EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 3, 2018, 682-701
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On the Irreducibility of Fourth Dimensional Tuba’s
Representation of the Pure Braid Group on Three

Strands

Hasan A. Haidar1, Mohammad N. Abdulrahim1,∗

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

Abstract. We consider Tuba’s representation of the pure braid group, P3, given by the map
φ : P3 −→ GL(4, F ), where F is an algebraically closed field. After, specializing the indeterminates
used in defining the representation to non- zero complex numbers, we find sufficient conditions that
guarantee the irreducibility of Tuba’s representation of the pure braid group P3 with dimension
d = 4. Under further restriction for the complex specialization of the indeterminates, we get a
necessary and sufficient condition for the irreducibility of φ.

2010 Mathematics Subject Classifications: 20F36

Key Words and Phrases: Braid group, pure braid group, irreducible

1. Introduction

Let Bn be the braid group on n strands. There exists a surjective group homomor-
phism π : Bn −→ Sn. The kernel of π is referred to as the pure braid group Pn with
n(n−1)

2 generators. In 2001, a representation of B3 was defined by I. Tuba and H. Wenzl ,
namely ρ : B3 −→ GL(V ), which is irreducible on the dimensional vector space V over an
algebraically closed field F . A complete classification of irreducible representations of the
braid group B3 was given by Tuba and Wenzl, for dimensions d ≤ 5 (see [7]). This was
done by assuming a certain triangular form of the matrices of the generators of B3. Albev-
erio has found a class of representations of B3 in every dimension n, which depends on n
parameters [1]. The author in that work uses a deformation of pascal’s triangle connected
with qshifted factorials to get the representations, and this generalizes the work of Tuba
and Wenzl who classified all irreducible representations of B3 for dimensions d ≤ 5 [7] .
This is also a generalization of the results of Humphries, who constructed the representa-
tions of the braid group B3 in arbitrary dimension using the classical pascal triangle [3]. Le

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3273

Email addresses: hah339@student.bau.edu.lb (H. A. Haidar),
mna@bau.edu.lb (M. N. Abdulrahim)

http://www.ejpam.com 682 c© 2018 EJPAM All rights reserved.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 683

Bruyn in [4] proved that all the components of n-dimensional irreducible representations
of B3 are densely parametrized by rational quiver varieties and the explicit parametriza-
tions are given for n < 12. Then Le Bruyn in [5] extended all this by establishing such
parametrizations for all finite dimensions n, which also generalizes the work of Tuba and
Wenzl. Also, researchers gave a great value for representations of the pure braid group
Pn, the normal subgroup of Bn.
Recently, N. Maanna and M. Abdulrahim gave a necessary and sufficient condition for the
irreducibility of the Tuba’s representation of pure braid group P3 for dimensions 2 and 3
(see [6]). In our work, we mainly consider the irreducibility criteria of Tuba’s represen-
tation of the pure braid group P3, with dimension four. Our main result is Theorem 11,
which determines sufficient conditions for the irreducibility of Tuba’s representation of P3

with dimension d = 4. Under further restriction on the indeterminates used in defining
Tuba’s representation of dimension 4, we get a necessary and sufficient condition for the
irreducibility of the representation. This will be corollary 12.

2. Preliminaries

Definition 1. [2] The braid group on n strings, Bn, is the abstract group with presentation
Bn ={σ1, ..., σn−1; σiσi+1σi = σi+1σiσi+1, for i = 1, 2, ..., n−2, σiσj = σjσi if |i− j| � 1}.

The generators σ1, ..., σn−1 are called the standard generators of Bn.

Definition 2. [2] The pure braid group, Pn, is defined as the kernel of the homomorphism
Bn −→ Sn, defined by σi −→ (i, i+ 1), 1 ≤ i ≤ n− 1. It has the following generators:

Aij = σj−1σj−2...σi+1σ
2
i σ
−1
i+1...σ

−1
j−2σ

−1
j−1, 1 ≤ i, j ≤ n

Definition 3. A representation is a map γ : G −→ GL(V ), where G is a group and
GL(V ) is the group of n× n invertible matrices over the algebraically closed field V .

Definition 4. A representation γ : G −→ GL(V ) is said to be irreducible if it has no non
trivial proper invariant subspaces.

3. Tuba’s Representation of B3

Imre Tuba and Hans Wenzl gave a complete classification of all simple representations
of B3 with dimensions d ≤ 5 by assuming a certain triangular form for the invertible d×d
matrices A and B of the generators of B3 that satisfy ABA = BAB. In particular, they
proved that a simple d− dimensional representation ϕ : B3 −→ GL(V ) is determined, up
to isomorphism, by the eigenvalues λ1, ...λd of the images of the generators σ1 and σ2 of
B3. For more details, see [7]. Below,we write the explicit matrices in the case d = 4.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 684

Proposition 1. [7, p.500] Tuba
′
s representaion of B3 of dimension d = 4 is defind as

follows:

σ1−→


λ1 (1 +D−1 +D−2)λ2 (1 +D−1 +D−2)λ3 λ4
0 λ2 (1 +D−1)λ3 λ4
0 0 λ3 λ4
0 0 0 λ4

 ,

σ2 −→


λ4 0 0 0
−λ3 λ3 0 0
Dλ2 −(D + 1)λ2 λ2 0
−D3λ1 (D3 +D2 +D)λ1 −(D2 +D + 1)λ1 λ1

 , where

λ1, λ2, λ3, and λ4 are indeterminates and D =
√

λ2λ3
λ1λ4

.

