Representations of 3-Dimensional Simple Multiplicative Hom-Lie Algebras

Abstract

The aim of this paper is to study representations of 3-dimensional simple multiplicative Hom-Lie algebras (𝔤;[⋅,⋅];𝛼) (whose structure is of A₁-type). In this paper we can see that a finite dimensional representation of (𝔤;[⋅,⋅];𝛼) is not always completely reducible, and a representation of (𝔤;[⋅,⋅];𝛼) 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 𝛼=id, 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 𝔤 makes this topic interesting and complicated.

Thanks to the relationship between multiplicative Hom-Lie algebras with 𝛼 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 ℂ [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 𝐴₁-type. In Section 3, the representation (𝜌,𝑉,𝛽) of a multiplicative Hom-Lie algebra with 𝛼 invertible is investigated and shows that when 𝛽 is invertible, (𝜌,𝑉,𝛽) 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 (𝔤,[⋅,⋅],𝛼). In this section we can see that a finite dimensional representation of (𝔤,[⋅,⋅],𝛼) is not always completely reducible and a representation of (𝔤,[⋅,⋅],𝛼) 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 ℂ.

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 (𝔤,[⋅,⋅],𝛼) consisting of a vector space 𝔤 over ℂ, a linear sef-map 𝛼, and a bilinear map [⋅,⋅]∶𝔤×𝔤→𝔤 such that 𝛼(𝑥),+𝛼𝑦,[+𝛼(𝑧),𝑥,𝑦=−𝑦,𝑥,∀𝑥,𝑦∈𝔤,𝑦,𝑧𝑧,𝑥]𝑥,𝑦=0,∀𝑥,𝑦,𝑧∈𝔤. (1)(2)

A Hom-Lie algebra (𝔤,[⋅,⋅],𝛼) is said to be multiplicative if 𝛼([𝑥,𝑦])=[𝛼(𝑥),𝛼(𝑦)] for any 𝑥,𝑦∈𝔤; see [7].

Definition 2. Let (𝔤,[⋅,⋅],𝛼) be a Hom-Lie algebra. If there exists a Lie algebra (𝔤,[⋅,⋅]^′) such that [𝑥,𝑦]=𝛼([𝑥,𝑦]^′)=[𝛼(𝑥),𝛼(𝑦)]^′,forall𝑥,𝑦∈𝔤, then (𝔤,[⋅,⋅],𝛼) is said to be of Lie-type, and (𝔤,[⋅,⋅]^′) is called the compatible Lie algebra of (𝔤,[⋅,⋅],𝛼). Furthermore, if the compatible Lie algebra of (𝔤,[⋅,⋅],𝛼) is an 𝐴 (or 𝐵, 𝐶, 𝐷)-type Lie algebra, one calls (𝔤,[⋅,⋅],𝛼) an 𝐴 (or 𝐵, 𝐶, 𝐷)-type Hom-Lie algebra.

A subspace 𝔤₁ of 𝔤 is called an ideal of (𝔤,[⋅,⋅],𝛼) if 𝛼(𝔤₁)⊆𝔤₁,[𝔤₁,𝔤]⊆𝔤₁ are satisfied. A center of (𝔤,[⋅,⋅],𝛼) is defined as 𝐶(𝔤)={𝑥∈𝔤∣[𝑥,𝔤]=0;[𝛼(𝑥),𝔤]=0}. (3)

A Hom-Lie algebra (𝔤,[⋅,⋅],𝛼) is called simple if it has no nontrivial ideals and [𝔤,𝔤]=𝔤.

Lemma 3. Let (𝔤,[⋅,⋅],𝛼) be a multiplicative Hom-Lie algebra with 𝛼 invertible. Then (𝔤,[⋅,⋅],𝛼) is Lie-type with the compatible Lie algebra (𝔤,[⋅,⋅]^′), where [⋅,⋅]^′ is defined by [𝑥,𝑦]^′=𝛼⁻¹([𝑥,𝑦]),𝑓𝑜𝑟𝑎𝑙𝑙𝑥,𝑦∈𝔤.

Proof. Let [𝑥,𝑦]^′=𝛼⁻¹([𝑥,𝑦]) for any 𝑥,𝑦∈𝔤. Since (𝔤,[⋅,⋅],𝛼) is multiplicative, we have 𝛼𝑥,𝑦=𝛼(𝑥),𝛼𝑦⟹𝛼²𝑥,𝑦^′=𝛼𝛼(𝑥),𝛼𝑦^′, (4) and thus 𝛼([𝑥,𝑦]^′)=[𝛼(𝑥),𝛼(𝑦)]^′ follows.
In the following we will show that (𝔤,[⋅,⋅]^′) is a Lie algebra. First it is obvious that [⋅,⋅]^′ is skew-symmetric. Next ∀𝑥,𝑦,𝑧∈𝔤, ↺_{𝑥,𝑦,𝑧}𝑥,𝑦,𝑧^′′=↺_{𝑥,𝑦,𝑧}𝛼⁻¹𝑥,𝛼⁻¹𝑦,𝑧=↺_{𝑥,𝑦,𝑧}𝛼⁻¹𝛼⁻¹𝛼(𝑥),𝛼⁻¹𝑦,𝑧=↺_{𝑥,𝑦,𝑧}𝛼⁻²𝛼(𝑥),𝑦,𝑧=0, (5) where ↺_{𝑥,𝑦,𝑧} denotes a summation over the cyclic permutation on 𝑥,𝑦,𝑧. Now it follows that (𝔤,[⋅,⋅]^′) is a Lie algebra.

