LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02

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Prefix insertion rewrites a trivial braid word into combing factors

Statement

Assume $n \geq 2$ and let $W = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{m}}^{ε_{m}}$ be a word in the Artin letters and their inverses whose image under the published surjection $φ$ of The Artin presentation surjects onto the geometric braid group is the trivial geometric braid. Write $W_{k} : = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{k}}^{ε_{k}}$ for $k = 0, \dots, m$ , and for $k = 0, \dots, m$ let $j_{k} : = π (φ (W_{k}))^{- 1} (n) \in {1, \dots, n}$ be the position at the bottom of the sub-braid $φ (W_{k})$ of the point that sits at position $n$ at its top, where $π ⁣ : G_{n} \to S_{n}$ is the endpoint permutation homomorphism; here $s_{r}$ denotes the transposition of $r$ and $r + 1$ . Then $j_{0} = n$ , the recursion $j_{k} = s_{i_{k}} (j_{k - 1}) (k = 1, \dots, m)$ holds, and $j_{m} = n$ because $φ (W) = 1$ . Equivalently, $j_{k}$ is the position of the point that starts at position $n$ at the top of $W$ after it has passed the first $k$ letters (the library's stacking puts the first letter of a word on top, so this point meets the letters in word order).

Using only free insertions of the words $α_{j} α_{j}^{- 1}$ and free deletions of cancelling pairs (no braid relation), $W$ is equivalent in the group $B_{n}$ of The braid group by Artin presentation to the product of combing factors $W \equiv \prod_{k = 1}^{m} (α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}}),$ where the words $α_{1}, \dots, α_{n}$ are those of The Zariski combing words alpha_i and x_i in the Artin presentation. The $k$ -th factor is the $k$ -th letter decorated by its two connectors: read bottom to top (the library's stacking puts the first letter of a word on top, so the last block of the factor is met first), the tracked point that starts at position $n$ travels through $α_{j_{k}}$ to position $j_{k}$ , is exchanged by the letter $σ_{i_{k}}^{ε_{k}}$ to position $j_{k - 1}$ when $j_{k} \in {i_{k}, i_{k} + 1}$ (and is fixed otherwise), and is carried back to position $n$ by $α_{j_{k - 1}}^{- 1}$ ; equivalently, in the source's bottom-up reading the point has position $j_{k}$ below the letter and $j_{k - 1}$ above it. It lies in the letter's support precisely when $j_{k} \in {i_{k}, i_{k} + 1}$ ; otherwise $j_{k - 1} = j_{k}$ and it remains fixed outside that support during the letter. In every case the recursion above is exactly the interchange rule $j_{k - 1} = s_{i_{k}} (j_{k})$ used by the six-case analysis, the empty word is allowed ( $m = 0$ , where $W$ is empty and $j_{0} = j_{m} = n$ ), and nothing but $φ (W) = 1$ is assumed about the geometric braid.

Facts & Assumptions

Given: An integer $n \geq 2$ , a word $W = σ_{i_{1}}^{ε_{1}} \dots σ_{i_{m}}^{ε_{m}}$ with $φ (W) = 1$ , its prefixes $W_{k}$ , and the words $α_{j}$ of The Zariski combing words alpha_i and x_i in the Artin presentation.

[F1]

$α_{j} = σ_{j} σ_{j + 1} \dots σ_{n - 1}$ for $1 \leq j \leq n - 1$ and $α_{n} = 1$ is the empty word; all these are words in the generators and their inverses (The Zariski combing words alpha_i and x_i in the Artin presentation).

[F2]

$B_{n}$ is the quotient of the free group on $σ_{1}, \dots, σ_{n - 1}$ by the normal closure of the two relation families; consequently words differing by insertions or deletions of adjacent inverse pairs $w^{\pm 1} w^{\mp 1}$ represent the same element, and a product of words telescopes whenever adjacent connector words cancel (The braid group by Artin presentation, The Zariski combing words alpha_i and x_i in the Artin presentation).

[F3]

The assignment $φ (σ_{i}) = [σ_{i}]$ extends to a surjective homomorphism $φ ⁣ : B_{n} \to G_{n}$ (The Artin presentation surjects onto the geometric braid group).

[F4]

The endpoint permutation $π ⁣ : G_{n} \to S_{n}$ is a homomorphism, the class of the half twist $σ_{r}$ has $π ([σ_{r}]) = s_{r}$ , and under the stacking convention of Stacking of geometric braids is a well-defined associative operation on isotopy classes the class of a word $u_{1} \dots u_{l}$ satisfies $π (φ (u_{1} \dots u_{l})) = π (φ (u_{1})) \circ \dots \circ π (φ (u_{l}))$ as functions, the first factor applied last (The elementary geometric half twist, its support disc, and its opposite, The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism, Stacking of geometric braids is a well-defined associative operation on isotopy classes).

