Exponent sum and writhe of a braid

For $n\ge 2$ let $B_n$ be the braid group of the Artin presentation (The braid group by Artin presentation), with generators ${\sigma }_1,\dots ,{\sigma }_{n-1}$. The assignment

${\sigma }_i⟼1(1\le i\le n-1)$

has equal values on the two sides of every defining relator of that presentation: a braid relator ${\sigma }_i{\sigma }_{i+1}{\sigma }_i={\sigma }_{i+1}{\sigma }_i{\sigma }_{i+1}$ has three letters on each side, and a distant-commutativity relator ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i$ has two letters on each side. Hence Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group gives a unique homomorphism

$w_n\text{\hspace{0.17em}⁣}:B_n⟶\mathbb{Z},w_n({\sigma }_i^{\pm 1})=\pm 1.$

It is the exponent sum, or writhe, of a braid. If $\beta ={\sigma }_{i_1}^{{ϵ}_1}\cdots {\sigma }_{i_k}^{{ϵ}_k}$ is an Artin word for $\beta$ with ${ϵ}_l\in \{\pm 1\}$, then $w_n(\beta )={\sum }_l{ϵ}_l$: the number of positive letters minus the number of negative letters. For $n=0,1$ set $w_n=0$ on the trivial group $B_n$.

The homomorphisms $w_n$ assemble to a function $w\text{\hspace{0.17em}⁣}:{⨆}_{n\ge 0}{B}_n\to \mathbb{Z}$. Let ${\iota }_n\text{\hspace{0.17em}⁣}:B_n\to B_{n+1}$, ${\iota }_n({\sigma }_i)={\sigma }_i$, be the standard inclusion, which is well defined because the defining relators of $B_n$ are among those of $B_{n+1}$. Then

$w_{n+1}({\iota }_n(\beta ))=w_n(\beta ),w_{n+1}({\iota }_n(\beta ){\sigma }_n^{\pm 1})=w_n(\beta )\pm 1,$

for all $n\ge 1$ and $\beta \in B_n$: the first identity holds because both sides are additive over an Artin word for $\beta$ and agree on generators, and the second adds the single letter ${\sigma }_n^{\pm 1}$. Consequently $w$ is invariant under conjugation,

$w(\gamma \beta {\gamma }^{-1})=w(\beta )$

for all braids $\gamma ,\beta$ for which the product is defined, since $w(\gamma \beta {\gamma }^{-1})=w(\gamma )+w(\beta )-w(\gamma )$.

The exponent sum is invariant under conjugation, while under stabilization it shifts by $\pm 1$; it is not a complete invariant of braids. For instance in $B_3$ the braids ${\sigma }_1{\sigma }_2$ and ${\sigma }_2{\sigma }_1$ have equal exponent sum $2$ and are distinct: their images under ${\sigma }_i\mapsto (ii+1)$ are the two different three-cycles. This generator assignment respects the Artin relations by direct permutation multiplication, so it extends by Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group.