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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 . Put , with the convention of Artin automorphisms of the free group. If the left middle letter is cancelled first, set ; if the right middle letter is cancelled first, set . In both cases is peripheral-boundary-preserving and has a representation whose total reduced conjugator length is strictly less than that of .
More precisely, its two new conjugators can be taken as in the left case, and in the right case; all other conjugators are unchanged. A representation of minimum total length therefore satisfies .
Facts & Assumptions
Given: The reduced expressions, the adjacent pair selected in the cancellation dichotomy, and the frozen Artin substitution .
The dichotomy gives in the left case and in the right case, as reduced concatenations (Artin's product-cancellation dichotomy).
The automorphism sends to and to ; its inverse sends them to and respectively, fixing all other basis letters (Artin automorphisms of the free group).
Peripheral-boundary-preserving means permutation of positive basis conjugacy classes and exact preservation of (Peripheral-boundary-preserving automorphisms of F_n).
Proof
Left middle letter. Here . Postcomposition gives and , where and . The first image has conjugator around , so and . Reducing gives conjugators of total length at most , whereas the old pair has length . The length falls by at least one.
Right middle letter. Here . Postcomposition with the inverse gives and . The second image has conjugator around . Thus the new conjugators are , of total length at most , while . Again the length falls by at least one; all other images are unchanged in either case.
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 and fix , since and the inverse fixes the same word. Consequently fixes and is peripheral-boundary-preserving. In the left case ; in the right case . The strict pair inequalities of steps 1.1 and 1.2 prove the total decrease; when the original representation is minimal they give . 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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Used by
Dependency tree · two levels
8 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, proof of Theorem 16, cases 2(a) and 2(b), printed pp. 114-115 (standard reference, not scraped)
- Juan Gonzalez-Meneses, Basic results on braid groups, section 1.6, printed pp. 9-10 (standard reference, not scraped)