TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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Positive braids have left and right gcds and lcms

Statement

Let $n \in N$ , let $B_{n}^{+}$ be the positive braid monoid of Positive braid monoid with its homogeneous length $ℓ$ (Positive artin relations preserve homogeneous length), its half twist $Δ$ (The Garside half twist and simple positive braids) and its divisibility orders $≼_{L}, ≼_{R}$ with the lcm and gcd notation $\lor_{L}, \land_{L}, \lor_{R}, \land_{R}$ of Left and right divisibility for positive braids. Then, for all $a, b \in B_{n}^{+}$ :

(a) Left join. The left-lcm $a \lor_{L} b$ exists: there is a common left multiple of $a$ and $b$ that left-divides every common left multiple of $a$ and $b$ . It is unique, and for words $u, v$ with $[u] = a$ , $[v] = b$ it is given by the reversing complement, $a \lor_{L} b = [u Θ (u, v)] = [v Θ (v, u)],$ where $Θ$ is the right complement of Artin right complements and word reversing.

(b) Left meet. The left-gcd $a \land_{L} b$ exists and is unique: there is a common left divisor of $a$ and $b$ that is a left multiple of every common left divisor of $a$ and $b$ .

(c) Right-hand versions. The right-lcm $a \lor_{R} b$ and the right-gcd $a \land_{R} b$ exist and are unique.

(d) Finite families. Every nonempty finite subset of $B_{n}^{+}$ has a left-lcm, a left-gcd, a right-lcm and a right-gcd; in particular the two orders are lattices on $B_{n}^{+}$ .

The lcm of (a) is computed by the finite reversing algorithm of Artin positive word reversing is complete, and no choice principle is used. For $n \leq 1$ the monoid is trivial and all these elements are $1$ .

Facts & Assumptions

Given: A natural number $n$ , the positive braid monoid $B_{n}^{+}$ with its length $ℓ$ and divisibility orders $≼_{L}, ≼_{R}$ , the half twist $Δ$ , and the partial map $Θ$ .

[F1]

$a ≼_{L} b ⟺ \exists c (b = a c)$ , $a ≼_{R} b ⟺ \exists c (b = c a)$ ; both are partial orders with $ℓ (a) \leq ℓ (b)$ when $a ≼_{L} b$ or $a ≼_{R} b$ ; there are at most $∣ Σ_{n} ∣^{k}$ elements of $B_{n}^{+}$ of length $k$ , where $Σ_{n}$ is the actual alphabet (empty for $n \leq 1$ and of size $n - 1$ for $n \geq 2$ ); and $1$ is a left and right divisor of every element (Left and right divisibility for positive braids, Positive artin relations preserve homogeneous length).

[F2]

Common multiples exist. For all $a, b \in B_{n}^{+}$ there is $m \in N$ with both $a ≼_{L} Δ^{m}$ and $b ≼_{L} Δ^{m}$ , and likewise for $≼_{R}$ ; in particular every pair has a common left multiple in the sense of the order $≼_{L}$ (Every positive braid divides a power of the half twist on both sides).

[F3]

Reversing criterion. For positive words $u, v$ the elements $[u], [v]$ admit a common left multiple if and only if right-reversing of the signed word $u^{- 1} v$ terminates, and then $[u Θ (u, v)] = [v Θ (v, u)]$ is the least common left multiple: every common left multiple of $[u], [v]$ is a left multiple of it, and it is itself a common left multiple. This is part (e) of Artin positive word reversing is complete, where the element is called the right-lcm because it is obtained by extending $u$ on the right; in the notation of Left and right divisibility for positive braids it is the join $[u] \lor_{L} [v]$ , since $[u] ≼_{L} [u Θ (u, v)]$ and $[v] ≼_{L} [v Θ (v, u)]$ hold by the definition of $≼_{L}$ (Left and right divisibility for positive braids).

[F4]

Reversal. The word reversal induces an involutive anti-automorphism $ρ$ of $B_{n}^{+}$ , and it exchanges the two orders: $a ≼_{L} b ⟺ ρ (a) ≼_{R} ρ (b)$ , $a ≼_{R} b ⟺ ρ (a) ≼_{L} ρ (b)$ (The positive braid monoid is left and right cancellative, Left and right divisibility for positive braids).

[F5]

Uniqueness of least elements. If a subset of a partially ordered set has a greatest element, it is unique (Left and right divisibility for positive braids for antisymmetry of the two orders).

Proof

technique · direct

1.1

Every pair has a common left multiple. Let $a, b \in B_{n}^{+}$ . By [F2] there are $m, m^{'}$ with $a ≼_{L} Δ^{m}$ and $b ≼_{L} Δ^{m^{'}}$ ; putting $M : = \max (m, m^{'})$ and writing $Δ^{m} = a c$ gives $Δ^{M} = Δ^{m} Δ^{M - m} = a (c Δ^{M - m})$ , so $a ≼_{L} Δ^{M}$ , and symmetrically $b ≼_{L} Δ^{M}$ .

F1F2

2.1

The left-lcm exists for every pair. Let $u, v$ be positive words with $[u] = a$ , $[v] = b$ . By step 1.1 the classes admit a common left multiple, so by the reversing criterion [F3] the class $[u Θ (u, v)] = [v Θ (v, u)]$ is a common left multiple of $a$ and $b$ that left-divides every common left multiple of $a$ and $b$ ; this is exactly $a \lor_{L} b$ , and it is unique by [F5]. This is (a).