Proposition 2. [7, p.503] Tuba’s representation of B3 of dimension four is irreducible if
and only if
−γ−2

(
λ2r + γ2

) (
λ2s + γ2

) (
γ2 + λrλk + λsλl

)
(γ2 + λrλl + λsλk) 6= 0,

where {r, s, k, l} = {1, 2, 3, 4} , and γ2 is a square root of the det(σ1).

A similar result is obtained for the pure braid group P3, the normal subgroup of the
braid group B3.

Proposition 3. [6] Tuba’s representation of P3 is irreducible if and only if
i) λ1 6= −λ2 and λ21 − λ1λ2 + λ22 6= 0 for dimension d = 2,
ii) λi 6= −λj and (λ2m + λkλn)(λ2n + λkλm) 6= 0, for dimension d = 3.
Here i 6= j, m 6= n 6= k, and i, j,m, n, k ∈ {1, 2, 3}.

4. Tuba’s Representation of P3

Let P3 be the pure braid group on three strings. Applying Tuba’s representation on the
normal subgroup of the braid group , namely the pure braid group, we get the following
representation of dimension d = 4.

Definition 5. Tuba’s representation of the pure braid group P3 of dimension d = 4 is
defined as follows:

A12 =


λ21 Jλ2 (λ1 + λ2) Jλ3[λ1 + λ3 + Iλ2] λ4[λ1 + λ4 + J (λ2 + λ3)]
0 λ22 Iλ3 (λ2 + λ3) λ4[λ2 + λ4 + Iλ3]
0 0 λ23 λ4 (λ3 + λ4)
0 0 0 λ24

 ,

A23 =


λ24 0 0 0

−λ3 (λ3 + λ4) λ23 0 0
λ2[Dλ4 + (D + 1)λ3 +Dλ2] −(D + 1)λ2 (λ2 + λ3) λ22 0

L K M λ21

 ,



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 685

where

I = 1 +D−1,

J = 1 +D−1 +D−2,

K = λ1(D
2 +D + 1)[D(λ1 + λ2 + λ3) + λ2],

L = λ1[−D3 (λ4 + λ1)− (D3 +D2 +D)(λ3 + λ2)],

M = −(D2 +D + 1)λ1(λ1 + λ2).

As for A13 = σ2σ
2
1σ
−1
2 , we will not need it in the proof of Theorem 11.

5. Irreducibility of Tuba’s Representation of the Pure Braid Group P3

with Dimension d=4

We specialize the indeterminates λ1, λ2, λ3, and λ4 to non zero complex numbers. Then
we find sufficient conditions for the irreducibility of the complex specialization of Tuba’s
representation of the pure braid group P3 with dimension d = 4.

Definition 6. Principal square root function is defined as follows:

For z = (1, θ),
√
z = e

θ
2
i, where −π ≺ θ ≤ π. Since θ ∈ (−π, π], it follows that

√
z2 = z

for any complex number z.

In what follows, we take
√
z2 = z for any complex number z.

Lemma 1. Let ϕ : P3 −→ GL4(C) be the complex specialization of Tuba’s representation
of the pure braid group P3. Hence, the following are true:

i) D + 1 = 0 if and only if γ2 + λ1λ4 = 0.

ii) Dλn + λm = 0 if and only if γ2 + λ2n = 0, where {m,n} = {2, 3}.

iii) 1 +D +D2 = 0 implies that γ2 + λ1λ4 + λ2λ3 = 0.

Proof. The proof of (i) follows from the fact that det(σ1) = λ1λ2λ3λ4 and γ2 = λ1λ4D.

The proof of (ii) follows from the fact that γ2 = λ2λ3D
−1.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 686

To prove (iii) : If 1 +D +D2 = 0 then D3 − 1 = 0. This implies that

(λ1λ4D)3 − (λ1λ4)
3 = 0, which is equivalent to γ6 −(λ1λ4)

3 = 0. Hence

(γ2 − λ1λ4)(γ4 + λ1λ4γ
2 + (λ1λ4)

2) = 0

In the case γ2 − λ1λ4 = 0, we get λ1λ4D − λ1λ4 = 0. This implies that D = 1,

a contradiction. Thus γ4 + λ1λ4γ
2 + (λ1λ4)

2 = 0. It follows that

λ2λ3γ
4 + λ1λ2λ3λ4γ

2 + λ21λ2λ3λ
2
4 = 0. This implies that λ2λ3γ

4 + γ6 + λ1λ4γ
4 = 0.

Thus λ2λ3 + γ2 + λ1λ4 = 0.

Theorem 1. Tuba’s representation ϕ : P3 −→ GL4(C) is irreducible if the following hold
true:

(1) λi 6= −λj , where i, j ∈ {1, 2, 3, 4}
(2) γ2 + λlλ4 6= 0, where l ∈ {1, 2, 3}
(3) γ2(λi + λ3 + λ4) + λiλjλ3 + λiλjλ4 + λjλ3λ4 6= 0, where {i, j} ∈ {1, 2}
(4) (γ2 + λ2r)(γ

2 + λrλl + λsλk) 6= 0, where {r, s, l, k} = {1, 2, 3, 4}

Proof. To get contradiction, suppose that this representation ϕ : P3 −→ GL4(C) is
reducible .That is, there exists a proper non-zero invariant subspace S, of dimension 1 , 2
or 3.We consider 15 cases. We use e1, e2, e3, and e4 as the canonical basis of C4. Let α, β
and δ be non-zero complex numbers.

