Hecke Markov Traces and Polynomial Link Invariants

Well-definedness. Write the target additively, so the identity of $\mathbb{Z}$ is $0$. Define $u$ on generators by $u({\sigma }_i):=1$. At the braid relator both sides receive $1+1+1=3$; at each far-commutation relator both sides receive $1+1=2$. Hence every defining relator of [F1] is sent to $0$, and [F2] produces a unique homomorphism $e_n:B_n\to \mathbb{Z}$ with $e_n({\sigma }_i)=1$. For $n\le 1$ the group $B_n$ is trivial by [F1] and the unique homomorphism to $\mathbb{Z}$ is the trivial one; this is the case $n=1$ of the statement.

Values on words and inverses. Let $\beta ={\sigma }_{i_1}^{{\epsilon }_1}\cdots {\sigma }_{i_k}^{{\epsilon }_k}$ be an Artin word. By [F3] applied successively, $e_n(\beta )={\sum }_{r=1}^k{\epsilon }_r{e}_n({\sigma }_{i_r})={\sum }_{r=1}^k{\epsilon }_r$; in particular the value does not depend on the word chosen to represent the element $\beta$, because it equals the value of the well-defined map $e_n$ at $\beta$. Taking $k=1$, $e_n({\sigma }_i^{-1})=-1=-e_n({\sigma }_i)$, so the same computation gives $e_n({\beta }^{-1})=-e_n(\beta )$ for every Artin word by [F3] and the multiplicativity of group homomorphisms.

Surjectivity. For $n\ge 2$ the element $e_n({\sigma }_1)=1$ generates the additive group $\mathbb{Z}$, so $e_n$ is surjective; for $n=1$ the domain $B_1$ is trivial, and the exponent sum is the (trivial, hence not surjective) map into $\mathbb{Z}$. This proves all claims of the definition and completes the construction of the unique homomorphism with the prescribed values.

The link exterior and its retraction. AC supplies the countable-choice hypothesis of [F2] by [A1]. Apply [F2] to each $K_i$. Equip its normal bundle with the metric induced by the standard metric on $S^3$ (identify the quotient normal fibres with the orthogonal complements of the tangent lines). Compactness of the zero section and a finite bundle trivialization cover give a radius ${\epsilon }_i>0$ whose closed fibre disks lie inside the tubular domain. Shrink these finitely many radii until their images $N_i$ are pairwise disjoint; this is possible since the $K_i$ are disjoint compact sets. Each $N_i$ is a compact smooth disk bundle with smooth boundary, without needing a global product trivialization. Put $N={\bigcup }_i{N}_i$ and $M_0=S^3\setminus \mathrm{int}N$; local fibre-boundary charts show that $M_0$ is a compact smooth $3$-manifold with boundary $\mathrm{\partial }N$. In normalized disk-bundle coordinates $0<\mathrm{\mid }v\mathrm{\mid }\le 1$ define $H_u(x,v)=(x,((1-u)+u\mathrm{/}\mathrm{\mid }v\mathrm{\mid })v)$, for $0\le u\le 1$. Its radius is $(1-u)\mathrm{\mid }v\mathrm{\mid }+u$, so it stays in the punctured disk bundle, equals the identity at $u=0$, reaches the fibre boundary at $u=1$, and fixes that boundary at every $u$. This formula is independent of local trivializations and glues with the identity on $M_0$. It is a strong deformation retraction $X_L\to M_0$.

Local structure. $X_L$ is the complement in the smooth $3$-manifold $S^3$ of the closed subset $L$, hence is an open subset of $S^3$, and by [F3] it carries a canonical smooth structure making it a smooth $3$-manifold without boundary. In particular every point of $X_L$ has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^3$: such a set is locally path-connected, and an open Euclidean ball about a point is simply connected, so the point has arbitrarily small simply connected neighbourhoods. Hence $X_L$ is locally path-connected and semilocally simply connected, and it is nonempty because $L\mathrm{\ne }{S}^3$ by [F1].

Connectedness. By [F5] applied to the ring $R=\mathbb{Z}$ and the compact weakly locally contractible set $L\subset S^3$ (nonempty and proper by [F1]), ${\tilde{H}}_0(X_L;\mathbb{Z})\cong {\tilde{H}}^2(L;\mathbb{Z})\mathrm{.}$ By [F4], $L$ is a finite CW complex of dimension $1$, so $H^2(L;\mathbb{Z})=0$; therefore ${\tilde{H}}_0(X_L;\mathbb{Z})=0$. The degree-zero homology theorem Zero-th singular homology is free on path components identifies this with the augmentation kernel of the free group on path components, so the nonempty $X_L$ has one path component and is connected.

Path-connectedness. A space that is connected and locally path-connected is path-connected: for $x\in X_L$ the set of points that can be joined to $x$ by a path is open (local path-connectedness) and closed (its complement is also open), hence equals all of the connected space $X_L$. Thus $X_L$ is path-connected.

