PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27

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The Artin presentation surjects onto the geometric braid group

Statement

Let $n \in N$ . Write $G_{n}$ for the geometric braid group of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, let $σ_{1}, \dots, σ_{n - 1}$ be the geometric half twists of The elementary geometric half twist, its support disc, and its opposite, with classes $[σ_{i}] \in G_{n}$ , and let $B_{n} = ⟨ x_{1}, \dots, x_{n - 1} ∣ R ⟩$ be the Artin braid group of The braid group by Artin presentation, whose generator written there as $σ_{i}$ is here written $x_{i}$ to keep it distinct from the geometric half twist. Then the assignment

$φ (x_{i}) : = [σ_{i}] (1 \leq i \leq n - 1)$

extends to a homomorphism $φ ⁣ : B_{n} \to G_{n}$ , it does so uniquely, and $φ$ is surjective. Consequently every element of $G_{n}$ is a finite product of the classes $[σ_{i}]^{\pm 1}$ of the half twists, and the composition of $φ$ with the endpoint permutation homomorphism $π ⁣ : G_{n} \to S_{n}$ is the permutation map $x_{i} \mapsto (i i + 1)$ of the presented group.

Only surjectivity is asserted. Nothing here shows that $φ$ is injective, that is, that the Artin relations are a complete set of relations for the geometric braid group; the presentation is shown to surject onto $G_{n}$ only. For $n \leq 1$ the presentation has no generator and $B_{n}$ and $G_{n}$ are both trivial, so the assertions are vacuous.

Facts & Assumptions

Given: A natural number $n$ , the geometric braid group $G_{n}$ of The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, the Artin presentation $B_{n} = ⟨ X ∣ R ⟩$ of The braid group by Artin presentation with $X = {σ_{1}, \dots, σ_{n - 1}}$ for $n \geq 2$ and $X = \emptyset$ for $n \leq 1$ , and the elementary half twists of The elementary geometric half twist, its support disc, and its opposite.

[F1]

For $n \geq 2$ the group $B_{n}$ is the quotient of the free group on $X = {σ_{1}, \dots, σ_{n - 1}}$ by the normal closure of the relations $σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1}$ ( $1 \leq i \leq n - 2$ ) and $σ_{i} σ_{j} = σ_{j} σ_{i}$ ( $∣ i - j ∣ > 1$ ), interpreted in the sense of Group presentation by generators and relations and Relators and relations; finitely generated, finitely related, and finite presentations; for $n = 0$ and $n = 1$ there are no generators and $B_{n}$ is the trivial group of the empty presentation (The braid group by Artin presentation, Group presentation by generators and relations, Free group on a set of generators, Group and abelian group).

[F2]

In the presentation $⟨ X ∣ R ⟩$ of [F1] an equation $u = v$ is recorded by the relator $u^{- 1} v$ in the sense of the free group on $X$ (Relators and relations; finitely generated, finitely related, and finite presentations, Free group on a set of generators, Group presentation by generators and relations).

[F3]

Let $⟨ X ∣ R ⟩$ be a presentation, $H$ a group, and $u ⁣ : X \to H$ a function. If the evaluation of every $r \in R$ under $u$ is $e_{H}$ , then there is a unique homomorphism $\overline{u} ⁣ : ⟨ X ∣ R ⟩ \to H$ with $\overline{u} ([x]) = u (x)$ for every $x \in X$ ; moreover $\overline{u}$ is surjective if and only if $u (X)$ generates $H$ (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

[F4]

The half twist $σ_{i}$ and its opposite $σ_{i}^{-}$ are braids based at $Q$ with classes $[σ_{i}], [σ_{i}]^{- 1} \in G_{n}$ , and their endpoint permutations are the transposition of $i$ and $i + 1$ ; the endpoint permutation is a homomorphism $π ⁣ : G_{n} \to S_{n}$ (The elementary geometric half twist, its support disc, and its opposite, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism).

[F5]

For $∣ i - j ∣ > 1$ the half twists satisfy $[σ_{i}] [σ_{j}] = [σ_{j}] [σ_{i}]$ in $G_{n}$ , and for $n \leq 3$ there are no such pairs of indices, so the assertion is vacuous (Far commutativity of elementary geometric half twists).

[F6]

For $1 \leq i \leq n - 2$ the half twists satisfy $[σ_{i}] [σ_{i + 1}] [σ_{i}] = [σ_{i + 1}] [σ_{i}] [σ_{i + 1}]$ in $G_{n}$ , and for $n \leq 2$ there is no such index, so the assertion is vacuous (The geometric three strand braid relation).

