# A question of braid words

Let $(W,S)$ be a Coxeter group, let $B(W,S)$ be the corresponding braid or Artin-Tits group. Set $S=\{s_1,\dots, s_n\}$ and denote by $\bf{S}=\{\sigma_1,\dots, \sigma_n\}$ the corresponding generators of $B(W)$.

I would like to know if the following is true. Let $\beta\in B(W,S)$. Assume that $\beta$ has two reduced braid words (i.e., words of minimal length with respect to the set $\bf{S}\cup\bf{S}^{-1}$) of the form $\sigma_{i_1}^{\varepsilon_1} \sigma_{i_2}^{\varepsilon_2}\cdots \sigma_{i_k}^{\varepsilon_k}$ and $\sigma_{i_1}^{\delta_1} \sigma_{i_2}^{\delta_2}\cdots \sigma_{i_k}^{\delta_k}$ respectively, where $i_j\in\{1,\dots,n\}$, $\varepsilon_{j},\delta_j\in\{\pm 1\}$. Then $\epsilon_j=\delta_j$ for all $j$.

By induction on the word length this can obviously be reduced to showing that $\varepsilon_1=\delta_1$. If the image of the word in the Coxeter group stays reduced, then it is easy (for example by expanding the image of the two words in the standard basis of the Hecke algebra one sees that the first exponents must be equal), but in general I have no idea. I would be very grateful if anyone had a hint towards an answer or a counterexample.

• Artin-Tits groups associated to general coxeter groups are called "generalized braid groups". The special terminology "braid groups" is reserved for the Artin-Tits group associated to a finite symmetric group $S_n$. – Lee Mosher Sep 5 '15 at 15:48