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Garside Structure, Normal Forms, and the Center — Examples

1 · Prerequisites

2 · Summary

These four worked entries make the Garside structure of the companion page concrete, all of them inside $B_{3}$ , where the half twist is $Δ = T_{1} T_{2} = σ_{1} (σ_{2} σ_{1}) = σ_{1} σ_{2} σ_{1}$ of length $N = 3$ and the braid relation also gives $Δ = σ_{2} σ_{1} σ_{2}$ . The first example lists the six simple braids and computes their relevant meets and joins in the positive divisibility lattice. There are exactly $3! = 6$ simple braids, namely $1, σ_{1}, σ_{2}, σ_{1} σ_{2}, σ_{2} σ_{1}, Δ$ , and their images under the permutation homomorphism $π ⁣ : B_{3}^{+} \to S_{3}$ are the six distinct elements $i d, (1 2), (2 3), (1 2 3), (1 3 2), (1 3)$ , which both proves the listing and exhibits the bijection of the theory. The two atoms are incomparable on the left with join their braid word, $σ_{1} \land_{L} σ_{2} = 1$ and $σ_{1} \lor_{L} σ_{2} = σ_{1} σ_{2} σ_{1} = Δ$ . For $a = σ_{1} σ_{2}$ and $b = σ_{1}$ the left meet is $a \land_{L} b = σ_{1}$ , computed from the divisors of an atom and of a two-letter element, while the right meet is $a \land_{R} b = 1$ : the two divisibility orders are therefore already different on the four-element subfamily ${1, σ_{1}, σ_{2}, σ_{1} σ_{2}}$ , and the balancedness of $Δ$ — its left and right divisor sets coincide — does not collapse them into one order.

The second example computes a left Garside normal form in full. For $x = σ_{1}^{- 1} σ_{2}$ the atom complement $σ_{1}^{- 1} = Δ^{- 1} σ_{1} σ_{2}$ gives the positive description $x = Δ^{- 1} σ_{1} σ_{2}^{2}$ , and the scaling identity for meets yields $Δ \land_{L} σ_{1} σ_{2}^{2} = (σ_{1} σ_{2}) (σ_{1} \land_{L} σ_{2}) = σ_{1} σ_{2}$ , so $Δ ̸ ≼_{L} σ_{1} σ_{2}^{2}$ and the normal form is $x = Δ^{- 1} \cdot (σ_{1} σ_{2}) \cdot σ_{2}$ with $p (x) = - 1$ , $A (x) = σ_{1} σ_{2}^{2}$ and two proper simple factors $a_{1} = σ_{1} σ_{2}$ , $a_{2} = σ_{2}$ , of permutations $s_{1} s_{2}$ and $s_{2}$ . The pair is left weighted, $(a_{1} a_{2}) \land_{L} Δ = a_{1}$ , and the negative exponent correctly reports that $x$ is not a positive braid.

The third example computes the full twist. The braid relation gives $(σ_{1} σ_{2})^{3} = σ_{1} σ_{2} σ_{1} \cdot σ_{2} σ_{1} σ_{2} = Δ^{2}$ , that is, the square of the half twist is the class of a positive word of length $6$ ; applying the sliding identity $σ_{i} Δ = Δ σ_{3 - i}$ twice shows that each atom commutes with $Δ^{2}$ , hence so does every element of $B_{3}$ and $Z (B_{3}) = ⟨ Δ^{2} ⟩$ is infinite cyclic by the center theorem for $n = 3 > 2$ . The half twist itself is not central only because $Δ σ_{1} = σ_{2} Δ$ and the atoms $σ_{1} \neq σ_{2}$ are distinct: the passage from $Δ$ to its square in the center is genuine, not an artifact of commuting generators.

The fourth entry is the counterexample. The exponent sum $ε ⁣ : B_{n} \to Z$ , $σ_{i}^{\pm 1} \mapsto \pm 1$ , is a well-defined homomorphism: every three-term braid relator and every far-commutation relator has equal exponent sum on both sides — and the permutation homomorphism $π ⁣ : B_{3} \to S_{3}$ is likewise well defined, with $π (σ_{1}) = (1 2) \neq (2 3) = π (σ_{2})$ , so $σ_{1} \neq σ_{2}$ in $B_{3}$ . Yet $ε (σ_{1}) = ε (σ_{2}) = 1$ , and both elements have the same coarse normal-form data: left normal form $Δ^{0} \cdot σ_{i}$ with exponent $p = 0$ and a single proper simple factor. The exponent sum is therefore not a complete braid normal form, and not even the triple $(ε (x), p (x), r (x))$ determines a braid; what separates the two witnesses is the factor itself, $σ_{1}$ against $σ_{2}$ , i.e. the permutation data that the left Garside normal form retains. Nothing in these four entries uses a choice principle.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-27Open item page →

The simple braids and divisibility lattice for b three

Example

Take $n = 3$ , so that $B_{3}^{+}$ has the two atoms $σ_{1}, σ_{2}$ and the half twist $Δ = σ_{1} σ_{2} σ_{1}$ of length $N = 3$ . The example verifies:

the six simple braids of $B_{3}$ are $1, σ_{1}, σ_{2}, σ_{1} σ_{2}, σ_{2} σ_{1}, Δ$ , with the six distinct endpoint permutations $i d, (1 2), (2 3), (1 2 3), (1 3 2), (1 3)$ ;
$σ_{1} \land_{L} σ_{2} = 1$ while $σ_{1} \lor_{L} σ_{2} = σ_{1} σ_{2} σ_{1} = Δ$ ;
for $a : = σ_{1} σ_{2}$ and $b : = σ_{1}$ one has $a \land_{L} b = σ_{1}$ but $a \land_{R} b = 1$ ;
consequently the left and the right divisibility orders on $B_{3}^{+}$ are different, even though by Simple positive braids are indexed by permutations the two orders have the same divisor set on $Δ$ .

