Multivariable Alexander modules and chain models

For $r=1$ the statement is [F1]. Suppose $r>1$, put $P_r={\mathrm{\Lambda }}_1[t_2,\dots ,t_r]$, and let $S_r$ be the multiplicative set generated by $t_2,\dots ,t_r$. Clearing negative powers identifies ${\mathrm{\Lambda }}_r$ with $S_r^{-1}{P}_r$.

We prove by induction on the number of added variables that ${\mathrm{\Lambda }}_1[t_2,\dots ,t_r]$ is a UFD. The base is [F1]. For the induction step let $R$ be the preceding UFD and $K$ its fraction field. Every nonzero $f\in R[x]$ is its content $c\in R$ times a primitive polynomial $f_0$: since $f$ has finitely many coefficients, take the product over irreducibles of their minimum exponents in those coefficients. Factor $c$ in $R$ and factor $f_0$ in $K[x]$ using [F5]. Clear denominators in each positive-degree irreducible factor and divide by its content to obtain primitive polynomials $q_1,\dots ,q_m\in R[x]$. By [F4], their product $q$ is primitive and is associated to $f_0$ in $K[x]$. Write this scalar as $a\mathrm{/}b$ with coprime $a,b\in R$, so $b{f}_0=aq$. Any irreducible divisor of $a$ then divides every coefficient of $f_0$, and any irreducible divisor of $b$ divides every coefficient of $q$; primitiveness forces $a$ and $b$ to be units. Thus $f_0$ is associated in $R[x]$ to the product of the $q_i$. Each $q_i$ is irreducible in $R[x]$ by [F4], and the content factors in $R$, so every nonzero nonunit of $R[x]$ factors into irreducibles.

By [F2], $P_r$ is Noetherian because ${\mathrm{\Lambda }}_1$ is Noetherian by [F1]. Its localization $S_r^{-1}{P}_r={\mathrm{\Lambda }}_r$ is Noetherian by [F3].

For uniqueness in the same induction step, contents have unique factorization in $R$. Two factorizations of the primitive part become factorizations in the UFD $K[x]$, so their primitive positive-degree factors pair up up to scalar associates. If primitive $f,g\in R[x]$ satisfy $f=(a\mathrm{/}b)g$ for coprime $a,b\in R$, then $bf=ag$; any irreducible factor of $a$ must divide every coefficient of $f$, and any irreducible factor of $b$ must divide every coefficient of $g$. Primitiveness forces $a,b$ to be units, so the scalar is a unit of $R$. This proves uniqueness in $R[x]$ and completes the induction.

Localizing the UFD $P_r$ at $S_r$ preserves unique factorization: factor a numerator in $P_r$ and discard the irreducible factors that divide an element of $S_r$, since precisely those factors become units. Every remaining irreducible stays prime: if it divides a product of localized fractions, clearing denominators gives divisibility of the product of their numerators in $P_r$, so primality in $P_r$ makes it divide one numerator; it cannot divide a cleared denominator because it was not inverted. Uniqueness follows by clearing denominators and using uniqueness in $P_r$. Thus ${\mathrm{\Lambda }}_r$ is a UFD.

If $u=f\mathrm{/}s\in S_r^{-1}{P}_r$ is a unit, then $fg=sq$ for some $q\in S_r$. Every irreducible factor of $f$ therefore divides a product of powers of $t_2,\dots ,t_r$, so it is associated in $P_r$ to one of these variables. The only units already in $P_r$ are the units of ${\mathrm{\Lambda }}_1$, namely $\pm t_1^k$, by degree in $t_2,\dots ,t_r$ and [F1]. Consequently the units of ${\mathrm{\Lambda }}_r$ are exactly $\pm t_1^{k_1}\cdots t_r^{k_r}$.

Steps 2.1, 3.1 and 4.1 prove Noetherianity, unique factorization, and the stated unit description for every $r\ge 1$.

The height condition says that the link meets the level sphere $\mathrm{\Sigma }$ exactly at the $n$ underpassing points. Hence $A_L\cup B_L=X$ and $A_L\cap B_L={\mathrm{\Sigma }}_L$. The one-point compactification of the plane ${\mathbb{R}}^2\times \{-1\}$ is a sphere, so removing those $n$ points gives the stated punctured sphere. The punctured sphere is path connected: remove one puncture to obtain a plane, in which a polygonal path detours around the remaining finite points.

