# Representations of 3-Dimensional Simple Multiplicative Hom-Lie Algebras

Authors: Xiuxian Li
Journal: Advances in Mathematical Physics
 (2013)

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## 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.


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## Body

## 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 withα 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 Section2 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 fieldℂ.


## 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
(1)[x,y]=-[y,x],∀x,y∈𝔤,(2)[α(x),[y,z]]+[α(y),[z,x]]+[α(z),[x,y]]=0,∀x,y,z∈𝔤.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 subspace𝔤1 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
(3)C(𝔤)={x∈𝔤∣[x,𝔤]=0;[α(x),𝔤]=0}.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
(4)α([x,y])=[α(x),α(y)]⟹α2([x,y]′)=α([α(x),α(y)]′),
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,
(5)↺x,y,z[x,[y,z]′]′=↺x,y,zα-1[x,α-1([y,z])]=↺x,y,zα-1[α-1α(x),α-1([y,z])]=↺x,y,zα-2([α(x),[y,z]])=0,
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
(6)(𝔤,[·,·],α)={h,e,f∣[h,e]=2ae,[h,f]=-2a-1f,[e,f]=h},α(h,e,f)=(h,e,f)(1000a000a-1)0≠a∈ℂ.Proof.
IfKer(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
(7)[α(x),y]=[α(x),α(y′)]=α([x,y′])=0.
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
(8)[𝔤,𝔤]=α-1([𝔤,𝔤]′)=[α-1(𝔤),α-1(𝔤)]′=[𝔤,𝔤]′≠𝔤,
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=expadkh,k∈ℂ. Thus
(9)α(h,e,f)=(h,e,f)(1000exp2k000exp-2k).
Now it follows that
(10)[h,e]=α([h,e]′)=2exp2ke,[h,f]=α([h,f]′)=-2exp-2kf,[e,f]=α([e,f]′)=h.
Leta=exp2k≠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
(11)ρ([x,y])β(v)=ρ(α(x))ρ(y)v-ρ(α(y))ρ(x)v,∀x,y∈𝔤,v∈V,(12)3β(ρ(x)v)=ρ(α(x))β(v),∀x∈𝔤,v∈V,
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⊕⋯⊕Vs, where β(Vi)⊆Vi(i=1,…,s) and (Vi,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
(13)ρ(α(x))=βρ(x)β-1,∀x∈𝔤.
Equation (11) is equivalent to
(14)ρ[x,y]=βρ(x)β-1ρ(y)β-1-βρ(y)β-1ρ(x)β-1,∀x∈𝔤,v∈V.
Denote that byα(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
(15)ρ(α-1(x))=β-1ρ(x)β,∀x∈𝔤.
On vector spaceβ-1ρ(𝔤), for all x,y∈𝔤, define a commutator bracket [·,·]′ as
(16)[β-1ρ(x),β-1ρ(y)]′=β-1ρ(x)β-1ρ(y)-β-1ρ(y)β-1ρ(x)=[ρ′(x),ρ′(y)]′.
Clearly,(β-1ρ(𝔤),[·,·]′) is a Lie algebra.
On the other hand, for allx,y∈𝔤,
(17)ρ′([x,y]𝔤′)=ρ′(α-1([x,y]))=β-1ρ([α-1(x),α-1(y)])=β-1(βρ(α-1(x))β-1ρ(α-1(y))β-1-βρ(α-1(y))β-1ρ(α-1(x))β-1)=β-1ρ(x)ββ-1β-1ρ(y)ββ-1-β-1ρ(y)ββ-1β-1ρ(x)ββ-1=β-1ρ(x)β-1ρ(y)-β-1ρ(y)β-1ρ(x)=[ρ′(x),ρ′(y)]′.
Then the result follows easily.From Proposition6, 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
(18)β(ρ′(x)v)=ρ′(α(x))β(v),∀x∈𝔤,v∈V,
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 ofβ and (12), we can get (18) easily.
(2)forallx∈𝔤,v∈V, (12) follows from (18) easily. forallx,y∈𝔤,v∈V, we have
(19)ρ([x,y])β(v)=β(ρ′(α([x,y]′))β(v))=β((ρ′(α(x))ρ′(α(y))-ρ′(α(y))ρ′(α(x)))β(v))=β(ρ′(α(x))β(ρ′(y)v)-ρ′(α(y))β(ρ′(x)v))=ρ(α(x))ρ(y)v-ρ(α(x))ρ(y)v.
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.
for
all
v
∈
