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Geometric braid classes and the unordered configuration fundamental group

Statement

Fix $n \in N$ and the explicit geometric base tuple $Q = (q_{1}, \dots, q_{n})$ from Geometric braids in the disc with setwise endpoints. In The configuration braid group $B_{n}^{c o n f}$ as the fundamental group of an unordered configuration space, take its parameterized base configuration to be this same tuple $q = Q$ . Thus $B_{n}^{c o n f} = π_{1} (C_{n} (D^{2}), [Q]) .$ Let $G_{n}$ be the group of geometric braid-isotopy classes based at $Q$ , with the stacking product $[γ] [β] = [γ ⋆ β]$ . Let $ι^{C} : C_{n} (int D^{2}) ↪ C_{n} (D^{2})$ be the inclusion and let $ι_{*}^{C}$ be its induced homomorphism at $[Q]$ . For a geometric braid $β$ , let $S (β)$ be its unordered configuration slice. Then raw slicing induces a bijection $S : G_{n} ⟶ π_{1} (C_{n} (int D^{2}), [Q]), [β] ⟼ [S (β)],$ and reverses products. Consequently $Φ : G_{n} ⟶ B_{n}^{c o n f}, [β] ⟼ (ι_{*}^{C} [S (β)])^{- 1}$ is a group isomorphism. Relating this group to one based at another configuration requires choosing a connecting path; no Artin-presentation completeness claim is made here.

Facts & Assumptions

Given: $n \in N$ , the specified tuple $Q$ , its geometric braid group $G_{n}$ , the open-to-closed configuration inclusion, and the slicing map.

[L1]

The geometric motion definition fixes the same explicit tuple $Q$ and specifies that the configuration braid group is based at its orbit $[Q]$ (Based motions of an unordered point configuration).

[L2]

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

[L3]

Slicing and tracing are well-defined mutually inverse bijections between geometric braid-isotopy classes at $Q$ and based path-homotopy classes in $C_{n} (int D^{2})$ at $[Q]$ (Tracing and slicing are inverse on relative classes).

[L4]

Stacking and slicing satisfy $[S (γ ⋆ β)] = [S (β)] [S (γ)]$ with the library's first-loop-then-second loop product (Raw slicing reverses geometric stacking products).

[L5]

At every $q \in F_{n} (int D^{2})$ , the inclusion-induced map $ι_{*}^{C} : π_{1} (C_{n} (int D^{2}), [q]) ⟶ π_{1} (C_{n} (D^{2}), [q])$ is a group isomorphism (The interior-disc and closed-disc configuration spaces are homotopy equivalent).

[L6]

The configuration braid group is $B_{n}^{c o n f} : = π_{1} (C_{n} (D^{2}), [q])$ for the chosen base configuration $q$ (The configuration braid group $B_{n}^{c o n f}$ as the fundamental group of an unordered configuration space).

[L7]

For a pointed continuous map $f$ , the induced map sends $[α]$ to $[f \circ α]$ and is a group homomorphism (The homomorphism on fundamental groups induced by a pointed continuous map).

[L8]

The loop product $[α] [η] = [α * η]$ traverses $α$ first and $η$ second (Based loops and the fundamental group).

[L9]

The target fundamental-group classes form a group with two-sided inverses and an identity element (Loop classes form the group $π_{1} (X, x_{0})$ under concatenation).

[L10]

A group homomorphism preserves products: $f (x y) = f (x) f (y)$ (Monoid homomorphism and group homomorphism).

[L11]

A function is bijective when it is both injective and surjective, and a two-sided inverse certifies those properties (Injection, surjection, bijection).

[L12]

A group isomorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L13]

For $n = 0$ there is exactly one empty geometric braid; for $n = 1$ the geometric braid condition imposes no collision restriction (Geometric braids in the disc with setwise endpoints).

[L14]

For $n = 0$ , both configuration spaces are one-point spaces and there is one based motion; for $n = 1$ , the configuration space is canonically the disk and there is no collision condition (Based motions of an unordered point configuration).

The tuple $Q$ and every braid coordinate path are specified. Tracing uses the unique lift from this $Q$ , and no arbitrary order, representative, or connecting path is chosen. No Axiom of Choice is used.

Proof

technique · direct

1.1L1L5L6L7

Fix the common basepoint. By [L1], the definition of $B_{n}^{c o n f}$ is instantiated at $q = Q$ , so the open-to-closed inclusion is based at the same orbit $[Q]$ on both sides. By [L5] and [L7], $ι_{*}^{C}$ is a group isomorphism from the open-disk fundamental group onto $B_{n}^{c o n f}$ .

1.2L3

Raw slicing is a bijection. By [L3], the class map $S : [β] \mapsto [S (β)]$ is well defined and has tracing as its inverse. It is therefore bijective, including the zero- and one-strand cases.

2.1L2L3L4L8step 1.2

Raw slicing reverses stacking. For $[γ], [β] \in G_{n}$ , [L2] defines their product by $[γ ⋆ β]$ , and [L4] gives $S ([γ] [β]) = [S (γ ⋆ β)] = [S (β)] [S (γ)] .$ This is the anti-homomorphism identity for the first-then-second loop product of [L8]. Since $S$ is bijective by step 1.2, it is an anti-isomorphism.

3.1L2L4L7L9L10step 2.1

The inverse-loop map is multiplicative. Write $a : = ι_{*}^{C} [S (γ)], b : = ι_{*}^{C} [S (β)] .$ By step 2.1 and the homomorphism property [L7], $Φ ([γ] [β]) = (b a)^{- 1} .$ In the group $B_{n}^{c o n f}$ , the element $a^{- 1} b^{- 1}$ is a two-sided inverse of $b a$ : associativity gives $(b a) (a^{- 1} b^{- 1}) = b (a a^{- 1}) b^{- 1} = 1$ and $(a^{- 1} b^{- 1}) (b a) = a^{- 1} (b^{- 1} b) a = 1$ . Uniqueness of inverses therefore gives $(b a)^{- 1} = a^{- 1} b^{- 1}$ . Hence $Φ ([γ] [β]) = (ι_{*}^{C} [S (γ)])^{- 1} (ι_{*}^{C} [S (β)])^{- 1} = Φ ([γ]) Φ ([β]),$ so $Φ$ is a group homomorphism by [L10].

4.1L3L5L9L11L12step 1.1step 1.2step 3.1

The map is bijective and hence an isomorphism. The formula for $Φ$ is the composite of the bijection $S$ from step 1.2, the isomorphism $ι_{*}^{C}$ from step 1.1, and inversion in $B_{n}^{c o n f}$ . Inversion is a bijection because applying it twice returns the original element. Thus $Φ$ is bijective by [L11]; together with step 3.1 it is a group isomorphism by [L12]. This also proves the stated anti-isomorphism property of raw slicing.

5.1

Boundary cases. When $n = 0$ , [L13] gives the unique empty braid and [L14] gives singleton configuration spaces and one based motion; by [L3] and [L5], the class sets and inclusion map are also singletons. When $n = 1$ , [L13]–[L14] give no collision condition; the same tracing/slicing bijection, based inclusion isomorphism, and product calculation above apply. These cases require no additional choice or basepoint path. [L3, L5, L9, L13, L14, step 1.1, step 1.2, step 2.1, step 3.1, step 4.1] $□$

Depends on

Used by

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Sources

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