The closure of a geometric braid

Let $n\ge 0$ and let $\beta =(z_1,\dots ,z_n)$ be a geometric braid with the fixed real basepoints $q_j=((2j-n-1)\mathrm{/}(4(n+1)),0)$ and endpoint permutation $\pi$, as in Geometric braids in the disc with setwise endpoints. Identify the disk with $\{x\in \mathbb{C}:\mathrm{\mid }x\mathrm{\mid }<1\}$ and put $V=(D^{\circ }\times [0,1])\mathrm{/}((x,1)\sim (x,0))\mathrm{.}$ The quotient and its topology use The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space and the circle $\mathbb{R}\mathrm{/}\mathbb{Z}$ of The circle as $S^1=\mathbb{R}\mathrm{/}\mathbb{Z}$ with basepoint $[0]$. The interval images $s_j=\{[(z_j(t),t)]:0\le t\le 1\},\mathrm{cl}(\beta )={\bigcup }_{j=1}^n{s}_j$ are the closure in the open solid torus. In general an individual $s_j$ is an interval image, not a closed circle: its last point is the first point of $s_{\pi (j)}$.

Components and orientation. For a cycle $(j,\pi (j),\dots ,{\pi }^{k-1}(j))$ concatenate these $k$ strand maps in that order, with parameter in $[0,k]$. Their endpoints agree in $V$, so the concatenation factors through $\mathbb{R}\mathrm{/}k\mathbb{Z}$. It is injective on this circle: points at distinct nonintegral heights can agree only when their parameters have the same fractional part, and the distinct strands at that height have distinct disk coordinates; at integral heights only the specified adjacent endpoints agree. Thus each cycle gives an embedded circle. Different cycles give disjoint circles, and every strand belongs to exactly one cycle. These circles are precisely the connected components of the closure. The orientation is increasing concatenation parameter. In particular the number of components equals the number of cycles of $\pi$.

The standard axis and fixed framing. In $S^3=\{(z,w)\in {\mathbb{C}}^2:\mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2=1\}$, as in Euclidean spheres and closed balls as subspaces of ${\mathbb{R}}^n$, set $A=\{(0,w):\mathrm{\mid }w\mathrm{\mid }=1\},P_{\theta }=\{(z,w)\in S^3:z\mathrm{\ne }0,\arg z=\theta \}\mathrm{.}$ Fix the following particular diffeomorphism, including its disk framing: $\phi ([(x,t)])=(\sqrt{1-\mathrm{\mid }x{\mathrm{\mid }}^2}{e}^{2\pi it},x):V⟶S^3\setminus A\mathrm{.}$ Its inverse has disk coordinate $x=w$ and circle coordinate $[t]=[\arg z\mathrm{/}(2\pi )]$. Every $P_{\theta }$ is consequently an open disk. The raw topological oriented closure is ${\widehat{\beta }}_{\mathrm{t}\mathrm{o}\mathrm{p}}=\phi (\mathrm{cl}(\beta ))\mathrm{.}$ This explicit $\phi$ is part of the construction; the definition does not allow an arbitrary page-preserving change of framing.

Page intersections and smooth representatives. The whole closure meets every page in exactly $n$ points. A component belonging to a $\pi$-cycle of length $k$ meets each page in exactly $k$ points: at any fractional height it uses exactly those $k$ distinct strands. The page coordinate of its concatenation is $[t]$, for $t\in \mathbb{R}\mathrm{/}k\mathbb{Z}$, so its oriented degree is $+k$. If all disk-coordinate strand maps are smooth and have all positive-order derivatives zero at their endpoints, their cycle concatenations have matching jets at every seam. The page coordinate has nonzero derivative, so the result is a smooth embedding of disjoint oriented circles, in the sense of Smooth embeddings. For an endpoint-flat smooth braid, its literal $\phi$-image is therefore a smooth oriented link and is denoted $\widehat{\beta }$.

Closure in the smooth oriented-link category. The raw image need not be smooth, even if individual strands are smooth but their endpoint jets fail to match at the cycle seams. For an arbitrary continuous braid, assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) and choose a smooth representative ${\beta }_s$ by Every geometric braid is braid-isotopic to a smooth braid. Make it endpoint-flat by composing all disk-coordinate strands with a fixed smooth nondecreasing $\psi :I\to I$, equal to $0$ near $0$ and $1$ near $1$. Interpolation of height maps $(1-u)t+u\psi (t)$ is an endpoint-fixed braid isotopy, because every strand is evaluated at the same height. The smooth-category closure of $\beta$ is the oriented-link isotopy class of this chosen model's literal $\phi$-image; when a subset is needed, use that chosen closed-braid representative. Independence of the representative is proved by The closure depends only on the braid isotopy class ↗. Page and cycle properties refer to the chosen constructed closed braid, with the same number of strands and endpoint permutation as $\beta$.

If $\beta$ already has smooth strands with matching cycle-seam jets, in particular if it is endpoint-flat, its literal raw image is a smooth link and is retained as $\widehat{\beta }$. The raw quotient, fixed $\phi$, finite cycles, page counts and trivial closures remain choice-free. Finite elementary words have explicit smooth endpoint-flat models, so forming their literal closures also requires no choice assumption; ${\mathrm{A}\mathrm{C}}_{\omega }$ is used for the general continuous-representative convention and its independence theorem.

The trivial closure. For the constant braid its circles are $\{(\sqrt{1-q_j^2}{e}^{2\pi it},q_j):[t]\in \mathbb{R}\mathrm{/}\mathbb{Z}\}$. They are latitudes of the sphere $S=\{(z,w)\in S^3:\mathrm{Im}w=0\}$, with $\mathrm{Re}w=q_j$. Each northern cap $C_j=\{p\in S:\mathrm{Re}w(p)\ge q_j\}$ is a smooth disk: stereographic coordinates on $S$ give radius $\sqrt{(1-q_j)\mathrm{/}(1+q_j)}$ for this cap. To make their spanning disks disjoint, let $e$ be the unit vector in the $\mathrm{Im}w$ direction, set $\epsilon =\pi \mathrm{/}16$ and $f_j(p)=\epsilon (\mathrm{Re}w(p)-q_j)$ on $C_j$, and use $F_j(p)=\cos (f_j(p))p+\sin (f_j(p))e\mathrm{.}$ The boundary is fixed. The tubular parametrization $(p,s)\mapsto \cos (s)p+\sin (s)e$ is injective for $\mathrm{\mid }s\mathrm{\mid }<\pi \mathrm{/}2$, and $0\le f_j<\pi \mathrm{/}8$. On overlapping caps, if $q_i<q_j$ then $f_i-f_j=\epsilon (q_j-q_i)>0$, so their graphs are disjoint. Thus these are pairwise disjoint smooth spanning disks. The trivial $n$-braid closes to the oriented $n$-component unlink; for $n=1$ this is the unknot.

At $n=0$ all strand and cycle unions are empty, every page has zero intersections, and the closure is the empty link. The basepoint list and disk constructions above then have no entries.