# Post-Lie Algebra Structures on the Lie Algebragl(2,C)

Authors: Yuqiu Sheng; Xiaomin Tang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The post-Lie algebra is an enriched structure of the Lie algebra. We give a complete classification of post-Lie algebra structures on the Lie algebragl(2,C) up to isomorphism.


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## Body

## 1. Introduction

Post-Lie algebras were introduced around 2007 by Vallette [1], who found the structure in a purely operadic manner as the Koszul dual of a commutative trialgebra. Moreover, Vallette [1] proves that post-Lie algebras have the important algebraic property of being Koszul. This property is shared by many other important algebras, such as Lie algebras, associative algebras, commutative algebras, pre-Lie algebras, LR-algebras, and dendriform algebras, see [2, 3]. Recently, many authors study some post-Lie algebras and post-Lie algebra structures [4–8]. We recall the definition of the post-Lie algebra (structure) as follows, see [1, 8].Definition 1.
A (left) post-LieC-algebra is a C-vector space L with two binary operations ∘ and [,] which satisfy the following relations:
(1)[x,y]=-[y,x],(2)[[x,y],z]+[[z,x],y]+[[y,z],x]=0,(3)z∘(y∘x)-y∘(z∘x)+(y∘z)∘x-(z∘y)∘x+[y,z]∘x=0,(4)z∘[x,y]-[z∘x,y]-[x,z∘y]=0.
Let (σ(L),[,]) denote the Lie algebra defined by (1) and (2). Call (L,[,],∘) a post-Lie algebra on (σ(L),[,]).Remark 2.
Suppose(g,[,]) is a Lie algebra. Two post-Lie algebra (g,[,],∘) and (g,[,],⋆) on the Lie algebra g are called isomorphic on the Lie algebra (g,[,]) if there is an automorphism φ of the Lie algebra (g,[,]) such that
(5)φ(x∘y)=φ(x)⋆φ(y),∀x,y∈g.One of the key problems in the study of post-Lie algebras is to find the post-Lie algebra structures on the (given) Lie algebras. In [8], the authors determined all isomorphic classes of post-Lie algebra structures on (sl(2,C),[,]), the special linear Lie algebra of order 2. They use an important fact that the derivation of a semisimple Lie algebra is inner. But for the nonsemisimple Lie algebra, this fact dos not hold. So that we must find another way to study such problem for nonsemisimple Lie algebra. The purpose of this paper is to give a complete classification of post-Lie algebra structures on nonsemisimple Lie algebra gl(2,C), the general linear Lie algebra of order 2, up to isomorphism. Now, we recall the above two Lie algebras.Denote(6)u1:=12[01-10],u2:=12-1[0110],u3:=12-1[100-1],u4:=12[1001].
It is obvious that the previous four matrices form a C-linear basis of gl(2,C) and determine the Lie algebra (gl(2,C),[,]) through the Lie product
(7)[u1,u2]=u3,[u2,u3]=u1,[u3,u1]=u2,[ui,u4]=0,i=1,2,3.
It is also well known that u1,u2,u3 form a C-linear basis of sl(2,C) and determine the Lie algebra (sl(2,C),[,]) through the relations (7). The authors in [8] got the following classification theorem, which will be used in our proof.Theorem 3 (see [8]).
The following is a complete set of representatives for the isomorphic classes of post-Lie algebra(sl(2,C),[,],∘) on the Lie algebra (sl(2,C),[,]).(1)
ui∘uj=0,i,j=1,2,3;(2)
ui∘uj=[ui,uj],i,j=1,2,3;(3)
u1∘ui=[-u1,ui],u2∘ui=[-((1+-1)/2)u2+((-1-1)/2)u3,ui], u3∘ui=[-((1+-1)/2)u2+((-1-1)/2)u3,ui],i=1,2,3;(4)
u1∘ui=[(-1-(1/2))u1+(1-(-1/2))u2,ui], u2∘ui=[(1+(-1/2))u1-(-1+(1/2))u2,ui], u3∘ui=0,i=1,2,3;(5)
u1∘ui=[ku1,ui], u2∘ui=[-(1/2)u2+(-1/2)u3,ui], u3∘ui=[-(-1/2)u2-(1/2)u3,ui],i=1,2,3,k∈C*.


## 2. Equations from Post-Lie Algebras

From (4), we obtain u1∘u2=u1∘[u3,u1]=[u1∘u3,u1]+[u3,u1∘u1]∈sl(2,C) and u4∘u1=u4∘[u2,u3]=[u4∘u2,u3]+[u2,u4∘u3]∈sl(2,C). Similarly, we have ui∘uj∈sl(2,C) for any i∈{1,2,3,4} and j∈{1,2,3}. Thus, we get the following.Lemma 4.
(
s
l
(
2
,
C
)
,
[
,
]
,
∘
) is a post-Lie subalgebra of (gl(2,C),[,],∘).Proposition 5.
Let(gl(2,C),[,],∘) and (gl(2,C),[,],⋆) be post-Lie algebras on the Lie algebra gl(2,C), they are isomorphic through automorphism φ of the Lie algebra gl(2,C). Then, sl(2,C) is a φ-subspace of gl(2,C), and φ|sl(2,C) is an isomorphism from (sl(2,C),[,],∘) to (sl(2,C),[,],⋆).Proof.
SupposeQ=(qij)4×4 is the matrix of φ with respect to the basis {u1,u2,u3,u4}; that is,
(8)φ(ui)=∑j=13‍qijuj,i=1,2,3,4.
From φ([ui,u4])=φ(0)=0,i=1,2,3,4, we obtain φ(u4)=q44u4. Moreover, φ(u1)=φ([u2,u3])=[φ(u2),φ(u3)]∈sl(2,C) means q41=0. Similarly, we have q42=q43=0. Now, we see that Q has the form
(9)Q=[q11q21q310q12q22q320q13q23q330000q44].
From this we can easily get the conclusion.Proposition 6.
Let(gl(2,C),[,],∘) be post-Lie algebra on the Lie algebra gl(2,C).(1)
There exists a linear mapf:sl(2,C)→sl(2,C) such that
(10)x∘y=[f(x),y],∀x,y∈sl(2,C).(2)
There existb1,b2,b3,b4∈C such that ui∘u4=biu4, i=1,2,3,4.(3)
There existl12,l13,l23∈C such that
(11)(u4∘u1,u4∘u2,u4∘u3)=(u1,u2,u3)[0l12l13-l120l23-l13-l230].Proof.
(1) The conclusion is given by [8].
(2) For anyx,z∈gl(2,C), we have [x,z∘u4]=z∘[x,u4]-[z∘x,u4]=0 from (4) and (7). Thus, z∘u4 is in the center of gl(2,C), and so z∘u4=h(z)u4, where h is a linear map from gl(2,C) to C. Let bi=h(ui). The conclusion of (2) is proved.
(3) Letz=u4, {x,y}={u1,u2},{u2,u3}, and {u3,u1} in (4), one can get the conclusion (3) by a simple computation.Definition 7.
SupposeA is the matrix of f (from Proposition 6) with respect to the basis {u1,u2,u3}; that is, f(ui)=∑j=13‍aijuj,i=1,2,3. Denote
(12)L=[0l12l13-l120l23-l13-l230],β=[b1,b2,b3,b4].
The matrix set {A,L,β} is unique for a given ∘. On the other hand, ∘ is defined uniquely by the matrix set {A,L,β}. Because of their uniqueness, the matrix set {A,L,β} is called the matrix set of the post-Lie algebra (gl(2,C),[,],∘) and is also denoted by {A∘,L∘,β∘}.Proposition 8.
Suppose that{A∘,L∘,β∘} and {A⋆,L⋆,β⋆} are the matrix sets of post-Lie algebras (gl(2,C),[,],∘) and (gl(2,C),[,],⋆), respectively. φ is an isomorphic map from (gl(2,C),[,],∘) to (gl(2,C),[,],⋆). Then,(1)
(13)φ(u1,u2,u3,u4)=(u1,u2,u3,u4)Q,Q=[Q100q44],whereQ1∈SO(3,C), the group of all3×3complex orthogonal matrices whose determinants are(1);(2)
Q1A∘Q1-1=A⋆;(3)
β∘=β⋆Q;(4)
Q1L∘Q1-1=q44L⋆.Proof.
(1) The conclusion is given by Proposition5 and [8].
(2) It is given by [8].
(3) Letβ∘=(b1,b2,b3,b4) and β⋆=(b1′,b2′,b3′,b4′). Note that
(14)φ(u1∘u4,u2∘u4,u3∘u4,u4∘u4)=φ(b1u4,b2u4,b3u4,b4u4)=q44(b1u4,b2u4,b3u4,b4u4),(φ(u1)⋆φ(u4),φ(u2)⋆φ(u4),φ(u3)⋆φ(u4),φ(u4)⋆φ(u4))=q44(φ(u1)⋆u4,φ(u2)⋆u4,φ(u3)⋆u4,φ(u4)⋆u4)=q44(u1⋆u4,u2⋆u4,u3⋆u4,u4⋆u4)Q=q44(b1′u4,b2′u4,b3′u4,b4′u4)Q.
So, we have by φ(ui∘uj)=φ(ui)⋆φ(uj) that q44(b1′,b2′,b3′,b4′)Q=q44(b1,b2,b3,b4), that is, β∘=β⋆Q.
(4) Since(15)φ(u4∘u1,u4∘u2,u4∘u3)=φ(u1,u2,u3)L∘=(u1,u2,u3)Q1L∘,(16)(φ(u4)⋆φ(u1),φ(u4)⋆φ(u2),φ(u4)⋆φ(u3))=q44(u4⋆φ(u1),u4⋆φ(u2),u4⋆φ(u3))=q44(u4⋆u1,u4⋆u2,u4⋆u3)Q1=q44(u1,u2,u3)L⋆Q1=(u1,u2,u3)Q1L∘,
we deduce by φ(ui∘uj)=φ(ui)⋆φ(uj) that Q1L∘Q1-1=q44L⋆, the proof is completed.Proposition 9.
Suppose that{A,L,β} is the matrix set of post-Lie algebra (gl(2,C),[,],∘), then we have the following equations:
(17)l12b2+l13b3+b1b4=0,l12b1-l23b3-b2b4=0,l13b1+l23b2-b3b4=0,i.e.,(b4I3+L)[b1b2b3]=0,(18)(a11+a22+1)b3-a13b1-a23b2=0,(a22+a33+1)b1-a21b2-a31b3=0,(a11+a33+1)b2-a12b1-a32b3=0,i.e.,((trA+1)I3-A)[b1b2b3]=0,(19)(b1-a23)l12+(a11-a33)l13+a12l23=0,(a12+a21)l12+(a13+a31)l13-b1l23=0,(a12+a21)l12-b2l13-(a23+a32)l23=0,(a11-a22)l12-a23l13+(b2-a31)l23=0,(a22-a11)l12+(a32+b1)l13+a13l23=0,(b2+a13)l12+a21l13+(a22-a33)l23=0,-a31l12+(b3-a12)l13+(a33-a22)l23=0,-a32l12+(a11-a33)l13+(b3+a21)l23=0,b3l12+(a13+a31)l13+(a23+a32)l23=0.Proof.
We consider (3) and (10). Let y=x=u4 and z=u1,u2, and u3, respectively, we get (17); Let x=u4, {z,y}={u2,u1}, {u3,u2} and {u3,u1}, respectively, we get (18); let z=u4 and {y,x}={u1,u2}, {u2,u3}, {u1,u1}, {u2,u1}, {u3,u3}, and {u3,u1}, respectively, we get (19).


## 3. Classification of Post-Lie Algebra(gl(2,C),[,],∘)

Lemma 10 (see [9]).
Suppose thatA is a complex skewsymmetric 3×3 matrix, there exists T∈O(3,C) such that
(20)TAT-1=[01+-120-1--120-1+-1201--120]or[0a0-a00000].
Based on Definition 7, we get the main result in this paper as follows.Theorem 11.
The following is a complete set of matrix sets of representatives for the isomorphic classes of post-Lie algebra(gl(2,C),[,],∘) on the Lie algebra (gl(2,C),[,])(21)(1)A=0,L=[01+-120-1--120-1+-1201--120],β=[0,0,0,b4],b4=0or1;(2)A=0,L=[0a0-a00000],β=[0,0,0,b4],a∈C,b4=0or1;(3)A=-I3,L=[01+-120-1--120-1+-1201--120],β=[0,0,0,b4],b4=0or1;(4)A=-I3,L=[0a0-a00000],β=[0,0,0,b4],a∈C,b4=0or1;(5)A=[-1000-1--12-1--120-1+-12-1+-12],L=0,β=[b1,-1b3,b3,0],b1,b3∈C;(6)A=[-1000-1--12-1--120-1+-12-1+-12],L=0,β=[0,0,0,1];(7)A=[k000-12--120-12-12],L=0,β=[b1,0,0,0],k≠-1,b1∈C*;(8)A=[k000-12--120-12-12],L=[00000a0-a0],β=[0,0,0,b4],k≠-1,a∈C,b4=0or1;(9)A=[-1000-12--120-12-12],L=[00000a0-a0],β=[0,0,0,1],a∈C;(10)A=[-1000-12--120-12-12],L=0,β=[b1,-1b3,b3,0],b1,b3∈C;(11)A=[-1000-12--120-12-12],L=[0000010-10],β=0;(12)A=[-1-121+-1201--12--1-120000],L=0,β=[0,0,0,1];(13)A=[-1-121+-1201--12--1-120000],L=0,β=[0,0,b3,0],b3∈C;(14)A=[-1-121+-1201--12--1-120000],L=0,β=[-1b2,b2,b3,0],b2∈C*,b3∈C.Proof.
Case  1.
A
=
0. By (18), we obtain b1=b2=b3=0; that is, β=[0,0,0,b4]. By Lemma 10 we have
(22)L=[01+-120-1--120-1+-1201--120]or[0a0-a00000].
Case  1.1. Consider
(23)L=[01+-120-1--120-1+-1201--120].
If b4≠0, let
(24)Q1=[b42+12b40-1(b42-1)2b4010-1(1-b42)2b40b42+12b4],(25)Q=[Q100b4];
then
(26)Q1AQ1-1=A,Q1LQ1-1=b4L,β=[0,0,0,1]Q.
In view of Proposition 8, we can suppose that
(27)A=0,L=[01+-120-1--120-1+-1201--120],β=[0,0,0,b4],b4=0or1.
Case  1.2. Consider L=[0a0-a00000]. If b4≠0, let
(28)Q1=I3,Q=[I300b4];
then
(29)Q1AQ1-1=A,Q1[0a0-a00000]Q1-1=b4[0ab4-10-ab4-100000],β=[0,0,0,1]Q.
Thus, we can suppose that A=0,L=[0a0-a00000],andβ=[0,0,0,b4],a∈C,b4=0 or 1.
Case  2. Consider A=-I3. By Lemma 10 we have
(30)L=[01+-120-1--120-1+-1201--120]or[0a0-a00000].
By (18), we obtain β=[0,0,0,b4]. In a similar way with the proof of Case 1, we can suppose that β=[0,0,0,b4],b4=0 or 1.
Case  3. Consider
(31)A=[-1000-1--12-1--120-1+-12-1+-12].
By (18), we obtain b2=-1b3. Then, by (19), we get
(32)l23=0,l12=l13,b2l12=b3l12=(b1-i)l12=0.
Therefore, by (17), we can have
(33)l12=l13=0,b1b4=b2b4=b3b4=0.
Hence, L=0. If b4≠0, then b1=b2=b3=0. In a similar way with the proof of Case 1, we can suppose that β=[0,0,0,1]. Else, if b4=0, then β=[b1,-1b3,b3,0],b1,b3∈C.
Case  4. Consider
(34)A=[k000-12--120-12-12].
By (18), we obtain
(35)b2=b3=0,k≠-1,b2=-1b3,k=-1.
Case  4.1. When k≠-1, b2=b3=0. By (17), we obtain b1b4=b1l12=b1l13=0. Then, by (19), we get l12=l13=b1l23=0.
Ifb1≠0, then b4=l23=0.
Therefore,(36)L=0,β=[b1,0,0,0],b1∈C*orL=[00000a0-a0],β=[0,0,0,b4],a,b4∈C.
In a similar way with the proof of Case 1, we can suppose thatb4=0 or 1.
Case  4.2. When k=-1, b2=-1b3. By (19), we obtain
(37)bil12=bil13=bil23=0i=2,3,b1l23=0,l13=-1l12,(b1--1)l13=0.
This, together with (17), implies that l12=l13=0,bib4=0,andi=1,2,3.
Ifb4≠0, then bi=0,andi=1,2,3. Hence, in a similar way with the proof of Case 1, we can suppose that
(38)L=[00000a0-a0],β=[0,0,0,1].
If b4=0 and l23=0, then
(39)L=0,β=[a,-1c,c,0],a,c∈C.
If b4=0 and l23≠0, then bi=0,i=1,2,3. Hence,
(40)L=[00000a0-a0],β=0,a=l23∈C*.
Let
(41)Q1=I3,Q=[I300a];
then
(42)Q1AQ1-1=A,Q1[00000a0-a0]Q1-1=a[0000010-10],0=β=βQ.
Thus, we can suppose thatL=[0000010-10],β=0.
Case  5. Consider
(43)A=[-1-121+-1201--12--1-120000].
By (18), we obtain b1=-1b2. Then, by (19), we get
(44)-1b2l12+(-1-12)l13+(1--12)l23=0,(45)2l12--1b2l23=0,(46)2l12-b2l13=0,(47)(b3-1+-12)l13+(12+-1)l23=0,(48)(-12+-1)l13+(b3+1+-12)l23=0.
From (45) and (46), we have b2(l13--1l23)=0.
Case  5.1. When b2=0, b1=0. From (45), we have l12=0. Then by (44), (47), and (48) we obtain l13=l23=0. Therefore, by (17), we get b3b4=0. So, L=0,β=[0,0,b3,b4],b3b4=0. In a similar way with the proof of Case 1, we can suppose that
(49)L=0,β=[0,0,0,1]orβ=[0,0,b3,0],b3∈C.
Case  5.2. When b2≠0, l13=-1l23. From (44), we have b2l12=-1l23=l13. Then, by (45) and (46), we get 2-1l23=2b2l12=-1b22l23,2l13=2b2l12=b22l13 and 2l12=b2l13=b22l12; that is, (b22-2)l23=(b22-2)l13=(b22-2)l12=0.
Ifb22≠2, then l23=l13=l12=0.
Else, ifb22=2, from (44) we obtain that -1b22l12+(-1-(1/2))b2l13+(1-(-1/2))b2l23=0; that is, 2-1l12+(-1-(1/2))2l12--1(1-(-1/2))-1b2l23=2-1l12+(-1-(1/2))2l12--1(1-(-1/2))2l12=0; thus, l12=0. Therefore, by (45) and (46), we conclude that l13=l23=0.
Therefore, we can getb4=0 by (17). So,
(50)L=0,β=[-1b2,b2,b3,0],b2∈C*,b3∈C.
The sufficiency of Theorem11 is obvious from the proof of the necessity.



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