Oriented Links, Braid Closures, and Markov Equivalence — Examples

The literal word model and component count. Write $R=\sqrt{1-r^2}$. The finite signed half-turn word has disk strands $\pm r{e}^{i\mathrm{\Theta }(t)}$, where $\mathrm{\Theta }$ is smooth, $\mathrm{\Theta }(0)=0$, $\mathrm{\Theta }(1)=m\pi$ and all positive-order endpoint derivatives vanish. Take $\mathrm{\Theta }=0$ at $m=0$; concatenating the finitely many flattened half-turn angles gives such a $\mathrm{\Theta }$ for every other $m$. The two points stay distinct. Its literal smooth closure has $z=R{e}^{2\pi it}$, $w=\pm r{e}^{i\mathrm{\Theta }(t)}$ by [F1]. The endpoint permutation is $(12{)}^m$, so there is one component when $m$ is odd and two when $m$ is even, including $m=0$. This is $\gcd (2,m)$, with the positive gcd convention for negative $m$.

An explicit ambient isotopy to the uniform torus model. Set $\mathrm{\Delta }(t)=m\pi t-\mathrm{\Theta }(t)$. Its values at both ends are zero, its first derivatives at both ends are $m\pi$, and all higher endpoint derivatives agree; hence it is a smooth function on the page circle. Choose a fixed smooth radial cutoff $\chi (\mathrm{\mid }x{\mathrm{\mid }}^2)$ equal to one near $r^2$ and zero near $\mathrm{\mid }x\mathrm{\mid }=1$. On $V$ use $H_s([(x,t)])=[(e^{is\chi (\mathrm{\mid }x{\mathrm{\mid }}^2)\mathrm{\Delta }(t)}x,t)]$. Its inverse rotates by the negative angle, since $\mathrm{\mid }x\mathrm{\mid }$ is unchanged. It is a smooth isotopy, and through the explicit $\phi$ it extends by identity across the axis because the cutoff is zero near the disk boundary. It preserves orientation as a smooth path of diffeomorphisms starting at identity. At $s=1$ the curve is exactly $z=R{e}^{2\pi it}$, $w=\pm r{e}^{im\pi t}$. For odd $m$, concatenate its two strands with parameter $u\in \mathbb{R}\mathrm{/}2\mathbb{Z}$; this gives $z=R{e}^{2\pi iu}$, $w=r{e}^{im\pi u}$, with winding pair $(2,m)$ after parameter $u\mathrm{/}2$. For even $m$, each strand is a circle with winding pair $(1,m\mathrm{/}2)$ and the two are disjoint. These are the standard torus curves $T(2,m)$, including the stated component count. No generic continuous-braid smoothing theorem or Markov-preservation assumption is used.

The two one-crossing oriented unknots without Choice. For $m=ϵ=\pm 1$, use the chart of $S^3\setminus \{w=0\}$ with $\theta =\arg w$ and $\xi =z{e}^{-2ϵi\theta }$. Its inverse is $(\xi ,\theta )\mapsto (\xi e^{2ϵi\theta },\sqrt{1-\mathrm{\mid }\xi {\mathrm{\mid }}^2}{e}^{i\theta })$, so it is explicitly $D_{\xi }^{\circ }\times S^1$. The curve from step 2.1 has $\xi =R$ constant. Move this disk point along the real segment from $R$ to $0$ by finitely many small cutoff translations $F_s(x)=x+s\eta (x)v$, with compact support in $D_{\xi }^{\circ }$ and $\mathrm{\parallel }D\eta {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }v\mathrm{\mid }<1$. A fixed smooth bump and a sufficiently fine finite subdivision suffice because that compact segment has positive distance $1-R$ from the boundary. The inverse is the unique limit of $x_{k+1}=y-s\eta (x_k)v$, starting at $x_0=y$: successive differences are bounded by a geometric sequence with ratio at most $\mathrm{\parallel }D\eta {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }v\mathrm{\mid }<1$, so the explicit sequence converges and its fixed point is unique. The Jacobian has determinant $1+sD\eta \cdot v>0$; [F3] makes the already constructed inverse smooth locally and hence globally. Each map is identity outside its compact interior support, so gives a disk diffeomorphism. Compose these finitely many disk-map families with the same parameter $s$; the finite smooth composition starts at identity and at $s=1$ sends $R$ to $0$. Its product with the unchanged $\theta$ extends by identity across $\{w=0\}$, since $\mathrm{\mid }\xi \mathrm{\mid }\to 1$ there. This ambient isotopy takes the curve to $(0,e^{i\theta })$. The explicit unitary rotation $(z,w)\mapsto (\cos az+\sin aw,-\sin az+\cos aw)$, $0\le a\le \pi \mathrm{/}2$, then takes it to the trivial one-braid circle $(e^{i\theta },0)$. For $ϵ=1$ its orientation is already the positive page orientation. For $ϵ=-1$ reverse that circle's negative angular orientation using $(z,w)\mapsto (\bar{z},\bar{w})$; this map is the endpoint of the explicit $\pi$ rotation in the real plane of the two imaginary coordinates, an orientation-preserving path in $SO(4)$. Thus both signs give the oriented unknot. Every construction is explicit or a finite selection; no countable choice is used.