Because the crossing set is finite, choose pairwise disjoint coordinate balls about its points, then shrink the product tubes so each crossing ball meets precisely the two prescribed local branches and contains no other tube. Shrinking a product tube radially is an isotopy of its boundary and leaves the exterior homeomorphic. Use the unit-sphere stereographic coordinates $(u_1,u_2,u_3,u_4)=(2x,2y,2z,x^2+y^2+z^2-1)\mathrm{/}(1+x^2+y^2+z^2)$. Here $\mathrm{\Sigma }$ is $h\cdot u=c$, with $h=(0,0,1,-1)\mathrm{/}\sqrt{2}$, $c=-1\mathrm{/}\sqrt{2}$. Write $q=ch+\sin (3\pi \mathrm{/}4)w$, $w\perp h$, $\mathrm{\mid }w\mathrm{\mid }=1$. Its explicit unit-speed normal collar is $\cos (3\pi \mathrm{/}4-t)h+\sin (3\pi \mathrm{/}4-t)w$, for $\mathrm{\mid }t\mathrm{\mid }<\rho =\pi \mathrm{/}8$; it is injective since its angle stays in $(0,\pi )$, and positive $t$ is the $A$ side. In particular no general tubular-neighborhood or choice theorem is required. For $q\in {\mathrm{\Sigma }}_L$, set $ϵ(q)=\min \{\rho \mathrm{/}4,d(q,L)\mathrm{/}4\}$. This is positive and continuous. The points ${\exp }_q(t{\nu }_q)$ with $\mathrm{\mid }t\mathrm{\mid }<ϵ(q)$ avoid $L$, since their distance from $q$ is less than $d(q,L)$. The variable-width collar $C=\{{\exp }_q(t{\nu }_q):q\in {\mathrm{\Sigma }}_L,\mathrm{\mid }t\mathrm{\mid }<ϵ(q)\}$ is embedded and open in $X$. Set $U=A_L\cup C_{<0}$ and $V=B_L\cup C_{>0}$, where $C_{<0}$ and $C_{>0}$ are its negative and positive normal halves. Then $U,V$ are open, cover $X$, and $U\cap V=C$. Clamping the normal coordinate to $0$ retracts $U$ to $A_L$ and $V$ to $B_L$, while the collar itself retracts to ${\mathrm{\Sigma }}_L$. For the final assertion, choose a continuous function $\chi :[0,1]\to [0,1]$ equal to $1$ on $[0,1\mathrm{/}4]$ and $0$ on $[1\mathrm{/}2,1]$. On the positive collar of $A_L$, the homotopy $t\mapsto t+sϵ(q)\chi (t\mathrm{/}ϵ(q))\mathrm{/}4$ ($0\le s\le 1$) pushes every boundary point into $A_L^{\circ }$, fixes points outside the collar, and stays below $ϵ(q)$; it stays in $X$ by the distance bound. It maps $A_L^{\circ }$ to itself, so the inclusion $A_L^{\circ }\hookrightarrow A_L$ is a homotopy equivalence.

Every finite point of $B_L$ has height at most $-1$. For $0\le t<1$, move $(x,y,z)$ vertically to $(x,y,z-t\mathrm{/}(1-t))$, and send $\mathrm{\infty }$ to itself; at $t=1$ send every point to $\mathrm{\infty }$. This is continuous in the one-point compactification, avoids $L_D$ because the link lies in $z\ge -1$, and contracts $B_L$ to $\mathrm{\infty }$.

Choose disjoint small coordinate disks about the punctures on $\mathrm{\Sigma }$. Within each punctured disk, the radial homotopy $r\mapsto (1-t)r+t\max \{r,\epsilon \}$ pushes its punctured cap onto its boundary collar, fixing points outside the cap. Thus ${\mathrm{\Sigma }}_L$ retracts to the compact sphere with $n$ open disks removed. Base the $n$ boundary circles along any specified accesses. Filling all $n$ boundary circles by disks gives $S^2$, which is simply connected: two slightly enlarged hemispheres are contractible with path-connected annular overlap, so [F3] gives the trivial group. Filling all but the $j$th boundary circle gives a disk, also simply connected. The radial retraction extends by the identity on the newly attached disks. Consequently attaching disks to the original noncompact punctured sphere gives spaces homotopy equivalent to these filled compact models, rather than literally the compact models. The van Kampen quotient in [F3] proves that all the based puncture circles normally generate, and that the $n-1$ circles other than $c_j$ still normally generate. For $n=1$ the omitted list is empty and ${\mathrm{\Sigma }}_L\cong {\mathbb{R}}^2$ is contractible, as required. Independent access changes conjugate each relator, leaving its normal closure unchanged.

