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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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An extremal cancellation shortens an Artin substitution

Statement

In case 2 of Artin's product-cancellation dichotomy, use its chosen adjacent pair and notation a,b,U,V,R. Put s=ρ(σi), with the convention of Artin automorphisms of the free group. If the left middle letter is cancelled first, set A′=A∘s; if the right middle letter is cancelled first, set A′=A∘s−1. In both cases A′ is peripheral-boundary-preserving and has a representation whose total reduced conjugator length is strictly less than that of A.

More precisely, its two new conjugators can be taken as red⁡(RU),U in the left case, and V,red⁡(RV) in the right case; all other conjugators are unchanged. A representation of minimum total length therefore satisfies ℓ(A′)<ℓ(A).

Facts & Assumptions

Given: The reduced expressions, the adjacent pair selected in the cancellation dichotomy, and the frozen Artin substitution s.

[F1]

The dichotomy gives V=RaU in the left case and U=Rb−1V in the right case, as reduced concatenations (Artin's product-cancellation dichotomy).

[F2]

The automorphism s sends xi to xixi+1xi−1 and xi+1 to xi; its inverse sends them to xi+1 and xi+1−1xixi+1 respectively, fixing all other basis letters (Artin automorphisms of the free group).

[F3]

Peripheral-boundary-preserving means permutation of positive basis conjugacy classes and exact preservation of δ=x1⋯xn (Peripheral-boundary-preserving automorphisms of F_n).

Proof

1.1F1F2algebra

Left middle letter. Here V=RaU. Postcomposition gives A′(xi)=TiTi+1Ti−1 and A′(xi+1)=Ti, where Ti=U−1aU and Ti+1=V−1bV. The first image has conjugator VTi−1=RaU U−1a−1U=RU around b, so A′(xi)=(RU)−1b(RU) and A′(xi+1)=U−1aU. Reducing RU gives conjugators of total length at most ∣R∣+2∣U∣, whereas the old pair has length ∣U∣+∣V∣=∣R∣+1+2∣U∣. The length falls by at least one.

1.2F1F2algebra

Right middle letter. Here U=Rb−1V. Postcomposition with the inverse gives A′(xi)=Ti+1 and A′(xi+1)=Ti+1−1TiTi+1. The second image has conjugator UTi+1=Rb−1V V−1bV=RV around a. Thus the new conjugators are V,red⁡(RV), of total length at most ∣R∣+2∣V∣, while ∣U∣+∣V∣=∣R∣+1+2∣V∣. Again the length falls by at least one; all other images are unchanged in either case.

2.1F2F3step 1.1step 1.2algebra∎

Admissibility and recovery. The displayed images swap the two middle generators and retain conjugates of every other generator, so they still permute the positive peripheral classes. Both s and s−1 fix δ, since s(xixi+1)=(xixi+1xi−1)xi=xixi+1 and the inverse fixes the same word. Consequently A′ fixes δ and is peripheral-boundary-preserving. In the left case A=A′∘s−1; in the right case A=A′∘s. The strict pair inequalities of steps 1.1 and 1.2 prove the total decrease; when the original representation is minimal they give ℓ(A′)≤ℓ(A)−1. No choice principle is used.

Remarks

These are Artin's two length reductions in the proof of Theorem 16, printed pp. 114–115, translated to the authored generator convention. Both operations change source basis letters by postcomposition; they do not apply a substitution to every target word by precomposition.

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