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Left garside normal form is unique

Statement

Let $n \geq 2$ , let $B_{n}$ be the braid group of The braid group by Artin presentation, identified with the group of fractions of the positive braid monoid $B_{n}^{+}$ of Positive braid monoid by The group of fractions of the positive braid monoid is the Artin braid group, so that $B_{n}^{+}$ is a submonoid of $B_{n}$ ; let $Δ$ be the half twist of The Garside half twist and simple positive braids, and let $≼_{L}$ denote the group order of Left and right divisibility extend to lattice orders on the braid group, defined by $x ≼_{L} y ⟺ x^{- 1} y \in B_{n}^{+}$ . Recall that a simple braid is a left divisor of $Δ$ in $B_{n}^{+}$ (The Garside half twist and simple positive braids), and call a simple braid $a$ proper if $1 ≺_{L} a ≺_{L} Δ$ . Then:

(a) Maximal $Δ$ -exponent. For every $x \in B_{n}$ the set $P (x) : = {p \in Z : Δ^{p} ≼_{L} x}$ is nonempty and bounded above. Writing $p (x) : = \max P (x)$ and $A (x) : = Δ^{- p (x)} x$ , one has $A (x) \in B_{n}^{+}$ and $Δ ̸ ≼_{L} A (x)$ . Moreover $x = Δ^{p} A$ with $p \in Z$ , $A \in B_{n}^{+}$ , $Δ ̸ ≼_{L} A$ holds for exactly one pair $(p, A)$ , namely $(p (x), A (x))$ .

(b) The greedy factorisation. Let $A = A (x) = A_{0} \in B_{n}^{+}$ . If $A = 1$ put $r : = 0$ ; otherwise define, as long as $A_{i - 1} \neq 1$ , $a_{i} : = Δ \land_{L} A_{i - 1} \in B_{n}^{+}, A_{i - 1} = a_{i} A_{i} with A_{i} \in B_{n}^{+},$ the factor $A_{i}$ being unique. Then there is an $r \geq 1$ with $A_{r} = 1$ , and for every $1 \leq i \leq r$ : $a_{i}$ is a proper simple braid, $A_{i - 1} = a_{i} a_{i + 1} \dots a_{r}$ , and $Δ ̸ ≼_{L} A_{i}$ . In particular $A (x) = a_{1} a_{2} \dots a_{r}$ , and each $a_{i}$ is reduced in the sense of Simple positive braids are indexed by permutations: $a_{i} = \hat{σ}$ for a unique $σ \in S_{n} ∖ {i d, w_{0}}$ , and $ℓ (a_{i}) = inv (π (a_{i}))$ .

(c) Left normal form. Every $x \in B_{n}$ has a unique expression $x = Δ^{p} a_{1} a_{2} \dots a_{r}$ with $p \in Z$ , $r \in N$ , every $a_{i}$ a proper simple braid, and $a_{i} = Δ \land_{L} (a_{i} a_{i + 1} \dots a_{r}) (1 \leq i \leq r) .$ In such an expression necessarily $p = p (x)$ , $a_{1} \dots a_{r} = A (x)$ , $a_{1} = Δ \land_{L} A (x)$ , and $a_{i} = Δ \land_{L} ((a_{1} \dots a_{i - 1})^{- 1} A (x))$ for $i \geq 2$ . This is the left normal form of $x$ .

(d) Specialisations. In the left normal form of $x$ one has: (i) $r = 0$ if and only if $x \in ⟨ Δ ⟩ = {Δ^{p} : p \in Z}$ ; (ii) $p (x) \geq 0$ if and only if $x \in B_{n}^{+}$ ; thus a positive braid has normal form $x = Δ^{p} a_{1} \dots a_{r}$ with $p = p (x) \geq 0$ , and $p = 0$ exactly when $Δ ̸ ≼_{L} x$ , so for instance $p (x) = 0$ always holds at $x = 1$ (where $r = 0$ ); (iii) for $n = 2$ there is no proper simple braid at all, and the left normal form of every $x \in B_{2}$ is $Δ^{p (x)}$ with $r = 0$ .

