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Geometric Braids and Artin Generators — Examples

1 · Prerequisites

2 · Summary

These four worked entries make the abstract items of the companion page concrete, and each one is a computation or a witness rather than a restatement. The first example handles the case $n = 2$ completely without assuming completeness of the Artin presentation: for $h = \frac{1}{12}$ and $q_{1} = (- \frac{1}{12}, 0)$ , $q_{2} = (\frac{1}{12}, 0)$ , the relative motion $w = z_{1} - z_{2}$ of a two-strand braid is a nowhere-zero path in $R^{2} ∖ {0}$ ending at one of $\pm w (0)$ , whose argument class lifts uniquely to $θ ⁣ : I \to R$ with $θ (0) = \frac{1}{2}$ , and $k (β) : = 2 θ (1) - 1$ is an integer. The example proves that $k$ is a braid-isotopy invariant, that $k (γ ⋆ β) = k (γ) + k (β)$ , that $k (e) = 0$ , $k (σ_{1}) = 1$ and $k (σ_{1}^{-}) = - 1$ , and hence that $[β] = [σ_{1}]^{k (β)}$ : every two-strand braid is braid-isotopic to exactly one integer twist, $G_{2} ≅ Z$ via $k$ , and the twist exponent is the total argument change of the relative motion divided by $π$ . The parity of $k$ records the endpoint permutation, which is exactly the ingredient that makes additivity of $k$ under stacking work.

The second example takes $n = 3$ , $i = 1$ and $h = \frac{1}{16}$ , so that the three base points are $- \frac{1}{8}, 0, \frac{1}{8}$ on the horizontal axis, and writes out the two words $W_{0} = σ_{1} ⋆ (σ_{2} ⋆ σ_{1})$ and $W_{1} = σ_{2} ⋆ (σ_{1} ⋆ σ_{2})$ strand by strand, in the six explicit windows dictated by the stacking formula; the windows glue at $u = \frac{1}{4}, \frac{1}{2}$ and both words end at $(q_{3}, q_{2}, q_{1})$ , giving the endpoint permutation $(1 3)$ . The example checks the two products in $S_{3}$ , $(1 2) (2 3) (1 2) = (2 3) (1 2) (2 3) = (1 3)$ , computes the separation $2 ∥ ρ (v) ∥_{2} \geq \sqrt{2} h = \frac{\sqrt{2}}{16}$ of the two moving strands and the clearance $2 h = \frac{1}{8}$ from the frozen base point, exhibits the rotation braid $r o t$ and the linear interpolation between $W_{0}$ and $r o t$ , whose slice at $\frac{1}{2}$ is $((h, - h), (- h, 0), (0, h))$ , and concludes $[σ_{1}] [σ_{2}] [σ_{1}] = [σ_{2}] [σ_{1}] [σ_{2}]$ in $G_{3}$ .

The two counterexamples record the two places where the geometric definitions could be misread. Setwise endpoints do not make a braid pure: the half twist $σ_{1}$ on two strands has top endpoint set ${q_{1}, q_{2}}$ but ends at $(q_{2}, q_{1})$ , so no strand returns to its own starting point, and since the endpoint permutation is an isotopy invariant $σ_{1}$ is not braid-isotopic to the trivial braid. And an arbitrary isotopy of arcs with fixed endpoints need not be a braid isotopy: the explicit family $α_{s} (u) = ((\frac{λ (s)}{8} w (u), 0), u + λ (s) w (u))$ with $λ (s) = \frac{1}{4} \min (2 s, 2 - 2 s)$ and the piecewise linear $w$ with $w (\frac{1}{4}) = 1, w (\frac{3}{4}) = - 1$ consists of simple arcs with the fixed endpoints $(q_{1}, 0)$ and $(q_{1}, 1)$ whose boundary arcs are the trivial braid, yet at $s = \frac{1}{2}$ the arc passes through the two distinct points $((\pm \frac{1}{32}, 0), \frac{1}{2})$ at the single height $\frac{1}{2}$ , so it is not the graph of a strand of any braid; the one-point-per-height requirement in the definition of braid isotopy is therefore not redundant. Nothing in these four entries uses a choice principle.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Geometric two strand braids are integer twists

Example

Let $n = 2$ , let $Q = (q_{1}, q_{2})$ with $h = \frac{1}{12}$ , $q_{1} = (- \frac{1}{12}, 0)$ and $q_{2} = (\frac{1}{12}, 0)$ , let $β = (z_{1}, z_{2})$ be a braid based at $Q$ (Geometric braids in the disc with setwise endpoints), and let $σ_{1}$ and $σ_{1}^{-}$ be the two half twists at $i = 1$ (The elementary geometric half twist, its support disc, and its opposite), with classes $[σ_{1}], [σ_{1}^{-}] \in G_{2}$ (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism). Write

$w : = z_{1} - z_{2} ⁣ : I ⟶ R^{2} ∖ {0}$

for the relative motion of the two strands, and write $σ_{1}^{m}$ for the stacking of $m$ copies of $σ_{1}$ when $m > 0$ , of $∣ m ∣$ copies of $σ_{1}^{-}$ when $m < 0$ , and for the trivial braid $e$ when $m = 0$ .

The invariant. There is a unique continuous $θ ⁣ : I \to R$ with $θ (0) = \frac{1}{2}$ whose class $[θ (t)] \in R / Z$ is the argument class of $w (t)$ , that is, the unique $s$ with $H ([s]) = w (t) / ∥ w (t) ∥_{2}$ , where $H ([s]) = (\cos 2 π s, \sin 2 π s)$ ( $[t] \mapsto (\cos 2 π t, \sin 2 π t)$ is a homeomorphism from $R / Z$ to the unit circle). The number

$k (β) : = 2 (θ (1) - θ (0)) = 2 θ (1) - 1$

is then an integer, and it is an invariant of the braid isotopy class of $β$ . The example proves:

$k ⁣ : G_{2} \to Z$ is a group homomorphism, with $k (γ ⋆ β) = k (γ) + k (β)$ ;
$k (e) = 0$ , $k (σ_{1}) = 1$ and $k (σ_{1}^{-}) = - 1$ , hence $k (σ_{1}^{m}) = m$ for every $m \in Z$ ;
$[β] = [σ_{1}]^{k (β)}$ in $G_{2}$ . Consequently $k$ is a group isomorphism $G_{2} ≅ Z$ , every two-strand braid is braid-isotopic to exactly one of the integer twists $σ_{1}^{m}$ ( $m \in Z$ ), and two two-strand braids are braid-isotopic if and only if they have the same invariant $k$ .

