arXiv:1707.06864v1 [] 21 Jul 2017

Interval structures for braid groups B(e, e, n)
Georges Neaime
Laboratoire de Mathématiques Nicolas Oresme
Université de Caen Normandie
georges.neaime@unicaen.fr
March 18, 2018
Abstract
Complex braid groups are a generalization of Artin-Tits groups. The general
goal is to extend what is known for Artin-Tits groups to other complex braid
groups. We are interested in Garside structures that derive from intervals. Actually, we construct intervals in the complex reflection group G(e, e, n) which
gives rise to Garside groups. Some of these groups correspond to the complex
braid group B(e, e, n). For the other Garside groups that appear, we give some
of their properties in order to understand these new structures.

Contents
1 Introduction

2

2 The groups G(e, e, n) and B(e, e, n)

6

3 Reduced words in G(e, e, n)

9

4 Balanced elements of maximal length

15

5 Interval structures

20

6 About the interval structures

24

1

1

Introduction

A reflection is an element s of GLn (C) with n ≥ 1 such that ker(s− 1) is a hyperplane
and s2 = 1. Relaxing the last condition to s of finite order defines the notion of pseudoreflection. Let W be a finite subgroup of GLn (C) and R be the set of reflections of
W . We say that W is a complex reflection group if W is generated by R. Since
every complex reflection group is a direct product of irreducible ones, we restrict our
work to irreducible complex reflection groups. These groups have been classified by
Shephard and Todd [17] in 1954. The classification consists of two families and 15
exceptions. For e, n ≥ 1, the first family is denoted by G(e, e, n) and defined as the
group of n × n matrices consisting of
• monomial matrices (each row and column has a unique nonzero entry),
• with all nonzero entries lying in µe , the group of e-th roots of unity, and
• for which the product of the nonzero entries is 1.
The second family is denoted by G(2e, e, n) and defined as the group of n × n matrices
consisting of monomial matrices, with all nonzero entries lying in µ2e , and for which
the product of the nonzero entries is 1 or −1. For the definition of the exceptional
groups, the reader may check [17].
For every complex reflection group W , there exists a corresponding hyperplane
arrangement
and hyperplane complement: A = {Ker(s − 1) | s ∈ R} and X =
S
Cn \ A. The pure braid group is defined as P := π1 (X) and the braid group (or
complex braid group) as B := π1 (X/W ). Note that we have the short exact sequence:
1 −→ P −→ B −→ W −→ 1.
If W = G(de, e, n) with d = 1 or 2, then we denote by B(de, e, n) the associated
braid group. This construction of the braid group is also valid for finite complex
pseudo-reflection groups. However, using the classification of Shephard and Todd for
irreducible complex pseudo-reflection groups and case-by-case results of [3], we may
restrict our work to complex reflection groups as far as group-theoretic properties of
the braid group are concerned.
The previous definitions are a generalization of the well-known Coxeter and ArtinTits groups that we recall now. One way of defining a finite Coxeter group W is by
a presentation with
• a generating set S and
• relations:
– quadratic relations: s2 = 1 for all s ∈ S and
– braid relations: sts
| {z· ·}· for s 6= t ∈ S where mst is the order of
| {z· ·}· = tst
mst

mst

st ∈ W .

2

The Artin-Tits group B(W ) corresponding to W is the group of fractions of the
monoid B + (W ) defined by a presentation with generators: a set Se in bijection with
e
the set S and the relations are only the braid relations: s̃|t̃s̃{z· ·}· = |t̃s̃{z
t̃ · ·}· for s̃ 6= t̃ ∈ S.
mst

mst

The seminal example of these groups is when W = Sn , the symmetric group with
n ≥ 2. It has a presentation with generators s1 , s2 , · · · , sn−1 and relations:
• s2i = 1 for 1 ≤ i ≤ n − 1,
• si si+1 si = si+1 si si+1 for 1 ≤ i ≤ n − 2, and
• si sj = sj si for |i − j| > 1.
The Artin-Tits group associated with Sn is the ‘classical’ braid group denoted by Bn .
The following diagram presentation encodes all the information about the generators
and relations of the presentation of Bn .
s̃1

s̃n−2 s̃n−1

s̃2

Figure 1: Diagram for the presentation of Bn .
The link with the first definitions is as follows. Consider W a real reflection group
meaning that W < GLn (R) < GLn (C). By a theorem of Coxeter, every real reflection group corresponds to a Coxeter group. Furthermore, by a theorem of Brieskorn,
the Artin-Tits group corresponding to W is the braid group π1 (X/W ) attached to W .
It is widely believed that complex braid groups share similar properties with ArtinTits groups. One would like to extend what is known for Artin-Tits groups to other
complex braid groups. For instance, it is shown by Bessis and Corran [1] in 2006,
and by Corran and Picantin [6] in 2009 that the complex braid group B(e, e, n) admits some Garside structures. We are interested in constructing Garside structures
for B(e, e, n) that derive from intervals in the associated complex reflection group
G(e, e, n). For instance, the complex braid group B(e, e, n) admits interval structures
that derive from the non-crossing partitions of type (e, e, n). See [1] for a detailed
study. The aim of this paper is to provide new interval structures for B(e, e, n). In
Section 3, we compute the length of all elements of G(e, e, n) over an appropriate
generating set. This allows us to construct intervals in the complex reflection group
G(e, e, n) and show that they are lattices which automatically gives rise to Garside
structures. This is done in Theorem 5.15. In the last section, we provide group presentations for these Garside structures which we denote by B (k) (e, e, n) for 1 ≤ k ≤ e − 1
and identify which of them correspond to B(e, e, n). For the other Garside structures
that appear, we give some of their properties in order to understand them. One of
the important results obtained is Theorem 6.14:
B (k) (e, e, n) is isomorphic to B(e, e, n) if and only if k ∧ e = 1.
In the remaining part of this section, we include the necessary preliminaries to
accurately describe these Garside structures.
3

1.1

Garside monoids and groups

In his PhD thesis, defended in 1965 [12], and in the article that followed [13], Garside
solved the Conjugacy Problem for the classical braid group Bn by introducing a submonoid Bn+ of Bn and an element ∆n of Bn+ that he called fundamental, and then
showing that there exists a normal form for every element in Bn . In the beginning of
the 1970’s, it was realized by Brieskorn and Saito [2] and Deligne [11] that Garside’s
results extend to all Artin-Tits groups. At the end of the 1990’s, after listing the
abstract properties of Bn+ and the fundamental element ∆n , Dehornoy and Paris [10]
defined the notion of Gaussian groups and Garside groups which leads, in “a natural,
but slowly emerging program” as stated in [8], to Garside theory. For a complete
study about Garside structures that is still far from complete, we refer the reader to
[8].
We start by defining Garside monoids and groups. Let M be a monoid. Under
some assumptions about M , more precisely the assumptions 1 and 2 of Definition 1.2,
one can define a partial order relation on M as follows.
Definition 1.1. Let f, g ∈ M . We say that f left-divides g or simply f divides g when
there is no confusion, written f  g, if f g ′ = g holds for some g ′ ∈ M . Similarly, we
say that f right-divides g, written f r g, if g ′ f = g holds for some g ′ ∈ M .
We are ready to define Garside monoids and groups.
Definition 1.2. We say that M is a Garside monoid if
1. M is cancellative, that is f g = f h =⇒ g = h and gf = hf =⇒ g = h for
f, g, h ∈ M ,
2. there exists λ : M −→ N s.t. λ(f g) ≥ λ(f ) + λ(g) and g 6= 1 =⇒ λ(g) 6= 0,
3. any two elements of M have a gcd and an lcm for  and r , and
4. there exists an element ∆ ∈ M such that the set of its left divisors coincides
with the set of its right divisors, generate M , and is finite.
The element ∆ is called a Garside element of M and the divisors of ∆ are called
the simples of the Garside structure.
Assumptions 1 and 3 of Definition 1.2 ensure that Ore’s conditions are satisfied.
Hence there exists a group of fractions of the monoid M in which it embeds. This
allows us to give the following definition.
Definition 1.3. A Garside group is the group of fractions of a Garside monoid.
Note that one of the important aspects of a Garside structure is the existence of
a normal form for all elements of the Garside group. Furthermore, many problems
like the Word and Conjugacy problems can be solved in Garside groups which makes
their study interesting.

4

1.2

Interval structures

Let G be a finite group generated by a finite set S. There is a way to construct
Garside structures from intervals in G.
We start by defining a partial order relation on G.
Definition 1.4. Let f, g ∈ G. We say that g is a divisor of f or f is a multiple of
g, and write g  f , if f = gh with h ∈ G and ℓ(f ) = ℓ(g) + ℓ(h), where ℓ(f ) is the
length over S of f ∈ G.
Definition 1.5. For w ∈ G, define a monoid M ([1, w]) by the presentation of monoid
with
• generating set P in bijection with the interval
[1, w] := {f ∈ G | 1  f  w} and
• relations: f g = h if f, g, h ∈ [1, w], f g = h, and f  h, that is ℓ(f ) + ℓ(g) = ℓ(h).
Similarly, one can define the partial order relation on G
g r f if and only if ℓ(f g −1 ) + ℓ(g) = ℓ(f ),
then define the interval [1, w]r and the monoid M ([1, w]r ).
Definition 1.6. Let w be in G. We say that w is a balanced element of G if [1, w] =
[1, w]r .
We have the following theorem (see Section 10 of [15] for a proof).
Theorem 1.7. If w ∈ G is balanced and both posets ([1, w], ) and ([1, w]r , r ) are
lattices, then M ([1, w]) is a Garside monoid with Garside element w and with simples
[1, w].
The previous construction gives rise to an interval structure. The interval monoid
is M ([1, w]). When M ([1, w]) is a Garside monoid, its group of fractions exists and
is denoted by G(M ([1, w])). We call it the interval group. We will give a seminal
example of this structure. It shows that Artin-Tits groups admit interval structures.
Example 1.8. Let W be a finite coxeter group
W =< S | s2 = 1, sts
| {z· ·}· = tst
| {z· ·}· for s 6= t ∈ S > .
mst

mst

Take G = W and g = w0 the longest element over S in W . We have [1, w0 ] = W .
Construct the interval monoid M ([1, w0 ]). We have M ([1, w0 ]) is the Artin-Tits
monoid B + (W ). Hence B + (W ) is generated by a copy W of W with f g = h if
f g = h and ℓ(f ) + ℓ(g) = ℓ(h); f, g, and h ∈ W . We have the following result.
Theorem 1.9. B + (W ) is a Garside monoid with Garside element w0 and with simples W .
5

The groups G(e, e, n) and B(e, e, n)

2

In this section, we recall some presentations by generators and relations of G(e, e, n)
and B(e, e, n) based on the results of [3] and [6].