Theorem 4. Let (𝔤,[⋅,⋅],𝛼) be a 3-dimensional simple multiplicative Hom-Lie algebra; then (𝔤,[⋅,⋅],𝛼) is 𝐴₁-type and ([])=[]𝔤,⋅,⋅,𝛼ℎ,𝑒,𝑓∣ℎ,𝑒=2𝑎𝑒,ℎ,𝑓=−2𝑎⁻¹,𝛼=𝑓,𝑒,𝑓=ℎℎ,𝑒,𝑓ℎ,𝑒,𝑓1000𝑎000𝑎⁻¹_{0≠𝑎∈ℂ}. (6)

Proof. If Ker(𝛼)≠0, then 𝛼(Ker(𝛼))=0, 𝛼([𝔤,Ker(𝛼)])=0; that is, Ker(𝛼) is a nontrivial ideal of (𝔤,[⋅,⋅],𝛼), which is a contradiction to the simplicity of (𝔤,[⋅,⋅],𝛼). So 𝛼 is invertible. Now by Lemma 3, we have that (𝔤,[⋅,⋅],𝛼) is Lie-type with the 3-dimensional Hom-Lie admissible algebra (𝔤,[⋅,⋅]^′)=(𝔤,𝛼⁻¹([⋅,⋅])).
If (𝔤,[⋅,⋅]^′) is an abelian Lie algebra, then we can deduce that (𝔤,[⋅,⋅],𝛼) is also abelian, which is absurd.
Suppose that (𝔤,[⋅,⋅]^′) has a 1-dimensional center ℂ𝑥. Note that [𝑥,𝔤]=𝛼([𝑥,𝔤]^′)=0, and forall𝑦∈𝔤,∃𝑦^′∈𝔤 such that 𝑦=𝛼(𝑦^′), we have 𝛼(𝑥),𝑦=𝛼(𝑥),𝛼𝑦^′=𝛼𝑥,𝑦^′=0. (7)
That is, (𝔤,[⋅,⋅],𝛼) has a 1-dimensional center ℂ𝑥, which is a contradiction to the simplicity of (𝔤,[⋅,⋅],𝛼).
If [𝔤,𝔤]^′≠𝔤, then [𝔤,𝔤]=𝛼⁻¹[𝔤,𝔤]^′=𝛼⁻¹(𝔤),𝛼⁻¹(𝔤)^′=[𝔤,𝔤]^′≠𝔤, (8) which is impossible.
Now we can get that [𝔤,𝔤]^′=𝔤. By Lie theory, (𝔤,[⋅,⋅]^′) is an 𝐴₁-type Lie algebra with a basis {ℎ,𝑒,𝑓} and a bracket [ℎ,𝑒]^′=2𝑒,[ℎ,𝑓]^′=−2𝑓,[𝑒,𝑓]^′=ℎ. On one hand, by the proof of Lemma 3, we have that 𝛼 is an automorphism of (𝔤,[⋅,⋅]^′). On the other hand, by Lie theory, the automorphism of 𝐴₁ has the form 𝛼=exp^{𝑎𝑑𝑘ℎ},𝑘∈ℂ. Thus 𝛼ℎ,𝑒,𝑓=ℎ,𝑒,𝑓1000exp^2𝑘000exp^−2𝑘. (9) Now it follows that [ℎ,𝑒]=𝛼[ℎ,𝑒]^′=2exp^2𝑘𝑒,ℎ,𝑓=𝛼ℎ,𝑓^′=−2exp^−2𝑘𝑓,𝑒,𝑓=𝛼𝑒,𝑓^′=ℎ. (10)
Let 𝑎=exp^2𝑘≠0. 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 (𝔤,[⋅,⋅],𝛼) be a multiplicative Hom-Lie algebra, 𝑉 a finite dimensional vector space, and 𝛽∈gl(𝑉). If a linear map 𝜌∶𝔤→gl(𝑉) satisfies 𝜌𝛽(𝑣)(𝛼(𝑥𝜌𝑦𝛼𝑦𝜌(𝑥)𝜌(𝑥)𝑣(𝛼(𝑥𝛽(𝑣)𝑥,𝑦=𝜌))𝑣−𝜌𝑣,∀𝑥,𝑦∈𝔤,𝑣∈𝑉,3𝛽=𝜌)),∀𝑥∈𝔤,𝑣∈𝑉, (11)(12) then (𝜌,𝑉,𝛽) is called a representation of (𝔤,[⋅,⋅],𝛼), and (𝑉,𝛽) is called a Hom-𝔤-module via the action 𝑥𝑣=𝜌(𝑥)𝑣,forall𝑥∈𝔤,𝑣∈𝑉.