Thus a two-strand braid is exactly an integer number of half twists, counted with sign, and the composition of such twists adds the numbers. The calculation is independent of any presentation of $G_{2}$ : it uses only the generation of $G_{2}$ by $[σ_{1}]$ from The Artin presentation surjects onto the geometric braid group together with the argument lift constructed below.

Facts & Assumptions

Given: The natural number $2$ , the base configuration $Q = (q_{1}, q_{2})$ with $h = \frac{1}{12}$ , $q_{1} = (- \frac{1}{12}, 0)$ , $q_{2} = (\frac{1}{12}, 0)$ , two-strand braids $β = (z_{1}, z_{2})$ , $β^{'}$ and $γ$ based at $Q$ , and the half twists $σ_{1}, σ_{1}^{-}$ based at $Q$ .

[F1]

A braid based at $Q$ is a pair $(u_{1}, u_{2})$ of continuous maps $u_{j} ⁣ : I \to D^{\circ}$ with $u_{1} (t) \neq u_{2} (t)$ for all $t$ , $u_{j} (0) = q_{j}$ and ${u_{1} (1), u_{2} (1)} = {q_{1}, q_{2}}$ , with endpoint permutation $π (u) \in S_{2}$ defined by $u_{j} (1) = q_{π (u) (j)}$ ; here $q_{1} - q_{2} = (- 2 h, 0)$ and $q_{2} - q_{1} = (2 h, 0)$ , and the two-element group $S_{2}$ consists of the identity and the transposition of $1$ and $2$ ; a braid isotopy from $β$ to $β^{'}$ is a pair of jointly continuous maps $Z_{j} ⁣ : I \times I \to D^{\circ}$ whose every slice $Z (s, \cdot)$ is such a braid based at $Q$ and whose boundary slices are $β$ and $β^{'}$ (Geometric braids in the disc with setwise endpoints, Braid isotopy relative to the top and bottom endpoints, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Continuity of a map of topological spaces at a point and globally, 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).

[F2]

The half twists are $(σ_{1})_{1} = m_{1} + ρ$ , $(σ_{1})_{2} = m_{1} - ρ$ and $(σ_{1}^{-})_{1} = m_{1} + ρ^{-}$ , $(σ_{1}^{-})_{2} = m_{1} - ρ^{-}$ with $m_{1} = \frac{q_{1} + q_{2}}{2} = (0, 0)$ , where $ρ (t) = (2 t h - h, - 2 t h)$ for $t \leq \frac{1}{2}$ and $ρ (t) = (2 t h - h, 2 t h - 2 h)$ for $t \geq \frac{1}{2}$ , and $ρ^{-} (t) = (ρ_{1} (t), - ρ_{2} (t))$ ; both are braids based at $Q$ , $π (σ_{1}) = π (σ_{1}^{-})$ is the transposition of $1$ and $2$ , and $[σ_{1}^{-}] = [σ_{1}]^{- 1}$ in $G_{2}$ (The elementary geometric half twist, its support disc, and its opposite, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

[F3]

Stacking is $(γ ⋆ β)_{j} (t) = z_{j} (2 t)$ for $t \leq \frac{1}{2}$ and $= w_{π (β) (j)} (2 t - 1)$ for $t \geq \frac{1}{2}$ , where $z_{j}, w_{j}$ are the strands of $β, γ$ ; $[γ ⋆ β] = [γ] [β]$ and $π$ is constant on braid isotopy classes, so $π (β) \in S_{2}$ is an invariant of the class $[β]$ (Stacking of geometric braids is a well-defined associative operation on isotopy classes, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

[F4]

Every element of $G_{2}$ is a finite product of the elements $[σ_{1}]$ and $[σ_{1}]^{- 1}$ ; more precisely the classes $[σ_{1}], \dots, [σ_{n - 1}]$ generate $G_{n}$ for every $n$ (The Artin presentation surjects onto the geometric braid group).

[F5]

$p ⁣ : R \to R / Z$ is a covering map; for a covering $p ⁣ : E \to B$ , a path $α ⁣ : I \to B$ and $e_{0} \in E$ with $p (e_{0}) = α (0)$ there is a unique path lift $\tilde{α} ⁣ : I \to E$ with $\tilde{α} (0) = e_{0}$ and $p \circ \tilde{α} = α$ , and for a homotopy $H ⁣ : I \times I \to B$ together with a lift ${\tilde{H}}_{0} ⁣ : I \to E$ of $H (\cdot, 0)$ there is a unique lift $\tilde{H} ⁣ : I \times I \to E$ of $H$ with $\tilde{H} (s, 0) = {\tilde{H}}_{0} (s)$ for all $s$ ; consequently two lifts of one path into $R / Z$ to continuous maps $I \to R$ differ by a constant integer, since their difference is continuous and takes values in $Z$ ( $p : R \to R / Z$ is a covering map with translated interval sheets, Existence and uniqueness of path lifts through a covering map, Existence and uniqueness of homotopy lifts through a covering map).

[F6]

The map $H ⁣ : R / Z \to S^{1}$ , $H ([s]) = (\cos 2 π s, \sin 2 π s)$ , is a homeomorphism onto the unit circle $S^{1} = {(x, y) \in R^{2} : ∥ (x, y) ∥_{2} = 1}$ , and $H ([\frac{1}{2}]) = (- 1, 0)$ ; moreover $H ([s + \frac{1}{2}]) = - H ([s])$ for every real $s$ , because $\cos (x + π) = - \cos x$ and $\sin (x + π) = - \sin x$ ( $[t] \mapsto (\cos 2 π t, \sin 2 π t)$ is a homeomorphism from $R / Z$ to the unit circle, Quarter-turn values and shifts by pi/2 and pi, Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[F7]

The radial normalisation $r ⁣ : R^{2} ∖ {0} \to S^{1}$ , $r (x) = x / ∥ x ∥_{2}$ , is continuous, and $r (- u) = - r (u)$ for every $u \neq 0$ ; so the composite $c : = H^{- 1} \circ r ⁣ : R^{2} ∖ {0} \to R / Z$ is continuous, and $c (- u) = c (u) + [\frac{1}{2}]$ for every $u \neq 0$ by [F6] (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ , Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[F8]

Sums, differences and scalar multiples of continuous maps are continuous, composites of continuous maps are continuous, continuity of a map into $R^{2}$ may be checked on the coordinate functions, and the continuous image of a connected subset is connected; since $I$ is connected and the connected subsets of $R$ are the intervals, a continuous map from $I$ into $Z$ is constant (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally, A continuous image of a connected space is connected, and connectedness is a topological property, The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ", Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

The relative motion and its endpoint values. Assume $n = 2$ and let $β = (z_{1}, z_{2})$ be a braid based at $Q$ ; then $w = z_{1} - z_{2}$ is continuous and $w (t) \neq 0$ 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 $w (1) = z_{1} (1) - z_{2} (1) \in {q_{1} - q_{2}, q_{2} - q_{1}} = {(- 2 h, 0), (2 h, 0)}$ , and by the convention $z_{j} (1) = q_{π (β) (j)}$ of [F1] the value $w (1) = (- 2 h, 0)$ occurs exactly when $π (β)$ is the identity, while $w (1) = (2 h, 0)$ occurs exactly when $π (β)$ is the transposition of $1$ and $2$ .

