Twisted Alexander polynomials and Reidemeister torsion — Examples and Counterexamples

Under [A1], realize $S^3$ as $\{(z,w)\in {\mathbb{C}}^2:\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2=1\}$ and its standard unknot as $K=\{w=0\}$. Choose $0<a<1\mathrm{/}2$ and the normal-disk tube $N(K)=\{\mathrm{\mid }w\mathrm{\mid }\le a\}$, parametrized by $(v,w)\mapsto (\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}v,w)$ for $v\in S^1$. By [F3], its exterior is $\{\mathrm{\mid }w\mathrm{\mid }\ge a\}$. Put $b=\sqrt{1-a^2}$. The map $D_b^2\times S^1\to E_K$ given by $(z,u)\mapsto (z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}u)$ has inverse $(z,w)\mapsto (z,w\mathrm{/}\mathrm{\mid }w\mathrm{\mid })$, exhibiting a solid torus. Contracting the disk factor retracts it onto its core; a positive normal-disk meridian winds once around the circle factor, oriented accordingly. For a finite simple-spine certificate, triangulate the disk as a cone on a polygon. Collapse its triangles along their outer edges, then its radial edges at their outer vertices, leaving the center. In the product CW structure, perform each disk collapse first over the circle's $1$-cell and then over its vertex. These elementary cellular collapses leave the core circle with one vertex and edge $y$, representing the positive meridian of [F4].

Attaching a $1$-cell $x$ at that vertex and a $2$-cell with attaching word $x$ is an elementary expansion: the new $1$-cell is a free face of the new $2$-cell, so collapsing this pair returns the circle spine. Therefore $⟨x,y\mid x⟩$ is a valid based deficiency-one spine of the same exterior, in the sense of [F1]. The relation kills $x$ in the fundamental group, so [F4] makes the full abelianization send $x$ to $1$ and the positive meridian $y$ to its generator $t$.

The Fox derivatives of $r=x$ are $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }x=1$ and $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }y=0$, so the row Fox matrix is $J=[10]$. By [F2], the $d_1$ blocks are $0$ for $x$ and $t-1$ for $y$, making the localized complex $0\to \mathbb{Q}(t)\stackrel{(1,0)}{\to }\mathbb{Q}(t{)}^2\stackrel{(0,t-1{)}^{\mathsf{T}}}{\to }\mathbb{Q}(t)\to 0$ acyclic. Deleting the $x$ column leaves the zero $1\times 1$ minor and gives denominator $1-1=0$, hence [F1] declares that quotient undefined. Deleting the $y$ column leaves minor $[1]$ and denominator $t-1$, so its quotient is $1\mathrm{/}(t-1)$. This establishes the stated boundary case.

Under [A1], let $A$ be the annulus left on the Heegaard torus after removing an open annular neighborhood of the $(2,3)$ curve. The torus-knot model identifies its core with the common source circle $c$ in the finite spine $X=M_2{\cup }_c{M}_3$, where $M_2$ and $M_3$ are mapping cylinders of the degree-two and degree-three maps of a circle and their target core circles are $a$ and $b$. By [F7], the compact exterior is the mapping cylinder of an explicit map from its boundary torus to $X$. Choose finite cell structures compatible with this map and the two degree maps, with circle vertices $v_c,v_a,v_b$ and cylinder edges $e_a:v_c\to v_a$, $e_b:v_c\to v_b$; take $v_c$ as basepoint. In the resulting exterior mapping-cylinder CW structure, each prism over a boundary-torus cell $\sigma$ pairs with its unglued face $\sigma \times \{0\}$; removing these pairs in decreasing cell dimension gives a finite sequence of elementary collapses onto $X$, fixing $X$ and $v_c$. Thus this specific exterior-to-$X$ equivalence is simple. The natural CW structure on $X$ has these three vertices, edges $a,b,c,e_a,e_b$, and $2$-cells with based words $e_a{a}^2{e}_a^{-1}{c}^{-1}$ and $e_b{b}^3{e}_b^{-1}{c}^{-1}$. The two connecting edges form a maximal tree. Contract it to obtain a based presentation complex $\bar{X}$: this contraction is simple, as can be seen by moving the loops and attaching paths at each terminal vertex across its connecting edge by cellular edge slides, then collapsing that edge with the now free terminal vertex. Each cellular slide is realized by an elementary expansion followed by an elementary collapse; the root $v_c$ is fixed. After the two contractions, $\bar{X}$ has one vertex, edges $a,b,c$, and $2$-cells ${\lambda }_2,{\lambda }_3$ attached by $r_2=a^2{c}^{-1}$ and $r_3=b^3{c}^{-1}$. Work in $\bar{X}$ for the following further simple changes. Slide ${\lambda }_3$ over the oppositely oriented ${\lambda }_2$, replacing its attaching word by $r_3{r}_2^{-1}=(b^3{c}^{-1})(a^2{c}^{-1}{)}^{-1}=b^3{a}^{-2}\mathrm{.}$ This elementary based $2$-cell slide removes $c$ from that attaching map. The edge $c$ now occurs exactly once in the boundary of ${\lambda }_2$ and nowhere else, so $(c,{\lambda }_2)$ is a free $1$- and $2$-cell pair and collapses. Reversing the orientation of the remaining $2$-cell gives the attaching word $a^2{b}^{-3}$. Thus the resulting one-vertex complex with core edges $a,b$ and one $2$-cell is based simple-homotopy equivalent to the chosen compact exterior and supplies the required certificate.

