---
author:
- 'Filippo Callegaro & Ivan Marin'
bibliography:
- 'biblio.bib'
date: 'November 18, 2010'
title: '[**Homology computations for complex braid groups**]{}'
---

Scuola Normale Superiore\
Piazza dei Cavalieri, 7\
56126 Pisa\
Italy\
—\
Institut de Mathématiques de Jussieu\
Université Paris 7\
175 rue du Chevaleret\
75013 Paris\
France

[**Abstract.**]{} Complex braid groups are the natural generalizations of braid groups associated to arbitrary (finite) complex reflection groups. We investigate several methods for computing the homology of these groups. In particular, we get the Poincaré polynomial with coefficients in a finite field for one large series of such groups, and compute the second integral cohomology group for all of them. As a consequence we get non-isomorphism results for these groups.

[**MSC 2010 : 20J06, 20F36, 20F55**]{}

Complexes from Garside theory
=============================

We recall a few homological constructions from the theory of Garside monoids and groups. Recall that a Garside group $G$ is the group of fractions of a Garside monoid $M$, where Garside means that $M$ satisfy several conditions for which we refer to [@dehpar]. In particular, $M$ admits (left) lcm’s, and contains a special element, called the Garside element. We denote $\mathcal{X}$ the set of atoms in $M$, assumed to be finite. The homology of $G$ coincides with the homology of $M$. Garside theory provides two useful resolutions of $\Z$ by free $\Z M$-modules.

The first one was defined in [@CMW]. Another one, with more complicated differential but a smaller number of cells, has been defined in [@DL].

The Dehornoy-Lafont complex
---------------------------

Let $M$ be a Garside monoid with a finite set of atoms $\mathcal{X}$. We choose an arbitrary linear order $<$ on $\mathcal{X}$. For $m \in M$, denote $\md(m)$ denote the smaller element in $\mathcal{X}$ which divides $m$ on the right ($m = a\md(m)$ for some $a \in M$). Recall that $\lcm(x,y)$ for $x,y \in M$ denotes the least common multiple on the left, that is $v = gx = h y$ implies $v = j \lcm(x,y)$ for some $j \in M$. If $A = (x,B)$ is a list of elements in $M$ we define inductively $\lcm(A) = \lcm(x, \lcm(B))$.

A $n$-cell is a $n$-tuple $[x_1,\dots,x_n]$ of elements in $\mathcal{X}$ such that $x_1 < \dots < x_n$ and $x_i
= \md (\lcm(x_i,x_{i+1},\dots,x_n))$. Let $\mathcal{X}_n$ denote the set of all such $n$-cells. By convention $\mathcal{X}_0
= \{ [\emptyset] \}$. The set $C_n$ of $n$-chains is the free $\Z M$-module with basis $\mathcal{X}_n$. A differential $\partial_n : C_n \to C_{n-1}$ is defined recursively through two auxiliary $\Z$-module homomorphisms $s_n : C_n \to C_{n+1}$ and $r_n : C_n \to C_n$. Let $[\alpha,A]$ be a $(n+1)$-cell, with $\alpha \in \mathcal{X}$ and $A$ a $n$-cell. We let $\alpha_{/A}$ denote the unique element in $M$ such that $(\alpha_{/A}) lcm(A) = lcm(\alpha,A)$. The defining equations for $\partial$ and $r$ are the following ones. $$\partial_{n+1}[\alpha,A] = \alpha_{/A} [A] - r_n(\alpha_{/A} [A]), \ \ r_{n+1} = s_n \circ
\partial_{n+1}, \ \ r_0(m[\emptyset]) = [\emptyset].$$ In order to define $s_n$, we say that $x [A]$ for $x \in M$ and $A$ a $n$-cell is *irreducible* if $x = 1$ and $A = \emptyset$, or if $\alpha = \md(x \lcm(A))$ coincides with the first coefficient in $A$. In that case, we let $s_n(x [A]) = 0$, and otherwise $$s_n(x[A]) = y[\alpha,A] + s_n(y r_n(\alpha_{/A} [A]))$$ with $x = y \alpha_{/A}$.

The Charney-Meyer-Wittlesey complex
-----------------------------------

Let again $G$ denote the group of fractions of a Garside monoid $M$, with Garside element $\Delta$. Let $\mathcal{D}$ denote the set of simple elements in $M$, namely the (finite) set of proper divisors of $\Delta$. We let $\mathcal{D}_n$ denote the set of $n$-tuples $[\mu_1|\dots|\mu_n]$ such that each $\mu_i$ as well as the product $\mu_1 \dots \mu_n$ lie in $\mathcal{D}$. The differential from the free $\Z M$-modules $\Z M \mathcal{D}_n$ to $\Z M \mathcal{D}_{n-1}$ is given by $$\partial_n [\mu_1|\dots|\mu_n] = \mu_1 [\mu_2|\dots|\mu_n]
+ \sum_{i=1}^{n-1} (-1)^i [\mu_1,\dots,\mu_i \mu_{i+1},\dots,\mu_n]
+ (-1)^n [\mu_1|\dots|\mu_{n-1}]$$ This complex in general has larger cells than the previous one. Its main advantage for us is that the definition of the differential is simpler, and does not involve many recursion levels anymore.

Application to the exceptional groups
-------------------------------------

When $W$ is well-generated, meaning that it can be generated by $n$ reflections, where $n$ denotes the rank of $W$, then $B$ is the group fractions of (usually) several Garside monoids that generalize the Birman-Ko-Lee monoid of the usual braid groups. These monoids have been introduced by D. Bessis in [@BESSISKPI1] and call there dual braid monoids. They are determined by the choice of a so-called Coxeter element $c$. Such an element is regular, meaning that it admits only one eigenvalue different from 1 with the corresponding eigenvector outside the reflection hyperplanes. A Coxeter element is a regular element with eigenvalue $\exp(2 \ii \pi/h)$, where $h$ denotes the (generalized) Coxeter number for $W$, namely its highest degree as a reflection group.

The corresponding Garside monoid $M_c$ is then generated by some set $R_c$ of braided reflections with relations of the form $r r' = r' r''$ (see [@BESSISKPI1] for more details). The above complexes for these monoids have been implemented by Jean Michel and the second author, using the (development version of) the CHEVIE package for GAP3. The chosen Coxeter element are indicated in Table \[coxeterdualmonoids\], in terms of the usual presentations of these groups (see [@BMR] for an explanations of the diagrams).

= lcircle10 scaled 1950 = lcircle10 scaled 3 = lcircle10 scaled 2 = lcircle10 \#1\#2[[-0.4pt\_[\#2]{}-8.6pt[\#1]{}2.3pt]{}]{} \#1pt[[ width\#1pt height4pt depth-3pt]{}]{} \#1[[-0.6pt\_[\#1]{}-1pt]{}]{} |\#1pt[[width\#1pt height3pt depth-2pt]{}]{} \#1[[-0.4pt-7pt[\#1]{}2.6pt]{}]{} \#1pt[[ width\#1pt height5pt depth-4pt]{}]{} \#1\#2[-1.5pt\^[\#1]{}-2pt]{} \#1\#2\#3[ \#1|14pt-13pt7.5pt \#2 -29pt9.5pt -2pt 17.5pt -1pt9.5pt ]{} \#1\#2[ 19.4pt]{} \#1\#2 \#1\#2\#3

$$\begin{array}{c||c|c|c|c|c}
\mbox{Group} & G_{24} & G_{27} & G_{29} & G_{33} & G_{34} \\
\hline
\mbox{Diagram} &  \nnode s\rlap{\kern 2.5pt{\raise 6pt\hbox{$\triangle$}}}
 \dbar14pt\nnode t
 \kern-26.2pt\raise8.5pt\hbox{$\diagup$}
 \kern -3.1pt
 \raise16.7pt\hbox{$\node$\rlap{\raise 2pt\hbox{$\kern 1pt\scriptstyle u$}}}
 \kern -3pt\raise8.3pt\hbox{$\diagdown$}
 \kern -7.6pt\raise9pt\hbox{$\diagdown$}
 \kern 4pt &  \nnode s\rlap{\kern 2.5pt{\raise 6pt\hbox{$\triangle$}}}
 \dbar14pt\nnode t
 \rlap{\kern-6pt\raise11pt\hbox{$\scriptstyle 5$}}
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 \kern -2pt
 \raise16.5pt\hbox{$\node$\rlap{\raise 2pt\hbox{$\kern 1pt\scriptstyle u$}}}
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 \kern 4pt &  \nnode s\bar10pt
 \nnode t\rlap{\kern 1pt\raise9pt\hbox{$\underleftarrow{}$}}
        \rlap{\kern 6.5pt\vrule width1pt height16pt depth-11pt
\kern 1pt \vrule width1pt height16pt depth-11pt}
\dbar14pt\nnode u
 \kern-27pt\raise9.5pt\hbox{$\diagup$}
 \kern -3pt\raise18.5pt
\hbox{$\node$\rlap{\raise 2pt\hbox{$\kern 1pt\scriptstyle v$}}}
 \kern -2pt\raise9.5pt\hbox{$\diagdown$} & \nnode s\bar10pt\trianglerel tvu\bar10pt\nnode w & 
\nnode s\bar10pt\trianglerel tvu\bar10pt\nnode w\bar10pt\nnode x \\
\hline
\mbox{Coxeter\ element} & stu & uts & stvu & wvtsu & xwuvts \\
\hline
\end{array}$$

Using the HAP package for GAP4 we then obtained the homologies described in Table \[tableexc\] (we recall in Table \[tableexcCox\] the ones obtained earlier by Salvetti for the Coxeter groups) except for the groups $G_{12}, G_{13}, G_{22}, G_{31}$, which are not well-generated, as well as the $H_3(B,\Z)$ of type $G_{33}$. When $W$ has type $G_{13}$, the group $B$ is the same as when $W$ has Coxeter type $I_2(6)$, and the result is known. For $G_{12}$ and $G_{22}$ one can use Garside monoids introduced by M. Picantin in [@THESEPIC].

A complex for $G_{31}$ can be obtained from the theory of Garside categories by considering it as the the centralizer of some regular element in the Coxeter group $E_8$. This viewpoint was used in [@BESSISKPI1] in order to prove that the corresponding spaces $X$ and $X/W$ are $K(\pi,1)$. More precisely, a simplicial complex (reminiscent from the Charney-Meyer-Wittlesey complex) is constructed in [@BESSISKPI1], which is homotopically equivalent to $X/W$. From this construction, we got a complex from an implementation by Jean Michel in CHEVIE.

However, for $G_{31}$, $G_{33}$ and $G_{34}$, the complexes obtained are too large to be dealt with completely through usual computers and software. The one missing for $G_{31}$ and $G_{33}$ are the middle homology $H_2(B,\Z)$ for $G_{31}$ and $H_3(B,\Z)$ for $G_{33}$. For $G_{33}$ the Dehornoy-Lafont complex for $G_{33}$ is however computable in reasonable time, and its small size enables to compute the whole homology by standard methods. For $G_{31}$, for which there is so far no construction analogous to the Dehornoy-Lafont complex, we used the following method for computing $H_2(B,\Z)$.