F1F8

2.1

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

F5F6F7F8step 1.1

3.1

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

F1F6F7step 1.1step 2.1

4.1

Isotopy invariance. Let $Z$ be a braid isotopy from $β$ to $β^{'}$ and put $W (s, t) : = Z_{1} (s, t) - Z_{2} (s, t)$ , a continuous and nowhere vanishing map $I \times I \to R^{2} ∖ {0}$ by [F1] and [F8]; then $c \circ W$ is a homotopy $I \times I \to R / Z$ by [F7] and [F8], and $W (s, 0) = z_{1} (0) - z_{2} (0) = q_{1} - q_{2} = (- 2 h, 0)$ 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 $Θ ⁣ : I \times I \to R$ of $c \circ W$ with $Θ (s, 0) = \frac{1}{2}$ 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 $t \mapsto Θ (s, t)$ 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 $2 Θ (s, 1) - 1 \in Z$ for every $s$ . The map $s \mapsto Θ (s, 1)$ is continuous, so $s \mapsto 2 Θ (s, 1) - 1$ is a continuous map from the connected interval $I$ into $Z$ and is constant by [F8]; moreover $Θ (0, \cdot) = θ$ by the uniqueness in [F5] applied to $c \circ W (0, \cdot) = c \circ w$ and step 2.1, and $Θ (1, \cdot)$ is the corresponding lift for the relative motion $W (1, \cdot)$ of $β^{'}$ ; therefore $k (β) = 2 Θ (0, 1) - 1 = 2 Θ (1, 1) - 1 = k (β^{'})$ , and $k$ is constant on braid isotopy classes.

F1F5F6F7F8step 2.1step 3.1

4.2

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

F1F3F5F6F7F8step 2.1step 3.1

5.1

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 $(- 2 h, 0)$ , 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 $β = σ_{1}$ the relative motion is $2 ρ$ , that is $w (t) = (4 t h - 2 h, - 4 t h)$ for $t \leq \frac{1}{2}$ and $w (t) = (4 t h - 2 h, 4 t h - 4 h)$ for $t \geq \frac{1}{2}$ , by [F2], and it vanishes nowhere: on the first half $w (t)$ runs through the closed third quadrant from $(- 2 h, 0)$ to $(0, - 2 h)$ and on the second half through the closed fourth quadrant from $(0, - 2 h)$ to $(2 h, 0)$ , in each case with a direction that turns strictly monotonically; hence the lift with value $\frac{1}{2}$ at $0$ satisfies $θ (\frac{1}{2}) = \frac{3}{4}$ and $θ (1) = 1$ , and $k (σ_{1}) = 2 (1 - \frac{1}{2}) = 1$ . For $β = σ_{1}^{-}$ 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 $θ (\frac{1}{2}) = \frac{1}{4}$ and $θ (1) = 0$ , so $k (σ_{1}^{-}) = 2 (0 - \frac{1}{2}) = - 1$ . By the additivity of step 4.2 and induction on $∣ m ∣$ this gives $k (σ_{1}^{m}) = m$ for every $m \in Z$ .

F1F2F5F6step 4.2

6.1

Every two-strand braid is an integer twist. Let $β$ be any braid based at $Q$ ; by [F4] the class $[β] \in G_{2}$ is a finite product of the elements $[σ_{1}]$ and $[σ_{1}]^{- 1}$ , that is $[β] = [σ_{1}]^{m}$ 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 $β$ is $k (β) = k (σ_{1}^{m}) = m$ ; hence $[β] = [σ_{1}]^{k (β)}$ , and braid-isotopic braids have equal invariants by step 4.1. Consequently $k$ descends to a well-defined map $G_{2} \to Z$ by step 4.1, that map is a group homomorphism by step 4.2, it is surjective because $k (σ_{1}^{m}) = m$ for every $m \in Z$ by step 5.1, and it is injective because $k (β) = 0$ forces $[β] = [σ_{1}]^{0} = [e]$ by the identity above; so $G_{2} ≅ Z$ via $k$ , the twists $σ_{1}^{m}$ represent pairwise distinct classes, and each class of $G_{2}$ is exactly one of them. ∎

F3F4step 4.1step 4.2step 5.1

Remarks

The invariant is the total argument change of the relative motion, divided by $π$ : the lift $θ$ measures the angle of the vector from the second strand to the first in units of full turns, and the half twists contribute $+ 1$ and $- 1$ . The factor $2$ in $k = 2 (θ (1) - θ (0))$ converts turns into half turns. Step 3.1 also records the parity dictionary used in step 4.2: $k (β)$ is even exactly for the braids with $π (β)$ the identity, and odd exactly for those whose endpoint permutation is the transposition; this is what makes the correction $z = (k (β) - ϵ) / 2$ in the second half of a stacking an integer.
Only the relative motion of the pair is used, and the endpoint set condition makes its argument change an integer multiple of half a turn: a pure two-strand braid returns the two labels, so the vector comes back to itself after an even number of half turns, while a transposition reverses it after an odd number.
The example does not use any presentation of $B_{2}$ , and in particular it does not use The two-strand braid group is infinite cyclic: generation comes from The Artin presentation surjects onto the geometric braid group and completeness of the presentation is never assumed.

