EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 1, 2018, 215-237
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On the Irreducibility of Perron Representations of
Degrees 4 and 5

Malak M. Dally1, Mohammad N. Abdulrahim1,∗

1 Department of Mathematics, Faculty of Science, Beirut Arab University,
P.O. Box 11-5020, Beirut, Lebanon

Abstract. We consider the graph En+1,1 with (n+1) generators σ1, ..., σn, and δ, where σi has an
edge with σi+1 for i = 1, ..., n+1, and σ1 has an edge with δ. We then define the Artin group of the
graph En+1,1 for n = 3 and n = 4 and consider its reduced Perron’s representation of degrees four
and five respectively. After we specialize the indeterminates used in defining the representation
to non-zero complex numbers, we obtain necessary and sufficient conditions that guarantee the
irreducibility of the representations for n = 3 and 4 .

2010 Mathematics Subject Classifications: 20F36

Key Words and Phrases: Artin representation, braid group, Burau representation, graph,
irreducibility

1. Introduction

Let Γ be an undirected simple graph. The Artin group A is defined as an abstract
group whose generators are the vertices of Γ that satisfy the two relations: xy = yx for
vertices x and y that have no edge in common and xyx = yxy if the vertices x and y have
a common edge.

Having defined A, we consider the graph An having n vertices σi’s (1 ≤ i ≤ n ) in
which σi and σi+1 share a comon edge, where i = 1, 2, ..., n−1. Indeed, the Artin group of
An, denoted by A(An), is the braid group on n+1 strands, Bn+1. That is, A(An) = Bn+1.

From the graph An, we obtain the graph En+1,p by adding a vertex δ and an edge
connecting σp and δ. Here 1 ≤ p ≤ n. Clearly, the graph An embeds in the graph En+1,p.
Consequently, A(An) ⊂ A(En+1,p). As a result, a representation of A(En+1,p) yields a
representation of Bn+1.

Perron’s strategy is to begin with the reduced Burau representation of Bn+1 of degree
n and extend it to a representation of Bn+1 of degree 2n. The representation obtained is
referred to as Burau bis representation. Next, Perron constructs for each λ = (λ1, . . . , λn)

∗Corresponding author.

Email addresses: malakdally@hotmail.com (M. Dally), mna@bau.edu.lb (M. Abdulrahim)

http://www.ejpam.com 215 c© 2018 EJPAM All rights reserved.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 216

a representation ψλ : A(En+1,p) → GL2n(Q(t, d1, . . . , dn)), where t, d1, . . . , dn λ1, . . . , λn
are indeterminates.

In [3], we determined necessary and sufficient condition that guarantees the irreducibil-
ity of the representation ψλ for n = 2. In our work, we extend our work to n = 3 and
n = 4. We reduce the complex specialization of the representation ψλ to representations
of A(E4,1) and A(E5,1) of degrees 4 and 5 respectively. In each case, a necessary and
sufficient condition which guarantees the irreducibility of the considered representation is
obtained. The obtained conditions are similar to the condition obtained in the case n = 2,
which was studied in [3].

2. Burau bis Representation

The Burau Bis representation is a representation of Bn+1 of degree 2n.
It is defined as follows:

ψ : Bn+1 → Gl2n(Z[t, t−1])

ψ(σi) =

(
In 0
Ri Ji

)
, 1 ≤ i ≤ n

Here, Ri denotes an n × n block of zeros with a t placed in the (i, i) th position and
In denotes the n× n identity matrix.

This representation was constructed by Perron by extending the reduced Burau rep-
resentation of degree n to a representation of Bn+1 of degree 2n.

The reduced Barau representation Bn+1 → GLn(Z[t, t−1]) is defined as follows:

σi → Ji =


Ii−2 0 0

0
1 0 0
t −t 1
0 0 1

0

0 0 In−i−1

 ,

where Ik stands for the k × k identity matrix. Here, i = 2, . . . , n− 1.

σ1 → J1 =

 −t 1
0 1

0

0 In−2



σn → Jn =

 In−2 0

0
1 0
t −t





M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 217

For more details, see [2] and [5].

3. Perron Representation

The Burau bis representation extends to A(En+1,p) for all possible values of n and p
in the following way.
We define the following n× n matrices:

A = (λ1b, λ2b, . . . , λnb)

B = (0, . . . , 0, b, 0, . . . , 0)

C = (λ1d, λ2d, . . . , λnd)

D = (0, . . . , 0, d, 0, . . . , 0),

where 0 denotes a column of n zeros, b =

b1...
bn

 , d =

d1...
dn

 , and λ = (λ1, . . . , λn).

For each i = 1, . . . , n, we have that bi satisfies the following conditions

tbi = −tdi−1 + (1 + t)di − di+1, i 6= p,

tbp = −tdp−1 + (1 + t)dp − dp+1 + t,
n∑
i=1

λibi = −(1 + dp + t),

setting any undefined dj equal zero.

