LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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The Artin automorphisms satisfy the braid relations

Statement

For the automorphisms of Artin automorphisms of the free group: if $∣ i - j ∣ > 1$ then $ρ (σ_{i}) ρ (σ_{j}) = ρ (σ_{j}) ρ (σ_{i})$ ; if $∣ i - j ∣ = 1$ then $ρ (σ_{i}) ρ (σ_{j}) ρ (σ_{i}) = ρ (σ_{j}) ρ (σ_{i}) ρ (σ_{j})$ .

Facts & Assumptions

Given: $n \in N$ , the free group $F_{n} = ⟨ x_{1}, \dots, x_{n} ⟩$ , and the automorphisms $ρ (σ_{i})$ , $1 \leq i \leq n - 1$ , of Artin automorphisms of the free group, with $ρ (σ_{i}) (x_{i}) = x_{i} x_{i + 1} x_{i}^{- 1}, ρ (σ_{i}) (x_{i + 1}) = x_{i}, ρ (σ_{i}) (x_{j}) = x_{j} (j \notin {i, i + 1}),$ $ρ (σ_{i})^{- 1} (x_{i}) = x_{i + 1}, ρ (σ_{i})^{- 1} (x_{i + 1}) = x_{i + 1}^{- 1} x_{i} x_{i + 1} .$

[F1]

The elements $x_{1}, \dots, x_{n}$ form a free basis of $F_{n}$ ; two endomorphisms agree if they agree on a free basis, and equality of elements is decided by equality of reduced words (Free group on a set of generators, Reduced words form the free group on an alphabet).

Proof

technique · direct computation on the basis

1.1F1given

Far commutation. Let $∣ i - j ∣ > 1$ . The two substitutions involve disjoint pairs of letters. For $k \notin {i, i + 1, j, j + 1}$ both composites fix $x_{k}$ ; for $k \in {i, i + 1}$ both send $x_{k}$ to $ρ (σ_{i}) (x_{k})$ , because $ρ (σ_{j})$ fixes every letter of that word, and similarly for $k \in {j, j + 1}$ . Thus the composites agree on every basis letter and are equal by [F1].

1.2given

Adjacent case, the composite $ρ (σ_{i}) ρ (σ_{i + 1}) ρ (σ_{i})$ . Let $∣ i - j ∣ = 1$ ; after swapping the names of $i, j$ if necessary this is the triple $(x_{i}, x_{i + 1}, x_{i + 2})$ , and the composite fixes every other basis letter. Composing the displayed substitutions (the rightmost letter acts first) gives $ρ (σ_{i}) ρ (σ_{i + 1}) ρ (σ_{i}) (x_{i}) = x_{i} x_{i + 1} x_{i + 2} x_{i + 1}^{- 1} x_{i}^{- 1},$ $ρ (σ_{i}) ρ (σ_{i + 1}) ρ (σ_{i}) (x_{i + 1}) = x_{i} x_{i + 1} x_{i}^{- 1},$ $ρ (σ_{i}) ρ (σ_{i + 1}) ρ (σ_{i}) (x_{i + 2}) = x_{i} .$

2.1givenstep 1.2

The other composite has the same values. Apply $ρ (σ_{i + 1})$ , then $ρ (σ_{i})$ , then $ρ (σ_{i + 1})$ . The successive images of $x_{i}$ are $x_{i}$ , $x_{i} x_{i + 1} x_{i}^{- 1}$ , and $x_{i} x_{i + 1} x_{i + 2} x_{i + 1}^{- 1} x_{i}^{- 1}$ . Those of $x_{i + 1}$ are $x_{i + 1} x_{i + 2} x_{i + 1}^{- 1}$ , $x_{i} x_{i + 2} x_{i}^{- 1}$ , and $x_{i} x_{i + 1} x_{i}^{- 1}$ ; those of $x_{i + 2}$ are $x_{i + 1}$ , $x_{i}$ , and $x_{i}$ . Every other basis letter is fixed. These are the values of step 1.2.

3.1F1step 1.2step 2.1

Comparison. A direct reduction using the formulas confirms the identity of the two triples of reduced words of steps 1.2 and 2.1: both composite automorphisms send $x_{i} \mapsto x_{i} x_{i + 1} x_{i + 2} x_{i + 1}^{- 1} x_{i}^{- 1}, x_{i + 1} \mapsto x_{i} x_{i + 1} x_{i}^{- 1}, x_{i + 2} \mapsto x_{i} .$

4.1F1step 1.1step 3.1∎

Conclusion. Steps 1.1 and 3.1 show that the two composites agree on every basis element in the far and the adjacent case respectively; by [F1] they are equal as automorphisms, which is the asserted braid relations. The computation used the displayed formulas only and made no case distinction beyond the two stated.

Remarks

No choice principle is used; the verificaton is finite and effective.

Depends on

Used by

The Artin representation on a free group Definition

Dependency tree · two levels

9 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

Emil Artin, Theory of Braids, Annals of Mathematics 48 (1947), pp. 101-126, equations (14)-(15) and relations (18)-(19), printed pp. 113-115 (standard reference, not scraped)
Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, printed pp. 9-10 (the braid-relation check) (standard reference, not scraped)