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The three strand geometric braid relation

Example

Take $n = 3$ and $i = 1$ , so that $h = \frac{1}{4 (3 + 1)} = \frac{1}{16}$ and the base configuration is

$q_{1} = (- 2 h, 0) = (- \frac{1}{8}, 0), q_{2} = (0, 0), q_{3} = (2 h, 0) = (\frac{1}{8}, 0),$

with $m : = m_{1} = q_{1} + (h, 0) = (- \frac{1}{16}, 0)$ and $m^{'} : = m_{2} = q_{2} + (h, 0) = (\frac{1}{16}, 0)$ , and with $c : = q_{2} = (0, 0)$ and $p_{k} : = q_{k}$ for $k = 1, 2, 3$ . Let $σ_{1}, σ_{2}$ be the elementary half twists (The elementary geometric half twist, its support disc, and its opposite) and set

$W_{0} : = σ_{1} ⋆ (σ_{2} ⋆ σ_{1}), W_{1} : = σ_{2} ⋆ (σ_{1} ⋆ σ_{2})$

with respect to the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes. Write $R_{θ} (a, b) : = (a \cos θ - b \sin θ, a \sin θ + b \cos θ)$ for the rotation of the plane about the origin, and let $r o t$ be the three-strand tuple whose $k$ -th strand is at height $u$ at the point $c + R_{π u} (p_{k})$ , the strands outside ${1, 2, 3}$ being constant (here $n = 3$ , so there are none). The example verifies:

$W_{0}$ and $W_{1}$ are braids based at $Q$ whose strand coordinates are the explicit windows displayed below, and both have endpoint permutation the transposition $(1 3)$ ;
$r o t$ is a braid based at $Q$ with endpoint permutation $(1 3)$ , and its strands move through the explicit positions $(\pm 2 h \cos π u, \pm 2 h \sin π u)$ and $(0, 0)$ ;
the linear interpolation $Z_{k} (s, u) : = (1 - s) (W_{0})_{k} (u) + s {r o t}_{k} (u)$ has bottom value $Q$ and top value $(q_{3}, q_{2}, q_{1})$ for every $s$ , and its slices at $u \in {0, \frac{1}{2}, 1}$ are collision-free with the displayed values;

consequently, by the isotopies exhibited in The geometric three strand braid relation, the two words are braid-isotopic and

$[σ_{1}] [σ_{2}] [σ_{1}] = [σ_{2}] [σ_{1}] [σ_{2}]$

holds in $G_{3}$ .

Facts & Assumptions

Given: The natural number $3$ , the index $i = 1$ , the base configuration $Q = (q_{1}, q_{2}, q_{3})$ with $h = \frac{1}{16}$ , the half twists $σ_{1}, σ_{2}$ based at $Q$ , and the words $W_{0} = σ_{1} ⋆ (σ_{2} ⋆ σ_{1})$ and $W_{1} = σ_{2} ⋆ (σ_{1} ⋆ σ_{2})$ .

[F1]

For $n \geq 3$ and $1 \leq i \leq n - 2$ one has $σ_{i} ⋆ σ_{i + 1} ⋆ σ_{i} \sim σ_{i + 1} ⋆ σ_{i} ⋆ σ_{i + 1}$ ; the proof exhibits the intermediate braid $r o t$ , the rotation of $q_{i}, q_{i + 1}, q_{i + 2}$ about $q_{i + 1}$ by the angle $π u$ at height $u$ with the remaining strands fixed, and shows that the bracketing $σ_{i} ⋆ (σ_{i + 1} ⋆ σ_{i})$ is braid-isotopic to $r o t$ , which is a braid based at $Q$ with endpoint permutation the transposition of $i$ and $i + 2$ , while the point reflection $κ (w) = 2 q_{i + 1} - w$ followed by the relabelling of $i$ and $i + 2$ turns that bracketing into $σ_{i + 1} ⋆ (σ_{i} ⋆ σ_{i + 1})$ , so that bracketing is braid-isotopic to $r o t$ as well (The geometric three strand braid relation).

[F2]

The base points are $q_{j} = ((2 j - n - 1) h, 0)$ , here $q_{1} = (- 2 h, 0)$ , $q_{2} = (0, 0)$ , $q_{3} = (2 h, 0)$ ; the half twist at $k$ is $(σ_{k})_{k} = m_{k} + ρ$ , $(σ_{k})_{k + 1} = m_{k} - ρ$ and $(σ_{k})_{j} = q_{j}$ otherwise, where $m_{k} = q_{k} + (h, 0)$ and the diamond path satisfies $ρ (0) = (- h, 0)$ , $ρ (\frac{1}{2}) = (0, - h)$ , $ρ (1) = (h, 0)$ ; $π (σ_{k})$ is the transposition of $k$ and $k + 1$ ; stacking places the right factor below: $(γ ⋆ β)_{j} (t) = z_{j} (2 t)$ for $t \leq \frac{1}{2}$ and $= w_{π (β) (j)} (2 t - 1)$ for $t \geq \frac{1}{2}$ , with $π (γ ⋆ β) = π (γ) \circ π (β)$ and $[γ ⋆ β] = [γ] [β]$ in $G_{n}$ (Geometric braids in the disc with setwise endpoints, The elementary geometric half twist, its support disc, and its opposite, Stacking of geometric braids is a well-defined associative operation on isotopy classes, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

[F3]

A braid based at $Q$ is a tuple of continuous maps $u_{k} ⁣ : I \to D^{\circ}$ with pairwise distinct values, $u_{k} (0) = q_{k}$ and ${u_{1} (1), u_{2} (1), u_{3} (1)} = {q_{1}, q_{2}, q_{3}}$ , its endpoint permutation being the unique $π$ with $u_{k} (1) = q_{π (k)}$ ; a braid isotopy is a jointly continuous family whose every slice is such a braid and whose boundary slices are the two given braids; the group $S_{3}$ is written in cycle notation with $(π \circ σ) (j) = π (σ (j))$ (Geometric braids in the disc with setwise endpoints, Braid isotopy relative to the top and bottom endpoints, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

[F4]

$\sin 0 = 0$ , $\cos 0 = 1$ , $\cos \frac{π}{2} = 0$ , $\sin \frac{π}{2} = 1$ , $\cos π = - 1$ , $\sin π = 0$ , and $\sin^{2} θ + \cos^{2} θ = 1$ for every real $θ$ ; hence $R_{0}$ is the identity, $R_{π} (a, b) = (- a, - b)$ , and $∥ R_{θ} (a, b) ∥_{2} = ∥ (a, b) ∥_{2}$ for all $a, b$ (The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ).