For any choice λ = (λ1, . . . , λn) , we get a linear representation

ψλ : A(En+1,p)→ Gl2n(R),

where R is the field of rational fractions in n+1 indeterminates Q(t, d1, ..., dn).

ψλ(σi)→
(
In 0
Ri Ji

)
,

ψλ(δ)→
(
In +A B
C In +D

)
.

For more details, see [2].



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 218

4. Reducibility of ψλ : A(E4,1) → GL6(C)

Having defined Perron’s representation, we set n = 3 and p = 1 to get the following

vectors. b =

b1b2
b3

 , d =

d1d2
d3

 , and λ = (λ1, λ2, λ3).

After we specialize the indeterminate d3 to −t(1+t+t2)
1+t , we get the following 3 × 3

matrices:

A =

λ1b1 λ2b1 λ3b1
λ1b2 λ2b2 λ3b2
λ1b3 λ2b3 λ3b3

 ,

B =

b1 0 0
b2 0 0
b3 0 0

 ,

C =

 λ1d1 λ2d1 λ3d1
λ1d2 λ2d2 λ3d2

−t(1+t+t2)
1+t λ1

−t(1+t+t2)
1+t λ2

−t(1+t+t2)
1+t λ3

 ,

and

D =

 d1 0 0
d2 0 0

−t(1+t+t2)
1+t 0 0

 .

Simple computations show that the parameters satisfy the following equations:

• tb2 = −td1 + (1 + t)d2 + t(1+t+t2)
1+t

• tb3 = −td2 − t(1 + t+ t2)

• tb1 = (1 + t)d1 − d2 + t

• λ1b1 + λ2b2 + λ3b3 = −(1 + t+ d1)

Having defined the 3 × 3 matrices A, B, C and D, we obtain the multiparameter
representation A(E4,1). This representation is of degree 6. We specialize the parameters
λ1, λ2, λ3, b1, b2, b3, d1, d2, t to values in C − {0}. We further assume that t 6= −1. The
representation ψλ : A(E4,1)→ GL6(C) is defined as follows:



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 219

ψλ(σ1) =



1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
t 0 0 −t 1 0
0 0 0 0 1 0
0 0 0 0 0 1

 ,

ψλ(σ2) =



1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 t 0 t −t 1
0 0 0 0 0 1

 ,

ψλ(σ3) =



1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 t 0 t −t

 ,

and

ψλ(δ) =



1 + λ1b1 λ2b1 λ3b1 b1 0 0
λ1b2 1 + λ2b2 λ3b2 b2 0 0
λ1b3 λ2b3 1 + λ3b3 b3 0 0
λ1d1 λ2d1 λ3d1 1 + d1 0 0
λ1d2 λ2d2 λ3d2 d2 1 0

−t(1+t+t2)
1+t λ1

−t(1+t+t2)
1+t λ2

−t(1+t+t2)
1+t λ3

−t(1+t+t2)
1+t 0 1


.

The graph E4,1 has 4 vertices σ1, σ2, σ3 and δ. Since p = 1, it follows that the vertex
δ has a common edge with σp = σ1. Therefore, the following relations are satisfied.

σ1σ2σ1 = σ2σ1σ2 (4.1)

σ2σ3σ2 = σ3σ2σ3 (4.2)

σ1σ3 = σ3σ1 (4.3)

σ2δ = δσ2 (4.4)



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 220

σ3δ = δσ3 (4.5)

σ1δσ1 = δσ1δ (4.6)

We note that relations (4.1),(4.2), and (4.3) are actually Artin’s braid relation of the
classical braid group, B4 having σ1, σ2, and σ3 as standard generators. This assures that
a representation of A(E4,1) yields a representation of B4.
For more details, see [1] and [4].

Lemma 1. The representation ψλ : A(E4,1)→ GL6(C) is reducible.

Proof. For simplicity, we write σi instead of ψλ(σi) .The subspace

S =
〈
e1 + b2

b1
e2 + b3

b1
e3, e4, e5, e6

〉
is an invariant subspace of dimension 4.

To see this:

(i) σ1(e1 + b2
b1
e2 + b3

b1
e3) = e1 + b2

b1
e2 + b3

b1
e3 + te4 ∈ S

(ii) σ2(e1 + b2
b1
e2 + b3

b1
e3) = e1 + b2

b1
e2 + b3

b1
e3 + t b2b1 e5 ∈ S

(iii) σ3(e1 + b2
b1
e2 + b3

b1
e3) = e1 + b2

b1
e2 + b3

b1
e3 + t b3b1 e6 ∈ S

(iv) δ(e1 + b2
b1
e2 + b3

b1
e3) = (1 + λ1b1 + λ2b2 + λ3b3)(e1 + b2

b1
e2 + b3

b1
e3) + d1

b1
(λ1b1 + λ2b2 +

λ3b3)e4 + d2
b1

(λ1b1 + λ2b2 + λ3b3)e5 + −(1+t+t2)
b1(1+t)