Facts & Assumptions

Given: The natural number $3$ , the monoid $B_{3}^{+}$ with atoms $σ_{1}, σ_{2}$ , half twist $Δ = σ_{1} σ_{2} σ_{1}$ of length $3$ , and the elements $a = σ_{1} σ_{2}$ , $b = σ_{1}$ .

[F1]

$B_{3}^{+}$ is generated by the atoms, $ℓ ([w]) = ∣ w ∣$ is additive and $ℓ (x) = 0$ only for $x = 1$ ; the orders are $x ≼_{L} y ⟺ y = x c$ and $x ≼_{R} y ⟺ y = c x$ for some $c \in B_{3}^{+}$ , with length monotone, so a proper divisor of an atom or of $1$ or $2$ letters has strictly smaller length (Positive braid monoid, Positive artin relations preserve homogeneous length, Left and right divisibility for positive braids).

[F2]

$Δ = σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ by the braid relation, and the simple braids — the left divisors of $Δ$ — are in bijection with $S_{3}$ : they are exactly $\hat{i d} = 1$ , $\hat{s_{1}} = σ_{1}$ , $\hat{s_{2}} = σ_{2}$ , $\hat{s_{1} s_{2}} = σ_{1} σ_{2}$ , $\hat{s_{2} s_{1}} = σ_{2} σ_{1}$ , $\hat{w_{0}} = Δ$ (Simple positive braids are indexed by permutations, The Garside half twist and simple positive braids).

[F3]

The homomorphism $π ⁣ : B_{3}^{+} \to S_{3}$ sends $σ_{i}$ to $s_{i}$ and products to products; $s_{1} = (1 2)$ , $s_{2} = (2 3)$ , $s_{1} s_{2} = (1 2 3)$ , $s_{2} s_{1} = (1 3 2)$ , $s_{1} s_{2} s_{1} = (1 3)$ in the composition convention of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition (Reduced adjacent-transposition words have well-defined positive lifts).

[F4]

For adjacent atoms, $σ_{1} \lor_{L} σ_{2} = σ_{1} σ_{2} σ_{1}$ is the least common left multiple and also the least common right multiple, and $σ_{1} \land_{L} σ_{2}$ exists and is unique; more generally every nonempty finite subset of $B_{3}^{+}$ has unique left and right gcds and lcms (Artin atoms have explicit left and right lcms and complements, Positive braids have left and right gcds and lcms).

Verification

technique · direct

1.1

The six simple braids and their permutations. By [F2] there are exactly $3! = 6$ simple braids. Applying [F3] to the six displayed elements gives the images $i d$ , $s_{1} = (1 2)$ , $s_{2} = (2 3)$ , $s_{1} s_{2} = (1 2 3)$ , $s_{2} s_{1} = (1 3 2)$ and $s_{1} s_{2} s_{1} = (1 3)$ in this order; these are the six elements of $S_{3}$ , so they are pairwise distinct, and since $π$ is a function the six braids $1, σ_{1}, σ_{2}, σ_{1} σ_{2}, σ_{2} σ_{1}, Δ$ are pairwise distinct and exhaust the left divisors of $Δ$ . In particular $σ_{1} σ_{2}$ and $σ_{2} σ_{1}$ are proper simple braids, because $σ_{1} σ_{2} \cdot σ_{1} = Δ$ and $σ_{2} σ_{1} \cdot σ_{2} = Δ$ by [F2].

F2F3

1.2

The atom pair $(σ_{1}, σ_{2})$ . The left divisors of $σ_{1}$ are $1$ and $σ_{1}$ , since a divisor $d$ of $σ_{1}$ has $ℓ (d) \leq 1$ , so either $d = 1$ or $d$ is an atom equal to $σ_{1}$ ; the same holds for $σ_{2}$ , and the distinct atoms are incomparable, so the common left divisors of $σ_{1}$ and $σ_{2}$ are just $1$ : thus $σ_{1} \land_{L} σ_{2} = 1$ . Their least common left multiple is $σ_{1} σ_{2} σ_{1}$ by [F4], which is $Δ$ by [F2], so $σ_{1} \lor_{L} σ_{2} = Δ$ .

F1F2F4

1.3

A pair with different left and right meets. For $a = σ_{1} σ_{2}$ and $b = σ_{1}$ : $σ_{1}$ left-divides $b$ trivially and left-divides $a = σ_{1} σ_{2}$ , so $σ_{1}$ is a common left divisor; every common left divisor $d$ of $a$ and $b$ divides $b = σ_{1}$ , hence is $1$ or $σ_{1}$ , so $d ≼_{L} σ_{1}$ : therefore $a \land_{L} b = σ_{1}$ . On the right: the right divisors of $b = σ_{1}$ are $1$ and $σ_{1}$ by [F1], while the right divisors of $a = σ_{1} σ_{2}$ are $1, σ_{2}, σ_{1} σ_{2}$ — indeed if $a = d c$ then $ℓ (c) \leq 2$ ; for $ℓ (c) = 2$ one has $ℓ (d) = 0$ , so $d = 1$ and $c = a$ ; for $ℓ (c) = 1$ the element $c$ is an atom, and $d$ then has length $1$ and is an atom as well, so $a$ must be a product of two atoms, the four candidates being $σ_{1} σ_{1}, σ_{1} σ_{2}, σ_{2} σ_{1}, σ_{2} σ_{2}$ , which are distinct because every defining relation has length three; only $σ_{1} σ_{2} = a$ occurs, so the only one-letter right divisor is $c = σ_{2}$ . Hence the common right divisors of $a$ and $b$ are ${1, σ_{2}, σ_{1} σ_{2}} \cap {1, σ_{1}} = {1}$ , and $a \land_{R} b = 1$ .