[F5]

Sums, scalar multiples and composites of continuous maps are continuous, and a function on the interval $I = [0, 1]$ whose restrictions to the finitely many closed pieces ${u \leq \frac{1}{4}}$ , ${\frac{1}{4} \leq u \leq \frac{1}{2}}$ , ${\frac{1}{2} \leq u \leq 1}$ are continuous is continuous; the same pasting applies in the isotopy parameter (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

The explicit strand windows of the two words. Applying the stacking formula of [F2] twice, with $m = (- \frac{1}{16}, 0)$ , $m^{'} = (\frac{1}{16}, 0)$ and the transposition couplings $π (σ_{1}) = (1 2)$ , $π (σ_{2}) = (2 3)$ , gives for $W_{0}$ the strands $(m + ρ (4 u), m - ρ (4 u), q_{3})$ for $0 \leq u \leq \frac{1}{4}$ , $(m^{'} + ρ (4 u - 1), q_{1}, m^{'} - ρ (4 u - 1))$ for $\frac{1}{4} \leq u \leq \frac{1}{2}$ and $(q_{3}, m + ρ (2 u - 1), m - ρ (2 u - 1))$ for $\frac{1}{2} \leq u \leq 1$ , and for $W_{1}$ the strands $(q_{1}, m^{'} + ρ (4 u), m^{'} - ρ (4 u))$ for $0 \leq u \leq \frac{1}{4}$ , $(m + ρ (4 u - 1), q_{3}, m - ρ (4 u - 1))$ for $\frac{1}{4} \leq u \leq \frac{1}{2}$ and $(m^{'} + ρ (2 u - 1), m^{'} - ρ (2 u - 1), q_{1})$ for $\frac{1}{2} \leq u \leq 1$ . The two windows of $W_{0}$ agree at $u = \frac{1}{4}$ , where the first gives $(m + ρ (1), m - ρ (1), q_{3}) = ((0, 0), (- 2 h, 0), (2 h, 0))$ and the second gives $(m^{'} + ρ (0), q_{1}, m^{'} - ρ (0)) = ((0, 0), (- 2 h, 0), (2 h, 0))$ , and at $u = \frac{1}{2}$ , where the second gives $(m^{'} + ρ (1), q_{1}, m^{'} - ρ (1)) = ((2 h, 0), (- 2 h, 0), (0, 0))$ and the third gives $(q_{3}, m + ρ (0), m - ρ (0)) = ((2 h, 0), (- 2 h, 0), (0, 0))$ ; the same two checks apply verbatim to the three windows of $W_{1}$ , so by [F5] the formulas define continuous tuples $W_{0}, W_{1} ⁣ : I \to (R^{2})^{3}$ , and each value of each of the six windows is one of $q_{1}, q_{2}, q_{3}$ or one of $m \pm ρ (v), m^{'} \pm ρ (v)$ .

F2F3F5

1.2

The rotation braid. By [F4] the motion $u \mapsto R_{π u} (p_{k})$ is continuous for each $k$ and satisfies $R_{π u} (p_{k}) = p_{k}$ at $u = 0$ and $R_{π u} (p_{k}) = - p_{k}$ at $u = 1$ , so the strands of $r o t$ run from $p_{1} = (- 2 h, 0), p_{2} = (0, 0), p_{3} = (2 h, 0)$ to $- p_{1} = (2 h, 0), - p_{2} = (0, 0), - p_{3} = (- 2 h, 0)$ , that is from $Q$ to $(q_{3}, q_{2}, q_{1})$ ; at height $u$ the three positions are $R_{π u} (p_{k})$ , whose mutual distances are those of the distinct points $p_{1}, p_{2}, p_{3}$ because $R_{π u}$ preserves the norm and is linear and injective; and every value lies in $D^{\circ}$ , since $∥ R_{π u} (p_{k}) ∥_{2} = ∥ p_{k} ∥_{2} \leq 2 h = \frac{1}{8} < 1$ ; hence $r o t$ is a braid based at $Q$ with endpoint permutation $(1 3)$ .

F3F4F5

2.1

Collision bounds for the windows. For $v \in I$ the diamond path satisfies $∥ ρ (v) ∥_{2}^{2} = h^{2} (8 v^{2} - 4 v + 1)$ for $v \leq \frac{1}{2}$ and $∥ ρ (v) ∥_{2}^{2} = h^{2} ((2 v - 1)^{2} + (2 v - 2)^{2})$ for $v \geq \frac{1}{2}$ ; the first is $\geq h^{2} \cdot \frac{1}{2}$ with equality at $v = \frac{1}{4}$ and the second is $\geq h^{2} \cdot \frac{1}{2}$ with equality at $v = \frac{3}{4}$ , and both are $\leq h^{2}$ . Hence in each window the two moving strands, which are $m + ρ (v)$ and $m - ρ (v)$ (or $m^{'} + ρ (v))$ and $m^{'} - ρ (v)$ ), are separated by $2 ∥ ρ (v) ∥_{2} \geq \sqrt{2} h = \frac{\sqrt{2}}{16}$ , 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 $3 h$ from that window's midpoint and therefore at distance at least $3 h - h = 2 h = \frac{1}{8}$ from each moving point; moreover every window value has norm at most $∥ m ∥_{2} + h = ∥ m^{'} ∥_{2} + h = 2 h = \frac{1}{8} < 1$ , 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 $Q$ .

F2F3F4step 1.1

2.2

The endpoint permutations. By [F2] and step 1.1, $π (W_{0}) = π (σ_{1}) \circ π (σ_{2} ⋆ σ_{1}) = π (σ_{1}) \circ π (σ_{2}) \circ π (σ_{1})$ and $π (W_{1}) = π (σ_{2}) \circ π (σ_{1}) \circ π (σ_{2})$ ; evaluating the first composite, $(1 2) \circ (2 3) \circ (1 2)$ , at $1, 2, 3$ gives $1 \mapsto (1 2) (3) = 3$ , $2 \mapsto (1 2) (1) = 2$ and $3 \mapsto (1 2) (2) = 1$ , that is the transposition $(1 3)$ , and evaluating the second, $(2 3) \circ (1 2) \circ (2 3)$ , 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 $(1 3)$ ; 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.