Finite CW type. By step 1.1, $M_0$ is a compact smooth $3$-manifold with boundary and $X_L\simeq M_0$. By the literature input [F6], $M_0$ admits a finite triangulation, hence is homeomorphic to a finite simplicial complex, and a finite simplicial complex is a finite CW complex. Therefore $M_0$, and with it $X_L$, has the homotopy type of a finite CW complex.

Conclusion. Steps 1.2 and 2.1 show that $X_L$ is nonempty, path-connected, locally path-connected and semilocally simply connected, so the hypotheses of the covering-space classification [F7] are satisfied; step 1.2 shows that $X_L$ is a smooth $3$-manifold without boundary, and step 2.2 gives its finite CW homotopy type. This proves every assertion of the statement.

The total linking homomorphism. The composite $\phi$ of [F2] is a homomorphism $H_1(X_L;\mathbb{Z})\to \mathbb{Z}$; writing it on ${\pi }_1$ through the Hurewicz abelianization surjection of [F4] gives the same homomorphism on abelianisations. Since the sum map ${⨁}_i{H}^1(K_i)\to \mathbb{Z}$ is surjective, so is $\phi$. By [F7] each positive meridian maps to its component coordinate, so $\phi ({\mu }_i)=1$. In particular $K=\ker \phi$ is a normal subgroup of $G$ with $G\mathrm{/}K\cong \mathbb{Z}$.

The cover and its deck group. By [F1] the base $X_L$ is nonempty, path-connected, locally path-connected and semilocally simply connected, so [F3] applies to the subgroup $K$ and provides a connected covering $p:\widetilde{X_L}\to X_L$ with $p_{\ast }{\pi }_1(\widetilde{X_L})=K$. As $K$ is normal by step 1.1, this covering is regular with deck group $\mathrm{Deck}(\widetilde{X_L}\mathrm{/}{X}_L)\cong G\mathrm{/}K\cong \mathbb{Z}$, generated by the deck transformation $\tau$ corresponding to $1\in \mathbb{Z}$.

Module structure. By [F6] the deck transformation $\tau$ acts as a group automorphism ${\tau }_{\ast }$ of $A_L=H_1(\widetilde{X_L};\mathbb{Z})$; since $\tau$ is invertible with inverse given by the deck transformation for $-1$, ${\tau }_{\ast }$ is an automorphism of the abelian group $A_L$. Sending $t\mapsto {\tau }_{\ast }$ therefore defines a group homomorphism $\mathbb{Z}\to \mathrm{Aut}(A_L)$, which by the universal property of the localisation in [F5] is exactly a $\mathrm{\Lambda }$-module structure on $A_L$ with $t$ acting as ${\tau }_{\ast }$. This completes the construction of $A_L$ and of its module structure.

Identification with the Fitting indexing. Given a finite presentation of $M$ with presentation matrix $A$ of size $n\times m$, [F1] defines $E_k(M)$ as the ideal generated by the $(n-k)\times (n-k)$ minors of $A$, with the values $R$ for $n-k\le 0$ and $0$ when $n-k$ exceeds the number $m$ of columns. This is exactly the ideal $I_{n-k}(A)$ of [F2] for the same presentation (whose index set $J$ is finite of size $m$), including both conventions; hence $E_k(M)=I_{n-k}(A)={\mathrm{Fitt}}_k(M)$ for every finite presentation of $M$.

Independence and isomorphism invariance. By [F2] the value $I_{n-k}(A)$ is independent of the presentation, so by step 1.1 $E_k(M)$ is independent of the chosen finite presentation. If $u:M\to M^{\mathrm{\prime }}$ is an isomorphism, [F4] transports any finite presentation of $M$ to one of $M^{\mathrm{\prime }}$ with the same matrix, so the two ideals agree. This proves the statement.

Well-formedness. The localisation $\mathrm{\Lambda }[z^{-1},u^{\pm 1},s]$ of the polynomial ring $\mathrm{\Lambda }[s,u]$ at the multiplicative subset generated by $z$ and $u$ is a commutative ring with unit by [F2], and by [F1] the images $v,z,u$ are units (for $v$ because $v$ is already a unit of $\mathrm{\Lambda }$). Forming the quotient by the ideal generated by the two relations gives a commutative ring $R$ with unit by [F3], and the displayed relations hold in it by construction. Since $s^2=v$ and $v$ is a unit, $s$ is a unit with $s^{-1}=s{v}^{-1}$, so $s^{-1}$ exists and $m=s-s^{-1}=s^{-1}(s^2-1)=s^{-1}(v-1)$ as displayed.

