On the Irreducibility of Perron Representations of Degrees 4 and 5

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1 EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. No ISSN Published by New York Business Global On the Irreducibility of Perron Representations of Degrees 4 and 5 Malak M. Dally Mohammad N. Abdulrahim Department of Mathematics Faculty of Science Beirut Arab University P.O. Box -52 Beirut Lebanon Abstract. We consider the graph E n+ with (n+) generators σ... σ n and δ where σ i has an edge with σ i+ for i =... n+ and σ has an edge with δ. We then define the Artin group of the graph E n+ 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. 2 Mathematics Subject Classifications: 2F36 Key Words and Phrases: irreducibility Artin representation braid group Burau representation graph. 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 A n having n vertices σ i s ( i n ) in which σ i and σ i+ share a comon edge where i = 2... n. Indeed the Artin group of A n denoted by A(A n ) is the braid group on n+ strands B n+. That is A(A n ) = B n+. From the graph A n we obtain the graph E n+p by adding a vertex δ and an edge connecting σ p and δ. Here p n. Clearly the graph A n embeds in the graph E n+p. Consequently A(A n ) A(E n+p ). As a result a representation of A(E n+p ) yields a representation of B n+. Perron s strategy is to begin with the reduced Burau representation of B n+ of degree n and extend it to a representation of B n+ of degree 2n. The representation obtained is referred to as Burau bis representation. Next Perron constructs for each λ = (λ... λ n ) Corresponding author. addresses: malakdally@hotmail.com (M. Dally) mna@bau.edu.lb (M. Abdulrahim) 25 c 28 EJPAM All rights reserved.

2 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) a representation ψ λ : A(E n+p ) GL 2n (Q(t d... d n )) where t d... d n λ... λ n are indeterminates. In [3] we determined necessary and sufficient condition that guarantees the irreducibility of the representation ψ λ for n = 2. In our work we extend our work to n = 3 and n = 4. We reduce the complex specialization of the representation ψ λ to representations of A(E 4 ) and A(E 5 ) 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 B n+ of degree 2n. It is defined as follows: ψ : B n+ Gl 2n (Z[t t ]) ( ) In ψ(σ i ) = i n R i J i Here R i denotes an n n block of zeros with a t placed in the (i i) th position and I n denotes the n n identity matrix. This representation was constructed by Perron by extending the reduced Burau representation of degree n to a representation of B n+ of degree 2n. The reduced Barau representation B n+ GL n (Z[t t ]) is defined as follows: I i 2 σ i J i = t t I n i where I k stands for the k k identity matrix. Here i = 2... n. σ J = t I n 2 σ n J n = I n 2 t t

3 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) For more details see [2] and [5]. 3. Perron Representation The Burau bis representation extends to A(E n+p ) for all possible values of n and p in the following way. We define the following n n matrices: where denotes a column of n zeros b = A = (λ b λ 2 b... λ n b) B = (... b... ) C = (λ d λ 2 d... λ n d) D = (... d... ). b n d = d. d n and λ = (λ... λ n ). For each i =... n we have that b i satisfies the following conditions tb i = td i + ( + t)d i d i+ i p tb p = td p + ( + t)d p d p+ + t n λ i b i = ( + d p + t) i= setting any undefined d j equal zero. For any choice λ = (λ... λ n ) we get a linear representation ψ λ : A(E n+p ) Gl 2n (R) where R is the field of rational fractions in n+ indeterminates Q(t d... d n ). ( ) In ψ λ (σ i ) R i J i For more details see [2]. ( ) In + A B ψ λ (δ). C I n + D