Case 1: Let e1 ∈ S, it follows that

A23e1 − λ24e1 ∈ S, then


0
R2

R3

R4

 ∈ S.

Here, the constants are given by

R2 = −λ3(λ3 + λ4),

R3 = λ2[Dλ4 + (D + 1)λ3 +Dλ2],

R4 = λ1[−D3 (λ4 + λ1)− (D3 +D2 +D)(λ3 + λ2)].



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 687

We have A23(R2e2 +R3e3 +R4e4)− λ23(R2e2 +R3e3 +R4e4) ∈ S.

Then


0
0
P3

P4

 ∈ S, where

P3 = −R2(D + 1)λ2(λ2 + λ3) +R3(λ
2
2 − λ23),

P4 = KR2 −R3(D
2 +D + 1)λ1(λ1 + λ2) +R4(λ

2
1 − λ23).

Also, we have A23(P3e3 + P4e4)− λ22(P3e3 + P4e4) ∈ S. Then


0
0
0
T

 ∈ S.
Here,

T = −P3(D
2 +D + 1)λ1(λ1 + λ2) + P4(λ

2
1 − λ22).

If T 6= 0, then e4 ∈ S. This implies that P3e3 ∈ S.

In the case P3 = 0, We get −R2(D + 1)λ2 +R3(λ2 − λ3) = 0,

and so (D + 1)λ2λ3(λ3 + λ4) + (λ2 − λ3)λ2[Dλ4 + (D + 1)λ3 +Dλ2] = 0.

Thus λ2(λ3 + Dλ2)(λ4 + λ2) = 0, which implies that λ3 + Dλ2 = 0. This is equiva-
lent to γ2 + λ22 = 0 (Lemma 10), a contradiction.

In the case P3 6= 0, we get e3 ∈ S. But we have R2e2 +R3e3 +R4e4 ∈ S,
and R2 = −λ3(λ3 + λ4) 6= 0. So e2 ∈ S, and also e1 ∈ S.
Thus S = C4, a contradiction.

Therefore, we have T = 0.

This implies that λ2λ3 +D3λ21 +Dλ1λ2 +Dλ1λ3 +D2λ1λ2 +D2λ1λ3 = 0.
We get λ1λ

2
4(λ2λ3 +D3λ21 +Dλ1λ2 +Dλ1λ3 +D2λ1λ2 +D2λ1λ3) = 0.

Thus γ2[γ2(λ2 + λ3 + λ4) + λ1λ2λ3 + λ1λ2λ4 + λ1λ3λ4] = 0, a contradiction.

Case 2: Let e2 ∈ S, it follows that



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 688

A12e2 − λ22e2 ∈ S . Then


(1 +D−1 +D−2)λ2 (λ1 + λ2)

0
0
0

 ∈ S.
But (1 +D−1 +D−2)λ2 (λ1 + λ2) 6= 0, a contradiction (Case 1).

Case 3: Let e4 ∈ S, it follows that

1
λ4

[A12e4 − λ24e4] ∈ S. Then


N1

N2

N3

0

 ∈ S.

Here, the constants are given by

N1 = λ1 + λ4 + (1 +D−1 +D−2) (λ2 + λ3) ,
N2 = λ2 + λ4 + (1 +D−1)λ3,
N3 = λ3 + λ4.

We have A12(N1e1 +N2e2 +N3e3)− λ23(N1e1 +N2e2 +N3e3) ∈ S.

This implies that


M1

M2

0
0

 ∈ S,

where

M1 = N1(λ
2
1 − λ23) +N2Jλ2 (λ1 + λ2) +N3Jλ3[λ1 + λ3 + (1 +D−1)λ2],

M2 = N2(λ
2
2 − λ23) +N3Iλ3 (λ2 + λ3) .

Also, we have A12(M1e1 +M2e2)− λ22(M1e1 +M2e2) ∈ S.

This implies that


T
0
0
0

 ∈ S, where

T = M1(λ
2
1 − λ22) +M2Jλ2 (λ1 + λ2) .

If T 6= 0, then we get a contradiction (Case 1).

If T = 0. Then λ2λ3 +D3λ21 +Dλ1λ2 +Dλ1λ3 +D2λ1λ2 +D2λ1λ3 = 0.
Thus γ2[γ2(λ2 + λ3 + λ4) + λ1λ2λ3 + λ1λ2λ4 + λ1λ3λ4] = 0, a contradiction.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 689

Case 4: Let e3 ∈ S, it follows that A23e3 − λ22e3 ∈ S.

So


0
0
0

−(D2 +D + 1)λ1 (λ1 + λ2)

 ∈ S.

But −(D2 +D + 1)λ1 (λ1 + λ2) 6= 0, a contradiction (Case 3).

Case 5: Let e1 + αe2 ∈ S, it follows that

A12(e1 + αe2)− λ22(e1 + αe2) ∈ S. This implies that


T
0
0
0

 ∈ S, where

T = λ21 − λ22 + αJλ2 (λ1 + λ2) .