(λ1b1 + λ2b2 + λ3b3)e6 ∈ S

(v) σ1e4 = −te4 ∈ S

(vi) σ2e4 = e4 + te5 ∈ S

(vii) σ3e4 = e4 ∈ S

(viii) δe4 = b1(e1 + b2
b1
e2 + b3

b1
e3) + (1 + d1)e4 + d2e5

−t(1+t+t2)
1+t e6 ∈ S

(ix) σ1e5 = e4 + e5 ∈ S

(x) σ2e5 = −te5 ∈ S

(xi) σ3e5 = e5 + te6 ∈ S

(xii) δe5 = e5 ∈ S

(xiii) σ1e6 = e6 ∈ S

(xiv) σ2e6 = e5 + e6 ∈ S

(xv) σ3e6 = −te6 ∈ S



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 221

(xvi) δe6 = e6 ∈ S

5. On the Irreducibility of ψ′
λ : A(E4,1) → GL4(C)

We consider the representation ψλ : A(E4,1)→ GL6(C) restricted to the basis e1, e2,
e1 + b2

b1
e2 + b3

b1
e3, e4, e5, and e6. The matrix of σ1 becomes

ψλ(σ1) =



1 0 0 t 0 0
0 1 0 0 0 0
0 0 1 t 0 0
0 0 0 −t 0 0
0 0 0 1 1 0
0 0 0 0 0 1

 .

We reduce our representation to a 4-dimensional one by considering the sub-basis
e1 + b2

b1
e2 + b3

b1
e3, e4, e5, and e6 to get ψ′λ : A(E4,1) → GL4(C). The representation is

defined as follows:

ψ′λ(σ1) =


1 t 0 0
0 −t 0 0
0 1 1 0
0 0 0 1

 ,

ψ′λ(σ2) =


1 0 tb2

b1
0

0 1 t 0
0 0 −t 0
0 0 1 1

 ,

ψ′λ(σ3) =


1 0 0 tb3

b1
0 1 0 0
0 0 1 t
0 0 0 −t

 ,

and

ψ′λ(δ) =



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 222


1 +

∑3
i=1 λibi

d1
b1

(
∑3

i=1 λibi)
d2
b1

(
∑3

i=1 λibi)
−t(1+t+t2)
b1(1+t)

(
∑3

i=1 λibi)

b1 1 + d1 d2 −t(1 + t+ t2)
0 0 1 0
0 0 0 1

 .

We then diagonalize the matrix corresponding to ψ′λ(σ1) by an invertible matrix, say
T , and conjugate the matrices of ψ′λ(σ2), ψ

′
λ(σ3), and ψ′λ(δ) by the same matrix T . The

invertible matrix T is given by

T =


0 0 1 t
0 0 0 −1− t
0 1 0 1
1 0 0 0

 .

In fact, a computation shows that

T−1ψ′λ(σ1)T =


1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −t

 .

After conjugation, we get

T−1ψ′λ(σ2)T =



1 1 0 1

0 −t2
1+t 0 −(1+t+t2)

1+t

0 t(b2+b1t+b2t)
b1(1+t)

1 t(b2+b1t+b2t)
b1(1+t)

0 −t
1+t 0 1

1+t


,

T−1ψ′λ(σ3)T =


−t 0 0 0
t 1 0 0
tb3
b1

0 1 0

0 0 0 1

 ,

and



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 223

T−1ψ′λ(δ)T =

1 0 0 0
−t(1+t+t2)

(1+t)2
1 + d2

1+t
b1
1+t

t
1+t

−t(1+t+t2)
b1(1+t)

k d2
b1
k 1 + k

(−d1(1+t)+d2+b1t)((
∑3

i=1 λibi)(1+t)+b1t)
b1(1+t)

t(1+t+t2)
1+t

−d2
1+t

−b1
1+t

1
1+t


.

where k = b1t
1+t +

∑3
i=1 λibi.

The entries of the matrices T−1ψ′λ(σ2)T and T−1ψ′λ(δ)T are well-defined since we assume
in our work that t 6= −1. For simplicity, we denote T−1ψ′λ(σ1)T by ψ′λ(σ1), T

−1ψ′λ(σ2)T
by ψ′λ(σ2), T

−1ψ′λ(σ3)T by ψ′λ(σ3), and T−1ψ′λ(δ)T by ψ′λ(δ).

We now prove some lemmas and propositions to determine a sufficient and necessary
condition for irreducibility of ψ′λ : A(E4,1)→ GL4(C).

Lemma 2. The proper subspace S =
〈
e1, e4, e2 + b3

b1
e3

〉
is not invariant if and only if

t4 + t3 + t2 + t+ 1 6= 0.