F1F3F5step 1.1

3.1

The left-gcd exists. Let $D : = {d \in B_{n}^{+} : d ≼_{L} a and d ≼_{L} b}$ be the set of common left divisors. It contains $1$ by [F1], and it is finite: every $d \in D$ satisfies $ℓ (d) \leq ℓ (a)$ by monotonicity, and there are only finitely many elements of each length at most $ℓ (a)$ by [F1]. Let $d_{1}, \dots, d_{r}$ be an enumeration of $D$ and define $δ_{1} : = d_{1}$ , $δ_{j + 1} : = δ_{j} \lor_{L} d_{j + 1}$ for $j < r$ . Each step is legitimate: if $δ_{j} \in D$ then $δ_{j} ≼_{L} a$ and $d_{j + 1} ≼_{L} a$ , so $a$ is a common left multiple of the pair and step 2.1 provides the join, which by leastness satisfies $δ_{j + 1} ≼_{L} a$ and $δ_{j + 1} ≼_{L} b$ , so $δ_{j + 1} \in D$ . Thus $δ_{r} \in D$ and $d_{j} ≼_{L} δ_{r}$ for every $j$ by construction; so $δ_{r}$ is a common left divisor of $a, b$ that is a left multiple of every common left divisor, that is, $a \land_{L} b = δ_{r}$ exists and is unique by [F5].

F1F5step 2.1

4.1

The right-hand versions. Apply the anti-automorphism $ρ$ of [F4] to step 2.1: if $u, v$ are words, then $ρ ([u]), ρ ([v])$ have the join $ρ ([u]) \lor_{L} ρ ([v])$ , and by the exchange of orders [F4] the element $ρ (ρ ([u]) \lor_{L} ρ ([v]))$ is the right-lcm of $[u], [v]$ , since $ρ$ carries $≼_{L}$ to $≼_{R}$ bijectively and preserves leastness; it is unique by [F5]. The same transport of step 3.1 gives the right-gcd, and the transport of steps 2.1 and 3.1 also supplies the common right multiples needed, since $ρ$ is a bijection, so no separate existence proof is needed. This is (c).

F4F5step 2.1step 3.1

5.1

Finite families. If $x_{1}, \dots, x_{r} \in B_{n}^{+}$ with $r \geq 1$ , then $x_{1} \lor_{L} \dots \lor_{L} x_{r} : = (x_{1} \lor_{L} \dots \lor_{L} x_{r - 1}) \lor_{L} x_{r}$ is defined by induction on $r$ using step 2.1 and is the least common left multiple, and similarly for $\land_{L}$ using step 3.1 and for the right-hand pair using step 4.1; the case $r = 1$ is $x_{1}$ itself, and the case $r = 0$ is excluded because the family is required to be nonempty. This is (d).

F1step 2.1step 3.1step 4.1

6.1

Assembly. Part (a) is step 2.1 including the computation $a \lor_{L} b = [u Θ (u, v)]$ , part (b) is step 3.1, part (c) is step 4.1 and part (d) is step 5.1. The hypothesis that makes the reversing criterion applicable is exactly step 1.1: the conditional form of Artin positive word reversing is complete is upgraded to an unconditional existence statement by the $Δ$ -power multiples of Every positive braid divides a power of the half twist on both sides, so no common-multiple hypothesis survives in the conclusion. For $n \leq 1$ the monoid is trivial, so all four elements are $1$ ; all arguments are finite and no choice principle is used. ∎

step 1.1step 2.1step 3.1step 4.1step 5.1

Remarks

Terminology. The source GM writes $≼$ for the prefix order and states "we will also have $x d = x a \land x b$ and $x m = x a \lor x b$ for every $x \in B_{n}^{+}$ ". In Dehornoy et al. one speaks of right-lcms and right-gcds, because the multiples are generated by extending words on the right. This item uses the letter convention of Left and right divisibility for positive braids: $a \lor_{L} b$ is the least common upper bound for $≼_{L}$ , which is the element called the right-lcm in Artin positive word reversing is complete. The dictionary is stated in [F3] and used in step 2.1, so the two vocabularies cannot be silently interchanged.
Where the $Δ$ -power hypothesis enters. The reversing criterion alone is conditional: it computes the lcm only when a common left multiple exists. Step 1.1 removes that hypothesis, and this is the only place where the half twist is used. The proof therefore follows the plan of GM's Section 4 ("as every two elements have a common multiple, induction on length gives unique lcms and gcds") but supplies the missing explicit common multiple before invoking the criterion.
Effectivity. Step 2.1 is effective: the complement $Θ (u, v)$ is computed by finitely many recursion steps from the displayed words, and the gcd of step 3.1 is the join of a finite explicitly bounded list (all common left divisors of $a$ and $b$ , enumerated by length and lexicographically within each finite level). This is what the word-problem corollary The braid group word problem is decidable by garside normal form uses.
Nothing here uses a choice principle: the enumerations are of finite sets of words and all joins are determined, not chosen.

Depends on

Used by

Dependency tree · two levels

17 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

J. Gonzalez-Meneses, Basic results on braid groups, Section 4, printed pp. 26-27 (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter II, Section 4, printed pp. 63-83 (standard reference, not scraped)