4 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) Reducibility of ψ λ : A(E 4 ) GL 6 (C) Having defined Perron s representation we set n = 3 and p = to get the following vectors. b = b 2 d = d 2 and λ = (λ λ 2 λ 3 ). b 3 d 3 d After we specialize the indeterminate d 3 to t(+t2 ) we get the following 3 3 matrices: λ λ 2 λ 3 A = λ b 2 λ 2 b 2 λ 3 b 2 λ b 3 λ 2 b 3 λ 3 b 3 B = b 2 b 3 λ d λ 2 d λ 3 d C = λ d 2 λ 2 d 2 λ 3 d 2 t(+t 2 ) t(+t λ 2 ) t(+t λ 2 ) 2 λ 3 and d D = d 2. t(+t 2 ) Simple computations show that the parameters satisfy the following equations: tb 2 = td + ( + t)d 2 + t(+t2 ) tb 3 = td 2 t( + t + t 2 ) t = ( + t)d d 2 + t λ + λ 2 b 2 + λ 3 b 3 = ( + t + d ) Having defined the 3 3 matrices A B C and D we obtain the multiparameter representation A(E 4 ). This representation is of degree 6. We specialize the parameters λ λ 2 λ 3 b 2 b 3 d d 2 t to values in C {}. We further assume that t. The representation ψ λ : A(E 4 ) GL 6 (C) is defined as follows:

5 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) ψ λ (σ ) = t t ψ λ (σ 2 ) = t t t ψ λ (σ 3 ) = t t t and + λ λ 2 λ 3 λ b 2 + λ 2 b 2 λ 3 b 2 b 2 λ b 3 λ 2 b 3 + λ 3 b 3 b 3 ψ λ (δ) = λ d λ 2 d λ 3 d + d. λ d 2 λ 2 d 2 λ 3 d 2 d 2 t(+t 2 ) t(+t λ 2 ) t(+t λ 2 ) t(+t 2 λ 2 ) 3 The graph E 4 has 4 vertices σ σ 2 σ 3 and δ. Since p = it follows that the vertex δ has a common edge with σ p = σ. Therefore the following relations are satisfied. σ σ 2 σ = σ 2 σ σ 2 (4.) σ 2 σ 3 σ 2 = σ 3 σ 2 σ 3 (4.2) σ σ 3 = σ 3 σ (4.3) σ 2 δ = δσ 2 (4.4)

6 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) σ 3 δ = δσ 3 (4.5) σ δσ = δσ δ (4.6) We note that relations (4.)(4.2) and (4.3) are actually Artin s braid relation of the classical braid group B 4 having σ σ 2 and σ 3 as standard generators. This assures that a representation of A(E 4 ) yields a representation of B 4. For more details see [] and [4]. Lemma. The representation ψ λ : A(E 4 ) GL 6 (C) is reducible. Proof. For simplicity we write σ i instead of ψ λ (σ i ).The subspace S = e + b 2 e 3 e 4 e 5 e 6 is an invariant subspace of dimension 4. To see this: (i) σ (e + b 2 e 3 ) = e + b 2 e 3 + te 4 S (ii) σ 2 (e + b 2 e 3 ) = e + b 2 e 3 + t b 2 e 5 S (iii) σ 3 (e + b 2 e 3 ) = e + b 2 e 3 + t b 3 e 6 S (iv) δ(e + b 2 e 3 ) = ( + λ + λ 2 b 2 + λ 3 b 3 )(e + b 2 e 3 ) + d (λ + λ 2 b 2 + λ 3 b 3 )e 4 + d 2 (λ + λ 2 b 2 + λ 3 b 3 )e 5 + (+t2 ) (λ + λ 2 b 2 + λ 3 b 3 )e 6 S (v) σ e 4 = te 4 S (vi) σ 2 e 4 = e 4 + te 5 S (vii) σ 3 e 4 = e 4 S () (viii) δe 4 = (e + b 2 e 3 ) + ( + d )e 4 + d 2 e 5 t(+t 2 ) e 6 S (ix) σ e 5 = e 4 + e 5 S (x) σ 2 e 5 = te 5 S (xi) σ 3 e 5 = e 5 + te 6 S (xii) δe 5 = e 5 S (xiii) σ e 6 = e 6 S (xiv) σ 2 e 6 = e 5 + e 6 S (xv) σ 3 e 6 = te 6 S

