The full twist in b three

The half twist and its two words. By [F1], $\mathrm{\Delta }={\sigma }_1{\sigma }_2{\sigma }_1$ has length $3$ and the braid relation gives the second triangular word $\mathrm{\Delta }={\sigma }_2{\sigma }_1{\sigma }_2$; by [F3] the monoid $B_3^{+}$ sits inside the group $B_3$, so $\mathrm{\Delta }$ may be inverted and cancelled there.

Both generators commute with ${\mathrm{\Delta }}^2$. Applying the identity ${\sigma }_i\mathrm{\Delta }=\mathrm{\Delta }{\sigma }_{3-i}$ of [F2] twice, ${\sigma }_1{\mathrm{\Delta }}^2=({\sigma }_1\mathrm{\Delta })\mathrm{\Delta }=(\mathrm{\Delta }{\sigma }_2)\mathrm{\Delta }=\mathrm{\Delta }({\sigma }_2\mathrm{\Delta })=\mathrm{\Delta }(\mathrm{\Delta }{\sigma }_1)={\mathrm{\Delta }}^2{\sigma }_1$, and likewise ${\sigma }_2{\mathrm{\Delta }}^2=({\sigma }_2\mathrm{\Delta })\mathrm{\Delta }=(\mathrm{\Delta }{\sigma }_1)\mathrm{\Delta }=\mathrm{\Delta }({\sigma }_1\mathrm{\Delta })=\mathrm{\Delta }(\mathrm{\Delta }{\sigma }_2)={\mathrm{\Delta }}^2{\sigma }_2$.

The cube of ${\sigma }_1{\sigma }_2$ is the square of the half twist. Using associativity and the two words for $\mathrm{\Delta }$ from step 1.1, $({\sigma }_1{\sigma }_2{)}^3={\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2{\sigma }_1{\sigma }_2=({\sigma }_1{\sigma }_2{\sigma }_1)({\sigma }_2{\sigma }_1{\sigma }_2)=\mathrm{\Delta }\cdot \mathrm{\Delta }={\mathrm{\Delta }}^2$.

Centrality of the full twist. By [F4] every $z\in B_3$ is a finite product of the letters ${\sigma }_1^{\pm 1},{\sigma }_2^{\pm 1}$. Step 1.2 gives ${\mathrm{\Delta }}^2{\sigma }_i={\sigma }_i{\mathrm{\Delta }}^2$ for $i=1,2$, and this passes to inverses: from ${\mathrm{\Delta }}^{2}g=g{\mathrm{\Delta }}^2$ one gets $g^{-1}{\mathrm{\Delta }}^2={\mathrm{\Delta }}^2{g}^{-1}$ by multiplying the identity on the left and on the right by $g^{-1}$. Induction on the number of letters therefore gives ${\mathrm{\Delta }}^{2}z=z{\mathrm{\Delta }}^2$ for every $z\in B_3$, that is ${\mathrm{\Delta }}^2\in Z(B_3)$.

The centre is infinite cyclic on the full twist. By [F5] with $n=3$, $Z(B_3)=⟨{\mathrm{\Delta }}^2⟩=\{{\mathrm{\Delta }}^{2k}:k\in \mathbb{Z}\}$, the element ${\mathrm{\Delta }}^2$ is not $1$, and $k\mapsto {\mathrm{\Delta }}^{2k}$ is an isomorphism $\mathbb{Z}\to Z(B_3)$; combining this with step 2.1, the centre is infinite cyclic and generated by the full twist ${\mathrm{\Delta }}^2=({\sigma }_1{\sigma }_2{)}^3$, consistently with the local centrality verification of step 2.2.

The half twist alone is not central. By [F2] with $i=1$ one has $\mathrm{\Delta }{\sigma }_1={\sigma }_2\mathrm{\Delta }$. If $\mathrm{\Delta }$ were central then also $\mathrm{\Delta }{\sigma }_1={\sigma }_1\mathrm{\Delta }$, so ${\sigma }_2\mathrm{\Delta }={\sigma }_1\mathrm{\Delta }$ and right cancellation by the invertible element $\mathrm{\Delta }$ in $B_3$ [F3] gives ${\sigma }_2={\sigma }_1$, contradicting $\pi ({\sigma }_2)=s_2\mathrm{\ne }{s}_1=\pi ({\sigma }_1)$ [F6]. Hence $\mathrm{\Delta }\mathrm{\notin }Z(B_3)$, although ${\mathrm{\Delta }}^2\in Z(B_3)$ by step 2.2: the full twist is the square of the half twist and is not the half twist itself.

Conclusion. In $B_3$ the full twist satisfies ${\mathrm{\Delta }}^2=({\sigma }_1{\sigma }_2{)}^3$ by step 2.1, it is central by step 2.2, it generates the centre $Z(B_3)=⟨{\mathrm{\Delta }}^2⟩\cong \mathbb{Z}$ by step 3.1, and the half twist $\mathrm{\Delta }$ 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 $\mathrm{\Delta }$. ∎