Proof. First, we prove that proper subspace S =
〈
e1, e4, e2 + b3

b1
e3

〉
is not invariant if

t4 + t3 + t2 + t+ 1 6= 0.
Assume, for contradiction, that S is invariant.

We have ψ′λ(σ2)(e4) =



1

−(1+t+t2)
1+t

t(b2+b2t+b1t)
b1(1+t)

1
1+t


∈ S.

This implies that (1 + t+ t2)b3 = −t(b2 + tb2 + tb1).

By using the equations: tb2 = −td1+(1+t)d2+ t(1+t+t2)
1+t , tb3 = −td2−t(1+t+t2), and

tb1 = (1 + t)d1− d2 + t, simple computations give t4 + t3 + t2 + t+ 1 = 0, a contradiction.

On the other hand, we assume that t4 + t3 + t2 + t+ 1 = 0. We prove that the proper

subspace S =
〈
e1, e4, e2 + b3

b1
e3

〉
is invariant as follows:



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 224

(i) ψ′λσ1(e1) = e1 ∈ S.

(ii) ψ′λσ2(e1) = e1 ∈ S.

(iii) ψ′λσ3(e1) =


−t
t
tb3
b1

0

 ∈ S.

(iv) ψ′λδ(e1) =



1

−t(1+t+t2)
(1+t)2

−t2(1+t+t2)
(1+t)2

+ −t(1+t+t2)
b1(1+t)

(λ1b1 + λ2b2 + λ3b3)

−t(1+t+t2)
(1+t)2


= ae1+be4+c(e2+ b3

b1
e3).

Here, we have a = 1, b = −d3
1+t , c = d3

1+t , and
cb3
b1

= −t2(1+t+t2)
(1+t)2

+ −t(1+t+t2)
b1(1+t)

(λ1b1 + λ2b2 + λ3b3).

Thus, b3
b1

= b1t+(1+t)(λ1b1+λ2b2+λ3b3)
b1

. (5.1)

(v) ψ′λσ1(e4) = −te4 ∈ S.

(vi) ψ′λσ2(e4) =



1

−(1+t+t2)
1+t

t(b2+b2t+b1t)
b1(1+t)

1
1+t


= ae1 + be4 + c(e2 + b3

b1
e3).

Here, we have a = 1, b = 1
1+t , c = −(1+t+t2)

1+t , and cb3
b1

= t(b2+b2t+b1t)
b1(1+t)

.

Thus, −(1 + t+ t2) b3b1 = t(b2+b2t+b1t)
b1

. (5.2)

(vii) ψ′λσ3(e4) = e4 ∈ S.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 225

(viii) ψ′λδ(e4) =



0

t
1+t

(−d1(1+t)+d2+b1t)((λ1b1+λ2b2+λ3b3)(1+t)+b1t)
b1(1+t)

1
1+t


= ae1 + be4 + c(e2 + b3

b1
e3).

Here, we have a = 0, b = 1
1+t , c = t

1+t , and
cb3
b1

= (−d1(1+t)+d2+b1t)((λ1b1+λ2b2+λ3b3)(1+t)+b1t)
b1(1+t)

.

Thus, b3
b1

= (−d1(1+t)+d2+b1t)((λ1b1+λ2b2+λ3b3)(1+t)+b1t)
b1t

. (5.3)

(ix) ψ′λσ1(e2 + b3
b1
e3) = e2 + b3

b1
e3 ∈ S.

(x) ψ′λσ2(e2 + b3
b1
e3) =



0
−t2
1+t

t(b2+b2t+b1t)
b1(1+t)

+ b3
b1

−t
1+t


= ae1 + be4 + c(e2 + b3

b1
e3).

Here, we have a = 1, b = −t
1+t , c = −t2

1+t , and cb3
b1

= t(b2+b2t+b1t)
b1(1+t)

+ b3
b1

.

Thus, −(1 + t+ t2) b3b1 = t(b2+b2t+b1t)
b1

. (5.4)

(xi) ψ′λσ3(e2 + b3
b1
e3) = e2 + b3

b1
e3 ∈ S.

(xii) ψ′λδ(e2 + b3
b1
e3) =

0

1 + d2
1+t + b3

1+t

d2
b1

(λ1b1 + λ2b2 + λ3b3 + b1t
1+t) + b3

b1
(1 + λ1b1 + λ2b2 + λ3b3 + b1t

1+t)
−d2
1+t −

b3
1+t


= ae1 + be4 + c(e2 + b3

b1
e3).



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 226

Here, we have a = 0, b = −d2
1+t −

b3
1+t , c = 1 + d2

1+t + b3
1+t , and c b3b1 = d2

b1
(λ1b1 + λ2b2 +

λ3b3 + b1t
1+t) + b3

b1
(1 + λ1b1 + λ2b2 + λ3b3 + b1t

1+t).

