Every geometric braid is isotopic to a stacking of signed elementary half twists

The cases $n\le 1$. If $n\le 1$ there is no index $i$ with $1\le i\le n-1$, so the family of generators is empty and its empty word has value $e$; moreover every strand satisfies $z_j(1)=q_j$, because for $n=1$ the top set $\{z_1(1)\}=\{q_1\}$ forces $z_1(1)=q_1$ and for $n=0$ there is no strand; then $H_j(s,t):=(1-s)z_j(t)+s{q}_j$ is a braid isotopy from $\beta$ to $e$, since it is jointly continuous, every value is a convex combination of two points of the convex set $D^{\circ }$, each slice has bottom $q_j$ and top set $\{q_j\}$, and a slice has at most one strand, so no two strands of a slice can meet. Hence $\beta \sim e$, the conclusion holds with $M=0$, and for the rest of the proof $n\ge 2$.

Reduction to generic polygonal representatives, and the structure of their crossings. For $M\ge 0$, let $\mathrm{A}(M)$ assert that every generic polygonal braid based at $Q$ with exactly $M$ projected crossing heights is isotopic to the signed half-twist word read from those crossings as in the statement, with chronological first crossing on the right in stacking order. By [F5] there is a generic polygonal braid ${\beta }^{\mathrm{\prime }}$ based at $Q$ with ${\beta }^{\mathrm{\prime }}\sim \beta$, and it is enough to prove the assertion for ${\beta }^{\mathrm{\prime }}$, because $\sim$ is transitive (two braid isotopies that meet end to end paste to a jointly continuous family by [F8]) and because the class in $G_n$ is unchanged; for such a ${\beta }^{\mathrm{\prime }}$ the differences $\mathrm{\Re }{z}_i-\mathrm{\Re }{z}_j$ of first coordinates are continuous functions of the height, so on each of the intervals $[0,c_1),(c_1,c_2),\dots ,(c_{m^{\mathrm{\prime }}},1]$ the left-to-right order of the labels is constant; on $[0,c_1)$ that order is $1,2,\dots ,n$, because $\mathrm{\Re }{\beta }_1^{\mathrm{\prime }}(0)<\cdots <\mathrm{\Re }{\beta }_n^{\mathrm{\prime }}(0)$ by [F1]; at $t=c_1$ the two strands with equal first coordinate are therefore adjacent in that order, and because no third strand has that first coordinate at $c_1$ the two of them are precisely the strands labelled $p$ and $p+1$, where $p$ is one more than the number of strands whose first coordinate at $c_1$ is strictly smaller than the common first coordinate $\chi :=\mathrm{\Re }{\beta }_p^{\mathrm{\prime }}(c_1)$ of the pair.

Reparametrisation of heights is a braid isotopy. Let $\phi \text{\hspace{0.17em}⁣}:I\to I$ be continuous and nondecreasing with $\phi (0)=0$ and $\phi (1)=1$, and let $\delta$ be a braid based at $Q$; then ${\delta }_j^{\phi }(t):={\delta }_j(\phi (t))$ is a braid based at $Q$, and ${\delta }^{\phi }\sim \delta$: indeed ${\phi }_s(t):=(1-s)t+s\phi (t)$ is again continuous and nondecreasing with ${\phi }_s(0)=0$, ${\phi }_s(1)=1$, and $H_j(s,t):={\delta }_j({\phi }_s(t))$ is jointly continuous with every slice a braid, because its bottom values are ${\delta }_j(0)=q_j$, its top values are ${\delta }_j(1)$, and the values ${\delta }_j({\phi }_s(t))$ are pairwise distinct for each $s,t$ as they are values of $\delta$ at the single height ${\phi }_s(t)$.

