Braids as Fundamental Groups of Configuration Spaces — Examples

Compute the round relative path. The addition law [L4] and Euler's identity [L3] give $v(0)=-h$, $v(1)=h$, and $v(t)=-h\exp (i\pi t)$. For $0<t<1$, [L3] and [L5] give $\mathrm{Im}v(t)=-h\sin (\pi t)<0$, since $h>0$ and $0<\pi t<\pi$; the modulus formula gives $\mathrm{\mid }v(t)\mathrm{\mid }=h$. By [L6], $v$ is continuous. The piecewise formula [L2] gives $\mathrm{Im}\rho (t)=-2th<0$ for $0<t\le 1\mathrm{/}2$ and $\mathrm{Im}\rho (t)=2h(t-1)<0$ for $1\mathrm{/}2\le t<1$.

Check the unordered loop. The coordinates $x_j(t)$ are continuous by [L6], [L7], and [L9]. Their moving pair is distinct because $\mathrm{\mid }v(t)\mathrm{\mid }=h>0$ by [L1] and [L3]. Each moving point is at distance $h$ from $m_i$, so lies in $U_i\subset D^{\circ }$; every fixed $q_j$ lies in $D^{\circ }$ and, for $j\mathrm{\notin }\{i,i+1\}$, outside $U_i$ by [L1]. The tuple therefore lies in $F_n(D^{\circ })$ by [L8], and [L10] makes it continuous into that subspace. Its orbit is continuous by [L11]. Since $v(0)=-h$ and $v(1)=h$, the ordered tuple starts at $Q$ and ends with $q_i,q_{i+1}$ exchanged; both orbit endpoints are $[Q]$. Thus ${\alpha }_i$ is an interior based loop by [L12].

Interpolate in the lower half-plane. Set $w_s(t):=(1-s)\rho (t)+sv(t)$ for $s,t\in I$. This is jointly continuous by [L6], [L7], and [L9]. For $0<t<1$, both imaginary parts are strictly negative by step 1.1, so $w_s(t)\mathrm{\ne }0$; at $t=0,1$ the common values are $-h,+h$, also nonzero. The triangle inequality and [L2], [L3] give $\mathrm{\mid }{w}_s(t)\mathrm{\mid }\le (1-s)\mathrm{\mid }\rho (t)\mathrm{\mid }+s\mathrm{\mid }v(t)\mathrm{\mid }\le h<3h\mathrm{/}2$. The tuple with coordinates $x_{s,i}(t)=m_i+w_s(t)$, $x_{s,i+1}(t)=m_i-w_s(t)$, and $x_{s,j}(t)=q_j$ for the other labels stays collision-free and interior: the moving pair is distinct and stays in $U_i$, while all fixed points remain outside $U_i$ by [L1]. Its coordinate maps are continuous, and [L8]–[L10] give a continuous family in the ordered configuration subspace. Each slice is a geometric braid by [L13], and the family is a braid isotopy by [L17], since its bottom tuple is $Q$ and its top set is $Q$ for every $s$. At $s=0$ the braid is ${\sigma }_i$; at $s=1$ it is the round tuple of step 1.2.

Identify the exact traces. The ordered family in step 2.1 is continuous into $F_n(D^{\circ })$ by [L8]–[L10], so its quotient $H(s,t):=[(x_{s,1}(t),\dots ,x_{s,n}(t))]$ is jointly continuous by [L11]. Its endpoints satisfy $H(s,0)=H(s,1)=[Q]$ for every $s$, and every slice is an interior based motion by [L12]. The transpose $(t,u)\mapsto H(u,t)$ is a path homotopy rel endpoints under [L16]. Thus [L15] makes the traces of the two boundary loops braid-isotopic. By [L14], those traces are exactly ${\sigma }_i$ and the round tuple; the latter is the trace of ${\alpha }_i$ by step 1.2. This proves the example.

Put $\omega (t):=\exp (2\pi it)$. By [L2], $\omega$ is continuous and $\mathrm{\mid }\omega (t)\mathrm{\mid }=1$. Since $\exp (i\pi )=-1$ and the exponential addition law gives $\exp (2\pi i)=\exp (i\pi {)}^2=1$, one has $\omega (0)=\omega (1)=1$. Hence $\eta (t)=(-h,-h+2h\omega (t))$ starts and ends at $Q$. Its coordinate difference is $2h\omega (t)\mathrm{\ne }0$, and $\mathrm{\mid }-h+2h\omega (t)\mathrm{\mid }\le h+2h=3h=\frac{1}{4}<1,$ while $\mathrm{\mid }-h\mathrm{\mid }=h<1$. Thus its values are ordered configurations in $F_2(\mathrm{int}{D}^2)$; coordinate continuity and [L5] show that $\eta$ is a continuous ordered loop at $Q$.