The identity. Using $s^2=v$ and $vz{u}^2=z+1-v$, so that $u^{2}v=(z+1-v)z^{-1}$ and $1-u^{2}v=(v-1)z^{-1}$, one computes $l^{-1}-l=(us{)}^{-1}-us=(1-u^2{s}^2)(us{)}^{-1}=(1-u^{2}v)(us{)}^{-1}=(v-1)z^{-1}{u}^{-1}{s}^{-1}=s^{-1}(v-1)(uz{)}^{-1}=m(uz{)}^{-1}$, which is the displayed identity.

Universal property. By [F2] a unital homomorphism from $\mathrm{\Lambda }[z^{-1},u^{\pm 1},s]$ to $T$ is exactly a unital homomorphism $\mathrm{\Lambda }[s,u]\to T$ sending $z$ and $u$ to units, i.e. a choice of images $v_0$ (a unit, since $v$ is a unit of $\mathrm{\Lambda }$ and homomorphisms send units to units), $z_0,u_0\in T^{\times }$ and $s_0\in T$; by [F3] it factors through the quotient $R$ exactly when the two relations hold at the images, i.e. $s_0^2=v_0$ and $v_0{z}_0{u}_0^2=z_0+1-v_0$; uniqueness holds because the images of $v,z,u,s$ determine those of the inverted elements, and these elements and inverses generate $R$.

Noetherian. Every nonzero ideal of $\mathbb{Z}$ is generated by its least positive element, by integer division, and the zero ideal is generated by $0$; thus $\mathbb{Z}$ is Noetherian. By [F2] the polynomial ring $\mathbb{Z}[t]$ is Noetherian, and $\mathrm{\Lambda }=\mathbb{Z}[t{]}_S$ is a localisation of it; localisations of Noetherian rings are Noetherian by [F2]. Hence every ideal of $\mathrm{\Lambda }$ is finitely generated. In $\mathbb{Z}[t]$, if $t$ divides $fg$, evaluation at zero gives $f(0)g(0)=0$ in the domain $\mathbb{Z}$, so one constant term vanishes and $t$ divides that factor. Since $t$ is nonzero and a nonunit there, it is prime; localization then makes it a unit by [F1].

The integer-polynomial UFD. Write any nonzero integer polynomial as its integer content times a primitive polynomial. Factor the latter over $\mathbb{Q}[t]$ by [F3], and clear denominators and contents in each nonconstant factor to obtain primitive integer factors. Their product is primitive by Gauss' lemma. Two primitive integer polynomials related by a nonzero rational scalar differ only by sign: a reduced denominator would divide every coefficient of the first, and the scalar's numerator would divide every coefficient of the second. Hence the primitive polynomial is, up to sign, the product of these primitive factors, which are irreducible over $\mathbb{Z}$ by Gauss' lemma. Integer prime factors of the content complete the factorization. Uniqueness follows by comparing integer contents using integer factorization, then comparing the remaining factors in the UFD $\mathbb{Q}[t]$; primitive rational associates are integer associates by the same scalar argument. Thus $\mathbb{Z}[t]$ is a UFD, and its irreducibles are prime.

Surviving irreducibles. An irreducible $p\in \mathbb{Z}[t]$ becomes a unit in $\mathrm{\Lambda }$ exactly when it divides a power of $t$: a relation $(p\mathrm{/}1)(a\mathrm{/}{t}^N)=1$ is equivalent, by injectivity of the localisation map, to $pa=t^N$, and the converse gives an inverse. Suppose $p$ remains a nonunit and $p=xy$ in $\mathrm{\Lambda }$. Write $x=a\mathrm{/}{t}^N$ and $y=b\mathrm{/}{t}^M$. Then $p{t}^{N+M}=ab$ in $\mathbb{Z}[t]$. The UFD property of step 1.2 makes $p$ prime, so, after interchanging $x,y$, write $a=p{a}^{\mathrm{\prime }}$. Cancellation gives $a^{\mathrm{\prime }}b=t^{N+M}$, hence $y(a^{\mathrm{\prime }}\mathrm{/}{t}^N)=1$. Thus $p$ is irreducible in $\mathrm{\Lambda }$.

Existence and uniqueness. Every nonzero nonunit $x\in \mathrm{\Lambda }$ has the form $a\mathrm{/}{t}^N$ with $0\mathrm{\ne }a\in \mathbb{Z}[t]$. Factor $a$ using step 1.2 and absorb all factors associated to $t$ into a Laurent unit. Step 2.1 shows that all remaining factors are irreducible in $\mathrm{\Lambda }$, giving existence. In particular every irreducible of $\mathrm{\Lambda }$ is associate there to one of these surviving integer-polynomial primes: its factorization can contain only one nonunit factor. To compare two Laurent factorizations, replace their factors by these integer-polynomial primes and absorb their Laurent units, which are $\pm t^k$ by [F1]. Clearing powers of $t$ gives equality in $\mathbb{Z}[t]$. Uniqueness there matches all primes not associated to $t$ on the two sides, hence matches the original Laurent factors up to permutation and Laurent units. Together with step 1.1, this proves the statement.