7 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) (xvi) δe 6 = e 6 S 5. On the Irreducibility of ψ λ : A(E 4) GL 4 (C) We consider the representation ψ λ : A(E 4 ) GL 6 (C) restricted to the basis e e 2 e + b 2 e 3 e 4 e 5 and e 6. The matrix of σ becomes t ψ λ (σ ) = t t. We reduce our representation to a 4-dimensional one by considering the sub-basis e + b 2 e 3 e 4 e 5 and e 6 to get ψ λ : A(E 4) GL 4 (C). The representation is defined as follows: t ψ λ (σ ) = t tb 2 ψ λ (σ 2) = t t tb 3 ψ λ (σ 3) = t t and ψ λ (δ) =

8 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) i= λ d i i ( 3 i= λ ib i ) d 2 ( 3 i= λ t(+t ib i ) ) () ( 3 + d d 2 t( + t + t 2 ) i= λ ib i ). We then diagonalize the matrix corresponding to ψ λ (σ ) by an invertible matrix say T and conjugate the matrices of ψ λ (σ 2) ψ λ (σ 3) and ψ λ (δ) by the same matrix T. The invertible matrix T is given by t T = t. In fact a computation shows that T ψ λ (σ )T =. t After conjugation we get t 2 T ψ λ (σ 2)T = t(b 2 + t+b 2 t) () (+t 2 ) t(b 2 + t+b 2 t) () t t T ψ λ (σ 3)T = t tb 3 and

9 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) T ψ λ (δ)t = t(+t 2 ) + d () 2 2 t(+t 2 ) () k t(+t 2 ) d 2 k d 2 where k = t + 3 i= λ ib i. + k t ( d ()+d 2 + t)(( 3 i= λ ib i )()+ t) () The entries of the matrices T ψ λ (σ 2)T and T ψ λ (δ)t are well-defined since we assume in our work that t. For simplicity we denote T ψ λ (σ )T by ψ λ (σ ) T ψ λ (σ 2)T by ψ λ (σ 2) T ψ λ (σ 3)T by ψ λ (σ 3) and T ψ 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(E 4) GL 4 (C). Lemma 2. The proper subspace S = t 4 + t 3 + t 2 + t +. e e 4 e 3 Proof. First we prove that proper subspace S = t 4 + t 3 + t 2 + t +. Assume for contradiction that S is invariant. We have ψ λ (σ 2)(e 4 ) = (+t 2 ) t(b 2 +b 2 t+ t) () S. This implies that ( + t + t 2 )b 3 = t(b 2 + tb 2 + t ).. is not invariant if and only if e e 4 e 3 is not invariant if By using the equations: tb 2 = td +()d 2 + t(+t2 ) tb 3 = td 2 t(+t 2 ) and t = ( + t)d d 2 + t simple computations give t 4 + t 3 + t 2 + t + = a contradiction. On the other hand we assume that t 4 + t 3 + t 2 + t + =. We prove that the proper subspace S = e e 4 e 3 is invariant as follows:

10 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) (i) ψ λ σ (e ) = e S. (ii) ψ λ σ 2(e ) = e S. t (iii) ψ λ σ t 3(e ) = tb 3 S. (iv) ψ λ δ(e ) = t 2 (+t 2 ) () 2 + t(+t2 ) t(+t 2 ) () 2 () (λ + λ 2 b 2 + λ 3 b 3 ) t(+t 2 ) () 2 Here we have a = b = d 3 c = d 3 and cb 3 = t2 (+t 2 ) + t(+t2 ) () 2 () (λ + λ 2 b 2 + λ 3 b 3 ). Thus b 3 = t+()(λ +λ 2 b 2 +λ 3 b 3 ). (5.) (v) ψ λ σ (e 4 ) = te 4 S. (vi) ψ λ σ 2(e 4 ) = (+t 2 ) t(b 2 +b 2 t+ t) () = ae + be 4 + c( e 3 ). = ae +be 4 +c( e 3 ). Here we have a = b = c = (+t2 ) and cb 3 = t(b 2+b 2 t+ t) (). Thus ( + t + t 2 ) b 3 = t(b 2+b 2 t+ t). (5.2) (vii) ψ λ σ 3(e 4 ) = e 4 S.

11 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) t (viii) ψ λ δ(e 4) = = ae ( d ()+d 2 + t)((λ +λ 2 b 2 +λ 3 b 3 )()+ t) + be 4 + c( e 3 ). () Here we have a = b = c = t and cb 3 = ( d ()+d 2 + t)((λ +λ 2 b 2 +λ 3 b 3 )()+ t) (). Thus b 3 = ( d ()+d 2 + t)((λ +λ 2 b 2 +λ 3 b 3 )()+ t) t. (5.3) (ix) ψ λ σ ( e 3 ) = e 3 S. (x) ψ λ σ 2( e 3 ) = t 2 t(b 2 +b 2 t+ t) () + b 3 t = ae + be 4 + c( e 3 ). Here we have a = b = t c = t2 and cb 3 = t(b 2+b 2 t+ t) () + b 3. Thus ( + t + t 2 ) b 3 = t(b 2+b 2 t+ t). (5.4) (xi) ψ λ σ 3( e 3 ) = e 3 S. (xii) ψ λ δ( e 3 ) = + d 2 + b 3 d 2 (λ + λ 2 b 2 + λ 3 b 3 + t ) + b 3 ( + λ + λ 2 b 2 + λ 3 b 3 + t = ae + be 4 + c( e 3 ). d 2 b 3 )

12 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) Here we have a = b = d 2 b 3 c = + d 2 + b 3 and c b 3 = d 2 (λ + λ 2 b 2 + λ 3 b 3 + t ) + b 3 ( + λ + λ 2 b 2 + λ 3 b 3 + t ). Thus (+ d 2 (5.5) + b 3 ) b 3 = d 2 (λ +λ 2 b 2 +λ 3 b 3 + t )+ b 3 (+λ +λ 2 b 2 +λ 3 b 3 + t ). By simple computations we can verify that equations (5.) (5.2) (5.3) and (5.5) are clearly satisfied without any assumption of the indeterminates whereas equation (5.4) is satisfied only if t 4 + t 3 + t 2 + t + =. Lemma 3. Any proper subspace S containing the vector e i + ue j + ve k where i j k { 2 3 4} except possibly the subspace having the form e e 4 e 3 is not invariant. Proof. We consider all the subspaces containing the vector e i + ue j + ve k where i j k { 2 3 4} except possibly the subspace of the form e e 4 e 3. 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. Assume all the indeterminates used in defining Perron representation of degree 4 are non zero complex numbers. Let d 3 = t(+t2 ) and t. The representation ψ λ : A(E 4) GL 4 (C) is irreducible if and only if t 4 + t 3 + t 2 + t +. In the following sections we set n = 4 and p = and we study the irreducibility of the reduced representation of ψ λ : A(E 5 ) GL 8 (C). Indeed we obtain a sufficient and necessary condition that gauarantees the irreducibility of ψ λ : A(E 5) GL 5 (C). 6. Reducibility of ψ λ : A(E 5 ) GL 8 (C) Having defined Perron s representation we set n = 4 and p = to get the following d vectors. b = b 2 b 3 d = d 2 d 3 and λ = (λ λ 2 λ 3 λ 4 ). b 4 d 4

13 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) After we specialize the indeterminates d 2 and d 3 to ( + t + t 2 ) and t( + t) respectively we get the following 4 4 matrices: λ λ 2 λ 3 λ 4 A = λ b 2 λ 2 b 2 λ 3 b 2 λ 4 b 2 λ b 3 λ 2 b 3 λ 3 b 3 λ 4 b 3 λ b 4 λ 2 b 4 λ 3 b 4 λ 4 b 4 B = b 2 b 3 b 4 λ d λ 2 d λ 3 d λ 4 d C = ( + t + t 2 )λ ( + t + t 2 )λ 2 ( + t + t 2 )λ 3 ( + t + t 2 )λ 4 t( + t)λ t( + t)λ 2 t( + t)λ 3 t( + t)λ 4 λ d 4 λ 2 d 4 λ 3 d 4 λ 4 d 4 and d D = ( + t + t 2 ) t( + t). d 4 Simple computations show that the parameters satisfy the following equations: tb 2 = td ( + t)( + t + t 2 ) + t( + t) = td ( + t)( + t 2 ) tb 3 = t( + t + t 2 ) t( + t) 2 d 4 = t(2t 2 + 3t + 2) d 4 tb 4 = t 2 ( + t) + ( + t)d 4 t = ( + t)d + + 2t + t 2 λ + λ 2 b 2 + λ 3 b 3 + λ 4 b 4 = ( + t + d ) Having defined the 4 4 matrices A B C and D we obtain the multiparameter representation A(E 5 ). This representation is of degree 8. We specialize the parameters λ λ 2 λ 3 λ 4 b 2 b 3 b 4 d d 4 t to values in C {}. We further assume that t. The representation ψ λ : A(E 5 ) GL 8 (C) is defined as follows:

14 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) ψ λ (σ ) = t t ψ λ (σ 2 ) = t t t ψ λ (σ 3 ) = t t t ψ λ (σ 4 ) = t t and

15 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) ψ λ (δ)= + λ λ 2 λ 3 λ 4 λ b 2 + λ 2 b 2 λ 3 b 2 λ 4 b 2 b 2 λ b 3 λ 2 b 3 + λ 3 b 3 λ 4 b 3 b 3 λ b 4 λ 2 b 4 λ 3 b 4 + λ 4 b 4 b 4 λ d λ 2 d λ 3 d λ 4 d + d kλ kλ 2 kλ 3 kλ 4 k t( + t)λ t( + t)λ 2 t( + t)λ 3 t( + t)λ 4 t( + t) λ d 4 λ 2 d 4 λ 3 d 4 λ 4 d 4 d 4 where k = ( + t + t 2 ). The graph E 5 has 5 vertices σ σ 2 σ 3 σ 4 and δ. Since p = it follows that the vertex δ has a common edge with σ p = σ. Therefore the following relations are satisfied. σ σ 2 σ = σ 2 σ σ 2 (6.) σ 2 σ 3 σ 2 = σ 3 σ 2 σ 3 (6.2) σ 3 σ 4 σ 3 = σ 4 σ 3 σ 4 (6.3) σ σ 3 = σ 3 σ (6.4) σ σ 4 = σ 4 σ (6.5) σ 2 σ 4 = σ 4 σ 2 (6.6) σ 2 δ = δσ 2 (6.7) σ 3 δ = δσ 3 (6.8) σ 4 δ = δσ 4 (6.9) σ δσ = δσ δ (6.) We note that relations (6.)(6.2) (6.3) (6.4) (6.5) and (6.6) are actually Artin s braid relation of the classical braid group B 5 having σ σ 2 σ 3 and σ 4 as standard generators. This assures that a representation of A(E 5 ) yields a representation of B 5. For more details see [] and [4]. Lemma 4. The representation ψ λ : A(E 5 ) GL 8 (C) is reducible. Proof. For simplicity we write σ i instead of ψ λ (σ i ).The subspace S = e + b 2 e 3 + b 4 e 4 e 5 e 6 e 7 e 8 is an invariant subspace of dimension 5.

16 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) On the Irreducibility of ψ λ : A(E 5) GL 5 (C) We consider the representation ψ λ : A(E 5 ) GL 8 (C) restricted to the basis e e 2 e 3 e + b 2 e 3 + b 4 e 4 e 5 e 6 e 7 and e 8 to get the subrepresentation ψ λ : A(E 5) GL 5 (C) which is the representation restricted to the sub-basis e + b 2 e 3 + b 4 e 4 e 5 e 6 e 7. This representation is defined as follows: t ψ λ (σ t ) = tb 2 ψ λ (σ t 2) = t tb 3 ψ λ (σ 3) = t t tb 4 ψ λ (σ 4) = t t and ψ λ (δ) =

17 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) where r = 4 i= λ ib i. + r d r (+t 2 ) r t() r d 4 r + d ( + t + t 2 ) t( + t) d 4 We then diagonalize the matrix corresponding to ψ λ (σ ) by an invertible matrix say T and conjugate the matrices of ψ λ (σ 2) ψ λ (σ 3)ψ λ (σ 4) and ψ λ (δ) by the same matrix T. The invertible matrix T is given by t t T =. In fact a computation shows that T ψ λ (σ )T =. t After conjugation we get t 2 T ψ λ (σ 2)T = t(b 2 + t+b 2 t) () (+t 2 ) t(b 2 + t+b 2 t) () t