We first get $H_2(B,\Q) = 0$ by computing the second Betti number from the lattice. Indeed, recall from [@OT] (cor. 6.17, p.223) that the Betti numbers of $X/W$ can be in principle computed from the lattice of the arrangement. Precisely, the second Betti number of $X/W$ is given by $\sum_{Z \in T_2} |\mathcal{H}_Z/W_Z| -1$ where $T_i$ is a system of representatives modulo $W$ of codimension $i$ subspaces in the arrangement lattice ; for $Z$ such a subspace, $\mathcal{H}_Z = \{ H \in \mathcal{A} \ | \ 
H \supset Z \}$, $W_Z = \{ w \in W \ | \ w(Z) = Z \}$. More generally, the $i$-th Betti number is given by $$(-1)^i \sum_{Z \in T_i} \sum_{\sigma \in U_Z} (-1)^{d(\sigma)}$$ where $U_Z$ is the set of classes modulo $W$ of the set of simplices of the augmented Folkman complex of the lattice $\mathcal{A}_z$, and $d(\sigma)$ denotes the dimension of a cell. The Folkman complex of a lattice is defined (see [@OT]) as the complex of poset obtained by removing the minimal and maximal elements of the lattice ; when the maximal codimension of the lattice is 1, then the Folkman complex is empty. The augmented Folkman complex is defined by adding to the Folkman complex one $G$-invariant simplex of dimension $-1$. In the case of $G_{31}$ the computation of this formula is doable and we get 0 for the second Betti number.

We then reduce our original complex mod $p^r$, for $p^r$ small enough so that we can encode each matrix entry inside one byte. Then we wrote a C program to compute $H_2(B,\Z_4) = H_2(B,\Z_2) = \Z_2$, $H_2(B,\Z_9) = H_2(B,\Z_3) = \Z_3$ and $H_2(B,\Z_5) = 0$ (the matrix of $d_3$ has size $11065 \times 15300$). Since $G_{31}$ has order $2^{10} .3^2. 5$ and $H_*(P)$ is torsion-free, for $p \not\in \{ 2, 3, 5 \}$ we have $H_2(B,\Z_p) = H_2(P,\Z_p)^W = (H_2(P,\Z)^W) \otimes \Z_p$. But $0=H_2(B,\Q) = H_2(P,\Q)^W = H_2(P,\Z)^W \otimes \Q$ whence $H_2(P,\Z)^W = 0$ and $H_2(B,\Z_p) = 0$. Now $H_1(B,\Z) = \Z$ is torsion-free, hence $H_2(B,\Z_n) \simeq
H_2(B,\Z) \otimes \Z_n$ for any $n$ by the universal coefficients theorem. Since $H_2(B,\Z)$ is a $\Z$-module of finite type this yields $H_2(B,\Z) = \Z_6$ and completes the computation for $G_{31}$.

Embeddings between Artin-like monoids {#sectembed}
-------------------------------------

We end this section by proving a few lemmas concerning submonoids, which will be helpful in computing differentials in concrete cases.

We consider Garside monoids with set of generators $S$ and endowed with a length function, namely a monoid morphism $\ell : M \to \N = \Z_{\geq 0}$ such that $\ell(x) = 0 \Leftrightarrow
x = 1$ and $\ell(s) = 1$ for all $s \in S$. We consider the divisibility relation on the left (that is $U | V$ means $\exists m \ V = U m$) and recall that such a monoid admit lcm’s (on the left).

Let $M,N$ be two such monoids, and $\varphi : M \to N$ a monoid morphism such that

1.  $\forall s \in S \ | \ \varphi(s) \neq 1$

2.  $\forall s,t \in S \ \lcm(\varphi(s),\varphi(t)) = \varphi(\lcm(s,t))$

The following results on such morphisms are basically due to J. Crisp, who proved them in [@crisp] in the case of finite-type Artin groups.

Let $U,V \in M$. If $\varphi(U) | \varphi(V)$ then $U | V$.

By induction on $\ell(V)$. Since $\forall s \in S \ \ell(\varphi(s)) \geq 1 = \ell(s)$, we have $\ell(\varphi(U)) \geq \ell(U)$. Since $\varphi(U) | \varphi(V)$, we have $\ell(\varphi(U)) \leq \ell(\varphi(V))$ hence $\ell(U) \leq \ell(\varphi(V))$. Hence $\varphi(V) = 1$ implies $\ell(U) = 0$ and $U = 1$, which settles the case $\ell(V) = 0$.

We thus assume $\ell(V) \geq 1$. The case $U=1$ being clear, we can assume $U \neq 1$. Then there exists $s,t \in S$ with $s | U$ and $t|V$. It follows that $\varphi(t) | \varphi(V)$ and $\varphi(s) | \varphi(U) | \varphi(V)$, hence $\lcm(\varphi(s),\varphi(t)) | \varphi(V)$.

Now $\lcm(s,t) = t m$ for some $m \in M$ and $V = t V'$ for some $V' \in M$, hence $\varphi(t) \varphi(m) | \varphi(V) = \varphi(t) \varphi(V')$ and this implies $\varphi(m) | \varphi(V')$ by cancellability in $M$. Since $\ell(V') < \ell(V)$, from the induction assumption follows that $m | V'$ hence $tm | V$ that is $\lcm(s,t) | V$. In particular we get $s | V$. Writing $V = s V''$ and $U = s U'$ for some $V'',U' \in M$, the assumption $\varphi(U) | \varphi(V)$ implies $\varphi(U') | \varphi(V'')$ by cancellability, and then $U' | V''$ by the induction assumption. It follows that $U|V$ which proves the claim.

The lemma has the following consequence.

The morphism $\varphi : M \to N$ is injective. If $G_M$, $G_N$ denotes the group of fractions of $M,N$, then $\varphi$ can be extended to $\widetilde{\varphi} : G_M \into G_N$.

Let $U,V \in M$ with $\varphi(U) = \varphi(V)$. By the lemma we get $U|V$ and $V | U$. This implies $\ell(U) = \ell(V)$ hence $U = V$. Composing $\varphi : M \to N$ with the natural morphism $N \into G_N$ yields a monoid morphism $M \to G_N$. Since $G_N$ is a group this morphism factors through the morphism $M \to G_M$ and this provides $\widetilde{\varphi} : G_M \to G_N$. Let $g \in \Ker \widetilde{\varphi}$. Since $g \in G_M$ there exists $a,b \in M$ with $g = ab^{-1}$ hence $\varphi(a) = \varphi(b)$, $a=b$ and $g = 1$.

We consider the following extra assumption on $\varphi$. We assume that, for all $m \in M$ and $n \in N$, $n | m$ implies

We can now identify in this $M,N,G_M$ to subsets of $G_N$. We consider the following extra assumption. We assume that, for all $m \in M, n \in N$, if $n$ divides $m$ in $N$ then $n \in M$.

Under this assumption, $U,V$ in $M$ have the same lcm in $M$ and in $N$. Moreover, $M = N \cap G_M$.

Since $\lcm_M(U,V)$ divides $U,V$ in $N$, we have that $\lcm_N(U,V))$ divides $\lcm_M(U,V)$ in $N$. Conversely, since $\lcm_N(U,V)$ divides $U$ in $N$ and $U \in M$, by the assumption we get $\lcm_N(U,V) \in M$. From the lemma we thus get that $\lcm_N(U,V)$ divides $U$ and $V$ in $M$ hence $\lcm_M(U,V)$ divides $\lcm_N(U,V)$ in $N$. It follows that $\lcm_M(U,V) = \lcm_N(U,V)$.

We have $M \subset N \cap G_M$. Let $n \in N \cap G_M$. Since $n \in G_M$ there exists $a,b \in M$ with $n = ab^{-1}$, hence $nb = a \in M$. Hence $n \in N$ divides $a \in M$ in $M$. By the assumption we get $n \in M$ and the conclusion.

The groups ${{\mathrm{B}}}(e,e,r)$ {#s:beer}
==================================

The Corran-Picantin monoid
--------------------------

We denote ${{\mathrm{B}}}(e,e,r)$ for $e \geq 1$ and $r \geq 2$ the braid group associated to the complex reflection group $G(e,e,r)$. The ${{\mathrm{B}}}(e,e,r)$ are the group of fractions of a Garside monoid introduced by R. Corran and M. Picantin (see [@corpic]). This monoid, that we denote $M(e,e,r)$, has generators (atoms) $t_0,t_1,\dots,t_{e-1}, s_3, s_4, \dots, s_r$ and relations

1.  $t_{i+1} t_i = t_{j+1} t_j$, with the convention $t_e = t_0$,

2.  $s_3 t_i s_3 = t_i s_3 t_i$

3.  $s_k t_i  = t_i s_k$ for $k \geq 4$

4.  $s_k s_{k+1} s_k = s_{k+1} s_k s_{k+1} $ for $k \geq 3$

5.  $s_k s_l = s_l s_k$ when $| l-k | \geq 2$.

Link with the topological definition
------------------------------------

The connection between this monoid and the group ${{\mathrm{B}}}(e,e,r)$ defined as a fundamental group is quite indirect. In [@BMR] a first presentation is obtained by combining embeddings into usual braid groups, fibrations and coverings. The presentation used here is deduced from this one in a purely algebraic matter, by adding generators in order to get a Garside presentation. Although it is folklore the description of all generators as braided reflection does not appear in the literature (see however [@BESSISCORRAN] for a statement without proof in a related context).