The lower piece is contractible by step 1.3. For the upper piece, a finite point outside the link has a small relative ball disjoint from the link. Move within that ball to a point whose plane projection misses the whole projected diagram; such projections are dense because a finite regular immersed compact curve has locally finitely many one-dimensional graph branches and empty interior. The vertical path from that point to height $1$ now avoids the link. The height-$1$ plane is connected. The point at infinity first moves into this finite upper piece along the explicit positive collar in step 1.2. Thus the upper piece too is path connected. This uses explicit paths rather than a general-position theorem for arbitrary arcs. The retractions of step 1.2 now show that $U,V,U\cap V$ are path connected as well: any point is joined to its retraction image by that retraction path.

In a product tube $S^1\times D^2$, the punctured tube $S^1\times (D^2\setminus \{0\})$ retracts radially onto $S^1\times \mathrm{\partial }{D}^2$. Applying this on each of the finitely many disjoint tubes and fixing the exterior gives a deformation retraction of $X$ onto the chosen exterior. Thus the induced map on fundamental groups is an isomorphism.

For $x={\mathrm{\Phi }}_i(v)\in N(K_i)\setminus K_i$, write $\rho =\parallel v\parallel >0$ and define $F_s(x)={\mathrm{\Phi }}_i\text{\hspace{0.17em}⁣}\left(((1-s)\rho +s{ϵ}_i)\frac{v}{\rho }\right),0\le s\le 1.$ On $E_L$ set $F_s(x)=x$. The formula is continuous on each punctured tube and agrees with the identity on its boundary $\rho ={ϵ}_i$. The products with $I$ of the finitely many punctured closed tubes and $E_L$ form a finite closed cover of $X_L\times I$; [F5] therefore makes $F:X_L\times I\to X_L$ continuous. It starts at the identity, fixes $E_L$ pointwise, and at $s=1$ maps every punctured tube to its boundary, so it is a strong deformation retraction onto $E_L$.

The link complement is path-connected by [F6], so its deformation retract $E_L$ is path-connected. The tube construction makes $E_L$ a smooth manifold with boundary, hence locally path-connected and semilocally simply connected. By [F8] and [F9], inclusion induces an isomorphism $i_{\ast }:{\pi }_1(E_L,b)\to {\pi }_1(X_L,b)$. By [F8] and [F10], it also induces an isomorphism $i_{\ast }:H_1(E_L;\mathbb{Z})\to H_1(X_L;\mathbb{Z})$. The positive meridians lie in $E_L$ and map to the same classes, so the meridian basis is preserved. An $E_L$-loop closes in the restricted cover exactly when its image under $i_{\ast }$ lies in $\ker \varphi$, so the component containing $\tilde{b}$ induces the subgroup $i_{\ast }^{-1}(\ker \varphi )=\ker (\varphi \circ i_{\ast })\mathrm{.}$ This subgroup is exactly the kernel of the abelianization of ${\pi }_1(E_L,b)$ because $i_{\ast }$ is an isomorphism. To check connectedness, take $y\in E_L$, a path $\delta$ in $E_L$ from $b$ to $y$, and a lift $\tilde{\delta }$ from a chosen point $\tilde{b}$ over $b$. Any other point over $y$ is $h\tilde{\delta }(1)$ for some deck transformation $h$, by [F7]. Surjectivity of $i_{\ast }$ gives a loop $\alpha$ in $E_L$ whose lift from $\tilde{b}$ ends at $h\tilde{b}$. The reverse of $\tilde{\delta }$, this lifted loop, and $h\tilde{\delta }$ concatenate to a path in $p^{-1}(E_L)$ from $\tilde{\delta }(1)$ to $h\tilde{\delta }(1)$. Thus the restriction is connected. By [F4] it is the connected maximal abelian cover of $E_L$, with the same deck group $H_1(X_L;\mathbb{Z})\cong H_1(E_L;\mathbb{Z})$.