2.1

Presentations

Recall that for e, n ≥ 1, G(e, e, n) is the group of n × n matrices consisting of monomial matrices, with all nonzero entries lying in µe , the e-th roots of unity, and for
which the product of the nonzero entries is 1. Note that this family includes three
families of finite Coxeter groups: G(1, 1, n) is the symmetric group, G(e, e, 2) is the
dihedral group, and G(2, 2, n) is type Dn .
Define a group by a presentation with generators and relations that can be described by the following diagram.
t1

2
e

t0

s3

s4

sn−1

sn

2

2

2

2

2

Figure 2: Diagram for the presentation of BMR of G(e, e, n).
The generators of this presentation are t0 , t1 , s3 , s4 , · · · , sn−1 , sn and the relations
are:
• quadratic relations for all generators,
• the relations of the symmetric group for s3 , s4 , · · · , sn−1 ,
• the relation of the dihedral group I2 (e) for t0 and t1 : t0 t1 t0 · · · = t1 t0 t1 · · ·,
| {z } | {z }
e

e

• s3 ti s3 = ti s3 ti for i = 0, 1, and

• sj ti = ti sj for i = 0, 1 and 4 ≤ j ≤ n.
It is shown in [3] that this group is isomorphic to the complex reflection
group G(e,
e, n)



I
0
0
0
j−2
0 ζe−i
0
 0
0 1
0 


0
0  for i = 0, 1 and sj 7−→ sj := 
with ti 7−→ ti := ζei
 0
1 0
0 
0
0 In−2
0
0 0 In−j
for 3 ≤ j ≤ n, where Ik is the identity matrix with 1 ≤ k ≤ n.
This presentation is called the presentation of BMR (Broué Malle Rouquier) of
G(e, e, n).

6

By [3], removing the quadratic relations gives a presentation of the complex braid
group B(e, e, n) attached to G(e, e, n) with diagram as follows.
t̃1
s̃3

s̃n−1

s̃4

s̃n

e
t̃0

Figure 3: Diagram for the presentation of BMR of B(e, e, n).
Note that the set of generators of this presentation is in bijection with
{t0 , t1 , s3 , s4 , · · · , sn }. This presentation is called the presentation of BMR of B(e, e, n).

2.2

Other presentations

Consider the presentation of BMR for the complex braid group B(e, e, n). For e ≥ 3
and n ≥ 3, it is shown in [5], p.122, that the monoid defined by this presentation
fails to embed in B(e, e, n). Thus, this presentation does not give rise to a Garside
structure for B(e, e, n). Considering the interest of the Garside groups given earlier,
it is legitimate to look for a Garside structure for B(e, e, n).
Let t̃i := t̃i−1 t̃i−2 t̃−1
i−1 for 2 ≤ i ≤ e − 1. Consider the following diagram presentation (the kite).
t̃i

t̃2
t̃1

t̃e−1

t̃0

s̃3
s̃4

s̃n

···
s̃n−1

Figure 4: Diagram for the presentation of CP of B(e, e, n).
The generators of this presentation are t̃0 , t̃1 , · · · , t̃e−1 , s̃3 , s̃4 , · · · , s̃n−1 , s̃n and
the relations are:
• the relations of the symmetric group for s̃3 , s̃4 , · · · , s̃n−1 ,

7

• the relation of the ‘dual’ dihedral group for t̃0 , t̃1 , · · · , t̃e−1 : t̃i t̃i−1 = t̃j t̃j−1 for
i, j ∈ Z/eZ,
• s̃3 t̃i s̃3 = t̃i s̃3 t̃i for 0 ≤ i ≤ e − 1, and
• s̃j t̃i = t̃i s̃j for 0 ≤ i ≤ e − 1 and 4 ≤ j ≤ n.
It is shown in [6] that the group defined by this presentation is isomorphic to
B(e, e, n). We call it the presentation of CP (Corran Picantin) of B(e, e, n). It is also
shown in [6] that:
Proposition 2.1. The presentation of CP gives rise to a Garside structure for
B(e, e, n) with
t se · · · sen and
t1 e
t se · · · sen · · · se3 e
t e
t0 se3 e
Garside element: ∆ = e
t1 e
|
{z0 3
}
| 1{z0 }3
|{z}
∆3

∆2

∆n

simples: the elements of the form δ2 δ3 · · · δn where δi is a divisor of ∆i for
2 ≤ i ≤ n.
It is stated in [6] that if one adds the relations x2 = 1 for all generators x of the
presentation of CP, one obtains a presentation of a group isomorphic to G(e, e, n). It
is called the presentation of CP of G(e, e, n) with diagram as follows.
ti
t2

2

2

t1
te−1

t0

2

2
2
2

s3

2

s4

···

2

sn

2

sn−1

Figure 5: Diagram for the presentation of CP of G(e, e, n).
The generators of this presentation belong to the set X := {t0 , t1 , · · · , te−1 , s3 , · · · , sn }
and the diagram presentation encodes the same relations as the diagram of Figure 2.2
with quadratic relations for all the generators in X.


0 ζe−i
0
0
0 
The isomorphism with G(e, e, n) is given by ti 7−→ ti := ζei
0
0
I
n−2


Ij−2 0 0
0
 0
0 1
0 
 for 3 ≤ j ≤ n. Denote
for 0 ≤ i ≤ e − 1, and sj 7−→ sj := 
 0
1 0
0 
0
0 0 In−j
X the set {t0 , t1 , · · · , te−1 , s3 , · · · , sn }.
8

In the next section, we compute the length over X of each element w in G(e, e, n)
by providing a minimal word representative over X of w.

3

Reduced words in G(e, e, n)

In this section, we represent each element of G(e, e, n) by a reduced word over X.
After some preparations, this is done in Proposition 3.10. Also, we characterize all
the elements of G(e, e, n) that are of maximal length over X.
Recall that an element w ∈ X∗ is called word over X. We denote by ℓ(w) the
length over X of the word w.
Definition 3.1. Let w be an element of G(e, e, n). We define ℓ(w) to be the minimal
word length ℓ(w) of a word w over X that represents w. A reduced expression of w
is any word representative of w of word length ℓ(w).
We introduce an algorithm that produces a word RE(w) over X for a given matrix
w in G(e, e, n). Later on, we prove that RE(w) is a reduced expression over X of w.
Input : w, a matrix in G(e, e, n).
Output: RE(w), a word over X.
Local variables: w′ , RE(w), i, U , c, k.
Initialisation: U := [1, ζe , ζe2 , ..., ζee−1 ], s2 := t0 , s2 := t0 ,
RE(w) = ε: the empty word, w′ := w.
for i from n down to 2 do
c := 1; k := 0;
while w′ [i, c] = 0 do
c := c + 1;
end
Then w′ [i, c] is the root of unity on row i;
while U [k + 1] 6= w′ [i, c] do
k := k + 1;
end
Then w′ [i, c] = ζek .
if k 6= 0 then
w′ := w′ sc sc−1 · · · s3 s2 tk ; Then w′ [i, 2] = 1;
RE(w) := tk s2 s3 · · · sc RE(w);
c := 2;
end
w′ := w′ sc+1 · · · si ; Then w′ [i, i] = 1;
RE(w) := si · · · sc+1 RE(w);
end
Return RE(w);
Algorithm 1: A word over X corresponding to an element w ∈ G(e, e, n).
9



0
0
Example 3.2. We apply Algorithm 1 to w := 
0
1

0
ζ32
Step 1 (i = 4, k = 0, c = 1): w′ := ws2 s3 s4 = 
0
0

0 0
 0 ζ32
Step 2 (i = 3, k = 1, c = 2): w′ := w′ s2 = 
ζ3 0
0 0


0 0 1 0
1 0 0 0
′
′
, then w′ := w′ s3

then w := w t1 = 
0 1 0 0
0 0 0 1
Step 3 (i = 2, k = 0, c = 1): w′ := w′ s2 = I4 .
Hence RE(w) = s2 s3 t1 s2 s4 s3 s2 = t0 s3 t1 t0 s4 s3 t0 .

0 0
ζ32 0
0 ζ3
0 0
0
1
0
0
ζ3 0
0 
0
1 0
0 0
,
0 0
0 1

0
1
=
0
0


1
0
 ∈ G(3, 3, 4).
0
0

0
0
.
0
1

1 0
0 0
0 1
0 0


0
0
.
0
1

Let wn := w ∈ G(e, e, n). For i from n to 2,
 the i-th step of Algorithm 1
0
wi
into a block diagonal matrix
transforms the block diagonal matrix
0 In−i


wi−1
0
∈ G(e, e, n) with w1 = I1 . Actually, for 2 ≤ i ≤ n, there exists
0
In−i+1
a unique c with 1 ≤ c ≤ n such that wi [i, c] 6= 0. At each step i of Algorithm 1, if
wi [i, c] = 1, we shift it into the diagonal position [i, i] by right multiplication by transpositions of the symmetric group Sn . If wi [i, c] 6= 1, we shift it into the first column
by right multiplication by transpositions, transform it into 1 by right multiplication
by an element of {t0 , t1 , · · · , te−1 }, and then shift the 1 obtained into the diagonal
position [i, i]. The following lemma is straightforward.
Lemma 3.3. For 2 ≤ i ≤ n, the block wi−1 is obtained by
• removing the row i and the column c from wi , then by
• multiplying the first column of the new matrix by wi [i, c].
Example 3.4. Let w be as in Example 3.2. The block
 wn−1 isobtained by removing
0 0 1
the row n and the column 1 from wn = w to obtain ζ32 0 0, then by multiplying
0 ζ3 0
the first column of this matrix by 1. Check that the block wn−1 obtained in Example
3.2, after multiplying w by t0 s3 s4 , is the same as the one obtained here. The same
can be said for the other blocks wi with 2 ≤ i ≤ n − 1.
Definition 3.5. At each step i from n to 2,

10

• if wi [i, c] = ζek with k
si · · · s3 tk t0 s3 · · · sc
si · · · s3 tk t0
si · · · s3 tk

6= 0, we define REi (w) to be the word
if c ≥ 3,
if c = 2,
if c = 1, and

• if wi [i, c] = 1, we define REi (w) to be the word
si · · · sc+1 .
Remark that for 2 ≤ i ≤ n, the word REi (w) is either the empty word (when
wi [i, i] = 1) or a word that contains si necessarily but does not contain any of
si+1 , si+2 , · · · , sn .
Lemma 3.6. We have RE(w) = RE2 (w)RE3 (w) · · · REn (w).
Proof. The output RE(w) of Algorithm 1 is a concatenation of the words
RE2 (w), RE3 (w), · · · , and REn (w) obtained at each step i from n to 2 of Algorithm 1.
Example 3.7. If w is defined as in Example 3.2, we have
RE(w) = t0
s t t s s t .
|{z} |3 {z1 }0 |4 {z3 }0
RE2 (w) RE3 (w) RE4 (w)

Proposition 3.8. The word RE(w) given by Algorithm 1 is a word representative
over X of w ∈ G(e, e, n).
Proof. Algorithm 1 transforms the matrix w into In by multiplying it on the right
by elements of X and RE(w) is a concatenation (in reverse order) of elements of X
corresponding to the matrices of X used to transform w into In . Hence RE(w) is a
word representative over X of w ∈ G(e, e, n).
The following proposition will prepare us to prove that the output of Algorithm 1
is a reduced expression over X of a given element w ∈ G(e, e, n).
Proposition 3.9. Let w be an element of G(e, e, n). For all x ∈ X, we have
|ℓ(RE(xw)) − ℓ(RE(w))| = 1.
Proof. For 1 ≤ i ≤ n, there exists unique ci such that w[i, ci ] 6= 0. We denote w[i, ci ]
by ai .
Case 1: Suppose x = si for 3 ≤ i ≤ n.
Since s2i w = w, we can assume without restriction that ci−1 < ci .
Set w′ := si w. Since the left multiplication by the matrix x exchanges the rows i − 1
and i of w and the other rows remain the same, by Definition 3.5 and Lemma 3.3, we
have:
REi+1 (xw)REi+2 (xw) · · · REn (xw) = REi+1 (w)REi+2 (w) · · · REn (w) and
RE2 (xw)RE3 (xw) · · · REi−2 (xw) = RE2 (w)RE3 (w) · · · REi−2 (w).
Then, in order to prove our property, we should compare ℓ1 := ℓ(REi−1 (w)REi (w))