F1F2F4

1.4

Conclusion. Steps 1.1--1.3 give the six simple braids, $σ_{1} \land_{L} σ_{2} = 1$ , $σ_{1} \lor_{L} σ_{2} = Δ$ , and the pair $a, b$ with $a \land_{L} b = σ_{1} \neq 1 = a \land_{R} b$ . So already on the four-element subfamily ${1, σ_{1}, σ_{2}, σ_{1} σ_{2}}$ the two orders have different meets, even though on $Δ$ itself the left and right divisor sets coincide by [F2]; the balancedness established in Delta is the lcm of the artin atoms and has the same left and right divisors is therefore a property of $Δ$ and does not identify the two orders. No choice principle is used. ∎

step 1.1step 1.2step 1.3

Remarks

Reading the six permutations. In the convention of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition the product acts with the right factor first, so $σ_{1} σ_{2}$ maps to $s_{1} s_{2} = (1 2 3)$ and $σ_{2} σ_{1}$ to the inverse cycle $(1 3 2)$ : the order of the two words is visible in the orientation of the $3$ -cycle, and this is what separates the two length-two simple braids.
Why the right meet is the smaller one here. The right divisors of $σ_{1} σ_{2}$ are its suffixes $1, σ_{2}, σ_{1} σ_{2}$ , while those of $σ_{1}$ are $1, σ_{1}$ ; the overlap is trivial even though the overlap of the corresponding prefix sets is ${1, σ_{1}}$ . This is the smallest instance of the asymmetry between the two orders.
No choice principle is used; all computations are finite word computations.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-27Open item page →

A left garside normal form computation in b three

Example

Take $n = 3$ , so that $B_{3}$ has the two generators $σ_{1}, σ_{2}$ , the positive monoid $B_{3}^{+}$ has the two atoms $σ_{1}, σ_{2}$ , and the half twist is

$Δ = T_{1} T_{2} = σ_{1} (σ_{2} σ_{1}) = σ_{1} σ_{2} σ_{1},$

of length $N = 1 + 2 = 3$ . Let $x : = σ_{1}^{- 1} σ_{2} \in B_{3}$ . The example computes the left Garside normal form of $x$ of Left garside normal form is unique and checks:

$σ_{1}^{- 1} = Δ^{- 1} σ_{1} σ_{2}$ , hence $x = Δ^{- 1} σ_{1} σ_{2}^{2}$ with $σ_{1} σ_{2}^{2} \in B_{3}^{+}$ ;
$Δ \land_{L} σ_{1} σ_{2}^{2} = σ_{1} σ_{2}$ , so in particular $Δ ̸ ≼_{L} σ_{1} σ_{2}^{2}$ ;
the left normal form of $x$ is $x = Δ^{- 1} \cdot (σ_{1} σ_{2}) \cdot σ_{2},$ that is $p (x) = - 1$ , $A (x) = σ_{1} σ_{2}^{2}$ and the two factors $a_{1} = σ_{1} σ_{2}$ , $a_{2} = σ_{2}$ are proper simple braids with $π (a_{1}) = s_{1} s_{2}$ , $π (a_{2}) = s_{2}$ ;
the factor pair is left weighted: $(a_{1} a_{2}) \land_{L} Δ = a_{1}$ ;
consequently $x \notin B_{3}^{+}$ , in agreement with $p (x) = - 1 < 0$ .

Facts & Assumptions

Given: The natural number $3$ ; the braid group $B_{3} = ⟨ σ_{1}, σ_{2} ∣ σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2} ⟩$ of The braid group by Artin presentation; its positive monoid $B_{3}^{+}$ with atoms $σ_{1}, σ_{2}$ and length $ℓ$ ; the half twist $Δ = σ_{1} σ_{2} σ_{1} \in B_{3}^{+}$ of length $3$ ; and the element $x : = σ_{1}^{- 1} σ_{2} \in B_{3}$ .

[F1]

$B_{3}^{+}$ is generated by the atoms, $ℓ ([w]) = ∣ w ∣$ is additive, $ℓ (z) = 0$ forces $z = 1$ , and the braid relation gives $σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ , so both triangular words for $Δ$ agree: $Δ = σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ (Positive braid monoid, Positive artin relations preserve homogeneous length, The Garside half twist and simple positive braids).

[F2]

$a ≼_{L} b$ means $b = a c$ for some $c \in B_{3}^{+}$ , and then $ℓ (a) \leq ℓ (b)$ ; a left divisor of an atom has length $0$ or $1$ , so the left divisors of $σ_{1}$ are among $1, σ_{1}$ and those of $σ_{2}$ among $1, σ_{2}$ (Left and right divisibility for positive braids, Positive artin relations preserve homogeneous length).

[F3]

$π ⁣ : B_{3}^{+} \to S_{3}$ is a monoid homomorphism with $π (σ_{i}) = s_{i}$ ; in the composition convention of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition the adjacent transpositions $s_{1} = (1 2)$ and $s_{2} = (2 3)$ are distinct, so $σ_{1} \neq σ_{2}$ (Reduced adjacent-transposition words have well-defined positive lifts, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[F4]

$B_{3}^{+}$ is a submonoid of $B_{3}$ , and for $x \in B_{3}$ one has $1 ≼_{L} x$ if and only if $x \in B_{3}^{+}$ (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).

[F5]

Scaling identity. For $D, A, B \in B_{3}^{+}$ one has $D (A \land_{L} B) = (D A) \land_{L} (D B)$ , where the meet is the positive left-gcd (Left and right divisibility extend to lattice orders on the braid group).

[F6]

Every nonempty finite family in $B_{3}^{+}$ has a unique left-gcd and a unique left-lcm; in particular $Δ \land_{L} A$ exists for every $A \in B_{3}^{+}$ , and every common left divisor of $Δ$ and $A$ left-divides $Δ \land_{L} A$ (Positive braids have left and right gcds and lcms).

[F7]

Normal form. Every $x \in B_{3}$ has a unique expression $x = Δ^{p} a_{1} \dots a_{r}$ with $p \in Z$ , $r \in N$ , every $a_{i}$ a proper simple braid and $a_{i} = Δ \land_{L} (a_{i} \dots a_{r})$ ; in such an expression $p = p (x)$ is the largest $p$ with $Δ^{p} ≼_{L} x$ , the product $a_{1} \dots a_{r}$ equals $A (x) = Δ^{- p (x)} x$ , and $(p (x), A (x))$ is the unique pair with $x = Δ^{p} A$ , $A \in B_{3}^{+}$ and $Δ ̸ ≼_{L} A$ . Moreover $p (x) \geq 0$ if and only if $x \in B_{3}^{+}$ , and if $r \geq 2$ then $(a_{i} a_{i + 1}) \land_{L} Δ = a_{i}$ for $1 \leq i < r$ (Left garside normal form is unique).