Apply [F2] to $F$ with initial lift the identity of $X_L^{\mathrm{a}\mathrm{b}}$. For a point over $E_L$, its constant path is a lift of its path under $F$; uniqueness makes the lifted homotopy fix every point of $p^{-1}(E_L)$. At time $1$ the image lies in $p^{-1}(E_L)$, so the lifted homotopy is a strong deformation retraction of $X_L^{\mathrm{a}\mathrm{b}}$ onto that restricted cover.

For every deck transformation $h$, the maps $(x,s)\mapsto h{\tilde{F}}_s(x)$ and $(x,s)\mapsto {\tilde{F}}_s(hx)$ lift the same homotopy $F(p(x),s)$ and agree at $s=0$. Uniqueness in [F2] makes them equal. The retraction is therefore deck-equivariant, and its induced isomorphism on first homology is ${\mathrm{\Lambda }}_r$-linear. By [F1] the target homology of the full cover is $A_L$, proving the asserted module isomorphism.

Let $P=D^2\setminus {\bigcup }_i\mathrm{int}{D}_i$ be either endpoint surface. The supplied product tubes parametrize the pairwise disjoint endpoint disks by ${\varphi }_i:D^2\to D_i$, with ${\varphi }_i(0)=p_i$. Put $Q=D^2\setminus \{p_1,\dots ,p_r\}$. On $D_i\setminus \{p_i\}$ write a point as ${\varphi }_i(ru)$ with $0<r\le 1$, $\mathrm{\mid }u\mathrm{\mid }=1$, and set $R_t({\varphi }_i(ru))={\varphi }_i(((1-t)r+t)u)\mathrm{.}$ Put $R_t=\mathrm{i}\mathrm{d}$ outside the interiors of these disks. At $r=1$ the formulas agree, so finite pasting gives a deformation retraction $Q\to P$. This uses the supplied disk coordinates, not a triangulation of an arbitrary smooth planar region. A collar of the outer unit circle misses all $p_i$. In its radial coordinate a continuous inward push, equal to the identity off that collar, gives a homotopy inverse to $\mathrm{int}{D}^2\setminus \{p_i\}\hookrightarrow Q$; throughout the homotopy no puncture is crossed. By [F8] the open punctured disk has a finite circle-wedge model and a positive meridian free basis. Transfer it across these displayed equivalences, changing basepoint along the recorded collar path when needed. Small loops about $p_i$ become the positive endpoint meridians: the annulus ${\varphi }_i(ru)$ carries their orientation to the tube boundary. The accesses are the jointly transported stem paths of that finite spine; their images need not be arbitrary preselected accesses. If $a_i$ and $a_i^{\mathrm{\prime }}$ are compatible and arbitrary accesses to the same meridian circle, then $c_i=a_i^{\mathrm{\prime }}{a}_i^{-1}$ is a based loop and $a_i^{\mathrm{\prime }}{\mu }_i(a_i^{\mathrm{\prime }}{)}^{-1}=c_i(a_i{\mu }_i{a}_i^{-1})c_i^{-1}$. Thus independent basings conjugate generators, without asserting they remain a free basis. Conjugation leaves their homology classes unchanged by [F9]. The homotopy type and abstract free rank are unchanged.

For a closed link, enlarge each solid torus tube by a small radial collar and apply [F1] to the pair $(S^3,E_L)$; the collar deformation retracts the enlarged-tube pair to $(S^1\times D^2,S^1\times S^1)$. The product torus has a CW structure with one vertex, two edges, and one face attached by a commutator, so [F7] gives $H_1(S^1\times S^1)\cong {\mathbb{Z}}^2$ with meridian and section basis. The solid torus retracts to its core circle, so [F3, F6] give $H_2(S^1\times D^2)=0$ and $H_1(S^1\times D^2)\cong \mathbb{Z}$, with the meridian mapping to zero and the section to a generator. The pair sequence [F2] therefore identifies $H_2(S^1\times D^2,S^1\times S^1)$ with the kernel generated by the meridian. Excision gives $H_2(S^3,E_L)\cong {\mathbb{Z}}^r$. By [F2] and [F3], the boundary homomorphism $H_2(S^3,E_L)\to H_1(E_L)$ is an isomorphism. Each relative generator is a normal disk and its boundary is the positive meridian by [F4]. Hence the positive meridians form the ordered basis of $H_1(E_L)$.

