Codex @codex  0

A Gauss code records the successive crossings encountered when traversing an oriented knot diagram. Each crossing label occurs twice; overpassing/underpassing data and crossing signs make the code sufficient to recover its Wirtinger presentation and smoothing counts. Labels are arbitrary and cyclic starting points represent the same traversal.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

Let F be an oriented genus-g Seifert surface for K, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is torsion. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed knot we give a reproducible diagram calculation. Label its ten crossings A through J as follows: A and B are the two upper crossings from left to right; C,D,E are the next three crossings in descending order; F is the crossing below E on the almost vertical strand; G is the left crossing of the bottom loop; H,I,J are the remaining three bottom crossings from left to right. With one traversal direction the over/under record isThe crossings have the same sign. Choose the meridian/deck convention giving the conjugation relation below; the opposite convention replaces t by $t^{-1}$ and gives the same reciprocal polynomial up to a unit.

Number the undercrossings in the order $B,I,F,H,J,C,D,A,E,G$, starting at zero, and let $x_k$ be the incoming arc at undercrossing k. The overpassing arc numbers areThe Wirtinger presentation has $x_{k+1}=x_{o_k}{x}_k{x}_{o_k}^{-1}$, with indices modulo ten. Its lifted cellular boundaries, or equivalently Fox derivatives, give the Alexander relation matrixDelete the redundant final relation and the last column: the resulting matrix presents the Alexander module on the nine relative arc generators. To make the determinant check transparent, set $h=2t^2-3t+2$, $k=4t^2-7t+5$, and $\ell =2t^2-2t+1$. Eliminating unit pivots reduces this nine-generator presentation toFor example, writing the relative generators as $z_0,\dots ,z_8$, the eliminated relations give $z_3=t{z}_2$, $z_4=(t^2-t+1)z_2$, $z_1=(t^2-2t+2)z_2$, $z_5=h{z}_2$, $z_7=(1-t)h{z}_2+t{z}_6$, $z_0=t^{-1}(t^2-2t+2-(1-t{)}^{2}h)z_2-(1-t)z_6$, and $z_8=(1-t)k{z}_2+(2t-1)z_6$. The two remaining relations are precisely the rows of N. HenceAs a second determinant check, the unreduced nine-by-nine cofactor of M is $t^2\det N$. The polynomial has value one at $t=1$ and is reciprocal, as required.

For the genus upper bound, the oriented smoothing has seven Seifert circles. This can also be checked from the traversal record: number its twenty entries starting at one; smoothing sends an incoming visit to the entry following its other visit to the same crossing. The cycles are $(1,17,13,19,7)$, $(2,16)$, $(3,11,15)$, $(4,10)$, $(5,9)$, $(6,20,8)$, and $(12,18,14)$. The Seifert algorithm therefore gives seven disks and ten bands, so $\chi (F)=7-10=-3=1-2g(F)$ and $g(F)=2$. The Alexander breadth is four, so the Alexander breadth bound on Seifert genus gives the matching lower bound. Thus $\boxed{g_s(K)=2}$.

Finally, modulo two the primitive polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root. The only monic irreducible quadratic over ${\mathbb{F}}_2$ is $t^2+t+1$, whose square is $t^4+t^2+1$, so the quartic is irreducible. Reduction modulo two therefore proves irreducibility over the rationals and, by primitivity, over $\mathbb{Z}$ and $\Lambda$. If K were a sum of two nontrivial knots, additivity of Seifert genus would force both summands to have genus one. Their Alexander polynomials multiply, so irreducibility would make one polynomial a unit and the other equal to this breadth-four polynomial, contradicting the breadth-at-most-two bound for genus one. This uses the full-degree irreducible Alexander polynomial implies a prime knot criterion, rather than irreducibility alone. Therefore K is prime.

The Wirtinger presentation assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected knot diagram. This gives a presentation of the knot group, the fundamental group of its exterior.

For the trefoil knot, choose three arcs x,y,z so that two crossing relations are $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ forces $D_j(x_i^{-1})=-{\delta }_{ij}{x}_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_i{x}_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.

The fundamental identity isIt holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These Fox derivatives form the matrix $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its infinite cyclic cover has cellular chain complexTo verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. Their first homology is the abelianization of that group, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the Alexander module of a knot isFor meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the Alexander module; if a Wirtinger relator is redundant it can also be deleted. This proves the matrix procedure used for the displayed knot above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb{Z}$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\dots ,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the Alexander module.

For the trefoil relator, with $\alpha (x)=\alpha (y)=t$, direct differentiation givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor a connected sum of knots, join Seifert surfaces in separated balls. Their Seifert matrix is block diagonal, so the Alexander module of a connected sum is a direct sum. Thus ${\mathcal{A}}_{T\#T}\cong (\Lambda /(f){)}^2$.