The unlink, trefoils and conclusion. At $m=0$ step 1.1 is the literal trivial two-braid closure, whose explicit disjoint spanning disks are [F1]. At $m=\pm 3$, step 2.1 gives the standard $(2,\pm 3)$ torus knots, the trefoils named in the Statement. The map $(z,w)\mapsto (z,\bar{w})$ changes $m$ to $-m$ and reverses the ambient orientation, so the two are mirrors. Steps 1.1-3.1 give the component formula and both one-crossing unknots, and the coordinate curves verify $T(2,m)$ for every integer $m$.

The closures of the stabilizations. By the preliminary ambient positioning in [F1], the two fixed-framing closures have representatives which are the round unknot with respectively a positive and a negative kink. Both closures are one-component links by [F3], since the endpoint permutation of each stabilization of $e$ is the transposition of the two strands.

The R1 isotopies. The explicit isotopy for the positive sign is the R1 move that pulls the kink straight inside a small ball neighbourhood of the kink, keeping the rest of the closed braid fixed; for the negative sign the mirror isotopy unrolls the opposite kink. In both cases the R1 move is realized by an ambient isotopy by [F2], so the closure of each stabilization is equivalent to the closure of the trivial braid, the unknot.

Conclusion. Both stabilizations $e{\sigma }_1$ and $e{\sigma }_1^{-1}$ in $B_2$ close to the unknot, with the explicit R1 isotopies of the two signs; the example illustrates that stabilization preserves the closure in the simplest possible case.

The source's five-arc picture and six circle pairs. Number the four circles in the first sketch of Figure 4 by their northwest, northeast, southwest and southeast positions. The northwest and southeast arrows are counterclockwise; the northeast and southwest arrows clockwise. For side-by-side circles in $S^2$, coherence requires opposite planar orientations, since the common annulus is outside their two disk interiors. Thus exactly the two diagonal pairs are incoherent; each of the four side pairs is coherent, so $h=2$. The five positive arcs are the two between the top circles, one on each vertical side and one on the bottom side, as shown in that rendered source panel. Their signs and incident circles record the original five crossings by [F1]. The heavy ${\alpha }_1$ joins the northwest and southeast incoherent pair; [F2] licenses its first reduction.

The two reductions. Perform the reducing move along ${\alpha }_1$: it replaces the chosen incoherent pair by two coherent circles joined by two oppositely signed arcs, and by [F2] the height drops to $1$; the new picture has a remaining incoherent pair, and the source's second reducing arc ${\alpha }_2$ joins it. Performing the reducing move along ${\alpha }_2$ drops the height to $0$, and the resulting picture has all pairs of Seifert circles coherent. The two moves are Reidemeister II moves of the original diagram, so the represented knot is unchanged.

