Wirtinger presentations and knot groups — Examples and Counterexamples

The crossing-free code has one under-arc and no crossing relators, so [F1] gives $⟨x\mid ⟩$. This is the free group on one generator, isomorphic to $(\mathbb{Z},+)$ by exponent sum using [F5]; [F3] identifies it with the unknot group.

The trefoil panels give the crossing triples $(b,c,a)$, $(c,a,b)$, and $(a,b,c)$. Applying [F2] gives the three crossing relations in the statement. The crossing with overarc $a$ gives $c=ab{a}^{-1}$. Substituting this into $a=bc{b}^{-1}$ gives $a=bab{a}^{-1}{b}^{-1}$. Right multiplication by $ba$ yields $aba=bab$. Under that braid relation, $ca{c}^{-1}=aba{b}^{-1}{a}^{-1}=b$, so the third crossing equation follows from the first two. Thus eliminating $c$ gives exactly $⟨a,b\mid aba=bab⟩$.

By [F7], these local charts belong to the explicit figure-eight code, rather than merely prescribing four abstract crossing relations. Each panel is an oriented local crossing chart. In every panel [F2] gives $x_o{x}_L=x_R{x}_o$. The side triples $(o,L,R)$ are $(x_1,x_2,x_3)$, $(x_4,x_2,x_1)$, $(x_2,x_4,x_3)$, and $(x_3,x_4,x_1)$. Substitution, and reversal of the written equality when needed, gives exactly the four equations in the statement. The under-arrow direction does not select a different conjugation rule.

In $S_3$, the two words $(12)(23)(12)$ and $(23)(12)(23)$ both equal $(13)$, so the assignment respects the trefoil relation. The transpositions $(12)$ and $(23)$ generate all six permutations $e,(12),(23),(12)(23),(23)(12),(13)$, hence the homomorphism is surjective by [F4] and [F6]. Since $S_3$ is nonabelian, this group is nonabelian.

From $R_3$, left multiplication by $x_1$ gives $x_1{x}_3{x}_2=x_1{x}_2{x}_4$. Applying $R_1$ to the right side gives $x_3{x}_1{x}_4$, and $R_2$ then gives $x_3{x}_4{x}_2$. Right cancellation of $x_2$ yields $x_1{x}_3=x_3{x}_4$, which is $R_4$.

By [F1] and [F3], the crossing-free unknot group is $⟨x\mid ⟩$, the free group on one generator. By [F5] it is isomorphic to $\mathbb{Z}$. All commutators in $\mathbb{Z}$ are trivial by [F7], so $[\mathbb{Z},\mathbb{Z}]=1$ and [F8] identifies its abelianization with $\mathbb{Z}$.

A valid rational code for the displayed trefoil uses the two diagonals from $(\pm 1,3-j)$ to $(\mp 1,2-j)$ for $j=0,1,2$, subdivided at their crossing $(0,5\mathrm{/}2-j)$, with the right-starting branch over and both diagonals oriented downward. Close the left endpoints by $(-1,0)\to (-2,0)\to (-2,3)\to (-1,3)$ and the right endpoints by the reflected route. Each strip has just its one transverse crossing, distinct strip interiors and the two exterior closure routes are disjoint, and the three swaps trace one circle. Thus these integer and half-integer vertices give admissible input for [F1]; rounding the bends gives the displayed diagram. The trefoil's displayed crossing triples and [F2] give $b=ca{c}^{-1}$, $a=bc{b}^{-1}$, and $c=ab{a}^{-1}$. Substituting the last into the second yields $a=bab{a}^{-1}{b}^{-1}$; right multiplication by $ba$ gives $aba=bab$. Under this relation, $ca{c}^{-1}=aba{b}^{-1}{a}^{-1}=b$, so the first equation follows. Thus eliminating $c$ gives $⟨a,b\mid aba=bab⟩$, and [F3] identifies it with $G_T$.

In the abelian quotient of [F8], written additively, $aba=bab$ gives $2[a]+[b]=[a]+2[b]$, hence $[a]=[b]$ and the abelianization is generated by $[a]$. The map $G_T\to \mathbb{Z}$ sending $a,b$ to $1$ respects the relator and is surjective; by [F8] it factors through the abelianization. The map $\mathbb{Z}\to G_T^{\mathrm{a}\mathrm{b}}$ sending $1$ to $[a]$ is inverse, so $G_T^{\mathrm{a}\mathrm{b}}\cong \mathbb{Z}$.

The map $a\mapsto (12)$, $b\mapsto (23)$ respects $aba=bab$, since both sides equal $(13)$. Its image contains two adjacent transpositions and hence all six elements of $S_3$, so it is surjective by [F4] and [F6]. Since $S_3$ is nonabelian, $G_T$ is nonabelian; every quotient of the cyclic group $\mathbb{Z}$ is cyclic, so $G_T\cong ̸\mathbb{Z}\cong G_U$.

If $U$ and $T$ were ambient isotopic, [F9] would give a homeomorphism of their complements; [F10] and [F11] would induce a fundamental-group isomorphism. Step 2.2 rules this out. Their abelianizations nevertheless agree by steps 1.1 and 2.1, proving the stated counterexample.