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The group of fractions of the positive braid monoid is the Artin braid group

Statement

Let $n \in N$ , let $B_{n}^{+}$ be the positive braid monoid of Positive braid monoid with its atoms $σ_{1}, \dots, σ_{n - 1}$ , its length $ℓ$ , its half twist $Δ$ (The Garside half twist and simple positive braids) and its cancellation laws (The positive braid monoid is left and right cancellative), and let $B_{n}$ be the Artin braid group of The braid group by Artin presentation, with the same generators $σ_{1}, \dots, σ_{n - 1}$ and the same defining relations. Then:

(a) Ore condition. For all $a, b \in B_{n}^{+}$ there exist $c, d \in B_{n}^{+}$ with $a c = b d$ ; indeed one may take $c, d$ with $a c = b d = Δ^{2 r}$ for some $r \in N$ . Consequently $B_{n}^{+}$ is a cancellative Ore monoid.

(b) The group of fractions. There is a group $G$ together with an injective monoid homomorphism $η ⁣ : B_{n}^{+} \to G$ such that

(i) every element of $G$ has the form $η (a) η (b)^{- 1}$ with $a, b \in B_{n}^{+}$ , and in fact the stronger description $η (a) η (Δ)^{- 2 k}$ with $a \in B_{n}^{+}$ , $k \in N$ holds; (ii) universal property. for every group $H$ and every monoid homomorphism $f ⁣ : B_{n}^{+} \to H$ there is a unique group homomorphism $\overset{ˉ}{f} ⁣ : G \to H$ with $\overset{ˉ}{f} \circ η = f$ .

We call $G$ the group of fractions of $B_{n}^{+}$ ; it is determined up to a unique isomorphism compatible with $η$ .

(c) Identification with the braid group. The assignment $σ_{i} \mapsto η (σ_{i})$ extends to an isomorphism $B_{n} \to G$ . Consequently the canonical map $κ ⁣ : B_{n}^{+} \to B_{n}$ , $σ_{i} \mapsto σ_{i}$ , is an injective monoid homomorphism: $B_{n}^{+}$ is isomorphic to the submonoid of $B_{n}$ consisting of the elements that can be written as positive words, so the two meanings of "positive braid" agree.

No choice principle is used; the group $G$ is an explicit quotient of $B_{n}^{+} \times N$ .

Facts & Assumptions

Given: A natural number $n$ , the monoid $B_{n}^{+}$ with generators $σ_{i}$ , length $ℓ$ , half twist $Δ$ and cancellation laws, and the Artin group $B_{n}$ with its presentation.

[F1]

$B_{n}^{+}$ is generated as a monoid by the $σ_{i}$ , with product $[u] [v] = [u v]$ ; $ℓ (x) = 0$ only for $x = 1$ , so $x y = 1$ forces $x = y = 1$ (Positive braid monoid, Positive artin relations preserve homogeneous length).

[F2]

Cancellation and $Δ$ -powers. $x a = x b \Rightarrow a = b$ and $a x = b x \Rightarrow a = b$ ; every $a \in B_{n}^{+}$ satisfies $a ≼_{L} Δ^{m}$ and $a ≼_{R} Δ^{m}$ for some $m$ , and if $a ≼_{L} Δ^{m}$ then also $a ≼_{L} Δ^{m + 1}$ , since $Δ^{m + 1} = Δ^{m} Δ = a c Δ = a (c Δ)$ when $Δ^{m} = a c$ (The positive braid monoid is left and right cancellative, Every positive braid divides a power of the half twist on both sides).

[F3]

Centrality of $Δ^{2}$ . $w Δ^{2} = Δ^{2} w$ for every positive word $w$ ; hence $Δ^{2}$ commutes with every element of $B_{n}^{+}$ (Conjugation by the half twist reverses Artin generators).