11

and ℓ2 := ℓ(REi−1 (xw)REi (xw)).
Since ci−1 < ci , by Lemma 3.3, the rows i − 1 and i of the blocks wi and wi′ are of
the form:
.. c .. c′ .. i
bi−1
i−1
wi :
ai
i
.. c
wi′ :

i−1
i

.. c′ ..
ai

i

bi−1

with c < c′ and where we write bi−1 instead of ai−1 since ai−1 is likely to change
when applying Algorithm 1 if ci−1 = 1, that is ai−1 on the first column of w.
We will discuss different cases depending on the values of ai and bi−1 .
• Suppose ai = 1.
– If bi−1 = 1,
we have REi (w) = si · · · sc′ +2 sc′ +1 and REi−1 (w) = si−1 · · · sc+2 sc+1 .
Furthermore, we have REi (xw) = si · · · sc+2 sc+1
and REi−1 (xw) = si−1 · · · sc′ +1 sc′ .
It follows that ℓ1 = ((i − 1) − (c + 1) + 1) + (i − (c′ + 1) + 1) = 2i − c − c′ − 1
and ℓ2 = ((i − 1) − c′ + 1) + (i − (c + 1) + 1) = 2i − c − c′ hence ℓ2 = ℓ1 + 1.
– If bi−1 = ζek with 1 ≤ k ≤ e − 1,
we have REi (w) = si · · · sc′ +2 sc′ +1 and REi−1 (w) = si−1 · · · s3 tk t0 s3 sc .
Furthermore, we have REi (xw) = si · · · s3 tk t0 s3 sc and REi−1 (xw) = sc′ · · · si−1 .
It follows that ℓ1 = (((i−1)−3+1)+2+(c−3+1))+(i−(c′ +1)+1) = 2i+
c−c′ −3 and ℓ2 = ((i−1)−c′ +1)+((i−3+1)+2+(c−3+1)) = 2i+c−c′ −2
hence ℓ2 = ℓ1 + 1.
It follows that
if ai = 1, then ℓ(RE(si w)) = ℓ(RE(w)) + 1.

(a)

• Suppose now that ai = ζek with 1 ≤ k ≤ e − 1.
– If bi−1 = 1,
we have REi (w) = si · · · s3 tk t0 s3 sc′ and REi−1 (w) = si−1 · · · sc+1 .
Also, we have REi (xw) = si · · · sc+1 and
REi−1 (xw) = si−1 · · · s3 tk t0 s3 sc′ −1 .
It follows that ℓ1 = ((i − 1) − (c + 1) − 1) + ((i − 3 + 1) + 2 + (c′ − 3 + 1)) =
2i−c+c′ −5 and ℓ2 = (((i−1)−3+1)+2+((c′−1)−3+1))+(i−(c+1)−1) =
2i − c + c′ − 6 hence ℓ2 = ℓ1 − 1.

12

′

– If bi−1 = ζek with 1 ≤ k ′ ≤ e − 1,
we have REi (w) = si · · · s3 tk t0 s3 sc′ and
REi−1 (w) = si−1 · · · s3 tk′ t0 s3 sc .
Also, we have REi (xw) = si · · · s3 tk′ t0 s3 sc and
REi−1 (xw) = si−1 · · · s3 tk t0 s3 sc′ −1 .
It follows that ℓ1 = ((i−1)−3+1)+2+(c−3+1)+(i−3+1)+2+(c′−3+1) =
2i + c + c′ − 5 and ℓ2 = ((i − 1) − 3 + 1) + 2 + ((c′ − 1) − 3 + 1) + (i − 3 +
1) + 2 + (c − 3 + 1) = 2i + c + c′ − 6 hence ℓ2 = ℓ1 − 1.
It follows that
if ai 6= 1, then ℓ(RE(si w)) = ℓ(RE(w)) − 1.

(b)

Case 2: Suppose x = ti for 0 ≤ i ≤ e − 1.
Set w′ := ti w. By definition of the left multiplication by ti , we have that the last
n − 2 rows of w and w′ are the same. Hence, by Definition 3.5 and Lemma 3.3, we
have:
RE3 (xw)RE4 (xw) · · · REn (xw) = RE3 (w)RE4 (w) · · · REn (w). In order to prove our
property in this case, we should compare ℓ1 := ℓ(RE2 (w)) and ℓ2 := ℓ(RE2 (xw)).
• Consider the case where c1 < c2 .
Since c1 < c2 , by Lemma 3.3, the blocks w2 and w2′ are of the form:




0
ζe−i a2
b
0
and w2′ =
w2 = 1
with b1 instead of a1 since a1 is likely
ζei b1
0
0 a2
to change when applying Algorithm 1 if c1 = 1.
– Suppose a2 = 1,
we have b1 = 1 necessarily hence ℓ1 = 0. Since RE2 (xw) = ti , we have
ℓ2 = 1. It follows that when c1 < c2 ,
if a2 = 1, then ℓ(RE(ti w)) = ℓ(RE(w)) + 1.

(c)

– Suppose a2 = ζek with 1 ≤ k ≤ e − 1,
we have RE2 (w) = tk t0 . Thus ℓ1 = 2. We also have ℓ2 = 1 for any b1 . It
follows that when c1 < c2 ,
if a2 6= 1, then ℓ(RE(ti w)) = ℓ(RE(w)) − 1.

(d)

• Now, consider the case where c1 > c2 .
Since c1 > c2 , by Lemma 3.3, the blocks w2 and w2′ are of the form:



 −i
0 a1
ζ b
0
w2 =
with b2 instead of a2 since a2 is likely
and w2′ = e 2
b2 0
0
ζei a1
to change when applying Algorithm 1 if c2 = 1.

13

– Suppose a1 6= ζe−i ,
we have ℓ1 = 1 necessarily for any b2 , and since ζei a1 6= 1, we have ℓ2 = 2.
Hence when c1 > c2 ,
if a1 6= ζe−i , then ℓ(RE(ti w)) = ℓ(RE(w)) + 1.

(e)

– Suppose a1 = ζe−i ,
we have ℓ1 = 1 and ℓ2 = 0 for any b2 . Hence when c1 > c2 ,
if a1 = ζe−i , then ℓ(RE(ti w)) = ℓ(RE(w)) − 1.

(f )

This finishes our proof.
Proposition 3.10. Let w be an element of G(e, e, n). The word RE(w) is a reduced
expression over X of w.
Proof. We must prove that ℓ(w) = ℓ(RE(w)).
Let x1 x2 · · · xr be a reduced expression over X of w. Hence ℓ(w) = ℓ(x1 x2 · · · xr ) = r.
Since RE(w) is a word representative over X of w, we have ℓ(RE(w)) ≥ ℓ(x1 x2 · · · xr ) =
r. We prove that ℓ(RE(w)) ≤ r. Observe that we can write w as x1 x2 · · · xr where
x1 , x2 , · · · , xr are the matrices of G(e, e, n) corresponding to x1 , x2 , · · · , xr .
By Proposition 3.9, we have: ℓ(RE(w)) = ℓ(RE(x1 x2 · · · xr )) ≤ ℓ(RE(x2 x3 · · · xr )) +
1 ≤ ℓ(RE(x3 · · · xr )) + 2 ≤ · · · ≤ r. Hence ℓ(RE(w)) = r = ℓ(w) and we are done.
The following proposition is useful in the other sections. It summarizes the proof
of Proposition 3.9.
Proposition 3.11. Let w be an element of G(e, e, n). Denote by ai the unique
nonzero entry w[i, ci ] on the row i of w where 1 ≤ i, ci ≤ n.
1. For 3 ≤ i ≤ n, we have:
(a) if ci−1 < ci , then
ℓ(si w) = ℓ(w) − 1 if and only if ai 6= 1.
(b) if ci−1 > ci , then
ℓ(si w) = ℓ(w) − 1 if and only if ai−1 = 1.
2. If c1 < c2 , then ∀ 0 ≤ k ≤ e − 1, we have
ℓ(tk w) = ℓ(w) − 1 if and only if a2 6= 1.
3. If c1 > c2 , then ∀ 0 ≤ k ≤ e − 1, we have
ℓ(tk w) = ℓ(w) − 1 if and only if a1 = ζe−k .
Proof. The claim 1(a) is deduced from (a) and (b), 2 is deduced from (c) and (d), and
3 is deduced from (e) and (f ) where (a), (b), (c), (d), (e), and (f ) are given in the
proof of Proposition 3.9. Since s2i = 1, 1(b) can be deduced from 1(a).

14

The next proposition is about elements of G(e, e, n) that are of maximal length.
Proposition 3.12. The maximal length of an element of G(e, e, n) is n(n − 1). It
is realized for diagonal matrices w such that w[i, i] is an e-th root of unity different
from 1 for 2 ≤ i ≤ n. A minimal word representative of such an element is then of
the form tk2 t0 s3 tk3 t0 s3 · · · sn · · · s3 tkn t0 s3 · · · sn with 1 ≤ k2 , · · · , kn ≤ (e − 1), and
|
{z
}
| {z } | {z }
the number of elements that are of maximal length is (e − 1)(n−1) .
Proof. By Algorithm 1, an element w in G(e, e, n) is of maximal length when wi [i, i] =
ζek for 2 ≤ i ≤ n and ζek 6= 1. By Lemma 3.3, this condition is satisfied when w is a
diagonal matrix such that w[i, i] is an e-th root of unity different from 1 for 2 ≤ i ≤ n.
A minimal word representative given by Algorithm 1 for such an element is of the
form tk2 t0 s3 tk3 t0 s3 · · · sn · · · s3 tkn t0 s3 · · · sn (1 ≤ k2 , · · · , kn ≤ (e − 1)) which is of
| {z } | {z }
|
{z
}
length 2 + 4 + ... + 2(n − 1) = n(n − 1). The number of elements of this form is
(e − 1)(n−1) .




Example 3.13. Consider λ := 




(ζe−1 )(n−1)
ζe
..

.



 ∈ G(e, e, n). We have


ζe
RE(λ) = t1 t0 s3 t1 t0 s3 · · · sn · · · s3 t1 t0 s3 · · · sn . Hence ℓ(λ) = n(n − 1) which is the
|
{z
}
|{z} | {z }
maximal length of an element of G(e, e, n).

Note that λ is the image in G(e, e, n) of Λ, the Garside element of the presentation
of Corran and Picantin of the complex braid group B(e, e, n).
Once the length function over X is understood, we can construct intervals in
G(e, e, n) and characterize the balanced elements that are of maximal length.

