# Rota-Baxter Operators on 3-Dimensional Lie Algebras and the ClassicalR-Matrices

Authors: Linli Wu; Mengping Wang; Yongsheng Cheng
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

Our aim is to classify the Rota-Baxter operators of weight 0 on the 3-dimensional Lie algebra whose derived algebra’s dimension is 2. We explicitly determine all Rota-Baxter operators (of weight zero) on the 3-dimensional Lie algebrasg. Furthermore, we give the corresponding solutions of the classical Yang-Baxter equation in the 6-dimensional Lie algebras g⋉ad⁎g⁎ and the induced left-symmetry algebra structures on g.


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## Body

## 1. Introduction

In physics, the Yang-Baxter equation is a consistency equation which was first introduced in the field of statistical mechanics. It depends on the idea that, in some scattering situations, particles may preserve their momentum while changing their quantum internal states. Rota-Baxter algebra started with the probability study and has since found applications in many areas of mathematics and physics, such as quasi-symmetric functions, number theory, dendriform algebras, and Yang-Baxter equations.A Rota-Baxter operator (of weight zero) on an associative algebraA is defined to be a linear map P:g→g satisfying(1)PxPy=PPxy+xPy,∀x,y∈A. Rota-Baxter operators (on associative algebras) were introduced by Baxter to solve an analytic formula in probability [1–4]. It has been related to other areas in mathematics and mathematical physics [5–9]. A Rota-Baxter operator (of weight zero) on a Lie algebra (g,[·,·]) is a linear operator P:g→g such that(2)Px,Py=PPx,y+x,Py,∀x,y∈g.In fact, a Rota-Baxter operator is also called the operator form of the classical Yang-Baxter equation [10–13]. Let g be a Lie algebra and r=∑iai⊗bi∈g⊗g. r is called a classical R-matrix if it is a solution of the classical Yang-Baxter equation (CYBE) in g: that is,(3)r12,r13+r12,r23+r13,r23=0in U(g), where U(g) is the universal enveloping algebra of g and(4)r12=∑iai⊗bi⊗1,r13=∑iai⊗1⊗bi,r23=∑i1⊗ai⊗bi. Set r21=∑ibi⊗ai. It is easy to obtain that r is skew-symmetric if and only if r=-r21. Semenov-Tian-Shansky proved in [14] that r is skew-symmetric and there is a nondegenerate symmetric invariant bilinear form on Lie algebra g; relation (2) is equivalent to relation (3) when the weight is zero. Furthermore, Rota-Baxter operators of weights 0 and 1 on a Lie algebra g give rise to solutions of CYBE on the double Lie algebra g⋉ad∗g∗ over the direct sum g⊕g∗ of the Lie algebra g and its dual space g∗ [12, 15, 16]. Moreover, we can get some solutions of CYBE in g⋉ad∗g∗ Lie algebras through Rota-Baxter operators of any weight on g.In [12], the authors gave all Rota-Baxter operators (of weight zero) on 3-dimensional simple Lie algebra sl(2,C). The aim of this paper is to determine the Rota-Baxter operators (of weight zero) on the 3-dimensional Lie algebra which is not simple, and the dimension of its derived algebra is 2. We will determine the Rota-Baxter operators on the Lie algebra g and give a family of solutions of CYBE in g⋉ad∗g∗. This paper is organized as follows. In Section 2, we give the classification theorem of Rota-Baxter operators (of weight zero) on g. In Section 3, we give the corresponding solutions of CYBE in g⋉ad∗g∗. In Section 4, we give the corresponding left-symmetry structure on g.


## 2. The Rota-Baxter Operators ong (of Weight Zero)

### 2.1. Notations and the Classification Theorem

Letg be a 3-dimensional linear Lie algebra whose standard (Cartan-Weyl) basis consists of e1, e2, e3 over the field of complex numbers C with the following Lie brackets:(5)e1,e2=e1,e1,e3=0,e2,e3=e1+e3. Thus, a linear operator P:g→g is determined by (6)Pe1Pe2Pe3=b11b12b13b21b22b23b31b32b33e1e2e3,where bij∈C,1≤i,j≤3. P is a Rota-Baxter operator on g if the above matrix (bij)3×3 satisfies (2). Here is our main theorem.Theorem 1.
All Rota-Baxter operators of weight zero ong are listed in their matrices form with respect to the Cartan-Weyl basis below, where a, b, and c are nonzero complex numbers. (7)P1=000000a10,P2=0000a-a201-a,P3=000000010,P4=000aba-a2b1-a,P5=000a01000,P6=000100000,P7=000001000,P8=000000000,P9=000a0b100,P10=000a00100,P11=00000a100,P12=000000100,P13=000a1b000,P14=000a10000,P15=00001a000,P16=000010000,P17=001a0b00-1,P18=a0100ba2-a0a-1a≠12,P19=a01b00a2-a0a-1a≠12,P20=00100a00-1,P21=001a0000-1,P22=a01000a2-a0a-1a≠12,P23=00100000-1,P24=a01b0ca2-a0a-1a≠12,P25=120100a-140-12,P26=1201a00-140-12,P27=1201000-140-12,P28=1201a0b-140-12,P29=012a0a2a20-12-a,P30=a12a000-a2-12-a,P31=0100000-120,P32=a12a+babb-2b2-2ab-a2-12-a+b.