Verification

technique · direct

1.1

The half twist and the rewriting of $σ_{1}^{- 1}$ . By [F1], $Δ = σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ has length $3$ . In the group $B_{3}$ , in which $B_{3}^{+}$ sits as a submonoid by [F4], multiplying out gives $Δ^{- 1} σ_{1} σ_{2} = (σ_{1} σ_{2} σ_{1})^{- 1} σ_{1} σ_{2} = σ_{1}^{- 1} σ_{2}^{- 1} σ_{1}^{- 1} σ_{1} σ_{2} = σ_{1}^{- 1} σ_{2}^{- 1} σ_{2} = σ_{1}^{- 1}$ , so $σ_{1}^{- 1} = Δ^{- 1} σ_{1} σ_{2}$ ; substituting this identity into the given element $x = σ_{1}^{- 1} σ_{2}$ gives $x = Δ^{- 1} σ_{1} σ_{2}^{2}$ , and $σ_{1} σ_{2}^{2} \in B_{3}^{+}$ because it is a product of atoms.

F1F4givenalgebra

1.2

The atoms are coprime. Let $d \in B_{3}^{+}$ be a common left divisor of $σ_{1}$ and $σ_{2}$ . By [F2] write $σ_{1} = d c$ with $c \in B_{3}^{+}$ ; additivity of $ℓ$ [F1] gives $ℓ (d) + ℓ (c) = 1$ . If $ℓ (d) = 0$ then $d = 1$ [F1]; if $ℓ (d) = 1$ then $ℓ (c) = 0$ , so $c = 1$ and $d = σ_{1}$ . Since $d$ also left-divides $σ_{2}$ , write $σ_{2} = d c^{'}$ with a possibly different suffix $c^{'}$ . Length gives $ℓ (c^{'}) = 0$ , hence $c^{'} = 1$ and $σ_{2} = d = σ_{1}$ , contradicting $π (σ_{2}) = s_{2} \neq s_{1} = π (σ_{1})$ [F3]. So the only common left divisor of the two atoms is $1$ , and by [F6] their left-gcd is $σ_{1} \land_{L} σ_{2} = 1$ .

F1F2F3F6

2.1

The meet with $σ_{1} σ_{2}^{2}$ . Apply the scaling identity [F5] to $D : = σ_{1} σ_{2}$ , $A : = σ_{1}$ and $B : = σ_{2}$ ; by [F1] the two products are $D A = σ_{1} σ_{2} σ_{1} = Δ$ and $D B = σ_{1} σ_{2} σ_{2} = σ_{1} σ_{2}^{2}$ , so $Δ \land_{L} σ_{1} σ_{2}^{2} = (σ_{1} σ_{2} σ_{1}) \land_{L} (σ_{1} σ_{2} σ_{2}) = (σ_{1} σ_{2}) (σ_{1} \land_{L} σ_{2}) = (σ_{1} σ_{2}) \cdot 1 = σ_{1} σ_{2}$ , the third equality by step 1.2. Hence $Δ ̸ ≼_{L} σ_{1} σ_{2}^{2}$ : if $Δ$ were a common left divisor of $Δ$ and $σ_{1} σ_{2}^{2}$ , then by [F6] it would satisfy $Δ ≼_{L} Δ \land_{L} σ_{1} σ_{2}^{2} = σ_{1} σ_{2}$ , and monotonicity of $ℓ$ [F2] would force $3 = ℓ (Δ) \leq ℓ (σ_{1} σ_{2}) = 2$ , a contradiction.

F1F2F5F6step 1.2

3.1

The maximal $Δ$ -exponent is $- 1$ . The pair $(p, A) : = (- 1, σ_{1} σ_{2}^{2})$ satisfies $x = Δ^{p} A$ by step 1.1, $A \in B_{3}^{+}$ by step 1.1 and $Δ ̸ ≼_{L} A$ by step 2.1; by the uniqueness in [F7] of the pair $(p, A)$ it coincides with $(p (x), A (x))$ , hence $p (x) = - 1$ and $A (x) = σ_{1} σ_{2}^{2}$ .

F7step 1.1step 2.1

3.2

The permutation images and left weighting. Since $π$ is a homomorphism with $π (σ_{i}) = s_{i}$ [F3], the two factors have images $π (a_{1}) = s_{1} s_{2}$ and $π (a_{2}) = s_{2}$ , which are distinct; in particular $a_{1} \neq a_{2}$ , so the two-factor form is not a repetition of one simple braid. Left weighting is the instance $i = 1 < r = 2$ of the last assertion of [F7]: $(a_{1} a_{2}) \land_{L} Δ = a_{1}$ , which is literally the computation of step 2.1.

F3F7step 2.1

4.1

The two greedy factors. Put $a_{1} : = σ_{1} σ_{2}$ and $a_{2} : = σ_{2}$ , so that $A (x) = a_{1} a_{2}$ by step 3.1. Then $Δ \land_{L} A (x) = Δ \land_{L} σ_{1} σ_{2}^{2} = a_{1}$ by step 2.1, and the unique $A_{1}$ with $A (x) = a_{1} A_{1}$ is $A_{1} = σ_{2} = a_{2}$ , since $(σ_{1} σ_{2}) σ_{2} = σ_{1} σ_{2}^{2}$ [F1]; further $a_{2} ≼_{L} Δ$ because $Δ = σ_{2} σ_{1} σ_{2} = σ_{2} (σ_{1} σ_{2})$ [F1], hence $Δ \land_{L} A_{1} = Δ \land_{L} a_{2} = a_{2}$ [F6], and the unique $A_{2}$ with $A_{1} = a_{2} A_{2}$ is $A_{2} = 1$ . So the recursion of [F7] produces $r = 2$ with factors $a_{1}, a_{2}$ . Both are proper simple braids: $Δ = (σ_{1} σ_{2}) σ_{1}$ exhibits $a_{1} ≼_{L} Δ$ and $Δ = σ_{2} (σ_{1} σ_{2})$ exhibits $a_{2} ≼_{L} Δ$ [F1], while $a_{1} \neq 1, Δ$ and $a_{2} \neq 1, Δ$ because $ℓ (a_{1}) = 2$ and $ℓ (a_{2}) = 1$ lie in ${1, 2}$ and not in ${0, 3}$ [F1].