Thus, (1+ d2
1+t+

b3
1+t)

b3
b1

= d2
b1

(λ1b1+λ2b2+λ3b3+ b1t
1+t)+ b3

b1
(1+λ1b1+λ2b2+λ3b3+ b1t

1+t).
(5.5)

By simple computations, we can verify that equations (5.1), (5.2), (5.3) , and (5.5) are
clearly satisfied without any assumption of the indeterminates whereas equation (5.4) is
satisfied only if t4 + t3 + t2 + t+ 1 = 0.

Lemma 3. Any proper subspace S containing the vector ei + uej + vek, where i, j, k ∈
{1, 2, 3, 4}, except possibly the subspace having the form

〈
e1, e4, e2 + b3

b1
e3

〉
, is not invari-

ant.

Proof. We consider all the subspaces containing the vector ei + uej + vek, where

i, j, k ∈ {1, 2, 3, 4} except possibly the subspace of the form
〈
e1, e4, e2 + b3

b1
e3

〉
.

We then assume, for contradiction, that each considered subspace is invariant.
In each case, simple computations give a contradiction.

Thus, we have determined a necessary and sufficient condition for irreducibility.

Theorem 1. Assume all the indeterminates used in defining Perron representation of de-

gree 4 are non zero complex numbers. Let d3 = −t(1+t+t2)
1+t and t 6= −1. The representation

ψ′λ : A(E4,1)→ GL4(C) is irreducible if and only if t4 + t3 + t2 + t+ 1 6= 0.

In the following sections, we set n = 4 and p = 1 and we study the irreducibility of
the reduced representation of ψλ : A(E5,1)→ GL8(C). Indeed, we obtain a sufficient and
necessary condition that gauarantees the irreducibility of ψ′λ : A(E5,1)→ GL5(C).

6. Reducibility of ψλ : A(E5,1) → GL8(C)

Having defined Perron’s representation, we set n = 4 and p = 1 to get the following

vectors. b =


b1
b2
b3
b4

 , d =


d1
d2
d3
d4

 , and λ = (λ1, λ2, λ3, λ4).



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 227

After we specialize the indeterminates d2 and d3 to −(1 + t+ t2) and −t(1 + t) respec-
tively, we get the following 4× 4 matrices:

A =


λ1b1 λ2b1 λ3b1 λ4b1
λ1b2 λ2b2 λ3b2 λ4b2
λ1b3 λ2b3 λ3b3 λ4b3
λ1b4 λ2b4 λ3b4 λ4b4

 ,

B =


b1 0 0 0
b2 0 0 0
b3 0 0 0
b4 0 0 0

 ,

C =


λ1d1 λ2d1 λ3d1 λ4d1

−(1 + t+ t2)λ1 −(1 + t+ t2)λ2 −(1 + t+ t2)λ3 −(1 + t+ t2)λ4
−t(1 + t)λ1 −t(1 + t)λ2 −t(1 + t)λ3 −t(1 + t)λ4

λ1d4 λ2d4 λ3d4 λ4d4

 ,

and

D =


d1 0 0 0

−(1 + t+ t2) 0 0 0
−t(1 + t) 0 0 0

d4 0 0 0

 .

Simple computations show that the parameters satisfy the following equations:

• tb2 = −td1 − (1 + t)(1 + t+ t2) + t(1 + t) = −td1 − (1 + t)(1 + t2)

• tb3 = t(1 + t+ t2)− t(1 + t)2 − d4 = −t(2t2 + 3t+ 2)− d4

• tb4 = t2(1 + t) + (1 + t)d4

• tb1 = (1 + t)d1 + 1 + 2t+ t2

• λ1b1 + λ2b2 + λ3b3 + λ4b4 = −(1 + t+ d1)

Having defined the 4 × 4 matrices A, B, C and D, we obtain the multiparameter
representation A(E5,1). This representation is of degree 8. We specialize the parameters
λ1, λ2, λ3, λ4, b1, b2, b3, b4, d1, d4, t to values in C − {0}. We further assume that t 6= −1.
The representation ψλ : A(E5,1)→ GL8(C) is defined as follows:



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 228

ψλ(σ1) =



1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
t 0 0 0 −t 1 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1


,

ψλ(σ2) =



1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 t 0 0 t −t 1 0
0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 1


,

ψλ(σ3) =



1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 t 0 0 t −t 1
0 0 0 0 0 0 0 1


,

ψλ(σ4) =



1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 0 0 1 0 0 t −t


,

and



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 229

ψλ(δ)=

1 + λ1b1 λ2b1 λ3b1 λ4b1 b1 0 0 0
λ1b2 1 + λ2b2 λ3b2 λ4b2 b2 0 0 0
λ1b3 λ2b3 1 + λ3b3 λ4b3 b3 0 0 0
λ1b4 λ2b4 λ3b4 1 + λ4b4 b4 0 0 0
λ1d1 λ2d1 λ3d1 λ4d1 1 + d1 0 0 0
kλ1 kλ2 kλ3 kλ4 k 1 0 0

−t(1 + t)λ1 −t(1 + t)λ2 −t(1 + t)λ3 −t(1 + t)λ4 −t(1 + t) 0 1 0
λ1d4 λ2d4 λ3d4 λ4d4 d4 0 0 1


,

where k = −(1 + t+ t2).