In order to provide this connection, we need to recall the way these generators are constructed. For clarity, we stick to the notations of [@BMR] ; in this paper, the authors introduce 4 different spaces, $\mathcal{M}(r+1) = \{ (z_0,\dots,z_r) \in \C^{r+1} \ | \ z_i \neq z_j \}$, $\mathcal{M}^{\#}(m,r) = \{ (z_1,\dots,z_r) \in \C^r \ | \ z_i \neq 0,
z_i /z_j \not\in \mu_m \}$, $\mathcal{M}(e,r) = \{ (z_1,\dots,z_r) \in \C^r \
| \ z_i \not\in \mu_e z_j \}$, and $\mathcal{M}^{\#}(r) = \{ (z_1,\dots,z_r) \in \C^r \ | \ z_i \neq 0 \}$, where $\mu_n$ denotes the set of $n$-th roots of 1 in $\C$. We have a Galois covering $r : \mathcal{M}^{\#}(m,r)
\to \mathcal{M}^{\#}(r) = \mathcal{M}^{\#}(m,r) / (\mu_m)^r$, a locally trivial fibration $p : \mathcal{M}(r+1) \to \mathcal{M}^{\#}(r)$ with fiber $\C$ given by $(z_0,\dots,z_r) \mapsto (z_0 - z_1,\dots,
z_0-z_r)$, and a natural action of $\mathfrak{S}_r$ on $\mathcal{M}(r+1)$ that leaves the $(r+1)$-st coordinate fixed. We choose a fixed point $x \in \mathcal{M}(r+1)/\mathfrak{S}_r$, and a lift $\widetilde{p(x)}$ of $p(x) \in \mathcal{M}^{\#}(r)/\mathfrak{S}_r$ in $\mathcal{M}^{\#}(d,r)/G(d,1,r) = (\mathcal{M}(d,r)/(\mu_m)^r)/\mathfrak{S}_r$. We get an isomorphism $\psi : \pi_1(\mathcal{M}^{\#}(d,r)/G(d,1,r),
\widetilde{p(x)}) \to \pi_1(\mathcal{M}(r+1)/\mathfrak{S}_r,x)$ by composing the isomorphisms induced by $r$ and $p$. $$\xymatrix{
 \pi_1(\mathcal{M}^{\#}(d,r)/G(d,1,r), \widetilde{p(x)}) \ar[dr]_{r}^{\simeq} \ar[rr]^{\psi} & &
\ar[dl]^{p}_{\simeq} \pi_1(\mathcal{M}(r+1)/\mathfrak{S}_r, x) \\
 & \pi_1(\mathcal{M}^{\#}(r)/\mathfrak{S}_r, p(x)) & \\
}$$ Since $\pi_1(\mathcal{M}^{\#}(d,r)/G(d,1,r)) = {{\mathrm{B}}}(d,1,r)$, $\psi$ identifies the latter group with $\pi_1(\mathcal{M}(r+1)
/\mathfrak{S}_r)$. The generators of ${{\mathrm{B}}}(d,1,r)$ are then obtained in [@BMR] by taking the preimages under $\psi$ and the covering of $\mathcal{M}(r+1)/\mathfrak{S}_r \to \mathcal{M}(r+1)/\mathfrak{S}_{r+1}$. Note that this covering provides an injection between fundamental groups, hence an embedding $\tilde{\psi} : {{\mathrm{B}}}(d,1,r) \into {{\mathrm{Br}}}(r+1)$, where ${{\mathrm{Br}}}(r+1)$ denotes the usual braid group on $r+1$ strands. We choose for base point in $\mathcal{M}(r+1)$ the point $x = (0,x_1,\dots,x_r)$ with the $x_i \in \R$ and $x_{i+1} \ll x_i$, and for generators of the usual braid group $\mathcal{M}(r+1)/\mathfrak{S}_{r+1}$ the elements $\xi_0, \xi_1,\dots,\xi_{r-1}$ as described below :

![image](Artingen.pdf)

Then (see [@BMR]), the group $\pi_1(\mathcal{M}(r+1)/\mathfrak{S}_r)$ is generated by $\xi_0^2,\xi_1,\dots,\xi_{r-1}$. The element $\xi_0^2$ is the class in $\mathcal{M}(r+1)$ of the loop $(\frac{x_1}{2}(1-e^{2 \ii \pi t}), \frac{x_1}{2}( e^{2 \ii \pi t} +1),x_2,
\dots,x_r )$. Taking its image by $p$ provides a loop based at $(-x_1,-x_2,\dots,-x_r)$ described by $(-x_1 e^{2 \ii \pi t},
\frac{x_1}{2} (1 - e^{2 \ii \pi t}) -x_2,\dots,\frac{x_1}{2} (1 - e^{2 \ii \pi t}) - x_r)$. Since $|x_i | \ll |x_{i+1}|$, this path is homotopic to $(-x_1 e^{2 \ii \pi t}, -x_2, \dots, -x_r)$, both in $\mathcal{M}^{\#}(r)$ and in $\mathcal{M}^{\#}(r)/\mathfrak{S}_r$. Letting $a_i = -x_i$, we have $0 < a_1 < a_2 < \dots < a_r$, and we choose $y = \widetilde{p(x)}$ to be $y = (a_1^{\frac{1}{d}},\dots,
a_r^{\frac{1}{d}})$. The above loop thus lifts under $r$ to the path $(a_1 e^{2 \ii \pi t/d},a_2,\dots,a_r)$ in $\mathcal{M}^{\#}(d,r)$. By definition of $\psi$, the class of this path $\sigma = \psi^{-1}(\xi_0^2)$. Similarly, we can determine $\psi^{-1}(\xi_i)$ when $i \geq 1$ : the image of $\xi_i$ under $p$ is a path in $\mathcal{M}^{\#}(r)$ homotopic to

On the open cone described in the picture, the map $z \mapsto z^d$ is a positive homeomorphism, and this enables one to lift this path to

meaning that $\tau_i = \psi^{-1}(\xi_i)$ is the class of this path, from $(a_1^{1/d}, \dots, a_i^{1/d}, a_{i+1}^{1/d}, \dots, a_r^{1/d})$ to $(a_1^{1/d}, \dots, a_{i+1}^{1/d}, a_{i}^{1/d}, \dots, a_r^{1/d})$. We recall that ${{\mathrm{B}}}(de,e,r)$ is defined by $\pi_1( \mathcal{M}^{\#}(de,r)/G(de,e,r))$ when $d > 1$. If moreover $e=1$, then ${{\mathrm{B}}}(d,1,r)$ is generated by $\sigma,
\tau_1,\dots,\tau_{r-1}$ ; in general, it is generated by $\sigma^e,
\tau_1,\dots,\tau_{r-1}$. Now, the morphism $\tilde{\psi} : {{\mathrm{B}}}(d,1,r) \into
{{\mathrm{Br}}}(r+1) = \pi_1(\mathcal{M}(r+1)/\mathfrak{S}_{r+1})$ commutes with the natural morphisms $$\xymatrix{ {{\mathrm{B}}}(d,1,r) \ar[rr] \ar[d] & & {{\mathrm{Br}}}(r+1) \ar[d] \\
G(d,1,r) \ar[rr] \ar[dr] & & \mathfrak{S}_{r+1} \\ 
 & \mathfrak{S}_r \ar[ur] & }$$ Letting as in [@BMR] $\xi'_1 = \xi_0^2 \xi_1 \xi_0^{-2} \in \pi_1(
\mathcal{M}(r+1)/\mathfrak{S}_r, x)$, we have $\tau'_1 = \psi^{-1}(\xi'_1)
\in \pi_1(\mathcal{M}^{\#}(d,r) / G(d,1,r),y)$. As before we let $\zeta = \exp(2 \ii \pi/d)$ and $g_i \in G(d,1,r)$ being defined by $g_i.(z_1,\dots,z_r) = (z_1,
z_2,\dots,\zeta z_i, \dots, z_r)$. We let $b_i = a_i^{1/d}$. $$\xymatrix{
(b_1,\dots,b_r) \ar@/^1pc/[d]^{\xi_0^{-2}} \ar@/_5pc/@{-->}[dd]_{g_1^{-1} s_1}
\ar@/_10pc/@{-->}[ddd]_{g_1^{-1}s_1g_1}\ar@/^10pc/[ddd]_{\xi'_1}\\
(\zeta^{-1} b_1,b_2,\dots,b_r) \ar@/^1pc/[d]^{g_1^{-1}.\xi_1}\\
(\zeta^{-1} b_2, b_1,\dots,b_r) \ar@/^1pc/[d]^{g_1^{-1} s_1 . \xi_0^2}\\
(\zeta^{-1} b_2, \zeta b_1,\dots,b_r) }$$ In order to generate ${{\mathrm{B}}}(e,e,r) = \pi_1(\mathcal{M}(e,r)/\mathfrak{S}_r, x)$, and letting $e=d$, we only need to take the image of $\tau'_1,\tau_1,\dots,\tau_{r-1}$ under $i^*$ where $i : \mathcal{M}^{\#}(e,r) \to \mathcal{M}(e,r)$ is the natural inclusion. We will use the following definition.

Let $X$ be the complement of an hyperplane arrangement $\mathcal{A}$ in $\C^l$, and $v,v' \in X$. A line segment from $v$ to $v'$ is $t \mapsto (1-t) v + t v'$ for $t \in [0,1]$, If this line segment crosses exactly one hyperplane of $\mathcal{A}$ at one point, a positive *detour* from $v$ to $v'$ is a path of the form $\gamma(t) = (1-t) v + t v' + \ii t (1-t) (v-v') \eps$ for $\eps > 0$ small enough so that it and the similar paths $\gamma'$ for $0 < \eps' < \eps$ do not cross any hyperplane in $\mathcal{A}$. All such detours are clearly homotopic to each other. A negative detour is defined similarly with $\ii$ replaced by $- \ii$.

Note that, for $v \in \mathcal{M}(e,r)$ and $s$ a reflection in $G(e,e,r)$, if there exists a positive detour from the base point $\underline{b} = (b_1,\dots,b_r)$ to $w.\underline{b}$, then it provides a braided reflection around the hyperplane attached to $s$.

The elements $i^*(\tau_k)$ are now easy-to-describe braided reflections, as the positive detours from $\underline{b}$ to their images by the corresponding reflections. In case $e = 2$, the given monoid is then clearly the classical Artin monoid of type $D_r$, so we can assume $e \geq 3$. The paths corresponding to $\xi_0^{2}$ and to its translates are homotopic to a line segment in $\mathcal{M}(e,r)$. The fact that $\tau'_1$ is a braided reflection essentially amounts to the fact that $i^*(g_0^{-1} .\tau_1)$ is a braided reflection in $\pi_1(\mathcal{M}(e,r)/G(e,e,r), g_0^{-1} . y)$, and this holds true because $\tau_1$ is a braided reflection in $\mathcal{M}^{\#}(e,r)/G(e,e,r)$.