Let $\alpha (t):=p_2(\eta (t))$. By [L6], this is a continuous interior based loop at $[Q]$. Its unique lift from $Q$ is $\eta$ itself, so [L7] identifies its trace with the coordinate braid ${\beta }_{\eta }(t)=(-h,-h+2h\omega (t))$. Because both coordinates of $\eta$ return to their starting values, [L8] shows ${\beta }_{\eta }$ is pure and its endpoint permutation is the identity.

Use the stacking formula [L9] on two copies of ${\sigma }_1$. Since the endpoint permutation of the lower copy exchanges labels $1$ and $2$, the stacked coordinate pair is $(\rho (2t),-\rho (2t))(0\le t\le \frac{1}{2}),(-\rho (2t-1),\rho (2t-1))(\frac{1}{2}\le t\le 1)\mathrm{.}$ Its centre is $0$ and its second-minus-first coordinate is $D_0(t)=\{\begin{array}{ll}-2\rho (2t),& 0\le t\le \frac{1}{2},\\ 2\rho (2t-1),& \frac{1}{2}\le t\le 1.\end{array}$ Substitution of the two branches of $\rho$ from [L1] shows that $D_0(t)$ lies successively in the first, second, third, and fourth closed quadrants on the four quarter intervals. It never vanishes, and $\mathrm{\mid }{D}_0(t)\mathrm{\mid }\le 2h$ by [L1]. By [L2] and [L4], $D_1(t):=2h\omega (t)$ lies in the same respective closed quadrant, never vanishes, and has modulus $2h$.

For $s,t\in I$ set $D_s(t):=(1-s)D_0(t)+s{D}_1(t),H_1(s,t):=(-D_s(t)\mathrm{/}2,D_s(t)\mathrm{/}2)\mathrm{.}$ Each closed quadrant is convex and contains no pair of opposite nonzero vectors, so $D_s(t)\mathrm{\ne }0$ for every $(s,t)$. By [L3], $\mathrm{\mid }{D}_s(t)\mathrm{\mid }\le (1-s)\mathrm{\mid }{D}_0(t)\mathrm{\mid }+s\mathrm{\mid }{D}_1(t)\mathrm{\mid }\le 2h,$ so both coordinates of $H_1$ have modulus at most $h<1$. The formulas and [L2], [L3], [L9] give joint continuity. Both $D_0$ and $D_1$ equal $2h$ at $t=0,1$, so $H_1(s,0)=H_1(s,1)=Q$ for every $s$. Thus [L11] makes $H_1$ a braid isotopy from the stacked diamond braid to the centred round pair ${\beta }_{\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}(t)=(-h\omega (t),h\omega (t))$.

Put $C_s(t):=s(-h+h\omega (t))$ and define $H_2(s,t):=(C_s(t)-h\omega (t),C_s(t)+h\omega (t))\mathrm{.}$ The coordinate difference is $2h\omega (t)\mathrm{\ne }0$, and [L3] gives $\mathrm{\mid }{C}_s(t)\pm h\omega (t)\mathrm{\mid }\le \mathrm{\mid }{C}_s(t)\mathrm{\mid }+h\le 2h+h=3h<1.$ Both coordinates are therefore in the open disk and distinct at every height. The formula is jointly continuous by [L2], [L3]. Since $C_s(0)=C_s(1)=0$ and $\omega (0)=\omega (1)=1$, the endpoints are $Q$ for every $s$. Thus [L11] makes $H_2$ a braid isotopy from ${\beta }_{\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}$ to ${\beta }_{\eta }$.

The families $H_1$ and $H_2$ agree at their common braid ${\beta }_{\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}}$. Pasting $H_1(2s,t)$ for $s\le 1\mathrm{/}2$ to $H_2(2s-1,t)$ for $s\ge 1\mathrm{/}2$ gives a jointly continuous family by [L11]. Each slice is a braid and its endpoints remain $Q$, so this is a braid isotopy from ${\sigma }_1\star {\sigma }_1$ to ${\beta }_{\eta }$. By [L9] and [L10], $[{\beta }_{\eta }]=[{\sigma }_1\star {\sigma }_1]=[{\sigma }_1{]}^2\mathrm{.}$ Together with step 1.2, this proves that the trace of the stated ordered loop is the positive full twist and has identity endpoint permutation.