If T 6= 0, then e1 ∈ S, a contradiction (Case 1). If T = 0 then
λ1 − λ2 + αJλ2 = 0. It follows that Jλ2e1 + (λ2 − λ1)e2 ∈ S. Thus

A23[Jλ2e1 + (λ2 − λ1)e2]− λ24[Jλ2e1 + (λ2 − λ1)e2] ∈ S . Then


0
R2

R3

R4

 ∈ S.
Here, the constants are given by

R2 = −λ2Jλ3(λ4 + λ3) + (λ2 − λ1)(λ23 − λ24),

R3 = λ22J [Dλ4 + (D + 1)λ3 +Dλ2]− (λ2 − λ1)(D + 1)λ2(λ3 + λ2),

R4 = λ2JL+ (λ2 − λ1)K.

Since R2e2 +R3e3 +R4e4 ∈ S, it follows that

A23(R2e2 +R3e3 +R4e4)− λ23(R2e2 +R3e3 +R4e4) ∈ S. Thus


0
0
P3

P4

 ∈ S,
where

P3 = −R2(D + 1)λ2 (λ2 + λ3) +R3(λ
2
2 − λ23),

P4 = R2K +R3M +R4(λ
2
1 − λ23).



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 690

On the other hand, A23(P3e3 + P4e4)− λ22(P3e3 + P4e4) ∈ S. Then


0
0
0
T1

 ∈ S.
Here, the constant T1 is given by

T1 = −P3(D
2 +D + 1)λ1(λ1 + λ2) + P4(λ

2
1 − λ22).

If T1 6= 0, then e4 ∈ S, a contradiction (by Case 3).

If T1 = 0 then D2λ31 +D2λ1λ2λ4 +Dλ21λ2 +Dλ1λ2λ3 +Dλ22λ4 + λ1λ
2
2 = 0.

Hence, λ1λ4
λ2

(D2λ31 +D2λ1λ2λ4 +Dλ21λ2 +Dλ1λ2λ3 +Dλ22λ4 + λ1λ
2
2) = 0.

This implies that (γ2 + λ21)(γ
2 + λ1λ3 + λ2λ4) = 0, a contradiction.

Case 6: Let e1 + αe3 ∈ S, it follows that

A12(e1 + αe3)− λ23(e1 + αe3) ∈ S. Then


N1

N2

0
0

 ∈ S.
Here, the constants are given by

N1 = λ21 − λ23 + αJλ3[λ1 + λ3 + Iλ2],
N2 = αIλ3 (λ2 + λ3) .

We have N2 6= 0 (I 6= 0 by Lemma 10). If N1 = 0, then e2 ∈ S, a contradiction (Case 2).
If N1 6= 0, we get a contradiction (Case 5).

Case 7: Let e3 + αe4 ∈ S, it follows that

A23(e3 + αe4)− λ22(e3 + αe4) ∈ S. So


0
0
0
T

 ∈ S, where

T = −D2Jλ1(λ1 + λ2) + α(λ21 − λ22).

If T 6= 0, then e4 ∈ S, a contradiction.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 691

If T = 0 then −D2Jλ1 + α(λ1 − λ2) = 0.

On the other hand, we have (e3 + αe4)(λ1 − λ2) ∈ S.
It follows that (λ1 − λ2)e3 +D2Jλ1e4 ∈ S.

Hence A12[(λ1 − λ2)e3 +D2Jλ1e4]− λ24[(λ1 − λ2)e3 +D2Jλ1e4] ∈ S.

Thus


N1

N2

N3

0

 ∈ S.

Here, the constants are given by

N1 = (λ1 − λ2)Jλ3[λ1 + λ3 + Iλ2] +D2Jλ1λ4[λ1 + λ4 + J (λ2 + λ3)],
N2 = (λ1 − λ2)Iλ3 (λ2 + λ3) +D2Jλ1λ4[λ2 + λ4 + Iλ3],
N3 = (λ1 − λ2)(λ23 − λ24) +D2Jλ1λ4(λ3 + λ4).

Also, we have A12(N1e1 +N2e2 +N3e3)− λ23(N1e1 +N2e2 +N3e3) ∈ S.

Then


M1

M2

0
0

 ∈ S, where

M1 = N1(λ
2
1 − λ23) +N2Jλ2 (λ1 + λ2) +N3Jλ3[λ1 + λ3 + Iλ2],

M2 = N2(λ
2
2 − λ23) +N3Iλ3 (λ2 + λ3) .

If M1 6= 0 or M2 6= 0. Then we get a contradiction (Case 1, Case 2, and Case 5).

If M1 = 0 and M2 = 0. Then

λ2λ3(λ1 + λ4) +D2λ1λ3λ4 +Dλ1λ2λ3 +Dλ1λ2λ4 +Dλ2λ3λ4 = 0.

So D−1(λ1λ2λ3 + λ2λ3λ4 +D2λ1λ3λ4 +Dλ1λ2λ3 +Dλ1λ2λ4 +Dλ2λ3λ4) = 0.

This implies that γ2(λ1 + λ3 + λ4) + λ1λ2λ3 + λ1λ2λ4 + λ2λ3λ4 = 0,
a contradiction.

Case 8: Let e1 + αe4 ∈ S, it follows that A23(e1 + αe4)− λ24(e1 + αe4) ∈ S.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 692

Then


0
R2

R3

R4

 ∈ S.
Here, the constants are given by

R2 = −λ3 (λ3 + λ4) ,
R3 = λ2[Dλ4 + (D + 1)λ3 +Dλ2],
R4 = L+ α(λ21 − λ24).

