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Lower-rank Artin letters conjugate x-letters

Statement

Assume $n \geq 3$ , let $1 \leq i \leq n - 2$ and $1 \leq j \leq n - 1$ , and work in the group $B_{n}$ of The braid group by Artin presentation with the words $σ_{1}, \dots, σ_{n - 1}$ and $x_{1}, \dots, x_{n - 1}$ of The Zariski combing words alpha_i and x_i in the Artin presentation. Using only the two families of defining relations and free insertions and deletions of adjacent inverse letters:

(i) $σ_{i}^{- 1} x_{j} σ_{i}$ equals $x_{j}$ when $j < i$ or $j > i + 1$ , equals $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ when $j = i + 1$ , and equals $x_{i + 1}$ when $j = i$ ;

(ii) $σ_{i} x_{j} σ_{i}^{- 1}$ equals $x_{j}$ when $j < i$ or $j > i + 1$ , equals $x_{i}$ when $j = i + 1$ , and equals $x_{i}^{- 1} x_{i + 1} x_{i}$ when $j = i$ .

Consequently, for every pair of signs $ε, δ \in {\pm 1}$ there is a word $w$ in the letters $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ with $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε},$ so that in any word over the mixed alphabet $σ_{1}^{\pm 1}, \dots, σ_{n - 2}^{\pm 1}, x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ each occurrence of a lower-rank $σ$ -letter can be moved to the right of every $x$ -letter, the $x$ -letters changing only by further $x$ -letters and their inverses. All identities also hold in the geometric braid group under the published surjection $φ$ of The Artin presentation surjects onto the geometric braid group.

Facts & Assumptions

Given: Integers $n \geq 3$ , $1 \leq i \leq n - 2$ , $1 \leq j \leq n - 1$ , the group $B_{n}$ of The braid group by Artin presentation, and the elements $x_{1}, \dots, x_{n - 1} \in B_{n}$ of The Zariski combing words alpha_i and x_i in the Artin presentation.

[F1]

In $B_{n}$ the two defining families of relations hold: $σ_{r} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}$ for $1 \leq r \leq n - 2$ , and $σ_{r} σ_{s} = σ_{s} σ_{r}$ whenever $∣ r - s ∣ > 1$ ; words equal in the free group on $σ_{1}, \dots, σ_{n - 1}$ and their inverses represent the same element of $B_{n}$ , so adjacent inverse letters may be freely inserted and deleted (The braid group by Artin presentation, The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

For every $r$ with $1 \leq r \leq n - 1$ one has the two displayed words $x_{r} = σ_{n - 1}^{- 1} \dots σ_{r + 1}^{- 1} σ_{r}^{2} σ_{r + 1} \dots σ_{n - 1} = α_{r + 1}^{- 1} σ_{r}^{2} α_{r + 1}$ , and $x_{r}^{- 1} = σ_{n - 1}^{- 1} \dots σ_{r + 1}^{- 1} σ_{r}^{- 2} σ_{r + 1} \dots σ_{n - 1}$ ; also $α_{s} = σ_{s} σ_{s + 1} \dots σ_{n - 1}$ for $1 \leq s \leq n - 1$ and $α_{n} = 1$ (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F3]

The map $φ ⁣ : B_{n} \to G_{n}$ of The Artin presentation surjects onto the geometric braid group is a homomorphism with $φ (σ_{r}) = [σ_{r}]$ , and in the geometric braid group $G_{n}$ the two families of relations of [F1] hold: $[σ_{r}] [σ_{r + 1}] [σ_{r}] = [σ_{r + 1}] [σ_{r}] [σ_{r + 1}]$ and $[σ_{r}] [σ_{s}] = [σ_{s}] [σ_{r}]$ for $∣ r - s ∣ > 1$ (The geometric three strand braid relation, Far commutativity of elementary geometric half twists).

Proof

technique · direct

1.1F1

Two mixed forms of the braid relation. Let $1 \leq r \leq n - 2$ . Multiplying $σ_{r} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}$ on the left by $σ_{r}^{- 1}$ and on the right by $σ_{r + 1}^{- 1}$ gives $σ_{r}^{- 1} σ_{r + 1} σ_{r} = σ_{r + 1} σ_{r} σ_{r + 1}^{- 1},$ and multiplying the braid relation on the left by $σ_{r + 1}^{- 1}$ and on the right by $σ_{r}^{- 1}$ gives $σ_{r + 1}^{- 1} σ_{r} σ_{r + 1} = σ_{r} σ_{r + 1} σ_{r}^{- 1}$ , whose inverse is $σ_{r} σ_{r + 1}^{- 1} σ_{r}^{- 1} = σ_{r + 1}^{- 1} σ_{r}^{- 1} σ_{r + 1} .$ Both are consequences of the defining relations of [F1] alone.

