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The geometric three strand braid relation

Statement

Let $n \geq 3$ and let $i$ be an index with $1 \leq i \leq n - 2$ . Let $σ_{i}$ and $σ_{i + 1}$ be the elementary half twists of The elementary geometric half twist, its support disc, and its opposite based at $Q$ , and let $⋆$ be the stacking of Stacking of geometric braids is a well-defined associative operation on isotopy classes. Then

$σ_{i} ⋆ σ_{i + 1} ⋆ σ_{i} \sim σ_{i + 1} ⋆ σ_{i} ⋆ σ_{i + 1},$

and consequently, in the group $G_{n}$ of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism,

$[σ_{i}] [σ_{i + 1}] [σ_{i}] = [σ_{i + 1}] [σ_{i}] [σ_{i + 1}] .$

The proof exhibits an explicit intermediate braid: writing $κ$ for the point reflection $w \mapsto 2 q_{i + 1} - w$ and $r o t$ for the rotation of the three points $q_{i}, q_{i + 1}, q_{i + 2}$ about $q_{i + 1}$ by the angle $π u$ at height $u$ , with the remaining strands fixed, 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 $i + 2$ , and the reflection $κ$ , 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 two bracketings of each word are themselves braid-isotopic by the associativity of stacking. All constructions are explicit and no choice principle is used.

Facts & Assumptions

Given: A natural number $n \geq 3$ , an index $i$ with $1 \leq i \leq n - 2$ , the base configuration $Q = (q_{1}, \dots, q_{n})$ with $q_{j + 1} - q_{j} = (2 h, 0)$ and $h = \frac{1}{4 (n + 1)}$ , and the half twists $σ_{i}, σ_{i + 1}$ based at $Q$ .

[F1]

A braid based at $Q$ is a tuple $(u_{1}, \dots, u_{n})$ of continuous maps $u_{j} ⁣ : I \to D^{\circ}$ with $u_{k} (t) \neq u_{l} (t)$ for $k \neq l$ , $u_{k} (0) = q_{k}$ , and ${u_{1} (1), \dots, u_{n} (1)} = {q_{1}, \dots, q_{n}}$ ; its endpoint permutation $π (u)$ is the unique permutation with $u_{k} (1) = q_{π (u) (k)}$ ; the points $q_{i}, q_{i + 1}, q_{i + 2}$ are collinear and equally spaced, so in the coordinates centred at $q_{i + 1}$ they are $(- 2 h, 0), (0, 0), (2 h, 0)$ , and $∣ q_{j} - q_{i + 1} ∣ \geq 4 h$ for every $j \notin {i, i + 1, i + 2}$ (Geometric braids in the disc with setwise endpoints).

[F2]

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 $m_{k} = q_{k} + (h, 0)$ ; the diamond path $ρ$ satisfies $ρ (0) = (- h, 0)$ , $ρ (\frac{1}{2}) = (0, - h)$ , $ρ (1) = (h, 0)$ , $∥ ρ (v) ∥_{2} \leq h$ , and $ρ_{2} (v) \leq 0$ with $ρ_{2} (v) = 0$ only for $v \in {0, 1}$ and $ρ (v) \neq 0$ for every $v \in I$ ; and $π (σ_{k})$ is the transposition of $k$ and $k + 1$ , while the support disc contains $q_{k}, q_{k + 1}$ and no other base point (The elementary geometric half twist, its support disc, and its opposite).

[F3]

A braid isotopy is a tuple of jointly continuous maps $Z_{j} ⁣ : I \times I \to D^{\circ}$ whose every slice is a braid based at $Q$ and whose boundary slices are the two given braids (Braid isotopy relative to the top and bottom endpoints).

[F4]

Stacking places its right factor in the lower half of the height interval and its left factor in the upper half: writing $z_{j}$ for the strands of the right factor $β$ and $w_{j}$ for those of the left factor $γ$ , one has $(γ ⋆ β)_{j} (t) = z_{j} (2 t)$ for $t \leq \frac{1}{2}$ and $(γ ⋆ β)_{j} (t) = w_{π (β) (j)} (2 t - 1)$ for $t \geq \frac{1}{2}$ ; stacking of braids is a braid, it descends to isotopy classes, $π (γ ⋆ β) = π (γ) \circ π (β)$ , and the two bracketings of a threefold stacking are braid-isotopic, $(δ ⋆ γ) ⋆ β \sim δ ⋆ (γ ⋆ β)$ (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[F5]

$G_{n}$ is a group with operation $[γ] [β] = [γ ⋆ β]$ , so equal isotopy classes have equal products (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

[L6]

Composites of continuous maps are continuous and continuity pastes over the two closed halves of a square; the interval $I = [0, 1]$ carries the subspace topology in which the points $0, \frac{1}{2}, 1$ cut it into closed pieces (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).