(e) Left weighting. If $x = Δ^{p} a_{1} \dots a_{r}$ is the left normal form of (c) and $i < r$ , then $(a_{i} a_{i + 1}) \land_{L} Δ = a_{i}$ .

No choice principle is used: the exponent $p (x)$ is obtained from an explicitly rewritten word, and all minima and maxima that occur are taken over nonempty subsets of $Z$ or over finite sets of positive words, for which the elementary well-ordering and induction principles suffice.

Facts & Assumptions

Given: A natural number $n \geq 2$ , the monoid $B_{n}^{+}$ with atoms $σ_{1}, \dots, σ_{n - 1}$ , length $ℓ$ , divisibility orders and half twist $Δ$ of length $N = n (n - 1) / 2$ , and the braid group $B_{n}$ with its group order $≼_{L}$ .

[F1]

$B_{n}^{+}$ has the atoms as generators, $ℓ ([w]) = ∣ w ∣$ , $ℓ$ is additive, $ℓ (z) = 0$ forces $z = 1$ , and $ℓ (Δ^{m}) = m N$ for $m \geq 0$ (Positive braid monoid, Positive artin relations preserve homogeneous length, The Garside half twist and simple positive braids). The order $a ≼_{L} b$ on $B_{n}^{+}$ means $b = a c$ for some $c \in B_{n}^{+}$ , with unique witness $c$ (Left and right divisibility for positive braids).

[F2]

For every atom $σ_{i}$ there is $R_{i} \in B_{n}^{+}$ with $Δ = σ_{i} R_{i}$ , and $ℓ (R_{i}) = N - 1$ ; hence $σ_{i}^{- 1} = R_{i} Δ^{- 1}$ holds in the group and $σ_{i} ≼_{L} Δ$ (Each Artin atom is a left and right divisor of the half twist). Moreover $σ_{j} Δ = Δ σ_{n - j}$ for all $j$ , and more generally $w Δ = Δ τ (w)$ for every positive word $w$ , where $τ$ is the involutive automorphism of $B_{n}^{+}$ induced by $σ_{j} \mapsto σ_{n - j}$ (Conjugation by the half twist reverses Artin generators).

[F3]

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

[F4]

Meets and joins in the positive monoid. Every nonempty finite subset of $B_{n}^{+}$ has a left-gcd $\land_{L}$ and a left-lcm, unique, and a common left divisor of the family divides the gcd; in particular $Δ \land_{L} A$ exists for every $A \in B_{n}^{+}$ (Positive braids have left and right gcds and lcms).

[F5]

Passage to the group. $B_{n}^{+}$ is a submonoid of $B_{n}$ and, on positive elements, the group order $≼_{L}$ of Left and right divisibility extend to lattice orders on the braid group agrees with the monoid order: for $a, b \in B_{n}^{+}$ , $a ≼_{L} b$ in $B_{n}$ if and only if $b = a c$ with $c \in B_{n}^{+}$ ; the group order is defined by $x ≼_{L} y ⟺ x^{- 1} y \in B_{n}^{+}$ (The group of fractions of the positive braid monoid is the Artin braid group, Left and right divisibility extend to lattice orders on the braid group).

[F6]

Proper simple braids are reduced. An element $a \in B_{n}^{+}$ satisfies $1 ≺_{L} a ≺_{L} Δ$ if and only if $a = \hat{σ}$ for a unique $σ \in S_{n} ∖ {i d, w_{0}}$ , and then $ℓ (a) = inv (π (a))$ (Simple positive braids are indexed by permutations).

Proof

technique · direct

1.1

Every element is a $Δ$ -power times a positive braid. Let $x \in B_{n}$ and let $σ_{i_{1}}^{\pm 1} \dots σ_{i_{k}}^{\pm 1}$ be a word representing it. Replacing every negative letter by $σ_{i}^{- 1} = R_{i} Δ^{- 1}$ with $R_{i}$ as in [F2] turns it into a product of positive letters and of symbols $Δ^{- 1}$ . From $w Δ = Δ τ (w)$ and $τ^{2} = i d$ [F2] one obtains $w Δ^{- 1} = Δ^{- 1} τ (w)$ for every positive $w$ ; this identity moves each $Δ^{- 1}$ to the left past positive letters. Since $τ$ preserves positivity, induction on the number of $Δ^{- 1}$ symbols rewrites $x$ as $x = Δ^{- j} A$ with $j \geq 0$ and $A \in B_{n}^{+}$ . Hence $P (x) \neq \emptyset$ , since it contains $- j$ .

