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Conjugacy alone does not classify braid closures

Statement refuted

Assume AC. Two braids have equivalent oriented closures if and only if they are connected using conjugations alone, without changing strand number.

Facts & Assumptions

Given: AC, the braids $e \in B_{1}$ , $σ_{1} \in B_{2}$ , $σ_{1} σ_{2} \in B_{3}$ and the closure construction of The closure of a geometric braid.

[F1]

Stabilization replaces $β \in B_{n}$ by the product in $B_{n + 1}$ of the standard inclusion of $β$ with $σ_{n}^{\pm 1}$ , the new strand being added on the right; conjugacy always takes place inside a single group $B_{n}$ and never changes the number of strands (Markov conjugation and stabilization moves).

[F2]

The stabilizations $e σ_{1}$ and $e σ_{1}^{- 1}$ close to the unknot, with the explicit R1 isotopies, so the closure of the stabilization of the trivial one-strand braid is the unknot; the example item A Markov stabilization preserves the unknot closure records this for both signs.

[F3]

Markov moves preserve the oriented closure up to equivalence; in particular the closure of a stabilized braid is equivalent to the closure of the original (Markov moves preserve the oriented closure up to isotopy).

[F4]

By closure of a braid, the number of components is the number of cycles of the endpoint permutation; the permutation of $σ_{1} \in B_{2}$ is the transposition $(1 2)$ with one cycle, and the permutation of $σ_{1} σ_{2} \in B_{3}$ is a three-cycle, also with one cycle, so both closures have one component (The closure of a geometric braid, The braid group by Artin presentation).

Counterexample

Assume the Axiom of Choice. The braids $σ_{1} \in B_{2}$ and $σ_{1} σ_{2} \in B_{3}$ have equivalent closures — both the unknot — but are not conjugate to each other, since conjugacy preserves the braid group and $B_{2} \neq B_{3}$ . Hence conjugation alone, with the number of strands fixed, does not classify braid closures; at least one stabilization or destabilization is genuinely necessary.

1.1F2F3F4algebra

Both closures are unknots. The trivial braid $e \in B_{1}$ closes to a round unknot. Its positive stabilization is $σ_{1} \in B_{2}$ , whose closure is the unknot by [F2]; stabilizing $σ_{1}$ once more, adding the third strand on the right, gives the braid $σ_{1} σ_{2} \in B_{3}$ , whose closure is equivalent to the closure of $σ_{1}$ by [F3], hence also an unknot. The component count is one in both cases by [F4], so both closures are one-component links, and by [F2] and [F3] they are the unknot.

1.2F1algebra

They are not conjugate. Conjugation, by [F1], stays inside a fixed braid group $B_{n}$ and never changes the number of strands. The braid $σ_{1} \in B_{2}$ has two strands and $σ_{1} σ_{2} \in B_{3}$ has three; no sequence of conjugations within a single braid group can relate them, because such a sequence would have to identify an element of $B_{2}$ with an element of $B_{3}$ . Therefore conjugation alone does not classify braid closures: the two closures are equivalent oriented links (both the unknot) while the braids are not conjugate.

2.1F1F2F3step 1.1step 1.2∎

Conclusion. The braids $σ_{1} \in B_{2}$ and $σ_{1} σ_{2} \in B_{3}$ have equivalent closures but are not conjugate; the only difference between the two braids in Markov terms is the stabilization that changes the group from $B_{2}$ to $B_{3}$ , which is the strand-changing move that conjugacy cannot simulate. This verifies the counterexample and shows that at least one stabilization or destabilization is genuinely necessary for the classification. The statement retains AC because it consumes the countable-choice-stated closure-preservation lemma [F3], with countable choice following from AC by AC implies DC implies countable choice and the ambient isotopy theory behind [F2].

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Sources

Birman and Brendle, Braids: A Survey, Handbook of Knot Theory chapter, author manuscript; section 2.3, printed pp. 17-19 (standard reference, not scraped)
Traczyk, A new proof of Markov's braid theorem, Banach Center Publications 42 (1998), 409-419; section 1 (standard reference, not scraped)