We consider the plane $P$ defined by the equations $z_i = b_i$ for $i = 3,\dots,r$, and identify it to $\C^2$ through $(z_1,z_2)$. We let $P^0 = \C^2 \setminus \bigcup \{ z_2 = z_1 \eta \ | \ \eta \in \mu_e \}
= P \cap \mathcal{M}(e,r)$. Then $\tau_1,\tau'_1$ lie in the plane $P$, and $\tau'_1$ is homotopic in $P^0$ to

where the half-circle represents the positive detour from $(\zeta^{-1} b_1,b_2)$ to $(\zeta^{-1} b_2,b_1)$. We let now $t_0 = \tau_1$, $t_1 = \tau'_1$, $t_{i+1} = t_i^{-1} t_{i-1}t_i$ for $1 \leq i \leq e-2$. A way to understand paths in $P \simeq \C^2$ is to use the projection $\C^2 \to \mathbbm{P}^1(\C)$ given by $(z_1,z_2) \mapsto
z_2/z_1$. Note for example that two paths $\gamma_1,\gamma_2$ in $P$ with the same endpoints whose images are homotopic in $\mathbbm{P}^1(\C) \setminus
\mu_e$ are homotopic in $P^0$ as soon as, writing $\gamma_i(t)
=(x_i(t),y_i(t))$, the set $x_1([0,1]) \cup x_2([0,1])$ is contained in some simply connected subspace of $\C \setminus \{ 0 \}$. We let $\alpha = b_2/b_1 \gg 1$. Then the positive detour $t_0$ is mapped to a path from $\alpha$ to $\alpha^{-1}$ close to the line segment, with image in the positive half-plane. The line segments of the form $\gamma(t) = (z_1,z_2(t)$ are mapped to line segments, and lines form $\gamma(t) = (z_1(t),z_2)$ are mapped to images of a line under $z \mapsto 1/z$, which is the composite of the complex conjugation with the geometric inversion with respect to the unit circle ; they are thus mapped to a line if the original line passes through 0, and otherwise to a circle passing through the origin. The induced action of $G(e,e,r)$ is given by $s_1 : z \mapsto \frac{1}{z}$, $g_1 : z \mapsto \zeta^{-1} z$, $g_2 : z \mapsto \zeta z$. The images of $t_1$ and $t_2$ are depicted in figure \[figt1t2\]. The images of $t_2$ and of the positive detour from $(b_1,b_2)$ to $(\zeta^{-2} b_2,\zeta^2 b_1)$ are then clearly homotopic (see figure \[figt2detour\]), and the first coordinate of both paths is easily checked to remain in a simply connected region of $\C \setminus \{ 0 \}$. With the same argument, using the relation $t_{i+1} = t_i^{-1} t_0 t_1$ and possibly using $(z_1,z_2) \mapsto z_1/z_2$ instead of $(z_1,z_2)
\mapsto z_2/z_1$, we get that each $t_i$ is (homotopic to) the positive detour from $(b_1,b_2)$ to $(\zeta^{-i}b_2,\zeta^i b_1)$. We thus got the following

Let $\underline{b} = (b_1,\dots,b_r) \in \mathcal{M}(e,r)$ with $0 < b_1 \ll b_2 \ll \dots \ll b_r$. Then ${{\mathrm{B}}}(e,e,r) = \pi_1( \mathcal{M}(e,r)/G(e,e,r),\underline{b})$ is generated by braided reflections $t_0,\dots,t_{e-1},s_3,\dots,s_r$ which are positive detours from $\underline{b}$ to their images under the corresponding reflection. Under ${{\mathrm{B}}}(e,e,r) \onto
G(e,e,r)$, $t_i$ is mapped to $(z_1,z_2,\dots,z_r) \mapsto
(\zeta^{-i} z_2,\zeta^i z_1,\dots,z_r)$, and $t_0,s_3,\dots,
s_r$ are mapped to the successive transpositions of $\mathfrak{S}_r$ in that order. These generators provide a presentation of ${{\mathrm{B}}}(e,e,r)$ with the relations (2)-(5) of page , and with (1) replaced by $t_i t_{i+1} = t_j t_{j+1}$.

We notice that the slight change in the presentation is meaningless in monoid-theoretic terms, as both monoids are isomorphic under $t_i \mapsto t_{- i}$, but it is not in topological terms, as $t_1 t_0 t_1^{-1}$ is *not* homotopic to a detour from $\underline{b}$ to its image (see figure \[figt2plus\]).

Let $S_0 = \{ t_i,s_3,\dots,s_r \}$. The subgroup of ${{\mathrm{B}}}(e,e,r) = \pi_1(\mathcal{M}(e,r)/G(e,e,r)$ generated by $S_0$ is a parabolic subgroup in the sense of [@BMR], and can be naturally identified with the braid group on $r$ strands as the fundamental group of $\{ (z_1,\dots,z_r) \ | \ z_i \neq z_j, z_1+\dots+z_r = 0 \}/\mathfrak{S}_r$, with base point $(-\zeta^{-i}(b_1+b_2+\dots + b_r),b_2,\dots,b_r)$, in such a way that the elements of $S_0$ are identified with positive detours.

The parabolic subgroup of $G(e,e,r)$ defined as the fixer of $(\zeta^i, 1,1 , \dots, 1)$ is obviously conjugated to the one fixing $(1,1,\dots,1)$, the latter being the natural $\mathfrak{S}_r  \subset G(e,e,r)$. We thus need only consider the case $i = 0$. Let $\alpha = -(b_1+\dots+b_r) \ll 0$, $\underline{b}_0
= (\alpha,b_2,\dots,b_r)$, $X = \mathcal{M}(e,r)$ and $X_0 = \{  (z_1,\dots,z_r) \ | \ z_i \neq z_j, z_1+\dots+z_r = 0 \}$. By [@BMR] we get an embedding $\pi_1(X_0/\mathfrak{S}_r, 
\underline{b}_0) \into \pi_1(X/G(e,e,r),\underline{b})$, natural only up to the choice of a path from $\underline{b}$ to $\underline{b}_0$ in $Y$. The line segment $\gamma$ from $\underline{b}$ to $\underline{b}_0$ provides such a natural choice.

We now need to prove that composing the positive detours from $\underline{b}$ with this path provides the positive detours from $\underline{b}_0$, up to homotopy in $Y = \{  (z_1,\dots,z_r) \ | \ z_i \neq z_j \}$. For $s_3,\dots,s_r$ this is true because the first component of the first path can be homotoped to the second one in $(\R_{\leq b_1}, b_1)$. For $t_0$ we let $\sigma_0$ and $\sigma$ denote the positive detours in $P^0$ from $(b_1,b_2)$ to $(b_2,b_1)$ and from $(\alpha,b_2)$ to $(b_2,\alpha)$, respectively. Let $\gamma, \gamma'$ denote the line segments $(b_1,b_2) \to (\alpha,b_2)$ and $(b_2,\alpha) \to (b_2,b_1)$. We need to prove that $\sigma_0$ is homotopic to $\gamma' \sigma \gamma$ in $\{ (z_1,z_2) \in \C^2 \ | \ 
z_1 \neq z_2 \}$, the other coordinates $z_3,\dots,z_r$ being the same for both paths. Since $b_1 - b_2$ and $\alpha -b_2$ have the same (negative) sign, we can homotope $\gamma' \sigma \gamma$ to a path with the same real part (for both coordinates), and with imaginary part the same as $\sigma$, up to possibly diminishing the chosen factor $\eps$ in the definition of the detours. Choosing then an homotopy in $\R_{\leq b_2}^2$ between the real parts of these two paths, provides an homotopy between them in $Y$.

Parabolic submonoids
--------------------

We apply the results of section \[sectembed\] on submonoids to the monoid $N = M(e,e,r)$, with generators $S = \{ t_0,\dots,t_{e-1}, s_3,\dots,s_r \}$. Let $\mathcal{C} = \{ t_0 , \dots ,t_{e-1} \}$. For $S_0 \subset
S$, let $M(S_0)$ be the monoid generated by $S_0$ with the defining relations of $M(e,e,r)$ which involve only elements of $S_0$. We get a natural morphism $\varphi : M(S_0) \to M(S) = M(e,e,r)$. We ask for the following extra assumption on $S_0$ : $$S_0 \cap \mathcal{C} \in \{ \emptyset, \mathcal{C}, \{ t_i \} \} \mbox{ for some } i \in \{0,1,\dots,e-1 \} .$$ In other terms, $S_0$ contains none, all or exactly one of the $t_i$’s. Note that all the corresponding monoids are known to be Garside and are endowed with a suitable length function.

This condition implies the extra condition on $\varphi$ in section \[sectembed\], namely that, if $n \in M(S)$ divides $\varphi(m)$ for some $m \in M(S_0)$, then $n \in \varphi(M(S_0))$. Indeed, if we have such $n,m$,then $n \in \varphi(M(S_0))$ unless $n$ can be written as a word containing some $x \in S \setminus S_0$. But in that case $\varphi(m)$ can also be written as a word in $S$ containing $x$. Now note that the defining relations involving such a $x$ cannot make it disappear, except when $x \in \mathcal{C}$. By contradiction this settles the cases $S_0 \cap \mathcal{C} = \emptyset$ and $S_0
\supset \mathcal{C}$. In case $S_0 \cap \mathcal{C} = \{ t_{i} \}$, we can assume $x = t_j$ for $j \neq i$, and would get equality in $M(S)$ of two words on $S$, one involving $t_i$ and no other element of $\mathcal{C}$, and the other involving $t_j$. But we check on the defining relations that all relations involving $t_i$ either involve only $t_i$ and no other elements of $\mathcal{C}$ in which case they preserve that property and do not make the $t_i$’s disappear, or they involve several elements of $\mathcal{C}$ and cannot be applied to the first word. This leads to a contradiction, which proves this property.

This condition also implies the property (2) for $\varphi$. For this we need to compute the lcm’s between two elements $x,y$ of $S$. We need to prove the following in $M(S_0)$, for any $S_0 \subset S$ satisfying the above condition that contains $x$ and $y$.

-   $\lcm(s_i,s_j) = s_i s_j = s_j s_i$ if $|j-i| \geq 2$

-   $\lcm(s_i,s_{i+1}) = s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}$

-   $\lcm(t_i,t_j) = t_1 t_0 = t_i t_{i-1} = t_jt_{j-1}$

-   $\lcm(t_i ,s_3) = t_i s_3 t_i = s_3 t_i s_3 $

-   $\lcm(t_i, s_j) = t_is_j = s_j t_i $ if $j \geq 4$.

The identities with length two are clear, as the lcm exist and cannot have length 1. For the ones of length 3, namely $\{ x, y \} = \{ s_i,s_{i+1} \}$ and $\{ x, y \}  =  \{ t_i ,s_3 \}$, we use that $\{ x , y \} \subset S_0 \subset S$ satisfies our condition. Since the lcm of $x,y$ in $M(S_0)$ should divide $xyx=yxy$, it should then come from $M(\{ x, y \})$, meaning that is should be a word in $x$ and $y$, of length at most $2$. Thus only few possibilities remain, all of them easily excluded.

Using the previous section, we thus get injective monoid morphisms $M(S_0) \to M(S) = M(e,e,r)$. Let $B(S_0)$ the group of fractions of $M(S_0)$. It is proved in [@corpic] that $B(S) = {{\mathrm{B}}}(e,e,r)$. We call the $B(S_0)$ the parabolic submonoids of ${{\mathrm{B}}}(e,e,r)$. Crucial examples of such submonoids are described below.

### Second homology group

We choose on the atoms the ordering $s_r < s_{r-1} < \dots< s_3 < t_0 < t_1 < \dots < t_{e-1}$. By the above construction, the parabolic submonoid $M(e,e,r-1) = M(\{ s_{r-1},\dots,s_3,t_0,\dots,t_{e-1} \})$ is indeed a submonoid of $M(e,e,r)$, and the lcm of a family of elements in $M(e,e,r-1)$ is also its lcm in $M(e,e,r)$. The same holds true for the following submonoids :

-   the ones generated by $s_3,t_i$, which is an Artin monoid of type $A_2$ ;

-   the ones generated by $s_k,t_i$, $k \geq 4$, which is an Artin monoid of type $A_1 \times A_1$ ;

-   the ones generated by $s_4,s_3,t_i$, which is an Artin monoid of type $A_3$ ;

-   the ones generated by $s_k,s_3,t_i$, $k \geq 5$ which is an Artin monoid of type $A_1 \times A_2$ ;

-   the ones generated by $s_k,s_l,t_i$, $k \geq l+2$, $l \geq 4$, which is an Artin monoid of type $A_1 \times A_1 \times A_1$ ;

-   the ones generated by $s_k,s_l,s_r$, which is an Artin monoid of type given by the obvious subdiagram (of type $A_1 \times A_1 \times A_1$, $A_2 \times A_1$, $A_1 \times A_2$ or $A_3$).