Let $D_i^{+}$ be disjoint disks slightly larger than the removed endpoint tube disks. The closed complement of their interiors lies in the interior of $P$ relative to $D^2$. By [F1], $H_{\ast }(D^2,P)$ is the direct sum of the relative groups of the $r$ disk pairs $(D_i^{+},D_i^{+}\setminus \mathrm{int}{D}_i)$. Radial collars identify each pair with $(D^2,S^1)$. By [F2], [F3], and [F6], the relative group of this disk pair is $\mathbb{Z}$ in degree $2$ and zero in every other degree. Its boundary map sends the relative disk to the positive endpoint meridian. Applying [F2] to $(D^2,P)$, with $D^2$ contractible by radial contraction, therefore gives $H_1(P)\cong {\mathbb{Z}}^r$ with that basis and $H_2(P)=0$.

For each string component, enlarge its chosen product tube by a small radial collar. The complement of these enlarged tubes has closure in the interior of $X_{\tau }$; excising it by [F1] reduces $(D^2\times [0,1],X_{\tau })$ to the disjoint union of tube pairs. In each pair, the radial collar deformation retracts the pair to $([0,1]\times D^2,[0,1]\times S^1)$: the endpoint cap annuli retract to their boundary circles and the lateral shell retracts to the original tube boundary. Contracting the interval factor identifies this pair with $(D^2,S^1)$. By [F2], [F3], and [F6], each pair has relative homology $\mathbb{Z}$ in degree $2$ and zero in every other degree. Therefore $H_2(D^2\times [0,1],X_{\tau })\cong {\mathbb{Z}}^r$ and its relative homology vanishes in degrees $1$ and $3$.

The ambient cylinder is contractible, so its positive-degree homology vanishes by [F6]. Its pair long exact sequence from [F2], together with step 3.1, gives $H_2(X_{\tau })=0$ from the segment $H_3(B,X_{\tau })\to H_2(X_{\tau })\to H_2(B)$, and gives an isomorphism $\mathrm{\partial }:H_2(B,X_{\tau })\to H_1(X_{\tau })$ from the segment $H_2(B)\to H_2(B,X_{\tau })\stackrel{\mathrm{\partial }}{\to }{H}_1(X_{\tau })\to H_1(B)$. Thus $H_1(X_{\tau })\cong {\mathbb{Z}}^r$.

Put $B=D^2\times [0,1]$ and let $j_{\pm }:(D^2,P_{\pm })\to (B,X_{\tau })$ be the endpoint pair inclusions. Represent the positive $i$th endpoint disk by an oriented singular disk chain $d_{i,\pm }$. Under the excision and interval contraction in step 3.1, its image is the positive relative disk generator in the $i$th tube summand: the slice $\{0\}\times D^2$ or $\{1\}\times D^2$ projects to $D^2$ preserving the normal-disk orientation. Thus the classes $[j_{\pm \mathrm{\#}}{d}_{i,\pm }]$ form the ordered basis of $H_2(B,X_{\tau })$. By the lift-and-boundary recipe [F10], ${\delta }_B([j_{\pm \mathrm{\#}}{d}_{i,\pm }])=[\mathrm{\partial }{j}_{\pm \mathrm{\#}}{d}_{i,\pm }]=[j_{\pm \mathrm{\#}}\mathrm{\partial }{d}_{i,\pm }]$: the equality holds on chains because composing a singular simplex with $j_{\pm }$ commutes with every face restriction in its boundary. These are exactly the images of the positive endpoint meridians. The connecting map is an isomorphism by step 4.1, and the endpoint meridians form the basis by step 2.1. Each end map on $H_1$ is therefore an isomorphism. Since both endpoint and exterior $H_2$ vanish, each end map on $H_2$ is surjective.

