Cyclic branched covers and knot determinants

If $n=0$, the codomain and image are both the zero group, so the cokernel is trivial and has order $1=\mathrm{\mid }\det A\mathrm{\mid }$. Now suppose $n>0$. Multiplication on the left or right by an integer matrix with determinant $\pm 1$ preserves the isomorphism type of the cokernel and preserves the absolute determinant: left multiplication changes the basis of the target, and right multiplication changes the basis of the domain.

Whenever the current remaining square submatrix is nonzero, among all matrices obtained from it by such row and column operations choose one whose least nonzero absolute entry is as small as possible; by swaps and a sign change, place that positive entry $d$ in position $(1,1)$. If any entry in its first row or column is not divisible by $d$, Euclidean division and a row or column subtraction produce a nonzero remainder strictly between $0$ and $d$, which contradicts the minimality of $d$. Clear the first row and column by subtracting multiples of the first column and row. If an entry $b$ in the remaining submatrix were not divisible by $d$, add its row to the first row, write $b=kd+r$ with $0<r<d$, and subtract $k$ times the first column from its column; swapping that column with the first would again give a smaller nonzero pivot. Hence $d$ divides the remaining submatrix.

Apply the same finite pivot procedure to each remaining square submatrix. If a remaining submatrix is zero, stop and retain it unchanged as part of the zero block. Since every nonzero pivot removes one row and one column, induction gives a diagonal matrix $\mathrm{diag}(d_1,\dots ,d_r,0,\dots ,0)$ with each $d_i>0$, obtained by unimodular row and column operations. Its cokernel is ${⨁}_{i=1}^r\mathbb{Z}\mathrm{/}{d}_i\mathbb{Z}\oplus {\mathbb{Z}}^{n-r}$. Thus it is finite exactly when $r=n$, equivalently when $\det A\mathrm{\ne }0$; then its order is ${\prod }_i{d}_i=\mathrm{\mid }\det A\mathrm{\mid }$. The operations in step 1.1 preserve both conclusions for the original matrix.

In each chosen tubular neighborhood, normalize the normal-disk radius to one and set $H_t(s,v)=(s,((1-t)+t\mathrm{/}\mathrm{\mid }v\mathrm{\mid })v)$ for $0<\mathrm{\mid }v\mathrm{\mid }\le 1$ and $0\le t\le 1$. Its radius is $(1-t)\mathrm{\mid }v\mathrm{\mid }+t$, so this homotopy stays off $J$, fixes the disk boundary, and glues to the identity on $X$. It is a strong deformation retraction from $S^3\setminus J$ onto $X$; in particular $X$ is connected by [F2]. By [F1], Alexander duality identifies $H_1(S^3\setminus J;\mathbb{Z})$ with ${\tilde{H}}^1(J;\mathbb{Z})={⨁}_i{H}^1(J_i;\mathbb{Z})\cong {\mathbb{Z}}^n$. The locally constructed retraction makes the inclusion of $X$ a homotopy equivalence and hence a homology isomorphism. Thus $H_1(X;\mathbb{Z})\cong {\mathbb{Z}}^n$.

The boundary of a small positive normal disk to $J_i$ maps to the $i$-th Alexander-duality coordinate generator: it meets $J_i$ once positively and misses every other component. Hence $[{\mu }_1],\dots ,[{\mu }_n]$ are the coordinate basis, with the positive orientation normalized as in [F3]. The $i$-th coordinate of $[{\lambda }_i]$ is zero by its zero self-linking convention, and for $j\mathrm{\ne }i$ its $j$-th coordinate is the defined linking number ${\mathrm{lk}}_{S^3}(J_i,J_j)$. Therefore $[{\delta }_i]=[{\lambda }_i+f_i{\mu }_i]={\sum }_j{A}_{ij}[{\mu }_j]$.

In Mayer–Vietoris for the filling, the map $H_1(\mathrm{\partial }{V}_i)\to H_1(V_i)$ is onto and its kernel is generated by the meridian-disk boundary ${\delta }_i$. The map $H_0({⨆}_i\mathrm{\partial }{V}_i)\to H_0(X)\oplus H_0({⨆}_i{V}_i)$ is injective: in coordinates it sends $(a_1,\dots ,a_n)$ to $({\sum }_i{a}_i,-a_1,\dots ,-a_n)$. Exactness therefore gives $H_1(Y;\mathbb{Z})={\mathbb{Z}}^n\mathrm{/}A{\mathbb{Z}}^n$.

