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Delta is the lcm of the artin atoms and has the same left and right divisors

Statement

Let $n \geq 2$ , let $B_{n}^{+}$ be the positive braid monoid of Positive braid monoid with its atoms $σ_{1}, \dots, σ_{n - 1}$ , its divisibility orders $≼_{L}, ≼_{R}$ and their lcm and gcd notation (Left and right divisibility for positive braids), and let $Δ$ be the half twist of The Garside half twist and simple positive braids. Then:

(a) Left lcm. $Δ$ is a common left multiple of all atoms, and every common left multiple $m \in B_{n}^{+}$ of $σ_{1}, \dots, σ_{n - 1}$ satisfies $Δ ≼_{L} m$ . Equivalently, $Δ = σ_{1} \lor_{L} σ_{2} \lor_{L} \dots \lor_{L} σ_{n - 1}$ .

(b) Right lcm. $Δ$ is a common right multiple of all atoms, and every common right multiple $m \in B_{n}^{+}$ satisfies $Δ ≼_{R} m$ ; equivalently $Δ = σ_{1} \lor_{R} σ_{2} \lor_{R} \dots \lor_{R} σ_{n - 1}$ .

(c) The divisors coincide. An element $a \in B_{n}^{+}$ is a left divisor of $Δ$ if and only if it is a right divisor of $Δ$ , and this happens if and only if $a = \hat{σ}$ for a unique $σ \in S_{n}$ ; in particular there are exactly $n!$ simple braids, and the sets of left and of right divisors of $Δ$ both equal ${\hat{σ} : σ \in S_{n}}$ .

(d) Characterisation by the atoms. For $m \in B_{n}^{+}$ one has $Δ ≼_{L} m$ if and only if $σ_{i} ≼_{L} m$ for every $i \in {1, \dots, n - 1}$ , and analogously with $≼_{R}$ .

For $n \leq 1$ the alphabet is empty, the monoid is trivial, $Δ = 1$ and all assertions hold with $n! = 1$ (there is exactly one simple braid, namely $1$ ). No choice principle is used; the only infinite objects are the finitely many fixed-length positive words used to invoke the gcd/lcm theorem.

Facts & Assumptions

Given: A natural number $n \geq 2$ , the positive braid monoid $B_{n}^{+}$ with atoms $σ_{1}, \dots, σ_{n - 1}$ , divisibility orders $≼_{L}, ≼_{R}$ and half twist $Δ$ , and the bijection $σ \mapsto \hat{σ}$ from $S_{n}$ onto the set of left divisors of $Δ$ .

[F1]

Every atom is both a left and a right divisor of $Δ$ : for each $i$ there are $R_{i}, L_{i} \in B_{n}^{+}$ with $Δ = σ_{i} R_{i} = L_{i} σ_{i}$ (Each Artin atom is a left and right divisor of the half twist). The half twist is the class of the triangular word with $ℓ (Δ) = N = n (n - 1) / 2$ (The Garside half twist and simple positive braids).

[F2]

Every nonempty finite subset of $B_{n}^{+}$ has a left-lcm and a left-gcd and a right-lcm and a right-gcd, and these are unique; a common left divisor of a family divides its left-gcd, and a left-lcm divides every common left multiple (Positive braids have left and right gcds and lcms, Left and right divisibility for positive braids).

[F3]

Simple braids and descents (Simple positive braids are indexed by permutations). An element $a \in B_{n}^{+}$ is a left divisor of $Δ$ if and only if it is a right divisor of $Δ$ , if and only if $a = \hat{π (a)}$ ; the map $σ \mapsto \hat{σ}$ is a bijection from $S_{n}$ onto the left divisors of $Δ$ , so there are $n!$ simple braids. Moreover, if $b \in B_{n}^{+}$ satisfies $ℓ (b) = inv (π (b))$ and $σ_{i} ≼_{L} b$ for every $i$ , then $π (b) = w_{0}$ and $b = Δ$ .

[F4]

Reversal (The positive braid monoid is left and right cancellative, Conjugation by the half twist reverses Artin generators). Reversal of words induces an involutive anti-automorphism $ρ$ of $B_{n}^{+}$ with $ρ (σ_{i}) = σ_{i}$ and $ρ (Δ) = Δ$ , and it exchanges the two divisibility orders: $a ≼_{L} b ⟺ ρ (a) ≼_{R} ρ (b)$ and $a ≼_{R} b ⟺ ρ (a) ≼_{L} ρ (b)$ .