[F4]

The Artin group. $B_{n}$ is the quotient of the free group on $σ_{1}, \dots, σ_{n - 1}$ by the normal closure of the words $σ_{i} σ_{i + 1} σ_{i} (σ_{i + 1} σ_{i} σ_{i + 1})^{- 1}$ and $σ_{i} σ_{j} (σ_{j} σ_{i})^{- 1}$ ; consequently, for any group $H$ and any elements $x_{1}, \dots, x_{n - 1} \in H$ satisfying $x_{i} x_{i + 1} x_{i} = x_{i + 1} x_{i} x_{i + 1}$ and $x_{i} x_{j} = x_{j} x_{i}$ for $∣ i - j ∣ > 1$ , there is a unique group homomorphism $B_{n} \to H$ with $σ_{i} \mapsto x_{i}$ , and $B_{n}$ is generated by the $σ_{i}$ (The braid group by Artin presentation).

[F5]

The monoid universal property. For any monoid $M$ and elements $a_{i} \in M$ satisfying the same relations, there is a unique monoid homomorphism $B_{n}^{+} \to M$ with $σ_{i} \mapsto a_{i}$ (Positive braid monoid).

Proof

technique · direct

1.1

The relation $\approx$ . On the set $P : = B_{n}^{+} \times N$ define $(a, k) \approx (b, l)$ to mean $a Δ^{2 l} = b Δ^{2 k}$ in $B_{n}^{+}$ . This is an equivalence relation: reflexivity and symmetry are immediate from the symmetry of the defining equation, and if $a Δ^{2 l} = b Δ^{2 k}$ and $b Δ^{2 m} = c Δ^{2 l}$ , then $a Δ^{2 l} Δ^{2 m} = b Δ^{2 k} Δ^{2 m} = Δ^{2 k} b Δ^{2 m}$ by [F3], and $b Δ^{2 m} = c Δ^{2 l}$ gives $Δ^{2 k} b Δ^{2 m} = Δ^{2 k} c Δ^{2 l} = c Δ^{2 k} Δ^{2 l}$ ; hence $a Δ^{2 m} Δ^{2 l} = c Δ^{2 k} Δ^{2 l}$ and right cancellation [F2] yields $a Δ^{2 m} = c Δ^{2 k}$ , that is, $(a, k) \approx (c, m)$ .

F1F2F3

1.2

The product is well defined. Put $[a, k] \cdot [b, l] : = [a b, k + l]$ , where $[a, k]$ denotes the $\approx$ -class. If $(a, k) \approx (a^{'}, k^{'})$ , that is $a Δ^{2 k^{'}} = a^{'} Δ^{2 k}$ , then $a b Δ^{2 (k^{'} + l)} = a Δ^{2 k^{'}} b Δ^{2 l} = (a Δ^{2 k^{'}}) b Δ^{2 l} = (a^{'} Δ^{2 k}) b Δ^{2 l} = a^{'} b Δ^{2 (k + l)}$ using [F3] twice, so $(a b, k + l) \approx (a^{'} b, k^{'} + l)$ ; the verification in the second argument is the same computation with the factors interchanged, $a b Δ^{2 (k + l^{'})} = a Δ^{2 k} b Δ^{2 l^{'}} = a Δ^{2 k} b^{'} Δ^{2 l} = a b^{'} Δ^{2 (k + l)}$ . Hence the product is independent of the chosen representatives, and it is associative with two-sided identity $[1, 0]$ because these hold for the product and for addition in $N$ ; so the quotient $P / ⁣ \approx$ is a monoid, denoted $G$ .

F1F3

1.3

Every class has a right inverse. Given $(a, k)$ , choose $r \geq k$ with $a ≼_{L} Δ^{2 r}$ , which is possible by [F2] because $≼_{L} Δ^{m}$ implies $≼_{L} Δ^{m^{'}}$ for every $m^{'} \geq m$ . Write $Δ^{2 r} = a c$ with $c \in B_{n}^{+}$ . Then $(a, k) \cdot (c, r - k) = (a c, r) = (Δ^{2 r}, r)$ , and $(Δ^{2 r}, r) \approx (1, 0)$ because $Δ^{2 r} Δ^{0} = 1 \cdot Δ^{2 r}$ .

F1F2

2.1

$G$ is a group. By step 1.3 every element $x$ of the monoid $G$ has a right inverse $y$ : $x y = 1$ . Applying the same to $y$ gives $z$ with $y z = 1$ . Then $x = x \cdot 1 = x (y z) = (x y) z = 1 \cdot z = z$ , so $y x = y z = 1$ as well: $y$ is a two-sided inverse of $x$ . Hence every element of $G$ is invertible and $G$ is a group.