Since $A$ is invertible over $\mathbb{Q}$, for every $x\in {\mathbb{Z}}^n$ some positive integer $m$ clears the denominators of $A^{-1}x$, so $mx\in A{\mathbb{Z}}^n$ and each class is torsion. If $F,F^{\mathrm{\prime }}$ bound $mC,m^{\mathrm{\prime }}C$, pass to a common multiple $M$; the difference $(M\mathrm{/}m)F-(M\mathrm{/}{m}^{\mathrm{\prime }})F^{\mathrm{\prime }}$ is a 2-cycle $Z$. If $r[D]=0$, write $rD=\mathrm{\partial }G$; the chain boundary identity gives $r(Z\cdot D)=Z\cdot \mathrm{\partial }G=\mathrm{\partial }Z\cdot G=0$, hence $Z\cdot D=0$. Thus the quotient is independent of the chosen bounding chain and multiple modulo integers. Replacing either cycle representative by a homologous one adds an integer by the same boundary identity.

Let $B$ be the slope-coordinate matrix $[\delta ]=B[\mu ]$ in $H_1(X;\mathbb{Q})$; step 2.1 gives $B=A$. In the reverse filling, the slope-coordinate matrix is $B^{-1}$. Let ${\tilde{J}}_i$ denote the core of the attached solid torus, oriented so the boundary longitude ${\mathrm{\ell }}_i$ mapping to it satisfies ${\delta }_i\cdot {\mathrm{\ell }}_i=1$. Since ${\delta }_i\cdot {\mu }_i=-1$, the image of ${\mu }_i$ in that solid torus is $-[{\tilde{J}}_i]$. Define $H_{ij}={\mathrm{lk}}_Y({\tilde{J}}_i,{\tilde{J}}_j)$ for $i\mathrm{\ne }j$, and let $H_{ii}$ be the reverse-surgery framing determined by the slope ${\mu }_i$. For $i\mathrm{\ne }j$, clear denominators in $[{\mu }_i]={\sum }_k(B^{-1}{)}_{ik}[{\delta }_k]$ and cap each ${\delta }_k$ by its meridian disk. The chain in $X$ realizing this homology relation misses every core; with ${\delta }_k\cdot {\mathrm{\ell }}_k=+1$, the capped disk in $V_k$ meets the oriented core ${\tilde{J}}_j$ once positively if $k=j$ and not at all otherwise. Hence $(B^{-1}{)}_{ij}={\mathrm{lk}}_Y({\mu }_i,{\tilde{J}}_j)=-H_{ij}$. The reverse diagonal framing is the reverse slope coefficient divided by ${\delta }_i\cdot {\mu }_i=-1$, so $H_{ii}=-(B^{-1}{)}_{ii}$. Therefore $H=-B^{-1}=-A^{-1}$.

Each old meridian ${\mu }_i$ represents $-{\tilde{J}}_i$ in its attached solid torus. Hence for distinct indices their linking equals the core-linking entry $H_{ij}$; for a self-pairing, a parallel meridian maps to the parallel core specified by the reverse framing, also giving $H_{ii}$. Thus ${\lambda }_Y([{\mu }_i],[{\mu }_j])=-(A^{-1}{)}_{ij}$. The chain formula is additive in both cycles, by taking common multiples and summing bounding chains, so its matrix on the meridian basis gives ${\lambda }_Y([x],[y])=-x^T{A}^{-1}y(\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})$. If $x$ is replaced by $x+Az$, the value changes by $-z^{T}y$; if $y$ is replaced by $y+Az$, it changes by $-x^{T}z$, both integers because $A$ is symmetric and integral.

By [F2], inclusion of the compact exterior into the open complement induces an isomorphism on fundamental groups and identifies the positive meridian classes. The total-linking homomorphism in [F3] therefore corresponds to $\varphi$; by [F4], $A_K$ is isomorphic to $N^{\mathrm{a}\mathrm{b}}$ with the action induced by conjugation by a positive meridian, up to replacing that action by its inverse according to the left-versus-right lift convention.

Put $G_q=\ker (\varphi <mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q)$. For $g\in G_q$, write $\varphi (g)=qk$ and set $n=g{s}^{-qk}\in N$. Then $g=n{s}^{qk}$, and the expression is unique because $\varphi (s^{qk})=qk$. Thus $G_q$ is the split extension of $N$ by the infinite cyclic group generated by $s^q$, acting on $N$ by conjugation. By [F5] and [F1], $G_q$ is the fundamental group of $E_q$.

Abelianizing this split extension kills the commutators in $N$ and the relations $s^{q}n{s}^{-q}=n$ for $n\in N$. By [F3], those relations impose $(t^q-1)A_K=0$ on $N^{\mathrm{a}\mathrm{b}}$; if the inverse deck convention is used, $t^{-q}-1=-t^{-q}(t^q-1)$ generates the same submodule. The homomorphism $G_q\to \mathbb{Z}$, $g\mapsto \varphi (g)\mathrm{/}q$, splits by $1\mapsto s^q$, so $H_1(E_q;\mathbb{Z})\cong A_K\mathrm{/}(t^q-1)A_K\oplus \mathbb{Z}[s^q]$.