[F7]

Every braid class in $G_{n}$ is a finite product of the classes $[σ_{1}], \dots, [σ_{n - 1}]$ and their inverses; equivalently, the set ${[σ_{1}], \dots, [σ_{n - 1}]}$ generates $G_{n}$ in the sense of The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, and for $n \leq 1$ the empty family generates the trivial subgroup ${e}$ (Every geometric braid is isotopic to a stacking of signed elementary half twists, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

Proof

technique · direct

1.1

The relators of the Artin presentation evaluate to the identity under the half-twist assignment. Assume $n \geq 2$ , let $X = {σ_{1}, \dots, σ_{n - 1}}$ and let $u ⁣ : X \to G_{n}$ be the assignment $u (σ_{i}) : = [σ_{i}]$ ; the relators of [F1] are, by [F2], the words $(σ_{i} σ_{i + 1} σ_{i})^{- 1} (σ_{i + 1} σ_{i} σ_{i + 1})$ for $1 \leq i \leq n - 2$ and $(σ_{i} σ_{j})^{- 1} (σ_{j} σ_{i})$ for $∣ i - j ∣ > 1$ . Their evaluations are $([σ_{i}] [σ_{i + 1}] [σ_{i}])^{- 1} [σ_{i + 1}] [σ_{i}] [σ_{i + 1}] = e$ by [F6] and $([σ_{i}] [σ_{j}])^{- 1} [σ_{j}] [σ_{i}] = e$ by [F5].

F1F2F5F6

1.2

The cases $n \leq 1$ . For $n \leq 1$ the presentation has no generators and defines the trivial group $B_{n}$ by [F1], while $⟨ \emptyset ⟩ = {e}$ is the trivial subgroup of $G_{n}$ and [F7] says that this empty family generates $G_{n}$ , so $G_{n}$ is trivial as well; the unique map $B_{n} \to G_{n}$ is therefore a group homomorphism, it is the only homomorphism between these groups, and it is surjective because its codomain is trivial.

F1F3F7

2.1

Von Dyck extends the assignment to a homomorphism. By step 1.1 the hypothesis of [F3] is satisfied, so there is a unique homomorphism $φ ⁣ : B_{n} \to G_{n}$ with $φ (x_{i}) = [σ_{i}]$ for every generator, and $φ$ is surjective if and only if the set of these images generates $G_{n}$ .

F1F3step 1.1

3.1

Surjectivity. The images $φ ({σ_{1}, \dots, σ_{n - 1}}) = {[σ_{1}], \dots, [σ_{n - 1}]}$ generate $G_{n}$ by [F7], so the surjectivity criterion of [F3] applies to the homomorphism of step 2.1 and $φ$ is surjective; consequently every element of $G_{n}$ is a finite product of the classes $[σ_{i}]^{\pm 1}$ , and composing the unique homomorphism with the endpoint permutation homomorphism of [F4] gives the permutation map $x_{i} \mapsto (i i + 1)$ , because $π ([σ_{i}])$ is that transposition.

F3F4F7step 2.1

4.1

Conclusion. Steps 2.1 and 3.1 give, for $n \geq 2$ , a unique homomorphism $φ ⁣ : B_{n} \to G_{n}$ with $φ (x_{i}) = [σ_{i}]$ , and step 1.2 gives the same for $n \leq 1$ ; in both cases $φ$ is surjective, and no injectivity is claimed. ∎

F1F3step 2.1step 3.1step 1.2

Remarks

The proposition is exactly the surjectivity half of the classical statement that the Artin presentation presents the geometric braid group. Its content is that the geometric relations of Far commutativity of elementary geometric half twists and The geometric three strand braid relation satisfy the defining relators, and that the generic-crossing decomposition of Every geometric braid is isotopic to a stacking of signed elementary half twists reaches every braid class.
Injectivity requires the opposite direction: it needs an argument that no further relations hold between the half twists, which is not available on this page and is not assumed anywhere below it.

Depends on

Used by

Geometric two strand braids are integer twists Example

Dependency tree · two levels

43 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, sections 1.5 and 3.2, printed pp. 7-8 and 23-26 (standard reference, not scraped)
Joan S. Birman and Tara E. Brendle, Braids: A Survey, sections 1.2-1.3, author manuscript pp. 5-7 (standard reference, not scraped)