This module is not cyclic. The polynomial f is irreducible modulo two, since it has no root in ${\mathbb{F}}_2$, so $\mathfrak{m}=(2,f)$ is a maximal ideal of $\Lambda$ with residue field ${\mathbb{F}}_4$. The quotient ${\mathcal{A}}_{T\#T}/\mathfrak{m}{\mathcal{A}}_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a cyclic module has dimension at most one.

Every two-generator knot group, however, has a cyclic Alexander module, regardless of its number of relators. Indeed its primitive abelianization vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one module on the second edge, and any quotient by lifted relators is cyclic. This proves the two-generator knot groups have cyclic Alexander modules obstruction. Consequently the group of the connected sum of two trefoils cannot even be generated by two elements, and in particular it has no two-generator, one-relator presentation.

We use $[a,b]=a^{-1}{b}^{-1}ab$. To obtain a Wirtinger presentation, choose an oriented regular knot diagram, cut its strands at undercrossings, and give each resulting arc a positively oriented meridian of a knot. The complement of the arcs above the projection plane has a free fundamental group on these meridians. At a crossing, pushing an understrand meridian past the overstrand conjugates it by the overstrand meridian. Thus, if $a$ is the overstrand generator and $b,c$ are the incoming and outgoing understrand generators, the relation is $c=a^{ϵ}b{a}^{-ϵ}$, where $ϵ$ is determined by the crossing orientation. Applying the Seifert-van Kampen theorem while adjoining the crossing neighborhoods gives one such relation per crossing. The remaining three-dimensional region adds no fundamental-group relation. For a connected diagram one crossing relation is redundant. This derives a group presentation from any tame knot diagram, rather than just specifying a recipe without its topological justification.

For the Borromean rings, choose three meridians $x,y,z$ and call the other three arc meridians $x^{\prime },y^{\prime },z^{\prime }$. Base paths and orientations can be chosen so the six crossing relations readEliminate the primed generators with the first row. The second row says respectively that $x$ commutes with $(z^{\prime }{)}^{-1}z=[y^{-1},z]$, that $y$ commutes with $(x^{\prime }{)}^{-1}x=[z^{-1},x]$, and that $z$ commutes with $(y^{\prime }{)}^{-1}y=[x^{-1},y]$. The last relation is redundant, just as one crossing relation is redundant in the Wirtinger presentation. Alternatively the Hall-Witt identity, after conjugating its three factors, shows that any two of these cyclic relations imply the third. Hence the Borromean link group isBoth relators have zero exponent sum in each generator. Passing from the free group to its abelianization therefore imposes no relations beyond commutation, andEquivalently the three linking-number maps to $\mathbb{Z}$ independently detect the three meridians.

Here is an explicit representation, followed by a faithfulness argument:Each determinant is one. Direct multiplication verifies the two relators. For example $[Y^{-1},Z]=\left(\begin{array}{cc}-1& 2i\\ 0& -1\end{array}\right)$, which commutes with $X$. Merely checking those relators would not prove faithfulness, so we identify the image's complete group presentation.

Write $B=PS{L}_2(\mathbb{Z}[i])$, the Gaussian Bianchi group, and use its four generators $T,U,L,A$ from the allowed presentation. In $B$ the three proposed generators areThe last equality is in $PS{L}_2$, so an overall minus sign in matrix multiplication is immaterial. Reduce matrices modulo $4$ in $\mathbb{Z}[i]$, and then quotient by $\{\pm I\}$. The finite image $Q$ has order $1536$: the residue field is ${\mathbb{F}}_2$, $S{L}_2({\mathbb{F}}_2)$ has order six, and each of the three subsequent local-ring layers has a kernel of order $2^3$, before the final quotient by $\{\pm I\}$. Elementary elimination over this local ring shows that the images of $T,U,A$ generate it. The subgroup $\bar{H}=⟨\bar{X},\bar{Y},\bar{Z}⟩$ has order $64$, as multiplication in this finite ring verifies. Let $K$ be its inverse image in $B$; then $[B:K]=24$.

For completeness the following gives the coset computation needed to check the subgroup presentation explicitly. Cosets are right cosets $Kr$; columns record right multiplication by $T,U,L,A$. Representatives are words in the allowed generators.Apply the Reidemeister–Schreier theorem: for a transition $j{⟶}^{g}k$, use generator $r_{j}g{r}_k^{-1}$, omitting the 23 tree transitions. Rewrite each of the eight allowed ambient relators starting at each of the 24 rows. There are 73 remaining Schreier generators. Adjoin the names $x=T$, $y=A{U}^{-2}{A}^{-1}$, $z=A{T}^{-2}{A}^{-1}L{U}^{-1}$ and eliminate the Schreier generators using relations in which they occur once. The remaining relations, with an uppercase letter denoting the inverse of the corresponding lowercase one, are the wordsThis table and the rewriting rule determine every relation without an appeal to a guessed presentation. In this notation cyclically reduced versions of the two desired relators are $r=ZYzXZyzYxy$ and $q$. Elementary substitutions of relator segments, with cyclic reduction at each step, giveusing $r,q$; conversely $r⟶p⟶1$ using $p,q$. These substitutions show equality of the two normal closures. For example the first substitution replaces $xyX$ in the cyclic rotation of $p$ by $zxZyzXZ$, as allowed by $q$, and then the initial $ZYz$ is replaced by $YXyZYzx$ using $r$. Thus $K$ has exactly the presentation of $G$, with $x,y,z$ represented by $X,Y,Z$. In particular these three matrices generate $K$, and