One-crossing braids: the hypotheses and their endpoint configuration. Let $P$ abbreviate the following hypothesis on a braid $\alpha$ based at $Q$: there are $p\in \{1,\dots ,n-1\}$ and $c\in (0,1)$ such that (i) the horizontal order of labels for $t<c$ is $1,\dots ,n$; (ii) for $t>c$ it is the sequence $1,\dots ,p-1,p+1,p,p+2,\dots ,n$, omitting the left or right block if $p=1$ or $p=n-1$; and (iii) at $c$ only the pair $p,p+1$ has equal first coordinates, its points are distinct, and its common first coordinate lies strictly between those of the neighboring labels whenever those neighbors exist. Under $P$ the top configuration is forced: at height $1$ the left-to-right order of labels is the sequence in (ii), while that of $q_1,\dots ,q_n$ is $1,\dots ,n$ by [F1]. Thus ${\alpha }_j(1)=q_j$ for $j\mathrm{\notin }\{p,p+1\}$, ${\alpha }_{p+1}(1)=q_p$ and ${\alpha }_p(1)=q_{p+1}$; writing $\tau :=(pp+1)$, we have $\pi (\alpha )=\tau$ and $\alpha (1)=Q_{\tau }$, where $(Q_{\tau }{)}_j:=q_{\tau (j)}$.

One-crossing braids: straightening the motion on each side of the crossing. Assume $P$ of step 1.4, and define ${\mu }_j(t):=(1-t\mathrm{/}c)q_j+(t\mathrm{/}c){\alpha }_j(c)$ for $t\le c$ and ${\nu }_j(t):=\frac{1-t}{1-c}{\alpha }_j(c)+\frac{t-c}{1-c}{\alpha }_j(1)$ for $t\ge c$; for $s\in I$ let ${\alpha }_j^{(s)}(t):=(1-s){\alpha }_j(t)+s{\mu }_j(t)$ for $t\le c$ and ${\alpha }_j^{(s)}(t):=(1-s){\alpha }_j(t)+s{\nu }_j(t)$ for $t\ge c$. Each ${\alpha }^{(s)}$ is a braid based at $Q$: its values lie in the convex set $(D^{\circ }{)}^n$, it is continuous on each of the two closed pieces by [F8] and hence continuous, its bottom values are ${\alpha }_j^{(s)}(0)=q_j={\mu }_j(0)$ and ${\nu }_j(1)={\alpha }_j(1)=q_{\tau (j)}$, so its top set is $\{q_1,\dots ,q_n\}$; and collisions are impossible, since for $t<c$ the differences of first coordinates are $\mathrm{\Re }{\alpha }_j^{(s)}(t)-\mathrm{\Re }{\alpha }_i^{(s)}(t)=(1-s)(\mathrm{\Re }{\alpha }_j(t)-\mathrm{\Re }{\alpha }_i(t))+s(\mathrm{\Re }{\mu }_j(t)-\mathrm{\Re }{\mu }_i(t))$ with the first summand positive for $i<j$ by (i) and the second equal to $(1-t\mathrm{/}c)2h(j-i)+(t\mathrm{/}c)(\mathrm{\Re }{\alpha }_j(c)-\mathrm{\Re }{\alpha }_i(c))\ge 0$ by [F1] and (iii), while for $t>c$ and $i$ preceding $j$ in the order of (ii) the corresponding expression with $\nu$ has second summand $\frac{t-c}{1-c}(\mathrm{\Re }{q}_{\tau (j)}-\mathrm{\Re }{q}_{\tau (i)})>0$ by [F1] and (ii), and at $t=c$ both formulas give the collision-free configuration $\alpha (c)$. Hence $s\mapsto {\alpha }^{(s)}$ is a braid isotopy from $\alpha$ to the two-piece affine braid ${\alpha }^{(1)}$ with ${\alpha }_j^{(1)}(t)={\mu }_j(t)$ for $t\le c$ and ${\alpha }_j^{(1)}(t)={\nu }_j(t)$ for $t\ge c$, so $\alpha \sim {\alpha }^{(1)}$.