F1F2F5

1.2

The greedy step (b). Suppose $A_{i - 1} \in B_{n}^{+}$ with $A_{i - 1} \neq 1$ . Choosing a positive word $w$ with $[w] = A_{i - 1}$ and $∣ w ∣ = ℓ (A_{i - 1}) \geq 1$ , its first letter is an atom $σ_{j}$ with $σ_{j} ≼_{L} A_{i - 1}$ ; by [F2] $σ_{j} ≼_{L} Δ$ as well. Hence the gcd $a_{i} : = Δ \land_{L} A_{i - 1}$ [F4] satisfies $σ_{j} ≼_{L} a_{i}$ , so $a_{i} \neq 1$ . Since $a_{i} ≼_{L} A_{i - 1}$ while $Δ ̸ ≼_{L} A_{i - 1}$ , we have $a_{i} \neq Δ$ . So $1 ≺_{L} a_{i} ≺_{L} Δ$ : $a_{i}$ is a proper simple braid, and by [F6] $a_{i} = \hat{σ}$ for a unique $σ \in S_{n} ∖ {i d, w_{0}}$ with $ℓ (a_{i}) = inv (π (a_{i}))$ . Since $a_{i} ≼_{L} A_{i - 1}$ there is $A_{i} \in B_{n}^{+}$ with $A_{i - 1} = a_{i} A_{i}$ , unique by left cancellation [F3], and it satisfies $ℓ (A_{i}) = ℓ (A_{i - 1}) - ℓ (a_{i}) < ℓ (A_{i - 1})$ because $ℓ (a_{i}) \geq 1$ .

F1F2F3F4F6

2.1

The maximal exponent (a). $P (x)$ is downward closed: if $Δ^{- p} x = c \in B_{n}^{+}$ and $p^{'} \leq p$ , then $Δ^{- p^{'}} x = Δ^{p - p^{'}} c \in B_{n}^{+}$ . To bound $P (x)$ above, fix one decomposition $x = Δ^{- j} A$ with $A \in B_{n}^{+}$ and let $p \in P (x)$ , so $x = Δ^{p} c$ with $c \in B_{n}^{+}$ ; then $A = Δ^{p + j} c$ . If $p + j > 0$ additivity and [F1] give $ℓ (A) = (p + j) N + ℓ (c) \geq (p + j) N$ , while if $p + j \leq 0$ then $p \leq - j \leq ℓ (A) / N - j$ because $ℓ (A) \geq 0$ ; in both cases $p \leq ℓ (A) / N - j$ , so $P (x)$ is bounded above. Hence $p (x) : = \max P (x)$ exists by the well-ordering of the nonempty bounded-above subset $P (x) \subseteq Z$ , and $A (x) : = Δ^{- p (x)} x \in B_{n}^{+}$ by membership in $P (x)$ . If $Δ ≼_{L} A (x)$ , then $Δ^{- (p (x) + 1)} x = Δ^{- 1} A (x) \in B_{n}^{+}$ , i.e. $p (x) + 1 \in P (x)$ , contradicting maximality; hence $Δ ̸ ≼_{L} A (x)$ . Finally, if $x = Δ^{p} A = Δ^{p^{'}} A^{'}$ with $A, A^{'} \in B_{n}^{+}$ , $Δ ̸ ≼_{L} A$ , $Δ ̸ ≼_{L} A^{'}$ and $p < p^{'}$ , then $A = Δ^{p^{'} - p} A^{'} \in Δ B_{n}^{+}$ , i.e. $Δ ≼_{L} A$ , a contradiction; so $p = p^{'}$ and then $A = A^{'}$ .