F3step 1.1

2.3

The interpolation and its values at sample heights. Put $Z_{k} (s, u) : = (1 - s) (W_{0})_{k} (u) + s {r o t}_{k} (u)$ 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 $R_{π u}$ , the identities ${r o t}_{l} (u) - {r o t}_{k} (u) = 2 h (l - k) (\cos π u, \sin π u)$ and $(1 - s) > 0$ , to the statement that $(W_{0})_{k} (u) - (W_{0})_{l} (u)$ is a positive multiple of $(\cos π u, \sin π u)$ . At $u = 0$ the three strand positions of $W_{0}$ are $q_{1}, q_{2}, q_{3}$ , whose differences $(- 2 h, 0), (- 4 h, 0), (- 2 h, 0)$ 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 $(4 h, 0), (2 h, 0), (- 2 h, 0)$ are horizontal and nonzero while $(\cos \frac{π}{2}, \sin \frac{π}{2}) = (0, 1)$ is vertical, so no collision occurs, and $Z (\frac{1}{2}, \frac{1}{2}) = \frac{1}{2} ((2 h, 0), (- 2 h, 0), (0, 0)) + \frac{1}{2} ((0, - 2 h), (0, 0), (0, 2 h)) = ((h, - h), (- h, 0), (0, h))$ ; at $u = 1$ the positions of $W_{0}$ are $q_{3}, q_{2}, q_{1}$ , whose differences $(2 h, 0), (4 h, 0), (2 h, 0)$ are positive multiples of $(1, 0)$ while $(\cos π, \sin π) = (- 1, 0)$ , so no collision occurs, and $Z (s, 1) = (q_{3}, q_{2}, q_{1})$ for every $s$ .

F1F2F4F5step 1.1step 1.2

3.1

Conclusion. By [F1] the bracketing $W_{0} = σ_{1} ⋆ (σ_{2} ⋆ σ_{1})$ is braid-isotopic to $r o t$ , and the reflection $κ (w) = - w$ followed by the relabelling of $1$ and $3$ turns $W_{0}$ into $W_{1} = σ_{2} ⋆ (σ_{1} ⋆ σ_{2})$ , so $W_{1}$ is braid-isotopic to $r o t$ as well; step 1.2 identifies $r o t$ as a braid based at $Q$ , and step 2.2 gives $π (W_{0}) = π (W_{1}) = (1 3)$ ; step 2.3 exhibits the explicit intermediate values of the deformation, and steps 2.1 and 1.1 record the numerical facts $2 ∥ ρ (v) ∥_{2} \geq \frac{\sqrt{2}}{16}$ and $3 h - h = \frac{1}{8}$ behind its collision-freeness. Hence $W_{0} \sim W_{1}$ and, passing to isotopy classes in the group $G_{3}$ of [F2], $[σ_{1}] [σ_{2}] [σ_{1}] = [σ_{2}] [σ_{1}] [σ_{2}]$ . ∎

F1F2F3step 1.2step 2.2step 2.3

Remarks

The numbers are the smallest case of the relation: with $h = \frac{1}{16}$ the three base points are $- \frac{1}{8}, 0, \frac{1}{8}$ on the horizontal axis, so the local picture of the lemma is the picture of three points spaced $\frac{1}{8}$ apart, and the whole isotopy happens inside the closed ball of radius $\frac{1}{8}$ around the middle point, well inside $D^{\circ}$ .
The rotation $r o t$ is the geometric meaning of the relation: performing the three half twists on the outer pair and the middle pair alternately is the same as rotating the three-point configuration rigidly by the angle $π$ , and the point reflection in the middle point exchanges the two outer strands, which is why the two words have the same endpoint permutation $(1 3)$ .

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Setwise endpoints do not make a braid pure

Statement refuted

Refuted claim: a geometric braid based at $Q$ whose top endpoint set is ${z_{1} (1), \dots, z_{n} (1)} = {q_{1}, \dots, q_{n}}$ returns every strand to its own starting point, that is $z_{j} (1) = q_{j}$ for every $j$ ; in other words, the setwise endpoint condition of Geometric braids in the disc with setwise endpoints forces a braid to be pure.

The witness is the elementary half twist $σ_{1}$ on two strands (The elementary geometric half twist, its support disc, and its opposite, Geometric braids in the disc with setwise endpoints): its top endpoint set is ${q_{1}, q_{2}}$ , exactly the base configuration, but its first strand starts at $q_{1}$ and ends at $q_{2}$ , and its second strand starts at $q_{2}$ and ends at $q_{1}$ , so no strand returns to its own starting point and the endpoint permutation of $σ_{1}$ is the transposition of $1$ and $2$ , not the identity.

What is and is not claimed. What is refuted is only the implication "top endpoint set equal to $Q$ $\Rightarrow$ each label returns to its own starting point". Nothing here asserts that some other braid fails to be pure, and nothing here computes any invariant beyond the endpoint permutation. The example is the definitional point recorded in the definition of the endpoint permutation: labels are transported continuously from the bottom, so a braid may permute them, and the setwise condition is exactly the condition that this permutation be defined. It also shows that the failure is not an artefact of the choice of representative: since $π$ is constant along braid isotopies (Stacking of geometric braids is a well-defined associative operation on isotopy classes), $σ_{1}$ cannot be braid-isotopic to the trivial braid, whose endpoint permutation is the identity (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

Facts & Assumptions

Given: The natural number $2$ , the base configuration $Q = (q_{1}, q_{2})$ with $h = \frac{1}{12}$ , $q_{1} = (- \frac{1}{12}, 0)$ , $q_{2} = (\frac{1}{12}, 0)$ , and the elementary half twist $σ_{1}$ based at $Q$ .

[F1]

A braid based at $Q$ is a tuple $(u_{1}, u_{2})$ of continuous maps $u_{j} ⁣ : I \to D^{\circ}$ with $u_{1} (t) \neq u_{2} (t)$ , $u_{j} (0) = q_{j}$ and ${u_{1} (1), u_{2} (1)} = {q_{1}, q_{2}}$ ; its endpoint permutation is the unique $π \in S_{2}$ with $u_{j} (1) = q_{π (j)}$ , and a braid is called pure when this permutation is the identity; $q_{1} \neq q_{2}$ (Geometric braids in the disc with setwise endpoints, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Continuity of a map of topological spaces at a point and globally).

[F2]

The half twist is $(σ_{1})_{1} = m_{1} + ρ$ , $(σ_{1})_{2} = m_{1} - ρ$ with $m_{1} = \frac{q_{1} + q_{2}}{2} = (0, 0)$ and $ρ (0) = (- h, 0)$ , $ρ (1) = (h, 0)$ ; $σ_{1}$ is a braid based at $Q$ and $π (σ_{1})$ is the transposition of $1$ and $2$ (The elementary geometric half twist, its support disc, and its opposite, Geometric braids in the disc with setwise endpoints).