The induction base: braids with no crossings. Let $\delta$ be a generic polygonal braid based at $Q$ with $m^{\mathrm{\prime }}=0$, so that no two strands ever have equal first coordinates; then by [F1] and step 1.2 the left-to-right order of the labels is the constant order $1,\dots ,n$, so ${\delta }_j(1)=q_j$ for every $j$ and $H_j(s,t):=(1-s){\delta }_j(t)+s{q}_j$ is a braid isotopy from $\delta$ to $e$: it is jointly continuous, every value lies in the convex set $D^{\circ }$, each slice has bottom $q_j$ and top $q_j$, and $\mathrm{\Re }{H}_j(s,t)-\mathrm{\Re }{H}_i(s,t)=(1-s)(\mathrm{\Re }{\delta }_j(t)-\mathrm{\Re }{\delta }_i(t))+s2h(j-i)>0$ for $i<j$ and every $s$, so no slice has a collision. Hence $\delta \sim e$, the empty word, and assertion $\mathrm{A}(0)$ of step 1.2 holds.

One-crossing braids: the wall of configurations with the pair vertically aligned. Assume $P$ of step 1.4, write ${\alpha }_p(c)=(\chi ,y_1)$ and ${\alpha }_{p+1}(c)=(\chi ,y_2)$ with $y_1\mathrm{\ne }{y}_2$ and $\chi =\mathrm{\Re }{\alpha }_p(c)$, put $m_p:=q_p+(h,0)=((2p-n)h,0)$, and choose the sign $ϵ:=+1$ if $y_1<y_2$ and $ϵ:=-1$ if $y_1>y_2$; let $T^{ϵ}$ be the configuration with $T_j^{ϵ}:=q_j$ for $j\mathrm{\notin }\{p,p+1\}$, $T_p^{ϵ}:=m_p+(0,-ϵh)$ and $T_{p+1}^{ϵ}:=m_p+(0,ϵh)$, and put $P_s:=(1-s)\alpha (c)+s{T}^{ϵ}$ for $s\in I$. Each $P_s$ is a collision-free configuration whose first coordinates are increasing with the single tie $\mathrm{\Re }{P}_s[p]=\mathrm{\Re }{P}_s[p+1]$: the values lie in $(D^{\circ }{)}^n$ by [F7]; for $j<p$ one has $\mathrm{\Re }{P}_s[j]=(1-s)\mathrm{\Re }{\alpha }_j(c)+s\mathrm{\Re }{q}_j<(1-s)\chi +s\mathrm{\Re }{m}_p=\mathrm{\Re }{P}_s[p]$, and symmetrically $\mathrm{\Re }{P}_s[j]>\mathrm{\Re }{P}_s[p+1]$ for $j>p+1$, because both endpoint inequalities are strict, so the pair strands never meet the others; the other strands keep their strict relative order for the same reason; and $P_s[p]-P_s[p+1]=i((1-s)(y_1-y_2)-2ϵhs)$ is purely imaginary and nonzero for every $s$, because for $ϵ=+1$ both summands are $\le 0$ and vanish simultaneously only if $s=1$ and $y_1=y_2$ or $s=0$ and $y_1=y_2$, and symmetrically for $ϵ=-1$; consequently the family ${\beta }_s$ that equals the affine path from $Q$ to $P_s$ on $[0,c]$ and the affine path from $P_s$ to $Q_{\tau }$ on $[c,1]$ consists of braids depending jointly continuously on $(s,t)$: for $t<c$ the difference of the $j$-th and $i$-th first coordinates is $(1-t\mathrm{/}c)2h(j-i)+(t\mathrm{/}c)(\mathrm{\Re }{P}_s[j]-\mathrm{\Re }{P}_s[i])>0$ for $i<j$, for $t>c$ it is $\frac{1-t}{1-c}(\mathrm{\Re }{P}_s[j]-\mathrm{\Re }{P}_s[i])+\frac{t-c}{1-c}(\mathrm{\Re }{q}_{\tau (j)}-\mathrm{\Re }{q}_{\tau (i)})>0$ for $i$ preceding $j$ in the order of (ii), and at $t=c$ the configuration is $P_s$, collision-free by the above; since ${\beta }_0={\alpha }^{(1)}$ and ${\beta }_1$ is the affine two-piece path through $T^{ϵ}$, this gives ${\alpha }^{(1)}\sim {\beta }_1$.