For a Hom-𝔤-module (𝑉,𝛽), if a subspace 𝑉₁⊆𝑉 is invariant under 𝛽, then (𝑉₁,𝛽) is called a Hom-𝔤-submodule of (𝑉,𝛽). A Hom-𝔤-module (𝑉,𝛽) is called irreducible, if it has precisely two Hom-𝔤-submodules (itself and 0). A Hom-𝔤-module (𝑉,𝛽) is called completely reducible if 𝑉=𝑉₁⊕⋯⊕𝑉_𝑠, where 𝛽(𝑉_𝑖)⊆𝑉_𝑖(𝑖=1,…,𝑠) and (𝑉_𝑖,𝛽)(𝑖=1,…,𝑠) are irreducible Hom-𝔤-submodules.

Proposition 6. Let (𝔤,[⋅,⋅],𝛼) be a multiplicative Hom-Lie algebra with 𝛼 invertible, (𝜌,𝑉,𝛽) its representation with 𝛽 invertible, and (𝔤,[⋅,⋅]^′_𝔤) the compatible Lie algebra. Let 𝜌^′=𝛽⁻¹𝜌; then (𝜌^′,𝑉) is a representation of (𝔤,[⋅,⋅]^′_𝔤).

Proof. Equation (12) is equivalent to 𝜌(𝛼(𝑥))=𝛽𝜌(𝑥)𝛽⁻¹,∀𝑥∈𝔤. (13)
Equation (11) is equivalent to 𝜌𝑥,𝑦=𝛽𝜌(𝑥)𝛽⁻¹𝜌𝑦𝛽⁻¹−𝛽𝜌𝑦𝛽⁻¹𝜌(𝑥)𝛽⁻¹,∀𝑥∈𝔤,𝑣∈𝑉. (14)
Denote that by 𝛼(𝑥)=𝑥^′, (13) can be rewritten as 𝜌(𝛼⁻¹(𝑥^′))=𝛽⁻¹𝜌(𝑥^′)𝛽; by the arbitrary of 𝑥 and the invertibility of 𝛼 we have 𝜌𝛼⁻¹(𝑥)=𝛽⁻¹𝜌(𝑥)𝛽,∀𝑥∈𝔤. (15)
On vector space 𝛽⁻¹𝜌(𝔤), for all 𝑥,𝑦∈𝔤, define a commutator bracket [⋅,⋅]^′ as 𝛽⁻¹𝜌(𝑥),𝛽⁻¹𝜌𝑦^′=𝛽⁻¹𝜌(𝑥)𝛽⁻¹𝜌𝑦−𝛽⁻¹𝜌𝑦𝛽⁻¹𝜌(𝑥)=𝜌^′(𝑥),𝜌^′𝑦^′. (16)
Clearly, (𝛽⁻¹𝜌(𝔤),[⋅,⋅]^′) is a Lie algebra.
On the other hand, for all 𝑥,𝑦∈𝔤, 𝜌^′𝑥,𝑦^′_𝔤=𝜌^′𝛼⁻¹𝑥,𝑦=𝛽⁻¹𝜌𝛼⁻¹(𝑥),𝛼⁻¹𝑦=𝛽⁻¹𝛽𝜌𝛼⁻¹(𝑥)𝛽⁻¹𝜌𝛼⁻¹𝑦𝛽⁻¹−𝛽𝜌𝛼⁻¹𝑦𝛽⁻¹𝜌𝛼⁻¹(𝑥)𝛽⁻¹=𝛽⁻¹𝜌(𝑥)𝛽𝛽⁻¹𝛽⁻¹𝜌𝑦𝛽𝛽⁻¹−𝛽⁻¹𝜌𝑦𝛽𝛽⁻¹𝛽⁻¹𝜌(𝑥)𝛽𝛽⁻¹=𝛽⁻¹𝜌(𝑥)𝛽⁻¹𝜌𝑦−𝛽⁻¹𝜌𝑦𝛽⁻¹𝜌(𝑥)=𝜌^′(𝑥),𝜌^′𝑦^′. (17)
Then the result follows easily.

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

Let (𝜌,𝑉,𝛽) be a representation of a Lie-type Hom-Lie algebra. If 𝜌=𝛽𝜌^′, where (𝜌^′,𝑉) is a representation of the compatible Lie algebra, then (𝜌,𝑉,𝛽) is called a Lie-type representation. It is easy to know that the representation in Proposition 6 is Lie-type. In addition, suppose that (𝜌^′,𝑉) is an irreducible representation of the compatible Lie algebra; then (𝜌,𝑉,𝛽) is called a regular Lie-type representation.