4

Balanced elements of maximal length

The aim of this section is to prove that the only balanced elements of G(e, e, n) that
are of maximal length are λk with 1 ≤ k ≤ e − 1, where λ was given in Section 3. This
is done by characterizing the intervals of the elements that are of maximal length.
We start by recalling the partial order relations on G(e, e, n).
Definition 4.1. Let w, w′ ∈ G(e, e, n). We say that w′ is a divisor of w or w is
a multiple of w′ , and write w′  w, if w = w′ w′′ with w′′ ∈ G(e, e, n) and ℓ(w) =
ℓ(w′ ) + ℓ(w′′ ). This defines a partial order relation on G(e, e, n).
Similarly, we have another partial order relation on G(e, e, n).
Definition 4.2. Let w, w′ ∈ G(e, e, n). We say that w′ is a right divisor of w or w
is a left multiple of w′ , and write w′ r w, if there exists w′′ ∈ G(e, e, n) such that
w = w′′ w′ and ℓ(w) = ℓ(w′′ ) + ℓ(w′ ).
The following lemma is straightforward.
15

Lemma 4.3. Let w, w′ ∈ G(e, e, n) and let x1 x2 · · · xr be a reduced expression of
w′ over X. We have w′  w if and only if ∀ 1 ≤ i ≤ r, ℓ(xi xi−1 · · · x1 w) =
ℓ(xi−1 · · · x1 w) − 1.
Proof. On the one hand, we have w′ w′′ = w with w′′ = xr xr−1 · · · x1 w and the condition ∀ 1 ≤ i ≤ r, ℓ(xi xi−1 · · · x1 w) = ℓ(xi−1 · · · x1 w)−1 implies that ℓ(w′′ ) = ℓ(w) − r.
So we get ℓ(w′′ ) + ℓ(w′ ) = ℓ(w). Hence w′  w.
On the other hand, since x2 = 1 for all x ∈ X, we have ℓ(xw) = ℓ(w) ± 1 for all
w ∈ G(e, e, n). If there exists i such that ℓ(xi xi−1 · · · x1 w) = ℓ(xi−1 · · · x1 w) + 1
with 1 ≤ i ≤ r, then ℓ(w′′ ) = ℓ(xr xr−1 · · · x1 w) > ℓ(w) − r. It follows that
ℓ(w′ ) + ℓ(w′′ ) > ℓ(w). Hence w′  w.
Consider the homomorphism − : X∗ −→ G(e, e, n) : x 7−→ x := x ∈ X. If
RE(w) = x1 x2 · · · xr with w ∈ G(e, e, n) and x1 , x2 , · · · , xr ∈ X, then RE(w) =
x1 x2 · · · xr = w where x1 , x2 , · · · , xr ∈ X.
Definition 4.4. We define D to be the set
o
n
w ∈ G(e, e, n) s.t. REi (w)  REi (λ) f or 2 ≤ i ≤ n .
Proposition 4.5. The set D consists of elements w of G(e, e, n) such that for all
2 ≤ i ≤ n, REi (w) is one of the following words
si · · · s3 t1 t0 s3 · · · si′ with 3 ≤ i′ ≤ i,
si · · · s3 t1 t0 ,
si · · · s3 tk
with 0 ≤ k ≤ e − 1, and
with 3 ≤ i′ ≤ i.
si · · · si′
Proof. We have REi (λ) = si · · · s3 t1 t0 s3 · · · si . Let w ∈ G(e, e, n). Note that REi (w)
is necessarily one of the words given in the the first column of the following table. For each REi (w), there exists unique w′ ∈ G(e, e, n) with RE(w′ ) given in the
second
column,
such that REi (w)w′ = REi(λ). In the last column, we compute



ℓ REi (w) + ℓ(w′ ) that is equal to ℓ REi (λ) = 2(i − 1) only for the first four cases.
The result follows immediately.
REi (w)
si · · · s3 t1 t0 s3 · · · si′ with 3 ≤ i′ ≤ i
si · · · s3 t1 t0
si · · · s3 tk with 0 ≤ k ≤ e − 1
si · · · si′ with 3 ≤ i′ ≤ i
si · · · s3 tk t0 with 2 ≤ k ≤ e − 1
si · · · s3 tk t0 s3 · · · si′ with 2 ≤ k ≤ e − 1,
and 3 ≤ i′ ≤ i

RE(w′ )
si′ +1 · · · si
s3 · · · si
tk−1 s3 · · · si
si′ −1 · · · s3 t1 t0 s3 · · · si
t0 tk−1 s3 · · · si
si′ · · · s3 t0 tk−1 s3 · · · si

The next proposition characterizes the divisors of λ in G(e, e, n).
Proposition 4.6. The set D is equal to the interval [1, λ] where
[1, λ] = {w ∈ G(e, e, n) s.t. 1  w  λ}.
16

2(i − 1)
2(i − 1)
2(i − 1)
2(i − 1)
2i
2(i − 1)+
2(i′ − 1)

Proof. Let w ∈ G(e, e, n). We have RE(w) = RE2 (w)RE3 (w) · · · REn (w). Let w ∈ X∗
− ∈ X∗ the word obtained by reading w
be a word representative of w. Denote by ←
w
←−−−−−−
←−−−−−
from right to left. For 3 ≤ i ≤ n, we denote by αi the element REi−1 (w) · · · RE2 (w) in
G(e, e, n).
Suppose that w ∈ D. We apply Lemma 4.3 to prove that w  λ. Fix 2 ≤ i ≤ n. By
Proposition 4.5, we have four different possibilities for REi (w).
First, consider the cases REi (w) = si · · · s3 t1 t0 s3 · · · si′ , si · · · s3 t1 t0 , or si · · · si′ with
←−−−−
3 ≤ i′ ≤ i. Hence REi (w) = si′ · · · s3 t0 t1 s3 · · · si , t0 t1 s3 · · · si , or si′ · · · si , respectively.
Note that the left multiplication of the matrix λ by αi produces permutations only
in the block consisting of the first i − 1 rows and the first i − 1 columns of λ. Since
λ[i, i] = ζe (6= 1), by 1(a) of Proposition 3.11, the left multiplication of αi λ by
s3 · · · si decreases the length. Also, by 2 of Proposition 3.11, the left multiplication
of s3 · · · si αi λ by t1 decreases the length. Note that by these left multiplications,
λ[i, i] = ζe is shifted to the first row then transformed to ζe ζe−1 = 1. Hence, by 1(b) of
Proposition 3.11, the left multiplication of t1 s3 · · · si αi λ by si′ · · · s3 t0 decreases the
length, as desired.
←−−−−
Suppose that REi (w) = si · · · s3 tk with 0 ≤ k ≤ e − 1. We have REi (w) = tk s3 · · · si .
Since λ[i, i] = ζe (6= 1), by 1(a) of Proposition 3.11, the left multiplication of αi λ
by s3 · · · si decreases the length. By 2 of Proposition 3.11, the left multiplication of
s3 · · · si αi λ by tk also decreases the length. Hence, applying Lemma 4.3, we have
w  λ.
Conversely, suppose that w ∈
/ D, we prove that w  λ. If RE(w) = x1 · · · xr ,
by Lemma 4.3, we show that there exists 1 ≤ i ≤ r such that ℓ(xi xi−1 · · · x1 λ) =
ℓ(xi−1 · · · x1 λ) + 1. Since w ∈
/ D, by Proposition 4.5, we may consider the first
REi (w) that appears in RE(w) = RE2 (w) · · · REn (w) such that REi (w) = si · · · s3 tk t0
←−−−−−
or si · · · s3 tk t0 s3 · · · si′ with 2 ≤ k ≤ e − 1 and 3 ≤ i′ ≤ i. Thus, we have RE i (w) =
t0 tk s3 · · · si or si′ · · · s3 t0 tk s3 · · · si , respectively. Since λ[i, i] = ζe (6= 1), by 1(a) of
Proposition 3.11, the left multiplication of αi λ by s3 · · · si decreases the length. By
2 of Proposition 3.11, the left multiplication of s3 · · · si αi λ by tk also decreases the
length. Note that by these left multiplications, λ[i, i] = ζe is shifted to the first row
then transformed to ζe ζe−k = ζe1−k . Since 2 ≤ k ≤ e − 1, we have ζe1−k 6= 1. By 3 of
Proposition 3.11, it follows that the left multiplication of tk s3 · · · si αi λ by t0 increases
the length. Hence w  λ.
We want to recognize if an element w ∈ G(e, e, n) is in the set D directly from its
matrix form. For this purpose, we start by describing a matrix form for each element
w ∈ G(e, e, n). Since w is a monomial matrix, there exists nonzero entries that we
refer to as bullets such that the upper left-hand side of w with respect to the bullets,
that we denote by Z(w), have all its entries zero. We denote the other side of the
matrix by Z ′ (w).

17



0

0

ζ32 0

0





 0 0 0 0 ζ 

3



Example 4.7. Let w = 
 0 0 ζ3 0 0  ∈ G(3, 3, 5). The bullets are the


 1 0 0 0 0


2
0 ζ3 0 0 0
encircled elements and the drawn path separates Z(w) from Z ′ (w).
Remark 4.8. Let w[i, c] be one of the bullets of w ∈ G(e, e, n). We have
w[i − 1, c] ∈ Z(w) and w[i, c − 1] ∈ Z(w).
Also the bullets are the only nonzero entries of w that satisfy this condition.
The following proposition gives a nice description of the divisors of λ in G(e, e, n).
Proposition 4.9. Let w ∈ G(e, e, n). We have that w ∈ D if and only if all nonzero
entries of Z ′ (w) are either 1 or ζe .
Proof. Let w ∈ D and let w[i, c] be a nonzero entry of Z ′ (w). Since w ∈ D, by
Proposition 4.5, we have REi (w) = si · · · s3 t1 t0 , si · · · s3 t1 t0 s3 · · · si′ , or si · · · si′ with
3 ≤ i′ ≤ i. By Lemma 3.3, we have wi [i, i′ ] = w[i, c] for 1 < i′ ≤ i. Hence, by
Definition 3.5, we have w[i, c] = 1 or ζe .
Conversely, let w[i, c] ∈ Z ′ (w). We have w[i, c] = 1 or ζe . Then we have again
REi (w) = si · · · s3 t1 t0 , si · · · s3 t1 t0 s3 · · · si′ , or si · · · si′ for 3 ≤ i′ ≤ i. If w[i, c] is a
bullet of w, by Lemma 3.3, we have wi [i, 1] = ζek for some 0 ≤ k ≤ e − 1, for which
case REi (w) = si · · · s3 tk . Hence, by Proposition 4.5, we have w ∈ D.


ζ32 0 0

0


 0 0 ζ 0
3

Example 4.10. Let w = 
 0 ζ3 0 0

0 0 0 ζ32
it follows immediately that w ∈
/ D.






 ∈ G(3, 3, 4). Since Z ′ (w) contains ζ32 ,



Our description of the interval [1, λ] allows us to prove easily that λ is balanced.
Let us recall the definition of a balanced element.
Definition 4.11. A balanced element in G(e, e, n) is an element w such that w′  w
holds precisely when w r w′ .
The next lemma is obvious. It will be useful later.
Lemma 4.12. Let g be a balanced element and let w, w′ ∈ [1, g]. If w′  w, then
(w′ )−1 w  g.
In order to prove that λ is balanced, we first check the following.
Lemma 4.13. If w ∈ D, we have w−1 λ ∈ D and λw−1 ∈ D.

18

Proof. We show that w−1 λ = tw λ ∈ D and λw−1 = λtw ∈ D, where tw is the complex
conjugate of the transpose tw of the matrix w. We use the matrix form of an element
of D. Actually, by Remark 4.8, the bullets of tw correspond to the complex conjugate
of the bullets of w. Also, all nonzero entries of w−1 , apart from its bullets, are in
{1, ζe−1 }. Multiplying w−1 by λ, we have that all nonzero entries of w−1 λ and of
λw−1 , apart from their bullets, are in {ζe , 1}. Hence w−1 λ ∈ D and λw−1 ∈ D.
Example 4.14. We illustrate the idea of the proof of Lemma

0

 0



Consider w ∈ D as follows and show that tw λ ∈ D: w =  0


 ζ3

0





0 0 0 ζ32 0
0 0 0 ζ3 0












0
0
0
0
1
 0 0 0 0 1 





 tw λ 
 tw 


2


 0 0 ζ3 0 0  −→  0 0 ζ3 0 0  −→ 








 1 0 0 0 0 
 1 0 0 0 0 








2
0 ζ3 0 0 0
0 ζ3 0 0 0

4.13.
0 0

1

0

0

0

0

ζ3

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

1

ζ32

0

0

0

1

0

0




ζ3 


 tw
0  −→


0 

0

1 0


0 ζ3 


0 0 .