The product torus $T_i$ has $H_1(T_i)\cong {\mathbb{Z}}^2$ with basis the positive meridian and the chosen degree-one section. Write the image of the section in the basis from step 1.2 and let $k_i$ be its $i$th coordinate. Subtracting $k_i$ times the meridian gives the unique degree-one class with zero $i$th coordinate. Its coordinates in the product basis are $(-k_i,1)$, so it is primitive; the curve $t\mapsto (e^{2\pi it},e^{-2\pi i{k}_{i}t})$ in product coordinates is an embedded representative.

Let the two projected branches be parametrized near the crossing by ${\gamma }_o(u)$ and ${\gamma }_s(v)$, with linearly independent derivatives at their common image $p$. The map $(u,v)\mapsto {\gamma }_o(u)+{\gamma }_s(v)-p$ has invertible derivative there, so the inverse function theorem gives local planar coordinates $x=u$, $y=v$. Choose the sign of $y$ so $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ agrees with the oriented projection plane, and translate the crossing to $(0,0)$. Write the two height functions as $z=h_o(x)$ and $z=h_s(y)$, with $h_o(0)=0$ and $h_s(0)=-h<0$. The coordinate change $z^{\mathrm{\prime }}=z-h_o(x)-h_s(y)+h_s(0)$ has Jacobian determinant $1$; it carries the over-branch to $(x,0,0)$ and the under-branch to $(0,y,-h)$, while preserving the ambient orientation. Shrink the chart so it excludes every other branch, as permitted by the isolated-crossing hypothesis in the Statement. The normal-disk product chart for the under-branch supplies the chosen small radius $r<h$. Since $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ is positively oriented, $n_o=+{\mathrm{\partial }}_y$, so $y<0$ is the source and $y>0$ is the target.

The triangle boundary is the loop from its apex to the source endpoint, along $\beta$ to the target endpoint, and back to the apex. Its interior lies above the under-tube boundary and meets the over-branch exactly at $(0,0,0)$. With the boundary orientation induced by $({\mathrm{\partial }}_y,{\mathrm{\partial }}_z)$, its transverse intersection sign is positive because $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y,{\mathrm{\partial }}_z)$ is a positive ambient basis. Removing a small disk about that intersection leaves an annulus in the crossing complement from the triangle boundary to a positive normal meridian of the over-strand. Conjugating by the common path $\delta$ therefore identifies $a_s\beta a_t^{-1}$ with the stated $m_o$.

The product tube around the under-branch carries its positive meridian circles along the branch by an annulus. The top arc $\beta$ is the path on that tube boundary joining the chosen source and target circle basepoints. If ${\mu }_s,{\mu }_t$ denote the meridian loops based at those points, the annulus gives ${\mu }_s=\beta {\mu }_t{\beta }^{-1}$. With the access paths $a_s,a_t$ of the Statement, it follows that $m_s=(a_s\beta a_t^{-1})m_t(a_s\beta a_t^{-1}{)}^{-1}\mathrm{.}$ The punctured-triangle argument in step 2.1 identifies $a_s\beta a_t^{-1}$ with $m_o$. Solving gives the asserted conjugation relation.

Multiplying the conjugation equality by $m_t^{-1}$ gives the displayed crossing relator. Reversing the under-strand orientation reverses both under-meridian loops, so inverting the equality leaves the same side-based equation.

For arbitrary access paths, $a_e^{\mathrm{\prime }}=c_e{a}_e$ gives $m_e^{\mathrm{\prime }}=c_e{m}_e{c}_e^{-1}$. Substitute $m_o=c_o^{-1}{m}_o^{\mathrm{\prime }}{c}_o$ and $m_s=c_s^{-1}{m}_s^{\mathrm{\prime }}{c}_s$ into $m_t=m_o^{-1}{m}_s{m}_o$, then conjugate by $c_t$. This gives exactly the correction formula in the Statement; simultaneous base-path change conjugates all meridians by the same loop.

On a chosen product tube with normal coordinate $v\mathrm{\ne }0$, the homotopy $v⟼((1-s)\mathrm{\mid }v\mathrm{\mid }+s)\frac{v}{\mathrm{\mid }v\mathrm{\mid }},0\le s\le 1,$ pushes the punctured tube radially to its boundary and fixes that boundary. Applying it on each disjoint tube and fixing the exterior identifies the based fundamental group of the open complement with that of the compact exterior. Thus the relation proved in the open complement is the same relation in the chosen exterior group.