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Every positive braid divides a power of the half twist on both sides

Statement

Let $n \in N$ , let $B_{n}^{+}$ be the positive braid monoid of Positive braid monoid with its homogeneous length $ℓ$ , its negation free half twist $Δ = Δ_{n}$ of The Garside half twist and simple positive braids (of length $N = n (n - 1) / 2$ ), and its divisibility orders $≼_{L}, ≼_{R}$ (Left and right divisibility for positive braids). Then:

(a) Left divisibility into a $Δ$ -power. For every $a \in B_{n}^{+}$ there exist $k \in N$ and $c \in B_{n}^{+}$ with $Δ^{k} = a c$ ; equivalently $a ≼_{L} Δ^{k}$ .

(b) Right divisibility into a $Δ$ -power. For every $a \in B_{n}^{+}$ there exist $k^{'} \in N$ and $c^{'} \in B_{n}^{+}$ with $Δ^{k^{'}} = c^{'} a$ ; equivalently $a ≼_{R} Δ^{k^{'}}$ .

(c) Common $Δ$ -power multiples. For all $a, b \in B_{n}^{+}$ there is $m \in N$ such that $Δ^{m}$ is both a common left multiple and a common right multiple of $a$ and $b$ ; more precisely, if $a ≼_{L} Δ^{k}$ and $b ≼_{L} Δ^{k^{'}}$ , then $a ≼_{L} Δ^{m}$ and $b ≼_{L} Δ^{m}$ for every $m \geq \max (k, k^{'})$ , and the analogous statement holds for $≼_{R}$ . In particular every pair of positive braids admits a common right multiple, so the right complement $Θ$ of Artin right complements and word reversing is defined on every pair of positive words (Artin positive word reversing is complete).

For $n \leq 1$ the monoid is trivial, $Δ = 1$ , and the assertions hold with $k = k^{'} = m = 0$ . The proof is effective in the sense that a dividing power is produced by reading a word for $a$ from left to right; no search over words is performed and no choice principle is used.

Facts & Assumptions

Given: A natural number $n$ , the monoid $B_{n}^{+}$ with its atoms $σ_{i}$ and length $ℓ$ , the half twist $Δ$ with blocks $T_{k}$ and the index-reversal automorphism $τ$ ( $σ_{j} \mapsto σ_{n - j}$ ), and the orders $≼_{L}, ≼_{R}$ .

[F1]

$B_{n}^{+}$ is generated as a monoid by the atoms; $ℓ$ is additive, $ℓ (x) = 0$ only for $x = 1$ , and $ℓ (σ_{i}) = 1$ (Positive braid monoid, Positive artin relations preserve homogeneous length).

[F2]

$a ≼_{L} b ⟺ \exists c (b = a c)$ and $a ≼_{R} b ⟺ \exists c (b = c a)$ ; both relations are partial orders, the left order is preserved by left multiplication and the right order by right multiplication, and each divisibility witness is unique by cancellation (Left and right divisibility for positive braids, The positive braid monoid is left and right cancellative).

[F3]

The half twist identities (Conjugation by the half twist reverses Artin generators, The Garside half twist and simple positive braids): $Δ w \equiv^{+} τ (w) Δ$ and $w Δ \equiv^{+} Δ τ (w)$ for every positive word $w$ , where $τ$ is the index-reversal automorphism $σ_{j} \mapsto σ_{n - j}$ ; $τ (Δ) = Δ$ ; and $τ$ is an automorphism of $B_{n}^{+}$ because it permutes the defining relations.

[F4]

Atoms divide the half twist (Each Artin atom is a left and right divisor of the half twist): for every $i$ there is $R_{i} \in B_{n}^{+}$ with $Δ = σ_{i} R_{i}$ ; in particular $Δ = σ_{i} R_{i}$ holds for every atom and every $n \geq 1$ for which the atom exists (for $n \leq 1$ there is no atom and $Δ = 1$ ).

[F5]

Reversal (The positive braid monoid is left and right cancellative): the word reversal $w \mapsto w^{r e v}$ induces an involutive anti-automorphism $ρ$ of $B_{n}^{+}$ with $ρ (x y) = ρ (y) ρ (x)$ and $ρ (σ_{j}) = σ_{j}$ ; it satisfies $ρ (Δ) = Δ$ (Conjugation by the half twist reverses Artin generators), and it exchanges the two divisibility orders: $a ≼_{L} b ⟺ ρ (a) ≼_{R} ρ (b)$ and $a ≼_{R} b ⟺ ρ (a) ≼_{L} ρ (b)$ (Left and right divisibility for positive braids).

Proof

technique · direct

1.1

The extension step. Let $a \in B_{n}^{+}$ , $k \in N$ with $a ≼_{L} Δ^{k}$ , and let $σ_{i}$ be an atom; write $Δ^{k} = a c$ with $c \in B_{n}^{+}$ . Then associativity and the mirror identity [F3] for the positive word $c$ , followed by the atom factorization [F4], give the chain of equalities $Δ^{k + 1} = Δ^{k} Δ = a (c Δ) = a (Δ τ (c)) = a (σ_{i} R_{i} τ (c))$ , whose last factor $σ_{i} R_{i} τ (c)$ lies in $B_{n}^{+}$ . Hence $a σ_{i} ≼_{L} Δ^{k + 1}$ .

F1F2F3F4

2.1

Induction along a word. Every element $a \in B_{n}^{+}$ is the class of a positive word $w$ , and we prove by induction on $∣ w ∣$ that $[w] ≼_{L} Δ^{k}$ for some $k \in N$ : for $w = ε$ we have $[ε] = 1 = Δ^{0}$ , and if $w = w^{'} σ_{i}$ with $[w^{'}] ≼_{L} Δ^{k}$ then $[w] = [w^{'}] σ_{i} ≼_{L} Δ^{k + 1}$ by step 1.1. This proves (a).

F1F2step 1.1

3.1

The right-hand version. Let $a \in B_{n}^{+}$ and apply step 2.1 to $ρ (a)$ : there is $k$ with $ρ (a) ≼_{L} Δ^{k}$ , say $Δ^{k} = ρ (a) c$ with $c \in B_{n}^{+}$ . Applying the anti-automorphism $ρ$ and using $ρ (Δ) = Δ$ , $ρ \circ ρ = i d$ and $ρ (x y) = ρ (y) ρ (x)$ [F5] gives $Δ^{k} = ρ (Δ^{k}) = ρ (c) ρ (ρ (a)) = ρ (c) a$ , so $a ≼_{R} Δ^{k}$ . This is (b).

F2F5step 2.1

4.1

Common multiples. Let $a, b \in B_{n}^{+}$ . By (a) and (b), choose four exponents $k_{a}, k_{b}, r_{a}, r_{b}$ such that $a ≼_{L} Δ^{k_{a}}$ , $b ≼_{L} Δ^{k_{b}}$ , $a ≼_{R} Δ^{r_{a}}$ and $b ≼_{R} Δ^{r_{b}}$ , and put $m : = \max (k_{a}, k_{b}, r_{a}, r_{b})$ . If $Δ^{k_{a}} = a c$ , then $Δ^{m} = Δ^{k_{a}} Δ^{m - k_{a}} = a (c Δ^{m - k_{a}})$ exhibits $a ≼_{L} Δ^{m}$ , and the same computation applies to $b$ . If $Δ^{r_{a}} = c^{'} a$ , then $Δ^{m} = Δ^{m - r_{a}} Δ^{r_{a}} = (Δ^{m - r_{a}} c^{'}) a$ exhibits $a ≼_{R} Δ^{m}$ , and likewise for $b$ . Thus the same power is a common multiple on both sides.

F1F2step 2.1step 3.1

5.1

Totality of the right complement. If $u, v$ are positive words then $[u]$ and $[v]$ admit the common right multiple $Δ^{m}$ produced in step 4.1. By the conditional termination criterion of Artin positive word reversing is complete, right-reversing of $u^{- 1} v$ therefore reaches a terminal positive--negative path. Reversing, say, the leftmost negative--positive adjacent pair at each stage gives a fixed finite algorithm for its terminal complement pair $Θ (u, v), Θ (v, u)$ ; the right-complemented uniqueness lemma makes the output independent of that fixed schedule. Thus $Θ$ is total for this Artin presentation. Termination follows from the explicit common $Δ$ power and the conditional criterion, not from any bound by the input-word length.

F2step 4.1

6.1

Assembly. Part (a) is step 2.1, part (b) is step 3.1, and part (c) is step 4.1 together with step 5.1; for $n \leq 1$ there are no atoms, $Δ = 1$ and $k = k^{'} = m = 0$ work. Every induction is on the length of an explicit word, all products are finite, and no inverse, no group and no choice principle occur. ∎

step 2.1step 3.1step 4.1step 5.1

Remarks

Why the induction multiplies on the right. Step 1.1 appends the atom $σ_{i}$ to $a$ on the right and increases the power of $Δ$ by one; the mechanism is that $Δ$ commutes with every element up to the index-reversal automorphism $τ$ (that is the content of $Δ w = τ (w) Δ$ ), and that $Δ$ itself begins with any prescribed atom $σ_{i}$ with complement $R_{i}$ . The mirror identity is used exactly once in step 1.1, for the word $c$ , and the atom factorization is used once, for the atom through which the new letter enters. Comparing with GM Section 4, this is the sentence "by induction on the length, for every $a \in B_{n}^{+}$ one has $a ≼ Δ^{m}$ and $Δ^{m} ≽ a$ for some $m$ ".
What is not used. The least common multiple theorem (Positive braids have left and right gcds and lcms) is not used; only the conditional direction "a common right multiple exists $\Rightarrow$ the reversing of the pair terminates" of Artin positive word reversing is complete enters, in step 5.1, and it is used only to record that the common multiples produced here are the ones that make right-reversing total. In particular the argument is not circular: it produces common multiples of a very special shape before any general lcm theory is available.
Conventions. For $n = 0, 1$ the notation $Δ^{k}$ for $k = 0$ is $1$ and no atom occurs; the statements of (a) and (b) are then satisfied by $k = k^{'} = 0$ . For $n = 2$ every positive braid is a power of the single atom $σ_{1} = Δ$ , so the dividing power is $k = ℓ (a)$ .
Nothing here uses a choice principle: the word induction is finite and the exponents are natural numbers computed from a word for $a$ .

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4, printed p. 28 (a ≼ Δ^m and Δ^m ≽ a for some m) (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter IX, Section 1.3, printed pp. 439-440 (standard reference, not scraped)