F1F6F7step 2.1step 3.1

5.1

The normal form conditions hold. For $i = 2$ , step 4.1 gives $Δ \land_{L} a_{2} = a_{2} = Δ \land_{L} (a_{2})$ ; for $i = 1$ , step 2.1 gives $Δ \land_{L} (a_{1} a_{2}) = Δ \land_{L} σ_{1} σ_{2}^{2} = a_{1}$ . So $x = Δ^{- 1} a_{1} a_{2}$ with both $a_{i}$ proper simple and $a_{i} = Δ \land_{L} (a_{i} \dots a_{r})$ for $i = 1, 2$ ; by the uniqueness in [F7] this is the left normal form of $x$ , and [F7] identifies its data as $p (x) = - 1$ and $A (x) = a_{1} a_{2} = σ_{1} σ_{2}^{2}$ , in agreement with step 3.1.

F7step 2.1step 4.1

6.1

Conclusion. The element $x = σ_{1}^{- 1} σ_{2}$ of $B_{3}$ has left Garside normal form $x = Δ^{- 1} (σ_{1} σ_{2}) σ_{2}$ with $p (x) = - 1$ and $A (x) = σ_{1} σ_{2}^{2}$ , with proper simple factors $a_{1} = σ_{1} σ_{2}$ and $a_{2} = σ_{2}$ whose images in $S_{3}$ are $s_{1} s_{2}$ and $s_{2}$ , and with the factor pair left weighted by step 3.2. Moreover $x \notin B_{3}^{+}$ , because $p (x) = - 1 < 0$ while $p \geq 0$ characterises positivity by [F7]; this is the qualitative content of the computation, since $x$ is visibly written with an inverse letter. No choice principle is used, the only selections being the explicit words displayed above. ∎

F7step 1.1step 2.1step 3.1step 5.1step 3.2

Remarks

Reading off the algorithm. In the notation of Left garside normal form is unique the recursion runs $A_{0} = σ_{1} σ_{2}^{2}$ , $a_{1} = Δ \land_{L} A_{0} = σ_{1} σ_{2}$ , $A_{1} = σ_{2}$ , $a_{2} = Δ \land_{L} A_{1} = σ_{2}$ , $A_{2} = 1$ ; the factor $σ_{1} σ_{2}$ is the maximal simple prefix of $σ_{1} σ_{2}^{2}$ because $Δ$ does not divide $σ_{1} σ_{2}^{2}$ on the left, and the remainder $σ_{2}$ is already simple.
Why the $Δ$ -exponent is negative. The computation above gives $p (x) = - 1$ ; by Left garside normal form is unique (d)(ii), this proves $σ_{1}^{- 1} σ_{2}$ is not positive. The calculation exhibits the witness pair $(- 1, σ_{1} σ_{2}^{2})$ of that theorem's uniqueness clause rather than merely asserting it. Note that $σ_{1} σ_{2}^{2}$ has the same length $3$ as $Δ$ ; length alone therefore decides neither left divisibility nor the meet, and it is the scaling computation of step 2.1 that shows $Δ ̸ ≼_{L} σ_{1} σ_{2}^{2}$ .
Conventions. All products are read left to right as words in the generators, and the permutation images are those of the positive monoid map $π$ ; the cycle notation is that of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, so $s_{1} s_{2} = (1 2 3)$ . No choice principle is used: every object in the computation is an explicitly displayed finite word.

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The full twist in b three

Example

Take $n = 3$ , so that $B_{3}$ has the generators $σ_{1}, σ_{2}$ , the positive monoid $B_{3}^{+}$ has the two atoms $σ_{1}, σ_{2}$ , and the half twist is $Δ = σ_{1} σ_{2} σ_{1}$ of length $N = 3$ . The example verifies:

$(σ_{1} σ_{2})^{3} = σ_{1} σ_{2} σ_{1} \cdot σ_{2} σ_{1} σ_{2} = Δ^{2}$ , the full twist of The center of b n is generated by the full twist for n greater than two;
$σ_{i} Δ^{2} = Δ^{2} σ_{i}$ for $i = 1, 2$ , hence $Δ^{2}$ commutes with every element of $B_{3}$ , i.e. $Δ^{2} \in Z (B_{3})$ ;
together with the centre theorem for $n = 3 > 2$ this gives $Z (B_{3}) = ⟨ Δ^{2} ⟩$ , and that centre is infinite cyclic: the full twist $Δ^{2} = (σ_{1} σ_{2})^{3}$ is a free generator;
the half twist itself is not central: $Δ σ_{1} = σ_{2} Δ$ while $σ_{2} \neq σ_{1}$ , so $Δ \notin Z (B_{3})$ and the passage from $Δ$ to its square is genuine.

Facts & Assumptions

Given: The natural number $3$ ; the braid group $B_{3}$ with generators $σ_{1}, σ_{2}$ of The braid group by Artin presentation; its positive monoid $B_{3}^{+}$ with atoms $σ_{1}, σ_{2}$ , length $ℓ$ and half twist $Δ = σ_{1} σ_{2} σ_{1}$ of length $3$ .

[F1]

$B_{3}^{+}$ is generated by the atoms, $ℓ ([w]) = ∣ w ∣$ is additive, $ℓ (z) = 0$ forces $z = 1$ , the braid relation gives $σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ , and $Δ = T_{1} T_{2} = σ_{1} (σ_{2} σ_{1}) = σ_{1} σ_{2} σ_{1}$ has length $N = 1 + 2 = 3$ (Positive braid monoid, Positive artin relations preserve homogeneous length, The Garside half twist and simple positive braids).