The graph E5,1 has 5 vertices σ1, σ2, σ3, σ4 and δ. Since p = 1, it follows that the
vertex δ has a common edge with σp = σ1. Therefore, the following relations are satisfied.

σ1σ2σ1 = σ2σ1σ2 (6.1)

σ2σ3σ2 = σ3σ2σ3 (6.2)

σ3σ4σ3 = σ4σ3σ4 (6.3)

σ1σ3 = σ3σ1 (6.4)

σ1σ4 = σ4σ1 (6.5)

σ2σ4 = σ4σ2 (6.6)

σ2δ = δσ2 (6.7)

σ3δ = δσ3 (6.8)

σ4δ = δσ4 (6.9)

σ1δσ1 = δσ1δ (6.10)

We note that relations (6.1),(6.2), (6.3), (6.4), (6.5) and (6.6) are actually Artin’s braid
relation of the classical braid group, B5 having σ1, σ2, σ3, and σ4 as standard generators.
This assures that a representation of A(E5,1) yields a representation of B5.
For more details, see [1] and [4].

Lemma 4. The representation ψλ : A(E5,1)→ GL8(C) is reducible.

Proof. For simplicity, we write σi instead of ψλ(σi) .The subspace

S =
〈
e1 + b2

b1
e2 + b3

b1
e3 + b4

b1
e4, e5, e6, e7, e8

〉
is an invariant subspace of dimension 5.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 230

7. On the Irreducibility of ψ′
λ : A(E5,1) → GL5(C)

We consider the representation ψλ : A(E5,1)→ GL8(C) restricted to the basis e1, e2, e3,
e1+ b2

b1
e2+ b3

b1
e3+ b4

b1
e4, e5, e6, e7, and e8 to get the subrepresentation ψ′λ : A(E5,1)→ GL5(C)

which is the representation restricted to the sub-basis e1 + b2
b1
e2 + b3

b1
e3 + b4

b1
e4, e5, e6, e7.

This representation is defined as follows:

ψ′λ(σ1) =


1 t 0 0 0
0 −t 0 0 0
0 1 1 0 0
0 0 0 1 0
0 0 0 0 1

 ,

ψ′λ(σ2) =


1 0 tb2

b1
0 0

0 1 t 0 0
0 0 −t 0 0
0 0 1 1 0
0 0 0 0 1

 ,

ψ′λ(σ3) =


1 0 0 tb3

b1
0

0 1 0 0 0
0 0 1 t 0
0 0 0 −t 0
0 0 0 1 1

 ,

ψ′λ(σ4) =


1 0 0 0 tb4

b1
0 1 0 0 0
0 0 1 0 0
0 0 0 1 t
0 0 0 0 −t

 ,

and
ψ′λ(δ) =



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 231



1 + r d1
b1
r −(1+t+t2)

b1
r −t(1+t)

b1
r d4

b1
r

b1 1 + d1 −(1 + t+ t2) −t(1 + t) d4
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1


,

where r =
∑4

i=1 λibi.

We then diagonalize the matrix corresponding to ψ′λ(σ1) by an invertible matrix, say
T , and conjugate the matrices of ψ′λ(σ2), ψ

′
λ(σ3),ψ

′
λ(σ4) and ψ′λ(δ) by the same matrix T .

The invertible matrix T is given by

T =


0 0 0 1 t
0 0 0 0 −1− t
0 0 1 0 1
0 1 0 0 0
1 0 0 0 0

 .

In fact, a computation shows that

T−1ψ′λ(σ1)T =


1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 −t

 .

After conjugation, we get

T−1ψ′λ(σ2)T =



1 0 0 0 0
0 1 1 0 1

0 0 −t2
1+t 0 −(1+t+t2)

1+t

0 0 t(b2+b1t+b2t)
b1(1+t)

1 t(b2+b1t+b2t)
b1(1+t)

0 0 −t
1+t 0 1

1+t


,



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 232

T−1ψ′λ(σ3)T =


1 1 0 0 0
0 −t 0 0 0
0 t 1 0 0

0 tb3
b1

0 1 0

0 0 0 0 1

 ,

T−1ψ′λ(σ4)T =


−t 0 0 0 0
t 1 0 0 0
0 0 1 0 0
tb4
b1

0 0 1 0

0 0 0 0 1

 ,

and
T−1ψ′λ(δ)T =

1 0 0 0 0
0 1 0 0 0
d4
1+t −t 1 + −(1+t+t2)

1+t
b1
1+t

t
1+t

d4
b1
w −t(1+t)

b1
w −(1+t+t2)

b1
w 1 + w (−d1(1+t)−(1+t+t2)+b1t)

b1
w

−d4
1+t t 1+t+t2

1+t
−b1
1+t

1
1+t


,

where w =
∑4

i=1 λibi + b1t
1+t .