Also, we have A23(R2e2 +R3e3 +R4e4)− λ23(R2e2 +R3e3 +R4e4) ∈ S,

then


0
0
P3

P4

 ∈ S.
Here, P3 = −R2(D + 1)λ2(λ2 + λ3) +R3(λ

2
2 − λ23),

P4 = R2K +R3M +R4(λ
2
1 − λ23).

If P3 6= 0 or P4 6= 0, then we get a contradiction (Case 3, Case 4, and Case 7).

Otherwise, if P3 = 0 then λ2 (λ2 + λ4) (λ2 + λ3) (Dλ2 + λ3) = 0.

Hence Dλ2 + λ3 = 0, which is equivalent to γ2 + λ22 = 0 (Lemma 10), a contradiction.

Case 9: Let e2 + αe3 ∈ S, it follows that

A12(e2 + αe3)− λ23(e2 + αe3) ∈ S. Then


N1

N2

0
0

 ∈ S.
Here, the constants are given by

N1 = Jλ2 (λ1 + λ2) + αJλ3[λ1 + λ3 + Iλ2],
N2 = λ22 − λ23 + αIλ3 (λ2 + λ3) .

If N1 6= or N2 6= 0, then we get a contradiction (Case 1, Case 2, and Case 5).

Otherwise, N2 = 0 and so

1
D (λ2 + λ3) (Dλ2 −Dλ3 + αλ3 + αDλ3) = 0.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 693

This implies that Dλ2 −Dλ3 + α(λ3 +Dλ3) = 0. (1)

Also, we have N1 = 0. It follows that

1
D3D

2J(Dλ22 +Dλ1λ2 + αλ2λ3 + αDλ23 + αDλ1λ3 + αDλ2λ3) = 0.

Hence Dλ22 +Dλ1λ2 + αλ2λ3 + αDλ23 + αDλ1λ3 + αDλ2λ3 = 0. (2)

Now, after subtracting equation (2) from equation λ2(1), we get
Dλ2λ3 +Dλ1λ2 + αDλ23 + αDλ1λ3 = 0. This implies that
D (λ1 + λ3) (λ2 + αλ3) = 0. Thus αλ3 = −λ2. Substituting αλ3 = −λ2 in (1).
We get −Dλ3 − λ2 = 0, which is equivalent to γ2 + λ23 = 0 (Lemma 10), a contradiction.

Case 10: Let e2 + αe4 ∈ S, it follows that

A23(e2 + αe4)− λ23(e2 + αe4) ∈ S. Then


0
0
P3

P4

 ∈ S.
Here, the constants are given by

P3 = −(D + 1)λ2 (λ2 + λ3) ,
P4 = λ1(D

2 +D + 1)[D(λ1 + λ2 + λ3) + λ2] + α(λ21 − λ23).

If P3 6= 0 or P4 6= 0, then we get a contradiction (Case 3, Case 4, and Case 7).

Otherwise, if P3 = 0 then −(D + 1)λ2 (λ2 + λ3) = 0, a contradiction (Lemma 10).

Case 11: Let αe1 + βe2 + e3 ∈ S, it follows that

A12(αe1 + βe2 + e3)− λ23(αe1 + βe2 + e3) ∈ S.Then


N1

N2

0
0

 ∈ S.
Here, the constants are given by

N1 = α(λ21 − λ23) + βJλ2 (λ1 + λ2) + Jλ3[λ1 + λ3 + Iλ2],
N2 = β(λ22 − λ23) + Iλ3 (λ2 + λ3) .

If N1 6= 0 or N2 6= 0, then we get a contradiction (Case 1, Case 2, and Case 5).

Otherwise, N2 = 0 and so β(λ2 − λ3) + Iλ3 = 0.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 694

In the case λ2 − λ3 = 0, we get Iλ3 = 0, a contradiction (Lemma 10). Hence β = Iλ3
λ3−λ2 .

Substituting, β = Iλ3
λ3−λ2 in the equation (λ3 − λ2)N1 = 0, we get

α
(
λ21 − λ23

)
(λ3 − λ2) + IJλ3λ2 (λ1 + λ2) + Jλ3[λ1 + λ3 + Iλ2](λ3 − λ2) = 0.

Then α
(
λ21 − λ23

)
(λ3 − λ2) + IJλ2λ3(λ1 + λ3) + λ3J (λ3 − λ2) (λ1 + λ3) = 0.

This implies that α (λ1 − λ3) (λ3 − λ2) + IJλ2λ3 + λ3J (λ3 − λ2) = 0.

Hence α (λ1 − λ3) (λ3 − λ2) + λ3(1 +D−1 +D−2)(λ3 +D−1λ2) = 0.

If (λ1 − λ3) (λ2 − λ3) = 0, then λ3(1 +D−1 +D−2)(λ3 +D−1λ2) = 0.

So λ3 +D−1λ2 = 0, which is equivalent to γ2 + λ23 = 0 (Lemma 10), a contradiction.

That is (λ1 − λ3) (λ2 − λ3) 6= 0. Thus α = λ3(1+D−1+D−2)(λ3+D−1λ2)
(λ1−λ3)(λ2−λ3) .

On the other hand, A23(αe1 + βe2 + e3)− λ24(αe1 + βe2 + e3) ∈ S. It follows that


0
R2

R3

R4

 ∈ S.
Here, the constants are given by

R2 = −αλ3 (λ4 + λ3) + β(λ23 − λ24),
R3 = αλ2[Dλ4 + (D + 1)λ3 +Dλ2]− β(D + 1)λ2 (λ2 + λ3) + λ22 − λ24,
R4 = αL+ βK +M.