18 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) T ψ λ (σ t 3)T = t tb 3 t T ψ λ (σ t 4)T = tb 4 and T ψ λ (δ)t = d 4 t + (+t2 ) d 4 w t() w t w + w ( d () (+t 2 )+ t) w (+t 2 ) d 4 +t t 2 where w = 4 i= λ ib i + t. The entries of the matrices T ψ λ (σ 2)T T ψ λ (σ 3)T T ψ λ (σ 4)T and T ψ λ (δ)t are well-defined since we assume in our work that t. For simplicity we denote T ψ λ (σ )T by ψ λ (σ ) T ψ λ (σ 2)T by ψ λ (σ 2) T ψ λ (σ 3)T by ψ λ (σ 3) T ψ λ (σ 4)T by ψ λ (σ 4) and T ψ 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(E 5) GL 5 (C). Lemma 5. Except possibly the subspaces having the forms e e 3 e 5 e 2 + ue 4 and e 2 e 3 e 5 e + ue 4 where u C every proper subspace is not invariant.

19 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) Proof. We assume for contradiction that every subspace except those having the forms e e 3 e 5 e 2 + ue 4 and e 2 e 3 e 5 e + ue 4 is invariant. We then study each possible form. In each case simple computations give a contradiction. Lemma 6. If t 3 then the subspaces e e 3 e 5 e 2 + ue 4 and e 2 e 3 e 5 e + ue 4 are not invariant. Proof. First we assume for contradiction that S = e e 3 e 5 e 2 + ue 4 is invariant. t t ψ λ σ 4(e ) = tb 4 = ae + be 3 + ce 5 + d(e 2 + ue 4 ) which implies that u = b 4. (7.) ψ λ σ 2(e 3 ) = t 2 t( t+b 2 +b 2 t) () t u = t(t+b 2 +b 2 t) (). (7.2) = ae + be 3 + ce 5 + d(e 2 + ue 4 ) which implies that ψ λ σ 3(e 2 + ue 4 ) = t t u + tb 3 u = tb 3 (). (7.3) = ae + be 3 + ce 5 + d(e 2 + ue 4 ) which implies that Since equations (7.) and (7.3) are equal we have ( + t + t 2 )d 4 = t 2 ( + t + t 2 ). Thus d 4 = t 2. Moreover equations (7.2) and (7.3) are equal. This implies that d 4 = ( 2 ) 2 +t 2. By substituting d 4 = t 2 we get t 4 + t 3 + t + = (t + )(t 3 + ) = a contradiction. Now we assume for contradiction that S = e 2 e 3 e 5 e + ue 4 is invariant.

20 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) ψ λ σ 2(e 3 ) = t 2 t( t+b 2 +b 2 t) () t = ae 2 + be 3 + ce 5 + d(e + ue 4 ). This implies that t( t + b 2 + b 2 t) =. Simple computations give ( + t + t 2 ) 2 + t( + t) 2 + t 2 =. Thus t 4 + t 3 + t + = 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 S = e e 3 e 5 e 2 + ue 4 and S 2 = e 2 e 3 e 5 e + ue 4 cannot be both invariant. Proof. Assume that S is invariant. This implies that ψ λ σ 2(e 3 ) and ψ λ σ 3(e 2 + ue 4 ) S. Simple computations give t + b 2 + b 2 t = b 3. Assume for contradiction that S 2 is invariant. This implies that ψ λ σ 2(e 3 ) S 2. Simple computations give t + b 2 + b 2 t = a contradiction. Lemma 8. If t 3 = then the subspace S = e 2 e 3 e 5 e + ue 4 is invariant. Proof. ψ λ σ (e 2 )=ψ λ σ 2(e 2 )=ψ λ σ 4(e 2 )=e 2 S. t ψ λ σ 3(e 2 ) = t tb 3 = ae 2 + be 3 + ce 5 + d(e + ue 4 ) if u = tb 3. (7.4)