The entries of the matrices T−1ψ′λ(σ2)T ,T−1ψ′λ(σ3)T , T−1ψ′λ(σ4)T and T−1ψ′λ(δ)T
are well-defined since we assume in our work that t 6= −1. For simplicity, we denote
T−1ψ′λ(σ1)T by ψ′λ(σ1), T

−1ψ′λ(σ2)T by ψ′λ(σ2), T
−1ψ′λ(σ3)T by ψ′λ(σ3),

T−1ψ′λ(σ4)T by ψ′λ(σ4), and T−1ψ′λ(δ)T by ψ′λ(δ).

We now prove some lemmas and propositions to determine a sufficient and necessary
condition for irreducibility of ψ′λ : A(E5,1)→ GL5(C).

Lemma 5. Except possibly the subspaces having the forms 〈e1, e3, e5, e2 + ue4〉 and
〈e2, e3, e5, e1 + ue4〉, where u ∈ C∗, every proper subspace is not invariant.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 233

Proof. We assume, for contradiction, that every subspace, except those having the
forms 〈e1, e3, e5, e2 + ue4〉 and 〈e2, e3, e5, e1 + ue4〉, is invariant.
We then study each possible form. In each case, simple computations give a contradiction.

Lemma 6. If t3 6= −1, then the subspaces 〈e1, e3, e5, e2 + ue4〉 and
〈e2, e3, e5, e1 + ue4〉 are not invariant.

Proof. First, we assume, for contradiction, that S = 〈e1, e3, e5, e2 + ue4〉 is invariant.

• ψ′λσ4(e1) =



−t
t
0
tb4
b1

0

 = ae1 + be3 + ce5 + d(e2 + ue4) which implies that u = b4
b1

.

(7.1)

• ψ′λσ2(e3) =



0
1
−t2
1+t

t(b1t+b2+b2t)
b1(1+t)
−t
1+t


= ae1 + be3 + ce5 + d(e2 + ue4) which implies that

u = t(b1t+b2+b2t)
b1(1+t)

. (7.2)

• ψ′λσ3(e2 + ue4) =



1
−t
t

u+ tb3
b1

0

 = ae1 + be3 + ce5 + d(e2 + ue4) which implies that

u = −tb3
b1(1+t)

. (7.3)

Since equations (7.1) and (7.3) are equal, we have (1 + t + t2)d4 = −t2(1 + t + t2).
Thus, d4 = −t2.

Moreover, equations (7.2) and (7.3) are equal. This implies that d4 = −(1+t2)2+t2.
By substituting d4 = −t2, we get t4 + t3 + t+1 = (t+1)(t3 +1) = 0, a contradiction.

Now, we assume, for contradiction, that S = 〈e2, e3, e5, e1 + ue4〉 is invariant.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 234

• ψ′λσ2(e3) =



0
1
−t2
1+t

t(b1t+b2+b2t)
b1(1+t)
−t
1+t


= ae2 + be3 + ce5 + d(e1 + ue4).

This implies that t(b1t+ b2 + b2t) = 0.

Simple computations give −(1+ t+ t2)2 + t(1+ t)2 + t2 = 0. Thus, t4 + t3 + t+1 = 0,
a contradiction.

We now determine conditions under which one of the subspaces mentioned in Lemma
7 is invariant. But first we write down the following lemma.

Lemma 7. The proper subspaces S1 = 〈e1, e3, e5, e2 + ue4〉 and
S2 = 〈e2, e3, e5, e1 + ue4〉 cannot be both invariant.

Proof. Assume that S1 is invariant. This implies that ψ′λσ2(e3) and ψ′λσ3(e2 + ue4) ∈
S1. Simple computations give b1t+ b2 + b2t = −b3 6= 0.

Assume, for contradiction, that S2 is invariant. This implies that ψ′λσ2(e3) ∈ S2. Simple
computations give b1t+ b2 + b2t = 0, a contradiction.

Lemma 8. If t3 = −1, then the subspace S = 〈e2, e3, e5, e1 + ue4〉 is invariant.

Proof.

• ψ′λσ1(e2)=ψ′λσ2(e2)=ψ′λσ4(e2)=e2 ∈ S.

• ψ′λσ3(e2) =



1
−t
t
tb3
b1

0

 = ae2 + be3 + ce5 + d(e1 + ue4),

if u = tb3
b1

. (7.4)



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 235

• ψ′λδ(e2) =



0
1
d3
1+t

d3t
1+t + d3

b1
(
∑4

i=1 λibi)

−d3
1+t


= ae2 + be3 + ce5 + d(e1 + ue4),

if t
1+t =

−
∑4

i=1 λibi
b1

. (7.5)

• ψ′λσ1(e3)=ψ′λσ3(e3)=ψ′λσ4(e3)=e3 ∈ S.

