Bridge presentations and unknotting number — Examples and Counterexamples

By [F2], each of the $k$ moves preserves the represented knot type and adds exactly one crossing, while disjoint supports ensure that no move changes crossings created in another disk. Induction on the finite number of moves gives $c(D_k)=k$ and shows that $D_k$ still represents $U$.

By [F1], $u(U)=0$ and $b(U)=1$. Since $k\ge 1$, step 1.1 yields $c(D_k)=k\mathrm{\ne }0=u(U)$, proving the counterexample.

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 $u(T)=1$.