Bridge and unknotting numbers of the trefoil

Apply [F4] to the three stated crossing triples, and rearrange using the group laws in [F9]. The resulting relations are $a=bc{b}^{-1}$, $b=ca{c}^{-1}$ and $c=ab{a}^{-1}$.

Write $z_1=x_1+i{x}_2$ and $z_2=x_3+i{x}_4$ on the unit sphere $S^3\subset {\mathbb{C}}^2$, so the fixed height is $h=\mathrm{Im}{z}_2$. The curve $\gamma (t)=2^{-1\mathrm{/}2}(e^{2it},e^{3it})$, $0\le t<2\pi$, lies on the standard torus $\mathrm{\mid }{z}_1\mathrm{\mid }=\mathrm{\mid }{z}_2\mathrm{\mid }=2^{-1\mathrm{/}2}$. If two parameter values give the same point, then both $e^{2i(t-s)}$ and $e^{3i(t-s)}$ equal $1$, whence $e^{i(t-s)}=1$; its derivative is nowhere zero, so it is an embedded circle. As $\arg z_1$ makes one turn, $\arg z_2$ advances by $3\pi$, so the two points in each $z_1$-meridian disk exchange by three positive half-twists; following the single closed component for two turns gives exactly the closure of the stated two-strand braid. Let $Q=2^{-1\mathrm{/}2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$ and $U_s=Q\mathrm{diag}(1,e^{i\pi s})Q^{\ast }$ for $0\le s\le 1$. Each $U_s$ is unitary and hence preserves the norm and has positive real determinant $\mathrm{\mid }{\det }_{\mathbb{C}}{U}_s{\mathrm{\mid }}^2=1$; $U_0=I$ and $U_1(z_1,z_2)=(z_2,z_1)$. Thus [F5] makes this path an orientation-preserving ambient isotopy carrying $\gamma$ to $\tilde{\gamma }(t)=2^{-1\mathrm{/}2}(e^{3it},e^{2it})$. On $\tilde{\gamma }$, $h=2^{-1\mathrm{/}2}\sin (2t)$ has derivative $h^{\mathrm{\prime }}(t)=\sqrt{2}\cos (2t)$ and second derivative $h^{\mathrm{\prime }\mathrm{\prime }}(t)=-2\sqrt{2}\sin (2t)$. Its critical parameters are $\pi \mathrm{/}4,3\pi \mathrm{/}4,5\pi \mathrm{/}4,7\pi \mathrm{/}4$; the second derivative is negative at $\pi \mathrm{/}4,5\pi \mathrm{/}4$ and positive at the other two. Hence it is a Morse representative with exactly two maxima and [F1] gives $b(T)\le 2$.

Switch the middle crossing of the braid diagram. Its braid word becomes ${\sigma }_1{\sigma }_1^{-1}{\sigma }_1$. The adjacent inverse crossings cancel by Reidemeister II, and the remaining one-crossing closure is a Reidemeister-I curl, which deletes to the standard unknot. Thus one crossing change suffices and $u(T)\le 1$.

Substitute $c=ab{a}^{-1}$ into $a=bc{b}^{-1}$ to obtain $a=bab{a}^{-1}{b}^{-1}$, which after right multiplication by $ba$ is $aba=bab$. Conversely, under this relation, $ca{c}^{-1}=ab{a}^{-1}aa{b}^{-1}{a}^{-1}=aba{b}^{-1}{a}^{-1}=bab{b}^{-1}{a}^{-1}=b$, so the second crossing relation follows. Eliminating $c$ therefore gives the trefoil group presentation $⟨a,b\mid aba=bab⟩$.

The assignments $a\mapsto (12)$ and $b\mapsto (23)$ respect $aba=bab$, since both sides map to $(13)$. The two transpositions generate $S_3$, so this presentation surjects onto the nonabelian group $S_3$ and is itself nonabelian.

The crossing-free code of the standard unknot has one generator and no relators by [F3], so its group is cyclic and therefore abelian. By [F5], ambient isotopy to the unknot would give an isomorphism of this group with the nonabelian group from step 3.1. Thus $T$ is nontrivial.

A Morse representative is parametrized by the standard unit circle $S^1=\{(x,y):x^2+y^2=1\}$ from [F11]. This set is nonempty, bounded because $x^2+y^2=1$, and closed as the inverse image of $\{1\}$ under the continuous polynomial $(x,y)\mapsto x^2+y^2$; [F12] makes it compact in its Euclidean metric. The height restricted to the circle is continuous, so [F10] gives a global, and therefore local, maximum; thus $b(T)\ge 1$. By [F7] and step 4.1, $b(T)\mathrm{\ne }1$; step 1.2 gives $b(T)\le 2$. Therefore $b(T)=2$.

A sequence with zero crossing changes would leave the represented knot type unchanged by [F6], so it cannot carry the nontrivial knot from step 4.1 to the standard unknot. Hence $u(T)\ge 1$; step 1.3 gives the reverse inequality and $u(T)=1$.