# Presentation of the pure Artin groups

Let $W$ be a Coxeter group attached to a Coxeter matrix with entries $m_{ij}$ . The presentation of $W$ is given by $$W= < T_1, \dots, T_n | T_i^2=1, T_iT_jT_i \ldots = T_jT_iT_j \ldots, i \neq j>$$ where each side of the second equation has $m_{ij}$ terms.

Suppose $B_{W}$ is the corresponding braid group (aka Artin group) obtained by removing the relations $T_i^2=1$: $$B_{W}=< T_1, \dots, T_n | T_iT_jT_i \cdots = T_jT_iT_j \cdots>.$$

Then there is a canonical surjective homomorphism $B_{W} \to W$, the kernel of this homomorphism is called pure Artin group.

Does anyone know the presentation of the pure Artin group?

Thank you!

• Since Jim Humphreys thinks the question is stated loosely, could I try to rephrase? Let's suppose that $W$ is the Coxeter group attached to a Coxeter matrix with entries $m_{ij}$, with generators $T_i$ and relations $T_{i}^2 = 1$ in addition to the relations $(T_i T_j)^{m_{ij} = 1$ for $i \neq j$, and suppose $B_W$ is the corresponding braid group obtained by removing the relations $T_{i}^2 = 1$. Then there is a canonical surjective homomorphism $B_W \to W$, and you want a presentation of the kernel of this homomorphism. Is that the question? – Todd Trimble Nov 9 '11 at 2:27
• Thank you, Todd! That's exactly my question, and I edited my question according to your rephrasing. Thank you again! – Yaping Yang Nov 9 '11 at 3:31
• I took the liberty of changing the form of the presentation of the first by replacing relations $(T_iT_j)^{m_{ij}} = 1$ by $T_iT_jT_i\ldots = T_jT_iT_j\ldots$, because if we simply remove the equations $T_i^2 = 1$ as stated, the group presented by $(T_iT_j)^{m_{ij}} = 1$ is different from the group presented by $T_iT_jT_i\ldots = T_jT_iT_j\ldots$. (I think this was my fault, not yours.) – Todd Trimble Nov 9 '11 at 13:40
• Yes, Todd! You are right. Thank you for correcting! – Yaping Yang Nov 10 '11 at 16:20

• You can find In this paper of Enriquez a presentation for the pure Artin group of type B: http://arxiv.org/abs/math/0408035 Proposition 1.1 (by setting $N=2$ is the formulaes)
• For all the infinite families, the corresponding pure braid groups are iterated semi-direct products of free groups. (For all the families but the $D_n$ one, it follows from the fact that the corresponding hyperplane arrangements are of fiber type, hence these are even almost-direct product). It should leads quite easily to a nice presentation of these groups.
Edit: You may also be interested by this paper of Crips and Paris: http://arxiv.org/abs/math/0210438. Recall that $W_{B_n}=(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ and that $W_{D_n}=(\mathbb{Z}/2\mathbb{Z})^{n-1}\rtimes S_n$. It is proved in this paper that $B_{B_n}=F_n \rtimes B_n$ and that $B_{D_n}=F_{n-1}\rtimes B_n$ where $B_n$ is the braid group of type $A$ and $F_n$ is a free group, and that the canonical projection from the braid group to the Coxeter group is compatible with these decomposition (in the sense that in both cases it restricts to the obvious projections $B_n \rightarrow S_n$ and $F_k \rightarrow (\mathbb{Z}/2\mathbb{Z})^k$).