The boundary of $E_q$ is a torus cover with basis the lift $\tilde{\mu }$ of ${\mu }^q$ and a lifted longitude. Thicken $E_q$ and the filling solid torus by collars to form an open cover. By [F6], its Mayer–Vietoris map presents $H_1({\mathrm{\Sigma }}_q(K))$ as $H_1(E_q)$ modulo $[\tilde{\mu }]$: the longitude lift maps to the core generator of the attached solid torus and eliminates that generator, while the filling meridian maps to zero there. The class $[\tilde{\mu }]$ is $[s^q]$, the generator of the displayed free summand in step 3.1. Quotienting by it gives the stated isomorphism.

Use the isotopy-track handle model supplied by [F5], not a cut of the codimension-two branch surface itself. By [F1] and [F2], the connected surface has one disk and $2g$ bands. In the model, the disk-face gluings combine the $q$ cut-open $4$-balls into a $4$-ball with one $0$-handle. The remaining face gluings give, for each band and each adjacent-copy layer $k=1,\dots ,q-1$, a $2$-handle whose attaching circle joins that band's two face-core arcs through the common disk region, with the carried band framing. Hence there are $2g(q-1)$ $2$-handles and no other handles. The identification of this model with the actual cyclic branched cover, including the legitimacy of omitting the terminal gluing, is the external input [F5]; the local linking-entry computation below uses its attaching-circle description.

For band $a_i$ and seam $k$, the attaching circle $C_{i,k}$ is obtained by joining its oriented core arc in the positive face of copy $k$ to the core arc in the negative face of copy $k+1$, with endpoints joined through the common disk $0$-handle. In the collar of $F$, closing either face arc through that disk gives the corresponding positive or negative push-off of the closed core cycle $a_i$. Simultaneously moving $a_i^{-}$ and $a_j$ through the product collar to $a_i$ and $a_j^{+}$ preserves linking, so symmetry of linking and [F3] give $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)=V_{ji}$. For two handles at the same seam, the positive-face and negative-face contributions are therefore $V_{ij}$ and $V_{ji}$, giving $V_{ij}+V_{ji}$; for the self-entry this is framing $2V_{ii}$. For handles at consecutive seams, only their common sheet contributes, and the reflection reverses one normal direction, giving $-\mathrm{lk}(a_j^{+},a_i)=-V_{ji}$ in the upper block; symmetry gives $-V_{ij}$ in the lower block. If the seams are separated by at least one unused seam, the attaching circles lie in disjoint sheet neighborhoods, so their linking is zero. These cases exhaust all pairs and give exactly $M_q(V)$.

Apply the branched covering map on the actual cover $W_q(F)$ identified with the handle model by [F5]. Its normal-disk map is $w\mapsto w^q$; this is not a claim about the unmatched terminal banks of the handle model. A positive meridian away from the branch locus moves cyclically through the $q$ sheets. Restriction to the boundary therefore gives the connected exterior cover classified by ${\pi }_1(E_K)\to \mathbb{Z}\mathrm{/}q$, $\mu \mapsto 1$, and the lift of ${\mu }^q$ bounds the normal filling disk. This is exactly the boundary restriction in [F5] and the exterior/filling definition [F4]. Thus the boundary with the marking induced by the actual branched covering map is ${\mathrm{\Sigma }}_q(K)$. No canonical covering map on the unclosed raw handle model is being asserted.

For $q=2$ there is one seam, so the block matrix has only its diagonal block $V+V^T$, as stated.

The handle decomposition in [F2] has cellular chain groups $C_2(W;\mathbb{Z})={\mathbb{Z}}^m$, $C_1(W;\mathbb{Z})=0$, and $C_0(W;\mathbb{Z})=\mathbb{Z}$. Hence $H_2(W;\mathbb{Z})={\mathbb{Z}}^m$, $H_1(W;\mathbb{Z})=0$, and $H^2(W;\mathbb{Z})={\mathbb{Z}}^m$. By [F3], $H_2(W,Y;\mathbb{Z})={\mathbb{Z}}^m$ as well.

In the handle bases, the map $j:H_2(W;\mathbb{Z})\to H_2(W,Y;\mathbb{Z})$ sends each absolute handle class to its intersection coordinates with the relative dual handle classes. A diagonal coordinate is the handle framing, and an off-diagonal coordinate is the linking of the corresponding attaching circles; by [F2] this matrix is $M_q(V)$.

