On the Irreducibility of Artin ’ s Group of Graphs

We consider the graph E3,1 with three generators σ1, σ2, δ, where σ1 has an edge with each of σ2 and δ. We then define the Artin group of the graph E3,1 and consider its reduced Perron representation of degree three. After we specialize the indeterminates used in defining the representation to non-zero complex numbers, we obtain a necessary and sufficient condition that guarantees the irreducibility of the representation.


Introduction
To any undirected simple graph T , we introduce the Artin group, A, which is defined as an abstract group with vertices of Γ as its generators and 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.
Let A n be the graph having n vertices σ i 's (1 ≤ i ≤ n ) in which σ i and σ i+1 share a comon edge, where i = 1, 2, ..., n − 1.We notice that the Artin group of A n is the braid group on n + 1 strands.That is, A(A n ) = B n+1 (J.S. Birman, 1975).
Having defined A n , we consider E n+1,p , which is the graph obtained from A n by adding a vertex δ and an edge connecting σ p and δ.Here 1 ≤ p ≤ n.It is easy to see that the graph A n embeds in the graph E n+1,p .That is, A(A n ) ⊂ A (E n+1,p ).This induces an injection on B n+1 to A(E n+1,p ).In other words, a representation of A(E n+1,p ) yields a representation of B n+1 .
Knowing the reduced Burau representation of B n+1 of degree n, Perron extends such a representation to a representation of B n+1 of degree 2n.The representation obtained is referred to as Burau bis representation.Next, Perron constructs for each λ = (λ 1 , . . ., λ n ) a representation ψ λ : A(E n+1,p ) → GL 2n (Q(t, d 1 , . . ., d n )), where t, d 1 , . . ., d n λ 1 , . . ., λ n are indeterminates.We specialize t, d 1 , . . ., d n to non zero complex numbers, and we study this representation explicitly in the case n = 2 and p = 1.We then reduce the complex specialization of the representation ψ λ to a representation of degree 3, namely A(E 3,1 ) → GL 3 (C).A necessary and sufficient condition which guarantees its irreducibility is obtained in that case.

Burau bis Representation
Perron's strategy is to begin with the Burau representation of the braid group and extend it to a representation of A(E n+1,p ).He begins with the reduced Burau representation: B n+1 → GL n (Z[t, t −1 ]) defined as follows: , where I k stands for the k × k identity matrix.Here, i = 2, . . ., n − 1.
Knowing that this representation is of degree n, Perron extends it to a representation of B n+1 of degree 2n.Let R i denote an n × n block of zeros with a t placed in the (i, i) th position, and let I n denote the n × n identity matrix.
The obtained representation is referred to as the Burau bis representation.It is defined as follows: For more details, see (T.E.Brendle, 2002, B.Perron, 1999).

Perron Representation
The Burau bis representation extends to A(E n+1,p ) for all possible values of n and p in the following way.
We define the following n × n matrices: For each i = 1, . . ., n, we have that b i satisfies the following conditions setting any undefined d j equal zero.
For any choice λ = (λ 1 , . . ., λ n ) , we get a linear representation where R is the field of rational fractions in n+1 indeterminates ) .
We get the following 2 × 2 matrices A = ) .
Simple computations show that the parameters satisfy the following equations: Having defined the 2 × 2 matrices A, B, C and D, we obtain the multiparameter representation A(E 3,1 ).This representation is of degree 4. We specialize the parameters We further assume that t −1 and d 2 = −t.The representation ψ λ : A(E 3,1 ) → GL 4 (C) is defined as follows: The graph E 3,1 has 3 vertices σ 1 , σ 2 and δ.Since p = 1, it follows that the vertex δ has a common edge with σ p = σ 1 .Therefore, the following relations are satisfied.
We note that relation ( 1) is actually Artin's braid relation of the classical braid group, B 3 having σ 1 and σ 2 as standard generators.This assures that a representation of A(E 3,1 )yields a representation of B 3 .
Proof.For simplicity, we write σ i instead of ψ λ (k) , where k is a generator of A(E 3,1 ).The subspace ⟩ is an invariant subspace of dimension 3. To see this: b 1 e 2 , e 3 , and e 4 .The matrix of σ 1 becomes Similarly, we determine the matrices of σ 2 and δ.It is easy to see that the first column of the matrices of all generators is (1, 0, 0, 0, 0) T , where T is the transpose.We thus reduce our representation to a 3-dimensional one by deleting the first row and the first column to get ψ ′ λ : A(E 3,1 ) → GL 3 (C).The representation is defined as follows: We then diagonalize the matrix corresponding to ψ ′ λ (σ 1 ) by an invertible matrix, say T , and conjugate the matrices of ψ ′ λ (σ 2 ) and ψ ′ λ (δ) by the same matrix T .The invertible matrix T is given by In fact, a computation shows that After conjugation, we get The entries of the matrices T −1 ψ ′ λ (σ 2 )T and T −1 ψ ′ λ (δ)T are well-defined since we assume in our work that t −1.For simplicity, we denote T −1 ψ ′ λ (σ 1 )T by ψ ′ λ (σ 1 ), T −1 ψ ′ λ (σ 2 )T by ψ ′ λ (σ 2 ), and T −1 ψ ′ λ (δ)T by ψ ′ λ (δ).We now prove some propositions to determine a sufficient condition for irreducibility of ψ ′ λ : Proof.The proof easily follows by considering the following relations: Proposition 3 The two expressions 1 + t + t 2 and b 1 t + b 2 t + b 2 cannot be both equal to zeros.
We use Proposition 2, Proposition 3 and Proposition 4 to prove the following Lemma.We recall that all the indeterminates used in defining the representations are specialized to non zero complex numbers and, in addition, the complex number associated with t is not equal to −1.
Lemma 5 If t ±i, then any non zero subspace S , which is invariant under the action of the representation ψ ′ λ : A(E 3,1 ) → Gl 3 (C) containing the standard unit vector e 3 , must be the whole space C 3 .
Proof.We have that ψ Moreover, This also implies that Having proved that 1 + t + t 2 and b 1 t + b 2 t + b 2 can't both be zeros, we consider the following cases: By Proposition 3 and (1), we get that e 2 ∈ S .By Proposition 4 and (2), we get that e 1 ∈ S .Thus, S is the whole space.
Case 2. 1 + t + t 2 0 Let us multiply (1) by −t + b 1 t − d 1 (1 + t) which is proved not to be zero in Proposition 4. We also multiply (2) by 1 + t + t 2 0. If we add the obtained equations, we get By Proposition 2, we have that t By Proposition 4 and by (3), we get e 2 ∈ S .
Thus, S is the whole space C 3 .
Next, we present the following theorem which gives a sufficient condition for irreducibility of ψ ′ λ : A(E 3,1 ) → GL 3 (C).
Theorem 6 If t ±i, then the representation ψ ′ λ : A(E 3,1 ) → GL 3 (C) is irreducible.Proof.Let S be a non zero proper subspace of C 3 , which is invariant under the action of ψ ′ λ .By Lemma 5, we have that e 3 S .
Then S is one of the following subspaces: ∈ S .This gives a contradiction because t 0.
Therefore, we conclude that the representation is irreducible because there is no proper non zero invariant subspace under the action of ψ ′ λ .We now give a necessary condition for irreducibility.