One-crossing braids: the end of the family is a signed half twist. In the notation of step 3.1, let $\phi \text{\hspace{0.17em}⁣}:I\to I$ be the nondecreasing piecewise affine map with $\phi (t)=t\mathrm{/}(2c)$ for $t\le c$ and $\phi (t)=\frac{1}{2}+\frac{t-c}{2(1-c)}$ for $t\ge c$; comparing the formulas of [F4] with the two-piece affine motion of step 3.1, whose pair moves affinely from $(q_p,q_{p+1})$ to $(m_p+(0,-ϵh),m_p+(0,ϵh))$ and then affinely to $(q_{p+1},q_p)$ while every other strand stays at its base point, gives ${\beta }_{1,j}(t)={\sigma }_{p,j}^{ϵ}(\phi (t))$ for every $j$ and $t$: for $ϵ=+1$ the pair motions of ${\beta }_1$ are those of $m_p\pm \rho$ on the two halves reparametrised by $\phi$, and for $ϵ=-1$ those of $m_p\pm {\rho }^{-}$. Hence ${\beta }_1={\sigma }_p^{ϵ}\circ \phi$ in the sense of step 1.3 and therefore ${\beta }_1\sim {\sigma }_p^{ϵ}$.

The one-crossing claim. Under hypothesis $P$ of step 1.4 the braid $\alpha$ is braid-isotopic to ${\sigma }_p^{ϵ}$, where $ϵ=+1$ when the strand labelled $p$ has the smaller second coordinate at the crossing height $c$ and $ϵ=-1$ otherwise: this is $\alpha \sim {\alpha }^{(1)}$ from step 2.1, ${\alpha }^{(1)}\sim {\beta }_1$ from step 3.1 and ${\beta }_1\sim {\sigma }_p^{ϵ}$ from step 4.1, composed with the transitivity of braid isotopy.

The induction step: cutting at a height between the two lowest crossings. Let ${\beta }^{\mathrm{\prime }}$ be generic polygonal with $m^{\mathrm{\prime }}\ge 1$ crossing heights $c_1<\cdots <c_{m^{\mathrm{\prime }}}$, let $p_1$ be as in step 1.2 for the crossing $c_1$, and let ${ϵ}_1=+1$ when the strand labelled $p_1$ has the smaller second coordinate at height $c_1$ and ${ϵ}_1=-1$ otherwise; if $m^{\mathrm{\prime }}=1$ then ${\beta }^{\mathrm{\prime }}$ satisfies hypothesis $P$ of step 1.4 with $p:=p_1$ and $c:=c_1$, so step 5.1 gives ${\beta }^{\mathrm{\prime }}\sim {\sigma }_{p_1}^{{ϵ}_1}$ and $\mathrm{A}(1)$ holds. If $m^{\mathrm{\prime }}\ge 2$, choose heights ${\theta }_1<\theta <{\theta }_2$ with $c_1<{\theta }_1<\theta <{\theta }_2<c_2$, let $l_1,\dots ,l_n$ be the labels in left-to-right order at height $\theta$ (constant on $(c_1,c_2)$ and hence on $[{\theta }_1,{\theta }_2]$), and let $Q_{\sigma }$ be the configuration with $(Q_{\sigma }{)}_{l_k}:=q_k$ for every $k$, so that $Q_{\sigma }$ has the same left-to-right label order as ${\beta }^{\mathrm{\prime }}(\theta )$ and both lie in the convex set $C:=\{u:\mathrm{\Re }{u}_{l_1}<\cdots <\mathrm{\Re }{u}_{l_n}\}$; replacing ${\beta }^{\mathrm{\prime }}$ on $[{\theta }_1,{\theta }_2]$ by the two-piece affine path through $Q_{\sigma }$, by the same interpolation as in steps 2.1 and 3.1, gives a braid isotopy from ${\beta }^{\mathrm{\prime }}$ to a generic polygonal braid ${\beta }^{\ast }$ with ${\beta }^{\ast }(\theta )=Q_{\sigma }$, because the interpolation of two points of the convex set $C\cap (D^{\circ }{)}^n$ stays in it, so no slice acquires a collision and the values stay in the disc; then ${\alpha }_j(u):={\beta }_j^{\ast }(\theta u)$ and ${\gamma }_k(v):={\beta }_{l_k}^{\ast }(\theta +v(1-\theta ))$ are braids based at $Q$, because ${\alpha }_j(0)=q_j$ and ${\gamma }_k(0)=q_k$, while their top configurations are permutations of $Q$; the braid $\alpha$ satisfies hypothesis $P$ with the crossing $c_1\mathrm{/}\theta$ and the same index $p_1$, since below that crossing the order is the base order and above it the order is the transposed order of (ii); and ${\beta }^{\ast }=\gamma \star \alpha$ up to the monotone reparametrisation $\phi (t)=t\mathrm{/}(2\theta )$ for $t\le \theta$, $\phi (t)=\frac{1}{2}+\frac{t-\theta }{2(1-\theta )}$ for $t\ge \theta$ of step 1.3, so ${\beta }^{\mathrm{\prime }}\sim \gamma \star \alpha$; finally $\gamma$ is generic polygonal with the $m^{\mathrm{\prime }}-1$ crossing heights $(c_r-\theta )\mathrm{/}(1-\theta )$ for $r\ge 2$. Relabelling the upper braid by its current horizontal rank does not alter the rank pair or the vertical sign at any upper crossing, so its crossing word is exactly the suffix of the crossing word of ${\beta }^{\ast }$.