[F3]

The endpoint permutation is constant along braid isotopies, so equal values of $π$ are necessary for two braids to be braid-isotopic; in particular the transposition of $1$ and $2$ differs from the identity permutation of $S_{2}$ , and the trivial braid has the identity endpoint permutation (Stacking of geometric braids is a well-defined associative operation on isotopy classes, Braid isotopy relative to the top and bottom endpoints, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

Counterexample

technique · direct

1.1

The witness and its endpoint values. Take $n = 2$ and $β : = σ_{1}$ ; by [F2] its strands are $(σ_{1})_{1} (t) = m_{1} + ρ (t)$ and $(σ_{1})_{2} (t) = m_{1} - ρ (t)$ , so $(σ_{1})_{1} (0) = m_{1} + ρ (0) = (- h, 0) = q_{1}$ , $(σ_{1})_{2} (0) = m_{1} - ρ (0) = (h, 0) = q_{2}$ , and at the top $(σ_{1})_{1} (1) = m_{1} + ρ (1) = (h, 0) = q_{2}$ while $(σ_{1})_{2} (1) = m_{1} - ρ (1) = (- h, 0) = q_{1}$ , since $m_{1} = (0, 0)$ and $ρ (0) = (- h, 0)$ , $ρ (1) = (h, 0)$ by [F2].

F2F1

2.1

The setwise condition holds. The top endpoint set of $σ_{1}$ is ${(σ_{1})_{1} (1), (σ_{1})_{2} (1)} = {q_{2}, q_{1}} = {q_{1}, q_{2}}$ by step 1.1, so $σ_{1}$ satisfies the hypothesis of the refuted claim; the endpoint permutation of $σ_{1}$ is the unique $π \in S_{2}$ with $(σ_{1})_{j} (1) = q_{π (j)}$ for $j = 1, 2$ , which by step 1.1 is the transposition $π (1) = 2$ , $π (2) = 1$ of [F2].

F1F2step 1.1

2.2

The pointwise conclusion fails. By step 1.1 the first strand ends at $q_{2} \neq q_{1}$ and the second strand ends at $q_{1} \neq q_{2}$ , since $q_{1} \neq q_{2}$ by [F1]; hence $(σ_{1})_{j} (1) \neq q_{j}$ for both labels $j$ , so the conclusion of the refuted claim fails for this braid.

F1step 1.1

3.1

The failure is isotopy invariant. By step 2.1 the endpoint permutation of $σ_{1}$ is the transposition of $1, 2$ , which is not the identity permutation of $S_{2}$ , whereas the trivial braid has the identity endpoint permutation; by [F3] the endpoint permutation is constant along braid isotopies, so $σ_{1}$ is not braid-isotopic to the trivial braid, and in particular it is not pure in the sense of [F1].

F3step 2.1

4.1

Conclusion. Steps 1.1, 2.1 and 2.2 exhibit a braid whose top endpoint set equals the base configuration while no strand returns to its own starting point, so the refuted claim is false; step 3.1 shows moreover that this braid is not braid-isotopic to the trivial braid. ∎

step 2.1step 2.2step 3.1

Remarks

The distinction is exactly the one the definition records: the top matching of a braid is an arbitrary permutation of the labels, the setwise condition only says that this matching is defined at all, and the pure braids are the special case in which the matching is the identity.
The witness is minimal: with two strands the only non-identity permutation is the transposition, and the half twist realises it with the smallest possible support, the disc $U_{1}$ containing exactly the two base points.

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An arbitrary isotopy of arcs need not be a braid isotopy

Statement refuted

Refuted claim: the requirement in Braid isotopy relative to the top and bottom endpoints that every slice of the family be a braid is redundant, that is: if $(α_{s})_{s \in I}$ is a continuous family of injective parameterisations of arcs in $D^{\circ} \times I$ whose endpoints $α_{s} (0)$ and $α_{s} (1)$ are held fixed at the bottom and top points of the base configuration throughout, then the images $A_{s} : = α_{s} [I]$ are strands of geometric braids, so that the family is a braid isotopy as soon as its two boundary arcs are braids.

The witness deforms the single strand of the trivial one-strand braid. At the middle of the deformation its height coordinate has a horizontal shelf: two distinct points at one and the same height, so its image meets a horizontal slice in two points and is therefore not the strand of any braid. Since reparametrising an arc does not change its image, no choice of parameterisation repairs this; the family is a legitimate continuous deformation of arcs with fixed endpoints, but it is not a braid isotopy, and the definition's one-point-per-height condition is not redundant.

What is and is not claimed. The two boundary arcs $A_{0}$ and $A_{1}$ of the exhibited family are equal (both are the trivial braid), so nothing here asserts that two braids fail to be braid-isotopic; what is refuted is only the claim that an arbitrary arc deformation with braid boundary arcs is itself a braid isotopy. Nothing is claimed about closed links or about the classification of knots. The height function of the middle arc is not strictly increasing; the definition of braid isotopy requires each intermediate object to be a braid, which for a single strand means exactly that it meets each horizontal slice once.

Facts & Assumptions

Given: The natural number $1$ , so that $h = \frac{1}{4 (1 + 1)} = \frac{1}{8}$ and the base configuration of Geometric braids in the disc with setwise endpoints is $Q = (q_{1})$ with $q_{1} = (0, 0)$ , and the family of arc parameterisations $α_{s} ⁣ : I \to D^{\circ} \times I$ , $s \in I$ , defined by

$λ (s) : = \frac{1}{4} \min (2 s, 2 - 2 s), w (u) : = {\begin{cases} 4 u, & 0 \leq u \leq \frac{1}{4}, \\ 2 - 4 u, & \frac{1}{4} \leq u \leq \frac{3}{4}, \\ 4 u - 4, & \frac{3}{4} \leq u \leq 1, \end{cases} α_{s} (u) : = ((\frac{λ (s)}{8} w (u), 0), u + λ (s) w (u)) .$

[F1]

For $n = 1$ a braid based at $Q$ is a continuous map $z_{1} ⁣ : I \to D^{\circ}$ with $z_{1} (0) = q_{1}$ and ${z_{1} (1)} = {q_{1}}$ , and its strand is the graph ${(z_{1} (t), t) : t \in I} \subseteq D^{\circ} \times I$ , which by construction meets each horizontal slice $D^{\circ} \times {t}$ , $t \in I$ , in exactly one point; the trivial braid has the constant strand $z_{1} (t) = q_{1}$ , whose graph is ${(0, 0)} \times I$ , and a braid isotopy from $β$ to $β^{'}$ is a family of braids $Z (s, \cdot)$ depending jointly continuously on $s$ , with each slice a braid (Geometric braids in the disc with setwise endpoints, Braid isotopy relative to the top and bottom endpoints, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[F2]

The closed unit disc is $D = {w \in R^{2} : ∥ w ∥_{2} \leq 1}$ with interior $D^{\circ}$ ; the set $D^{\circ} \times I$ carries the product topology, and its points are written as $(spatial point, height)$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ , 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).

