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The fundamental group of the figure-eight knot complement has the displayed two-generator one-relator presentation. It follows from the Artin automorphism of a braid for $({\sigma }_1{\sigma }_2^{-1}{)}^2$ by eliminating the third meridian from its fixed-generator presentation.

Use the standard closed three-strand braid $B=({\sigma }_1{\sigma }_2^{-1}{)}^2$, a four-crossing diagram of the figure-eight knot. The diagram has two positive and two negative crossings, so $w(B)=0$.

We calculate its Kauffman bracket, making the inversion symmetry explicit rather than merely asserting amphichirality.

In the Temperley-Lieb diagram algebra on three strands put $u=e_1$, $v=e_2$. The relations are $u^2=\delta u$, $v^2=\delta v$, $uvu=u$, $vuv=v$. The bracket image of one two-crossing block isClosing planar diagrams gives the bracket trace on the three-strand Temperley-Lieb algebra:These values follow because the closures have respectively three, two and one circles. Expanding and reducing $H^2$ givesTaking its trace and substituting $\delta =-A^2-A^{-2}$ yieldsThere is no writhe correction, soThe equality is immediate from the five displayed coefficients. The same braid model supplies the meridians used in part (d).

Label the four crossings $A,B,C,D$ in top, left-middle, right-middle, bottom order. Label the bounded regions $d$ (left), $e$ (right), $a$ (upper middle), $b$ (central bigon), $c$ (lower middle), and the unbounded region $o$. The drawing preserves the crossings and the orientation of the supplied figure-eight knot diagram. Its overpassing strands are the outer-left/upper-right strand at $A$, the upper-left/lower-central strand at $B$, the upper-central/lower-right strand at $C$, and the lower-left/outer-right strand at $D$.

Here is the Dehn presentation Heegaard diagram specified by this drawing. Replace the underlying four-valent planar graph $G$ by a small three-dimensional regular neighborhood $N(G)$ and put $\Sigma =\partial N(G)$. Since $G$ has four vertices and eight edges, $\Sigma$ has genus $8-4+1=5$. On $\Sigma$, take the five $\alpha$ curves bordering the bounded faces, pushed slightly onto the boundary of $N(G)$. They bound a complete disk system in the complementary handlebody $S^3\setminus \mathrm{int}N(G)$; their labels are ${\alpha }_d,{\alpha }_e,{\alpha }_a,{\alpha }_b,{\alpha }_c$.

The four $\beta$ curves bound the disks dual to the four short crossing tunnels in $N(G)\setminus \mathrm{int}\nu K_8$. Lift the crossing circles shown in red onto the upper and lower sheets of $\Sigma$, routing them along the corresponding face boundaries; the crossing's overpassing strand determines the lift. Orient the $\alpha$ disk normals and the $\beta$ curves so that their signed successive intersections with the $\alpha$ curves areAt $A,D$ the missing fourth corner is the unbounded region $o$, whose generator is omitted. This sheet-lifting prescription, together with the projected graph, specifies the curves on the Heegaard surface; the red circles alone are only their local projections.

To recover the knot exterior from $\Sigma \times I$, attach five two-handles along the $\alpha$ curves on one side and cap the resulting sphere with a three-handle. On the other side attach four two-handles along the $\beta$ curves. Compressing that side leaves the boundary torus of the knot exterior, which is retained. In particular, the counts are five $\alpha$ compressions and four $\beta$ compressions, rather than two complete genus-five disk systems.

Keep all four relators and all five generators, in the orders $(r_A,r_B,r_C,r_D)$ and $(d,e,a,b,c)$. Under abelianization, these generators become $(t,t,t^2,t^3,t^2)$. The Fox calculus rules areFor example, $r_B=a{d}^{-1}c{b}^{-1}$ gives the nonzero Fox derivative entries $-a{d}^{-1}$, $1$, $-a{d}^{-1}c{b}^{-1}$, $a{d}^{-1}$ in columns $d,a,b,c$, respectively. After abelianization these are $-t,1,-1,t$. The full Alexander matrix isThe Fox calculus identity supplies a useful check:Deleting the $e$ column, adjacent to the unbounded region across the marked meridian of a knot, givesThus $\det B$ is a representative of the Alexander polynomial of a knot of the figure-eight knot, with its customary ambiguity of multiplication by $\pm t^n$.

The component knot types can change under this Conway mutation. To construct an example, place a knotted arc carrying a trefoil knot summand in one strand of $D_1$ and a knotted arc carrying a figure-eight knot summand in one strand of $D_2$, leaving the other two strand portions trivial. Choose the outside pairing so that before rotation the two knotted portions lie on different components, while after rotation they lie on the same component. Then the component multisets arerespectively. A link isotopy preserves the unordered multiset of component knot types, so these links are not isotopic, although part (a) shows that suitable orientations give them the same Jones polynomial.