F1step 1.2step 1.3

2.2

$η$ is injective. Define $η (a) : = [a, 0]$ . It is a monoid homomorphism by step 1.2: $η (a b) = [a b, 0] = [a, 0] [b, 0] = η (a) η (b)$ because $0 + 0 = 0$ . If $η (a) = η (b)$ , then $(a, 0) \approx (b, 0)$ , that is $a Δ^{0} = b Δ^{0}$ , so $a = b$ ; hence $η$ is injective.

F1step 1.2

3.1

The shape of the elements of $G$ . By step 1.3, applied to $(Δ^{2 k}, 0)$ , the class $[Δ^{2 k}, 0]$ is invertible with $[Δ^{2 k}, 0]^{- 1} = [1, k]$ ; and $[a, k] = [a, 0] \cdot [1, k]$ because $(a \cdot 1, 0 + k) = (a, k)$ . Hence every element of $G$ has the form $η (a) η (Δ)^{- 2 k}$ ; taking $b : = Δ^{2 k}$ and using $[Δ^{2 k}, 0] = η (b)$ this is $η (a) η (b)^{- 1}$ , which is (b)(i).

F1step 1.2step 1.3step 2.1

4.1

The universal property. Let $f ⁣ : B_{n}^{+} \to H$ be a monoid homomorphism into a group. Define $\overset{ˉ}{f} ([a, k]) : = f (a) f (Δ)^{- 2 k}$ . This is well defined: if $a Δ^{2 l} = b Δ^{2 k}$ then applying $f$ and multiplying by $f (Δ)^{- 2 k - 2 l}$ gives $f (a) f (Δ)^{- 2 k} = f (b) f (Δ)^{- 2 l}$ . It is a homomorphism: $\overset{ˉ}{f} ([a, k] [b, l]) = \overset{ˉ}{f} ([a b, k + l]) = f (a) f (b) f (Δ)^{- 2 k - 2 l}$ , while $\overset{ˉ}{f} ([a, k]) \overset{ˉ}{f} ([b, l]) = f (a) f (Δ)^{- 2 k} f (b) f (Δ)^{- 2 l}$ , and these agree because $f (Δ)^{2} = f (Δ^{2})$ lies in the centre of the image of $f$ by [F3], so that $f (Δ)^{- 2 k} f (b) = f (b) f (Δ)^{- 2 k}$ . Finally $\overset{ˉ}{f} (η (a)) = \overset{ˉ}{f} ([a, 0]) = f (a)$ , and $\overset{ˉ}{f}$ is unique with this property because every element of $G$ is a product of elements $η (a)$ and inverses $η (Δ)^{- 1}$ , as shown in step 3.1, so a group homomorphism out of $G$ is determined by its values on the $η (a)$ .

F1F3step 2.2step 3.1

5.1

Identification with $B_{n}$ . The assignment $σ_{i} \mapsto η (σ_{i})$ satisfies the defining relations of [F4] because they hold in $B_{n}^{+}$ and $η$ is a homomorphism, so [F4] gives a group homomorphism $ψ ⁣ : B_{n} \to G$ with $ψ (σ_{i}) = η (σ_{i})$ . In the other direction, the relations hold in $B_{n}$ itself, so [F5] gives a monoid homomorphism $κ ⁣ : B_{n}^{+} \to B_{n}$ with $κ (σ_{i}) = σ_{i}$ , and step 4.1 applied to $f : = κ$ gives a group homomorphism $φ ⁣ : G \to B_{n}$ with $φ \circ η = κ$ and hence $φ (η (σ_{i})) = σ_{i}$ . Then $φ \circ ψ$ and ${i d}_{B_{n}}$ are group endomorphisms of $B_{n}$ agreeing on the generators $σ_{i}$ , which generate $B_{n}$ by [F4], so $φ \circ ψ = {i d}_{B_{n}}$ ; similarly $ψ \circ φ$ and ${i d}_{G}$ are group endomorphisms of $G$ agreeing on the $η (σ_{i})$ , and these generate $G$ as a group because the $σ_{i}$ generate $B_{n}^{+}$ by [F1] so every element of $G$ is a product of elements $η (σ_{i})^{\pm 1}$ by step 3.1; hence $ψ \circ φ = {i d}_{G}$ . Thus $ψ$ is an isomorphism with inverse $φ$ , and $κ = φ \circ η$ is injective with image $φ (η (B_{n}^{+}))$ , the set of classes of positive words. This is (c).

