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The free-kernel words for three-strand braid combing

Example

Assume AC for the free-kernel basis. For $n = 3$ the combing words of The Zariski combing words alpha_i and x_i in the Artin presentation are $x_{2} = σ_{2}^{2}, x_{1} = σ_{2}^{- 1} σ_{1}^{2} σ_{2} .$ With the standard pure braids $A_{23} = σ_{2}^{2}$ and $A_{13} = σ_{2} σ_{1}^{2} σ_{2}^{- 1}$ of Standard geometric pure braid generators A_ij, free cancellation gives $x_{2} = A_{23}, x_{1} = A_{23}^{- 1} A_{13} A_{23},$ which is the combing identity of The combed geometric decomposition is unique at $n = 3$ . Hence the free kernel of the forgetting map $P B_{3} \to P B_{2}$ , freely generated by $A_{13}, A_{23}$ through the batch-21 identification, is equally freely generated by $x_{1}, x_{2}$ ; under the identification of that kernel with the fundamental group of the twice-punctured disc fibre of the forgetting map, the standard generators correspond to clockwise based meridians $μ_{1}, μ_{2}$ of the two punctures, the inverses of the positively oriented meridians specified by The $A_{i n}$ are meridian generators of the forgetful free kernel. The combing basis corresponds to $μ_{2}^{- 1} μ_{1} μ_{2}, μ_{2}$ : its first element is a conjugate of the first standard meridian, rather than the same based class for the original stem.

Facts & Assumptions

Given: The group $B_{3} = ⟨ σ_{1}, σ_{2} ⟩$ of The braid group by Artin presentation with its defining Artin braid relation and permitted free insertions and deletions of adjacent inverse letters, the combing words $α_{j}, x_{j}$ of The Zariski combing words alpha_i and x_i in the Artin presentation for $n = 3$ , the standard pure braids $A_{i j}$ of Standard geometric pure braid generators A_ij for $1 \leq i < j \leq 3$ , and the surjection $φ_{3} ⁣ : B_{3} \to G_{3}$ of The Artin presentation surjects onto the geometric braid group.

[F1]

For $n = 3$ the combing words are $α_{3} = 1$ , $α_{2} = σ_{2}$ , $α_{1} = σ_{1} σ_{2}$ and $x_{2} = α_{3}^{- 1} σ_{2}^{2} α_{3} = σ_{2}^{2}$ , $x_{1} = α_{2}^{- 1} σ_{1}^{2} α_{2} = σ_{2}^{- 1} σ_{1}^{2} σ_{2}$ (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

The standard pure braid generators of Standard geometric pure braid generators A_ij are the classes of the words $W_{i j} = σ_{j - 1} \dots σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} \dots σ_{j - 1}^{- 1}$ ; for $n = 3$ this gives $A_{23} = σ_{2}^{2}$ and $A_{13} = σ_{2} σ_{1}^{2} σ_{2}^{- 1}$ , in the geometric group and, through the published isomorphism $Ψ_{3}$ , in $P B_{3}$ .

[F3]

Assume AC. Then AC implies dependent choice and countable choice (The Axiom of Choice, AC implies DC implies countable choice), the forgetting map $P B_{3} \to P B_{2}$ has free kernel of rank $2$ , and under the fiber-inclusion identification the elements $A_{13}, A_{23}$ are a free basis of that kernel (The Fadell-Neuwirth short exact sequence for pure braids, The $A_{i n}$ are meridian generators of the forgetful free kernel); this is the batch-21 identification referred to in the statement.

[F4]

At $n = 3$ the uniqueness lemma says: if $W_{1}$ is a word in $x_{1}^{\pm 1}, x_{2}^{\pm 1}$ and $W_{2}$ a word in $σ_{1}^{\pm 1}$ with $φ_{3} (W_{1} W_{2}) = 1$ , then (i) $Ψ_{3} (φ_{3} (x_{1})) = A_{23}^{- 1} A_{13} A_{23}$ and $Ψ_{3} (φ_{3} (x_{2})) = A_{23}$ , and $Ψ_{3} (φ_{3} (x_{1})), Ψ_{3} (φ_{3} (x_{2}))$ form a free basis of the kernel; (ii) $φ_{3} (W_{1}) = 1$ with $W_{1}$ freely trivial; (iii) $φ_{2} (W_{2}) = 1$ (The combed geometric decomposition is unique).

Verification

technique · direct

1.1F1

The combing words. By [F1], $x_{2} = σ_{2}^{2}$ and $x_{1} = α_{2}^{- 1} σ_{1}^{2} α_{2} = σ_{2}^{- 1} σ_{1}^{2} σ_{2}$ ; both displays are literal word computations from the definition, with $α_{3} = 1$ the empty word.

1.2F2

The standard generators. By [F2], $A_{23} = [W_{23}]$ with $W_{23} = σ_{2}^{2}$ (both outer blocks of the display empty) and $A_{13} = [W_{13}]$ with $W_{13} = σ_{2} σ_{1}^{2} σ_{2}^{- 1}$ .