F1F5step 1.1

3.1

The invariant $Δ ̸ ≼_{L} A_{i}$ . Suppose $A_{i - 1} \neq 1$ , $A_{i - 1} = a_{i} A_{i}$ and $Δ ̸ ≼_{L} A_{i - 1}$ ; assume for contradiction $A_{i} = Δ c$ with $c \in B_{n}^{+}$ . Then, using $a_{i} Δ = Δ τ (a_{i})$ with $τ (a_{i}) \in B_{n}^{+}$ [F2], $A_{i - 1} = a_{i} Δ c = Δ τ (a_{i}) c \in Δ B_{n}^{+}$ , i.e. $Δ ≼_{L} A_{i - 1}$ , contradiction. Hence $Δ ̸ ≼_{L} A_{i}$ , and induction on $i$ gives $Δ ̸ ≼_{L} A_{i}$ for all $i \geq 0$ starting from $Δ ̸ ≼_{L} A_{0} = A (x)$ of [step 2.1]. Consequently the recursion never produces $a_{i + 1} = Δ$ : if $A_{i} \neq 1$ then $a_{i + 1} = Δ \land_{L} A_{i} ≼_{L} A_{i}$ while $Δ ̸ ≼_{L} A_{i}$ .

F2F4step 2.1

4.1

Termination and the factorisation (b). The recursion of step 1.2 either stops at $A_{i} = 1$ or produces a strictly decreasing sequence $ℓ (A_{0}) > ℓ (A_{1}) > \dots$ in $N$ , which cannot be infinite; so there is a least $r \geq 0$ with $A_{r} = 1$ . If $r \geq 1$ , then $A_{i - 1} = a_{i} A_{i}$ for $i = 1, \dots, r$ , and unfolding the recursion gives $A_{i - 1} = a_{i} a_{i + 1} \dots a_{r}$ for every $1 \leq i \leq r$ ; at $i = 1$ this is $A (x) = a_{1} \dots a_{r}$ . By step 1.2 each $a_{i}$ is a proper simple braid and by step 3.1 each $Δ ̸ ≼_{L} A_{i}$ , and [F6] gives the reduced-lift description of the $a_{i}$ stated in (b).

F1F3step 1.2step 3.1

5.1

Uniqueness of the left normal form (c). Let $x = Δ^{p} a_{1} \dots a_{r} = Δ^{p^{'}} b_{1} \dots b_{s}$ be two decompositions as in (c), and put $A : = a_{1} \dots a_{r}$ , $B : = b_{1} \dots b_{s}$ . If $r = 0$ , then $A = 1$ and $Δ ̸ ≼_{L} A$ , since $ℓ (Δ) = N > 0$ . If $r \geq 1$ and $Δ ≼_{L} A$ , then $Δ$ is a common left divisor of $Δ$ and $A$ , so $Δ ≼_{L} Δ \land_{L} A = a_{1}$ ; because $a_{1}$ is simple, $a_{1} ≼_{L} Δ$ as well, and antisymmetry gives $a_{1} = Δ$ , contradicting properness. Thus $Δ ̸ ≼_{L} A$ in either case, and likewise $Δ ̸ ≼_{L} B$ . By the uniqueness in (a), proved in step 2.1, we get $p = p^{'}$ and $A = B$ . If $A = 1$ , additivity of positive length and $ℓ (a_{i}), ℓ (b_{j}) \geq 1$ force $r = s = 0$ , so the two lists agree. Otherwise $r, s \geq 1$ , and their greedy conditions give $a_{1} = Δ \land_{L} A = Δ \land_{L} B = b_{1}$ . Cancelling $a_{1} = b_{1}$ [F3] gives $a_{2} \dots a_{r} = b_{2} \dots b_{s}$ . The defining greedy conditions pass unchanged to these tails, so induction on their length gives $r = s$ and $a_{i} = b_{i}$ for every $i$ . The identified $p$ and $A$ are $p (x)$ and $A (x)$ by step 2.1; when $r \geq 1$ , the factor identities $a_{1} = Δ \land_{L} A (x)$ and $a_{i} = Δ \land_{L} (a_{i} \dots a_{r}) = Δ \land_{L} ((a_{1} \dots a_{i - 1})^{- 1} A (x))$ follow from the defining conditions. Existence is step 4.1.