We first compute the differentials of the top cell for the corresponding Artin monoids (see Table \[typesA\]), and then use this remark for computing the differentials of the 2-cells and 3-cells. We let $d_n = \partial_n \otimes_{\Z M} \Z :  C_n \otimes_{\Z M} \Z \to C_{n-1} \otimes_{\Z M} \Z$ denote the differential with trivial coefficients.

The 2-cells are the following : $[t_0,t_{i}]$ for $1<i < e$, $[s_3,t_i]$ , $[s_k,t_i]$ for $k \geq 4$ and $[s_k,s_l]$ for $k<l$. From Table \[typesA\] we get $$\begin{array}{lcll}
d_2 [t_0,t_i] &=& [t_i] + [t_{i+1}] - [t_0] - [t_1] \\
d_2 [s_3, t_i] &=& [t_i] - [s] \\
d_2 [s_k, t_i] &=& 0 \mbox{ if } k \geq 4 \\
d_2 [s_k,s_l] &=& 0 \mbox{ if } l> k+2 \geq 4 \\
d_2 [s_k,s_{k+1}] &=& [s_{k+1}] - [s_k] 
\end{array}$$ We let $\delta_2 = t_1 t_0$ denote the Garside element of $M(e,e,2)$ and we assume $e > 1$. For the 3-cells, we also need to compute

$$\begin{array}{lcl}
\partial_3 [s,t_0,t_j] &=& (s\delta_2 s - t_{j+2} t_{j+1} s +t_{j+2} s)[t_0,t_j]
-t_{j+2}st_{j+1}[s,t_j] \\ & & +(t_{j+2}-st_{j+2})[s,t_{j+1}] + (s-t_{j+2}s-1)[t_0,t_{j+1}]
+(st_2-t_2)[s,t_1] \\ & & + (t_2 s +1 -s)[t_0,t_1]
  +[s,t_{j+2}] +t_2 s t_1[s,t_0]
-[s,t_2]
\end{array}$$ when $j \not\equiv -1 \mod e$, and $$\begin{array}{lcl}
\partial_3 [s,t_0,t_{-1}] &=& (s\delta_2 s -t_1t_0s +t_1s)[t_0,t_{-1}]
 -t_{1}st_{0}[s,t_{-1}]  + (1-t_2+st_2)[s,t_1] \\ & & + (1+t_2 s - s)[t_0,t_1]
+(t_1 - st_1)[s,t_0]
 +t_2 s t_1[s,t_0]
-[s,t_2]
\end{array}$$ This means $d_3 [s,t_0,t_{-1}] = [t_0,t_{-1}]
 -[s,t_{-1}]  + [s,t_1] + [t_0,t_1]
 +[s,t_0]
-[s,t_2]
$ and $d_3 [s,t_0,t_j] = [t_0,t_j]
-[s,t_j] -[t_0,t_{j+1}]
 + [t_0,t_1]
  +[s,t_{j+2}] +[s,t_0]
-[s,t_2]
$ for $j \not\equiv 1 \mod e$.

$$\begin{array}{|c|c|c|lcl|}
\hline
\mathrm{Type} & \mathrm{Atoms} & \mathrm{Relations} & & & \mathrm{Differential\ of\ top\ cell} \\
\hline
M(e,e,2) & t_0 < \dots < t_r & t_i t_{i+1} = t_j t_{j+1} & \partial_2 [t_0,t_i] & = & t_{i+1} [t_i] + [t_{i+1}] - t_1[t_0] - [t_1] \\
\hline
M(e,e,2) \times A_1 & s < t_0 < \dots < t_r & t_i t_{i+1} = t_j t_{j+1} & \partial_2 [s,t_0,t_i,] & = &(s-1)[t_0,t_i] - t_{i+1} [s,t_i]  \\
 & & t_i s = st_i &  & +  &  t_1 [s,t_0] - [s,t_{i+1}] + [s,t_1] \\
\hline
A_2 & s < t & sts = tst & \partial_2 [s,t] & =& (ts+1-s)[t] + (t-st-1)[s] \\
\hline
A_1 \times A_1 & s<u & su=us & \partial_2 [s,u] &=& (s-1)[u] - (u-1)[s] \\
\hline
A_3 & s<t<u & sts = tst & \partial_3 [s,t,u] &=& (u+stu-tu-1)[s,t] - [s,u] \\
 &  & su=us & & + & (su-u-s+1-tsu)t[s,u] \\
 &  & tut = utu & & + & (s-1-ts+uts)[t,u] \\
\hline
A_2 \times A_1 & s<t<u & tu = ut & \partial_3 [s,t,u] &= & (1-s+ts)[t,u] \\
 &  & su = us & & + & (t-1-st)[s,u] \\
 &  & sts = tst & & + &  (u-1)[s,t]\\
\hline
A_1 \times A_2 & s < t < u & st=ts & \partial_3 [s,t,u] &=& (1+tu-u)[s,t]\\ 
 & & su = us & & + &(t-1-ut)[s,u] \\
 & & tut=utu & & + & (s-1)[t,u] \\
\hline
A_1 \times A_1 \times A_1 & s<t<u & su = us & \partial_3[s,t,u] &=&  (1-t)[s,u] \\
& & st = ts & & + & (u-1) [s,t]\\
& & tu = ut & & + & (s-1)[t,u] \\
\hline
\end{array}$$

We now compute the second homology group, starting with $\Ker d_2$. Let $$v_i = [t_0,t_i] + [s,t_0] + [s,t_1] - [s,t_i] - [s,t_{i+1}] \in \Ker d_2$$ for $1 \leq i \leq e-1$. Let $K_1$ denote the submodule of $\Ker d_2$ spanned by the $v_i$. It is easy to show that $K_1$ is free on the $v_i$, and $K_1 = \Ker d_2$ for $r = 3$ ; if $r > 3$ we have $\Ker d_2 = K_1 \oplus K_2$ where $K_2$ is the free $\Z$-module with basis the $[s_k,t_i]$ for $k \geq 4$ and, if $r \geq 5$, the $[s_l,s_k]$ for $l \geq k+2$, $k \geq 3$.

Now decompose $\Z \mathcal{X}_3 = C_1 \oplus C_2$ where $C_1$ has for basis the $[s_3,t_0,t_i]$ and $C_2$ has for basis the other 3-cells. By the above computations we get $d_3(C_1) \subset K_1$, and $d_3(C_2) \subset K_2$. Thus $H_2(B,\Z) = (K_1/d_3(C_1)) \oplus (K_2 / d_3(C_2))$. We first compute $K_1/d_3(C_1)$. We have $$\begin{array}{lclr}
d_3[s_3,t_0,t_j] & = & v_j - v_{j+1} + v_1 & \mbox{ if } 0<j<e-1 \\
d_3[s_3,t_0,t_{e-1}] & = & v_{e-1} + v_{1} + v_1 
\end{array}$$ We denote $u_{i} = [s_3,t_0,t_i]$ for $1 \leq i \leq e-1$, and let $w_i = u_i + u_{i+1} + \dots + u_{e-1}$. Then $d_3 w_i = v_i + (e-i) v_1$. Written on the $\Z$-basis $(w_i)$ and $(v_i)$, $d_3$ is in triangular form, and the only diagonal coefficient that differs from $1$ is $e$, since $d_3 w_1 = e v_1$. It follows that $K_1/d_3(C_1) \simeq \Z_e$. Since $H_2(B,\Z) = K_1/d_3(C_1)$ for $r = 3$, we can now assume $r \geq 4$.

First assume $r=4$. In $K_2/d_3(C_2)$ we have $2[s_4,t_i] \equiv 0$, because $d_3[s_4,s_3,t_i] = -2[s_4,t_i]$. Since $d_3[s_4,t_0,t_i] = -[s_4,t_i]+[s_4,t_0]-[s_4,t_{i+1}]+[s_4,t_1]$. we get $[s_4,t_i] + [s_4,t_{i+1}] \equiv [s_4,t_0]+[s_4,t_1]$ when $i>0$. In particular, $[s_4,t_i] + [s_4,t_{i+1}] \equiv [s_4,t_{i+1}] + [s_4,t_{i+2}]$ that is $[s_4,t_i] \equiv [s_4,t_{i+2}]$, at least if $0<i<e-1$. From $d_3[s_4,t_0,t_1]\equiv 0$ we deduce $[s_4,t_2] \equiv [s_4,t_0]$, and from $d_3[s_4,t_0,t_{e-1}] \equiv 0$ we deduce $[s_4,t_{e-1}] \equiv [s_4,t_1]$. Thus $[s_4,t_i] \equiv [s_4,t_{i+2}]$ for every $i$. When $e$ is odd, $K_2/d_3(C_3)$ is then spanned by the class of $[s_4,t_0]$. From the other relations one easily gets that this class is nonzero, and since $2[s_4,t_i] \equiv 0$ we get $K_2/d_3(C_2) \simeq \Z_2$. When $e$ is even, this quotient is spanned by the classes of $[s_4,t_0]$ and $[s_4,t_1]$, and we get similarly $K_2/d_3(C_2) \simeq \Z_2^2$.

We now assume $r \geq 5$. Then $d_3([s_5,s_3,t_i]) = [s_5,s_3]-[s_5,t_i]$ whence $a := [s_5,t_0] \equiv [s_5,t_i]$ for all $i$, regardless whether $e$ is even or odd. From $d_3[s_5,s_4,t_i] = [s_4,t_i] - [s_5,t_i]$ we get $[s_4,t_i] \equiv a$ and from $[s_{k+1},s_k,t_i] = [s_k,t_i] - [s_{k+1},t_i]$ we deduce by induction $[s_k,t_i] \equiv a$. The only remaining relation involving $a$ is then as before $2a \equiv 0$.

On the other hand, we have $[s_5,s_3] \equiv a$. Assume we have $[s_l,s_k] \equiv a$ for some $l,k$ with $l \geq k+2$. From $d_3[s_l,s_{k+1},s_k] = [s_l,s_{k+1}] - [s_l,s_k]$ for $l \geq k+3$ we get $[s_l,s_{k'}] \equiv a$ for all $l-2 \geq k' \geq 3$, and then that $[s_{l'},s_{k'}] \equiv a$ for all $l'-2 \geq k' \geq 3$. We thus get $K_2/d_3 C_2 \simeq \Z/2\Z$.

As a consequence, we get the following result.

\[H2BEER\] Let $B = {{\mathrm{B}}}(e,e,r)$ with $r \geq 3$ and $e \geq 2$.

-   When $r = 3$, $H_2(B,\Z) \simeq \Z_e$

-   When $r = 4$ and $e$ is odd, $H_2(B,\Z) \simeq \Z_e \times \Z_2 \simeq \Z_{2e}$

-   When $r = 4$ and $e$ is even, $H_2(B,\Z) \simeq \Z_e \times \Z_2^2$

-   When $r \geq 5$, $H_2(B,\Z) \simeq \Z_e \times \Z_2$

The case $r=2$ is when $W$ is a dihedral group, and this case is known by [@SALVETTI] : we have $H_2(B,\Z) = 0$ if $e$ is odd, $H_2(B,\Z) = \Z$ if $e$ is even.

Low-dimensional homology {#s:low}
========================

The second homology group {#ss:H2}
-------------------------

The computations above provide the second integral homology group $H_2(B,\Z)$. In the case of the finite group $W$, the group $H_2(W,\Z)$ can be identified with the Schur multiplier $H^2(W,\C^{\times})$, which is relevant for dealing with projective representations. We use the determination of the $H_2(B,\Z)$ to show a direct connection between the two groups $H^2(B,\C^{\times})$ and $H^2(W,\C^{\times})$. We first start with a lemma.

Let $W$ be an irreducible finite complex 2-reflection group, and $B$ the associated braid group. The inflation morphism $H^2(W,\C^{\times}) \to H^2(B,\C^{\times})$ is into.

The Hochschild-Serre exact sequence associated to $1 \to P \to B \to W \to 1$ is $$0 \to H^1(W,\C^{\times}) \to H^1(B,\C^{\times}) \to
H^1(P,\C^{\times})^W \to H^2(W,\C^{\times}) \to H^2(B,\C^{\times}).$$ Now $H^1(P,\C^{\times}) = \operatorname{\mathrm{Hom}}(P^{ab},\C^{\times})^W
= \operatorname{\mathrm{Hom}}((P^{ab})^W ,\C^{\times})$ and $H^1(B,\C^{\times})
= \operatorname{\mathrm{Hom}}(B^{ab},\C^{\times})$. Now $P^{ab} = H_1(P,\Z)$ and $B^{ab}$ (see [@BMR] thm. 2.17) are torsion-free, with $B^{ab}
\simeq \Z^r$ where $r$ denotes the number of hyperplane orbits, and $(P^{ab})^W$ can be identified with $(2 \Z)^r$. The induced map $\operatorname{\mathrm{Hom}}(\Z^r,\C^{\times})\to \operatorname{\mathrm{Hom}}((2 \Z)^r,\C^{\times}) $ is then onto, since $\C$ is algebraically closed. By the Hochschild-Serre exact sequence above the conclusion follows.

[**Remark.**]{} Another proof of the lemma can be given using projective representations instead of the Hochschild-Serre exact sequence. Let $\alpha \in Z^2(W,\C^{\times})$ with zero image in $H^2(B,\C^{\times})$, choose some projective representation $R$ of $W$ with 2-cocycle $\alpha$, and consider its lift $\tilde{R}$ to $B$. By assumption, it is linearizable into some linear representation $\tilde{S}$. Choosing one generator of the monodromy $\sigma_i$ in $X/W$ for each hyperplane orbit (see [@BMR] appendix A) we find that $\tilde{S}(\sigma_i^2) = {\lambda}_i \in
\C^{\times}$. By [@BMR] Theorem 2.17 there exists a morphism $\varphi : B \to \C^{\times}$ with $\varphi(\sigma_i) = 1/{\lambda}_i$, and then $\tilde{T} = \tilde{S} \circ \varphi$ is a linear representation of $B$ that factors through $W$ and linearizes $R$, thus proving that $\alpha$ has zero image in $H^2(W,\C^{\times})$.

It is known by work of Read [@READ] and van der Hout [@HOUT] that $H^2(W,\C^{\times}) \simeq H_2(W,\Z)$ is a free $\Z_2$-module in all cases. A nice property that follows from our computation is that the part of $H^2(B,\C^{\times})$ that comes from $H^2(W,\C^{\times})$ is exactly the 2-torsion (except for 2 exceptional cases). Indeed, since $H_1(B,\Z)$ is torsion-free and $\C^{\times}$ is divisible, by the Universal Coefficients Theorem we get $H^2(B,\C^{\times}) \simeq \operatorname{\mathrm{Hom}}(H_2 B, \C^{\times})$ and the proposition below is a consequence of our computation of $H_2(B,\Z)$ (see Table \[tableH2\] for the exceptional groups, Theorems \[H2B2EER\] and \[H2BEER\] for the $G(2e,e,r)$ and the $G(e,e,r)$) and of the works of Read and van der Hout on $W$. We recall their computation of $H_2(W,\Z)$ in Table \[tableH2\] for the exceptional groups, and the rank over $\Z_2$ for the other ones in Table \[schurgen\].

$$\begin{array}{|c|l|c|c|}
\hline
r & e& G(e,e,r) & G(2e,e,r) \\
\hline
2 & \mbox{odd} & 0 & 1 \\
 & \mbox{even} & 1 & 2 \\
\hline
3 & \mbox{odd} & 0  & 2 \\
 & \mbox{even} & 1 & 2 \\
\hline
4 & \mbox{odd} & 1 & 3 \\
 & \mbox{even} & 3 & 4 \\
\hline
5 & \mbox{odd} & 1 & 3 \\
 & \mbox{even} & 2 & 3 \\
\hline
\end{array}$$

Except for $W = G_{33}$ or $W = G_{34}$, $H^2(W,\C^{\times})$ coincides with the 2-torsion of $H^2(B,\C^{\times})$.

$$\begin{array}{|c|c|c||c|c|c|}
\hline
W & H_2 W & H_2 B & W & H_2 W & H_2 B \\
\hline
G_{12} & 0 & 0 & G_{30} & \Z_2 & \Z_2 \\
G_{13} & \Z_2 & \Z & G_{31} & \Z_2 & \Z_6 \\
G_{22} & 0 & 0 & G_{33} & 0 & \Z_6 \\
G_{23} & \Z_2 & \Z & G_{34} & 0 & \Z_6 \\
G_{24} & \Z_2 & \Z & G_{35} & \Z_2 & \Z_2 \\
G_{27} & \Z_2 & \Z_3 \times \Z & G_{36} & \Z_2 & \Z_2 \\
G_{28} & (\Z_2)^2 & \Z^2 & G_{37} & \Z_2 & \Z_2 \\
G_{29} & (\Z_2)^2 & \Z_2 \times \Z_4 & & & \\
\hline
\end{array}$$

First homology in the sign representation {#ss:sign}
-----------------------------------------

If $r = |\mathcal{A}/W|$ denotes the number of hyperplane classes, the abelianization $B_{ab}$ is isomorphic to $\Z^r$. There are thus $2^r-1$ nonzero morphisms $B \onto \Z_2$, which define $2^r-1$ subgroups of even braids. When $r = 1$, there is only one such morphism $\eps : B \to \Z_2$ and group $B^{(2)} = \Ker \eps$. We investigate here two abelian invariants of $B$ which are naturally attached to this group : the abelianization $B^{(2)}_{ab}$ of $B^{(2)}$ and $H_1(B,\Z_{\eps})$.

Let $u \in B \setminus B^{(2)}$. The group $H_1(B,\Z_{\eps})$ is isomorphic to the quotient of $B^{(2)}_{ab}$ by the relations $[u^2] \equiv 0$ and $[h^u] \equiv
-[h]$ for $h \in B^{(2)}_{ab}$, where $h^u = u^{-1} h u$.

We start from the bar resolution $C_2 \to C_1 \to C_0$, where $C_i$ is a free $\Z B$-module with basis the $[g_1,\dots,g_i]$ for $g_i \in B$, we have $d_1([g]) = (g-1)[\emptyset]$, $d_2([g_1,g_2]) = g_1 [g_2] - [g_1 g_2] + [g_1]$. Denoting $d_i^{\eps}$ the differential with coefficients in $\Z_{\eps}$ and $C_i^{\eps} = C_i \otimes_{\Z B} \Z_{\eps}$ with $\Z$-basis the $[g_1,\dots,g_i]$, we get that $\Ker d_1^{\eps}$ is the direct sum $\Z B^{(2)} \oplus I$ where $I = \{ \sum_{g \not\in B^{(2)}} x_g [g] \ | \ x_g \in \Z, \sum x_g = 0 \}$. Choose some $u \in B \setminus B^{(2)}$. The image of $d_2^{\eps}$ is spanned by the $[g_1g_2] - \eps(g_1)
[g_2] - [g_1]$. Among them we find

1.  $[u^2] + [u] - [u] = [u^2]$

2.  $[h_1 h_2] - [h_1] - [h_2]$, for $h_1,h_2$ in $B^{(2)}$

3.  $[uh] - [u] + [h]$ for $h \in B^{(2)}$

4.  $[h^u] + [h]$ for $h \in B^{(2)}$.

Indeed, the element (4) is the difference of two elements clearly in $\mathrm{Im} d_2^{\eps}$, $[hu] - [u] - [h]$ and $[u h^u] + [h^u] - [u]$, where $h^u = u^{-1} h u$, since $u h^u = hu$. By (3), and since $I$ is spanned by the $[hu] - [u]$ for $h \in B^{(2)}$, we see that $H_1(B,\Z_{\eps})$ is generated by the images of the $[h]$ for $h \in B^{(2)}$. It is easy to check that the relations of the form $d_2^{\eps}([g_1,g_2]) \equiv 0$ are consequences of (1-4), hence $H_1(B,\Z_{\eps})$ is the quotient of $ B^{(2)}_{ab}$ by the relations $(1)$ and $(4)$.

The computation of $B^{(2)}$ can be done for exceptional groups by using the Reidemeister-Schreier method (see [@MKS]) and the presentations of [@BMR] and [@bessismichel]. Note that they are known to provide presentations of $B$ for all groups but $G_{31}$, for which our results as well will be conjectural. We start from one of these standard presentation of $B$ by braided reflections $\sigma_1,\dots,\sigma_{n}$ and use $\{ 1, \sigma_1\}$ for Schreier transversal. Then generators for $B^{(2)}$ are given by $\sigma_1^2,\sigma_1 \sigma_2,
\sigma_1 \sigma_3, \dots,\sigma_1 \sigma_n$ and $\sigma_2 \sigma_1^{-1}$, $\sigma_3 \sigma_1^{-1},\dots ,\sigma_n \sigma_1^{-1}$. We then apply the Reidemeister-Schreier process and find a presentation of $B^{(2)}$ from the relations $R$, $\sigma_1 R \sigma_1^{-1}$ where $R$ runs among the relations for $B$. The presentations obtained for exceptional groups are tabulated in figure \[figpreseven\] (the column ‘ST’ refers to the Shephard-Todd number of the group). It is then easy to abelianize these relations. We choose $u = \sigma_1$.

In order to get $H_1(B,\Z_{\eps})$ from $B^{(2)}_{ab}$ we start by adding the relation $[\sigma_1^2] \equiv 0$. Note that $\sigma_1 (\sigma_i \sigma_1^{-1})  \sigma_1^{-1}
= (\sigma_1 \sigma_i) (\sigma_1^{-2})$ hence $-[\sigma_i \sigma_1^{-1}]
 \equiv [\sigma_1 \sigma_i] - [\sigma_1^2] \equiv [\sigma_1 \sigma_i]$, and that $\sigma_1 (\sigma_1 \sigma_i)\sigma_1^{-1} = \sigma_1^2.
\sigma_i \sigma_1^{-1}$ hence $-[\sigma_1 \sigma_i] \equiv [\sigma_i \sigma_1^{-1}]$. The relations defining $H_1(B,\Z_{\eps})$ from $B^{(2)}_{ab}$ thus boil down to $-[\sigma_1 \sigma_i] \equiv [\sigma_i \sigma_1^{-1}]$ and $[\sigma_1^2] \equiv 0$.

ß

In order to get $H_1(B,\Z_{\eps})$ for the groups $G(*e,e,r)$, instead of using the complicated presentations of $B$ afforded by [@BMR], we use the semidirect product decomposition described in section \[s:iso\]. Recall that $B = \Z \ltimes \tilde{A}$ where we denote by $A$ the affine Artin group of type $\tilde{A}_{r-1}$. Then $A$ has Artin generators $\ss_1,\dots,\ss_r$ and the semidirect product is defined by $\tau \ss_i \tau^{-1} = \ss_{i+e}$ where addition is considered modulo $r$. From the split exact sequence $1 \to A \to B \to \Z \to 1$ we get the Hochschild-Serre short exact sequence $$0 = H_2(\Z,H_0(A,\Z_{\eps})) \to H_0(\Z,H_1(A,\Z_{\eps})) \to H_1(B,\Z_{\eps}) \to H_1(\Z,H_0(A,\Z_{\eps})) \to 0$$ with $H_2(\Z,H_0(A,\Z_{\eps})) = 0$ since $\Z$ has homological dimension 1. Since $A$ acts on $\Z_{\eps}$ through $\ss_i \mapsto -1$ we have $H_0(A,\Z_{\eps}) = \Z/2\Z = \Z_2$ ; since $\tau$ acts trivially on $H_0(A,\Z_{\eps})$ we thus get $H_1(\Z,H_0(A,\Z_{\eps})) \simeq H_1(\Z,\Z_2) \simeq \Z_2$. The short exact sequence thus boils down to $0 \to H_0(\Z,H_1(A,\Z_{\eps})) \to
H_1(B,\Z_{\eps}) \to \Z_2 \to 0$ and our task is reduced to computing $H_1(A,\Z_{\eps})$ while keeping track of the action of $\tau$.

In order to compute $H_1(A,\Z_{\eps})$ we apply the above process. Generators for $A^{(2)}$ are given by $u = \ss_1^2$, $x_i = \ss_1 \ss_i$ and $y_i = \ss_i \ss_1^{-1}$ for $2 \leq i \leq r$, and relations are given by rewriting $R$ and $\ss_1 R \ss_1^{-1}$ with $R$ running along the braid relations for $A$. These braid relations are the following (where $|j-i| \geq 2$ actually mean that $j,i$ are not connected in the braid diagram) $$\begin{array}{lllcl}
(R) & 1 \not\in \{ i, i+1\} & \ss_i \ss_{i+1} \ss_i \ss_{i+1}^{-1} \ss_i^{-1} \ss_{i+1}^{-1} & \rightsquigarrow & 
y_i  x_{i+1}  y_i  y_{i+1}^{-1}  x_i^{-1} y_{i+1}^{-1}
\\
& |j-i| \geq 2, 1 \not\in \{ i,j \}  & \ss_i \ss_j \ss_i^{-1} \ss_j^{-1} & \rightsquigarrow & 
y_i x_j x_i^{-1}y_j^{-1}\\
& & \ss_1 \ss_2 \ss_1 \ss_2^{-1} \ss_1^{-1} \ss_2^{-1} & \rightsquigarrow &
x_2 y_2^{-1}u^{-1} y_2^{-1} \\
& & \ss_1 \ss_r \ss_1 \ss_r^{-1} \ss_1^{-1} \ss_r^{-1} & \rightsquigarrow & 
x_r  y_r^{-1} u^{-1}y_r^{-1}\\
& i \not\in \{2,r \} & \ss_1 \ss_i \ss_1^{-1} \ss_i^{-1} & \rightsquigarrow &  x_i  u^{-1} y_i^{-1}\\
(\ss_1 R \ss_1^{-1} ) & 1 \not\in \{ i, i+1\} & \ss_1\ss_i \ss_{i+1} \ss_i \ss_{i+1}^{-1} \ss_i^{-1} \ss_{i+1}^{-1}\ss_1^{-1} & \rightsquigarrow & 
x_i y_{i+1}x_i  x_{i+1}^{-1} y_i^{-1} x_{i+1}^{-1} \\
& |j-i| \geq 2, 1 \not\in \{ i,j \}  & \ss_1\ss_i \ss_j \ss_i^{-1} \ss_j^{-1} \ss_1^{-1}& \rightsquigarrow & 
x_i y_j y_i^{-1}x_j^{-1} 
\\
& & \ss_1\ss_1 \ss_2 \ss_1 \ss_2^{-1} \ss_1^{-1} \ss_2^{-1}\ss_1^{-1} & \rightsquigarrow &
u y_2 u x_2^{-1} x_2^{-1}
 \\
& & \ss_1\ss_1 \ss_r \ss_1 \ss_r^{-1} \ss_1^{-1} \ss_r^{-1}\ss_1^{-1} & \rightsquigarrow & 
u y_ru  x_r^{-1} x_r^{-1}
\\
& i \not\in \{2,r \} &\ss_1 \ss_1 \ss_i \ss_1^{-1} \ss_i^{-1}\ss_1^{-1} & \rightsquigarrow & 
u  y_i   x_i^{-1}
\\ 
\end{array}$$ Abelianizing and dividing out by the relations $y_i = -x_i$ yields an abelian presentation for $H_1(A,\Z_{\eps})$ by generators $u, x_i$ for $2 \leq i \leq r$ and relations $$\begin{array}{lll}
 1 \not\in \{ i, i+1\} &  3 x_{i+1} =3 x_i
\\
|j-i| \geq 2, 1 \not\in \{ i,j \}  &  
 2 x_j =2 x_i \\
 & 
3x_2 =0 \\
 & 
3x_r =0\\
 i \not\in \{2,r \} & 2 x_i = 0\\
\end{array}$$ Thus, for $r = 3$, $H_1(A,\Z_{\eps}) = <x_2,x_3 | 3x_2 = 3x_3 = 0> = \Z_3 x_2 \oplus \Z_3 x_3 \simeq \Z_3^2$, for $r = 4$, $$H_1(A,\Z_{\eps}) =  < x_2,x_3,x_4 | 3x_2 = 3x_4 = 0, 2x_3 = 0 , 2x_2 =2x_4, 3x_3 = 3x_2=3x_4 >$$ hence $H_1(A,\Z_{\eps} ) =  < x_2, x_4 | 3x_2 = 3 x_4 = 0 , x_2 = x_4 >  = \Z_3 x_2 \simeq \Z_3$. When $r \geq 5$ , $H_1(A,\Z_{\eps})$ is generated by $x_2,\dots,x_r$, and we have $3 x_2 = 3 x_r= 0$. We have $2 < 3 < r-1 < r$. Then $2x_3 = 2x_{r-1} = 0$ but $0= 3x_2 = 3x_3$ and $0 = 3x_r = 3x_{r-1}$. It follows that $x_3 = 0$ and $x_{r-1} = 0$. Since $2x_3 = 2x_r$ and $2x_2 = 2x_{r-1}$ we get $x_2 = x_{r-1}$ and $x_3 = x_{r}$ hence $x_i = 0$ for all $i$ and $H_1(A,\Z_{\eps}) = 0$.

For $r \in \{ 3 , 4 \}$ it remains to compute the action of $\tau$ on $H_1(A,\Z_{\eps})$. We have $\tau.\ss_i = \ss_{i+e}$ hence $\tau.(\ss_1 \ss_i) = \ss_{1+e} \ss_{i+e}
= \ss_{1+e} \ss_1^{-1} \ss_1 \ss_{i+e}$. For $e \equiv  0 \mod r$ we have $\tau.x_i = x_i$ and $H_0(\Z,H_1(A,\Z_{\eps})) \simeq H_1(A,\Z_{\eps})$. For $r = 3$, $e \equiv 1 \mod 3$, $\tau.x_2 = \ss_2 \ss_3 
\equiv  y_2 + x_3 \equiv -x_2 + x_3  $ and $\tau.x_3 = \ss_2\ss_1 
= \ss_2 \ss_1^{-1} \ss_1^2 \equiv -x_2$. It follows that $H_0(\Z,H_1(A,\Z_{\eps})) \simeq \Z_3$. For $r=4$, $e \equiv 1 \mod 4$, $\tau.x_2 = \ss_2 \ss_3 = \ss_2 \ss_1^{-1} \ss_1 \ss_3
\equiv y_2 + x_3 \equiv -x_2 + x_3 \equiv -x_2$ hence $H_0(\Z,H_1(A,\Z_{\eps}))=0$. For $r=4$, $e \equiv 2 \mod 4$, $\tau.x_2 = \ss_3 \ss_4 = \ss_3 \ss_1^{-1} \ss_1 \ss_4 \equiv  x_2$ hence $H_0(\Z,H_1(A,\Z_{\eps}))= H_1(A,\Z_{\eps})$. Altogether, this yields

For $B = {{\mathrm{B}}}(*e,e,r)$, and $r \geq 3$, $$\begin{array}{ll}
H_1(B,\Z_{\eps}) \simeq \Z_2 & \mbox{ for } r \geq 5 \\
H_1(B,\Z_{\eps}) \simeq  \Z_6 & \mbox{ for } r=4, e \equiv 0,2 \mod 4 \\
H_1(B,\Z_{\eps}) \simeq  \Z_2 &  \mbox{ for } r=4, e \equiv 1 \mod 4 \\
H_1(B,\Z_{\eps}) \simeq  \Z_3 \oplus \Z_3 \oplus \Z_2 & \mbox{ for } r=3, e \equiv 0 \mod 4 \\
H_1(B,\Z_{\eps}) \simeq \Z_6 & \mbox{ for } r=3, e \equiv 1 \mod 4  \\
\end{array}$$