Since ${\beta }_{\eta }$ is pure by step 1.2, the exact ordered representative of its class under the pure-braid isomorphism [L12] is $\mathrm{\Psi }([{\beta }_{\eta }])=({\iota }_{\ast }^F[\eta ]{)}^{-1}\in P{B}_2\mathrm{.}$ Thus the displayed collision-free ordered loop records this positive full twist under the common basepoint convention, including the inverse in the published identification. [L12, step 1.2, step 3.1] $\mathrm{\square }$

Track the ordered pair. The midpoint of $q_1$ and $q_2$ is $0$, so the published half-twist formula in [L4] gives the ordered path ${\tilde{\sigma }}_1(t)=(\rho (t),-\rho (t))\mathrm{.}$ Its coordinates lie in $D^{\circ }$ because $\parallel \rho (t)\parallel \le h<1$, and they are distinct because their difference is $2\rho (t)\mathrm{\ne }0$. Thus this is a path in $F_2(D^{\circ })$, with endpoints ${\tilde{\sigma }}_1(0)=(q_1,q_2)=Q,{\tilde{\sigma }}_1(1)=(q_2,q_1)\mathrm{.}$

Since $q_1\mathrm{\ne }{q}_2$, the terminal ordered tuple $(q_2,q_1)$ is not $Q$. By [L7], ${\tilde{\sigma }}_1$ is a path in the ordered configuration space but is not a based loop at $Q$.

Forget the labels. By [L2], the endpoint tuples have the same orbit: $[(q_1,q_2)]=[(q_2,q_1)]=[Q]\mathrm{.}$ By [L3], the quotient projection is continuous, so $\alpha (t):=[{\tilde{\sigma }}_1(t)]$ is a continuous path in $C_2(D^{\circ })$ with $\alpha (0)=\alpha (1)=[Q]$. By [L5, L6], this is the slice path $S({\sigma }_1)$ of the positive geometric half twist, so it is the unordered based loop promised by the counterexample.

The ordered coordinate path fails to close at $Q$, while its unordered image closes at $[Q]$. Hence forgetting labels can close a nonlooping ordered coordinate path, refuting the claim in the statement.

Construct the folded arc and vertical strand. Let $a=h\mathrm{/}4$. In ${\mathbb{R}}^2\times I$, define $P_0=(-h,0,0),P_1=(-h+a,0,3\mathrm{/}4),P_2=(-h+2a,a,1\mathrm{/}4),P_3=(-h,0,1)\mathrm{.}$ Let $A$ be the polygonal path $P_0{P}_1\cup P_1{P}_2\cup P_2{P}_3$, and let $B=\{(h,0,t):t\in I\}$. The endpoint sets of $A\cup B$ at heights $0$ and $1$ are both $\{q_1,q_2\}$. The height of $A$ has a local maximum $3\mathrm{/}4$ at $P_1$ and a local minimum $1\mathrm{/}4$ at $P_2$, so $A$ is not height-monotone.

Verify that these are disjoint embedded arcs in the cylinder. The segment $P_0{P}_1$ has spatial coordinate $y=0$, while $P_1{P}_2$ has $y>0$ except at $P_1$ and $P_2{P}_3$ has $y>0$ except at $P_3$. The only possible intersection of $P_0{P}_1$ and $P_2{P}_3$ at $y=0$ would be $P_3$, whose height is $1$ while $P_0{P}_1$ has height at most $3\mathrm{/}4$, so those segments are disjoint. Along $P_1{P}_2$ the spatial coordinates satisfy $x+h-y=a$; along $P_2{P}_3$ they satisfy $x+h=2y$. A common point must therefore have $y=a$, which occurs on both segments only at $P_2$. Thus the three segments form an embedded arc. Its spatial points lie in the convex ball of radius $\sqrt{5}a$ about $q_1$, since that ball contains all four vertices. As $\sqrt{5}a<3a=3h\mathrm{/}4<2h=\mathrm{\mid }{q}_2-q_1\mathrm{\mid }$, the arc misses $B$. Also $\mathrm{\mid }{q}_1\mathrm{\mid }+\sqrt{5}a<h+3a=7h\mathrm{/}4=7\mathrm{/}48<1$, so $A$ lies in $D^{\circ }\times I$; $B$ lies there because $\mathrm{\mid }{q}_2\mathrm{\mid }=h<1$.

Compute the slice at height $1\mathrm{/}2$. The height coordinate is strictly monotone on each segment of $A$, so each segment meets that plane once. Linear interpolation gives the three spatial points $(-h+2a\mathrm{/}3,0),(-h+3a\mathrm{/}2,a\mathrm{/}2),(-h+4a\mathrm{/}3,2a\mathrm{/}3)\mathrm{.}$ The first has $y=0$ and the other two have different positive $y$ coordinates; their $x$ coordinates are also different, so these three points are distinct. The strand $B$ contributes the fourth point $(h,0)=q_2$, which is not on $A$ by step 2.1. The slice therefore contains four distinct points.

By [L2], a value of $C_2(D^{\circ })$ must have exactly two points, whereas the horizontal slice at $1\mathrm{/}2$ has four. Hence this embedded picture does not define a path $I\to C_2(D^{\circ })$ by slicing, despite having the same endpoint set $\{q_1,q_2\}$ at heights $0$ and $1$. Its folded arc also fails the one-point-per-height condition in [L3], so the level-preserving braid slicing lemma [L4] does not apply.