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The induced permutation does not determine a braid

Statement refuted

In $B_{3}$ , the identity and the pure braid word $σ_{1}^{2}$ induce the same permutation of the punctures (the trivial one), but they act differently on $F_{3}$ and are distinct braids. Hence the endpoint permutation of a braid does not determine the braid. The argument uses no choice principle.

Facts & Assumptions

Given: the Artin braid group $B_{3}$ on generators $σ_{1}, σ_{2}$ , the free group $F_{3} = ⟨ x_{1}, x_{2}, x_{3} ⟩$ , the automorphisms $ρ (σ_{i})$ of Artin automorphisms of the free group, the homomorphism $ρ : B_{3} \to Aut (F_{3})$ of The Artin representation on a free group, and the permutation homomorphism $π_{3} : B_{3} \to S_{3}$ of The braid group surjects onto the symmetric group.

[F1]

The substitutions. $ρ (σ_{1}) (x_{1}) = x_{1} x_{2} x_{1}^{- 1}$ , $ρ (σ_{1}) (x_{2}) = x_{1}$ , $ρ (σ_{1}) (x_{3}) = x_{3}$ , and $ρ$ is a well-defined homomorphism, so $ρ (σ_{1}^{2}) = ρ (σ_{1}) \circ ρ (σ_{1})$ and $ρ (1) = id$ ; a homomorphism of $F_{3}$ is determined by its values on the basis, and two endomorphisms agree as soon as they agree on the basis. (Artin automorphisms of the free group, The Artin representation on a free group, The braid group by Artin presentation.)

[F2]

The endpoint permutation. $π_{3}$ is a homomorphism with $π_{3} (σ_{i}) = (i i + 1)$ for $i = 1, 2$ , and it assigns to each braid its endpoint permutation of the three strands (The braid group surjects onto the symmetric group).

[F3]

Reduced words. Reduced words in the free basis represent the same element of $F_{3}$ only if they are equal (Reduced words form the free group on an alphabet, Free group on a set of generators).

Counterexample

The two braids compared are the identity $1 \in B_{3}$ and the pure braid word $σ_{1}^{2} \in B_{3}$ ; they are shown to induce the same permutation and different automorphisms of $F_{3}$ .

1.1F1algebra

The value of $ρ (σ_{1}^{2})$ on $x_{1}$ . By [F1], $ρ (σ_{1}) (x_{1}) = x_{1} x_{2} x_{1}^{- 1}$ and $ρ (σ_{1}) (x_{2}) = x_{1}$ , so, using that $ρ (σ_{1})$ is an automorphism, $ρ (σ_{1}^{2}) (x_{1}) = ρ (σ_{1}) (x_{1} x_{2} x_{1}^{- 1}) = ρ (σ_{1}) (x_{1}) ρ (σ_{1}) (x_{2}) ρ (σ_{1}) (x_{1})^{- 1} = x_{1} x_{2} x_{1} x_{2}^{- 1} x_{1}^{- 1} .$

1.2F1F3algebra

The two automorphisms differ. The words $x_{1} x_{2} x_{1} x_{2}^{- 1} x_{1}^{- 1}$ and $x_{1}$ are both reduced; they differ, so by [F3] they represent different elements of $F_{3}$ . Hence $ρ (σ_{1}^{2}) (x_{1}) \neq x_{1} = ρ (1) (x_{1})$ , so $ρ (σ_{1}^{2}) \neq ρ (1)$ . Since $ρ$ is a well-defined function on $B_{3}$ with $ρ (1) = id$ , the word $σ_{1}^{2}$ does not represent the trivial braid: $σ_{1}^{2} \neq 1$ in $B_{3}$ .

1.3F2algebra

The two endpoint permutations agree. By [F2], $π_{3} (σ_{1}) = (1 2)$ , so $π_{3} (σ_{1}^{2}) = (1 2)^{2} = 1 = π_{3} (1)$ : the braid $σ_{1}^{2}$ and the identity braid induce the same trivial permutation of the three strands.

2.1step 1.2step 1.3∎

Conclusion. Steps 1.2 and 1.3 exhibit the distinct braids $1$ and $σ_{1}^{2}$ in $B_{3}$ with equal endpoint permutation; the word $σ_{1}^{2}$ is moreover pure. Therefore the endpoint permutation of a braid does not determine the braid. The computation used finitely many substitutions and the choice-free suppliers [F1]-[F3], so no choice principle is used.

Remarks

The faithfulness theorem thm-the-artin-representation-is-faithful is not needed: the difference is already visible at the level of the well-defined homomorphism $ρ$ , since $ρ (1) = id$ is known without injectivity.
The braid $σ_{1}^{2}$ generates the kernel of $π_{3}$ on two strands; the example is the first nontrivial instance of the fact that the pure braid group is strictly larger than the center.

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, printed pp. 8-10 (the action of sigma_i and the pure braid sigma_i^2) (standard reference, not scraped)
Emil Artin, Theory of Braids, Annals of Mathematics 48 (1947), pp. 101-126, equations (14)-(15), printed pp. 113-114 (standard reference, not scraped)