The pair sequence [F4] contains $H_2(W;\mathbb{Z})\stackrel{j}{\to }{H}_2(W,Y;\mathbb{Z})⟶H_1(Y;\mathbb{Z})⟶H_1(W;\mathbb{Z})=0$. By [F1], $Y=\mathrm{\partial }W={\mathrm{\Sigma }}_q(K)$. Exactness and step 2.1 therefore give $H_1({\mathrm{\Sigma }}_q(K);\mathbb{Z})\cong \mathrm{coker}(M_q(V))$ in the disk-band basis.

For any other ordered integral basis, [F5] gives $V^{\mathrm{\prime }}=P^{T}VP$ with $P$ unimodular. Let $P_q=\mathrm{diag}(P,\dots ,P)$ with $q-1$ diagonal blocks. Blockwise substitution in the definition shows $M_q(V^{\mathrm{\prime }})=P_q^T{M}_q(V)P_q$. Left and right multiplication by unimodular integer matrices preserve the isomorphism class of the cokernel, so the presentation holds in the stated basis. For $q=2$ the block matrix has the single block $V+V^T$, and for $g=0$ it is empty.

The block indices in $M_q(V)$ run from $1$ to $q-1$. At $q=2$ there is only one index, so there are no adjacent blocks and $M_2(V)$ consists of the single diagonal block $V+V^T$.

Substitute $q=2$ in [F2]. By [F1], the resulting manifold is the double branched cover ${\mathrm{\Sigma }}_2(K)$; hence its first homology is the cokernel of $V+V^T$. If $g=0$, the block has size zero and the cokernel is the zero group by the empty-matrix convention. The required AC hypothesis is inherited through [F1].

Quotienting the module presentation in [F2] by $(t^q-1)A_K$ gives $A_K\mathrm{/}(t^q-1)A_K\cong \mathrm{coker}(V-t{V}^T:R_q^{2g}\to R_q^{2g})$ as abelian groups. The group $R_q^{2g}$ is free abelian of rank $2gq$ on $t^k{a}_i$ for $0\le k<q$ and $1\le i\le 2g$, so this map is represented by a square integer matrix $Q_q$.

After tensoring with $\mathbb{C}$, multiplication by $t$ on $R_q\otimes \mathbb{C}$ is diagonalizable: its eigenvalues are the $q$ distinct roots ${\zeta }_q^j$ for $0\le j<q$, each with one-dimensional eigenspace. On the corresponding $2g$-dimensional summand, $Q_q$ acts by $V-{\zeta }_q^j{V}^T$. Therefore $\det Q_q={\prod }_{j=0}^{q-1}\det (V-{\zeta }_q^j{V}^T)$. The $j=0$ factor is $\det (V-V^T)=\pm 1$ by [F3]. Thus $\det Q_q\mathrm{\ne }0$ exactly when every factor with $1\le j<q$ is nonzero, and its absolute value is the product of the absolute values of those factors.

By [F1], the group in question is $\mathrm{coker}(Q_q)$. Applying [F5] shows it is finite exactly when $\det Q_q\mathrm{\ne }0$, with order $\mathrm{\mid }\det Q_q\mathrm{\mid }$; if any nontrivial-root factor vanishes, the square integer cokernel has positive free rank and is infinite.

By [F4], $D(t)=\pm t^k{\mathrm{\Delta }}_K(t)$ for some integer $k$. At every root of unity, $\mathrm{\mid }\pm ({\zeta }_q^j{)}^k\mathrm{\mid }=1$, so $\mathrm{\mid }D({\zeta }_q^j)\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Delta }}_K({\zeta }_q^j)\mathrm{\mid }$. Hence the determinant-root criterion and product from step 3.1 give exactly the Alexander-polynomial criterion and order in the Statement, including the empty-matrix case $g=0$.

By [F1], the double-cover homology is presented by the square integer matrix $V+V^T=V-(-1)V^T$. Its determinant is $D(-1)$. By [F2], the homology is finite exactly when $D(-1)\mathrm{\ne }0$, and then its order is $\mathrm{\mid }D(-1)\mathrm{\mid }$. The hypothesis needed by [F1] is the Axiom of Choice in [A1].

By [F3], $D(-1)=\pm (-1{)}^k{\mathrm{\Delta }}_K(-1)$ for some integer $k$, so $\mathrm{\mid }D(-1)\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Delta }}_K(-1)\mathrm{\mid }$ and the zero tests agree. This proves the finiteness criterion and order formula.

Write ${\mathrm{\Delta }}_K(t)={\sum }_{r=m}^n{c}_r{t}^r$ with integers $c_r$ and finitely many terms. Since $(-1{)}^r\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$ for every integer $r$, including negative $r$, [F4] gives ${\mathrm{\Delta }}_K(-1)\equiv {\sum }_r{c}_r={\mathrm{\Delta }}_K(1)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Thus ${\mathrm{\Delta }}_K(-1)$ is odd and nonzero; step 1.1 then gives finiteness for every knot.