0 0 

0 0

Proposition 4.15. We have that λ is balanced.
Proof. Suppose that w  λ. We have w ∈ D then λw−1 ∈ D by Lemma 4.13. Hence
λ = (λw−1 )w satisfies ℓ(λw−1 ) + ℓ(w) = ℓ(λ), namely w r λ. Conversely, suppose
that w r λ. We have λ = w′ w with w′ ∈ G(e, e, n) and ℓ(w′ ) + ℓ(w) = ℓ(λ). It
−1
−1
follows that w′ ∈ D then w′ λ ∈ D by Lemma 4.13. Since w = w′ λ, we have
w ∈ D, namely w  λ.
We state without proof a similar result for the powers of λ in G(e, e, n).
Definition 4.16. For 1 ≤ k ≤ e − 1, let Dk be the set of w ∈ G(e, e, n) such that all
nonzero entries of Z ′ (w) are either 1 or ζek .
Proposition 4.17. For 1 ≤ k ≤ e − 1,

0 0 0

 0 0 0


Example 4.18. Let w =  0 0 ζ3

 1 0 0

0 ζ32 0

we have [1, λk ] = Dk and λk is balanced.

1 0

0 1

0 0
 ∈ G(3, 3, 5). We have w ∈ [1, λ2 ].

0 0

0 0

19

More generally, let w ∈ G(e, e, n) be an element of maximal length, namely by
Proposition 3.12, a diagonal matrix such that for 2 ≤ i ≤ n, w[i, i] is an e-th root
of unity different from 1. As previously, we can prove that a divisor w′ of w satisfies
that for all 2 ≤ i ≤ n, if w′ [i, c] 6= 0 is not a bullet of w′ , then w′ [i, c] = 1 or w[i, i].
Suppose that w is of maximal length such that w[i, i] 6= w[j, j] with 2 ≤ i 6= j ≤ n.
We have (i, j)  w where (i, j) is the transposition matrix. Hence w′ := (i, j)−1 w =
(i, j)w r w. We have w′ [j, i] = w[i, i]. Thus, w′ [j, i] that is not a bullet of w′ is
different from 1 and w[j, j], since it is equal to w[i, i]. Hence w′  w. We thus get
the following.
Proposition 4.19. The balanced elements of G(e, e, n) that are of maximal length
over X are precisely λk where 1 ≤ k ≤ e − 1.
We are ready to study the interval structures associated with the intervals [1, λk ]
where 1 ≤ k ≤ e − 1.

5

Interval structures

In this section, we construct the monoid M ([1, λk ]) associated with each of the intervals [1, λk ] constructed in Section 4 with 1 ≤ k ≤ e − 1. By Proposition 4.19, λk
is balanced. Hence, by Theorem 1.7, in order to prove that M ([1, λk ]) is a Garside
monoid, it remains to show that both posets ([1, λk ], ) and ([1, λk ], r ) are lattices.
This is stated in Corollary 5.13. The interval structure is given in Theorem 5.15.
Let 1 ≤ k ≤ e − 1 and let w ∈ [1, λk ]. For each 1 ≤ i ≤ n there exists a unique
ci such that w[i, ci ] 6= 0. We denote w[i, ci ] by ai . We apply Lemma 4.3 to prove the
following lemmas. The reader is invited to write the matrix form of w to illustrate
each step of the proof.
Lemma 5.1. Let ti  w where i ∈ Z/eZ.
• If c1 < c2 , then tk t0  w and
• if c2 < c1 , then tj  w for all j with j 6= i.
Hence if ti  w and tj  w with i, j ∈ Z/eZ and i 6= j, then tk t0  w.
Proof. Suppose c1 < c2 .
Since w ∈ [1, λk ], by Proposition 4.9, a2 = 1 or ζek . Since ti  w, we have ℓ(ti w) =
ℓ(w) − 1. Hence by 2 of Proposition 3.11, we get a2 6= 1. Hence a2 = ζek . Again by 2
of Proposition 3.11, since a2 6= 1, we have ℓ(tk w) = ℓ(w) − 1. Let w′ := tk w. We have
w′ [1, c2 ] = ζe−k a2 = ζe−k ζek = 1. Hence by 3 of Proposition 3.11, ℓ(t0 w′ ) = ℓ(w′ ) − 1.
It follows that tk t0  w.
Suppose c2 < c1 .
Since ti  w, we have ℓ(ti w) = ℓ(w) − 1. Hence by 3 of Proposition 3.11, we have
a1 = ζe−i . If there exists j ∈ Z/eZ with j 6= i such that tj  w, then ℓ(tj w) = ℓ(w)−1.
Again by 3 of Proposition 3.11, we have a1 = ζe−j . Thus, i = j which is not allowed.
The last statement of the lemma follows immediately.

20

Lemma 5.2. If ti  w with i ∈ Z/eZ and s3  w, then s3 ti s3 = ti s3 ti  w.
Proof. Set w′ := s3 w and w′′ := ti w.
Suppose c1 < c2 .
Since w ∈ [1, λk ], by Proposition 4.9, we have a2 = 1 or ζek . Since ti  w, we have
ℓ(ti w) = ℓ(w) − 1. Thus, by 2 of Proposition 3.11, we get a2 6= 1. Hence a2 = ζek .
Suppose that c3 < c2 . Since s3  w, we have ℓ(s3 w) = ℓ(w) − 1. Hence by 1(b)
of Proposition 3.11, a2 = 1 which is not allowed. Then, assume c2 < c3 . Since
w ∈ [1, λk ], we have a3 = 1 or ζek . By 1(a) of Proposition 3.11, we have a3 6= 1.
Hence a3 = ζek . Now, we prove that s3 ti s3  w by applying Lemma 4.3. Indeed, since
a3 6= 1, by 2 of Proposition 3.11, ℓ(ti w′ ) = ℓ(w′ ) − 1, and since a2 6= 1, by 1(a) of
Proposition 3.11, we have ℓ(s3 w′′ ) = ℓ(w′′ ) − 1.
Suppose c2 < c1 .
Since ℓ(ti w) = ℓ(w) − 1, by 3 of Proposition 3.11, we have a1 = ζe−i .
• Assume c2 < c3 .
Since ℓ(s3 w) = ℓ(w) − 1, by 1(a) of Proposition 3.11, we have a3 6= 1. We have
ℓ(ti w′ ) = ℓ(w′ ) − 1 for both cases c1 < c3 and c3 < c1 . Actually, if c1 < c3 , since
a3 6= 1, by 2 of Proposition 3.11, we have ℓ(ti w′ ) = ℓ(w′ ) − 1, and if c3 < c1 ,
since a1 = ζe−i , by 3 of Proposition 3.11, ℓ(ti w′ ) = ℓ(w′ ) − 1. Now, since
ζei a1 = ζei ζe−i = 1, by 1(b) of Proposition 3.11, we have ℓ(s3 w′′ ) = ℓ(w′′ ) − 1.
• Assume c3 < c2 .
Since a1 = ζe−i , by 3 of Proposition 3.11, ℓ(ti w′ ) = ℓ(w′ ) − 1. Since ζei a1 = 1,
by 1(b) of Proposition 3.11, we have ℓ(s3 w′′ ) = ℓ(w′′ ) − 1.

Lemma 5.3. If ti  w with i ∈ Z/eZ and sj  w with 4 ≤ j ≤ n, then ti sj = sj ti  w.
Proof. We distinguish four different cases: case 1: c1 < c2 and cj−1 < cj , case 2:
c1 < c2 and cj < cj−1 , case 3: c2 < c1 and cj−1 < cj , and case 4: c2 < c1 and
cj < cj−1 . The proof is similar to the proofs of Lemmas 5.1 and 5.2 so we prove that
sj ti  w only for the first case. Suppose that c1 < c2 and cj−1 < cj . Since ti  w,
we have ℓ(ti w) = ℓ(w) − 1. Hence, by 2 of Proposition 3.11, we have a2 6= 1. Also,
since sj  w, we have ℓ(sj w) = ℓ(w) − 1. Hence, by 1(a) of Proposition 3.11, we have
aj 6= 1. Set w′ := sj w. Since a2 6= 1, then ℓ(ti w′ ) = ℓ(w′ ) − 1. Hence sj ti  w.
The proof of the following lemma is similar to the proofs of Lemmas 5.2 and 5.3
and is left to the reader.
Lemma 5.4. If si  w and si+1  w for 3 ≤ i ≤ n−1, then si si+1 si = si+1 si si+1  w,
and if si  w and sj  w for 3 ≤ i, j ≤ n and |i − j| > 1, then si sj = sj si  w.
The following proposition is a direct consequence of all the preceding lemmas.
Proposition 5.5. Let x, y ∈ X = {t0 , t1 , · · · , te−1 , s3 , · · · , sn }. The least common
multiple in ([1, λk ], ) of x and y, denoted by x∨y, exists and is given by the following
identities:

21

• ti ∨ tj = tk t0 = ti ti−k = tj tj−k for i 6= j ∈ Z/eZ,
• ti ∨ s3 = ti s3 ti = s3 ti s3 for i ∈ Z/eZ,
• ti ∨ sj = ti sj = sj ti for i ∈ Z/eZ and 4 ≤ j ≤ n,
• si ∨ si+1 = si si+1 si = si+1 si si+1 for 3 ≤ i ≤ n − 1, and
• si ∨ sj = si sj = sj si for 3 ≤ i 6= j ≤ n and |i − j| > 1.
We have a similar result for ([1, λk ], r ).
Proposition 5.6. Let x, y ∈ X. The least common multiple in ([1, λk ], r ) of x and
y, denoted by x ∨r y, exists and is equal to x ∨ y.
Proof. Define an antihomomorphism φ : G(e, e, n) −→ G(e, e, n) : ti 7−→ t−i , sj 7−→ sj
with i ∈ Z/eZ and 3 ≤ j ≤ n. It is obvious that φ2 is the identity. Also, if
w ∈ G(e, e, n), we have ℓ(φ(w)) = ℓ(w). Let x, y ∈ X and w ∈ [1, λk ]. We prove that
if x r w and y r w, then x ∨ y r w. Actually, we have w = vx and w = v ′ y
with v, v ′ ∈ G(e, e, n). Thus, φ(w) = φ(x)φ(v) and φ(w) = φ(y)φ(v ′ ). Since φ respects the length over X in G(e, e, n), we have φ(x)  φ(w) and φ(y)  φ(w). Hence
φ(x) ∨ φ(y)  φ(w). Moreover, one can check that φ(x) ∨ φ(y) = φ(x ∨ y) for all
x, y ∈ X. Thus, φ(x ∨ y)  φ(w), that is φ(w) = φ(x ∨ y)u for some u ∈ G(e, e, n).
We get w = φ(u)(x ∨ y) and since φ respects the length function, we have x ∨ y r w.
Note that Propositions 5.5 and 5.6 are important to prove that both posets
([1, λk ], ) and ([1, λk ], r ) are lattices. Actually, they will make possible an induction proof of Proposition 5.12 below. For now, let us state some general properties
about lattices that will be useful in our proof. Let (S, ) be a finite poset.
Definition 5.7. Say that (S, ) is a meet-semilattice if and only if f ∧ g := gcd(f, g)
exists for any f, g ∈ S.
V
Equivalently, (S, ) is a meet-semilattice if and only if T exists for any finite
nonempty subset T of S.
Definition 5.8. Say that (S, ) is a join-semilattice if and only if f ∨ g := lcm(f, g)
exists for any f, g ∈ S.
W
Equivalently, (S, ) is a join-semilattice if and only if T exists for any finite
nonempty subset T of S.
Proposition 5.9. (S, ) is a meet-semilattice if and only if for any f, g ∈ S, either
f ∨ g exists, or f and g have no common multiples.
Proof. Let f, g ∈ S and suppose that f and g have at least one common multiple.
Let A := {h ∈ S | f  h and g  h} be the set of the common multiples
V of f and g.
Since
S
is
finite,
A
is
also
finite.
Since
(S,
)
is
a
meet-semilattice,
A exists and
V
A = lcm(f, g) = f ∨ g.
Conversely, let f, g ∈ S and let B := {h ∈ S | h  f and h  g} be the set ofWall
common divisors ofWf and g. Since all elements of B have common multiples, B
exists and we have B = gcd(f, g) = f ∧ g.
22

Definition 5.10. The poset (S, ) is a lattice if and only if it is both a meet- and
join- semilattice.
The following lemma is a consequence of Proposition 5.9.
W
Lemma 5.11. If (S, ) is a meet-semilattice such that S exists, then (S, ) is a
lattice.
We will prove that ([1, λk ], ) is a meet-semilattice by applying Proposition 5.9.
On occasion, for 1 ≤ m ≤ ℓ(λk ) with ℓ(λk ) = n(n − 1), we introduce
([1, λk ])m := {w ∈ [1, λk ] s.t. ℓ(w) ≤ m}.
Proposition 5.12. Let 0 ≤ k ≤ e − 1. For 1 ≤ m ≤ n(n − 1) and u, v in ([1, λk ])m ,
either u∨v exists in ([1, λk ])m , or u and v do not have common multiples in ([1, λk ])m .
Proof. Let 1 ≤ m ≤ n(n − 1). We make a proof by induction on m. By Proposition
5.5, our claim holds for m = 1. Suppose m > 1. Assume that the claim holds for
all 1 ≤ m′ ≤ m − 1. We want to prove it for m′ = m. The proof is illustrated in
Figure 6 below. Let u, v be in ([1, λk ])m such that u and v have at least one common
multiple in ([1, λk ])m which we denote by w. Write u = xu1 and v = yv1 such that
x, y ∈ X and ℓ(u) = ℓ(u1 ) + 1, ℓ(v) = ℓ(v1 ) + 1. By Proposition 5.5, x ∨ y exists.
Since x  w and y  w, x ∨ y divides w. We can write x ∨ y = xy1 = yx1 with
ℓ(x ∨ y) = ℓ(x1 ) + 1 = ℓ(y1 ) + 1. By Lemma 4.12, we have x1 , v1 ∈ [1, λk ]. Also, we
have ℓ(x1 ) < m, ℓ(v1 ) < m and x1 , v1 have a common multiple in ([1, λk ])m−1 . Thus,
by the induction hypothesis, x1 ∨ v1 exists in ([1, λk ])m−1 . Similarly, y1 ∨ u1 exists in
([1, λk ])m−1 . Write x1 ∨v1 = v1 x2 = x1 v2 with ℓ(x1 ∨v1 ) = ℓ(v1 )+ℓ(x2 ) = ℓ(v2 )+ℓ(x1 )
and write y1 ∨ u1 = u1 y2 = y1 u2 with ℓ(y1 ∨ u1 ) = ℓ(y1 ) + ℓ(u2 ) = ℓ(u1 ) + ℓ(y2 ). By
Lemma 4.12, we have u2 , v2 ∈ [1, λk ]. Also, we have ℓ(u2 ) < m, ℓ(v2 ) < m and u2 , v2
have a common multiple in ([1, λk ])m−1 . Thus, by the induction hypothesis, u2 ∨ v2
exists in ([1, λk ])m−1 . Write u2 ∨ v2 = v2 u3 = u2 v3 with ℓ(u2 ∨ v2 ) = ℓ(v2 ) + ℓ(u3 ) =
ℓ(u2 ) + ℓ(v3 ). Since uy2 v3 = vx2 u3 is a common multiple of u and v that divides
every common multiple w of u and v, we deduce that u ∨ v = uy2 v3 = vx2 u3 and we
are done.

y
x

y1

u1

u2
y2

v1
x1

v2

x2
u3

v3

w

Figure 6: The proof of Proposition 5.12.
23

Similarly, applying Proposition 5.6, we obtain the same results for ([1, λk ], r ).
We thus proved the following.
Corollary 5.13. Both posets ([1, λk ], ) and ([1, λk ], r ) are lattices.
Proof. Applying Proposition 5.9, ([1,Wλk ], ) is a meet-semilattice. Also, by definition of the interval [1, λk ], we have [1, λk ] = λk . Thus, applying Proposition 5.9,
([1, λk ], ) is a lattice. The same can be done for ([1, λk ], r ).
We are ready to define the interval monoid M ([1, λk ]).
Definition 5.14. Let Dk be a set in bijection with Dk = [1, λk ] with
[1, λk ] −→ Dk : w 7−→ w.
We define the monoid M ([1, λk ]) by the following presentation of monoid with
• generating set: Dk (a copy of the interval [1, λk ]) and
• relations: w = w′ w′′ whenever w, w′ and w′′ ∈ [1, λk ], w = w′ w′′ and ℓ(w) =
ℓ(w′ ) + ℓ(w′′ ).
We have that λk is balanced. Also, by Corollary 5.13, both posets ([1, λk ], ) and
([1, λk ], r ) are lattices. Hence, by Theorem 1.7, we have:
Theorem 5.15. M ([1, λk ]) is an interval monoid with Garside element λk and with
simples Dk . Its group of fractions exists and is denoted by G(M ([1, λk ])).
These interval structures have been implemented into GAP3, Contribution to the
chevie package (see [16]). The next section is devoted to the study of these interval
structures.

6

About the interval structures

In this section, we provide a new presentation of the interval monoid M ([1, λk ]).
Furthurmore, we prove that G(M ([1, λk ])) is isomorphic to the complex braid group
B(e, e, n) if and only if k ∧ e = 1. When k ∧ e 6= 1, we describe these new structures
and show some of their properties.

6.1

Presentations

Our first aim is to prove that the interval monoid M ([1, λk ]) is isomorphic to the
monoid B ⊕k (e, e, n) defined as follows.
Definition 6.1. For 1 ≤ k ≤ e − 1, we define the monoid B ⊕k (e, e, n) by a presentation of monoid with
e = {t̃0 , t̃1 , · · · , t̃e−1 , s̃3 , · · · , s̃n } and
• generating set: X
24

• relations:

for
for
for
for
for

s̃i s̃j s̃i = s̃j s̃i s̃j
s̃i s̃j = s̃j s̃i
s̃3 t̃i s̃3 = t̃i s̃3 t̃i
s̃j t̃i = t̃i s̃j
t̃i t̃i−k = t̃j t̃j−k

|i − j| = 1,
|i − j| > 1,
i ∈ Z/eZ,
i ∈ Z/eZ and 4 ≤ j ≤ n, and
i, j ∈ Z/eZ.

Note that the monoid B ⊕1 (e, e, n) is the monoid B ⊕ (e, e, n) of Corran and Picantin introduced in Section 2.
The following result is similar to Matsumoto’s property in the case of real reflection
groups.
Proposition 6.2. There exists a map F : [1, λk ] −→ B ⊕k (e, e, n) defined by
F (w) = x̃1 x̃2 · · · x̃r whenever x1 x2 · · · xr is a reduced expression over X of w, where
e for 1 ≤ i ≤ r.
x̃i ∈ X
B

w2 if w2 is obtained from w1 by
Proof. Let w1 and w2 be in X∗ . We write w1
applying only the relations of the presentation of B ⊕k (e, e, n) where we replace t̃i by
ti and s̃j by tj for all i ∈ Z/eZ and 3 ≤ j ≤ n.
Let w be in [1, λk ] and suppose that w1 and w2 are two reduced expressions over X
B

of w. We prove that w1
w2 by induction on ℓ(w1 ).
The result holds vacuously for ℓ(w1 ) = 0 and ℓ(w1 ) = 1. Suppose that ℓ(w1 ) > 1.
Write
w1 = x1 w′ 1 and w2 = x2 w′ 2 , with x1 , x2 ∈ X.
If x1 = x2 , we have x1 w1′ = x2 w2′ in G(e, e, n) from which we get w1′ = w2′ . Then, by
B

B

the induction hypothesis, we have w′ 1
w′ 2 . Hence w1
w2 .
If x1 6= x2 , since x1  w and x2  w, we have x1 ∨ x2  w where x1 ∨ x2 is the lcm
of x1 and x2 in ([1, λk ], ) given in Proposition 5.5. Write w = (x1 ∨ x2 )w′ . Also,
B

write x1 ∨ x2 = x1 v1 and x1 ∨ x2 = x2 v2 where we can check that x1 v1
x2 v2 for
all possible cases for x1 and x2 . All the words x1 w′ 1 , x2 w′ 2 , x1 v1 w′ , and x2 v2 w′
represent w. In particular, x1 w′ 1 and x1 v1 w′ represent w. Hence w1′ = v1 w′ and by
the induction hypothesis, we have w′ 1

B

x1 w′ 1

v1 w′ . Thus, we have
B

x1 v1 w′ .

Similarly, since x2 w′ 2 and x2 v2 w′ represent w, we get
x2 v2 w′
B

Since x1 v1

B

x2 w′ 2 .

x2 v2 , we have
x1 v1 w′

B

x2 v2 w′ .

x1 v1 w′

B

x2 v2 w′

We obtain:
w1
Hence w1

B

B

x1 w′ 1

B

w2 and we are done.

25

B

x2 w′ 2

B

w2 .

