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The center of b n is generated by the full twist for n greater than two

Statement

Let $n > 2$ , let $B_{n}$ be the braid group of The braid group by Artin presentation with its generators $σ_{1}, \dots, σ_{n - 1}$ and its half twist $Δ$ (The Garside half twist and simple positive braids). Call $Δ^{2}$ the full twist of $B_{n}$ . Then the center of $B_{n}$ is $Z (B_{n}) : = {x \in B_{n} : x g = g x for all g \in B_{n}} = ⟨ Δ^{2} ⟩ = {Δ^{2 k} : k \in Z},$ and it is infinite cyclic: $Δ^{2} \neq 1$ and the map $Z \to Z (B_{n})$ , $k \mapsto Δ^{2 k}$ , is a group isomorphism.

The statement is false for $n = 2$ , where the center is all of $B_{2} = ⟨ Δ ⟩$ : that exception is The center of b two is all of b two. The proof is choice free.

Facts & Assumptions

Given: A natural number $n > 2$ , the braid group $B_{n}$ with generators $σ_{1}, \dots, σ_{n - 1}$ , the positive braid monoid $B_{n}^{+}$ with half twist $Δ$ of length $N = n (n - 1) / 2$ , and a group element $x \in B_{n}$ .

[F1]

The square of the half twist is central. $σ_{i} Δ^{2} = Δ^{2} σ_{i}$ for every $i$ ; moreover $σ_{i} Δ = Δ σ_{n - i}$ (Conjugation by the half twist reverses Artin generators).

[F2]

Central positive braids. If $z \in B_{n}^{+}$ is central in $B_{n}$ , then $z = Δ^{2 s}$ for some $s \geq 0$ (A central positive braid is a power of delta squared for n greater than two).

[F3]

$Δ$ -power divisibility. Every positive braid $b \in B_{n}^{+}$ is a right divisor of some power of $Δ$ : there is $k \geq 0$ with $Δ^{k} = c b$ for some $c \in B_{n}^{+}$ ; exponents may be enlarged, so an exponent of the form $2 r$ may be chosen (Every positive braid divides a power of the half twist on both sides).

[F4]

Description by fractions. Every element of $B_{n}$ has the form $a b^{- 1}$ with $a, b \in B_{n}^{+}$ , the positive monoid being regarded as a submonoid of $B_{n}$ through its embedding (The group of fractions of the positive braid monoid is the Artin braid group).

[F5]

Length and torsion freeness. On $B_{n}^{+}$ the length is additive, $ℓ (w) = 0$ only for $w = 1$ , and $ℓ (Δ^{m}) = m N$ for $m \geq 0$ ; the group $B_{n}$ is torsion free (Positive artin relations preserve homogeneous length, Braid groups are torsion free by the garside lattice).

Proof

technique · direct

1.1

$⟨ Δ^{2} ⟩ \subseteq Z (B_{n})$ . By [F1] $Δ^{2}$ commutes with every generator $σ_{i}$ ; multiplying the identity $σ_{i} Δ^{2} = Δ^{2} σ_{i}$ on both sides by $σ_{i}^{- 1}$ shows that $Δ^{2}$ commutes with $σ_{i}^{- 1}$ as well. Every element of $B_{n}$ is a product of generators and their inverses (it is a class of a word in the $σ_{i}^{\pm 1}$ ), so induction on the number of letters of such a word shows that $Δ^{2} g = g Δ^{2}$ for every $g \in B_{n}$ ; hence $Δ^{2} \in Z (B_{n})$ and every power $Δ^{2 k}$ , $k \in Z$ , is central.

F1F4

1.2

Enlarging the exponent to an even one. Let $b \in B_{n}^{+}$ . By [F3] there is $k \geq 0$ and $c \in B_{n}^{+}$ with $Δ^{k} = c b$ ; choose $r$ with $2 r \geq k$ (and $2 r \geq 0$ ). Then $Δ^{2 r} = Δ^{2 r - k} Δ^{k} = (Δ^{2 r - k} c) b$ with $Δ^{2 r - k} c \in B_{n}^{+}$ , so $Δ^{2 r} = c^{'} b$ for some $c^{'} \in B_{n}^{+}$ : $b$ is a right divisor of an even power of $Δ$ .

F3F5

2.1

Every central element is a power of $Δ^{2}$ . Let $x \in Z (B_{n})$ . By [F4] write $x = a b^{- 1}$ with $a, b \in B_{n}^{+}$ , and by step 1.2 choose $r \geq 0$ and $c \in B_{n}^{+}$ with $Δ^{2 r} = c b$ ; thus $b^{- 1} = Δ^{- 2 r} c$ and, since $Δ^{2 r}$ is central by step 1.1, $z : = Δ^{2 r} x = Δ^{2 r} a b^{- 1} = a Δ^{2 r} b^{- 1} = a c b b^{- 1} = a c \in B_{n}^{+}$ . The element $z$ is central: it is the product of the central elements $Δ^{2 r}$ and $x$ . Hence by [F2] $z = Δ^{2 s}$ with $s \geq 0$ , and therefore $x = Δ^{- 2 r} z = Δ^{2 (s - r)} = (Δ^{2})^{s - r} \in ⟨ Δ^{2} ⟩$ .

F2F4step 1.1step 1.2

3.1

Infinite cyclic order. By steps 1.1 and 2.1, $Z (B_{n}) = ⟨ Δ^{2} ⟩$ , and $Z \to Z (B_{n})$ , $k \mapsto Δ^{2 k}$ , is a surjective homomorphism. It is injective: if $Δ^{2 k} = 1$ with $k > 0$ then $Δ^{2}$ is a nonidentity torsion element, because $Δ^{2} \neq 1$ — indeed $ℓ (Δ^{2}) = 2 N > 0 = ℓ (1)$ and $ℓ (w) = 0$ forces $w = 1$ in $B_{n}^{+}$ — contradicting the torsion freeness of $B_{n}$ . Hence $Z (B_{n}) ≅ Z$ is infinite cyclic, generated by the full twist $Δ^{2}$ .

F5step 1.1step 2.1

4.1

Assembly. The inclusion is step 1.1, the reverse inclusion is step 2.1 and the cyclic description is step 3.1. The key use of the hypothesis $n > 2$ is in [F2], where an odd exponent of $Δ$ is excluded by the distinct atoms $σ_{1} \neq σ_{n - 1}$ ; for $n = 2$ the argument fails exactly because $σ_{1} = σ_{n - 1}$ , and the center is larger. All steps are algebraic; no geometric model of braids is used and no choice principle is used. ∎

step 1.1step 1.2step 2.1step 3.1

Remarks

The full twist. The generator of the center is the square of the half twist, $Δ^{2} = (σ_{1} σ_{2} \dots σ_{n - 1})^{n}$ in the classical notation for type $A$ ; here only the description $Δ^{2}$ in terms of the triangular word of The Garside half twist and simple positive braids is used, so the identity with $(σ_{1} \dots σ_{n - 1})^{n}$ is not needed.
Why centrality of squares helps. The two ingredients are structural: an arbitrary group element can be shifted into the positive monoid by a central even power of $Δ$ (step 1.2, applied in step 2.1), and central positive braids are even $Δ$ -powers (the preceding lemma). The same two ingredients give Garside's theorem in J. González-Meneses, Basic results on braid groups, Theorem 4.2, printed pp. 30--31.
Comparison with $n = 2$ . For $n = 2$ the conclusion is false: $B_{2}$ is abelian, so $Z (B_{2}) = B_{2} = ⟨ Δ ⟩$ (The center of b two is all of b two). Both statements together give the complete description of the center of $B_{n}$ for every $n \geq 2$ .
Nothing here uses the Axiom of Choice or any weaker choice principle.

Depends on

Used by

The full twist in b three Example
The center of b two is all of b two Proposition

Dependency tree · two levels

26 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, Theorem 4.2, printed pp. 30-31 (standard reference, not scraped)
J. Birman and T. Brendle, Braids: A Survey, Section 5.2 (standard reference, not scraped)