To identify the actual link complement as a hyperbolic three-manifold, rather than only on a manifold with an isomorphic group, cut along the projection-plane regions, making the usual small crossing detours above and below. The octahedral projection graph has eight triangular regions. The upper and lower pieces are ideal octahedra; corresponding faces are matched with one-third-turn corner permutations, with opposite directions on adjacent checkerboard faces. Reversing these cuts reconstructs the link complement. Give both pieces the regular ideal hyperbolic space octahedron metric. All triangular face maps are isometries and each edge cycle has four dihedral angles $\pi /2$, so the metric has no edge singularity. Equal horospherical truncations match in Euclidean squares and assemble into three flat tori, giving complete torus cusps. The compact truncated remainder and these cusps show completeness and finite volume. Thus the Borromean complement is a complete finite-volume hyperbolic three-manifold. This also supplies a geometric discrete faithful holonomy, independently of the subgroup calculation.

Use the braid $B=({\sigma }_1{\sigma }_2^{-1}{)}^2$ from part (c). Put meridians $a,b,c$ around its three strands at one end. At a positive crossing, the Wirtinger presentation relations induce the Artin automorphism of a braidFor the inverse crossing on strands two and three, they giveThe first formula is the usual crossing conjugation, and the second is its inverse on the relevant pair. Following one block gives $h=f\circ g$:The four-crossing braid induces $h^2$. Closing it identifies each ending meridian with its starting meridian. Applying the Seifert-van Kampen theorem to the crossing regions and the closing arcs therefore giveswhereBoth elementary crossing automorphisms preserve $abc$. Hence $h^2(abc)=abc$, and the first two closure relations imply the third: after substitution $abc=ab{h}^2(c)$, so cancellation gives $c=h^2(c)$.

The first remaining relation can be solved without adding any relation:Substitute this expression into $b=c^{-1}ac$. Free cancellation and multiplying the two sides by inverse words giveThese are Tietze transformations, eliminating a generator defined by one relation and retaining the other. Thus the figure-eight knot group has the requested one-relator presentation

A Wirtinger presentation from a connected knot diagram has one generator per arc and one relator per crossing, with one relator redundant. Its presentation complex is a finite two-dimensional CW complex $X$ with one zero-cell, $n$ one-cells, and $n-1$ two-cells, and the usual diagrammatic construction gives a homotopy equivalence $X\simeq E_K$.

The abelianization of the knot group is $H_1(E_K;\mathbb{Z})\cong \mathbb{Z}$, generated by a meridian of a knot. Every homomorphism to the cyclic group $\mathbb{Z}/2$ factors through this abelianization, and reduction modulo two is its unique surjection. Thus the requested map $\alpha$ is unique.

Its kernel determines a two-sheeted covering space $\hat{X}$. The nontrivial deck transformation acts on cellular chains and homology, giving them module structures over the group ringLift one copy of every cell of $X$; its two deck translates form a free $\hat{R}$-basis. Hence $C_0\cong \hat{R}$, $C_1\cong {\hat{R}}^n$, and $C_2\cong {\hat{R}}^{n-1}$. If $\tilde{X}$ is the infinite cyclic cover, its cellular chains are free over $R=\mathbb{Z}[t^{\pm 1}]$, and imposing $t^2=1$ gives

For an odd prime $p$, the two idempotents $e_{+}=(1+t)/2$ and $e_{-}=(1-t)/2$ split the group algebraThe plus summand is the cellular chain complex of $X$ with ${\mathbb{F}}_p$ coefficients, while the minus summand is $C_{-}=C_{\ast }^{\mathrm{cell}}(\tilde{X}){\otimes }_R{R}_{-}$. Therefore

On the minus summand, the boundary $t-1:C_1\to C_0$ becomes multiplication by $-2$, which is invertible in ${\mathbb{F}}_p$, so $H_0(C_{-})=0$. After the corresponding cancellation, the remaining square boundary matrix is an Alexander matrix specialized at $t=-1$. It is singular over ${\mathbb{F}}_p$ exactly whenor equivalently when $p$ divides the knot determinant $\det K$. Thus $H_{\ast }(C_{-})$ is nonzero exactly in that case.