The source word and its product convention. The final sketch has four concentric counterclockwise circles. Number them outermost to innermost and read from twelve o'clock counterclockwise. The ordered signed events are $(2,+),(1,-),(2,+),(3,-),(2,+),(1,+),(2,+),(3,+),(2,+)$, giving the displayed chronological list $X$. For this counterclockwise motion use the transverse frame $(-e_r,+e_Z)$; together with tangent $+e_{\theta }$ it preserves ambient orientation. Its positive half twist has the first outer indexed point go through negative physical depth, so the source's crossing signs agree with the fixed signed generators. By [F7] the actual geometric element of this chronological list is $\mathrm{rev}(X)$, not automatically $X$. We verify both closures by an explicit word comparison. Write $s={\sigma }_1$, $t={\sigma }_2$, $r={\sigma }_3$. The Artin relations [F4] give $X=t{s}^{-1}t{r}^{-1}tstrt=t{s}^{-1}t{r}^{-1}stsrt=t{s}^{-1}ts{r}^{-1}trst=t{s}^{-1}tstr{t}^{-1}st\mathrm{.}$ The successive operations are $tst=sts$, commuting $s$ with $r$, and $r^{-1}tr=tr{t}^{-1}$. Put $A=t{s}^{-1}tst$, $B=t^{-1}st$, so $X=ArB$. By the interior insertion sequence [F5], this is related by one positive ordinary destabilization and conjugations to the three-braid $Y=AB=t{s}^{-1}tst{t}^{-1}st$. Inverse cancellation gives $Y=Z=t{s}^{-1}t{s}^{2}t$, a six-letter three-braid. Word reversal is an anti-automorphism because the Artin relations are palindromic or far commutations. It carries a right stabilization to a left one of the same sign, which cyclic conjugation converts to a right stabilization; hence the reversed identities likewise give $\mathrm{rev}(X)\leftrightarrow \mathrm{rev}(Z)$ by ordinary Markov moves. Finally put $C=t{s}^{2}t$. Directly $sC=st{s}^{2}t=(sts)st=(tst)st=ts(tst)=ts(sts)=t{s}^{2}ts=Cs\mathrm{.}$ Therefore $t^{-1}Zt=s^{-1}Ct=C{s}^{-1}t=\mathrm{rev}(Z)$. Combining the two destabilization comparisons with this old-strand conjugation gives an explicit Markov sequence between $X$ and $\mathrm{rev}(X)$. By [F6], both have the oriented closure of the source diagram. The algorithm itself yields the nine-letter four-braid; $Y$ is its eight-letter three-braid after destabilization and $Z$ is a strictly shorter word for that same three-braid.

Conclusion. The verified two reductions give the source's four-strand, nine-crossing closed braid. Step 3.1 proves that its actual chronological interpretation and the commissioned word $X$ have the same oriented closure, and explicitly gives the eight-letter and six-letter three-braid words. The existence of the three-braid proves that the initial four-braid uses more strands than necessary; the six-letter representative proves that the eight-letter word is longer than necessary. These are separate index and length comparisons, with no assertion that the three-braid has nonminimal braid index. AC supplies the reducing/height-zero constructions and the countable choice in [F6].

Both closures are unknots. The trivial braid $e\in B_1$ closes to a round unknot. Its positive stabilization is ${\sigma }_1\in B_2$, whose closure is the unknot by [F2]; stabilizing ${\sigma }_1$ once more, adding the third strand on the right, gives the braid ${\sigma }_1{\sigma }_2\in B_3$, whose closure is equivalent to the closure of ${\sigma }_1$ by [F3], hence also an unknot. The component count is one in both cases by [F4], so both closures are one-component links, and by [F2] and [F3] they are the unknot.

They are not conjugate. Conjugation, by [F1], stays inside a fixed braid group $B_n$ and never changes the number of strands. The braid ${\sigma }_1\in B_2$ has two strands and ${\sigma }_1{\sigma }_2\in B_3$ has three; no sequence of conjugations within a single braid group can relate them, because such a sequence would have to identify an element of $B_2$ with an element of $B_3$. Therefore conjugation alone does not classify braid closures: the two closures are equivalent oriented links (both the unknot) while the braids are not conjugate.

Conclusion. The braids ${\sigma }_1\in B_2$ and ${\sigma }_1{\sigma }_2\in B_3$ have equivalent closures but are not conjugate; the only difference between the two braids in Markov terms is the stabilization that changes the group from $B_2$ to $B_3$, which is the strand-changing move that conjugacy cannot simulate. This verifies the counterexample and shows that at least one stabilization or destabilization is genuinely necessary for the classification. The statement retains AC because it consumes the countable-choice-stated closure-preservation lemma [F3], with countable choice following from AC by AC implies DC implies countable choice and the ambient isotopy theory behind [F2].