#### Abstract

The aim of this paper is to study representations of 3-dimensional simple multiplicative Hom-Lie algebras (whose structure is of *A*_{1}-type). In this paper we can see that a finite dimensional representation of is not always completely reducible, and a representation of is irreducible if and only if it is a regular Lie-type representation.

#### 1. Introduction

In 2006, Hartwig, Larsson, and Silvestrov introduced the notion of a Hom-Lie algebra [1], which is a generalization of the notion of a Lie algebra. In particular, if , then a Hom-Lie algebra is exactly a Lie algebra.

Because the Hom-Lie algebras are closely related to discrete and deformed vector fields, differential calculus [2, 3], and mathematical physics [4, 5], the Hom-Lie algebras have attracted more and more attention and become an active topic in recent years [6–8].

The representation theory plays an important role in Lie theory [9–11]. By means of the representation theory, we would be more aware of the corresponding algebras. Thus it is meaningful to obtain more information about the representations of Hom-Lie algebras.

In [7] the author defined the representations of Hom-Lie algebras and the corresponding Hom-cochain complexes, and studied the cohomologies associated with the adjoint representation and the trivial representation. As is known, specific calculations about the representations of Hom-Lie algebras are still not solved. The diversity of the twist map of makes this topic interesting and complicated.

Thanks to the relationship between multiplicative Hom-Lie algebras with invertible and Lie algebras (Lemma 3), the representation theory of Lie algebras can be a reference to what is considered. The representation of a 3-dimensional simple Lie algebra plays a crucial role in the representation theory of semisimple Lie algebras over [9]. By the same reason, in this paper, we study the representations of 3-dimensional simple multiplicative Hom-Lie algebras.

The paper is organized as follows. In Section 2 we study the structures of 3-dimensional simple multiplicative Hom-Lie algebra and show that 3-dimensional simple multiplicative Hom-Lie algebras are of -type. In Section 3, the representation of a multiplicative Hom-Lie algebra with invertible is investigated and shows that when is invertible, is of Lie-type, which makes it convenient to study representations of multiplicative Hom-Lie algebras. In Section 4, we study regular Lie-type representations of 3-dimensional simple multiplicative Hom-Lie algebras and reflect the existence and irreducibility of representations of this type. In Section 5, we work over finite dimensional representations of 3-dimensional simple multiplicative Hom-Lie algebras . In this section we can see that a finite dimensional representation of is not always completely reducible and a representation of is irreducible if and only if it is a regular Lie-type one.

Throughout this paper, unless otherwise stated, all algebras are finite dimensional and over the complex field .

#### 2. The Structures of 3-Dimensional Simple Multiplicative Hom-Lie Algebras

First we give some important definitions on Hom-Lie algebras.

*Definition 1 (see [1]). *A Hom-Lie algebra is a triple consisting of a vector space over , a linear sef-map , and a bilinear map such that

A Hom-Lie algebra is said to be multiplicative if for any ; see [7].

*Definition 2. *Let be a Hom-Lie algebra. If there exists a Lie algebra such that , then is said to be of Lie-type, and is called the compatible Lie algebra of . Furthermore, if the compatible Lie algebra of is an (or , , )-type Lie algebra, one calls an (or , , )-type Hom-Lie algebra.

A subspace of is called an ideal of if are satisfied. A center of is defined as

A Hom-Lie algebra is called simple if it has no nontrivial ideals and .

Lemma 3. *Let be a multiplicative Hom-Lie algebra with invertible. Then is Lie-type with the compatible Lie algebra , where is defined by .*

*Proof. *Let for any . Since is multiplicative, we have
and thus follows.

In the following we will show that is a Lie algebra. First it is obvious that is skew-symmetric. Next ,
where denotes a summation over the cyclic permutation on . Now it follows that is a Lie algebra.

Theorem 4. *Let be a 3-dimensional simple multiplicative Hom-Lie algebra; then is -type and
*

*Proof. *If , then , ; that is, is a nontrivial ideal of , which is a contradiction to the simplicity of . So is invertible. Now by Lemma 3, we have that is Lie-type with the 3-dimensional Hom-Lie admissible algebra .

If is an abelian Lie algebra, then we can deduce that is also abelian, which is absurd.

Suppose that has a 1-dimensional center . Note that , and such that , we have

That is, has a 1-dimensional center , which is a contradiction to the simplicity of .

If , then
which is impossible.

Now we can get that . By Lie theory, is an -type Lie algebra with a basis and a bracket . On one hand, by the proof of Lemma 3, we have that is an automorphism of . On the other hand, by Lie theory, the automorphism of has the form . Thus
Now it follows that

Let . The result follows.

#### 3. The Representations of Multiplicative Hom-Lie Algebras

First we give the definition of the representations of multiplicative Hom-Lie algebras.

*Definition 5 (see [7]). *Let be a multiplicative Hom-Lie algebra, a finite dimensional vector space, and . If a linear map satisfies
then is called a representation of , and is called a Hom--module via the action .

For a Hom--module , if a subspace is invariant under , then is called a Hom--submodule of . A Hom--module is called irreducible, if it has precisely two Hom--submodules (itself and 0). A Hom--module is called completely reducible if , where and are irreducible Hom--submodules.

Proposition 6. *Let be a multiplicative Hom-Lie algebra with invertible, its representation with invertible, and the compatible Lie algebra. Let ; then is a representation of .*

* Proof. *Equation (12) is equivalent to

Equation (11) is equivalent to

Denote that by , (13) can be rewritten as ; by the arbitrary of and the invertibility of we have

On vector space , for all , define a commutator bracket as

Clearly, is a Lie algebra.

On the other hand, for all ,

Then the result follows easily.

From Proposition 6, we can get a method of computing representations of a multiplicative Hom-Lie algebra with invertible.

Let be a representation of a Lie-type Hom-Lie algebra. If , where is a representation of the compatible Lie algebra, then is called a Lie-type representation. It is easy to know that the representation in Proposition 6 is Lie-type. In addition, suppose that is an irreducible representation of the compatible Lie algebra; then is called a regular Lie-type representation.

Theorem 7. *Let be a Lie-type Hom-Lie algebra with the compatible Lie algebra .**(1) If ( invertible) is a representation of , then
**
where is a representation of the compatible Lie algebra.**(2) Suppose that is a representation of . If such that (18) is satisfied, let ; then is a representation of .*

*Proof. *(1) By the invertibility of and (12), we can get (18) easily.

(2) , (12) follows from (18) easily. , we have

Now we can get (11). Therefore is a representation of .

#### 4. Regular Lie-Type Representations of 3-Dimensional Simple Multiplicative Hom-Lie Algebras

Lemma 8. *Let be a representation of a multiplicative Hom-Lie algebra ; then is a Hom--submodule of .*

*Proof. *, by , it is easy to know that ; then the result follows easily.

Lemma 9. *Let be an irreducible or a completely reducible representation of a multiplicative Hom-Lie algebra ; then is invertible.*

*Proof. *By the reason of Lemma 8, if is an irreducible Hom- module, we have that is invertible. If is a completely reducible Hom- module, then , where () are irreducible Hom- submodules. By the irreducibility of , we have that () is invertible, so is an invertible linear map of .

Lemma 10. *Let with invertible be a nontrivial finite dimensional regular Lie-type representation of a Lie-type Hom-Lie algebra; then is an irreducible Hom--module.*

*Proof. *Suppose to the contrary that is reducible. Then we assume that is a nontrivial Hom--submodule of . Let be the representation of the compatible Lie algebra. Then

That is, is a nontrivial subrepresentation of the compatible Lie algebra, which is a contradiction. Therefore is an irreducible Hom--module.

It is natural to ask the question: are there nontrivial finite dimensional regular Lie-type representations in 3-dimensional simple multiplicative Hom-Lie algebras? Let us see the following theorem.