Theorem 7. Let (𝔤,[⋅,⋅],𝛼) be a Lie-type Hom-Lie algebra with the compatible Lie algebra (𝔤,[⋅,⋅]^′).
(1) If (𝜌,𝑉,𝛽) (𝛽 invertible) is a representation of (𝔤,[⋅,⋅],𝛼), then 𝛽𝜌^′(𝑥)𝑣=𝜌^′(𝛼(𝑥))𝛽(𝑣),∀𝑥∈𝔤,𝑣∈𝑉, (18) where (𝜌^′,𝑉)=(𝛽⁻¹𝜌,𝑉) is a representation of the compatible Lie algebra.
(2) Suppose that (𝜌^′,𝑉) is a representation of (𝔤,[⋅,⋅]^′). If  ∃𝛽∈gl(𝑉) such that (18) is satisfied, let 𝜌=𝛽𝜌^′; then (𝜌,𝑉,𝛽) is a representation of (𝔤,[⋅,⋅],𝛼).

Proof. (1) By the invertibility of 𝛽 and (12), we can get (18) easily.
(2) forall𝑥∈𝔤,𝑣∈𝑉, (12) follows from (18) easily. forall𝑥,𝑦∈𝔤,𝑣∈𝑉, we have 𝜌𝑥,𝑦𝛽(𝑣)=𝛽𝜌^′𝛼𝑥,𝑦^′𝛽(𝑣)=𝛽𝜌^′(𝛼(𝑥))𝜌^′𝛼𝑦−𝜌^′𝛼𝑦𝜌^′(𝛼(𝑥𝛽(𝑣)𝜌))=𝛽^′(𝛼(𝑥))𝛽𝜌^′𝑦𝑣−𝜌^′𝛼𝑦𝛽𝜌^′(𝑥)𝑣=𝜌(𝛼(𝑥))𝜌𝑦𝑣−𝜌(𝛼(𝑥))𝜌𝑦𝑣. (19)
Now we can get (11). Therefore (𝜌,𝑉,𝛽) is a representation of (𝔤,[⋅,⋅],𝛼).

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

Lemma 8. Let (𝜌,𝑉,𝛽) be a representation of a multiplicative Hom-Lie algebra (𝔤,[⋅,⋅],𝛼); then (Ker(𝛽),𝛽) is a Hom-𝔤-submodule of (𝑉,𝛽).

Proof. forall𝑣∈Ker(𝛽), by 𝛽(𝜌(𝑥)𝑣)=𝜌(𝛼(𝑥))𝛽(𝑣)=0,forall𝑥∈𝔤, it is easy to know that 𝜌(𝑥)Ker(𝛽)⊆Ker(𝛽); then the result follows easily.

Lemma 9. Let (𝜌,𝑉,𝛽) be an irreducible or a completely reducible representation of a multiplicative Hom-Lie algebra (𝔤,[⋅,⋅],𝛼); then 𝛽 is invertible.

Proof. By the reason of Lemma 8, if (𝑉,𝛽) is an irreducible Hom-𝔤 module, we have that 𝛽 is invertible. If (𝑉,𝛽) is a completely reducible Hom-𝔤 module, then 𝑉=𝑉₁⊕⋯⊕𝑉_𝑠, where (𝑉_𝑖,𝛽|_𝑉𝑖) (𝑖=1,…,𝑠) are irreducible Hom-𝔤 submodules. By the irreducibility of (𝑉_𝑖,𝛽|_𝑉𝑖), we have that 𝛽|_𝑉𝑖 (𝑖=1,…,𝑠) is invertible, so 𝛽 is an invertible linear map of 𝑉.

Lemma 10. Let (𝜌,𝑉,𝛽) with 𝛽 invertible be a nontrivial finite dimensional regular Lie-type representation of a Lie-type Hom-Lie algebra; then (𝑉,𝛽) is an irreducible Hom-𝔤-module.

Proof. Suppose to the contrary that (𝑉,𝛽) is reducible. Then we assume that (𝑉₁,𝛽) is a nontrivial Hom-𝔤-submodule of (𝑉,𝛽). Let (𝜌^′,𝑉)=(𝛽⁻¹𝜌,𝑉) be the representation of the compatible Lie algebra. Then 𝜌^′(𝑥)𝑉₁=𝛽⁻¹𝜌(𝑥)𝑉₁⊆𝛽⁻¹𝑉₁⊆𝑉₁. (20)
That is, (𝜌^′,𝑉₁) is a nontrivial subrepresentation of the compatible Lie algebra, which is a contradiction. Therefore (𝑉,𝛽) is an irreducible Hom-𝔤-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 (𝔤,[⋅,⋅],𝛼) be a 3-dimensional simple multiplicative Hom-Lie algebra; then there exist nontrivial finite dimensional regular Lie-type representations (𝜌,𝑉,𝛽), and these representations are irreducible. For every such representation, 𝜌(ℎ) is a semisimple linear transformation of 𝑉. In addition, there is a basis {𝑣₀,𝑣₁,…,𝑣_𝑚} of 𝑉 such that(a)𝜌(ℎ)𝑣_𝑖=(𝑚−2𝑖)𝑎^−𝑖𝑏₀𝑣_𝑖,𝛽(𝑣_𝑖)=𝑎^−𝑖𝑏₀𝑣_𝑖,𝑖=0,…,𝑚; (b)𝜌(𝑓)𝑣_𝑖=(𝑖+1)𝑎^−𝑖−1𝑏₀𝑣_𝑖+1,𝑖=0,…,𝑚−1;𝜌(𝑓)𝑣_𝑚=0; (c)𝜌(𝑒)𝑣₀=0,𝜌(𝑒)𝑣_𝑖=(𝑚−𝑖+1)𝑎^−𝑖+1𝑏₀𝑣_𝑖−1,𝑖=1,…,𝑚.

Proof. Let (𝜌^′,𝑉) be an 𝑚+1 dimensional irreducible representation of the compatible Lie algebra 𝐴₁. Take {𝑣₀,𝑣₁,…,𝑣_𝑚} as a basis of 𝑉, where 𝑣_𝑖=1/𝑖𝜌^′(𝑓)𝑣_𝑖−1(𝑖=1,…,𝑚) satisfies 𝜌^′(ℎ)𝑣_𝑖=(𝑚−2𝑖)𝑣_𝑖,𝜌^′(𝑒)𝑣_𝑖=(𝑚−𝑖+1)𝑣_𝑖−1(),𝜌𝑖=1,…,𝑚_′(𝑒)𝑣₀=0. (21)
Now we prove that when 𝛽∈gl(𝑉) is defined by 𝛽𝑣₀𝑣₁⋮𝑣_𝑚^𝑇=𝑣₀𝑣₁⋮𝑣_𝑚^𝑇𝑏₀𝑎⁻¹𝑏₀⋱𝑎^−𝑚𝑏₀,𝑏₀≠0, (22)
(18) is always satisfied. Because𝛽𝜌^′(ℎ)𝑣_𝑖=(𝑚−2𝑖)𝛽𝑣_𝑖=(𝑚−2𝑖)𝑎^−𝑖𝑏₀𝑣_𝑖,𝜌^′(𝛼(ℎ))𝛽𝑣_𝑖=𝜌^′(ℎ)𝑎^−𝑖𝑏₀𝑣_𝑖=𝑎^−𝑖𝑏₀(𝑚−2𝑖)𝑣_𝑖=𝛽𝜌^′(ℎ)𝑣_𝑖,𝛽𝜌𝑖=0,…,𝑚,_′𝑓𝑣_𝑖=(𝑖+1)𝛽𝑣_𝑖+1=(𝑖+1)𝑎^−𝑖−1𝑏₀𝑣_𝑖+1,𝜌^′𝛼𝑓𝛽𝑣_𝑖=𝑎⁻¹𝜌^′𝑓𝑎^−𝑖𝑏₀𝑣_𝑖=𝑎^−𝑖−1𝑏₀(𝑖+1)𝑣_𝑖+1=𝛽𝜌^′𝑓𝑣_𝑖𝛽𝜌,𝑖=0,…,𝑚,_′(𝑒)𝑣_𝑖=𝛽(𝑚−𝑖+1)𝑣_𝑖−1=(𝑚−𝑖+1)𝛽𝑣_𝑖−1=(𝑚−𝑖+1)𝑎^{−(𝑖−1)}𝑏₀𝑣_𝑖−1,𝜌^′(𝛼(𝑒))𝛽𝑣_𝑖=𝑎𝜌^′(𝑒)𝛽𝑣_𝑖=𝜌^′(𝑒)𝑎^{−(𝑖−1)}𝑏₀𝑣_𝑖=𝑎^{−(𝑖−1)}𝑏₀(𝑚−𝑖+1)𝑣_𝑖−1=𝛽𝜌^′(𝑒)𝑣_𝑖,𝑖=0,…,𝑚. (23)
Thus for 𝛽 defined by (22), (18) is always established. Let 𝜌=𝛽𝜌^′; then it follows from Theorem 7 (2) that (𝜌,𝑉,𝛽) is a nontrivial finite dimensional regular Lie-type representation of (𝔤,[⋅,⋅],𝛼). By Lemma 10, we have that (𝜌,𝑉,𝛽) is irreducible.
Furthermore, we have 𝜌(ℎ)𝑣_𝑖=𝛽𝜌^′(ℎ)𝑣_𝑖=(𝑚−2𝑖)𝑎^−𝑖𝑏₀𝑣_𝑖,𝑖=0,…,𝑚, (24) and thus we can get (a). For some fixed 𝑏₀, it is obvious that 𝜌(ℎ) is a semisimple linear transformation of 𝑉. Take 𝑥=𝑒,𝑓 in 𝜌(𝑥)𝑣_𝑖=𝛽(𝜌^′(𝑥)𝑣_𝑖)(𝑖=0,…,𝑚); 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, (𝔤,[⋅,⋅],𝛼) 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 (𝔤,[⋅,⋅],𝛼). However, are there other nontrivial irreducible representations of (𝔤,[⋅,⋅],𝛼)? Is any finite dimensional representation of (𝔤,[⋅,⋅],𝛼) completely reducible? We will study these questions in this section.

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

Let (𝜌,𝑉,𝛽) with 𝛽 invertible be a finite dimensional representation of (𝔤,[⋅,⋅],𝛼). By Proposition 6, we have that (𝜌,𝑉,𝛽) is of Lie-type, and (𝜌^′,𝑉)=(𝛽⁻¹𝜌,𝑉) is a finite dimensional representation of the compatible Lie algebra 𝐴₁. By Weyl theorem, we know that (𝜌^′,𝑉) is completely reducible. That is, 𝑉=𝑉₁⊕⋯⊕𝑉_𝑠, where 𝑉_𝑖(𝑖=1,…,𝑠) are irreducible 𝐴₁-modules. Suppose that dim𝑉_𝑖=𝑚_𝑖+1. Denote that a tuple 𝜏=(𝑚₁,…,𝑚_𝑠). By the representation theory of 𝐴₁, we have that 𝑉_𝑗(𝑗=1,…,𝑠) is a highest weight module with highest weight vectors 𝑣_𝑗0 and the highest weight 𝑚_𝑗, respectively. Take {𝑣_𝑗0,𝑣_𝑗1,…,𝑣_𝑗𝑚𝑗} as a basis of 𝑉_𝑗 satisfying 𝑣_𝑗𝑖=1𝑖𝜌^′𝑓𝑣_{𝑗,𝑖−1},𝑖=1,…,𝑚_𝑗,𝜌^′(ℎ)𝑣_𝑗𝑖=𝑚_𝑗−2𝑖𝑣_𝑗𝑖,𝑖=0,…,𝑚_𝑗,𝜌^′(𝑒)𝑣_𝑗𝑖=𝑚_𝑗−𝑖+1𝑣_{𝑗,𝑖−1},𝑖=1,…,𝑚_𝑗,𝜌^′(𝑒)𝑣₀=0. (25)

Theorem 12. When 𝜏=(𝑚₁,…,𝑚_𝑠), 𝑚_𝑖≠𝑚_𝑗 for 1≤𝑖≠𝑗≤𝑠, then(1)𝛽=diag(…,𝑏_𝑖0,𝑏_𝑖0/𝑎,…,𝑏_𝑖0/𝑎^𝑚𝑖,…),𝑖=1,…,𝑠; (2)(𝑉,𝛽) is a completely reducible Hom-𝔤-module with a decomposition 𝑉=𝑉₁⊕⋯⊕𝑉_𝑠, where (𝑉_𝑖,𝛽_𝑖),(𝑖=1,…,𝑠) are irreducible Hom-𝔤-submodules; here 𝛽_𝑖=𝛽|_𝑉𝑖.

Proof. (1) According to (18), we get the result.
(2) Because (𝑉_𝑗)⊆𝑉_𝑗,𝑗=1,…,𝑠, 𝜌(𝑥)𝑉_𝑗=𝛽𝜌^′(𝑥)𝑉_𝑗⊆𝑉_𝑗. (26)Let 𝛽_𝑗=𝛽|_𝑉𝑗, combined with Lemma 10; we have that (𝑉_𝑗,𝛽_𝑗)(𝑗=1,…,𝑠) are irreducible Hom-𝔤-submodules.

When 𝜏=(𝑚,…,𝑚), by (18) it can be checked that 𝐵𝛽=¹¹𝐵₁₂…𝐵_1𝑠𝐵₂₁𝐵₂₂…𝐵_2𝑠𝐵⋮⋮⋱⋮_𝑠1𝐵_𝑠2…𝐵_𝑠𝑠, (27) where 𝐵_𝑖𝑗=diag(𝑏_𝑖𝑗,𝑏_𝑖𝑗/𝑎,…,𝑏_𝑖𝑗/𝑎^𝑚).

Take 𝐵₁=𝑏₁₁𝑏₁₂…𝑏_1𝑠𝑏₂₁𝑏₂₂…𝑏_2𝑠𝑏⋮⋮⋱⋮_𝑠1𝑏_𝑠2…𝑏_𝑠𝑠. (28)

By the theory of linear algebra, there is an invertible matrix 𝑝𝑃=¹¹𝑝₁₂…𝑝_1𝑠𝑝₂₁𝑝₂₂…𝑝_2𝑠𝑝⋮⋮⋱⋮_𝑠1𝑝_𝑠2…𝑝_𝑠𝑠 (29) such that 𝑃⁻¹𝐵₁𝑃 is a Jordan canonical form; that is, 𝑃⁻¹𝐵₁𝐽𝑃=¹¹𝐽₂₂⋱𝐽_𝑡𝑡, (30) where 𝐽_𝑖𝑖=^{𝜆𝑖1𝜆𝑖}_{⋱⋱1𝜆𝑖}(𝑖=1,…,𝑡) are Jordan blocks.

Theorem 13. The condition is the same as the previous remark.(1)If 𝑡=1, then (𝑉,𝛽) is a reducible but not completely reducible Hom-𝔤-module.(2)If 𝑠>𝑡>1, then (𝑉,𝛽) is a reducible but not completely reducible Hom-𝔤-module.(3)If 𝑡=𝑠, then (𝑉,𝛽) is a completely reducible Hom-𝔤-module.