F1F4F5step 3.1step 4.1

6.1

Assembly. Part (a) is the existence statement of [F2] together with the common multiple $Δ^{2 r}$ of step 1.3 (take $c, d$ as there, for $a$ and $b$ , with a common even power). Part (b) comprises the construction of $G$ in steps 1.1--1.3, the group axioms in step 2.1, the injectivity of $η$ in step 2.2, the shape of the elements in step 3.1, and the universal property in step 4.1. Part (c) is step 5.1. No inverse is assumed in $B_{n}^{+}$ anywhere: the inverses live in the constructed quotient, and the only inputs about $Δ$ are the $Δ$ -power divisibility and the centrality of $Δ^{2}$ . All constructions are explicit, all exponents are natural numbers, and no choice principle is used. ∎

step 1.1step 1.2step 1.3step 2.1step 2.2step 3.1step 4.1step 5.1

Remarks

Why the construction uses $Δ^{2}$ and not $Δ$ . The relation defining $\approx$ compares $a Δ^{2 l}$ with $b Δ^{2 k}$ ; centrality of $Δ^{2}$ is what makes the product well defined and makes $f (Δ)^{- 2 k}$ commute with the image of $B_{n}^{+}$ in the universal property. Odd powers would only be central in the cases $n \leq 2$ : for $n \geq 3$ the element $Δ$ conjugates $σ_{i}$ to $σ_{n - i}$ rather than centralising it (Conjugation by the half twist reverses Artin generators), and the same construction with $2$ replaced by $1$ would fail to be well defined.
Comparison with GM. GM argue that "as every two elements have a common multiple (some power of $Δ$ ), and $B_{n}^{+}$ is cancellative, Ore's condition says that $B_{n}^{+}$ embeds in its group of fractions. This group of fractions, due to presentation (3.1), is precisely $B_{n}$ ." Steps 1.1--4.1 spell out the standard construction behind that sentence: the Ore condition is used only to find $r$ in step 1.3, cancellation only in steps 1.1 and 1.2 and in step 2.2, and the presentation comparison is step 4.1.
What the injectivity of $κ$ says. Since every element of $B_{n}$ is $κ (a) κ (b)^{- 1}$ , and $κ$ is injective, the usual abuse of notation is justified: from this point on a positive braid may be regarded as an element of $B_{n}$ , and the monoid orders $≼_{L}, ≼_{R}$ extend to $B_{n}$ (Left and right divisibility extend to lattice orders on the braid group). The statement that $B_{n}^{+}$ is not itself a group is Positive artin relations preserve homogeneous length (no nontrivial invertible element).
The group $G$ is presented by the same generators and the same relations as $B_{n}$ : this is what step 4.1 verifies, and it is the sense in which "the group of fractions is the Artin braid group" rather than merely a group containing $B_{n}^{+}$ .
Nothing here uses a choice principle: $G$ is a quotient of an explicit set, and the exponent $r$ of step 1.3 is bounded by a natural number read off from a word for $a$ .

Depends on

Used by

Exponent sum is not a complete braid normal form Counterexample
A left garside normal form computation in b three Example
The full twist in b three Example
Simple positive braids are indexed by permutations Lemma
Left and right divisibility extend to lattice orders on the braid group Theorem
Left garside normal form is unique Theorem
The center of b n is generated by the full twist for n greater than two Theorem

Dependency tree · two levels

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4, printed p. 28 (Ore's condition and the embedding of B_n^+) (standard reference, not scraped)
Patrick Dehornoy et al., Foundations of Garside Theory, Chapter IX, Lemma 1.22, printed p. 438 (standard reference, not scraped)