2.1F1F2step 1.1step 1.2

The two identities, by free cancellation. Substituting the words of steps 1.1 and 1.2, $A_{23}^{- 1} A_{13} A_{23} = σ_{2}^{- 2} (σ_{2} σ_{1}^{2} σ_{2}^{- 1}) σ_{2}^{2} \equiv σ_{2}^{- 1} σ_{1}^{2} σ_{2} = x_{1}$ and $A_{23} = x_{2}$ ; the only moves are deletions of the adjacent inverse pairs $σ_{2}^{- 1} σ_{2}$ and $σ_{2}^{- 1} σ_{2}$ , in the middle of the first display. This is the identity $x_{1} = P_{1}^{- 1} A_{13} P_{1}$ with $P_{1} = A_{23}$ of the combing lemma at $n = 3$ .

3.1F3F4step 2.1

The kernel is freely generated by the two combing words. Assume AC, so that the kernel $K$ of $P B_{3} \to P B_{2}$ is free with basis $A_{13}, A_{23}$ by [F3]. Identify $K$ with the abstract free group $F (a_{1}, a_{2})$ through $a_{1} \mapsto A_{13}$ , $a_{2} \mapsto A_{23}$ , and define the endomorphism $θ ⁣ : F \to F$ on that basis by $θ (a_{1}) : = a_{2} a_{1} a_{2}^{- 1}$ , $θ (a_{2}) : = a_{2}$ . Let $ρ ⁣ : F \to F$ be the homomorphism with $ρ (a_{1}) : = a_{2}^{- 1} a_{1} a_{2}$ and $ρ (a_{2}) : = a_{2}$ . Then $θ (ρ (a_{1})) = θ (a_{2})^{- 1} θ (a_{1}) θ (a_{2}) = a_{2}^{- 1} (a_{2} a_{1} a_{2}^{- 1}) a_{2} = a_{1}$ and $θ (ρ (a_{2})) = a_{2}$ , so $θ \circ ρ$ is the identity on the free basis and hence on $F$ ; therefore $ρ$ is injective and the elements $ρ (a_{1}) = A_{23}^{- 1} A_{13} A_{23}$ , $ρ (a_{2}) = A_{23}$ are a free basis of the subgroup they generate. By step 2.1 these are $Ψ_{3} (φ_{3} (x_{1}))$ and $Ψ_{3} (φ_{3} (x_{2}))$ under the identification of [F4], and they generate $K$ : each of $A_{13} = A_{23} ρ (a_{1}) A_{23}^{- 1}$ and $A_{23} = ρ (a_{2})$ lies in $⟨ ρ (a_{1}), ρ (a_{2}) ⟩$ , while both $ρ (a_{i})$ lie in $K$ . Hence $Ψ_{3} (φ_{3} (x_{1})), Ψ_{3} (φ_{3} (x_{2}))$ are a free basis of $K$ ; the argument is a free-group computation and uses no choice principle beyond the freeness of $K$ supplied by [F3].

4.1F3F4step 2.1step 3.1

Conclusion. Combining steps 2.1 and 3.1: the two combing words satisfy $x_{2} = A_{23}$ and $x_{1} = A_{23}^{- 1} A_{13} A_{23}$ by free cancellation, and the free kernel of $P B_{3} \to P B_{2}$ , freely generated by $A_{13}, A_{23}$ , is equally freely generated by $x_{1}, x_{2}$ . Under the fibre-inclusion identification, write $μ_{i}$ for the clockwise meridian corresponding to $A_{i 3}$ by [F3]. Step 2.1 gives the fibre classes $μ_{2}^{- 1} μ_{1} μ_{2}$ and $μ_{2}$ for $x_{1}$ and $x_{2}$ respectively. In the free group these first-meridian classes differ: the word $μ_{2}^{- 1} μ_{1} μ_{2}$ is reduced and is not $μ_{1}$ ; conjugation changes the based stem class. ∎

Remarks

The two free-cancellation displays of step 2.1 are choice-free; AC enters only through [F3], the batch-21 identification of the free kernel with basis $A_{13}, A_{23}$ , and through the uniqueness lemma [F4] that names the images of the combing words. The free-basis argument of step 3.1 is the $n = 3$ instance of the left-inverse argument of The combed geometric decomposition is unique: an endomorphism fixing the conjugated basis shows that conjugation by $A_{23}$ is injective on the free group.
The identification of the fibre with a twice-punctured disc and of $A_{13}, A_{23}$ with the clockwise based meridians is asserted here only as the reading of the batch-21 supplier statement (the formula $Ψ ([A_{i n}]) = (κ_{*} [γ_{i}])^{- 1}$ with $γ_{i}$ counterclockwise), read during this dispatch; the suppliers The Fadell-Neuwirth short exact sequence for pure braids and The $A_{i n}$ are meridian generators of the forgetful free kernel are in-run drafts, and their certification, in particular the meridian clause, is flagged for the owner rather than proved locally.

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, sections 2.1 and 3.1, printed pp. 11-13 and 19-22 (standard reference, not scraped)