Finally, for groups of type $G(e,e,r)$, we use the Dehornoy-Lafont complex associated to the Corran-Picantin monoid. The 1-cells $[s]$ are mapped to $(\eps(s) - 1)[\emptyset]
= -2 [\emptyset]$, hence the 1-cycles are spanned by the $[s]-[t]$ for $s,t$ two atoms. We have $d_{\eps}[s_j,s_i] = 2(s_j-s_i)$ when $|j-i| \geq 2$, $d_{\eps}[t_0,t_i] = - t_i + t_{i+1} + t_0 - t_1$, $d_{\eps}[s_3,t_i] = 3t_i - 3 s_3$, $d_{\eps}[s_i,t_0] = 2(s_i-t_0)$ for $i \geq 4$, and $d_{\eps} [s_{i+1},s_i] = 3 (s_{i+1}-s_i)$. Since a basis of the 1-cycles is given by the $t_i - t_0$, $t_0-s_3$, $s_3-s_4$, …, $s_{r-1}-s_r$, $H_1(B,\Z_{\eps})$ is spanned by $t_1-t_0, t_0-s_3,
\dots,s_{r-1}-s_r$, each of these elements being annihilated by 3. From $d_{\eps}[s_i,t_0] = 2(s_i-t_0)$ for $i \geq 4$ we get that $s_4 - s_3 \equiv
s_3 - t_0$, from $d_{\eps}[s_5,s_3] = 2(s_5-s_3)$ we get $s_5 - s_4 \equiv s_4 - s_3$, and so on. Finally, from $$d_{\eps}[t_1,s_4] = 2(t_1-s_4) = 2(t_1 - t_0) + 2(t_0-s_3) + 2(s_3-s_4)
\equiv 2(t-1-t_0) +(t_0-s_3)$$ we get that $t_1 - t_0 \equiv t_0-s_3$. It follows that, for $r \geq 4$, $H_1(B,\Z_{\eps})$ is generated by $t_1-t_0$ hence $H_1(B,\Z_{\eps}) \simeq \Z_3$ ; for $r = 3$, it is generated by $t_1 - t_0$ and $t_0 - s_3$ and $H_1(B,\Z_{\eps}) \simeq \Z_3\oplus \Z_3$ ; it is generated by $t_1-t_0$ for $r =2$.

The case $e=1$ (that is, of the usual braid group) follows the same pattern. On the whole, we get the following.

For the groups ${{\mathrm{B}}}(e,e,r)$ with $e \geq 2$, $H_1(B,\Z_{\eps}) \simeq \Z_3$ if $r \geq 4$. If $r = 3$ then $H_1(B,\Z_{\eps}) \simeq \Z_3 \oplus \Z_3$. If $r=2$ then $H_1(B,\Z_{\eps}) \simeq \Z$. When $e=1$, we have $H_1(B,\Z_{\eps}) = 0$ for $r = 2$ or $r \geq 5$, and $H_1(B,\Z_{\eps})=
\Z_3$ if $r=3$ or $r = 4$.

$$\begin{array}{|c|l|}
ST & \mbox{Presentations for the group of even braids} \\
\hline
12 & vbu=awv,uaw=vbua,vbu=buaw,uaw=wvb\\
13 & buaw=awv,wvb=vbua,vbua=buaw,uawv=wvbu\\
22 & vbua=awvb,uawv=vbuaw,vbua=buaw,uawv=wvbu\\
23 & auaua=vv,vvv=uauau,bu=w,w=ub,bvb=awa,waw=vbv\\ 
24 & vv=auau,uaua=vv,awaw=bvbv,vbvb=wawa,w=bub,ubu=ww, \\
 & auawvb=vbuaw,vvbuaw=uawvbu\\
27 & w=bub,ubu=ww,vv=auau,uaua=vv,awawa=bvbvb,vbvbv=wawaw, \\ & 
bvwaua=waubv,waubvv=ubvwau\\ 
28 & aua=v,vv=uau,bu=w,w=ub,bvbv=awaw,wawa=vbvb,cu=x,x=uc,cv=ax, \\ & 
cv=ax, xa=vc,cwc=bxb,xbx=wcw\\ 
29 & v=aua,uau=vv,axa=cvc,vcv=xax,bxb=cwc,wcw=xbx,awaw=bvbv, \\ & 
vbvb=wawa,w=bu,ub=w,x=cu,uc=x,cwaxbv=bvcwax,xbvcwa=waxbvc\\ 
30 & auaua=vv,vvv=uauau,bu=w,w=ub,bvb=awa,waw=vbv,cu=x,x=uc, \\ & 
cv=ax, xa=vc,cwc=bxb,xbx=wcw\\ 
31^* & x=cuc,ucu=xx,axa=cvc,vcv=xax,dwd=byb,yby=wdw,aya=dvd,vdv=yay, \\ & 
vb=aw,uaw=vbu,aw=bua,vbu=wv,y=du,ud=y,bx=cw,wc=xb,dx=cy,\\
 & yc=xd \\
33 & v=aua,uau=vv,bvb=awa,waw=vbv,cvc=axa,xax=vcv,cwc=bxb,xbx=wcw,\\ &
cyc=dxd,xdx=ycy,w=bu,ub=w,x=cu,uc=x,y=du,ud=y,ay=dv,vd=ya, \\ & 
by=dw,wd=yb,cvbxaw=bxawcv,xawcvb=wcvbxa\\ 
34 & v=aua,uau=vv,bvb=awa,waw=vbv,cvc=axa,xax=vcv,cwc=bxb,xbx=wcw, \\ & 
cyc=dxd,xdx=ycy,w=bu,ub=w,x=cu,uc=x,y=du,ud=y,ay=dv,vd=ya, \\ & 
by=dw,wd=yb,dzd=eye,yey=zdz,z=eu,ue=z,az=ev,ve=za,bz=ew,\\ & 
we=zb,cz=ex,xe=zc,cvbxaw=bxawcv,xawcvb=wcvbxa\\
 35 & au=v,v=ua,bub=w,ww=ubu,bv=aw,wa=vb,cu=x,x=uc,cvc=axa, \\ & 
xax=vcv,cwc=bxb,xbx=wcw,du=y,y=ud,dv=ay,ya=vd,dw=by,yb=wd, \\ & 
dxd=cyc,ycy=xdx,eu=z,z=ue,ev=az,za=ve,ew=bz,zb=we,\\ 
 & ex=cz,zc=xe,eye=dzd,zdz=yey \\
36 & au=v,v=ua,bub=w,ww=ubu,bv=aw,wa=vb,cu=x,x=uc,cvc=axa,\\ & 
xax=vcv,cwc=bxb,xbx=wcw,du=y,y=ud,dv=ay,ya=vd,dw=by, \\ & 
yb=wd,dxd=cyc,ycy=xdx,eu=z,z=ue,ev=az,za=ve,ew=bz,zb=we, \\ & 
ex=cz,zc=xe,eye=dzd,zdz=yey,fu=x_2,x_2=uf,fv=ax_2,x_2a=vf,\\ &
fw=bx_2,x_2b=wf,fx=cx_2,x_2c=xf,fy=dx_2,x_2d=yf,fzf=ex_2e,x_2ex_2=zfz\\ 
37 & au=v,v=ua,bub=w,ww=ubu,bv=aw,wa=vb,cu=x,x=uc,cvc=axa, \\ & 
xax=vcv,cwc=bxb,xbx=wcw,du=y,y=ud,dv=ay,ya=vd,dw=by,\\ & 
yb=wd,dxd=cyc,ycy=xdx,eu=z,z=ue,ev=az,za=ve,ew=bz,zb=we,\\ & 
ex=cz,zc=xe,eye=dzd,zdz=yey,fu=x_2, x_2=uf,fv=ax_2,x_2a=vf, \\ &
fw=bx_2,x_2b=wf,fx=cx_2,x_2c=xf,fy=dx_2,x_2d=yf,fzf=ex_2e,\\ & 
x_2ex_2=zfz,gu=y_2,y_2=ug,gv=ay_2,y_2a=vg,gw=by_2,y_2b=wg,gx=cy_2,\\ 
 & y_2c=xg,gy=dy_2,y_2d=yg,gz=ey_2,y_2e=zg,gx_2g=fy_2f,y_2fy_2=x_2gx_2 \\
\end{array}$$ $^*$ Provided that the presentation of $B$ suggested in [@BMR] for $G_{31}$ is correct.

$$\begin{array}{|c|ccccccc|}
 & H_0 & H_1 & H_2 & H_3 & H_4 & H_5 & H_6 \\
\hline
G_{12} & \Z & \Z & 0 & & & & \\ 
G_{13} & \Z & \Z^2 & \Z & & & & \\ 
G_{22} & \Z & \Z & 0 & & & & \\
G_{24} & \Z & \Z & \Z & \Z & & & \\
G_{27} & \Z & \Z & \Z_3 \times \Z & \Z & & & \\
G_{29} & \Z & \Z & \Z_2 \times \Z_4 & \Z_2 \times \Z & \Z & & \\
G_{31} & \Z & \Z & \Z_6 & \Z & \Z & & \\
G_{33} & \Z & \Z & \Z_6 & \Z_6 & \Z & \Z & \\
G_{34} & \Z & \Z & \Z_6 & ? & ? & ? & ? \\
\end{array}$$

$$\begin{array}{|l|ccccccccc|}
 & H_0 & H_1 & H_2 & H_3 & H_4 & H_5 & H_6 & H_7 & H_8\\
\hline
I_2(2m) & \Z & \Z^2 & \Z & & & & & & \\ 
I_2(2m+1) & \Z & \Z & 0  & & & & & & \\
H_3 = G_{23} & \Z & \Z & \Z & \Z & & & & & \\
H_4 = G_{30} & \Z & \Z & \Z_2   & \Z & \Z & & & & \\
F_4 = G_{28} & \Z & \Z^2 & \Z^2 & \Z^2 & \Z & & & & \\
E_6=G_{35} & \Z & \Z & \Z_2 & \Z_2 & \Z_6 & \Z_3 & 0 & & \\
E_7 = G_{36} & \Z & \Z & \Z_2 & \Z_2^2 & \Z_6^2 & \Z_3 \times \Z_6 & \Z & \Z & \\
E_8 = G_{37} & \Z & \Z & \Z_2 & \Z_2 & \Z_2 \times \Z_6 & \Z_3 \times \Z_6 & \Z_2 \times \Z_6 & \Z & \Z \\
\end{array}$$

$$\begin{array}{|c|c|c|}
ST & B^{(2)}_{ab} & H_1(B,\Z_{\eps}) \\
\hline
 12 & \Z_3\times \Z & \Z_3 \\ 13 & \Z\times \Z & \Z_2 \\ 22 & \Z & 0 \\  
  23 & \Z & 0 \\ 24 & \Z & 0 \\ 27 & \Z & 0 \\ 28 & \Z\times \Z & \Z_2 \\ 
  29 & \Z & 0 \\ 30 & \Z & 0 \\ 31^* & \Z & 0 \\ 33 & \Z & 0 \\ 34 & \Z & 0 \\ 
  35 & \Z & 0 \\ 36 & \Z & 0 \\ 37 & \Z & 0 \\ 
\end{array}$$