F1F3F4step 2.1step 4.1

6.1

Specialisations (d) and left weighting (e). (i): $r = 0$ means $A (x) = 1$ , i.e. $x = Δ^{p (x)} \in ⟨ Δ ⟩$ ; conversely if $x = Δ^{p}$ then $P (x) = {q : q \leq p}$ has maximum $p$ and $A (x) = 1$ . (ii): if $p (x) \geq 0$ then $x = Δ^{p (x)} A (x)$ is a product of positive elements, so $x \in B_{n}^{+}$ ; conversely if $x \in B_{n}^{+}$ then $Δ^{0} = 1 ≼_{L} x$ by [F5], so $0 \in P (x)$ and $p (x) \geq 0$ . (iii): for $n = 2$ the divisors of $Δ = σ_{1}$ have length $\leq 1$ , hence are $1$ and $Δ$ by [F1], so there is no proper simple braid and the normal form forced by (c) has $r = 0$ . (e): if $s$ is a common left divisor of $a_{i} a_{i + 1}$ and $Δ$ , then $s ≼_{L} a_{i} a_{i + 1} ≼_{L} a_{i} \dots a_{r}$ , so $s$ is a common left divisor of $a_{i} \dots a_{r}$ and $Δ$ , whence $s ≼_{L} Δ \land_{L} (a_{i} \dots a_{r}) = a_{i}$ ; therefore $a_{i}$ is the greatest common left divisor of $a_{i} a_{i + 1}$ and $Δ$ .

F1F4F5step 5.1

7.1

Assembly. Part (a) is step 2.1, part (b) is steps 1.2, 3.1 and 4.1, part (c) is steps 4.1 and 5.1, part (d) is step 6.1 and part (e) is step 6.1. The exponent extraction of step 1.1 uses only the atom factors $σ_{i}^{- 1} = R_{i} Δ^{- 1}$ and the index-reversal sliding $w Δ = Δ τ (w)$ ; the greedy recursion uses the left-gcd of the positive lattice, which is unconditional by [F4], and no appeal to the $Δ$ -divisibility of an arbitrary positive braid is made. For $n = 2$ the theorem reduces to the statement that every element of $B_{2}$ is a power of $Δ = σ_{1}$ , in accordance with the free-group description of $B_{2}$ ; the empty factor case $r = 0$ is the case $x \in ⟨ Δ ⟩$ . No choice principle is used anywhere. ∎

step 1.1step 2.1step 1.2step 3.1step 4.1step 5.1step 6.1

Remarks

Comparison with the source. The statement is the left normal form of Garside--Elrifai--Morton as presented in J. González-Meneses, Basic results on braid groups, Section 4.1, printed pp. 29--30: the source defines $A$ by maximality of $p$ , then sets $a_{1} = A \land Δ$ and $a_{i} = (a_{i - 1}^{- 1} \dots a_{1}^{- 1} A) \land Δ$ , and records the left-weighting $(a_{i} a_{i + 1}) \land Δ = a_{i}$ . Here the characterisation $a_{i} = Δ \land_{L} (a_{i} \dots a_{r})$ is used as the defining condition of the normal form, which is exactly what the greedy recursion produces; it implies the source's adjacent-pair weighting as part (e).
What uniqueness rests on. Only the uniqueness of the pair $(p, A)$ (pure positivity of $Δ$ -powers) and the determinism of the gcd $A \land_{L} Δ$ are used; no confluence property of a rewriting system and no injectivity of a geometric braid model is invoked.
Effective content. Every step of the recursion is a finite operation once the left-gcd $A_{i - 1} \land_{L} Δ$ is computable, and the first step (rewriting inverses to the left) uses the explicit factors $R_{i}$ of [F2]; the resulting algorithm is the subject of The braid group word problem is decidable by garside normal form.
Nothing here uses the Axiom of Choice or any weaker choice principle; the only maximum taken is that of a nonempty bounded-above set of integers.

Depends on

Used by

The braid group word problem is decidable by garside normal form Corollary
Exponent sum is not a complete braid normal form Counterexample
A left garside normal form computation in b three Example
A central positive braid is a power of delta squared for n greater than two Lemma

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4.1, printed pp. 28-30 (standard reference, not scraped)
J. Birman and T. Brendle, Braids: A Survey, Section 5.1 (standard reference, not scraped)