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The two-strand pure braid group is infinite cyclic

Example

Assume the Axiom of Choice and let $Q^{(2)} = (q_{1}, q_{2})$ be the canonical base configuration of Geometric braids in the disc with setwise endpoints, so that $P B_{2} = π_{1} (F_{2} (D^{2}), Q^{(2)})$ in the convention of The pure braid group $P B_{n}$ as the fundamental group of an ordered configuration space. Then $P B_{2} ≅ Z,$ generated by the geometrically positive two-strand full twist $A_{12} = σ_{1}^{2}$ of Standard geometric pure braid generators A_ij. Under the inverse-slicing identification of The $A_{i n}$ are meridian generators of the forgetful free kernel, the generator $A_{12}$ corresponds to the clockwise meridian of $q_{1}$ inside the once-punctured disc fibre, that is to the inverse of the counterclockwise meridian class; relative to the counterclockwise spine basis its winding is $- 1$ . The group $P B_{2}$ is infinite cyclic and, in particular, torsion-free.

Facts & Assumptions

Given: the Axiom of Choice, the canonical base configuration $Q^{(2)} = (q_{1}, q_{2})$ of interior points of the disc, and the truncation notation of The Fadell-Neuwirth short exact sequence for pure braids.

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

In ZF, AC implies DC (AC implies DC implies countable choice).

[F2]

Assume AC and let $n \geq 2$ . With the base configuration $Q^{(n)}$ and its truncation $q^{'}$ , the forgetting map $φ : P B_{n} \to P B_{n - 1}$ sits in the short exact sequence $1 \to F_{n - 1} \overset{κ}{\to} P B_{n} \overset{φ}{\to} P B_{n - 1} \to 1$ with $κ$ injective, $φ$ surjective and $im κ = \ker φ$ , and $P B_{1}$ is trivial; here $F_{n - 1} = π_{1} (int D^{2} ∖ {q_{1}, \dots, q_{n - 1}}, q_{n})$ is free on the $n - 1$ positively oriented meridian classes of the punctures $q_{1}, \dots, q_{n - 1}$ (The Fadell-Neuwirth short exact sequence for pure braids).

[F3]

Assume AC and let $n \geq 2$ . Under the identification $F_{n - 1} ≅ \ker φ$ given by $κ$ , the elements $A_{1 n}, \dots, A_{n - 1, n}$ of Standard geometric pure braid generators A_ij are a free basis of $\ker φ$ , and the $i$ -th of them corresponds to the clockwise meridian of the $i$ -th puncture, the inverse of the counterclockwise spine-basis class (The $A_{i n}$ are meridian generators of the forgetful free kernel).

[F4]

The free group on $X = {γ_{1}}$ has the reduced-word model, in which each element has a unique reduced word in $γ_{1}^{\pm 1}$ (Reduced words form the free group on an alphabet). Since there is only one generator, a reduced word contains only $γ_{1}$ or only $γ_{1}^{- 1}$ , so every element is uniquely $γ_{1}^{k}$ for some $k \in Z$ .

[F5]

Assume AC. For every $n \geq 0$ the pure braid group $P B_{n}$ is torsion-free (Pure braid groups are torsion-free).

Verification

technique · direct

1.1A1F1F2

The choice deduction and the exact sequence at $n = 2$ . By [F1] the Axiom of Choice [A1] yields DC, so the sequence of [F2] is available at $n = 2$ : $1 \to F_{1} \overset{κ}{\to} P B_{2} \overset{φ}{\to} P B_{1} \to 1$ with $κ$ injective, $im κ = \ker φ$ , and $F_{1} = π_{1} (int D^{2} ∖ {q_{1}}, q_{2})$ . Since $P B_{1}$ is trivial by [F2], $φ$ has trivial codomain, so $\ker φ = P B_{2}$ and $im κ = P B_{2}$ : the map $κ$ is a group isomorphism $F_{1} \to P B_{2}$ .

1.2F2F3

The fibre and its meridian basis. By [F2] the fibre group $F_{1} = π_{1} (int D^{2} ∖ {q_{1}}, q_{2})$ is free on the single positively oriented meridian class of $q_{1}$ , written $γ_{1}$ , and by the case $n = 2$ of [F3] the corresponding generator $A_{12}$ corresponds under $κ$ to the inverse class $γ_{1}^{- 1}$ , the clockwise meridian; that is, $κ (γ_{1}^{- 1}) = A_{12}$ and $κ (γ_{1}) = A_{12}^{- 1}$ .

2.1step 1.1step 1.2F2F4

$P B_{2}$ is infinite cyclic. By [F2] and [F4], every element of the free group $F_{1}$ on ${γ_{1}}$ is uniquely $γ_{1}^{k}$ for $k \in Z$ . Concatenation followed by free reduction adds exponents, so $ν : F_{1} \to (Z, +)$ , $ν (γ_{1}^{k}) = k$ , is a group isomorphism. Composing the negation automorphism $k \mapsto - k$ of $Z$ with $ν^{- 1}$ and then with $κ$ of step 1.1 gives the isomorphism $Z \to P B_{2}$ , $k \mapsto κ (γ_{1}^{- k}) = A_{12}^{k}$ , by step 1.2. Thus $P B_{2}$ is infinite cyclic, generated by $A_{12}$ , and its inverse sends $A_{12}$ to $1$ , as claimed.

3.1step 1.2step 2.1F5

The winding sign and torsion-freeness. By step 1.2 the generator $A_{12}$ corresponds under $κ$ to $γ_{1}^{- 1}$ , the clockwise meridian of $q_{1}$ , while the counterclockwise spine-basis class is $γ_{1}$ itself; the identification of step 1.2 therefore assigns to $A_{12}$ the winding $- 1$ relative to the counterclockwise basis, as claimed. The isomorphism in step 2.1 shows that $P B_{2}$ is infinite cyclic and generated by $A_{12}$ ; it is torsion-free by the general theorem [F5].

∎

Depends on

Used by

PB₃ as F₂ by Z, with its section action Example

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Joan S. Birman and Tara E. Brendle, Braids: A Survey, section 1.2, author manuscript p. 5 (the pure braid group P_2 is infinite cyclic and generated by A_{1,2}) (standard reference, not scraped)