Proof

technique · direct

1.1

(The two bracketings, in local coordinates.) Put $c : = q_{i + 1}$ , $p_{1} : = q_{i}$ , $p_{2} : = q_{i + 1}$ , $p_{3} : = q_{i + 2}$ , $m : = m_{i} = q_{i} + (h, 0)$ and $m^{'} : = m_{i + 1} = q_{i + 1} + (h, 0)$ , so that the centred vectors $p_{1} - c = (- 2 h, 0)$ , $p_{2} - c = (0, 0)$ , $p_{3} - c = (2 h, 0)$ , while $m - c = (- h, 0)$ and $m^{'} - c = (h, 0)$ ; since $π (σ_{i})$ is the transposition of $i, i + 1$ and $π (σ_{i + 1})$ that of $i + 1, i + 2$ by [F2] and $π$ multiplies by [F4], the composite $π (σ_{i} ⋆ σ_{i + 1}) = π (σ_{i}) \circ π (σ_{i + 1})$ is the 3-cycle $i \mapsto i + 1 \mapsto i + 2 \mapsto i$ and $π (σ_{i + 1} ⋆ σ_{i})$ is its inverse $i \mapsto i + 2 \mapsto i + 1 \mapsto i$ , so applying the stacking formula of [F4] twice shows that the strands $i, i + 1, i + 2$ of $W_{0} : = σ_{i} ⋆ (σ_{i + 1} ⋆ σ_{i})$ are $(m + ρ (4 u), m - ρ (4 u), p_{3})$ for $0 \leq u \leq \frac{1}{4}$ , $(m^{'} + ρ (4 u - 1), p_{1}, m^{'} - ρ (4 u - 1))$ for $\frac{1}{4} \leq u \leq \frac{1}{2}$ and $(p_{3}, m + ρ (2 u - 1), m - ρ (2 u - 1))$ for $\frac{1}{2} \leq u \leq 1$ , while the same computation with the roles of $i$ and $i + 1$ interchanged shows that the strands $i, i + 1, i + 2$ of $W_{1} : = σ_{i + 1} ⋆ (σ_{i} ⋆ σ_{i + 1})$ are $(p_{1}, m^{'} + ρ (4 u), m^{'} - ρ (4 u))$ for $0 \leq u \leq \frac{1}{4}$ , $(m + ρ (4 u - 1), p_{3}, m - ρ (4 u - 1))$ for $\frac{1}{4} \leq u \leq \frac{1}{2}$ and $(m^{'} + ρ (2 u - 1), m^{'} - ρ (2 u - 1), p_{1})$ for $\frac{1}{2} \leq u \leq 1$ , every remaining strand of either tuple being constantly at its base point; hence $W_{0}$ and $W_{1}$ are braids based at $Q$ by [F4], with $π (W_{0}) = (i i + 1) \circ (i + 1 i + 2) \circ (i i + 1)$ and $π (W_{1}) = (i + 1 i + 2) \circ (i i + 1) \circ (i + 1 i + 2)$ , both the transposition of $i$ and $i + 2$ , and by the associativity clause of [F4] $W_{0}$ is braid-isotopic to $σ_{i} ⋆ σ_{i + 1} ⋆ σ_{i}$ and $W_{1}$ to $σ_{i + 1} ⋆ σ_{i} ⋆ σ_{i + 1}$ .

F1F2F4L6

1.2

(The rotation braid.) Let $R_{θ}$ be the linear rotation of $R^{2}$ about the origin through angle $θ$ , and define $v_{k} : = q_{k} - c$ and ${r o t}_{k} (u) : = c + R_{π u} v_{k}$ for $k \in {i, i + 1, i + 2}$ ; put ${r o t}_{k} (u) : = q_{k}$ for all other labels. The motions are continuous, the two outer centred vectors are antipodal because $v_{i + 2} = - v_{i}$ , and $v_{i + 1} = 0$ , so the middle strand stays at $c$ and all three remain pairwise distinct. Their distance from the origin is at most $∥ c ∥_{2} + 2 h \leq (n + 1) h < 1$ ; every other base point is at distance at least $4 h$ from $c$ by [F1], so no moving strand meets a constant one. At $u = 0$ the triple has values $c + v_{k} = q_{k}$ , and at $u = 1$ it has values $c - v_{k} = q_{2 i + 2 - k}$ ; thus $r o t$ is based at $Q$ and has endpoint permutation $(i i + 2)$ , shared by $W_{0}$ and $W_{1}$ .