By the following proposition, we provide an alternative presentation of the interval
monoid M ([1, λk ]) given in Definition 5.14.
Proposition 6.3. The monoid B ⊕k (e, e, n) is isomorphic to M ([1, λk ]).
Proof. Consider the map ρ : Dk −→ B ⊕k (e, e, n) : w 7−→ F (w) where F is defined
in Proposition 6.2. Let w = w ′ w ′′ be a defining relation of M ([1, λk ]). Since ℓ(w) =
ℓ(w′ ) + ℓ(w′′ ), a reduced expression for w′ w′′ is obtained by concatenating reduced
expressions for w′ and w′′ . It follows that F (w′ w′′ ) = F (w′ )F (w′′ ). We conclude that
ρ has a unique extension to a monoid homomorphism M ([1, λk ]) −→ B ⊕k (e, e, n),
which we denote by the same symbol.
e −→ M ([1, λk ]) : x
Conversely, consider the map ρ′ : X
e 7−→ x. In order to prove
′
that ρ extends to a unique monoid homomorphism B ⊕k (e, e, n) −→ M ([1, λk ]), we
e1 = w
e2 of
have to check that w1 = w2 in M ([1, λk ]) for any defining relation w
B ⊕k (e, e, n). Given a relation w
e1 = w
e2 = x̃1 x̃2 · · · x̃r of B ⊕k (e, e, n), we have w1 =
w2 = x1 x2 · · · xr a reduced word over X. On the other hand, applying repeatedly
the defining relations in M ([1, λk ]) yields to w = x1 x2 · · · xr if w = x1 x2 · · · xr is a
reduced expression over X. Thus, we can conclude that w1 = w2 , as desired.
e −→
Hence we have defined two homomorphisms ρ : Dk −→ B ⊕k (e, e, n) and ρ′ : X
k
′
′
M ([1, λ ]) such that ρ ◦ ρ = idB ⊕k (e,e,n) and ρ ◦ ρ = idM([1,λk ]) . It follows that
B ⊕k (e, e, n) is isomorphic to M ([1, λk ]).
We deduce that B ⊕k (e, e, n) is a Garside monoid and we denote by B (k) (e, e, n)
its group of fractions.
Fix k such that 1 ≤ k ≤ e − 1. A diagram presentation for B (k) (e, e, n) is the same
as the diagram corresponding to the presentation of Corran and Picantin of B(e, e, n)
given in Figure 2.2, with a dashed edge between t̃i and t̃i−k and between t̃j and t̃j−k
for each relation of the form t̃i t̃i−k = t̃j t̃j−k , i, j ∈ Z/eZ. For example, the diagram
corresponding to B (2) (8, 8, 2) is as follows.

3
•

2
•

1
•
0
•

4
•
•
5

•
6

•
7

Figure 7: Diagram for the presentation of B (2) (8, 8, 2).

26

6.2

Identifying B(e, e, n)

Now, we want to check which of the monoids B ⊕k (e, e, n) are isomorphic to B ⊕ (e, e, n).
Assume there exists an isomorphism φ : B ⊕k (e, e, n) −→ B ⊕ (e, e, n) for a given k with
0 ≤ k ≤ e − 1. We start with the following lemma.
Lemma 6.4. The isomorphism φ fixes s̃3 , s̃4 , · · · , s̃n and permutes the t̃i where
i ∈ Z/eZ.
e ∗ . We have ℓ(f ) ≤ ℓ(φ(f )). Thus, we have ℓ(x̃) ≤ ℓ(φ(x̃)) for
Proof. Let f be in X
e Also, ℓ(φ(x̃)) ≤ ℓ(φ−1 (φ(x̃))) = ℓ(x). Hence ℓ(x̃) = ℓ(φ(x̃)) = 1. It follows
x̃ ∈ X.
that φ(x̃) ∈ {t̃0 , t̃1 , · · · , t̃e−1 , s̃3 , · · · , s̃n }.
Furthurmore, the only generator of B ⊕k (e, e, n) that commutes with all other generators except for one of them is s̃n . On the other hand, s̃n is the only generator of
B ⊕ (e, e, n) that satisfies the latter property. Hence φ(s̃n ) = s̃n . Next, s̃n−1 is the
only generator of B ⊕k (e, e, n) that does not commute with s̃n . The only generator
of B ⊕ (e, e, n) that does not commute with s̃n is also s̃n−1 . Hence φ(s̃n−1 ) = s̃n−1 .
Next, the only generator of B ⊕k (e, e, n) different from s̃n and that does not commute
with s̃n−1 is s̃n−2 . And so on, we get φ(s̃j ) = s̃j for 3 ≤ j ≤ n. It remains that
φ({t̃i | 0 ≤ i ≤ e − 1}) = {t̃i | 0 ≤ i ≤ e − 1}.
Proposition 6.5. The monoids B ⊕k (e, e, n) and B ⊕ (e, e, n) are isomorphic if and
only if k ∧ e = 1.
Proof. Assume there exists an isomorphism φ between the monoids B ⊕k (e, e, n) and
B ⊕ (e, e, n). By Lemma 6.4, we have φ(s̃j ) = s̃j for 3 ≤ j ≤ n and φ({t̃i | 0 ≤ i ≤ e − 1})
= {e
ti | 0 ≤ i ≤ e − 1}. A diagram presentation Γk of B ⊕k (e, e, 2) for 1 ≤ k ≤ e − 1
can be viewed as the same diagram presentation of B (k) (e, e, 2) given earlier. The
isomorphism φ implies that we have only one connected component in Γk or, in another words, a closed chain. Hence k is a generator of the cyclic group Z/eZ. Then
k satisfies the condition k ∧ e = 1.
Conversely, let 1 ≤ k ≤ e − 1 such that k ∧ e = 1. We define a map φ : B ⊕ (e, e, n) −→
B ⊕k (e, e, n) where φ(s̃j ) = s̃j for 3 ≤ j ≤ n and φ(t̃0 ) = t̃k , φ(t̃1 ) = t̃2k , φ(t̃2 ) = t̃3k ,
· · · , φ(t̃e−1 ) = e
tek . The map φ is a well-defined monoid homomorphism, which is both
surjective (as it corresponds a generator of B ⊕ (e, e, n) to a generator of B ⊕k (e, e, n))
and injective (as it is bijective on the relations). Hence φ defines an isomorphism of
monoids.
When k ∧ e = 1, since B ⊕k (e, e, n) is isomorphic to B ⊕ (e, e, n), we have the
following.
Corollary 6.6. B (k) (e, e, n) is isomorphic to the complex braid group B(e, e, n) for
k ∧ e = 1.
The reason that the proof of Proposition 6.5 fails in the case k ∧ e 6= 1 is that
we have more than one connected component in Γk that link t̃0 , t̃1 , · · · , and t̃e−1
together, as we can see in Figure 7. Actually, it is easy to check that the number of
connected components that link t̃0 , t̃1 , · · · , and t̃e−1 together is the number of cosets
of the subgroup of Z/eZ generated by the class of k, that is equal to k ∧ e, and each
of these cosets have e′ = e/k ∧ e elements. This will be useful in the next subsection.
27

6.3

New Garside groups

When k ∧ e 6= 1, we describe B (k) (e, e, n) as an amalgamated product of k ∧ e copies
of the complex braid group B(e′ , e′ , n) with e′ = e/e ∧ k, over a common subgroup
which is the Artin-Tits group B(2, 1, n − 1). This allows us to compute the center of
B (k) (e, e, n). Finally, using the Garside structure of B (k) (e, e, n), we compute its first
and second integral homology groups using the Dehornoy-Lafont complex [9] and the
method used in [4].
By an adaptation of the results of Crisp in [7] as in Lemma 5.2 of [4], we have the
following embedding. Let B := B(2, 1, n − 1) be the Artin-Tits group defined by the
following diagram presentation.
q1

q2

qn−2

q3

qn−1

Proposition 6.7. The group B injects in B (k) (e, e, n).
Proof. Define a monoid homomorphism φ : B + −→ B ⊕k (e, e, n) : q1 7−→ t̃i t̃i−k ,
q2 7−→ s̃3 , · · · , qn−1 7−→ s̃n . It is easy to check that for all x, y ∈ {q1 , q2 , · · · , qn−1 }, we
have lcm(φ(x), φ(y)) = φ(lcm(x, y)). Hence by applying Lemma 5.2 of [4], B(2, 1, n − 1)
injects in B (k) (e, e, n).
We construct B (k) (e, e, n) as follows.
Proposition 6.8. Let B(1) := B(e′ , e′ , n) ∗B B(e′ , e′ , n) be the amalgamated product
of two copies of B(e′ , e′ , n) over B = B(2, 1, n − 1) with e′ = e/e ∧ k. Define B(2) :=
B(e′ , e′ , n) ∗B (B(e′ , e′ , n) ∗B B(e′ , e′ , n)) and so on until defining B((e ∧ k) − 1).
We have B((e ∧ k) − 1) = B (k) (e, e, n).
Proof. Due to the presentation of B (k) (e, e, n) given in Definition 6.1 and to the
presentation of the amalgamated products (see Section 4.2 of [14]), one can deduce
that B((e ∧ k) − 1) is equal to B (k) (e, e, n).

28

B((e ∧ k) − 1) = B (k) (e, e, n)
B(e′ , e′ , n)
B(3)
B(2)
B(e′ , e′ , n)
B(e′ , e′ , n)

B(e′ , e′ , n)

Figure 8: The construction of B (k) (e, e, n).
Example 6.9. Consider the case of B (2) (6, 6, 3). It is an amalgamated product of
k ∧ e = 2 copies of B(e′ , e′ , 3) with e′ = e/(k ∧ e) = 3 over the Artin-Tits group
B(2, 1, 2). Consider the following diagram of this amalgamation.
The presentation of B(3, 3, 3) ∗ B(3, 3, 3) over B(2, 1, 2) is as follows:
• the generators are the union of the generators of the two copies of B(3, 3, 3),
• the relations are the union of the relations of the two copies of B(3, 3, 3) with
the additional relations s̃3 = s̃′3 and t̃2 t̃0 = t̃3 t̃1
This is exactly the presentation of B (2) (6, 6, 3) given in Definition 6.1.
s̃3

t̃3 t̃1

∼

s̃′3

t̃2 t̃0

−→

t̃5

t̃4

t̃3

t̃2
t̃0

t̃1

s̃′3

s̃3

B (2) (6, 6, 3)

Proposition 6.10. The center of B (k) (e, e, n) is infinite cyclic isomorphic to Z.
Proof. By Corollary 4.5 of [14] that computes the center of an amalgamated product,
the center of B (k) (e, e, n) is the intersection of the centers of B and B(e′ , e′ , n). Since
the center of B and B(e′ , e′ , n) is infinite cyclic [3], the center of B (k) (e, e, n) is infinite
cyclic isomorphic to Z.

29

Since the center of B(e, e, n) is also isomorphic to Z (see [3]), in order to distinguish these structures from the braid groups B(e, e, n), we compute their first and
second integral homology groups. We recall the Dehornoy-Lafont complex and follow
the method in [4] where the second integral homology group of B(e, e, n) is computed.
e by considering s̃n < s̃n−1 < · · · < s̃3 < t̃0 < t̃1 <
We order the elements of X
e which
· · · < t̃e−1 . For f ∈ B ⊕k (e, e, n), denote by d(f ) the smaller element in X
e
divides f on the right. An r-cell is an r-tuple [x1 , · · · , xr ] of elements in X such that
x1 < x2 < · · · < xr and xi = d(lcm(xi , xi+1 , · · · , xr )). The set Cr of r-chains is
er , the set of all r-cells with the convention
the free ZB ⊕k (e, e, n)-module with basis X
e0 = {[∅]}. We provide the definition of the differential ∂r : Cr −→ Cr−1 .
X
e and A an r-cell. Denote
Definition 6.11. Let [α, A] be an (r + 1)-cell, with α ∈ X
⊕k
α/A the unique element of B (e, e, n) such that (α/A )lcm(A) = lcm(α, A). Define
the differential ∂r : Cr −→ Cr−1 recursively through two Z-module homomorphisms
sr : Cr −→ Cr+1 and ur : Cr −→ Cr as follows.
∂r+1 [α, A] = α/A [A] − ur (α/A [A]),
with ur+1 = sr ◦ ∂r+1 where u0 (f [∅]) = [∅], for all f ∈ B ⊕k (e, e, n), and
sr ([∅]) = 0, sr (x[A]) = 0 if α := d(xlcm(A)) coincides with the first coefficient in A,
and otherwise sr (x[A]) = y[α, A] + sr (yur (α/A [A])) with x = yα/A .
We provide the final result of the computation of ∂1 , ∂2 , and ∂3 for all 1, 2, and
e we have
3-cells, respectively. For all x ∈ X,
∂1 [x] = (x − 1)[∅],

for all 1 ≤ i ≤ e − 1,
∂2 [t̃0 , t̃i ] = t̃i+k [t̃i ] − t̃k [t̃0 ] − [t̃k ] + [t̃i+k ],
e with xyx = yxy,
for x, y ∈ X

∂2 [x, y] = (yx + 1 − x)[y] + (y − xy − 1)[x], and

e with xy = yx,
for x, y ∈ X

∂2 [x, y] = (x − 1)[y] − (y − 1)[x].

