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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.

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4.1, printed pp. 29-30 (standard reference, not scraped)
J. Birman and T. Brendle, Braids: A Survey, Section 5.1 (standard reference, not scraped)