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A pure two-strand full twist as an ordered loop

Example

For n=2 with Q=(−h,h), the ordered path η(t)=(−h,−h+2hexp⁡(2πit)) stays collision-free in the open disk and closes at Q. Its traced geometric braid is the positive full twist [σ1]2 and has identity endpoint permutation.

Facts & Assumptions

Given: n=2, the published base tuple Q, spacing h, positive elementary half twist σ1, and its diamond relative path ρ.

[L1]

For n=2, h=1/(4(2+1))=1/12, Q=(−h,h), the midpoint is 0, and (σ1)1=ρ, (σ1)2=−ρ. The path ρ is continuous, has endpoints −h,+h, never vanishes, and satisfies ∣ρ(t)∣≤h. The positive convention is the published anticlockwise half twist (The elementary geometric half twist, its support disc, and its opposite, Geometric braids in the disc with setwise endpoints).

[L2]

The complex exponential is continuous, exp⁡(iθ)=cos⁡θ+isin⁡θ and ∣exp⁡(iθ)∣=1 for real θ, and exp⁡(iπ)=−1; also exp⁡(z+w)=exp⁡zexp⁡w (The complex exponential is entire and its complex derivative is itself, Complex differentiability at a point implies continuity there, exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0, exp⁡(z+w)=exp⁡z exp⁡w, and the complex exponential extends the real exponential).

[L3]

Complex modulus is definite and satisfies ∣z+w∣≤∣z∣+∣w∣; complex addition and scalar multiplication are continuous (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive, Vector addition and scalar multiplication are continuous in a normed space).

[L4]

On [0,π] cosine decreases through 0 at π/2 and sine is nonnegative; on [π,2π] cosine increases through 0 at 3π/2 and sine is nonpositive by its π-shift. Thus exp⁡(2πit) lies in the corresponding closed quadrant for t∈[0,1/4], [1/4,1/2], [1/2,3/4], and [3/4,1] (Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi, Pi is the first positive zero of sine).

[L6]

The orbit map p2:F2(X)→C2(X) is continuous, and an interior based motion is a continuous path with both endpoints [Q] (Unordered configuration spaces Cn(X), Based motions of an unordered point configuration).

[L7]

Every interior based configuration loop at [Q] has a unique ordered lift from Q, whose coordinate graphs are its geometric braid trace (An interior configuration loop traces a geometric braid).

[L8]

A geometric braid starts at the labelled tuple Q, remains collision-free in the open disk, and is pure when every labelled endpoint returns to its starting point (Geometric braids in the disc with setwise endpoints).

[L9]

In γ⋆β, the first half runs β and the second half runs γ with labels permuted by the lower braid, and the induced class product is [γ][β]=[γ⋆β] (Stacking of geometric braids is a well-defined associative operation on isotopy classes).

[L10]

The isotopy classes of geometric braids based at Q form a group with the stacking product (The isotopy classes of geometric braids based at Q form a group, and the endpoint permutation is a homomorphism).

[L11]

A braid isotopy is a jointly continuous family of braids with bottom tuple Q and top endpoint set Q at every isotopy parameter (Braid isotopy relative to the top and bottom endpoints); piecewise continuous maps on two closed sets covering the square paste continuously (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L12]

For a pure geometric braid with coordinate loop zβ at Q, the pure-braid isomorphism is Ψ([β])=(ι∗F[zβ])−1 (Pure geometric braids and ordered configuration loops).

The formulae below specify every path and homotopy. No lift, representative, or point of a nonempty set is chosen, so the Axiom of Choice is not used.

Proof

technique · direct
1.1L1L2L3L5

Put ω(t):=exp⁡(2πit). By [L2], ω is continuous and ∣ω(t)∣=1. Since exp⁡(iπ)=−1 and the exponential addition law gives exp⁡(2πi)=exp⁡(iπ)2=1, one has ω(0)=ω(1)=1. Hence η(t)=(−h,−h+2hω(t)) starts and ends at Q. Its coordinate difference is 2hω(t)≠0, and ∣−h+2hω(t)∣≤h+2h=3h=14<1, while ∣−h∣=h<1. Thus its values are ordered configurations in F2(int⁡D2); coordinate continuity and [L5] show that η is a continuous ordered loop at Q.

1.2L6L7L8

Let α(t):=p2(η(t)). By [L6], this is a continuous interior based loop at [Q]. Its unique lift from Q is η itself, so [L7] identifies its trace with the coordinate braid βη(t)=(−h,−h+2hω(t)). Because both coordinates of η return to their starting values, [L8] shows βη is pure and its endpoint permutation is the identity.

1.3L1L2L4L9

Use the stacking formula [L9] on two copies of σ1. Since the endpoint permutation of the lower copy exchanges labels 1 and 2, the stacked coordinate pair is (ρ(2t),−ρ(2t))(0≤t≤12),(−ρ(2t−1),ρ(2t−1))(12≤t≤1). Its centre is 0 and its second-minus-first coordinate is D0(t)={−2ρ(2t),0≤t≤12,2ρ(2t−1),12≤t≤1. Substitution of the two branches of ρ from [L1] shows that D0(t) lies successively in the first, second, third, and fourth closed quadrants on the four quarter intervals. It never vanishes, and ∣D0(t)∣≤2h by [L1]. By [L2] and [L4], D1(t):=2hω(t) lies in the same respective closed quadrant, never vanishes, and has modulus 2h.

2.1L1L2L3L9L11step 1.3

For s,t∈I set Ds(t):=(1−s)D0(t)+sD1(t),H1(s,t):=(−Ds(t)/2,Ds(t)/2). Each closed quadrant is convex and contains no pair of opposite nonzero vectors, so Ds(t)≠0 for every (s,t). By [L3], ∣Ds(t)∣≤(1−s)∣D0(t)∣+s∣D1(t)∣≤2h, so both coordinates of H1 have modulus at most h<1. The formulas and [L2], [L3], [L9] give joint continuity. Both D0 and D1 equal 2h at t=0,1, so H1(s,0)=H1(s,1)=Q for every s. Thus [L11] makes H1 a braid isotopy from the stacked diamond braid to the centred round pair βround(t)=(−hω(t),hω(t)).

2.2L1L2L3L11step 1.1

Put Cs(t):=s(−h+hω(t)) and define H2(s,t):=(Cs(t)−hω(t),Cs(t)+hω(t)). The coordinate difference is 2hω(t)≠0, and [L3] gives ∣Cs(t)±hω(t)∣≤∣Cs(t)∣+h≤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 Cs(0)=Cs(1)=0 and ω(0)=ω(1)=1, the endpoints are Q for every s. Thus [L11] makes H2 a braid isotopy from βround to βη.

3.1L9L10L11step 1.2step 2.1step 2.2

The families H1 and H2 agree at their common braid βround. Pasting H1(2s,t) for s≤1/2 to H2(2s−1,t) for s≥1/2 gives a jointly continuous family by [L11]. Each slice is a braid and its endpoints remain Q, so this is a braid isotopy from σ1⋆σ1 to βη. By [L9] and [L10], [βη]=[σ1⋆σ1]=[σ1]2. 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.

4.1

Since βη is pure by step 1.2, the exact ordered representative of its class under the pure-braid isomorphism [L12] is Ψ([βη])=(ι∗F[η])−1∈PB2. 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] □

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