For j 6= −k mod e, we have:
∂3 [s̃3 , t̃0 , t̃j ] = (s̃3 t̃k t̃0 s̃3 − t̃k t̃0 s̃3 + t̃j+2k s̃3 )[t̃0 , t̃j ] − t̃j+2k s̃3 t̃j+k [s̃3 , t̃j ] + (t̃j+2k −
s̃3 t̃j+2k )[s̃3 , t̃j+k ] + (s̃3 − t̃j+2k s̃3 − 1)[t̃0 , t̃j+k ] + (s̃3 t̃2k − t̃2k )[s̃3 , t̃k ] + (t̃2k s̃3 + 1 −
s̃3 )[t̃0 , t̃k ] + [s̃3 , t̃j+2k ] + t̃2k s̃3 t̃k [s̃3 , t̃0 ] − [s̃3 , t̃2k ] and
∂3 [s̃3 , t̃0 , t̃−k ] = (s̃3 t̃k t̃0 s̃3 − t̃k t̃0 s̃3 + t̃k s̃3 )[t̃0 , t̃−k ] − t̃k s̃3 t̃0 [s̃3 , t̃−k ] + (1 − t̃2k +
s̃3 t̃2k )[s̃3 , t̃k ] + (1 + t̃2k s̃3 − s̃3 )[t̃0 , t̃k ] + (t̃k − s̃3 t̃k + t̃2k s̃3 t̃k )[s̃3 , t̃0 ] − [s̃3 , t̃2k ].
Also, for 1 ≤ i ≤ e − 1 and 4 ≤ j ≤ n, we have:
30

∂3 [s̃j , t̃0 , t̃i ] = (s̃j − 1)[t̃0 , t̃i ] − t̃i+k [s̃j , t̃i ] + t̃k [s̃j , t̃0 ] − [s̃j , t̃i+k ] + [s̃j , t̃k ],
e with xyx = yxy, xz = zx, and yzy = zyz,
for x, y, z ∈ X

∂3 [x, y, z] =
(z + xyz − yz − 1)[x, y] − [x, z] + (xz − z − x + 1 − yxz)y[x, z] + (x − 1 − yx + zyx)[y, z],
e with xyx = yxy, xz = zx, and yz = zy,
for x, y, z ∈ X

∂3 [x, y, z] = (1 − x + yx)[y, z] + (y − 1 − xy)[x, z] + (z − 1)[x, y],

e with xy = yx, xz = zx, and yzy = zyz,
for x, y, z ∈ X

∂3 [x, y, z] = (1 + yz − z)[x, y] + (y − 1 − zy)[x, z] + (x − 1)[y, z], and

e with xy = yx, xz = zx, and yz = zy,
for x, y, z ∈ X

∂3 [x, y, z] = (1 − y)[x, z] + (z − 1)[x, y] + (x − 1)[y, z].

Let dr = ∂r ⊗ZB ⊕k (e,e,n) Z : Cr ⊗ZB ⊕k (e,e,n) Z −→ Cr−1 ⊗ZB ⊕k (e,e,n) Z be the
differential with trivial coefficients. For example, for d2 , we have: for all 1 ≤ i ≤ e − 1,
d2 [t̃0 , t̃i ] = [t̃i ] − [t̃0 ] − [t̃k ] + [t̃i+k ],
e with xyx = yxy,
for x, y ∈ X
e with xy = yx,
for x, y ∈ X

d2 [x, y] = [y] − [x], and

d2 [x, y] = 0.

The same can be done for ∂3 .
Definition 6.12. Define the integral homology group of order r to be
Hr (B (k) (e, e, n), Z) = ker(dr )/Im(dr+1 ).
We are ready to compute the first and second integral homology groups of B (k) (e, e, n).
Using the presentation of B (k) (e, e, n) given in 6.1, one can check that the abelianization of B (k) (e, e, n) is isomorphic to Z. Since H1 (B (k) (e, e, n), Z) is isomorphic to the
abalianization of B (k) (e, e, n), we deduce that H1 (B (k) (e, e, n), Z) is isomorphic to Z.
Since H1 (B(e, e, n), Z) is also isomorphic to Z (see [3]), the first integral homology
group does not give any additional information weither these groups are isomorphic
to some B(e, e, n) or not.
Recall that by [4], we have
• H2 (B(e, e, 3), Z) ≃ Z/eZ where e ≥ 2,
• H2 (B(e, e, 4), Z) ≃ Z/eZ × Z/2Z when e is odd and H2 (B(e, e, 4), Z) ≃ Z/eZ ×
(Z/2Z)2 when e is even, and
• H2 (B(e, e, n), Z) ≃ Z/eZ × Z/2Z when n ≥ 5 and e ≥ 2.
31

In order to compute H2 (B (k) (e, e, n), Z), we follow exactly the proof of Theorem 6.4
in [4]. Using the same notations as in [4], we set
vi = [t̃0 , t̃i ] + [s̃3 , t̃0 ] + [s̃3 , t̃k ] − [s̃3 , t̃i ] − [s̃3 , t̃i+k ] where 1 ≤ i ≤ e − 1,
and we also have
H2 (B (k) (e, e, n), Z) = (K1 /d3 (C1 )) ⊕ (K2 /d3 (C2 )).
We have d3 [s̃3 , t̃0 , t̃j ] = vj − vj+k + vk if j 6= −k, and d3 [s̃3 , t̃0 , t̃−k ] = v−k + vk . Denote
ui = [s̃3 , t̃0 , t̃i ] for 1 ≤ i ≤ e − 1. We define a basis of C2 as follows. For each coset of
the subgroup of Z/eZ generated by the class of k, say {t̃x , t̃x+k , · · · , t̃x−k } such that
1 ≤ x ≤ e − 1, we define wx+ik = ux+ik + ux+(i+1)k + · · · + ux−k for 0 ≤ i ≤ e − 1,
and when x = 0, we define wik = uik + u(i+1)k + · · · + u−k for 1 ≤ i ≤ e − 1. In
the Z-basis (wk , w2k , · · · , w−k , w1 , w1+k , · · · , w1−k , · · · , wx , wx+k , · · · , wx−k , · · · ) and
(vk , v2k , · · · , v−k , v1 , v1+k , · · · , v1−k , · · · , vx , vx+k , · · · , vx−k , · · · ), d3 is in triangular
form with (e ∧ k) − 1 diagonal coefficients that are zero, all other diagonal coefficients are equal to 1 except one of them that is equal to e′ = e/(e ∧ k). In this case,
we have H2 (B (k) (e, e, 3), Z) = Z(e∧k)−1 × Z/e′ Z. The rest of the proof is essentially
similar to the proof of Theorem 6.4 in [4]. When n = 4, we get [s̃4 , t̃i ] ≡ [s̃4 , t̃i+2k ]
for every i and since 2[s̃4 , t̃i ] ≡ 0 for every i, we get K2 /d3 (C2 ) ≃ (Z/2Z)c , where c is
the number of cosets of the subgroup of Z/eZ generated by the class of 2k.
In the following proposition, we provide the second integral homology group of
B (k) (e, e, n).
Proposition 6.13. Let n ≥ 3 and e ≥ 2.
• When n = 3, we have H2 (B (k) (e, e, 3), Z) ≃ Z(e∧k)−1 × Z/e′ Z,
• when n = 4, H2 (B (k) (e, e, 4), Z) ≃ Z(e∧k)−1 × Z/e′ Z × (Z/2Z)c , where c is the
number of cosets of the subgroup of Z/eZ generated by the class of 2k, and
• when n ≥ 5, H2 (B (k) (e, e, n), Z) ≃ Z(e∧k)−1 × Z/e′ Z × Z/2Z.
Comparing this result with H2 (B(e, e, n), Z), one can check that if k ∧ e 6= 1,
B (k) (e, e, n) is not isomorphic to a complex braid group of type B(d, d, n) with d ≥ 2.
Thus, we conclude by the following theorem.
Theorem 6.14. B (k) (e, e, n) is isomorphic to B(e, e, n) if and only if k ∧ e = 1.

Acknowledgements
I would like to thank my PhD supervisors, Eddy Godelle and Ivan Marin for fruitful
discussions. Also, I would like to thank all the members of Laboratoire de Mathématiques Nicols Oresme and Laboratoire Amiénois de Mathémathiques Fondamentales
et Appliquées for providing me the right environment to finish this work.

32

References
[1] D. Bessis and R. Corran. Non-crossing partitions of type (e, e, r). Adv. Math.,
202:1–49, 2006.
[2] E. Brieskorn and K. Saito. Artin-gruppen und coxeter-gruppen. Invent. Math.,
17:245–271, 1972.
[3] M. Broué, G. Malle, and R. Rouquier. Complex reflection groups, braid groups,
hecke algebras. J. Reine Angew. Math., 500:127–190, 1998.
[4] F. Callegaro and I. Marin. Homology computations for complex braid groups. J.
European Math. Soc., (16):103–164, 2014.
[5] R. Corran. On monoids related to braid groups. PhD thesis, University of Sydney,
May 2000.
[6] R. Corran and M. Picantin. A new garside structure for braid groups of type
(e, e, r). J. London Math. Soc., 84(3):689–711, 2011.
[7] J. Crisp. Injective maps between artin groups. “Geometric Group Theory Down
Under” ed. J. Cossey et al., de Gruyter Verlag, pages 119–137, 1999.
[8] P. Dehornoy, F. Digne, E. Godelle, D. Krammer, and J. Michel. Foundations of
Garside Theory, volume 22. EMS Tracts in Mathematics, 2015.
[9] P. Dehornoy and Y. Lafont. Homology of gaussian groups. Ann. Inst. Fourier,
53(2):1001–1052, 2003.
[10] P. Dehornoy and L. Paris. Gaussian groups and garside groups, two generalizations of artin-groups. Proc. London Math. Soc., 79(3):569–604, 1999.
[11] P. Deligne. Les immeubles des groupes de tresses généralisés. Invent. Math.,
17:273–302, 1972.
[12] F.A. Garside. The theory of knots and associated problems. PhD thesis, Oxford
University, 1965.
[13] F.A. Garside. The braid group and other groups. Quart. J. Math. Oxford,
20:235–254, 1969.
[14] A. Karrass, W. Magnus, and D. Solitar. Combinatorial Group Theory. Reprint
of the Interscience Publishers, New York, 1976.
[15] J. Michel. Groupes finis de réflexion. University notes, Université Paris Diderot,
2004. http://webusers.imj-prg.fr/ jean.michel/papiers/cours2004.pdf.
[16] J. Michel and G. Neaime. Contribution to the chevie package, cp.g.
http://webusers.imj-prg.fr/ jean.michel/gap3.

see

[17] G.C. Shephard and J.A. Todd. Finite unitary reflection groups. Canad. J. Math.,
6(2):274–304, 1954.

33