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 Lemma8, 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⊕⋯⊕Vs, where (Vi,β|Vi) (i=1,…,s) are irreducible Hom-𝔤 submodules. By the irreducibility of (Vi,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
(20)ρ′(x)V1=β-1(ρ(x)V1)⊆β-1(V1)⊆V1.
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,…,vm} of V such that(a)
ρ(h)vi=(m-2i)a-ib0vi,β(vi)=a-ib0vi,i=0,…,m;(b)
ρ(f)vi=(i+1)a-i-1b0vi+1,i=0,…,m-1;ρ(f)vm=0;(c)
ρ(e)v0=0,ρ(e)vi=(m-i+1)a-i+1b0vi-1,i=1,…,m.Proof.
Let(ρ′,V) be an m+1 dimensional irreducible representation of the compatible Lie algebra A1. Take {v0,v1,…,vm} as a basis of V, where vi=1/iρ′(f)vi-1(i=1,…,m) satisfies
(21)ρ′(h)vi=(m-2i)vi,ρ′(e)vi=(m-i+1)vi-1(i=1,…,m),ρ′(e)v0=0.
Now we prove that whenβ∈gl(V) is defined by
(22)β(v0v1⋮vm)T=(v0v1⋮vm)T(b0a-1b0⋱a-mb0),b0≠0,
(18) is always satisfied. Because(23)β(ρ′(h)vi)=(m-2i)β(vi)=(m-2i)a-ib0vi,ρ′(α(h))β(vi)=ρ′(h)a-ib0vi=a-ib0(m-2i)vi=β(ρ′(h)vi),ffffff0.ffffi=0,…,m,β(ρ′(f)vi)=(i+1)β(vi+1)=(i+1)a-i-1b0vi+1,ρ′(α(f))β(vi)=a-1ρ′(f)a-ib0vi=a-i-1b0(i+1)vi+1=β(ρ′(f)vi),i=0,…,m,β(ρ′(e)vi)=β((m-i+1)vi-1)=(m-i+1)β(vi-1)=(m-i+1)a-(i-1)b0vi-1,ρ′(α(e))β(vi)=aρ′(e)β(vi)=ρ′(e)a-(i-1)b0vi=a-(i-1)b0(m-i+1)vi-1=β(ρ′(e)vi),i=0,…,m.
Thus forβ 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(24)ρ(h)vi=β(ρ′(h)vi)=(m-2i)a-ib0vi,i=0,…,m,
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)vi=β(ρ′(x)vi)(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 Theorem11, 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 Lemma9, 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⊕⋯⊕Vs, where Vi(i=1,…,s) are irreducible A1-modules. Suppose that dimVi=mi+1. Denote that a tuple τ=(m1,…,ms). By the representation theory of A1, we have that Vj(j=1,…,s) is a highest weight module with highest weight vectors vj0 and the highest weight mj, respectively. Take {vj0,vj1,…,vjmj} as a basis of Vj satisfying
(25)vji=1iρ′(f)vj,i-1,(i=1,…,mj),ρ′(h)vji=(mj-2i)vji,(i=0,…,mj),ρ′(e)vji=(mj-i+1)vj,i-1,(i=1,…,mj),ρ′(e)v0=0.Theorem 12.
Whenτ=(m1,…,ms), mi≠mj for 1≤i≠j≤s, then(1)
β=diag(…,bi0,bi0/a,…,bi0/ami,…),i=1,…,s;(2)
(V,β) is a completely reducible Hom-𝔤-module with a decomposition V=V1⊕⋯⊕Vs, where (Vi,βi),(i=1,…,s) are irreducible Hom-𝔤-submodules; here βi=β|Vi.Proof.
(1) According to (18), we get the result.
(2) Because(Vj)⊆Vj,j=1,…,s,
(26)ρ(x)Vj=β(ρ′(x)Vj)⊆Vj.Let βj=β|Vj, combined with Lemma 10; we have that (Vj,βj)(j=1,…,s) are irreducible Hom-𝔤-submodules.Whenτ=(m,…,m), by (18) it can be checked that
(27)β=(B11B12…B1sB21B22…B2s⋮⋮⋱⋮Bs1Bs2…Bss),
where Bij=diag(bij,bij/a,…,bij/am).Take(28)B1=(b11b12…b1sb21b22…b2s⋮⋮⋱⋮bs1bs2…bss).By the theory of linear algebra, there is an invertible matrix(29)P=(p11p12…p1sp21p22…p2s⋮⋮⋱⋮ps1ps2…pss)
such that P-1B1P is a Jordan canonical form; that is,
(30)P-1B1P=(J11J22⋱Jtt),
where Jii=(λi1λi⋱⋱1λi)(i=1,…,t) are Jordan blocks.Theorem 13.
The condition is the same as the previous remark.(1)
Ift=1, then (V,β) is a reducible but not completely reducible Hom-𝔤-module.(2)
Ifs>t>1, then (V,β) is a reducible but not completely reducible Hom-𝔤-module.(3)