[F3]

The functions $s \mapsto λ (s)$ and $u \mapsto w (u)$ are continuous, being piecewise linear with continuous gluing, and sums, products and composites of continuous maps are continuous; continuity of a function on $I$ or on $I \times I$ follows from continuity on the finitely many closed pieces ${s \leq \frac{1}{2}}$ , ${s \geq \frac{1}{2}}$ and ${u \leq \frac{1}{4}}$ , ${\frac{1}{4} \leq u \leq \frac{3}{4}}$ , ${u \geq \frac{3}{4}}$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Continuity of a map of topological spaces at a point and globally, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

The family is jointly continuous with the prescribed fixed endpoints. By [F3] the maps $s \mapsto λ (s)$ and $u \mapsto w (u)$ are continuous, hence $(s, u) \mapsto α_{s} (u)$ is continuous by [F3]; at $u = 0$ one has $w (0) = 0$ , so $α_{s} (0) = ((0, 0), 0) = (q_{1}, 0)$ for every $s$ , and at $u = 1$ one has $w (1) = 0$ and $λ (s) w (1) = 0$ , so $α_{s} (1) = ((0, 0), 1) = (q_{1}, 1)$ for every $s$ ; thus the endpoints of the parameterised arcs are fixed at the bottom and top points of $Q$ .

F1F2F3

1.2

Every slice lies in $D^{\circ} \times I$ . For every $s \in I$ one has $0 \leq λ (s) \leq \frac{1}{4}$ , and $w$ takes the values $w (0) = 0$ , $w (\frac{1}{4}) = 1$ , $w (\frac{1}{2}) = 0$ , $w (\frac{3}{4}) = - 1$ , $w (1) = 0$ with $∣ w (u) ∣ \leq 1$ for all $u$ ; hence the spatial coordinate satisfies $∣ \frac{λ (s)}{8} w (u) ∣ \leq \frac{1}{32} < 1$ , so each spatial point $(\frac{λ (s)}{8} w (u), 0)$ lies in $D^{\circ}$ ; and the height $u + λ (s) w (u)$ lies in $I$ , because $w (u) \geq 0$ for $u \leq \frac{1}{2}$ gives $0 \leq u \leq u + λ (s) w (u) \leq \frac{1}{2} + \frac{1}{4} \leq 1$ there, while $w (u) \leq 0$ for $u \geq \frac{1}{2}$ gives $\frac{1}{2} - \frac{1}{4} \leq u + λ (s) w (u) \leq u \leq 1$ there.

F2F3

2.1

Each slice is a simple arc. Let $s \in I$ and suppose $α_{s} (u) = α_{s} (u^{'})$ ; comparing the spatial coordinates gives $λ (s) w (u) = λ (s) w (u^{'})$ and comparing the heights gives $u - u^{'} = λ (s) (w (u^{'}) - w (u)) = 0$ , so $u = u^{'}$ ; hence every $α_{s}$ is injective, and its image $A_{s} : = α_{s} [I]$ is a simple arc with the endpoints $(q_{1}, 0)$ and $(q_{1}, 1)$ .

step 1.1step 1.2

2.2

The middle slice is not the strand of a braid. At $s = \frac{1}{2}$ one has $λ (\frac{1}{2}) = \frac{1}{4}$ , and the two parameters $u = \frac{1}{4}$ and $u = \frac{3}{4}$ are distinct while $w (\frac{1}{4}) = 1$ , $w (\frac{3}{4}) = - 1$ give equal heights $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ and $\frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ and spatial coordinates $\frac{1}{32}$ and $- \frac{1}{32}$ ; hence $α_{1 / 2} (\frac{1}{4}) = ((\frac{1}{32}, 0), \frac{1}{2})$ and $α_{1 / 2} (\frac{3}{4}) = ((- \frac{1}{32}, 0), \frac{1}{2})$ are two distinct points of $A_{1 / 2}$ in the same horizontal slice $D^{\circ} \times {\frac{1}{2}}$ .

F2step 1.2

3.1

The boundary arcs are braids. For $s \in {0, 1}$ one has $λ (s) = \frac{1}{4} \min (0, 2) = 0$ , so $α_{s} (u) = ((0, 0), u)$ ; the image $A_{0} = A_{1} = {(0, 0)} \times I$ is exactly the strand of the trivial braid of [F1], hence both boundary slices are geometric braids based at $Q$ and the hypothesis of the refuted claim is satisfied.

F1step 1.1step 2.1

4.1

Conclusion. The family $(α_{s})$ satisfies the hypotheses of the refuted claim by steps 1.1, 1.2, 2.1 and 3.1, but its slice at $s = \frac{1}{2}$ meets the horizontal slice at height $\frac{1}{2}$ in two distinct points by step 2.2, whereas the strand of a braid based at $Q$ meets each horizontal slice in exactly one point by [F1]; since a reparametrisation does not change the image $A_{1 / 2}$ , no choice of parameterisation makes that slice a braid, so the family is not a braid isotopy. Hence the one-point-per-height requirement in the definition of braid isotopy is not redundant, and the refuted claim is false. ∎

F1step 2.1step 3.1step 2.2

Remarks

At $s = 1 / 2$ the height function is $2 u$ on $[0, 1 / 4]$ , constant $1 / 2$ on $[1 / 4, 3 / 4]$ , and $2 u - 1$ on $[3 / 4, 1]$ . This horizontal shelf gives many distinct points at height $1 / 2$ ; the height is nondecreasing, but not strictly increasing, and the arc is not a one-point-per-height graph.
The example is one-dimensional in the sense that a single strand suffices: no collision analysis between different strands arises, and the whole phenomenon is the failure of the height projection to restrict to a homeomorphism of the arc onto $I$ .
The two boundary arcs being equal is what makes the point sharp: the deformation does not change the isotopy class of anything, and yet it leaves the class of braid isotopies, because braid isotopy is a relation between braids (tuples of one-point-per-height strands) and not between arcs.

Geometric Braids and Artin Generators — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Geometric two strand braids are integer twists

Example

Facts & Assumptions

Verification

Remarks

The three strand geometric braid relation

Example

Facts & Assumptions

Verification

Remarks

Setwise endpoints do not make a braid pure

Statement refuted

Facts & Assumptions

Counterexample

Remarks

An arbitrary isotopy of arcs need not be a braid isotopy

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources