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Each Artin atom is a left and right divisor of the half twist

Statement

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

(a) Left divisibility. $σ_{i} ≼_{L} Δ$ , that is, there is $R_{i} \in B_{n}^{+}$ with $σ_{i} R_{i} = Δ$ .

(b) Right divisibility. $σ_{i} ≼_{R} Δ$ , that is, there is $L_{i} \in B_{n}^{+}$ with $L_{i} σ_{i} = Δ$ .

(c) Uniqueness and length. The complement $R_{i}$ of (a) and the complement $L_{i}$ of (b) are unique, and $ℓ (R_{i}) = ℓ (L_{i}) = N - 1$ ; in particular $R_{i} = 1$ holds if and only if $n = 2$ and $i = 1$ , and likewise for $L_{i}$ .

For $n \leq 1$ the alphabet is empty, $Δ = 1$ and the assertions are vacuous. The proof is effective: it exhibits $R_{i}$ and $L_{i}$ as classes of explicit positive words built from the recursion $Δ_{m} = Δ_{m - 1} T_{m - 1}$ and the sliding identity $σ_{j} T_{k} \equiv^{+} T_{k} σ_{j + 1}$ , and it does not use the future least-common-multiple theorem (Positive braids have left and right gcds and lcms). No choice principle is used.

Facts & Assumptions

Given: A natural number $n$ , the monoid $B_{n}^{+}$ with its atoms $σ_{i}$ , homogeneous length $ℓ$ , and the half twist $Δ = T_{1} \dots T_{n - 1}$ with blocks $T_{k} = σ_{k} σ_{k - 1} \dots σ_{1}$ .

[F1]

$B_{n}^{+} = Σ_{n}^{*} / ⁣ \equiv^{+}$ with $[u v] = [u] [v]$ , generated as a monoid by the ${\overline{σ}}_{i} = σ_{i}$ ; $ℓ ([w]) = ∣ w ∣$ is additive, $ℓ (x) = 0$ only for $x = 1$ , and $ℓ (σ_{i}) = 1$ (Positive braid monoid, Positive artin relations preserve homogeneous length).

[F2]

Sub-alphabet compatibility. Every defining pair of $R_{n - 1}$ is a defining pair of $R_{n}$ , because the pairs are indexed by relations on adjacent or distant indices and the index ranges for $n - 1$ are contained in those for $n$ . Hence the universal property of $B_{n - 1}^{+}$ (Positive braid monoid) gives a monoid homomorphism $B_{n - 1}^{+} \to B_{n}^{+}$ carrying the class of a word over ${σ_{1}, \dots, σ_{n - 2}}$ to its class in $B_{n}^{+}$ ; in particular the half twist $Δ_{n - 1} = T_{1} \dots T_{n - 2}$ of $B_{n - 1}^{+}$ maps to the class of $T_{1} \dots T_{n - 2}$ in $B_{n}^{+}$ , which is the element denoted $Δ_{n - 1}$ there (The Garside half twist and simple positive braids).

[F3]

Divisibility. $a ≼_{L} b ⟺ \exists c (b = a c)$ and $a ≼_{R} b ⟺ \exists c (b = c a)$ ; the witness is unique by cancellation, and $ℓ$ is additive over the witness (Left and right divisibility for positive braids, The positive braid monoid is left and right cancellative).

[F4]

The half twist and its identities (The Garside half twist and simple positive braids, Conjugation by the half twist reverses Artin generators): $Δ_{n} = Δ_{n - 1} T_{n - 1}$ for $n \geq 2$ , $ℓ (Δ) = N$ , the conjugation identity $σ_{i} Δ \equiv^{+} Δ σ_{n - i}$ holds for every $i$ , and the sliding identity $σ_{j} T_{k} \equiv^{+} T_{k} σ_{j + 1}$ holds whenever $1 \leq j \leq k - 1$ . Both are proved in the positive monoid, without inverting anything.

[F5]

Cancellation (The positive braid monoid is left and right cancellative): $x a = x b \Rightarrow a = b$ and $a x = b x \Rightarrow a = b$ in $B_{n}^{+}$ .

Proof

technique · direct

1.1

The atom $σ_{1}$ is a right divisor of $Δ$ . In the word $T_{1} T_{2} \dots T_{n - 1}$ the last letter is $σ_{1}$ , because $T_{n - 1} = σ_{n - 1} σ_{n - 2} \dots σ_{1}$ ends with $σ_{1}$ . Hence, putting $L^{(1)} : = [T_{1} \dots T_{n - 2} σ_{n - 1} σ_{n - 2} \dots σ_{2}]$ , the associativity of concatenation and multiplicativity of the quotient product give $Δ = [T_{1} \dots T_{n - 1}] = L^{(1)} σ_{1}$ , so $σ_{1} ≼_{R} Δ$ ; here $L^{(1)}$ is the empty product $1$ when $n = 2$ . Its length is $ℓ (L^{(1)}) = N - 1$ by additivity.

F1F3F4

2.1

Induction on the number of strands: every atom is a right divisor. We prove for every $m \geq 1$ : for every $1 \leq j \leq m - 1$ , the atom $σ_{j}$ of $B_{m}^{+}$ satisfies $σ_{j} ≼_{R} Δ_{m}$ . For $m = 1$ the range is empty. Assume the claim for $m - 1$ , where $m \geq 2$ , and let $1 \leq j \leq m - 1$ . If $j = 1$ , step 1.1 with $n = m$ gives the assertion. If $j \geq 2$ , then $j - 1 \leq m - 2$ , so the induction hypothesis in the sub-alphabet ${σ_{1}, \dots, σ_{m - 2}}$ gives $Δ_{m - 1} = L^{'} σ_{j - 1}$ for some $L^{'} \in B_{m - 1}^{+}$ , viewed inside $B_{m}^{+}$ by [F2]. Multiplying by $T_{m - 1}$ and using the recursion $Δ_{m} = Δ_{m - 1} T_{m - 1}$ gives $Δ_{m} = L^{'} σ_{j - 1} T_{m - 1} \equiv^{+} L^{'} T_{m - 1} σ_{j}$ , the last step by the sliding identity [F4] with $k = m - 1$ and $j - 1 \leq k - 1 = m - 2$ , which is exactly the hypothesis $j \leq m - 1$ . Since $L^{'} T_{m - 1} \in B_{m}^{+}$ , this says $σ_{j} ≼_{R} Δ_{m}$ , completing the induction.

F1F2F3F4step 1.1

3.1

Left divisors from right divisors. Fix $i \in {1, \dots, n - 1}$ . Since $n - i \in {1, \dots, n - 1}$ , step 2.1 with $m = n$ and $j = n - i$ gives $Δ = L σ_{n - i}$ for some $L \in B_{n}^{+}$ . The conjugation identity [F4] gives $σ_{i} Δ \equiv^{+} Δ σ_{n - i} = L σ_{n - i} σ_{n - i}$ , while associativity gives $σ_{i} Δ = σ_{i} (L σ_{n - i}) = (σ_{i} L) σ_{n - i}$ . Therefore $(σ_{i} L) σ_{n - i} = L (σ_{n - i} σ_{n - i})$ , and right cancellation [F5] yields $σ_{i} L = L σ_{n - i}$ . Substituting back, $Δ = L σ_{n - i} = σ_{i} L$ , so $σ_{i} ≼_{L} Δ$ with complement $L$ .

F1F3F4F5step 2.1

4.1

Uniqueness, length and the degenerate cases. Left cancellation [F5] gives the uniqueness of $R_{i}$ in $Δ = σ_{i} R_{i}$ and right cancellation gives the uniqueness of $L_{i}$ in $Δ = L_{i} σ_{i}$ : if $σ_{i} R_{i} = σ_{i} R_{i}^{'}$ then $R_{i} = R_{i}^{'}$ , and if $L_{i} σ_{i} = L_{i}^{'} σ_{i}$ then $L_{i} = L_{i}^{'}$ . For the lengths, additivity of $ℓ$ [F1, F3] and $ℓ (σ_{i}) = 1$ give $ℓ (Δ) = ℓ (σ_{i}) + ℓ (R_{i}) = 1 + ℓ (R_{i})$ and likewise for $L_{i}$ , so $ℓ (R_{i}) = ℓ (L_{i}) = N - 1$ ; and $R_{i} = 1$ happens if and only if $N - 1 = 0$ , that is $N = 1$ , that is $n = 2$ , in which case $i = 1$ . For $n \leq 1$ there is no $i$ in the range and $Δ = 1$ , so the assertions are vacuous.

F1F3F4F5step 1.1step 3.1

5.1

Assembly. Parts (a) and (b) are steps 3.1 and 2.1 respectively (the left divisors being transported from the right divisors by the conjugation identity), and part (c) is step 4.1. The proof never invokes a least common multiple, only the displayed recursion, the sliding identity and cancellation; all inductions are on natural numbers and all arguments are finite, so no choice principle is used. ∎

step 2.1step 3.1step 4.1

Remarks

What the construction exhibits. Combining the steps, the complements are the words obtained by the recursive recipe of step 2.1: the right complement of $σ_{j}$ in $Δ_{m}$ is (up to the sub-alphabet inclusion) the word $L^{'} T_{m - 1}$ whose factor $L^{'}$ is the right complement of $σ_{j - 1}$ in $Δ_{m - 1}$ , and the descent from $j$ to $j - 1$ is precisely one application of the sliding identity; the base case $j = 1$ is the trivial factorization read off from the last letter of $T_{m - 1}$ . The left complements are then obtained by conjugating indices, $R_{i} = L$ where $Δ = L σ_{n - i}$ . This is the elementary argument of GM Section 4 ("recall that for every $i$ one has $σ_{i} ≼ Δ$ "), made explicit; it is the reason why the later theorem that $Δ$ is the least common multiple of the atoms (Delta is the lcm of the artin atoms and has the same left and right divisors) is not needed here.
The hypothesis $j \leq k - 1$ of the sliding identity is met exactly once. Step 2.1 slides $σ_{j - 1}$ through the block $T_{m - 1}$ , which is legal precisely because $1 \leq j - 1 \leq (m - 1) - 1$ . Sliding the full block index $j - 1 = m - 1$ would, when the next generator exists, compare words with different generator supports and is false; the boundary case $j = m$ is therefore handled by the separate induction hypothesis (and, for $m = 2$ , by the base case $j = 1$ ).
No least common multiple and no group are used. All identities live in the monoid $B_{n}^{+}$ ; the conjugation identity of Conjugation by the half twist reverses Artin generators is the two-sided sliding $σ_{i} Δ = Δ σ_{n - i}$ , not a group conjugation. The complement $R_{i}$ is an element only of $B_{n}^{+}$ , and for $n = 2$ it is $1$ : the one atom of $B_{2}^{+}$ has complements of length $0$ , as $Δ = σ_{1}$ .
Nothing here uses a choice principle: the factorizations are read off from explicit words, and the only induction is on the number of strands.

Depends on

Used by

Dependency tree · two levels

13 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. 27-28 (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter IX, Lemma 1.22 and Section 1.3, printed pp. 438-440 (standard reference, not scraped)