Theorem 11. *Let be a 3-dimensional simple multiplicative Hom-Lie algebra; then there exist nontrivial finite dimensional regular Lie-type representations , and these representations are irreducible. For every such representation, is a semisimple linear transformation of . In addition, there is a basis of such that*(a)*;
*(b)*;
*(c)*. *

*Proof. *Let be an dimensional irreducible representation of the compatible Lie algebra . Take as a basis of , where satisfies

Now we prove that when is defined by

(18) is always satisfied. Because

Thus for defined by (22), (18) is always established. Let ; then it follows from Theorem 7 (2) that is a nontrivial finite dimensional regular Lie-type representation of . By Lemma 10, we have that is irreducible.

Furthermore, we have
and thus we can get (a). For some fixed , it is obvious that is a semisimple linear transformation of . Take in ; we can get (b) and (c) easily.

#### 5. Irreducible and Completely Reducible Representations of a 3-Dimensional Simple Multiplicative Hom-Lie Algebra

In this section, denotes a 3-dimensional simple multiplicative Hom-Lie algebra.

By Theorem 11, we know that there exist nontrivial finite dimensional irreducible regular Lie-type representations of . However, are there other nontrivial irreducible representations of ? Is any finite dimensional representation of completely reducible? We will study these questions in this section.

By Lemma 9, we only need to consider the case when is invertible.

Let with invertible be a finite dimensional representation of . By Proposition 6, we have that is of Lie-type, and is a finite dimensional representation of the compatible Lie algebra . By Weyl theorem, we know that is completely reducible. That is, , where are irreducible -modules. Suppose that . Denote that a tuple . By the representation theory of , we have that is a highest weight module with highest weight vectors and the highest weight , respectively. Take as a basis of satisfying

Theorem 12. *When , for , then*(1)*;
*(2)* is a completely reducible Hom--module with a decomposition , where are irreducible Hom--submodules; here .*

*Proof. *(1) According to (18), we get the result.

(2) Because ,
Let , combined with Lemma 10; we have that are irreducible Hom--submodules.

When , by (18) it can be checked that where .

Take

By the theory of linear algebra, there is an invertible matrix such that is a Jordan canonical form; that is, where are Jordan blocks.

Theorem 13. *The condition is the same as the previous remark.*(1)*If , then is a reducible but not completely reducible Hom--module.*(2)*If , then is a reducible but not completely reducible Hom--module.*(3)*If , then is a completely reducible Hom--module.*

*Proof. *Let
That is,
Then
where
here are matrices satisfying
Because is invertible, it is easy to check that the matrix is invertible; therefore is a basis of .

Let , because

By (32) and the representation of Lie algebra , we have
Through (38) it is easy to get that are -irreducible-modules.

(1) In this case, by (30), we have
Let
By (32), it is easy to check that
Then . Let ; then is an irreducible Hom--submodule, but is not completely reducible.

(2) Suppose (omit the order of )
here are nondiagonal Jordan blocks ().

Let

denoted by . It is easy to check that
By the statement of the proof and Lemma 10, we have that are irreducible Hom--submodules.

Let
Then
Let ; then are Hom--submodules. As (1) of the theorem, we can prove that are reducible but not completely reducible Hom--modules. We have the conclusion.

(3) In this case, by (30) and Theorem 12 (1) we get

Let

then are irreducible -modules and

Denoting by , thanks to Lemma 10 we have that are irreducible Hom--submodules. So is a completely reducible Hom--module.

When , suppose that the multiplicity of () is and . By (18) we can get as follows: where are matrices of the form (27),

Theorem 14. *The condition is as the previous remark. is completely reducible if there exist invertible matrices such that are diagonal matrices. Otherwise is reducible but not completely reducible.*

*Proof. *It can be got from Theorems 12 and 13 directly.

Proposition 15. *Let be a finite dimensional representation of ; then is not always completely reducible, and is irreducible if and only if it is of regular Lie-type.*

*Proof. *The claim follows from Theorems 11, 12, 13, and 14 directly.