• ψ′λσ2(e3) =



0
1
−t2
1+t

t(b1t+b2+b2t)
b1(1+t)
−t
1+t


= ae2 + be3 + ce5 + d(e1 + ue4),

if t(b1t+ b2 + b2t) = 0. (7.6)

• ψ′λδ(e3) =



0
0

1 + −(1+t+t2)
1+t

−t2(1+t+t2)
1+t + −(1+t+t2)

b1
(
∑4

i=1 λibi)
1+t+t2

1+t


= ae2 + be3 + ce5 + d(e1 + ue4),

if
∑4

i=1 λibi + b1t
1+t = 0. (7.7)

• ψ′λσ1(e5) = −te5 ∈ S.

• ψ′λσ3(e5)=ψ′λσ4(e5)=e5 ∈ S.



M. Dally, M. Abdulrahim / Eur. J. Pure Appl. Math, 11 (1) (2018), 215-237 236

• ψ′λσ2(e5) =



0
1

−(1+t+t2)
1+t

t(b1t+b2+b2t)
b1(1+t)

1
1+t


= ae2 + be3 + ce5 + d(e1 + ue4),

if t(b1t+ b2 + b2t) = 0. (7.8)

• ψ′λδ(e5) =



0
0
t

1+t

(−d1(1+t)−t(1+t+t2)+b2t)(
∑4

i=1 λibi(1+t)+b1t)
b1(1+t)

1
1+t


= ae2+be3+ce5+d(e1+ue4),

if (−d1(1 + t) + d2 + b2t)(
∑4

i=1 λibi(1 + t) + b1t) = 0. (7.9)

• ψ′λσ1(e1 + ue4)=ψ
′
λσ2(e1 + ue4)=ψ

′
λσ3(e1 + ue4)=e1 + ue4 ∈ S.

• ψ′λσ4(e1 + ue4) =


−t
t
0

u+ tb4
b1

0

 = ae2 + be3 + ce5 + d(e1 + ue4),

if u = −tb4
b1(1+t)

. (7.10)

• ψ′λδ(e1 + ue4) =



1
0

d4
1+t + b1u

1+t

d4
b1

(
∑4

i=1 λibi + b1t
1+t) + u(1 +

∑4
i=1 λibi + b1t

1+t)
−d4
1+t −

b1u
1+t


= ae2 + be3 +

ce5 + d(e1 + ue4),

if (
∑4

i=1 λibi + b1t
1+t)(

d4
b1

+ u) = 0. (7.11)



REFERENCES 237

Using the relations, we prove that equations (7.5), (7.7), (7.9), and (7.11) are clearly
satisfied.
Also, we verify that equations (7.4), (7.6), (7.8) and (7.10) are satisfied if −t(1 + t)2 =
−(1 + t+ t2)(1 + t2) + t2 which implies that t3 = −1.

Thus, we have determined a necessary and sufficient condition for irreducibility.

Theorem 2. Assume all the indeterminates used in defining Perron representation of
degree 5 are non zero complex numbers. Let d2 = −(1 + t + t2), d3 = −t(1 + t), and
t 6= −1. The representation ψ′λ : A(E5,1)→ GL5(C) is irreducible if and only if t3 6= −1.

Remark 1. • For n=2 and for t 6= −1, we proved that a complex specialization of the
representation ψ′λ : A(E3,1) → Gl3(C) is irreducible if and only if t2 6= −1 which is
equivalent to t3 + t2 + t+ 1 6= 0.

• For n=3 and for t 6= −1, we have proved that a complex specialization of the repre-
sentation ψ′λ : A(E4,1)→ Gl4(C) is irreducible if and only if t4 + t3 + t2 + t+ 1 6= 0.

• For n=4 and for t 6= −1, we have proved that a complex specialization of the rep-
resentation ψ′λ : A(E5,1) → Gl5(C) is irreducible if and only if t3 6= −1 which is
equivalent to t5 + t4 + t3 + t2 + t+ 1 6= 0.

References

[1] J. S. Birman, Braids, Links and Mapping Class Groups. Annals of Mathematical Stud-
ies. Princeton University Press, volume 82, New Jersey, 1975.

[2] T.E. Brendle, The Torelli Group and Representations of Mapping Class Groups. Doc-
toral Thesis, Columbia University, 2002.

[3] M. Dally and M. Abdulrahim, On the Irreducibility of Artin’s Group of Graphs. Journal
of Mathematics Research, volume 7, number 2, 2015.

[4] V.L. Hansen, Braids and Coverings. London Mathematical Society, Cambridge Uni-
versity Press, 1989.

[5] B. Perron, A linear representaton of a finite rank of the mapping class group of surfaces
(preprint). Laboratoire de Topologie, volume 99, number 204, 1999.