Geometric two strand braids are integer twists

The relative motion and its endpoint values. Assume $n=2$ and let $\beta =(z_1,z_2)$ be a braid based at $Q$; then $w=z_1-z_2$ is continuous and $w(t)\mathrm{\ne }0$ for every $t$ by [F1] and [F8], since distinct strands do not meet; the endpoint condition $\{z_1(1),z_2(1)\}=\{q_1,q_2\}$ of [F1] gives $w(1)=z_1(1)-z_2(1)\in \{q_1-q_2,q_2-q_1\}=\{(-2h,0),(2h,0)\}$, and by the convention $z_j(1)=q_{\pi (\beta )(j)}$ of [F1] the value $w(1)=(-2h,0)$ occurs exactly when $\pi (\beta )$ is the identity, while $w(1)=(2h,0)$ occurs exactly when $\pi (\beta )$ is the transposition of $1$ and $2$.

The argument class and its lift. By [F7] the class map $c=H^{-1}\circ r$ is continuous, so $c\circ w\text{\hspace{0.17em}⁣}:I\to \mathbb{R}\mathrm{/}\mathbb{Z}$ is continuous by [F8]; its value at $0$ is $c((-2h,0))=H^{-1}((-1,0))=[\frac{1}{2}]$ because $r((-2h,0))=(-1,0)$ and $H([\frac{1}{2}])=(-1,0)$ by [F6]; hence [F5] applied to the covering $p\text{\hspace{0.17em}⁣}:\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ and the path $c\circ w$ gives a unique continuous $\theta \text{\hspace{0.17em}⁣}:I\to \mathbb{R}$ with $\theta (0)=\frac{1}{2}$ and $[\theta (t)]=c(w(t))$, that is $H([\theta (t)])=w(t)\mathrm{/}\parallel w(t){\parallel }_2$ for every $t\in I$.

The invariant, and its parity against the endpoint permutation. By step 2.1 the class $[\theta (1)]$ equals $c(w(1))$, and by [F6] and [F7] one has $c((-2h,0))=H^{-1}((-1,0))=[\frac{1}{2}]$ and $c((2h,0))=H^{-1}((1,0))=[0]$; so step 1.1 gives $\theta (1)\in \frac{1}{2}+\mathbb{Z}$ when $\pi (\beta )$ is the identity and $\theta (1)\in \mathbb{Z}$ when $\pi (\beta )$ is the transposition. In the first case $2\theta (1)$ is odd and in the second it is even, so in both cases $k(\beta ):=2\theta (1)-1=2(\theta (1)-\theta (0))$ is an integer, and it is even exactly when $\pi (\beta )$ is the identity and odd exactly when $\pi (\beta )$ is the transposition.

Isotopy invariance. Let $Z$ be a braid isotopy from $\beta$ to ${\beta }^{\mathrm{\prime }}$ and put $W(s,t):=Z_1(s,t)-Z_2(s,t)$, a continuous and nowhere vanishing map $I\times I\to {\mathbb{R}}^2\setminus \{0\}$ by [F1] and [F8]; then $c\circ W$ is a homotopy $I\times I\to \mathbb{R}\mathrm{/}\mathbb{Z}$ by [F7] and [F8], and $W(s,0)=z_1(0)-z_2(0)=q_1-q_2=(-2h,0)$ for every $s$ by [F1], so the constant map $s\mapsto \frac{1}{2}$ is a continuous lift of $(c\circ W)(\cdot ,0)$; hence [F5] provides a unique lift $\mathrm{\Theta }\text{\hspace{0.17em}⁣}:I\times I\to \mathbb{R}$ of $c\circ W$ with $\mathrm{\Theta }(s,0)=\frac{1}{2}$ for all $s\in I$. For each $s$ the slice $Z(s,\cdot )$ is a braid based at $Q$ by [F1], so its relative motion is $W(s,\cdot )$, and $t\mapsto \mathrm{\Theta }(s,t)$ is the unique lift of $c\circ W(s,\cdot )$ with value $\frac{1}{2}$ at $t=0$; step 3.1 applied to that slice therefore gives $2\mathrm{\Theta }(s,1)-1\in \mathbb{Z}$ for every $s$. The map $s\mapsto \mathrm{\Theta }(s,1)$ is continuous, so $s\mapsto 2\mathrm{\Theta }(s,1)-1$ is a continuous map from the connected interval $I$ into $\mathbb{Z}$ and is constant by [F8]; moreover $\mathrm{\Theta }(0,\cdot )=\theta$ by the uniqueness in [F5] applied to $c\circ W(0,\cdot )=c\circ w$ and step 2.1, and $\mathrm{\Theta }(1,\cdot )$ is the corresponding lift for the relative motion $W(1,\cdot )$ of ${\beta }^{\mathrm{\prime }}$; therefore $k(\beta )=2\mathrm{\Theta }(0,1)-1=2\mathrm{\Theta }(1,1)-1=k({\beta }^{\mathrm{\prime }})$, and $k$ is constant on braid isotopy classes.

Additivity under stacking. Let $\beta ,\gamma$ be two-strand braids based at $Q$, let ${\theta }_{\beta },{\theta }_{\gamma }$ be their argument lifts of step 2.1, and let $ϵ:=0$ when $\pi (\beta )$ is the identity and $ϵ:=1$ when $\pi (\beta )$ is the transposition, so that $ϵ\equiv k(\beta )(\mathrm{m}\mathrm{o}\mathrm{d}2)$ by step 3.1; by [F3] the relative motion of the stacking is $w_{\gamma \star \beta }(t)=w_{\beta }(2t)$ for $t\le \frac{1}{2}$ and $w_{\gamma \star \beta }(t)=(-1{)}^{ϵ}{w}_{\gamma }(2t-1)$ for $t\ge \frac{1}{2}$. Let $\mathrm{\Theta }$ be the unique lift of $c\circ w_{\gamma \star \beta }$ with $\mathrm{\Theta }(0)=\frac{1}{2}$, granted by [F5]. On $[0,\frac{1}{2}]$ the map $t\mapsto {\theta }_{\beta }(2t)$ is a lift of $c\circ w_{\beta }(2t)$ with value $\frac{1}{2}$ at $t=0$, so $\mathrm{\Theta }(t)={\theta }_{\beta }(2t)$ there by uniqueness of path lifts, and $\mathrm{\Theta }(\frac{1}{2})={\theta }_{\beta }(1)=\frac{1}{2}+\frac{k(\beta )}{2}$, because ${\theta }_{\beta }(1)-{\theta }_{\beta }(0)=\frac{k(\beta )}{2}$ by step 3.1. On $[\frac{1}{2},1]$, since $c((-1{)}^{ϵ}u)=c(u)+ϵ[\frac{1}{2}]$ for $u\mathrm{\ne }0$ by [F6] and [F7], the maps $t\mapsto {\theta }_{\gamma }(2t-1)+\frac{ϵ}{2}$ and $t\mapsto \mathrm{\Theta }(t)$ are two lifts of the same path, so by [F5] the second is the first plus a constant integer: $\mathrm{\Theta }(t)={\theta }_{\gamma }(2t-1)+\frac{ϵ}{2}+z$ for some $z\in \mathbb{Z}$ and all $t\in [\frac{1}{2},1]$. Evaluating at $t=\frac{1}{2}$ gives $\frac{1}{2}+\frac{ϵ}{2}+z=\frac{1}{2}+\frac{k(\beta )}{2}$, that is $z=\frac{k(\beta )-ϵ}{2}$, which is an integer precisely because $ϵ\equiv k(\beta )(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Hence $\mathrm{\Theta }(1)={\theta }_{\gamma }(1)+\frac{ϵ}{2}+\frac{k(\beta )-ϵ}{2}={\theta }_{\gamma }(1)+\frac{k(\beta )}{2}$, and therefore $k(\gamma \star \beta )=2\mathrm{\Theta }(1)-1=2{\theta }_{\gamma }(1)-1+k(\beta )=k(\gamma )+k(\beta )$.

The values on the trivial braid and on the two half twists. The relative motion of the trivial braid $e$, whose strands are the constant maps $t\mapsto q_j$ by [F1], is the constant path $(-2h,0)$, whose argument lift with value $\frac{1}{2}$ at $0$ is the constant $\frac{1}{2}$; so $k(e)=2\cdot \frac{1}{2}-1=0$. For $\beta ={\sigma }_1$ the relative motion is $2\rho$, that is $w(t)=(4th-2h,-4th)$ for $t\le \frac{1}{2}$ and $w(t)=(4th-2h,4th-4h)$ for $t\ge \frac{1}{2}$, by [F2], and it vanishes nowhere: on the first half $w(t)$ runs through the closed third quadrant from $(-2h,0)$ to $(0,-2h)$ and on the second half through the closed fourth quadrant from $(0,-2h)$ to $(2h,0)$, in each case with a direction that turns strictly monotonically; hence the lift with value $\frac{1}{2}$ at $0$ satisfies $\theta (\frac{1}{2})=\frac{3}{4}$ and $\theta (1)=1$, and $k({\sigma }_1)=2(1-\frac{1}{2})=1$. For $\beta ={\sigma }_1^{-}$ the relative motion is the reflection in the horizontal axis of the previous one, running through the second and then the first quadrant, and the same computation gives $\theta (\frac{1}{2})=\frac{1}{4}$ and $\theta (1)=0$, so $k({\sigma }_1^{-})=2(0-\frac{1}{2})=-1$. By the additivity of step 4.2 and induction on $\mathrm{\mid }m\mathrm{\mid }$ this gives $k({\sigma }_1^m)=m$ for every $m\in \mathbb{Z}$.

Every two-strand braid is an integer twist. Let $\beta$ be any braid based at $Q$; by [F4] the class $[\beta ]\in G_2$ is a finite product of the elements $[{\sigma }_1]$ and $[{\sigma }_1{]}^{-1}$, that is $[\beta ]=[{\sigma }_1{]}^m$ for the integer $m$ which is the sum of the exponents of that product, and by steps 4.2 and 5.1 the invariant of $\beta$ is $k(\beta )=k({\sigma }_1^m)=m$; hence $[\beta ]=[{\sigma }_1{]}^{k(\beta )}$, and braid-isotopic braids have equal invariants by step 4.1. Consequently $k$ descends to a well-defined map $G_2\to \mathbb{Z}$ by step 4.1, that map is a group homomorphism by step 4.2, it is surjective because $k({\sigma }_1^m)=m$ for every $m\in \mathbb{Z}$ by step 5.1, and it is injective because $k(\beta )=0$ forces $[\beta ]=[{\sigma }_1{]}^0=[e]$ by the identity above; so $G_2\cong \mathbb{Z}$ via $k$, the twists ${\sigma }_1^m$ represent pairwise distinct classes, and each class of $G_2$ is exactly one of them. ∎