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Theorem 1.9 (Artin, 1925): which endomorphisms of FnF_nFn​ are braid automorphisms

Proved
BraidsLinksMCG.thm_1_9_artin_characterization

by Lucas · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

braid-groupsfree-groupsgroup-theory

Theorem 1.9 (Artin, 1925). Let Fn=⟨x1,…,xn⟩F_n = \langle x_1, \dots, x_n \rangleFn​=⟨x1​,…,xn​⟩ be free of rank nnn and let β\betaβ be an endomorphism of FnF_nFn​. Then β\betaβ belongs to Bn⊂Aut FnB_n \subset \mathrm{Aut}\,F_nBn​⊂AutFn​ — that is, β\betaβ is the automorphism induced by some braid under the Artin representation — if and only if

(xi)β=Ai xμi Ai−1(1≤i≤n),(x1x2⋯xn)β=x1x2⋯xn,(x_i)\beta = A_i\, x_{\mu_i}\, A_i^{-1} \quad (1 \le i \le n), \qquad (x_1x_2\cdots x_n)\beta = x_1x_2\cdots x_n,(xi​)β=Ai​xμi​​Ai−1​(1≤i≤n),(x1​x2​⋯xn​)β=x1​x2​⋯xn​,

where (μ1,…,μn)(\mu_1, \dots, \mu_n)(μ1​,…,μn​) is a permutation of (1,…,n)(1, \dots, n)(1,…,n) and each AiA_iAi​ is a word in the generators. In words: an endomorphism is a braid automorphism exactly when it permutes the generators up to conjugacy and fixes the product x1x2⋯xnx_1x_2\cdots x_nx1​x2​⋯xn​.

The representation ξ\xiξ is taken as a hypothesis: the statement is made for any homomorphism from BnB_nBn​ to Aut Fn\mathrm{Aut}\,F_nAutFn​ whose value on each σi\sigma_iσi​ is the endomorphism (1-14). Corollary 1.8.3 guarantees that at least one such ξ\xiξ exists, so the statement is not vacuous.

Preamble
import Mathlib
import Definitions.Def_BraidsLinksMCG_ArtinBraidGroup
import Definitions.Def_BraidsLinksMCG_ArtinEndo
Formal statement
namespace BraidsLinksMCG

theorem thm_1_9_artin_characterization (n : ℕ)
    (xi : ArtinBraidGroup n →* MulAut (FreeGroup (Fin n)))
    (hxi : ∀ i : Fin (n - 1), ∀ w : FreeGroup (Fin n), xi (sigma i) w = artinEndo n i w)
    (beta : FreeGroup (Fin n) →* FreeGroup (Fin n)) :
    (∃ b : ArtinBraidGroup n, ∀ w : FreeGroup (Fin n), xi b w = beta w) ↔
      ((∃ mu : Equiv.Perm (Fin n), ∃ A : Fin n → FreeGroup (Fin n),
            ∀ i : Fin n, beta (FreeGroup.of i) = A i * FreeGroup.of (mu i) * (A i)⁻¹) ∧
        beta (freeWordProd n) = freeWordProd n) := by sorry

end BraidsLinksMCG
Source
Joan S. Birman, *Braids, Links, and Mapping Class Groups*, Annals of Mathematics Studies 82, Princeton University Press, 1974, Chapter 1, p. 30, Theorem 1.9 [Artin, 1925], conditions (1-22) and (1-23)
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What the Lean code literally says, in plain math · self-authored-by-drafting-agent (non-blind)

Provenance: non-blind read-back. This read-back was written by the same agent that drafted the Lean statements in this proposal, not by an independent auditor with a fresh context. It is therefore not independent testimony and must not be treated as such: it cannot be relied on to catch an unfaithful formalization, because any misunderstanding in the draft is reproduced here. Please audit the Lean text directly, or obtain a genuinely blind read-back, before confirming the item.

Fix a natural number n≥0n \ge 0n≥0, let FnF_nFn​ be the free group on x0,…,xn−1x_0, \dots, x_{n-1}x0​,…,xn−1​, and let BnB_nBn​ be the presented group with generators σi\sigma_iσi​ and the braid relations.

Hypotheses. The statement is made for a given group homomorphism ξ:Bn→Aut(Fn)\xi : B_n \to \mathrm{Aut}(F_n)ξ:Bn​→Aut(Fn​) that satisfies, for every braid index iii and every w∈Fnw \in F_nw∈Fn​, ξ(σi)(w)=αi(w)\xi(\sigma_i)(w) = \alpha_i(w)ξ(σi​)(w)=αi​(w), where αi\alpha_iαi​ is Artin's endomorphism

xi↦xixi+1xi−1,xi+1↦xi,xj↦xj (j≠i,i+1);x_i \mapsto x_i x_{i+1} x_i^{-1}, \qquad x_{i+1} \mapsto x_i, \qquad x_j \mapsto x_j \ (j \neq i, i+1);xi​↦xi​xi+1​xi−1​,xi+1​↦xi​,xj​↦xj​ (j=i,i+1);

and for a given group endomorphism β:Fn→Fn\beta : F_n \to F_nβ:Fn​→Fn​ (a homomorphism, not assumed bijective).

Conclusion. The following two statements are equivalent.

Left side. There exists b∈Bnb \in B_nb∈Bn​ such that ξ(b)\xi(b)ξ(b) and β\betaβ agree at every element of FnF_nFn​.

Right side. Both of the following hold:

  1. there exist a permutation μ\muμ of {0,…,n−1}\{0, \dots, n-1\}{0,…,n−1} and a family of words A0,…,An−1∈FnA_0, \dots, A_{n-1} \in F_nA0​,…,An−1​∈Fn​ such that for every iii,
β(xi)  =  Ai xμ(i) Ai−1;\beta(x_i) \;=\; A_i\, x_{\mu(i)}\, A_i^{-1};β(xi​)=Ai​xμ(i)​Ai−1​;
  1. β\betaβ fixes the ordered product of all generators:
β(x0x1⋯xn−1)  =  x0x1⋯xn−1.\beta\bigl(x_0 x_1 \cdots x_{n-1}\bigr) \;=\; x_0 x_1 \cdots x_{n-1}.β(x0​x1​⋯xn−1​)=x0​x1​⋯xn−1​.

Exact logical strength. The equivalence is stated for the given ξ\xiξ, which is a hypothesis of the statement rather than a constructed object: if no homomorphism with the displayed values on the generators existed, the statement would be vacuously true. The conjugators AiA_iAi​ in (1) are arbitrary elements of FnF_nFn​ and may depend on iii; the permutation μ\muμ is arbitrary. Membership in the image of ξ\xiξ is expressed as pointwise agreement of functions on FnF_nFn​, not as equality of automorphisms in some quotient. No uniqueness of bbb, μ\muμ or the AiA_iAi​ is asserted.

Degenerate cases. n≤1n \le 1n≤1 is included: then BnB_nBn​ is trivial and the equivalence reduces to a statement about the identity endomorphism of a free group of rank at most one.

Human review
  • Endorsed by Shuze Chen · Sep 13, 2026

  • Endorsed by Lucas · Sep 13, 2026

    Confirmed by the mission captain (proposal self-audit).

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