[F2]

$σ_{i} Δ = Δ σ_{3 - i}$ for $i = 1, 2$ (Conjugation by the half twist reverses Artin generators).

[F3]

$B_{3}^{+}$ is a submonoid of $B_{3}$ and the elements of $B_{3}^{+}$ are multiplied in $B_{3}$ as in the monoid, so $Δ$ is an invertible element of the group $B_{3}$ and group cancellation is available (The group of fractions of the positive braid monoid is the Artin braid group).

[F4]

In the presentation of [F1], every element of $B_{3}$ is a finite product of the letters $σ_{1}^{\pm 1}, σ_{2}^{\pm 1}$ (The braid group by Artin presentation).

[F5]

The centre. For $n > 2$ one has $Z (B_{n}) = ⟨ Δ^{2} ⟩ = {Δ^{2 k} : k \in Z}$ , and this group is infinite cyclic: $Δ^{2} \neq 1$ and $k \mapsto Δ^{2 k}$ is an isomorphism $Z \to Z (B_{n})$ (The center of b n is generated by the full twist for n greater than two).

[F6]

$π ⁣ : B_{3}^{+} \to S_{3}$ is a monoid homomorphism with $π (σ_{i}) = s_{i}$ , and in the composition convention of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition the adjacent transpositions $s_{1} = (1 2)$ and $s_{2} = (2 3)$ are distinct, so $σ_{1} \neq σ_{2}$ (Reduced adjacent-transposition words have well-defined positive lifts, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

Verification

technique · direct

1.1

The half twist and its two words. By [F1], $Δ = σ_{1} σ_{2} σ_{1}$ has length $3$ and the braid relation gives the second triangular word $Δ = σ_{2} σ_{1} σ_{2}$ ; by [F3] the monoid $B_{3}^{+}$ sits inside the group $B_{3}$ , so $Δ$ may be inverted and cancelled there.

F1F3

1.2

Both generators commute with $Δ^{2}$ . Applying the identity $σ_{i} Δ = Δ σ_{3 - i}$ of [F2] twice, $σ_{1} Δ^{2} = (σ_{1} Δ) Δ = (Δ σ_{2}) Δ = Δ (σ_{2} Δ) = Δ (Δ σ_{1}) = Δ^{2} σ_{1}$ , and likewise $σ_{2} Δ^{2} = (σ_{2} Δ) Δ = (Δ σ_{1}) Δ = Δ (σ_{1} Δ) = Δ (Δ σ_{2}) = Δ^{2} σ_{2}$ .

F2algebra

2.1

The cube of $σ_{1} σ_{2}$ is the square of the half twist. Using associativity and the two words for $Δ$ from step 1.1, $(σ_{1} σ_{2})^{3} = σ_{1} σ_{2} σ_{1} σ_{2} σ_{1} σ_{2} = (σ_{1} σ_{2} σ_{1}) (σ_{2} σ_{1} σ_{2}) = Δ \cdot Δ = Δ^{2}$ .

F1step 1.1

2.2

Centrality of the full twist. By [F4] every $z \in B_{3}$ is a finite product of the letters $σ_{1}^{\pm 1}, σ_{2}^{\pm 1}$ . Step 1.2 gives $Δ^{2} σ_{i} = σ_{i} Δ^{2}$ for $i = 1, 2$ , and this passes to inverses: from $Δ^{2} g = g Δ^{2}$ one gets $g^{- 1} Δ^{2} = Δ^{2} g^{- 1}$ by multiplying the identity on the left and on the right by $g^{- 1}$ . Induction on the number of letters therefore gives $Δ^{2} z = z Δ^{2}$ for every $z \in B_{3}$ , that is $Δ^{2} \in Z (B_{3})$ .

F4step 1.2

3.1

The centre is infinite cyclic on the full twist. By [F5] with $n = 3$ , $Z (B_{3}) = ⟨ Δ^{2} ⟩ = {Δ^{2 k} : k \in Z}$ , the element $Δ^{2}$ is not $1$ , and $k \mapsto Δ^{2 k}$ is an isomorphism $Z \to Z (B_{3})$ ; combining this with step 2.1, the centre is infinite cyclic and generated by the full twist $Δ^{2} = (σ_{1} σ_{2})^{3}$ , consistently with the local centrality verification of step 2.2.

F5step 2.1step 2.2

3.2

The half twist alone is not central. By [F2] with $i = 1$ one has $Δ σ_{1} = σ_{2} Δ$ . If $Δ$ were central then also $Δ σ_{1} = σ_{1} Δ$ , so $σ_{2} Δ = σ_{1} Δ$ and right cancellation by the invertible element $Δ$ in $B_{3}$ [F3] gives $σ_{2} = σ_{1}$ , contradicting $π (σ_{2}) = s_{2} \neq s_{1} = π (σ_{1})$ [F6]. Hence $Δ \notin Z (B_{3})$ , although $Δ^{2} \in Z (B_{3})$ by step 2.2: the full twist is the square of the half twist and is not the half twist itself.

F2F3F6step 2.2

4.1

Conclusion. In $B_{3}$ the full twist satisfies $Δ^{2} = (σ_{1} σ_{2})^{3}$ by step 2.1, it is central by step 2.2, it generates the centre $Z (B_{3}) = ⟨ Δ^{2} ⟩ ≅ Z$ by step 3.1, and the half twist $Δ$ is not central by step 3.2. The four assertions of the Example section are therefore verified. No choice principle is used: the identities are finite word computations and the only inverse taken is the explicit inverse of the displayed word $Δ$ . ∎