1.2F1F2

The case $j > i + 1$ . Every letter $σ_{r}^{\pm 1}$ occurring in the displayed word for $x_{j}$ of [F2] has $r \geq j \geq i + 2$ , so $∣ i - r ∣ \geq 2$ and $σ_{i}$ commutes with that letter by the far-commutation relation of [F1]; repeating this letter by letter, $σ_{i}$ commutes with the whole word, so $σ_{i}^{\pm 1} x_{j} σ_{i}^{\mp 1} = x_{j}$ . In particular both assertions (i) and (ii) hold for $j > i + 1$ .

2.1F1F2step 1.1

The case $j < i$ : sliding $σ_{i}$ past $x_{j}$ . Decompose the word of [F2] for $x_{j}$ at the index $i$ , which satisfies $j < i$ , as $x_{j} = A σ_{i}^{- 1} K σ_{i} B, A : = σ_{n - 1}^{- 1} \dots σ_{i + 1}^{- 1}, K : = σ_{i - 1}^{- 1} \dots σ_{j + 1}^{- 1} σ_{j}^{2} σ_{j + 1} \dots σ_{i - 1}, B : = σ_{i + 1} \dots σ_{n - 1},$ where each of $A$ , $K$ , $B$ may be empty and the displayed equality is free cancellation in the two words of [F2]. Since $σ_{i}$ commutes with every letter of $A$ except $σ_{i + 1}^{- 1}$ , and $σ_{i} σ_{i + 1}^{- 1} σ_{i}^{- 1} = σ_{i + 1}^{- 1} σ_{i}^{- 1} σ_{i + 1}$ by step 1.1, one has $σ_{i} A σ_{i}^{- 1} = A σ_{i}^{- 1} σ_{i + 1}$ ; since $σ_{i + 1}$ commutes with every letter of $K$ (all its indices are at most $i - 1$ ), and since $B = σ_{i + 1} B^{'}$ with $B^{'} : = σ_{i + 2} \dots σ_{n - 1}$ , the computation $σ_{i} x_{j} = σ_{i} A σ_{i}^{- 1} K σ_{i} B = A σ_{i}^{- 1} σ_{i + 1} K σ_{i} B = A σ_{i}^{- 1} K σ_{i + 1} σ_{i} σ_{i + 1} B^{'} = A σ_{i}^{- 1} K σ_{i} σ_{i + 1} σ_{i} B^{'} = A σ_{i}^{- 1} K σ_{i} σ_{i + 1} B^{'} σ_{i} = x_{j} σ_{i}$ uses only far commutation, the braid relation, and the fact that $σ_{i}$ commutes with every letter of $B^{'}$ . Left-multiplying by $σ_{i}^{- 1}$ gives $σ_{i}^{- 1} x_{j} σ_{i} = x_{j}$ , and right-multiplying by $σ_{i}^{- 1}$ gives $σ_{i} x_{j} σ_{i}^{- 1} = x_{j}$ ; both assertions hold for $j < i$ .

2.2F1F2step 1.1

The case $j = i + 1$ . Put $C : = σ_{n - 1}^{- 1} \dots σ_{i + 2}^{- 1}$ and $D : = σ_{i + 2} \dots σ_{n - 1}$ , so that $C = D^{- 1}$ as words and, by [F2], $x_{i + 1} = C σ_{i + 1}^{2} D, x_{i} = C σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} D, x_{i + 1}^{- 1} = D^{- 1} σ_{i + 1}^{- 2} C^{- 1} .$ Since $σ_{i}$ commutes with every letter of $C$ and of $D$ , and using step 1.1, $σ_{i}^{- 1} x_{i + 1} σ_{i} = C σ_{i}^{- 1} σ_{i + 1}^{2} σ_{i} D = C (σ_{i}^{- 1} σ_{i + 1} σ_{i})^{2} D = C (σ_{i + 1} σ_{i} σ_{i + 1}^{- 1})^{2} D = C σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} D .$ Expanding the product $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ with the three displayed words and using $D C = D D^{- 1} = 1$ and $C^{- 1} = D$ gives $x_{i + 1} x_{i} x_{i + 1}^{- 1} = C σ_{i + 1}^{2} (D C) σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} (D D^{- 1}) σ_{i + 1}^{- 2} C^{- 1} = C σ_{i + 1} σ_{i}^{2} σ_{i + 1}^{- 1} D,$ so $σ_{i}^{- 1} x_{i + 1} σ_{i} = x_{i + 1} x_{i} x_{i + 1}^{- 1}$ .

3.1F1F2step 1.1

The case $j = i$ . With the same words $C, D$ of step 2.2, [F2] gives $x_{i} = C σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} D$ , and $σ_{i}$ commutes with every letter of $C$ and of $D$ , so $σ_{i}^{- 1} x_{i} σ_{i} = C σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i}^{2} σ_{i + 1} σ_{i} D = C (σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i}) (σ_{i} σ_{i + 1} σ_{i}) D .$ Inverting the first identity of step 1.1 gives $σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i} = σ_{i + 1} σ_{i}^{- 1} σ_{i + 1}^{- 1}$ , and the defining braid relation gives $σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1}$ ; substituting, $σ_{i}^{- 1} x_{i} σ_{i} = C σ_{i + 1} σ_{i}^{- 1} σ_{i + 1}^{- 1} σ_{i + 1} σ_{i} σ_{i + 1} D = C σ_{i + 1} σ_{i}^{- 1} σ_{i} σ_{i + 1} D = C σ_{i + 1}^{2} D = x_{i + 1},$ the penultimate equality deleting the adjacent inverse pairs and the last equality being [F2] again.

4.1step 2.2step 3.1

The second orientation for $j = i$ and $j = i + 1$ . The map $c_{i} ⁣ : B_{n} \to B_{n}$ , $c_{i} (g) : = σ_{i}^{- 1} g σ_{i}$ , is the conjugation automorphism by $σ_{i}$ , with inverse $c_{i}^{- 1} (g) = σ_{i} g σ_{i}^{- 1}$ . Steps 2.2 and 3.1 give $c_{i} (x_{i}) = x_{i + 1}$ and $c_{i} (x_{i + 1}) = x_{i + 1} x_{i} x_{i + 1}^{- 1}$ , hence $c_{i} (x_{i}^{- 1} x_{i + 1} x_{i}) = x_{i + 1}^{- 1} (x_{i + 1} x_{i} x_{i + 1}^{- 1}) x_{i + 1} = x_{i}$ ; therefore $c_{i}^{- 1} (x_{i}) = x_{i}^{- 1} x_{i + 1} x_{i}$ , that is $σ_{i} x_{i} σ_{i}^{- 1} = x_{i}^{- 1} x_{i + 1} x_{i}$ , and $c_{i}^{- 1} (x_{i + 1}) = x_{i}$ , that is $σ_{i} x_{i + 1} σ_{i}^{- 1} = x_{i}$ . This is assertion (ii) in the two remaining cases.

5.1F2step 1.2step 2.1step 2.2step 3.1step 4.1

All four signs. Let $ε, δ \in {\pm 1}$ . If $δ = 1$ , then $σ_{i}^{ε} x_{j} = (σ_{i}^{ε} x_{j} σ_{i}^{- ε}) σ_{i}^{ε}$ ; by steps 1.2, 2.1, 2.2, 3.1 and 4.1 the middle factor equals $x_{j}$ (for $j < i$ or $j > i + 1$ ), or $x_{i + 1}$ , $x_{i + 1} x_{i} x_{i + 1}^{- 1}$ , $x_{i}$ , $x_{i}^{- 1} x_{i + 1} x_{i}$ (for $j = i$ or $j = i + 1$ , according to the sign of $ε$ ), so in every case it is a word in $x_{1}^{\pm 1}, \dots, x_{n - 1}^{\pm 1}$ . If $δ = - 1$ , then $σ_{i}^{ε} x_{j}^{- 1} = (σ_{i}^{ε} x_{j} σ_{i}^{- ε})^{- 1} σ_{i}^{ε}$ , and the same case list applies with the inverse word. Hence for all signs $σ_{i}^{ε} x_{j}^{δ} = w σ_{i}^{ε}$ with $w$ a word in the $x$ -letters and their inverses, and a leftmost occurrence of $σ_{i}^{\pm 1}$ in any mixed word can therefore be moved one $x$ -letter at a time to the right of all $x$ -letters, only $x$ -letters changing.

6.1F3step 1.1step 5.1

Transfer to the geometric braid group. Since $φ$ is a homomorphism with $φ (σ_{r}) = [σ_{r}]$ by [F3], and since the relations used in steps 1.1-5.1 are exactly the two families of [F1], applying $φ$ to each identity yields the corresponding identity in $G_{n}$ between $φ (σ_{i})^{\pm 1}$ and $φ (x_{j})^{\pm 1}$ : the images satisfy the braid relation and far commutation by the published geometric lemmas, and the free cancellations map to cancellations in the group $G_{n}$ .

7.1step 1.2step 2.1step 2.2step 3.1step 4.1step 5.1step 6.1

Assertion (i) is steps 1.2, 2.1, 2.2 and 3.1, assertion (ii) is steps 1.2, 2.1 and 4.1, the collection statement is step 5.1, and the geometric transfer is step 6.1. ∎

Depends on

Used by

Combing a four-strand braid word Example
Every trivial braid word combs as W₁W₂ Lemma

Dependency tree · two levels

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, section 3.1, printed pp. 20-21 (standard reference, not scraped)