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The Artin representation is faithful

Statement

Assume AC. For every $n \geq 1$ the Artin representation $ρ : B_{n} \to Aut (F_{n})$ of The Artin representation on a free group is injective; equivalently, a braid word acts trivially on $F_{n}$ only if it represents the trivial braid.

Facts & Assumptions

Given: AC, the number $n \geq 1$ , the abstract braid group $B_{n}$ on $σ_{1}, \dots, σ_{n - 1}$ , the geometric braid group $G_{n}$ with the surjection $φ_{n} : B_{n} \to G_{n}$ and the isomorphism $Ψ : G_{n} ⟶ Mod (D^{2}, Q_{n}; \partial D^{2})$ of The Artin presentation surjects onto the geometric braid group and Braid group as boundary-fixed punctured-disk mapping classes, and the identification of $F_{n}$ with $π_{1} (D^{2} ∖ Q_{n}, d)$ by the standard meridians (The Artin representation on a free group, The braid group by Artin presentation).

[F1]

The geometric action is the Artin representation. Assume AC. For every braid word $β$ , the automorphism of $F_{n}$ induced by the mapping class $Ψ (φ_{n} (β))$ equals $ρ (β)$ ; in particular, if $ρ (β) = id$ then the homeomorphism representing $Ψ (φ_{n} (β))$ acts as the identity on $π_{1} (D^{2} ∖ Q_{n}, d)$ . (The geometric action on meridians is the Artin representation.)

[F2]

Trivial action implies isotopy to the identity. Assume AC. Let $h \in {Homeo}^{+} (D^{2}, \partial D^{2})$ preserve $Q_{n}$ setwise and act as the identity on $π_{1} (D^{2} ∖ Q_{n}, d)$ ; then $h$ is isotopic to ${i d}_{D^{2}}$ relative to $\partial D^{2}$ and $Q_{n}$ . Hence such an $h$ represents the identity element of $Mod (D^{2}, Q_{n}; \partial D^{2})$ . (A boundary-fixed punctured-disk homeomorphism acting trivially on the fundamental group is isotopic to the identity.)

[F3]

Completeness of the presentation. Assume AC. The published surjection $φ_{n}$ is injective, hence an isomorphism; that is, a braid word whose geometric braid is trivial represents the trivial element of $B_{n}$ . (The Artin presentation is complete for geometric braids.)

[F4]

The isomorphism $Ψ$ . $Ψ$ is a group isomorphism, so it is injective: if $Ψ (φ_{n} (β))$ is the identity mapping class, then $φ_{n} (β) = 1$ in $G_{n}$ . (Braid group as boundary-fixed punctured-disk mapping classes, AC used through the published isomorphism.)

Proof

technique · direct

1.1given

The case $n = 1$ . For $n = 1$ there is no generator, $B_{1}$ is the trivial group by The braid group by Artin presentation, and the unique map $ρ : B_{1} \to Aut (F_{1})$ is injective; the assertion holds vacuously.

1.2F1given

Assume a word acts trivially. Let $n \geq 2$ and let $β$ be a braid word with $ρ (β) = id$ . By [F1] the mapping class $Ψ (φ_{n} (β))$ induces the identity automorphism of $π_{1} (D^{2} ∖ Q_{n}, d)$ . Choose a homeomorphism $h$ representing this mapping class (for instance the homeomorphism attached to $β$ by the geometric construction underlying $φ_{n}$ ); then $h$ fixes $\partial D^{2}$ pointwise, preserves $Q_{n}$ setwise and acts as the identity on $π_{1} (D^{2} ∖ Q_{n}, d)$ .

2.1F2step 1.2

Trivial action forces the identity mapping class. By [F2], applied under the present assumption AC, the homeomorphism $h$ of step 1.2 is isotopic to the identity relative to $\partial D^{2}$ and $Q_{n}$ ; hence $Ψ (φ_{n} (β)) = 1$ in $Mod (D^{2}, Q_{n}; \partial D^{2})$ .

3.1F3F4step 2.1

Injectivity of the presentation. By [F4] the isomorphism $Ψ$ is injective, so step 2.1 gives $φ_{n} (β) = 1$ in $G_{n}$ . By [F3], $φ_{n}$ is injective, so the braid word $β$ represents the trivial element of $B_{n}$ .

4.1F1F2F3step 1.1step 3.1∎

Conclusion. Steps 1.1, 1.2, 2.1 and 3.1 show that every braid word acting trivially on $F_{n}$ represents the trivial braid, which is exactly the injectivity of $ρ$ . The converse (the trivial braid acts trivially) is immediate from $ρ$ being a homomorphism, so injectivity holds for every $n \geq 1$ . AC is used exactly through the three published or previously proved inputs [F1], [F2] and [F3], namely the geometric-action proposition, the isotopy-to-identity lemma and the completeness theorem, all of which assume AC; the final argument itself is elementary.

Remarks

The proof replaces Artin's original topological faithfulness argument by the route through the mapping class group: the geometric action identifies $ρ$ with the action of $Mod (D^{2}, Q_{n}; \partial D^{2})$ on $π_{1}$ , and a boundary-fixed homeomorphism acting trivially on $π_{1}$ is isotopic to the identity by induction on the number of punctures and the point-pushing kernel theorem. The last point-motion loop is detected by a compact tether square after filling that puncture; no Markov theorem or general arc-tameness theorem is used.
Consequently $ρ$ is an isomorphism onto its image, and $B_{n}$ is isomorphic to the Artin braid subgroup of $Aut (F_{n})$ characterized in thm-artins-characterization-of-the-braid-subgroup-of-aut-f-n.

Depends on

Used by

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, printed pp. 8-9 (Artin showed that rho is faithful by topological arguments) (standard reference, not scraped)
Emil Artin, Theory of Braids, Annals of Mathematics 48 (1947), pp. 101-126, Theorem 15 and the faithfulness statement, printed pp. 112-115 (standard reference, not scraped)