F1F2L6

2.1

(The interpolation family.) For $s \in I$ define $Z_{k} (s, u) : = (1 - s) (W_{0})_{k} (u) + s {r o t}_{k} (u)$ for every label $k$ , where $(W_{0})_{k}$ and ${r o t}_{k}$ are the motions of step 1.1 and step 1.2; each $Z_{k}$ is jointly continuous, being a sum of products of continuous functions, and satisfies $∥ Z_{k} (s, u) ∥_{2} \leq (1 - s) ∥ (W_{0})_{k} (u) ∥_{2} + s ∥ {r o t}_{k} (u) ∥_{2} < 1$ , so it maps $I \times I$ into $D^{\circ}$ ; at $u = 0$ one has $Z_{k} (s, 0) = q_{k}$ for every $s$ because $(W_{0})_{k} (0) = {r o t}_{k} (0) = q_{k}$ , and at $u = 1$ one has $Z_{k} (s, 1) = q_{π (k)}$ for every $s$ because $(W_{0})_{k} (1) = {r o t}_{k} (1) = q_{π (k)}$ for the transposition $π$ of $i$ and $i + 2$ shared by both braids; so each slice $Z (s, \cdot)$ satisfies the endpoint conditions of [F1].

F1F3step 1.1step 1.2L6

2.2

(The collision criterion.) Fix $s \in (0, 1)$ and $u \in I$ and let $k < l$ be two labels of the triple ${i, i + 1, i + 2}$ ; since ${r o t}_{k} (u) - {r o t}_{l} (u) = R_{π u} (q_{k} - q_{l})$ by step 1.2 and $q_{k} - q_{l} = - 2 h λ_{k l} (1, 0)$ with $λ_{k l} : = l - k > 0$ by [F1], the equality $Z_{k} (s, u) = Z_{l} (s, u)$ is equivalent to $(1 - s) ((W_{0})_{k} (u) - (W_{0})_{l} (u)) = 2 h s λ_{k l} (\cos π u, \sin π u)$ , hence to the statement that $(W_{0})_{k} (u) - (W_{0})_{l} (u)$ is a positive multiple of $(\cos π u, \sin π u)$ ; so a slice $Z (s, \cdot)$ with $s \in (0, 1)$ has no collision between two labels of the triple as soon as no difference $(W_{0})_{k} - (W_{0})_{l}$ with $k < l$ in the triple is a positive multiple of $(\cos π u, \sin π u)$ at any height $u$ .

F1F2step 1.1step 1.2

2.3

(Phase one: $0 \leq u \leq \frac{1}{4}$ , $v = 4 u$ .) In this and the next two phase calculations, a subscript $a \in {1, 2, 3}$ on $(W_{0})_{a}$ denotes the local position in the active triple, namely the global strand $(W_{0})_{i + a - 1}$ ; all other strands remain fixed. Here $(W_{0})_{1} - (W_{0})_{2} = 2 ρ (v)$ has second coordinate $\leq 0$ , while $(W_{0})_{1} - (W_{0})_{3} = (- 3 h, 0) + ρ (v)$ and $(W_{0})_{2} - (W_{0})_{3} = (- 3 h, 0) - ρ (v)$ have first coordinate at most $- 2 h < 0$ ; since $\cos (π u) \geq \cos (\frac{π}{4}) > 0$ and $\sin (π u) \geq 0$ on this phase, a positive multiple of $(\cos π u, \sin π u)$ has positive first coordinate, which excludes the pairs $(1, 3)$ and $(2, 3)$ , while for the pair $(1, 2)$ the second coordinates would have to agree, forcing $2 ρ_{2} (v) = λ \sin (π u) \geq 0$ and hence $ρ_{2} (v) = 0$ , that is $v \in {0, 1}$ ; at $v = 0$ one has $(W_{0})_{1} - (W_{0})_{2} = 2 ρ (0) = (- 2 h, 0)$ , a negative multiple of $(\cos 0, \sin 0) = (1, 0)$ , and at $v = 1$ one has $(W_{0})_{1} - (W_{0})_{2} = 2 ρ (1) = (2 h, 0)$ , which is not a multiple of $(\cos \frac{π}{4}, \sin \frac{π}{4})$ at all.

