The three strand geometric braid relation

The explicit strand windows of the two words. Applying the stacking formula of [F2] twice, with $m=(-\frac{1}{16},0)$, $m^{\mathrm{\prime }}=(\frac{1}{16},0)$ and the transposition couplings $\pi ({\sigma }_1)=(12)$, $\pi ({\sigma }_2)=(23)$, gives for $W_0$ the strands $(m+\rho (4u),m-\rho (4u),q_3)$ for $0\le u\le \frac{1}{4}$, $(m^{\mathrm{\prime }}+\rho (4u-1),q_1,m^{\mathrm{\prime }}-\rho (4u-1))$ for $\frac{1}{4}\le u\le \frac{1}{2}$ and $(q_3,m+\rho (2u-1),m-\rho (2u-1))$ for $\frac{1}{2}\le u\le 1$, and for $W_1$ the strands $(q_1,m^{\mathrm{\prime }}+\rho (4u),m^{\mathrm{\prime }}-\rho (4u))$ for $0\le u\le \frac{1}{4}$, $(m+\rho (4u-1),q_3,m-\rho (4u-1))$ for $\frac{1}{4}\le u\le \frac{1}{2}$ and $(m^{\mathrm{\prime }}+\rho (2u-1),m^{\mathrm{\prime }}-\rho (2u-1),q_1)$ for $\frac{1}{2}\le u\le 1$. The two windows of $W_0$ agree at $u=\frac{1}{4}$, where the first gives $(m+\rho (1),m-\rho (1),q_3)=((0,0),(-2h,0),(2h,0))$ and the second gives $(m^{\mathrm{\prime }}+\rho (0),q_1,m^{\mathrm{\prime }}-\rho (0))=((0,0),(-2h,0),(2h,0))$, and at $u=\frac{1}{2}$, where the second gives $(m^{\mathrm{\prime }}+\rho (1),q_1,m^{\mathrm{\prime }}-\rho (1))=((2h,0),(-2h,0),(0,0))$ and the third gives $(q_3,m+\rho (0),m-\rho (0))=((2h,0),(-2h,0),(0,0))$; the same two checks apply verbatim to the three windows of $W_1$, so by [F5] the formulas define continuous tuples $W_0,W_1\text{\hspace{0.17em}⁣}:I\to ({\mathbb{R}}^2{)}^3$, and each value of each of the six windows is one of $q_1,q_2,q_3$ or one of $m\pm \rho (v),m^{\mathrm{\prime }}\pm \rho (v)$.

The rotation braid. By [F4] the motion $u\mapsto R_{\pi u}(p_k)$ is continuous for each $k$ and satisfies $R_{\pi u}(p_k)=p_k$ at $u=0$ and $R_{\pi u}(p_k)=-p_k$ at $u=1$, so the strands of $\mathrm{r}\mathrm{o}\mathrm{t}$ run from $p_1=(-2h,0),p_2=(0,0),p_3=(2h,0)$ to $-p_1=(2h,0),-p_2=(0,0),-p_3=(-2h,0)$, that is from $Q$ to $(q_3,q_2,q_1)$; at height $u$ the three positions are $R_{\pi u}(p_k)$, whose mutual distances are those of the distinct points $p_1,p_2,p_3$ because $R_{\pi u}$ preserves the norm and is linear and injective; and every value lies in $D^{\circ }$, since $\parallel R_{\pi u}(p_k){\parallel }_2=\parallel p_k{\parallel }_2\le 2h=\frac{1}{8}<1$; hence $\mathrm{r}\mathrm{o}\mathrm{t}$ is a braid based at $Q$ with endpoint permutation $(13)$.

Collision bounds for the windows. For $v\in I$ the diamond path satisfies $\parallel \rho (v){\parallel }_2^2=h^2(8v^2-4v+1)$ for $v\le \frac{1}{2}$ and $\parallel \rho (v){\parallel }_2^2=h^2((2v-1{)}^2+(2v-2{)}^2)$ for $v\ge \frac{1}{2}$; the first is $\ge h^2\cdot \frac{1}{2}$ with equality at $v=\frac{1}{4}$ and the second is $\ge h^2\cdot \frac{1}{2}$ with equality at $v=\frac{3}{4}$, and both are $\le h^2$. Hence in each window the two moving strands, which are $m+\rho (v)$ and $m-\rho (v)$ (or $m^{\mathrm{\prime }}+\rho (v))$ and $m^{\mathrm{\prime }}-\rho (v)$), are separated by $2\parallel \rho (v){\parallel }_2\ge \sqrt{2}h=\frac{\sqrt{2}}{16}$, and the frozen base point of that window, namely $q_3$ in the first and third windows and $q_1$ in the second, is at distance exactly $3h$ from that window's midpoint and therefore at distance at least $3h-h=2h=\frac{1}{8}$ from each moving point; moreover every window value has norm at most $\parallel m{\parallel }_2+h=\parallel m^{\mathrm{\prime }}{\parallel }_2+h=2h=\frac{1}{8}<1$, so all values lie in $D^{\circ }$. Consequently each of the two window tuples is a collision-free tuple, and together with steps 1.1 and 1.2 this shows that $W_0$ and $W_1$ are braids based at $Q$.