### 2.2. Reduction to Quadratic Equations

In order to show thatP is a Rota-Baxter operator, we only check the following: (8)Pe1,Pe2=PPe1,e2+e1,Pe2,Pe2,Pe3=PPe2,e3+e2,Pe3,Pe1,Pe3=PPe1,e3+e1,Pe3. It follows from (5) and (6) that(9)Pe2,Pe3=b21b32-b22b31+b22b33-b23b32e1+b22b33-b23b32e3, while(10)PPe2,e3+e2,Pe3=Pb22-b31+b33e1-Pb22+b33e3=b22b11-b31b11+b33b11+b22b31+b33b31e1+b22b12-b31b12+b33b12+b22b32+b33b32e2+b22b13-b31b13+b33b13+b22b33+b33b33e3. Comparing the coefficients in (9) and (10), we have(11)b21b32-2b22b31+b22b33-b23b32-b22b11+b31b11-b33b11-b33b31=0,(12)b22b12-b31b12+b33b12+b22b32+b33b32=0,(13)b23b32+b22b13-b31b13+b33b13+b33b33=0. Similarly, from (14)Pe1,Pe2=PPe1,e2+e1,Pe2,Pe1,Pe3=PPe1,e3+e1,Pe3, we obtain the following six equations:(15)b11b11+b12b21-b13b11-b13b31-b12b23+b13b22=0,(16)b11b12+b22b12-b13b12-b13b32=0,(17)b11b13+2b22b13-b13b13-b13b33-b12b23=0,(18)b12b33-2b12b31-b13b32-b12b11=0,(19)b12b12+2b32b12=0,(20)b12b13+2b32b13=0.


### 2.3. Solving the Quadratic Equations

Equation (19) implies b12(b12+2b32)=0. To solve the quadratic equations (11), (12), (13), (15), (16), (17), (18), (19), and (20), we distinguish the following cases depending on whether b12=0 or not.Case 1.
b
12
=
0, b12+2b32≠0. That is, b12=0, b32≠0, taking b32=1. Equation (16) implies b13=0. Equation (15) implies b11=0. Equation (11) implies b21=b22b31+b222+b23. Equation (12) implies b33=-b22. Equation (13) implies b23=-b332=-b222. We obtain(21)P=000b22b31b22-b222b311-b22. Taking b22=0, b31=a, we obtain P1. Taking b22=a, b31=0, we obtain P2. Taking b22=0, b31=0, we obtain P3. Taking b22=a, b31=b, we obtain P4.Case 2.
Assumeb12=0, b12+2b32=0. That is, b12=0, b32=0. We distinguish the two cases depending on whether b13=0 or not.
Subcase 2.1.If b13=0, then (13) implies b33=0. Equation (15) implies b11=0. Equation (11) implies b22b31=0.
Subcase 2.1.1.If b22=0, b31=0, we obtain(22)P=000b210b23000. Taking b21=a, b23=1, we obtain P5. Taking b21=1, b23=0, we obtain P6. Taking b21=0, b23=1, we obtain P7. Taking b21=0, b23=0, we obtain P8.
Subcase 2.1.2.If b22=0, b31≠0, taking b31=1, we obtain(23)P=000b210b23100. Taking b21=a, b23=b, we obtain P9. Taking b21=a, b23=0, we obtain P10. Taking b21=0, b23=a, we obtain P11. Taking b21=0, b23=0, we obtain P12.
Subcase 2.1.3.If b22≠0, b31=0, taking b22=1, we obtain(24)P=000b211b23000. Taking b21=a, b23=b, we obtain P13. Taking b21=a, b23=a, we obtain P14. Taking b21=0, b23=a, we obtain P15. Taking b21=0, b23=0, we obtain P16.
Subcase 2.2.If b13≠0, taking b13=1, (17) implies b22=b33-b11+1/2. Then (15) implies b31=2b112-3b11+b33+1/2. Equation (13) implies b11+b33=b112-b332. That is, (b11+b33)(b11-b33-1)=0.
Subcase 2.2.1.If b11+b33≠0, b11-b33-1=0, (b11≠1/2), and then b22=0, b31=b112-b11, we obtain(25)P=b1101b210b23b112-b110b11-1,b11≠12.Taking b11=0,  b21=a, b23=b, we obtain P17. Taking b11=a, b21=0, b23=b  (a≠1/2), we obtain P18. Taking b11=a, b21=b, b23=0  (a≠1/2), we obtain P19. Taking b11=0, b21=0, b23=a, we obtain P20. Taking b11=0, b21=a, b23=0, we obtain P21. Taking b11=a, b21=0, b23=0  (a≠1/2), we obtain P22. Taking b11=0, b21=0, b23=0, we obtain P23. Taking b11=a, b21=b, b23=c  (a≠1/2), we obtain P24.
Subcase 2.2.2.If b11+b33=0, b11-b33-1≠0(b11≠1/2), and then b22=-2b11+1/2, b31=2b112-4b11+1/2, (3.4) implies 8b113-12b112+6b11-1=0. Then we have 8(b11-1/2)3=0, b11=1/2, giving a contradiction.
Subcase 2.2.3.If b11+b33=0, b11-b33-1=0, that is, b11=1/2, b33=-1/2, and then b22=0, b31=-1/4, we obtain(26)P=1201b210b23-140-12.Taking b21=0, b23=a, we obtain P25. Taking b21=a,  b23=0, we obtain P26. Taking b21=0, b23=0, we obtain P27. Taking b21=a, b23=b, we obtain P28.Case 3.
Assumeb12≠0,  (b12+2b32)=0. Taking b12=1, then b32=-1/2. Equation (16) implies b13=2b11+2b22. Equations (12) and (18) imply b31=b22+b33/2. Equation (17) implies b23=-2b112-2b11b22-2b11b33-2b22b33. Equation (15) implies b21=-b11b22-b11b33-b22b33-b112-b222. Equations (11), (15), and (17) imply b112+b222+b332+2b11b22+2b11b33+2b22b33=0. Then we have (b11+b22+b33)2=0. So b33=-(b11+b22), b21=b11b22, b23=-2b222-2b11b22, b31=-b11/2. We obtain(27)P=b1112b11+2b22b11b22b22-2b222-2b11b22-b112-12-b11+b22. Taking b11=0, b22=a, we obtain P29. Taking b11=a, b22=0, we obtain P30. Taking b11=0, b22=0, we obtain P31. Taking b11=a, b22=b, we obtain P32.


## 2.1. Notations and the Classification Theorem

Letg be a 3-dimensional linear Lie algebra whose standard (Cartan-Weyl) basis consists of e1, e2, e3 over the field of complex numbers C with the following Lie brackets:(5)e1,e2=e1,e1,e3=0,e2,e3=e1+e3. Thus, a linear operator P:g→g is determined by (6)Pe1Pe2Pe3=b11b12b13b21b22b23b31b32b33e1e2e3,where bij∈C,1≤i,j≤3. P is a Rota-Baxter operator on g if the above matrix (bij)3×3 satisfies (2). Here is our main theorem.Theorem 1.
All Rota-Baxter operators of weight zero ong are listed in their matrices form with respect to the Cartan-Weyl basis below, where a, b, and c are nonzero complex numbers. (7)P1=000000a10,P2=0000a-a201-a,P3=000000010,P4=000aba-a2b1-a,P5=000a01000,P6=000100000,P7=000001000,P8=000000000,P9=000a0b100,P10=000a00100,P11=00000a100,P12=000000100,P13=000a1b000,P14=000a10000,P15=00001a000,P16=000010000,P17=001a0b00-1,P18=a0100ba2-a0a-1a≠12,P19=a01b00a2-a0a-1a≠12,P20=00100a00-1,P21=001a0000-1,P22=a01000a2-a0a-1a≠12,P23=00100000-1,P24=a01b0ca2-a0a-1a≠12,P25=120100a-140-12,P26=1201a00-140-12,P27=1201000-140-12,P28=1201a0b-140-12,P29=012a0a2a20-12-a,P30=a12a000-a2-12-a,P31=0100000-120,P32=a12a+babb-2b2-2ab-a2-12-a+b.