We have A23(R2e2 +R3e3 +R4e4)− λ23(R2e2 +R3e3 +R4e4) ∈ S.

It follows that


0
0
P3

P4

 ∈ S.
Here,

P3 = −R2(D + 1)λ2(λ2 + λ3) +R3(λ
2
2 − λ23),

P4 = R2K +R3M +R4(λ
2
1 − λ23).



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 695

If P3 6= 0 or P4 6= 0, then we get a contradiction (Case 3, Case 4, and Case 7).

Thus P3 = 0 and P4 = 0.

But α = λ3(1+D−1+D−2)(λ3+D−1λ2)
(λ1−λ3)(λ2−λ3) , and β = (1+D−1)λ3

(λ3−λ2) .

After substituting the obtaining values of α and β in the equation P3 = 0, we get

(λ2+λ3)(λ2+Dλ3)(λ3+Dλ2)(λ2+λ4)
D3(λ2−λ3)(λ1−λ3) (D2(λ1λ2 + λ3λ4 − λ1λ4) +Dλ2λ3 + λ2λ3) = 0.

It follows that, either

(λ2 +Dλ3)(λ3 +Dλ2) = 0 or D2(λ1λ2 + λ3λ4 − λ1λ4) +Dλ2λ3 + λ2λ3 = 0.

If (λ2 + Dλ3)(λ3 + Dλ2) = 0, which is equivalent to (γ2 + λ23)(γ
2 + λ22) = 0 (Lemma

10), we get a contradiction,

If D2(λ1λ2 + λ3λ4 − λ1λ4) +Dλ2λ3 + λ2λ3 = 0, we get that
D(λ1λ4λ2λ3 +Dλ1λ4λ1λ2 +Dλ1λ4λ3λ4) = 0, which is equivalent to
Dγ2(γ2 + λ1λ2 + λ3λ4). This give a contradiction.

Case 12: Let e2 + αe3 + βe4 ∈ S, it follows that

A23(e2 + αe3 + βe4)− λ23(e2 + αe3 + βe4) ∈ S.
.

Then


0
0
P3

P4

 ∈ S.
Here, the constants are given by

P3 = −(D + 1)λ2 (λ2 + λ3) + α(λ22 − λ23),
P4 = K + αM + β(λ21 − λ23).

If P3 6= 0 or P4 6= 0, then we get a contradiction (Case 3, Case 4, and Case 7).

Otherwise, if P3 = 0 then −(D + 1)λ2 + α(λ2 − λ3) = 0.

In the case λ2 − λ3 = 0, we get −(D + 1)λ2 = 0, a contradiction (Lemma 10). Hence

α = (D+1)λ2
λ2−λ3 .



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 696

Substituting α = (D+1)λ2
λ2−λ3 in the equation P4(λ2 − λ3) = 0, we get

K(λ2 − λ3) + (D + 1)λ2M + β(λ21 − λ23) = 0.
Then −λ1(D2 +D + 1)(λ1 + λ3)(Dλ3 + λ2) + β(λ21 − λ23) = 0 and so
−λ1(D2 +D + 1)(Dλ3 + λ2) + β(λ1 − λ3) = 0.

If λ1 − λ3 = 0, then Dλ3 + λ2 = 0, which is equivalent to γ2 + λ23 = 0 (Lemma 10),
a contradiction.

Hence β = λ1(D2+D+1)(Dλ3+λ2)
λ1−λ3 .

On the other hand, A12(e2 + αe3 + βe4)− λ24(e2 + αe3 + βe4) ∈ S.

Then


N1

N2

N3

0

 ∈ S.
Here, the constants are given by

N1 = Jλ2 (λ1 + λ2) + αJλ3[λ1 + λ3 + Iλ2] + βλ4[λ1 + λ4 + J (λ2 + λ3)] ,
N2 = λ22 − λ24 + αIλ3 (λ2 + λ3) + βλ4[λ2 + λ4 + Iλ3],
N3 = α(λ23 − λ24) + βλ4 (λ3 + λ4) .

If N1 6= 0 or N2 6= 0 or N3 6= 0,
then we get a contradiction (Case 1,Case 2, Case 4, Case 5, Case 6, Case 9, and Case 11).

Otherwise, if N3 = 0 then βλ4 = α(λ4 − λ3).
Substituting βλ4 = α(λ4 − λ3) in the equation N2 = 0, we get

λ22 − λ24 + αIλ3 (λ2 + λ3) + α(λ4 − λ3)[λ2 + λ4 + Iλ3] = 0 (3).

Substituting α = (D+1)λ2
λ2−λ3 in (1), we get (λ2+λ4)(λ2+Dλ4)(λ3+Dλ2)D(λ2−λ3) = 0.

This implies that λ2 +Dλ4 = 0 or λ3 +Dλ2 = 0.

If λ2 + Dλ4 = 0 then D−1λ2λ3 + λ3λ4 = 0, which is equivalent to γ2 + λ3λ4 = 0, a
contradiction,

If λ3 +Dλ2 = 0 then γ2 + λ22 = 0 (Lemma 10), a contradiction.

Case 13 : Let αe1 + βe2 + e4 ∈ S, it follows that

A12(αe1 + βe2 + e4)− λ24(αe1 + βe2 + e4) ∈ S.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 697

Then


N1

N2

N3

0

 ∈ S.
Here, the constants are given by

N1 = α(λ21 − λ24) + βJλ2 (λ1 + λ2) + λ4[λ1 + λ4 + J (λ2 + λ3)],
N2 = β(λ22 − λ24) + λ4[λ2 + λ4 + Iλ3],
N3 = λ4 (λ3 + λ4) .

If N1 6= 0 or N2 6= 0 or N3 6= 0,
then we get a contradiction (Case 1, Case 2, Case 4, Case 5, Case 6, Case 9, and Case 11).

Otherwise, if N3 = 0 then λ4 (λ3 + λ4) = 0, a contradiction.

Case 14: Let e1 + αe3 + βe4 ∈ S, it follows that

A23(e1 + αe3 + βe4)− λ24(e1 + αe3 + βe4) ∈ S.

Then


0
R2

R3

R4

 ∈ S.
Here, the constants are given by

R2 = −λ3 (λ3 + λ4) ,
R3 = λ2[Dλ4 + (D + 1)λ3 +Dλ2] + α(λ22 − λ24),
R4 = L+ αM + β(λ21 − λ24).

If R2 6= 0 or R3 6= 0 or R4 6= 0 then we get a contradiction (Case 2, Case 3, Case 4,
Case 7,Case 9, Case 10, and Case 12).

Otherwise, if R2 = 0 then −λ3 (λ3 + λ4) = 0, a contradiction.

Case 15: Let αe1 + βe2 + δe3 + e4 ∈ S, it follows that

A12(αe1 + βe2 + δe3 + e4)− λ24(αe1 + βe2 + δe3 + e4) ∈ S.



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 698

.

Then


N1

N2

N3

0

 ∈ S.
Here, the constants are given by

N1 = α(λ21 − λ24) + βJλ2 (λ1 + λ2) + δJλ3[λ1 + λ3 + Iλ2] + λ4[λ1 + λ4 + J (λ2 + λ3)],
N2 = β(λ22 − λ24) + δIλ3 (λ2 + λ3) + λ4[λ2 + λ4 + Iλ3],
N3 = δ(λ23 − λ24) + λ4 (λ3 + λ4) .

If N1 6= 0 or N2 6= 0 or N3 6= 0 then we get a contradiction (Case 1, Case 2, Case 4,
Case 5, Case 6, Case 9, and Case 11).

Otherwise, if N3 = 0 then δ(λ3 − λ4) + λ4 = 0. This implies that δ = λ4
λ4−λ3 .

Substituting δ = λ4
λ4−λ3 in the equation N2(λ4 − λ3) = 0, we get

β(λ22 − λ24)(λ4 − λ3) + λ4Iλ3 (λ2 + λ3) + λ4(λ4 − λ3)[λ2 + λ4 + Iλ3] = 0.

So β(λ22 − λ24)(λ4 − λ3) + λ4Iλ3(λ2 + λ4) + λ4(λ4 − λ3)(λ2 + λ4) = 0.

This implies that (λ2 + λ4)[β(λ2 − λ4)(λ4 − λ3) + λ4(λ4 + λ3D
−1)] = 0.

Hence β(λ2 − λ4)(λ4 − λ3) + λ4(λ4 + λ3D
−1) = 0.

In the case (λ2 − λ4)(λ4 − λ3) = 0, we get λ4 + λ3D
−1 = 0.

This implies that λ2λ4 + λ2λ3D
−1 = 0, which is equivalent to γ2 + λ2λ4 = 0,

a contradiction. Hence β = λ4(λ4+λ3D−1)
(λ2−λ4)(λ3−λ4) .

Let us substitute δ = λ4
λ4−λ3 , and β = λ4(λ4+λ3D−1)

(λ2−λ4)(λ3−λ4) in the equation

N1(λ2 − λ4)(λ3 − λ4) = 0. It follows that

αf + λ4(λ4 + λ3D
−1)Jλ2 (λ1 + λ2)− λ4(λ2 − λ4)Jλ3[λ1 + λ3 + Iλ2] + g = 0.

Then

αf +D−3λ4 (λ1 + λ4) (λ2λ3 +D3λ24 +Dλ2λ4 +Dλ3λ4 +D2λ2λ4 +D2λ3λ4) = 0.(4)

Here, the constants are f and g are given by

f = (λ21 − λ24)(λ2 − λ4)(λ3 − λ4),



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 699

g = λ4[λ1 + λ4 + J (λ2 + λ3)](λ2 − λ4)(λ3 − λ4).

In the case f = 0, we get λ1 = λ4. Let us substitute λ1 = λ4 in (4), we getD−3λ4 (λ1 + λ4) (λ2λ3+
D3λ24 +Dλ2λ4 +Dλ3λ4 +D2λ2λ4 +D2λ3λ4) = 0.

This implies that λ4(λ2λ3 +D3λ24 +Dλ2λ4 +Dλ3λ4 +D2λ2λ4 +D2λ3λ4) = 0.

But D3 = Dλ2λ3
λ24

, for λ1 = λ4.

So λ2λ3 +Dλ2λ3 +Dλ2λ4 +Dλ3λ4 +D2λ2λ4 +D2λ3λ4 = 0.

Then (D + 1)(λ2λ3 +Dλ2λ4 +Dλ3λ4) = 0. Hence λ2λ3 +Dλ2λ4 +Dλ3λ4 = 0.

This implies that D−1λ2λ3 + λ2λ4 + λ3λ4 = 0.

It follows that γ2 + λ2λ4 + λ3λ4 = 0, a contradiction. Thus f 6= 0.

Now, (4) implies that α =
λ4(λ2λ3+D3λ24+Dλ2λ4+Dλ3λ4+D

2λ2λ4+D2λ3λ4)
D3(λ1−λ4)(λ2−λ4)(λ4−λ3) .

On the other hand, A23(αe1 + βe2 + δe3 + e4)− λ24(αe1 + βe2 + δe3 + e4) ∈ S.

Then


0
R2

R3

R4

 ∈ S.
Here, the constants are given by

R2 = −αλ3 (λ3 + λ4) + β(λ23 − λ24),
R3 = αλ2[Dλ4 + (D + 1)λ3 +Dλ2]− β(D + 1)λ2 (λ2 + λ3) + δ(λ22 − λ24),
R4 = αL+ βK + δM + λ21 − λ24.

If R2 6= 0 or R3 6= 0 or R4 6= 0 then we get a contradiction (Case 2, Case 3, Case 4,
Case 7, Case 9, Case 10, and Case 12).

Otherwise, if R2 = 0 then −αλ3 + β(λ3 − λ4) = 0.

Hence α = β(λ3−λ4)
λ3

, and we have β = λ4(λ4+λ3D−1)
(λ2−λ4)(λ3−λ4) .

Thus α = λ4(λ4+λ3D−1)
λ3(λ2−λ4) . Also, α =

λ4(λ2λ3+D3λ24+Dλ2λ4+Dλ3λ4+D
2λ2λ4+D2λ3λ4)

D3(λ1−λ4)(λ2−λ4)(λ4−λ3) .



H. A. Haidar, M. N. Abdulrahim / Eur. J. Pure Appl. Math, 11 (3) (2018), 682-701 700

So −λ4(λ2λ3+D3λ24+Dλ2λ4+Dλ3λ4+D
2λ2λ4+D2λ3λ4)

D3(λ1−λ4)(λ2−λ4)(λ4−λ3) + λ4(λ4+λ3D−1)
λ3(λ2−λ4) = 0.

This implies that

λ2λ23+D
3λ34+D

2λ1λ23−D3λ1λ24+D
2λ3λ24+Dλ

2
3λ4−D2λ1λ3λ4+D3λ1λ3λ4+(D+1)Dλ2λ3λ4

D3λ3(λ1−λ4)(λ2−λ4)(λ3−λ4) = 0.

It’s easy to see that, λ2λ
2
3 −D2λ1λ3λ4, so we get h = 0, where

h = D3λ34 +D2λ1λ
2
3 −D3λ1λ

2
4 +D2λ3λ

2
4 +Dλ23λ4 +D3λ1λ3λ4 + (D + 1)Dλ2λ3λ4.

Now, we multiply the equation h = 0 by λ1λ4
D , we get

λ2λ3λ
3
4 + λ1λ

2
3γ

2 − λ4γ4 + λ3λ
2
4γ

2 + λ1λ
2
3λ

2
4 + λ3γ

4 + λ2λ3λ4γ
2 + λ4γ

4 = 0.

Hence λ3(γ
2 + λ24)(γ

2 + λ1λ3 + λ2λ4) = 0, a contradiction.

Therefore, there is no non-zero invariant proper subspace. This implies that we have
determined sufficient condition under which the representation ϕ : P3 −→ GL4(C) is
irreducible.

If we require further the conditions λ1 = λ3 and λ2 = λ4, we get a necessary and
sufficient condition for the irreducibility of ϕ : P3 −→ GL4(C).

Corollary 1. Let ϕ : P3 −→ GL4(C) be the complex specialization of Tuba’s representa-
tion of the pure braid group P3. Assume that λ1 = λ3, λ2 = λ4 and λ1 6= −λ2.
Then ϕ is irreducible if and only if (γ2 + λ2r)(γ

2 + λrλl + λsλk) 6= 0, where
{r, s, l, k} = {1, 2, 3, 4}.

Proof. Let us show that if (γ2+λ2r)(γ
2+λrλl+λsλk) = 0, where {r, s, l, k} = {1, 2, 3, 4},

then the representation ϕ is reducible.

Assume that (γ2 + λ2r)(γ
2 + λrλl + λsλk) = 0, where {r, s, l, k} = {1, 2, 3, 4}, then the

reducibility on P3 follows from reducibility on B3 (see Proposition 6).

Now, let us show that if (γ2 + λ2r)(γ
2 + λrλl + λsλk) 6= 0, where {r, s, l, k} = {1, 2, 3, 4},

then ϕ is irreducible.

Given that λ1 = λ3, λ2 = λ4, and λ1 6= −λ2. In this case, γ2 = λ1λ2. Hence, it’s
easy to verify that all the conditions of Theorem 11 are satisfied: For instance, the second
condition of Theorem 11 is equivalent to λ1 6= −λ2. Also, the third condition is equivalent



REFERENCES 701

to λ1 6= −λ2.
Therefore, by Theorem 11, ϕ is irreducible.

Note that, provided that λ1 = λ3 and λ2 = λ4, we have to require λ1 6= −λ2 in order
for the matrices of the generators of B3 not to be constant matrices.

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