The abelianization of this spine has generators $[a],[b]$ and relation $2[a]-3[b]=0$. By [F6], the positive meridian class $[m]$ generates $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$. Write $[a]=p[m]$ and $[b]=q[m]$; the relation gives $2p=3q$, and $a,b$ generate the abelianization, so $\gcd (p,q)=1$. Hence $(p,q)=(3,2)$ or $(-3,-2)$; orient the two core generators consistently with the positive meridian to choose $(3,2)$. Thus the full positive-meridian deck map sends $a\mapsto t^3$ and $b\mapsto t^2$.

For $r=a^2{b}^{-3}$, the Fox product and inverse rules give $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }a=1+a$ and $\mathrm{\partial }r\mathrm{/}\mathrm{\partial }b=-a^2(b^{-1}+b^{-2}+b^{-3})$. Evaluating at $a=t^3$, $b=t^2$ gives $1+t^3$ and $-(t^4+t^2+1)$. The $a$ denominator is $t^3-1\mathrm{\ne }0$, so the deleted-$a$ quotient is $-(t^4+t^2+1)\mathrm{/}(t^3-1)$. Factoring $t^4+t^2+1=(t^2+t+1)(t^2-t+1)$ and $t^3-1=(t-1)(t^2+t+1)$ reduces it to $(t^2-t+1)\mathrm{/}(1-t)$.

In the cumulative basis, $g_1=x_1$ and $g_2=x_1{x}_2$. The positive Artin rule gives ${\sigma }_1(g_1)=g_2{g}_1^{-1}$ and fixes $g_2$. The Fox row of $g_2{g}_1^{-1}$ is $(-g_2{g}_1^{-1},1)$, which evaluates to $(-t,1)$ under the trivial representation and equal strand colors; the fixed boundary word contributes the second row $(0,1)$. Therefore the reduced block for ${\sigma }_1$ is $[-t]$. For automorphisms $\beta ,\gamma$ of the free group, the product and inverse recursions imply $\mathrm{\partial }(\beta (u))\mathrm{/}\mathrm{\partial }{g}_j={\sum }_k\beta (\mathrm{\partial }u\mathrm{/}\mathrm{\partial }{g}_k)\mathrm{\partial }(\beta (g_k))\mathrm{/}\mathrm{\partial }{g}_j$: it holds for generators and inverses, and the product rule extends it to every word by induction. After evaluation, the braid-preserved color map makes coefficient substitution by $\beta$ leave the coefficients unchanged, so the full Fox matrices compose by matrix multiplication; their block form has bottom row $(0,1)$, so the reduced scalar blocks multiply. The two strands close to one component, hence this coloring is preserved, and the reduced block for ${\sigma }_1^3$ is $[-t{]}^3=[-t^3]$.

The fraction-field $d_1$ has the invertible block $t^3-1$, so it is onto. The deleted Fox minor from step 3.1 is nonzero, so $d_2$ is injective; since the chain dimensions are $1,2,1$ and $d_1{d}_2=0$, its image is the one-dimensional kernel of $d_1$. The complex is acyclic. Moreover, any Laurent-ring $1$-cycle becomes a boundary over $\mathbb{Q}(t)$; clearing its denominator and using that the chain module is free over the domain $\mathbb{Q}[t^{\pm 1}]$ shows that the cycle's homology class is annihilated by a nonzero Laurent polynomial. Thus the degree-one Laurent-ring homology is torsion, and the hypotheses of the exterior-spine Wada theorem hold.

Apply [F2] to the certified spine and admissible $a$ column. Its Wada quotient from step 3.1 represents the twisted torsion class. This identifies the displayed rational function with the torsion computed from the torus-spine certificate.

The Burau numerator and boundary denominator are $N=\det ([-t^3]-[1])=-(t^3+1)$ and $D=t^2-1$. Hence $N\mathrm{/}D=-(t^3+1)\mathrm{/}(t^2-1)=-((t+1)(t^2-t+1))\mathrm{/}((t-1)(t+1))=(t^2-t+1)\mathrm{/}(1-t)$. This is exactly the quotient in step 3.1, so the finite determinant calculation equals the based-spine torsion value. If the compatible axis-filling certificate is supplied as well, [F5] identifies the same ratio with the procedure output.