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The center of b two is all of b two

Statement

Let $B_{2}$ be the braid group of The braid group by Artin presentation, with its single generator $σ_{1}$ , and let $Δ$ be the half twist of The Garside half twist and simple positive braids, so that $Δ = σ_{1}$ for $n = 2$ . Then $B_{2}$ is infinite cyclic, $B_{2} = {σ_{1}^{k} : k \in Z} = ⟨ Δ ⟩,$ and $B_{2}$ is abelian; consequently $Z (B_{2}) = B_{2} = ⟨ Δ ⟩ .$ This is the exceptional case $n = 2$ of the centre theorem: for $n > 2$ one has $Z (B_{n}) = ⟨ Δ^{2} ⟩ \neq B_{n}$ (The center of b n is generated by the full twist for n greater than two). The argument is choice free; it uses the free-group description of $B_{2}$ rather than the free-group reduced-word theorem in the form " $σ_{1} \neq 1$ ".

Facts & Assumptions

Given: The braid group $B_{2}$ with its single generator $σ_{1}$ and no relation, and the half twist $Δ = Δ_{2}$ .

[F1]

For $n = 2$ the presentation of The braid group by Artin presentation has the single generator $σ_{1}$ ; the braid relation $σ_{i} σ_{i + 1} σ_{i} = σ_{i + 1} σ_{i} σ_{i + 1}$ requires $1 \leq i \leq 0$ and the commutation relation requires a pair $i, j \in {1}$ with $∣ i - j ∣ > 1$ , so there is no relation at all. By Group presentation by generators and relations the presented group is the quotient $F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ with $X = {σ_{1}}$ and $R = \emptyset$ , and the normal closure of the empty set is trivial.

[F2]

The reduced words on $X ⊔ X^{- 1}$ form a group with multiplication given by concatenation followed by free reduction, and the one-letter words realise the universal property of the free group on $X$ (Reduced words form the free group on an alphabet).

[F3]

$Δ = Δ_{n} = T_{1} T_{2} \dots T_{n - 1}$ with $T_{k} = σ_{k} σ_{k - 1} \dots σ_{1}$ (The Garside half twist and simple positive braids); for $n = 2$ this is $Δ = T_{1} = σ_{1}$ .

Proof

technique · direct

1.1

$B_{2}$ is free on one generator. By [F1] $B_{2} = F ({σ_{1}}) / ⟨ ⁣ ⟨ \emptyset ⟩ ⁣ ⟩$ and the normal closure of the empty set is the trivial subgroup, so $B_{2} ≅ F ({σ_{1}})$ via the identity on the generator.

2.1

The elements of $B_{2}$ . By [F2] the elements of $F ({σ_{1}})$ are the freely reduced words on ${σ_{1}, σ_{1}^{- 1}}$ . A word on this two-letter alphabet is reduced exactly when it contains no adjacent pair $σ_{1} σ_{1}^{- 1}$ or $σ_{1}^{- 1} σ_{1}$ , i.e. exactly when it has the form $σ_{1}^{k}$ for a unique $k \in Z$ (with $k = 0$ for the empty word); for such a word no free reduction applies, so two of them represent different elements unless the exponents are equal. Hence $B_{2} = {σ_{1}^{k} : k \in Z}$ with $σ_{1}^{k} σ_{1}^{l} = σ_{1}^{k + l}$ , i.e. $B_{2}$ is infinite cyclic and abelian.

F2step 1.1

3.1

$Δ = σ_{1}$ and the centre. By [F3] with $n = 2$ , $Δ = T_{1} = σ_{1}$ , so $⟨ Δ ⟩ = ⟨ σ_{1} ⟩ = B_{2}$ by step 2.1. Since $B_{2}$ is abelian (step 2.1), every element commutes with every other, whence $Z (B_{2}) = B_{2} = ⟨ Δ ⟩$ .

F3step 2.1

4.1

Assembly and contrast with $n > 2$ . Steps 1.1, 2.1 and 3.1 give the infinite cyclic description and the centre. This is genuinely exceptional: for $n > 2$ the centre is the proper subgroup $⟨ Δ^{2} ⟩$ generated by the full twist, and $Δ \notin Z (B_{n})$ , as proved in The center of b n is generated by the full twist for n greater than two; the difference is that for $n = 2$ the two atoms $σ_{1}$ and $σ_{n - 1}$ coincide. The item uses only the empty presentation of $B_{2}$ and the free-group description of its elements; no geometric statement about two-strand braids and no choice principle is used. ∎

step 1.1step 2.1step 3.1

Remarks

A shortcut avoided. A tempting proof of the infinite order of $σ_{1}$ invokes torsion freeness of the free group together with the nonidentity of $σ_{1}$ ; the nonidentity is exactly what the algebraically presented free group gives by construction, and it is recorded here through the reduced-word description of $F ({σ_{1}})$ rather than through a separate torsion argument.
The two-strand exception. The centre theorem for $n > 2$ rules out odd powers of $Δ$ because $σ_{1}$ and $σ_{n - 1}$ are distinct atoms; for $n = 2$ there is only one atom, all powers of $Δ = σ_{1}$ are central, and the centre is the whole group.
Nothing here uses the Axiom of Choice or any weaker choice principle.

Depends on

Used by

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Dependency tree · two levels

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Sources

J. Gonzalez-Meneses, Basic results on braid groups, Section 4.3, printed p. 31 (standard reference, not scraped)