-abelian-cover convention. The Axiom of Choice; Twisted chain complex and based Reidemeister torsion for link exteriors

[F1]

The evaluated Fox block matrix of a based presentation complex is the cellular boundary matrix of its twisted chain complex, with the stated row-coefficient convention. The evaluated Fox Jacobian is the twisted cellular boundary

[F2]

Torsion degree-one homology, positive-rank deck map, and the stated deficiency-one cell counts imply acyclicity over the fraction field. Torsion first homology makes the localized exterior spine complex acyclic

[F3]

For an admissible deleted column of a certified deficiency-one spine, the Fox minor divided by its boundary determinant is the based twisted torsion class, and all admissible columns agree. The admissible Wada deleted minor equals based twisted torsion

[F4]

Changing lifts, orientations or the certified simple cellular model changes torsion only by an element of the determinant and deck-unit subgroup. Based twisted torsion changes only by determinant and deck units

[F5]

The twisted-chain definition gives the based fraction-field torsion and its exact unit quotient. Twisted chain complex and based Reidemeister torsion for link exteriors

Proof

1.1givenF1

The supplied simple-spine certificate identifies the cellular presentation complex with the selected knot exterior model and transports the based cells, representation, and deck map. Therefore [F1] identifies the evaluated Fox matrix with the actual twisted cellular boundary matrix of this model; no conclusion here follows merely from an isomorphism of presented groups.

1.2A1F2F5given

The spine has one 0-cell, n 1-cells and n−1 2-cells, and the deck map has positive rank. The given torsion hypothesis is exactly the degree-one module hypothesis in [F2]. Hence its fraction-field complex is acyclic, so the torsion class in [F5] is defined and nonzero.

2.1F1F3step 1.2

The chosen column is admissible by hypothesis. Apply [F3] to the based spine and the acyclic complex from step 1.2. It states that the deleted Fox minor divided by the selected d1 block determinant represents the torsion class in Q×/Uρ,ψ, and it proves the same conclusion for every other admissible column.

3.1F3F4F5step 2.1algebra∎

If a second certified deficiency-one spine is related to the first by the stated finite based simple cellular equivalence, [F4] makes its based torsion class equal in the same quotient; the representation and deck map are transported by that equivalence, so the subgroup Uρ,ψ is unchanged. Applying step 2.1 to the second spine shows its admissible Wada quotient has that same class. This proves the permitted presentation-change assertion without promoting a group isomorphism to a spine equivalence.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The colored twisted Burau procedure computes exterior torsion

Statement

Assume the Axiom of Choice convention of the twisted-chain definition. Let a finite colored geometric braid, a field with exact arithmetic, a representation of the closure exterior group, the full positive-meridian abelianization followed by a declared surjective color map, and compatible based spine and axis-filling certificates satisfy all input conditions of the colored twisted Burau procedure. Assume its designated denominator D=det⁡(ρ(gn)ψ(gn)−Id) is nonzero, so the procedure is applicable, and assume the representation and deck map extend over the filling with the axis meridian acting trivially. Then the procedure returns the class of the twisted Reidemeister torsion of the closure exterior. If the fraction-field complex is acyclic, the returned value is N/D, where N=det⁡(B‾ρ(β)−I(n−1)d); if it is nonacyclic, the procedure returns 0 and N=0, in agreement with the declared zero-torsion convention. The geometric certificates are essential hypotheses: matching a group presentation alone does not certify an exterior spine or the axis-filling decomposition.

Facts & Assumptions

Given: AC; all finite encoded procedure inputs; a nonzero designated denominator; the actual based spine and compatible axis-filling certificates; and the stated coefficient extension across filling.

[A1]

AC is inherited from the full abelianization and twisted-chain convention. The Axiom of Choice; Twisted chain complex and based Reidemeister torsion for link exteriors

[F1]

The finite procedure checks its declared input conditions, evaluates D before the acyclicity branch, returns zero for a nonacyclic complex, and otherwise returns N/D. Compute the twisted Burau determinant for a colored braid closure

[F2]

For the certified closure and axis exteriors with D≠0, the braid-axis torsion product identifies the exterior torsion with N/D on the acyclic branch and proves N=0 on the nonacyclic branch. The braid-axis torsion product gives the twisted Burau determinant

[F3]

The twisted torsion convention assigns zero to a nonacyclic fraction-field complex and takes acyclic torsion modulo the determinant and deck-unit subgroup. Twisted chain complex and based Reidemeister torsion for link exteriors

[F4]

The colored-braid algorithm forms the full positive-meridian abelianization before applying the declared color map. Compute the twisted Burau determinant for a colored braid closure

Proof

1.1A1givenF1F4

By [A1] and the given input hypotheses, all finite spine, gluing, color and representation checks in [F1] pass. Since D≠0, the designated deleted column is admissible and the procedure does not return not-applicable. The coefficient system is the full positive-meridian system followed by the declared color map, as required by [F4], and the axis meridian acts trivially by the extension hypothesis.

2.1F1F2F3step 1.1cases∎

If the fraction-field complex is acyclic, [F1] returns N/D and [F2] identifies this value with the twisted torsion class. If the complex is nonacyclic, [F1] returns 0, [F2] gives N=0, and [F3] assigns torsion 0. Thus in either branch the procedure returns the twisted Reidemeister torsion of the closure exterior, with precisely the stated spine and extension hypotheses.

4 · Examples, counterexamples and false statements

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