step 2.1step 2.2step 3.1step 3.2

Remarks

Why the name. Geometrically the half twist $Δ$ is the half turn of the $n$ strands, whose square is the full turn; the star of the example is the algebraic counterpart: $Δ^{2} = (σ_{1} σ_{2})^{3}$ is the full twist of $B_{3}$ , and The center of b n is generated by the full twist for n greater than two identifies it as the generator of the centre. The words $Δ = σ_{1} σ_{2} σ_{1}$ and $(σ_{1} σ_{2})^{3}$ have exponent sums $3$ and $6$ respectively, matching their homogeneous lengths $ℓ (Δ) = 3 = N$ and $ℓ (Δ^{2}) = 6 = 2 N$ .
The local and the global verification. Step 2.2 checks centrality directly from the two generator identities, using only that $B_{3}$ is generated by $σ_{1}$ and $σ_{2}$ ; the appeal to the centre theorem in step 3.1 is then used only to conclude that no other central elements exist. This matches the source's proof of the general theorem, where the same generator identities supply centrality of $Δ^{2}$ before the divisibility argument bounds the centre.
The asymmetry of the half twist. Centrality of $Δ^{2}$ and failure of centrality of $Δ$ are both visible in the index reversal $σ_{i} \mapsto σ_{3 - i}$ : conjugation by $Δ$ permutes the two generators instead of fixing them, whereas its square fixes both. No choice principle is used anywhere.

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Exponent sum is not a complete braid normal form

Statement refuted

Refuted claim. The exponent sum $ε ⁣ : B_{n} \to Z$ , defined on words by $σ_{i}^{\pm 1} \mapsto \pm 1$ , is a complete braid normal form: for all $x, y \in B_{n}$ , if $ε (x) = ε (y)$ then $x = y$ .

The witness is the pair $σ_{1}, σ_{2} \in B_{3}$ : both braids have exponent sum $1$ , and they are distinct. The refutation is robust — the two witnesses agree not only in $ε$ but also in the $Δ$ -exponent $p = 0$ and in the number $r = 1$ of simple factors of their left Garside normal forms — so no invariant of the coarse Garside data $(ε (x), p (x), r (x))$ distinguishes them either. In the left normal form of the theorem of Garside--Elrifai--Morton they differ only in which simple factor occurs, $σ_{1}$ against $σ_{2}$ , that is, in their permutations $s_{1} = (1 2)$ against $s_{2} = (2 3)$ .

What is and is not claimed. Nothing here is a claim against the usefulness of the exponent sum: for $n \geq 2$ it is a surjective homomorphism onto $Z$ (for $n \leq 1$ it is the zero homomorphism on the trivial group), it separates $σ_{1}$ from $σ_{1}^{2}$ , and for $n \geq 3$ it also separates $σ_{1}$ from $Δ$ ; being abelian it is insensitive to the braid relations. What is refuted is only completeness. Nor is it claimed that braid groups have no complete normal forms: the left Garside normal form is one, and it is precisely the additional simple-factor data (the first factor $σ_{1}$ versus $σ_{2}$ ) that separates the two witnesses below.

Facts & Assumptions

Given: The braid group $B_{3} = ⟨ σ_{1}, σ_{2} ∣ σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2} ⟩$ , that is, the case $n = 3$ of The braid group by Artin presentation; its positive monoid $B_{3}^{+}$ with the two atoms $σ_{1}, σ_{2}$ , the length $ℓ$ and the half twist $Δ = σ_{1} σ_{2} σ_{1}$ of length $3$ ; the symmetric group $S_{3}$ with adjacent transpositions $s_{i} = (i i + 1)$ ; and the two elements $σ_{1}, σ_{2} \in B_{3}$ .

[F1]

$B_{3} = ⟨ σ_{1}, σ_{2} ∣ R ⟩$ is a presented group in the sense of Group presentation by generators and relations, where $R$ consists of the single braid relation $σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ (The braid group by Artin presentation). If $u$ is a function on the generators whose evaluation sends every relator to the identity, then $u$ extends uniquely to a homomorphism of $B_{3}$ (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group, Group presentation by generators and relations).

[F2]

In $S_{n}$ the adjacent transpositions satisfy $s_{i}^{2} = i d$ , $s_{i} s_{j} = s_{j} s_{i}$ for $∣ i - j ∣ > 1$ , and $s_{i} s_{i + 1} s_{i} = s_{i + 1} s_{i} s_{i + 1}$ ; in the composition convention of The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition the transpositions $s_{1} = (1 2)$ and $s_{2} = (2 3)$ are distinct (Reduced adjacent-transposition words have well-defined positive lifts, The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[F3]

$B_{3}^{+}$ is generated by the atoms $σ_{1}, σ_{2}$ , $ℓ ([w]) = ∣ w ∣$ is additive with $ℓ (z) = 0$ only for $z = 1$ , and the half twist $Δ = T_{1} T_{2} = σ_{1} (σ_{2} σ_{1})$ of $B_{3}$ satisfies $Δ = σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ (Positive braid monoid, Positive artin relations preserve homogeneous length, The Garside half twist and simple positive braids).

[F4]

$a ≼_{L} b$ means $b = a c$ with $c \in B_{3}^{+}$ , and then $ℓ (a) \leq ℓ (b)$ ; in particular every left divisor of an atom has length $0$ or $1$ (Left and right divisibility for positive braids, Positive artin relations preserve homogeneous length).

[F5]

$B_{3}^{+}$ is a submonoid of $B_{3}$ (The group of fractions of the positive braid monoid is the Artin braid group).

[F6]

Every $x \in B_{3}$ has a unique left normal form $x = Δ^{p} a_{1} \dots a_{r}$ with $p \in Z$ , $r \in N$ , every $a_{i}$ a proper simple braid and $a_{i} = Δ \land_{L} (a_{i} \dots a_{r})$ ; in it $p = p (x)$ is the largest integer with $Δ^{p} ≼_{L} x$ , the product $a_{1} \dots a_{r}$ is $A (x) = Δ^{- p (x)} x$ , and $p (x) \geq 0$ exactly when $x \in B_{3}^{+}$ (Left garside normal form is unique).