Proof

technique · direct

1.1F1F3F4

The connectors move the $n$ -th point down to position $j$ . By [F3] and [F1], $φ (α_{j}) = φ (σ_{j}) \dots φ (σ_{n - 1}) = [σ_{j}] \dots [σ_{n - 1}]$ , so by [F4] the endpoint permutation is $π (φ (α_{j})) = s_{j} \circ s_{j + 1} \circ \dots \circ s_{n - 1}$ as a function. Evaluating on positions, this function sends $n \mapsto n - 1 \mapsto n - 2 \mapsto \dots \mapsto j$ and fixes every $x < j$ , so it is the cycle $t_{j} : = (j j + 1 \dots n)$ with $t_{j} (n) = j$ ; in particular a point starting at position $n$ ends at position $j$ , and $t_{j}^{- 1} (j) = n$ . For $j = n$ the word $α_{n}$ is empty and $π (φ (α_{n})) = id$ with $id (n) = n$ .

1.2F3F4

The recursion and its endpoints. For each $k$ , [F3] and [F4] give $π (φ (W_{k})) = π (φ (σ_{i_{1}}^{ε_{1}})) \circ \dots \circ π (φ (σ_{i_{k}}^{ε_{k}})) = s_{i_{1}} \circ \dots \circ s_{i_{k}}$ , because $s_{r}^{\pm 1} = s_{r}$ . Taking inverses, $π (φ (W_{k}))^{- 1} = s_{i_{k}} \circ \dots \circ s_{i_{1}}$ , so $j_{k} : = π (φ (W_{k}))^{- 1} (n)$ satisfies $j_{0} = n$ (the empty product) and $j_{k} = s_{i_{k}} (j_{k - 1})$ for $1 \leq k \leq m$ . Since $W_{m} = W$ and $φ (W) = 1$ , $π (φ (W_{m})) = id$ and therefore $j_{m} = id (n) = n$ .

2.1F1F2step 1.2

The telescoping insertion. For $k = 1, \dots, m - 1$ insert the word $α_{j_{k}} α_{j_{k}}^{- 1}$ between the $k$ -th and $(k + 1)$ -st letter of $W$ and bracket the result as $W = α_{j_{0}}^{- 1} σ_{i_{1}}^{ε_{1}} α_{j_{1}} \cdot α_{j_{1}}^{- 1} σ_{i_{2}}^{ε_{2}} α_{j_{2}} \dots α_{j_{m - 1}}^{- 1} σ_{i_{m}}^{ε_{m}} α_{j_{m}} .$ Each interior position contributes $α_{j_{k}} α_{j_{k}}^{- 1} = 1$ , which is a free cancellation by [F2], and by [F1] and step 1.2 the two end connectors are $α_{j_{0}} = α_{n} = 1$ and $α_{j_{m}} = α_{n} = 1$ ; expanding the displayed product therefore returns the original word $W$ by free cancellations alone, and conversely $W$ is obtained from the displayed product by the inverse free moves. No defining relation of the Artin presentation is used.

2.2F3F4step 1.1step 1.2

The bookkeeping inside a factor. Fix $k$ and read the factor $F_{k} = α_{j_{k - 1}}^{- 1} σ_{i_{k}}^{ε_{k}} α_{j_{k}}$ from the bottom upward, that is, starting from its last and lowest block $α_{j_{k}}$ and ending with its first and topmost block $α_{j_{k - 1}}^{- 1}$ (the word's first letter is the topmost layer in the stacking of [F4]). A point starting at position $n$ at the bottom of the factor is carried by the block $α_{j_{k}}$ to position $t_{j_{k}} (n) = j_{k}$ by step 1.1, then by the letter $σ_{i_{k}}^{ε_{k}}$ to position $s_{i_{k}}^{ε_{k}} (j_{k}) = s_{i_{k}} (j_{k}) = j_{k - 1}$ by [F4] and step 1.2 (if $j_{k} = j_{k - 1}$ the letter fixes it), and then by the block $α_{j_{k - 1}}^{- 1}$ to position $t_{j_{k - 1}}^{- 1} (j_{k - 1}) = n$ by step 1.1. Hence inside the $k$ -th factor the letter acts on the tracked point exactly through the interchange rule connecting the two connector positions $j_{k}$ (below the letter) and $j_{k - 1}$ (above it), and the tracked point returns to position $n$ at the top of every factor, as it must because at the top of $W$ it is again at position $n$ by $φ (W) = 1$ .

3.1F4step 1.2step 2.1step 2.2

Conclusion. Steps 1.2 and 2.1 establish the recursion, its endpoints, and the telescoping product. Step 2.2 gives the positions $j_{k}$ below and $j_{k - 1}$ above each letter. By the half-twist definition in [F4], the tracked point lies in the letter's support if $j_{k} \in {i_{k}, i_{k} + 1}$ ; otherwise it is a fixed base point outside that support. For $m = 0$ the product is empty. ∎

Remarks

The lemma is a pure bookkeeping statement: the group element is unchanged because each inserted connector is immediately cancelled, and the geometric input is only the published endpoint-permutation homomorphism, which fixes the positions $j_{k}$ by the triviality of $φ (W)$ .
In the source the same product is displayed with $α_{j_{0}} = α_{j_{m}} = 1$ , reading words bottom-up; the recursion $j_{k} = s_{i_{k}} (j_{k - 1})$ is identical in both conventions, and the six-case reduction of the next item depends only on this recursion and on the displayed shape of the factors.

Depends on

Used by

Dependency tree · two levels

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Sources

Juan Gonzalez-Meneses, Basic results on braid groups, section 3.1, printed pp. 19-20 (standard reference, not scraped)