21 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) d 3 ψ λ δ(e 2) = d 3 t + d 3 ( = ae 2 + be 3 + ce 5 + d(e + ue 4 ) 4 i= λ ib i ) d 3 t if = 4 i= λ ib i. (7.5) ψ λ σ (e 3 )=ψ λ σ 3(e 3 )=ψ λ σ 4(e 3 )=e 3 S. ψ λ σ 2(e 3 ) = t 2 t( t+b 2 +b 2 t) () t if t( t + b 2 + b 2 t) =. (7.6) = ae 2 + be 3 + ce 5 + d(e + ue 4 ) ψ λ δ(e 3) = if 4 i= λ ib i + t ψ λ σ (e 5 ) = te 5 S. + (+t2 ) t 2 (+t 2 ) + (+t2 ) ( 4 +t 2 =. (7.7) = ae 2 + be 3 + ce 5 + d(e + ue 4 ) i= λ ib i ) ψ λ σ 3(e 5 )=ψ λ σ 4(e 5 )=e 5 S.

22 M. Dally M. Abdulrahim / Eur. J. Pure Appl. Math () (28) ψ λ σ 2(e 5 ) = (+t 2 ) t( t+b 2 +b 2 t) () if t( t + b 2 + b 2 t) =. (7.8) ψ λ δ(e 5) = = ae 2 + be 3 + ce 5 + d(e + ue 4 ) t ( d () t(+t 2 )+b 2 t)( 4 i= λ ib i ()+ t) () if ( d ( + t) + d 2 + b 2 t)( 4 i= λ ib i ( + t) + t) =. (7.9) ψ λ σ (e + ue 4 )=ψ λ σ 2(e + ue 4 )=ψ λ σ 3(e + ue 4 )=e + ue 4 S. t ψ λ σ t 4(e + ue 4 ) = u + tb 4 = ae 2 + be 3 + ce 5 + d(e + ue 4 ) if u = tb 4 (). (7.) = ae 2 +be 3 +ce 5 +d(e +ue 4 ) ψ λ δ(e + ue 4 ) = ce 5 + d(e + ue 4 ) d 4 + u d 4 ( 4 i= λ ib i + t ) + u( + 4 i= λ ib i + t d 4 u ) = ae 2 + be 3 + if ( 4 i= λ ib i + t )( d 4 + u) =. (7.)

23 REFERENCES 237 Using the relations we prove that equations (7.5) (7.7) (7.9) and (7.) are clearly satisfied. Also we verify that equations (7.4) (7.6) (7.8) and (7.) are satisfied if t( + t) 2 = ( + t + t 2 )( + t 2 ) + t 2 which implies that t 3 =. 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 d 2 = ( + t + t 2 ) d 3 = t( + t) and t. The representation ψ λ : A(E 5) GL 5 (C) is irreducible if and only if t 3. Remark. For n=2 and for t we proved that a complex specialization of the representation ψ λ : A(E 3) Gl 3 (C) is irreducible if and only if t 2 which is equivalent to t 3 + t 2 + t +. For n=3 and for t we have proved that a complex specialization of the representation ψ λ : A(E 4) Gl 4 (C) is irreducible if and only if t 4 + t 3 + t 2 + t +. For n=4 and for t we have proved that a complex specialization of the representation ψ λ : A(E 5) Gl 5 (C) is irreducible if and only if t 3 which is equivalent to t 5 + t 4 + t 3 + t 2 + t +. References [] J. S. Birman Braids Links and Mapping Class Groups. Annals of Mathematical Studies. Princeton University Press volume 82 New Jersey 975. [2] T.E. Brendle The Torelli Group and Representations of Mapping Class Groups. Doctoral Thesis Columbia University 22. [3] M. Dally and M. Abdulrahim On the Irreducibility of Artin s Group of Graphs. Journal of Mathematics Research volume 7 number [4] V.L. Hansen Braids and Coverings. London Mathematical Society Cambridge University Press 989. [5] B. Perron A linear representaton of a finite rank of the mapping class group of surfaces (preprint). Laboratoire de Topologie volume 99 number

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