Proof. Let 𝑣^′_𝑖0=𝑝_1𝑖𝑣₁₀+𝑝_2𝑖𝑣₂₀+⋯+𝑝_𝑠𝑖𝑣_𝑠0𝑣,𝑖=1,…,𝑠,_′𝑖𝑗=1𝑗𝜌^′𝑓𝑣^′_{𝑖,𝑗−1},𝑖=1,…,𝑠;𝑗=1,…,𝑚. (31)(32) That is, 𝑣^′_𝑖𝑗=𝑝_1𝑖𝑣_1𝑗+𝑝_2𝑖𝑣_2𝑗+⋯+𝑝_𝑠𝑖𝑣_𝑠𝑗,𝑖=1,…,𝑠;𝑗=1,…,𝑚. (33) Then 𝑣^′₁₀,…,𝑣^′_1𝑚,…,𝑣^′_𝑠0,…,𝑣^′_𝑠𝑚=𝑣₁₀,…,𝑣_1𝑚,…,𝑣_𝑠0,…,𝑣_𝑠𝑚𝑄, (34) where 𝑄𝑄=¹¹𝑄₁₂…𝑄_1𝑠𝑄₂₁𝑄₂₂…𝑄_2𝑠𝑄⋮⋮⋱⋮_𝑠1𝑄_𝑠2…𝑄_𝑠𝑠; (35) here 𝑄_𝑖𝑗 are 𝑚×𝑚 matrices satisfying 𝑄_𝑖𝑗=diag𝑝_𝑖𝑗,…,𝑝_𝑖𝑗,𝑖,𝑗=1,…,𝑠. (36) Because 𝑃 is invertible, it is easy to check that the matrix 𝑄 is invertible; therefore {𝑣^′₁₀,…,𝑣^′_1𝑚,…,𝑣^′_𝑠0,…,𝑣^′_𝑠𝑚} is a basis of 𝑉.
Let 𝑉^′_𝑖=ℂ{𝑣^′_𝑖0,𝑣^′_𝑖1,…,𝑣^′_𝑖𝑚}, because 𝜌^′(ℎ)𝑣^′_𝑖0=𝜌^′(ℎ)𝑝_1𝑖𝑣₁₀+𝑝_2𝑖𝑣₂₀+⋯+𝑝_𝑠𝑖𝑣_𝑠0=𝑚𝑣_𝑖0. (37)
By (32) and the representation of Lie algebra 𝐴₁, we have 𝜌^′(ℎ)𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚=𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚𝑚⋱,𝜌𝑚−2−𝑚_′𝑓𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚=𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚,𝜌00…0010…0002…00⋮⋮⋱⋮⋮00…𝑚0_′(𝑒)𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚=𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚×.0𝑚0…000𝑚−1…0⋮⋮⋮⋱⋮000…1000…0 (38) Through (38) it is easy to get that 𝑉^′₁,…,𝑉^′_𝑠 are 𝐴₁-irreducible-modules.
(1) In this case, by (30), we have 𝛽𝑣^′₁₀,𝑣^′₂₀,…,𝑣^′_𝑠0=𝑣^′₁₀,𝑣^′₂₀,…,𝑣^′_𝑠0𝜆001𝜆⋱⋱1𝜆. (39) Let 𝑉^′_𝑖=ℂ𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚,𝑖=1,…,𝑠, (40) By (32), it is easy to check that 𝛽𝑉^′_𝑖⊆𝑉^′_𝑖+𝑉^′_𝑖+1,𝑖=1,…,𝑠−1;𝛽𝑉^′_𝑠⊆𝑉^′_𝑠. (41) Then 𝜌(𝑥)𝑉^′_𝑠=𝛽(𝜌^′(𝑥)𝑉^′_𝑠)⊆𝑉^′_𝑠. Let 𝛽_𝑠=𝛽|_𝑉′𝑠; then (𝑉^′_𝑠,𝛽_𝑠) is an irreducible Hom-𝔤-submodule, but (𝑉,𝛽) is not completely reducible.
(2) Suppose (omit the order of 𝑣^′₁₀,…,𝑣^′_𝑠0) 𝛽𝑣^′₁₀,𝑣^′₂₀,…,𝑣^′_𝑠0=𝑣^′₁₀,𝑣^′₂₀,…,𝑣^′_𝑠0𝜆₁⋱𝜆_𝑗𝐵_𝑗+1⋱𝐵_𝑡; (42) here 𝜆_𝑖∈ℂ,𝑖=1,…,𝑗,𝐵_𝑗+1,…,𝐵_𝑡 are 𝑘_𝑝×𝑘_𝑝 nondiagonal Jordan blocks (𝑘_𝑝∈ℤ⁺,𝑝=1,…,𝑡−𝑗).
Let 𝑉^′_𝑖=ℂ𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚,𝑖=1,…,𝑗, (43)
denoted by 𝛽_𝑖=𝛽|_𝑉′𝑖. It is easy to check that 𝛽_𝑖𝑉^′_𝑖⊆𝑉^′_𝑖,𝜌(𝑥)𝑉^′_𝑖=𝛽𝜌^′(𝑥)𝑉^′_𝑖⊆𝛽𝑉^′_𝑖⊆𝑉^′_𝑖,𝑖=1,…,𝑗. (44) By the statement of the proof and Lemma 10, we have that (𝑉^′_𝑖,𝛽_𝑖)(𝑖=1,…,𝑗) are irreducible Hom-𝔤-submodules.
Let 𝑉^′_𝑗+𝑝=ℂ𝑣^′_{𝑗+𝑘1+⋯+𝑘𝑝−1+1,0}𝑣,…,_{′𝑗+𝑘1+⋯+𝑘𝑝−1+1,𝑚}𝑣,…,_{′𝑗+𝑘1+⋯+𝑘𝑝,0}𝑣,…,_{′𝑗+𝑘1+⋯+𝑘𝑝,𝑚},𝑝=1,…,𝑡−𝑗. (45) Then 𝛽𝑉^′_𝑗+𝑝⊆𝑉^′_𝑗+𝑝,𝑝=1,…,𝑡−𝑗. (46) Let 𝛽_𝑖=𝛽|^′_𝑉𝑖,(𝑖=𝑗+1,…,𝑡); then (𝑉_𝑖,𝛽_𝑖)(𝑖=𝑗+1,…,𝑡) are Hom-𝔤-submodules. As (1) of the theorem, we can prove that (𝑉_𝑖,𝛽_𝑖)(𝑖=𝑗+1,…,𝑡) are reducible but not completely reducible Hom-𝔤-modules. We have the conclusion.
(3) In this case, by (30) and Theorem 12 (1) we get 𝛽𝑣^′₁₀…𝑣^′_1𝑚…𝑣^′_𝑠0…𝑣^′_𝑠𝑚^𝑡=𝑣^′₁₀…𝑣^′_1𝑚…𝑣^′_𝑠0…𝑣^′_𝑠𝑚^𝑡𝜆₁⋱𝜆₁𝑎^𝑚⋱𝜆_𝑠⋱𝜆_𝑠𝑎^𝑚. (47)
Let 𝑉^′_𝑖=ℂ𝑣^′_𝑖0,…,𝑣^′_𝑖𝑚,𝑖=1,…,𝑠; (48)
then 𝑉^′₁,…,𝑉^′_𝑠 are irreducible 𝐴₁-modules and 𝛽𝑉^′_𝑖⊆𝑉^′_𝑖𝜌(𝑥)𝑉,𝑖=1,…,𝑠,_′𝑖=𝛽𝜌^′(𝑥)𝑉^′_𝑖⊆𝑉^′_𝑖,𝑖=1,…,𝑠,∀𝑥∈𝔤. (49)
Denoting by 𝛽_𝑖=𝛽|_𝑉𝑖,𝑖=1,…,𝑠, thanks to Lemma 10 we have that (𝑉^′_𝑖,𝛽_𝑖)(𝑖=1,…,𝑠) are irreducible Hom-𝔤-submodules. So (𝑉,𝛽) is a completely reducible Hom-𝔤-module.