F2step 1.1

2.4

(Phase two: $\frac{1}{4} \leq u \leq \frac{1}{2}$ , $v = 4 u - 1$ .) Here $(W_{0})_{1} - (W_{0})_{2} = (2 h + 2 h v, - 2 h v)$ respectively $(2 h + 2 h v, 2 h v - 2 h)$ according to whether $v \leq \frac{1}{2}$ or $v \geq \frac{1}{2}$ , $(W_{0})_{1} - (W_{0})_{3} = 2 ρ (v)$ and $(W_{0})_{2} - (W_{0})_{3} = (2 h v - 4 h, - 2 h v)$ respectively $(2 h v - 4 h, 2 h v - 2 h)$ ; all three have second coordinate at most $0$ , while $\sin (π u) \geq \sin (\frac{π}{4}) > 0$ on this phase, so no one of them is a positive multiple of $(\cos π u, \sin π u)$ , whose second coordinate is positive.

F2step 1.1

2.5

(Phase three: $\frac{1}{2} \leq u \leq 1$ , $v = 2 u - 1$ .) Here $(W_{0})_{1} - (W_{0})_{2} = (3 h, 0) - ρ (v)$ and $(W_{0})_{1} - (W_{0})_{3} = (3 h, 0) + ρ (v)$ have first coordinate at least $2 h > 0$ , while $\cos (π u) \leq 0$ on this phase, so neither is a positive multiple of $(\cos π u, \sin π u)$ ; the remaining difference $(W_{0})_{2} - (W_{0})_{3} = 2 ρ (v)$ has second coordinate at most $0$ , and a positive multiple of $(\cos π u, \sin π u)$ has second coordinate $\geq 0$ because $\sin (π u) \geq 0$ on this phase, so a coincidence would force the common second coordinate to vanish, that is $ρ_{2} (v) = 0$ and $\sin (π u) = 0$ , which gives $v \in {0, 1}$ and $u \in {\frac{1}{2}, 1}$ ; with $v = 2 u - 1$ this leaves the two candidates $(u, v) = (\frac{1}{2}, 0)$ , where $(W_{0})_{2} - (W_{0})_{3} = (- 2 h, 0)$ is not a multiple of $(\cos \frac{π}{2}, \sin \frac{π}{2}) = (0, 1)$ , and $(u, v) = (1, 1)$ , where $(W_{0})_{2} - (W_{0})_{3} = (2 h, 0)$ is a negative multiple of $(\cos π, \sin π) = (- 1, 0)$ rather than a positive one.

F2step 1.1

3.1

(The isotopy from $W_{0}$ to $r o t$ .) By steps 2.2, 2.3, 2.4 and 2.5 no two labels of the triple can meet in any slice $Z (s, \cdot)$ with $s \in (0, 1)$ and any height $u$ , and at $s = 0$ the slice is $W_{0}$ and at $s = 1$ it is $r o t$ , which are braids with pairwise distinct strands by steps 1.1 and 1.2; the triple values $(W_{0})_{k} (u)$ and ${r o t}_{k} (u)$ all lie within distance $2 h$ of $c$ by steps 1.1 and 1.2 and [F2], so $∥ Z_{k} (s, u) - c ∥_{2} \leq (1 - s) ∥ (W_{0})_{k} (u) - c ∥_{2} + s ∥ {r o t}_{k} (u) - c ∥_{2} \leq 2 h$ puts every triple strand of every slice in the ball of radius $2 h$ about $c$ , while the remaining strands are constantly at base points of distance at least $4 h$ from $c$ by [F1], so no triple strand ever meets one of them; with the endpoint computations of step 2.1 this makes $Z$ a braid isotopy from $W_{0}$ to $r o t$ in the sense of [F3].