Proof

technique · direct

1.1

The left lcm (a). By [F1] $Δ$ is a common left multiple of the atoms. Let $m \in B_{n}^{+}$ be any common left multiple and put $d : = Δ \land_{L} m$ , which exists by [F2]. For every $i$ the atom $σ_{i}$ is a common left divisor of $Δ$ (by [F1]) and of $m$ (by hypothesis), hence $σ_{i} ≼_{L} d$ by the defining property of the gcd. In particular $d \neq 1$ unless $n = 1$ ; more importantly $d ≼_{L} Δ$ , so $d$ is a simple braid and therefore $d = \hat{π (d)}$ with $ℓ (d) = inv (π (d))$ by [F3]. Since every atom left-divides $d$ , [F3] applied to $b : = d$ gives $π (d) = w_{0}$ , hence $d = \hat{w_{0}} = Δ$ . Thus $Δ = d ≼_{L} m$ : $Δ$ left-divides every common left multiple of the atoms, so it is their left-lcm.

F1F2F3

2.1

The right lcm (b). By [F1] $Δ$ is a common right multiple. Let $m$ be any common right multiple of the atoms and apply the involutive anti-automorphism $ρ$ of [F4]: $σ_{i} ≼_{R} m$ is equivalent to $ρ (σ_{i}) = σ_{i} ≼_{L} ρ (m)$ , so $ρ (m)$ is a common left multiple of the atoms, whence $Δ ≼_{L} ρ (m)$ by step 1.1. Applying $ρ$ again and using $ρ (Δ) = Δ$ gives $Δ = ρ (Δ) ≼_{R} ρ (ρ (m)) = m$ . Hence $Δ$ right-divides every common right multiple of the atoms and is their right-lcm.

F1F4step 1.1

3.1

Divisors and the atom criterion (c), (d). Part (c) is [F3] restated: a left divisor of $Δ$ is the same as a right divisor, the common set is ${\hat{σ} : σ \in S_{n}}$ , and it has $n!$ elements. For (d): if $σ_{i} ≼_{L} m$ for every $i$ then $m$ is a common left multiple of the atoms, so $Δ ≼_{L} m$ by step 1.1; conversely $Δ ≼_{L} m$ implies $σ_{i} ≼_{L} m$ for every $i$ because $σ_{i} ≼_{L} Δ$ by [F1] and $≼_{L}$ is transitive. The right-handed statement is the same argument with step 1.1 replaced by step 2.1 and [F1]'s right divisibility.

F1F2F3step 1.1step 2.1

4.1

Assembly. Part (a) is step 1.1, part (b) is step 2.1, parts (c) and (d) are step 3.1. No use is made of an assumed lcm of the atoms before it is proved: the argument only uses the existence of the gcd $Δ \land_{L} m$ for two elements, which is supplied by [F2], and it identifies the gcd with $Δ$ by the descent criterion of [F3]. For $n \leq 1$ the alphabet is empty, $B_{n}^{+} = {1}$ , $Δ = 1$ , the only simple braid is $1$ , and all assertions are trivial. No choice principle is used. ∎

step 1.1step 2.1step 3.1

Remarks

No circularity. The plan of this page warns against using the future lcm claim Positive braids have left and right gcds and lcms to define $Δ$ : here the gcd/lcm machinery is applied to the pair $(Δ, m)$ , and the specific element $Δ$ is the independently defined triangular word of The Garside half twist and simple positive braids.
The source form. The left lcm statement is the algebraic content of Garside's observation $Δ = σ_{1} \lor \dots \lor σ_{n - 1}$ used in J. González-Meneses, Basic results on braid groups, Section 4, printed p. 27. The proof given here derives it from the permutation indexing of the divisors of $Δ$ (Garside's own argument compares the lengths of the divisors), so it does not presuppose the crossing number of a braid.
Consequences. Part (a) is used in A central positive braid is a power of delta squared for n greater than two to turn "every atom is a left divisor of $A$ " into " $Δ$ is a left divisor of $A$ ", and part (c) supplies the balanced divisor set used there and in Left garside normal form is unique.
Nothing here uses a choice principle: the gcd of two positive braids is obtained by a finite enumeration of the positive words of bounded length (Positive braids have left and right gcds and lcms).

Depends on

Used by

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4, printed pp. 27-28 (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter IX, printed pp. 433-438 (standard reference, not scraped)