When 𝜏=(𝑚₁,…,𝑚₁,…,𝑚_𝑖,…,𝑚_𝑖,𝑚_𝑖+1,…,𝑚_𝑙), suppose that the multiplicity of 𝑚_𝑡 (𝑡=1,…,𝑖) is 𝑘_𝑡 and 𝑘₁+⋯+𝑘_𝑖+𝑙−𝑖=𝑠. By (18) we can get 𝛽 as follows: 𝛽𝑣₁₀…𝑣_1𝑚1…𝑣_𝑠0…𝑣_𝑠𝑚𝑙=𝑣₁₀…𝑣_1𝑚1…𝑣_𝑠0…𝑣_𝑠𝑚𝑙𝐵₁⋱𝐵_𝑖𝐵_𝑖+1⋱𝐵_𝑙, (50) where 𝐵₁,…,𝐵_𝑖 are matrices of the form (27), 𝐵_𝑗=diag𝜆_𝑗,𝜆_𝑗𝑎𝜆,…,^𝑗𝑎^𝑚𝑗,𝑗=𝑖+1,…,𝑙. (51)

Theorem 14. The condition is as the previous remark. (𝑉,𝛽) is completely reducible if there exist invertible matrices 𝑃_𝑗(𝑗=1,…,𝑖) such that 𝑃_𝑗⁻¹𝐵_𝑗𝑃_𝑗(𝑗=1,…,𝑖) are diagonal matrices. Otherwise (𝑉,𝛽) is reducible but not completely reducible.

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

Proposition 15. Let (𝜌,𝑉,𝛽) be a finite dimensional representation of (𝔤,[⋅,⋅],𝛼); then (𝜌,𝑉,𝛽) is not always completely reducible, and (𝜌,𝑉,𝛽) is irreducible if and only if it is of regular Lie-type.

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