The endpoint permutations. By [F2] and step 1.1, $\pi (W_0)=\pi ({\sigma }_1)\circ \pi ({\sigma }_2\star {\sigma }_1)=\pi ({\sigma }_1)\circ \pi ({\sigma }_2)\circ \pi ({\sigma }_1)$ and $\pi (W_1)=\pi ({\sigma }_2)\circ \pi ({\sigma }_1)\circ \pi ({\sigma }_2)$; evaluating the first composite, $(12)\circ (23)\circ (12)$, at $1,2,3$ gives $1\mapsto (12)(3)=3$, $2\mapsto (12)(1)=2$ and $3\mapsto (12)(2)=1$, that is the transposition $(13)$, and evaluating the second, $(23)\circ (12)\circ (23)$, at $1,2,3$ gives $1\mapsto 1\mapsto 2\mapsto 3$, $2\mapsto 3\mapsto 3\mapsto 2$ and $3\mapsto 2\mapsto 1\mapsto 1$, which is again $(13)$; the same conclusion is read off at $u=1$, where the windows of step 1.1 give $(q_3,q_2,q_1)$ for both words.

The interpolation and its values at sample heights. Put $Z_k(s,u):=(1-s)(W_0{)}_k(u)+s{\mathrm{r}\mathrm{o}\mathrm{t}}_k(u)$ for $k=1,2,3$, a jointly continuous map by [F5] and step 1.1; for $k<l$ the collision equation $Z_k(s,u)=Z_l(s,u)$ with $s\in (0,1)$ is equivalent, by the linearity of $R_{\pi u}$, the identities ${\mathrm{r}\mathrm{o}\mathrm{t}}_l(u)-{\mathrm{r}\mathrm{o}\mathrm{t}}_k(u)=2h(l-k)(\cos \pi u,\sin \pi u)$ and $(1-s)>0$, to the statement that $(W_0{)}_k(u)-(W_0{)}_l(u)$ is a positive multiple of $(\cos \pi u,\sin \pi u)$. At $u=0$ the three strand positions of $W_0$ are $q_1,q_2,q_3$, whose differences $(-2h,0),(-4h,0),(-2h,0)$ are negative multiples of $(\cos 0,\sin 0)=(1,0)$, so no collision occurs for $s\in (0,1)$, and $Z(s,0)=(q_1,q_2,q_3)$; at $u=\frac{1}{2}$ the positions of $W_0$ are $q_3,q_1,q_2$, whose differences $(4h,0),(2h,0),(-2h,0)$ are horizontal and nonzero while $(\cos \frac{\pi }{2},\sin \frac{\pi }{2})=(0,1)$ is vertical, so no collision occurs, and $Z(\frac{1}{2},\frac{1}{2})=\frac{1}{2}((2h,0),(-2h,0),(0,0))+\frac{1}{2}((0,-2h),(0,0),(0,2h))=((h,-h),(-h,0),(0,h))$; at $u=1$ the positions of $W_0$ are $q_3,q_2,q_1$, whose differences $(2h,0),(4h,0),(2h,0)$ are positive multiples of $(1,0)$ while $(\cos \pi ,\sin \pi )=(-1,0)$, so no collision occurs, and $Z(s,1)=(q_3,q_2,q_1)$ for every $s$.

Conclusion. By [F1] the bracketing $W_0={\sigma }_1\star ({\sigma }_2\star {\sigma }_1)$ is braid-isotopic to $\mathrm{r}\mathrm{o}\mathrm{t}$, and the reflection $\kappa (w)=-w$ followed by the relabelling of $1$ and $3$ turns $W_0$ into $W_1={\sigma }_2\star ({\sigma }_1\star {\sigma }_2)$, so $W_1$ is braid-isotopic to $\mathrm{r}\mathrm{o}\mathrm{t}$ as well; step 1.2 identifies $\mathrm{r}\mathrm{o}\mathrm{t}$ as a braid based at $Q$, and step 2.2 gives $\pi (W_0)=\pi (W_1)=(13)$; step 2.3 exhibits the explicit intermediate values of the deformation, and steps 2.1 and 1.1 record the numerical facts $2\parallel \rho (v){\parallel }_2\ge \frac{\sqrt{2}}{16}$ and $3h-h=\frac{1}{8}$ behind its collision-freeness. Hence $W_0\sim W_1$ and, passing to isotopy classes in the group $G_3$ of [F2], $[{\sigma }_1][{\sigma }_2][{\sigma }_1]=[{\sigma }_2][{\sigma }_1][{\sigma }_2]$. ∎