Ift=s, then (V,β) is a completely reducible Hom-𝔤-module.Proof.
Let(31)vi0′=p1iv10+p2iv20+⋯+psivs0,i=1,…,s,(32)vij′=1jρ′(f)vi,j-1′,i=1,…,s;j=1,…,m.
That is,
(33)vij′=p1iv1j+p2iv2j+⋯+psivsj,i=1,…,s;j=1,…,m.
Then
(34)(v10′,…,v1m′,…,vs0′,…,vsm′)=(v10,…,v1m,…,vs0,…,vsm)Q,
where
(35)Q=(Q11Q12…Q1sQ21Q22…Q2s⋮⋮⋱⋮Qs1Qs2…Qss);
here Qij are m×m matrices satisfying
(36)Qij=diag(pij,…,pij),i,j=1,…,s.
Because P is invertible, it is easy to check that the matrix Q is invertible; therefore {v10′,…,v1m′,…,vs0′,…,vsm′} is a basis of V.
LetVi′=ℂ{vi0′,vi1′,…,vim′}, because
(37)ρ′(h)vi0′=ρ′(h)(p1iv10+p2iv20+⋯+psivs0)=mvi0.
By (32) and the representation of Lie algebra A1, we have
(38)ρ′(h)(vi0′,…,vim′)=(vi0′,…,vim′)(mm-2⋱-m),ρ′(f)(vi0′,…,vim′)=(vi0′,…,vim′)(00…0010…0002…00⋮⋮⋱⋮⋮00…m0),ρ′(e)(vi0′,…,vim′)=(vi0′,…,vim′)×(0m0…000m-1…0⋮⋮⋮⋱⋮000…1000…0).
Through (38) it is easy to get that V1′,…,Vs′ are A1-irreducible-modules.
(1) In this case, by (30), we have
(39)β(v10′,v20′,…,vs0′)=(v10′,v20′,…,vs0′)(λ001λ⋱⋱1λ).
Let
(40)Vi′=ℂ{vi0′,…,vim′},i=1,…,s,
By (32), it is easy to check that
(41)β(Vi′)⊆Vi′+Vi+1′,i=1,…,s-1;β(Vs′)⊆Vs′.
Then ρ(x)Vs′=β(ρ′(x)Vs′)⊆Vs′. Let βs=β|Vs′; then (Vs′,βs) is an irreducible Hom-𝔤-submodule, but (V,β) is not completely reducible.
(2) Suppose (omit the order ofv10′,…,vs0′)
(42)β(v10′,v20′,…,vs0′)=(v10′,v20′,…,vs0′)(λ1⋱λjBj+1⋱Bt);
here λi∈ℂ,i=1,…,j,Bj+1,…,Bt are kp×kp nondiagonal Jordan blocks (kp∈ℤ+,p=1,…,t-j).
Let(43)Vi′=ℂ{vi0′,…,vim′},i=1,…,j,
denoted byβi=β|Vi′. It is easy to check that
(44)βi(Vi′)⊆Vi′,ρ(x)Vi′=β(ρ′(x)Vi′)⊆β(Vi′)⊆Vi′,i=1,…,j.
By the statement of the proof and Lemma 10, we have that (Vi′,βi)(i=1,…,j) are irreducible Hom-𝔤-submodules.
Let(45)Vj+p′=ℂ{vj+k1+⋯+kp-1+1,0′,…,000000000vj+k1+⋯+kp-1+1,m′,…,000000000vj+k1+⋯+kp,0′,…,000000000vj+k1+⋯+kp,m′},p=1,…,t-j.
Then
(46)β(Vj+p′)⊆Vj+p′,p=1,…,t-j.
Let βi=β|Vi′,(i=j+1,…,t); then (Vi,βi)(i=j+1,…,t) are Hom-𝔤-submodules. As (1) of the theorem, we can prove that (Vi,βi)(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
(47)β(v10′…v1m′…vs0′…vsm′)t=(v10′…v1m′…vs0′…vsm′)t(λ1⋱λ1am⋱λs⋱λsam).
Let(48)Vi′=ℂ{vi0′,…,vim′},i=1,…,s;
thenV1′,…,Vs′ are irreducible A1-modules and
(49)β(Vi′)⊆Vi′,i=1,…,s,ρ(x)Vi′=β(ρ′(x)Vi′)⊆Vi′,i=1,…,s,∀x∈𝔤.
Denoting byβi=β|Vi,i=1,…,s, thanks to Lemma 10 we have that (Vi′,βi)(i=1,…,s) are irreducible Hom-𝔤-submodules. So (V,β) is a completely reducible Hom-𝔤-module.Whenτ=(m1,…,m1,…,mi,…,mi,mi+1,…,ml), suppose that the multiplicity of mt (t=1,…,i) is kt and k1+⋯+ki+l-i=s. By (18) we can get β as follows:
(50)β(v10…v1m1…vs0…vsml)=(v10…v1m1…vs0…vsml)(B1⋱BiBi+1⋱Bl),
where B1,…,Bi are matrices of the form (27),
(51)Bj=diag(λj,λja,…,λjamj),j=i+1,…,l.Theorem 14.
The condition is as the previous remark.(V,β) is completely reducible if there exist invertible matrices Pj(j=1,…,i) such that Pj-1BjPj(j=1,…,i) are diagonal matrices. Otherwise (V,β) is reducible but not completely reducible.Proof.
It can be got from Theorems12 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 Theorems11, 12, 13, and 14 directly.



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