F1F2F3step 1.1step 1.2step 2.1step 2.2step 2.3step 2.4step 2.5

4.1

(The reflection.) Let $κ (w) : = 2 c - w$ be the point reflection in $c$ and let $τ$ be the transposition of $i$ and $i + 2$ ; for $k \in {i, i + 1, i + 2}$ put $Z_{k}^{'} (s, u) : = κ (Z_{τ (k)} (s, u))$ and $Z_{k}^{'} (s, u) : = q_{k}$ for the remaining labels; each $Z_{k}^{'}$ is jointly continuous and, by step 3.1 and $∥ κ (w) - c ∥_{2} = ∥ w - c ∥_{2}$ , every value lies within distance $2 h$ of $c$ and hence in $D^{\circ}$ , because $∥ c ∥_{2} + 2 h \leq (n + 1) h < 1$ as in step 1.2; within the triple $κ$ is injective, so distinct reflected strands stay distinct, and they stay within distance $2 h$ of $c$ and hence at distance at least $2 h$ from the constant strands, which are at distance at least $4 h$ from $c$ by [F1]; so every slice $Z^{'} (s, \cdot)$ is a braid based at $Q$ ; at $s = 1$ one has $Z^{'} = r o t$ , because $κ ({r o t}_{τ (k)} (u)) = c - R_{π u} v_{τ (k)} = c + R_{π u} v_{k} = {r o t}_{k} (u)$ for the three points $p_{1}, p_{2}, p_{3}$ , whose labelling by $τ$ only exchanges the two outer centred vectors and satisfies $v_{τ (k)} = - v_{k}$ ; at $s = 0$ the reflected slice is $W_{1}$ , because the reflection identities $κ (m + ρ (v)) = m^{'} - ρ (v)$ , $κ (m - ρ (v)) = m^{'} + ρ (v)$ , $κ (p_{1}) = p_{3}$ and $κ (p_{3}) = p_{1}$ turn the window values of $W_{0}$ of step 1.1 into those of $W_{1}$ : on $[0, \frac{1}{4}]$ the reflected values are $(κ (p_{3}), κ (m - ρ (4 u)), κ (m + ρ (4 u))) = (p_{1}, m^{'} + ρ (4 u), m^{'} - ρ (4 u))$ , on $[\frac{1}{4}, \frac{1}{2}]$ they are $(κ (m^{'} - ρ (4 u - 1)), κ (p_{1}), κ (m^{'} + ρ (4 u - 1))) = (m + ρ (4 u - 1), p_{3}, m - ρ (4 u - 1))$ and on $[\frac{1}{2}, 1]$ they are $(κ (m - ρ (2 u - 1)), κ (m + ρ (2 u - 1)), κ (p_{3})) = (m^{'} + ρ (2 u - 1), m^{'} - ρ (2 u - 1), p_{1})$ , which are exactly the window values of $W_{1}$ ; hence $Z^{'}$ is a braid isotopy from $W_{1}$ to $r o t$ .

F1F2F3step 1.1step 1.2step 3.1L6

5.1

(Conclusion.) Steps 3.1 and 4.1 show that $W_{0}$ and $W_{1}$ are both braid-isotopic to $r o t$ , hence braid-isotopic to each other; step 1.1 identifies $W_{0}$ as the bracketing $σ_{i} ⋆ (σ_{i + 1} ⋆ σ_{i})$ and $W_{1}$ as the bracketing $σ_{i + 1} ⋆ (σ_{i} ⋆ σ_{i + 1})$ , and the associativity clause of [F4] shows that each is braid-isotopic to the corresponding word with the other bracketing, so that $σ_{i} ⋆ σ_{i + 1} ⋆ σ_{i} \sim σ_{i + 1} ⋆ σ_{i} ⋆ σ_{i + 1}$ ; passing to isotopy classes with [F5] gives $[σ_{i}] [σ_{i + 1}] [σ_{i}] = [σ_{i + 1}] [σ_{i}] [σ_{i + 1}]$ in $G_{n}$ . ∎

F4F5step 1.1step 3.1step 4.1

Remarks

The proof is local: only the three strands $i, i + 1, i + 2$ move, and all formulas are those of the three-strand picture with base points $(- 2 h, 0), (0, 0), (2 h, 0)$ spaced $2 h$ apart, which is why the lemma holds for every $n \geq 3$ and every $i$ with $1 \leq i \leq n - 2$ .
The moving strands stay within distance $2 h$ of $q_{i + 1}$ throughout the interpolation and the reflection, while every other base point is at distance at least $4 h$ from $q_{i + 1}$ ; this clearance is what makes the local computation an isotopy of $n$ -strand braids.
The braid relation is the geometric statement that three consecutive half twists of a triple can be deformed into the same three half twists performed by rotating the triple rigidly by $π$ about its middle point; the point reflection in that middle point exchanges the two outer strands and turns one word into the other.
The bracketing enters the formulas but not the conclusion: the proof computes the bracketings $σ_{i} ⋆ (σ_{i + 1} ⋆ σ_{i})$ and $σ_{i + 1} ⋆ (σ_{i} ⋆ σ_{i + 1})$ , and the associativity clause of Stacking of geometric braids is a well-defined associative operation on isotopy classes supplies the isotopy to the bracketings $(σ_{i} ⋆ σ_{i + 1}) ⋆ σ_{i}$ and $(σ_{i + 1} ⋆ σ_{i}) ⋆ σ_{i + 1}$ .

Depends on

Used by

The three strand geometric braid relation Example
The Artin presentation surjects onto the geometric braid group Proposition

Dependency tree · two levels

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, sections 1.5 and 3.2, printed pp. 7-8 and 23-26 (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 1.2, author manuscript pp. 5-6 (standard reference, not scraped)