The induction step concluded. Assume $m^{\mathrm{\prime }}\ge 2$ and the notation of step 6.1; by the induction hypothesis $\mathrm{A}(m)$ for every $m<m^{\mathrm{\prime }}$, applied to the generic polygonal braid $\gamma$ with its $m^{\mathrm{\prime }}-1<m^{\mathrm{\prime }}$ crossings, the braid $\gamma$ is isotopic to its exact signed crossing word $W$, and by step 5.1 the braid $\alpha$ satisfies $\alpha \sim {\sigma }_{p_1}^{{ϵ}_1}$; hence, using that stacking respects braid isotopy in each factor, ${\beta }^{\mathrm{\prime }}\sim \gamma \star \alpha \sim W\star {\sigma }_{p_1}^{{ϵ}_1}$, which is again a stacking of signed elementary half twists. Together with the case $m^{\mathrm{\prime }}=1$ of step 6.1 and the base case $\mathrm{A}(0)$ of step 2.2, the induction of [L9] gives $\mathrm{A}(m^{\mathrm{\prime }})$ for every $m^{\mathrm{\prime }}$.

Conclusion, and generation of $G_n$. For the original braid $\beta$ let ${\beta }^{\mathrm{\prime }}$ be the generic polygonal representative of [F5]; it has finitely many crossing heights, say $m^{\mathrm{\prime }}$, and $\beta \sim {\beta }^{\mathrm{\prime }}$; by $\mathrm{A}(m^{\mathrm{\prime }})$ of step 7.1 there are $M\ge 0$, indices $i_k$ and signs ${\epsilon }_k$ with ${\beta }^{\mathrm{\prime }}\sim {\sigma }_{i_1}^{{\epsilon }_1}\star \cdots \star {\sigma }_{i_M}^{{\epsilon }_M}$ and hence $\beta \sim {\sigma }_{i_1}^{{\epsilon }_1}\star \cdots \star {\sigma }_{i_M}^{{\epsilon }_M}$; moreover the more precise reading of the word, with $p_k$ and ${\epsilon }_k$ as in step 6.1 for each crossing, is exactly the one recorded in the statement. Passing to classes in $G_n$ with [F3] and [F6] gives $[\beta ]=[{\sigma }_{i_1}{]}^{{\epsilon }_1}\cdots [{\sigma }_{i_M}{]}^{{\epsilon }_M}$, using $[{\sigma }_i^{+1}]=[{\sigma }_i]$ and $[{\sigma }_i^{-1}]=[{\sigma }_i{]}^{-1}$ of [F4] and the empty word for $M=0$; since every element of $G_n$ is the class of a braid, the classes $[{\sigma }_1],\dots ,[{\sigma }_{n-1}]$ generate $G_n$, and for $n\le 1$ step 1.1 gives the trivial group on the empty family. ∎