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Pure braid groups are torsion-free

Statement

Assume the Axiom of Choice. For every $n \geq 0$ the pure braid group $P B_{n} = π_{1} (F_{n} (D^{2}), q)$ of The pure braid group $P B_{n}$ as the fundamental group of an ordered configuration space is torsion-free: if $g \in P B_{n}$ and $m \geq 1$ satisfy $g^{m} = e$ , then $g = e$ . Equivalently, every nonidentity element of $P B_{n}$ has infinite order.

Facts & Assumptions

Given: the Axiom of Choice and an integer $n \geq 0$ ; the pure braid groups $P B_{n}$ of the closed-disc convention, with the same base configuration $q$ fixed throughout.

[A1]

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

[F1]

$P B_{0}$ and $P B_{1}$ are the one-element groups, and the inclusion $F_{m} (int D^{2}) \to F_{m} (D^{2})$ induces an isomorphism $π_{1} (F_{m} (int D^{2}), q) \to P B_{m}$ at every configuration of interior points (The pure braid group $P B_{n}$ as the fundamental group of an ordered configuration space).

[F2]

Assume AC and let $n \geq 2$ . The last-coordinate forgetful map $φ : P B_{n} \to P B_{n - 1}$ fits into a 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 $im κ = \ker φ$ , where $F_{n - 1} = π_{1} (int D^{2} ∖ {q_{1}, \dots, q_{n - 1}}, q_{n})$ is a free group (The Fadell-Neuwirth short exact sequence for pure braids).

[F3]

Every free group is torsion-free: if $G$ is free, $x \in G$ and $m \geq 1$ satisfy $x^{m} = e$ , then $x = e$ (Free groups are torsion-free).

[F4]

In ZF, AC implies DC, so the choice hypothesis of [F2] is available under [A1] (AC implies DC implies countable choice).

Proof

technique · induction on $n$

1.1baseF1

The base cases. By [F1] the groups $P B_{0}$ and $P B_{1}$ are one-element groups, so their only element is the identity and is not a nonidentity element of finite order; hence $P B_{0}$ and $P B_{1}$ are torsion-free.

1.2F3

The torsion-freeness of the free fibre. By [F3] every free group is torsion-free; in particular this applies to the group $F_{n - 1}$ of the exact sequence in [F2], so that if $x \in F_{n - 1}$ and $x^{m}$ is the identity of $F_{n - 1}$ for some $m \geq 1$ , then $x$ is the identity.

1.3A1F2F4

The exact sequence used in the successor step. Let $n \geq 2$ . The Axiom of Choice [A1] holds, so by [F4] the choice hypothesis of [F2] is met and [F2] supplies the short exact sequence $1 \to F_{n - 1} \overset{κ}{\to} P B_{n} \overset{φ}{\to} P B_{n - 1} \to 1$ for the last-coordinate forgetful map $φ$ : the map $φ$ is a surjective homomorphism, $κ$ is an injective homomorphism, and $im κ = \ker φ$ . The Axiom of Choice is used only to invoke [F2]; no further choice is made in this proof.

1.4ih

Induction hypothesis. Fix $n \geq 2$ and assume that $P B_{n - 1}$ is torsion-free: every $y \in P B_{n - 1}$ with $y^{m} = e$ for some $m \geq 1$ equals $e$ .

2.1step 1.3step 1.4step 1.2

The successor step. Let $g \in P B_{n}$ and $m \geq 1$ satisfy $g^{m} = e$ , where $e$ denotes the identity of $P B_{n}$ . Since $φ$ is a homomorphism, $φ (g)^{m} = φ (g^{m}) = φ (e) = e$ in $P B_{n - 1}$ , so $φ (g) = e$ by the induction hypothesis of step 1.4. Hence $g \in \ker φ = im κ$ , and there is $x \in F_{n - 1}$ with $g = κ (x)$ . Then $κ (x^{m}) = κ (x)^{m} = g^{m} = e$ ; since $κ$ is injective, $x^{m}$ is the identity of $F_{n - 1}$ , so $x$ is the identity by step 1.2, and therefore $g = κ (x) = e$ . Thus every element of $P B_{n}$ of finite order is the identity, that is, $P B_{n}$ is torsion-free.

3.1step 1.1step 2.1discharge-induction

Induction conclusion. Step 1.1 establishes the statement for $n = 0$ and $n = 1$ , and step 2.1 proves the successor implication for every $n \geq 2$ ; by induction on $n$ , $P B_{n}$ is torsion-free for every $n \geq 0$ .

The argument applies only to the pure braid groups: torsion-freeness of the kernel $P B_{n}$ and finiteness of the quotient $S_{n}$ of the full braid group $B_{n}$ are not used to make any claim about torsion in $B_{n}$ itself. ∎

Depends on

Used by

The short exact sequence to Sₙ does not prove Bₙ torsion-free Counterexample
The two-strand pure braid group is infinite cyclic Example

Dependency tree · two levels

28 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

Juan Gonzalez-Meneses, Basic results on braid groups, section 2.1, printed pp. 11-14 (equation (2.2), the free-kernel pure braid tower, and Corollary 2.3) (standard reference, not scraped)