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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.

Depends on

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, sections 1.2-1.3, printed pp. 4-5 (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 1.1, author manuscript pp. 3-5 (standard reference, not scraped)