Counterexample

technique · direct

1.1

The exponent sum is a well-defined homomorphism. Define $u (σ_{i}) : = 1 \in Z$ on the generators of [F1]. The single relator $σ_{1} σ_{2} σ_{1} = σ_{2} σ_{1} σ_{2}$ evaluates to $1 + 1 + 1 - (1 + 1 + 1) = 0$ . By von Dyck [F1] there is therefore a unique homomorphism $ε ⁣ : B_{3} \to Z$ with $ε (σ_{i}) = 1$ for $i = 1, 2$ ; it satisfies $ε (σ_{i}^{- 1}) = - 1$ and $ε (x y) = ε (x) + ε (y)$ .

1.2

The permutation homomorphism exists. Define $v (σ_{i}) : = s_{i}$ . By the type-A relation of [F2], $v$ sends the three-term relator to $s_{1} s_{2} s_{1} (s_{2} s_{1} s_{2})^{- 1} = i d$ ; the further type-A relations $s_{i}^{2} = i d$ and the far commutation are properties of $S_{3}$ and impose no condition, because the corresponding words are not relators of [F1]. By von Dyck [F1] there is a homomorphism $π ⁣ : B_{3} \to S_{3}$ with $π (σ_{i}) = s_{i}$ .

F1F2

2.1

The two atoms are distinct. By step 1.2, $π (σ_{1}) = s_{1} = (1 2)$ and $π (σ_{2}) = s_{2} = (2 3)$ , and these are distinct by [F2]. If $σ_{1} = σ_{2}$ held in $B_{3}$ , then applying the function $π$ to the two sides would give $s_{1} = s_{2}$ , a contradiction; hence $σ_{1} \neq σ_{2}$ .

F2step 1.2

3.1

Their exponent sums agree. By step 1.1, $ε (σ_{1}) = 1 = ε (σ_{2})$ , and since $σ_{1} \neq σ_{2}$ by step 2.1, the assignment $ε$ is not injective: two distinct braids have equal exponent sum.

step 1.1step 2.1

3.2

Both witnesses are one-factor left normal forms with $p = 0$ . Fix $i \in {1, 2}$ . Since $σ_{i} \in B_{3}^{+}$ [F3], [F6] gives $p (σ_{i}) \geq 0$ ; also $Δ ̸ ≼_{L} σ_{i}$ , because a left divisor $d$ of the atom $σ_{i}$ has $ℓ (d) \leq 1$ by [F4] while $ℓ (Δ) = 3$ by [F3], so $ℓ (d) < ℓ (Δ)$ and $d \neq Δ$ . For every $p \geq 1$ , $Δ^{p} = Δ Δ^{p - 1}$ , so $Δ^{p} ≼_{L} σ_{i}$ would imply $Δ ≼_{L} σ_{i}$ by transitivity, which was just ruled out. Hence $0$ is the largest integer $p$ with $Δ^{p} ≼_{L} σ_{i}$ , that is $p (σ_{i}) = 0$ and $A (σ_{i}) = Δ^{0} σ_{i} = σ_{i}$ . Further $Δ = σ_{1} (σ_{2} σ_{1})$ for $i = 1$ and $Δ = σ_{2} (σ_{1} σ_{2})$ for $i = 2$ by [F3], so $σ_{i} ≼_{L} Δ$ , and the greedy first factor of [F6] is $a_{1} = Δ \land_{L} σ_{i} = σ_{i}$ , with remainder $σ_{i}^{- 1} σ_{i} = 1$ computed in the group $B_{3}$ , in which $B_{3}^{+}$ is a submonoid by [F5]; the factor is proper because $σ_{i} \neq 1$ and $σ_{i} \neq Δ$ by [F3], and it is simple because $σ_{i} ≼_{L} Δ$ . So the left normal form of each witness is $σ_{i} = Δ^{0} \cdot σ_{i}$ with $r = 1$ simple factor, and the two witnesses also agree in the data $p = 0$ and $r = 1$ .

F3F4F5F6step 2.1

4.1

Conclusion. The braids $σ_{1}$ and $σ_{2}$ are distinct by step 2.1, while $ε (σ_{1}) = ε (σ_{2}) = 1$ by step 3.1; moreover both have left normal form $Δ^{0} \cdot σ_{i}$ with one proper simple factor by step 3.2. The exponent sum therefore fails to be a complete normal form, and even the triple $(ε (x), p (x), r (x))$ fails to determine a braid. The refuted claim is false, already in $B_{3}$ . No choice principle is used: both homomorphisms are obtained from the explicit generator assignments $u (σ_{i}) = 1$ and $v (σ_{i}) = s_{i}$ , and every computation is a finite word computation. ∎

step 2.1step 3.1step 3.2

Remarks

Why the failure is minimal. The two witnesses are the two distinct generators of the smallest braid group that has more than one of them, $B_{3}$ ; on one generator there is no such pair to exhibit. The failure is not a defect of the presentation used to define $ε$ : the homomorphism is well defined by von Dyck's theorem, as step 1.1 checks.
What the exponent sum does see. It is additive, it takes the value $- 1$ on each inverse letter, it is surjective for $n \geq 2$ because $ε (σ_{1}) = 1$ , and for $n = 3$ it takes the value $3$ on the half twist $Δ$ and $6$ on the full twist $Δ^{2} = (σ_{1} σ_{2})^{3}$ . Being abelian data, it cannot see the non-abelian structure of the braid group, and the pair $σ_{1}, σ_{2}$ is the smallest instance of that insensitivity.
The extra data that does separate them. The two braids have the same $Δ$ -exponent and the same number of simple factors but different first (and only) simple factors, $σ_{1}$ against $σ_{2}$ , corresponding to the distinct permutations $s_{1}$ and $s_{2}$ . This is exactly the refinement supplied by the left Garside normal form of Left garside normal form is unique.
No choice principle is used; only the explicitly displayed generator assignments and finite word evaluations occur.

Garside Structure, Normal Forms, and the Center — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The simple braids and divisibility lattice for b three

Example

Facts & Assumptions

Verification

Remarks

A left garside normal form computation in b three

Example

Facts & Assumptions

Verification

Remarks

The full twist in b three

Example

Facts & Assumptions

Verification

Remarks

Exponent sum is not a complete braid normal form

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources