The Garside half twist and simple positive braids

Let $n\in \mathbb{N}$ and let $B_n^{+}$ be the positive braid monoid of Positive braid monoid with its homogeneous length $\mathrm{\ell }$ (Positive artin relations preserve homogeneous length) and its divisibility orders of Left and right divisibility for positive braids. For $1\le k\le n-1$ put

$T_k:={\sigma }_k{\sigma }_{k-1}\cdots {\sigma }_1\in B_n^{+},$

the word that moves the $k+1$-st strand across the first $k$ strands; for $n=1$ there is no $T_k$ and products below are empty. The Garside half twist (or fundamental element) of $B_n^{+}$ is

$\mathrm{\Delta }:={\mathrm{\Delta }}_n:=T_1{T}_2\cdots T_{n-1}={\sigma }_1({\sigma }_2{\sigma }_1)\cdots ({\sigma }_{n-1}{\sigma }_{n-2}\cdots {\sigma }_1),$

the class in $B_n^{+}$ of the displayed word. Equivalently, by the recursion ${\mathrm{\Delta }}_1:=1$ and ${\mathrm{\Delta }}_n:={\mathrm{\Delta }}_{n-1}{T}_{n-1}$ for $n\ge 2$, which expands to the same word. Its length is

$\mathrm{\ell }(\mathrm{\Delta })={\sum }_{k=1}^{n-1}k=\frac{n(n-1)}{2}=:N,$

since each block $T_k$ has length $k$ and the product of positive words has length equal to the sum of the lengths (Positive artin relations preserve homogeneous length). For $n=0$ or $n=1$ the alphabet is empty, $N=0$ and $\mathrm{\Delta }=1$.

The reversed triangular word. Reversing the displayed word gives

${\mathrm{\Delta }}^{\mathrm{r}\mathrm{e}\mathrm{v}}=({\sigma }_1{\sigma }_2\cdots {\sigma }_{n-1})({\sigma }_1\cdots {\sigma }_{n-2})\cdots {\sigma }_1,$

the product of the increasing blocks $U_k:={\sigma }_1{\sigma }_2\cdots {\sigma }_k$ in the order $U_{n-1}{U}_{n-2}\cdots U_1$. This is the word displayed in the plan of this page; that its class is again $\mathrm{\Delta }$ is not a formal triviality but a consequence of the braid relations, and it is proved together with the conjugation identity in Conjugation by the half twist reverses Artin generators, where reversal is also used. Until that point $\mathrm{\Delta }$ always denotes the class of the word $T_1\cdots T_{n-1}$ displayed above.

Simple positive braids. A simple positive braid is an element $s$ of $B_n^{+}$ that left-divides $\mathrm{\Delta }$ in the sense of Left and right divisibility for positive braids: $s{\preccurlyeq }_L\mathrm{\Delta }$, i.e. $\mathrm{\Delta }=sc$ for some $c\in B_n^{+}$. The set of simple positive braids is denoted ${\mathrm{D}\mathrm{i}\mathrm{v}}_L(\mathrm{\Delta })$. Since $\mathrm{\ell }$ is monotone for ${\preccurlyeq }_L$ and takes finitely many classes of words of length $\le N$, the set ${\mathrm{D}\mathrm{i}\mathrm{v}}_L(\mathrm{\Delta })$ is finite. The atoms ${\sigma }_1,\dots ,{\sigma }_{n-1}$ are the first examples of simple braids, and $\mathrm{\Delta }$ itself and $1$ are the largest and smallest; the identification of ${\mathrm{D}\mathrm{i}\mathrm{v}}_L(\mathrm{\Delta })$ with the symmetric group is Simple positive braids are indexed by permutations.

Balanced divisors. A divisor $s{\preccurlyeq }_L\mathrm{\Delta }$ is called balanced if it is also a right divisor of $\mathrm{\Delta }$, that is, if $\mathrm{\Delta }=ds$ for some $d\in B_n^{+}$; note that the complementary factor $c$ in $\mathrm{\Delta }=sc$ is a right divisor of $\mathrm{\Delta }$ for every left divisor $s$, since $\mathrm{\Delta }=sc$ exhibits $c$ as such, so the content of balancedness lies in the opposite divisibility of $s$ itself, not in that of $c$. The proof that the simple braids are exactly the balanced divisors of $\mathrm{\Delta }$ is Simple positive braids are indexed by permutations, and nothing on this page uses that equivalence before it. Where the distinction matters, a divisor of $\mathrm{\Delta }$ is called a left divisor or a right divisor of $\mathrm{\Delta }$ according to the side of $\mathrm{\Delta }$ on which it is written.

• Conventions: the blocks $T_k$ are written in decreasing index order, so that $T_k$ is the positive braid in which the $(k+1)$-st strand crosses the $k$-th, then the $(k-1)$-st, and so on. The product $T_1{T}_2\cdots T_{n-1}$ is the half turn of the $n$ strands read from the top strand downwards; the recursion ${\mathrm{\Delta }}_n={\mathrm{\Delta }}_{n-1}{T}_{n-1}$ is equation (1.6) of Dehornoy et al., Chapter I.
• Index reversal ${\sigma }_j\mapsto {\sigma }_{n-j}$ preserves the presentation and hence induces an automorphism $\tau$ of $B_n^{+}$, and similarly the reversal anti-automorphism $\rho$ of The positive braid monoid is left and right cancellative is available. The two words displayed above are reverses of one another as words: ${\mathrm{\Delta }}^{\mathrm{r}\mathrm{e}\mathrm{v}}=U_{n-1}{U}_{n-2}\cdots U_1$, with $U_k={\sigma }_1\cdots {\sigma }_k$; note that $\tau$ sends the block $T_k$ to ${\sigma }_{n-k}{\sigma }_{n-k+1}\cdots {\sigma }_{n-1}$, so $\tau$ does not simply exchange the two displayed words. That the classes of the two words agree, that $\tau (\mathrm{\Delta })=\mathrm{\Delta }$, and the conjugation identity ${\sigma }_i\mathrm{\Delta }=\mathrm{\Delta }{\sigma }_{n-i}$ are all proved in Conjugation by the half twist reverses Artin generators; until that point only the class of the $T$-word is called $\mathrm{\Delta }$.
• Nothing in this definition uses a choice principle: $\mathrm{\Delta }$ is the class of an explicit finite word, and the modularity of the recursion is a finite induction on $n$.