## 2.2. Reduction to Quadratic Equations

In order to show thatP is a Rota-Baxter operator, we only check the following: (8)Pe1,Pe2=PPe1,e2+e1,Pe2,Pe2,Pe3=PPe2,e3+e2,Pe3,Pe1,Pe3=PPe1,e3+e1,Pe3. It follows from (5) and (6) that(9)Pe2,Pe3=b21b32-b22b31+b22b33-b23b32e1+b22b33-b23b32e3, while(10)PPe2,e3+e2,Pe3=Pb22-b31+b33e1-Pb22+b33e3=b22b11-b31b11+b33b11+b22b31+b33b31e1+b22b12-b31b12+b33b12+b22b32+b33b32e2+b22b13-b31b13+b33b13+b22b33+b33b33e3. Comparing the coefficients in (9) and (10), we have(11)b21b32-2b22b31+b22b33-b23b32-b22b11+b31b11-b33b11-b33b31=0,(12)b22b12-b31b12+b33b12+b22b32+b33b32=0,(13)b23b32+b22b13-b31b13+b33b13+b33b33=0. Similarly, from (14)Pe1,Pe2=PPe1,e2+e1,Pe2,Pe1,Pe3=PPe1,e3+e1,Pe3, we obtain the following six equations:(15)b11b11+b12b21-b13b11-b13b31-b12b23+b13b22=0,(16)b11b12+b22b12-b13b12-b13b32=0,(17)b11b13+2b22b13-b13b13-b13b33-b12b23=0,(18)b12b33-2b12b31-b13b32-b12b11=0,(19)b12b12+2b32b12=0,(20)b12b13+2b32b13=0.


## 2.3. Solving the Quadratic Equations

Equation (19) implies b12(b12+2b32)=0. To solve the quadratic equations (11), (12), (13), (15), (16), (17), (18), (19), and (20), we distinguish the following cases depending on whether b12=0 or not.Case 1.
b
12
=
0, b12+2b32≠0. That is, b12=0, b32≠0, taking b32=1. Equation (16) implies b13=0. Equation (15) implies b11=0. Equation (11) implies b21=b22b31+b222+b23. Equation (12) implies b33=-b22. Equation (13) implies b23=-b332=-b222. We obtain(21)P=000b22b31b22-b222b311-b22. Taking b22=0, b31=a, we obtain P1. Taking b22=a, b31=0, we obtain P2. Taking b22=0, b31=0, we obtain P3. Taking b22=a, b31=b, we obtain P4.Case 2.
Assumeb12=0, b12+2b32=0. That is, b12=0, b32=0. We distinguish the two cases depending on whether b13=0 or not.
Subcase 2.1.If b13=0, then (13) implies b33=0. Equation (15) implies b11=0. Equation (11) implies b22b31=0.
Subcase 2.1.1.If b22=0, b31=0, we obtain(22)P=000b210b23000. Taking b21=a, b23=1, we obtain P5. Taking b21=1, b23=0, we obtain P6. Taking b21=0, b23=1, we obtain P7. Taking b21=0, b23=0, we obtain P8.
Subcase 2.1.2.If b22=0, b31≠0, taking b31=1, we obtain(23)P=000b210b23100. Taking b21=a, b23=b, we obtain P9. Taking b21=a, b23=0, we obtain P10. Taking b21=0, b23=a, we obtain P11. Taking b21=0, b23=0, we obtain P12.
Subcase 2.1.3.If b22≠0, b31=0, taking b22=1, we obtain(24)P=000b211b23000. Taking b21=a, b23=b, we obtain P13. Taking b21=a, b23=a, we obtain P14. Taking b21=0, b23=a, we obtain P15. Taking b21=0, b23=0, we obtain P16.
Subcase 2.2.If b13≠0, taking b13=1, (17) implies b22=b33-b11+1/2. Then (15) implies b31=2b112-3b11+b33+1/2. Equation (13) implies b11+b33=b112-b332. That is, (b11+b33)(b11-b33-1)=0.
Subcase 2.2.1.If b11+b33≠0, b11-b33-1=0, (b11≠1/2), and then b22=0, b31=b112-b11, we obtain(25)P=b1101b210b23b112-b110b11-1,b11≠12.Taking b11=0,  b21=a, b23=b, we obtain P17. Taking b11=a, b21=0, b23=b  (a≠1/2), we obtain P18. Taking b11=a, b21=b, b23=0  (a≠1/2), we obtain P19. Taking b11=0, b21=0, b23=a, we obtain P20. Taking b11=0, b21=a, b23=0, we obtain P21. Taking b11=a, b21=0, b23=0  (a≠1/2), we obtain P22. Taking b11=0, b21=0, b23=0, we obtain P23. Taking b11=a, b21=b, b23=c  (a≠1/2), we obtain P24.
Subcase 2.2.2.If b11+b33=0, b11-b33-1≠0(b11≠1/2), and then b22=-2b11+1/2, b31=2b112-4b11+1/2, (3.4) implies 8b113-12b112+6b11-1=0. Then we have 8(b11-1/2)3=0, b11=1/2, giving a contradiction.
Subcase 2.2.3.If b11+b33=0, b11-b33-1=0, that is, b11=1/2, b33=-1/2, and then b22=0, b31=-1/4, we obtain(26)P=1201b210b23-140-12.Taking b21=0, b23=a, we obtain P25. Taking b21=a,  b23=0, we obtain P26. Taking b21=0, b23=0, we obtain P27. Taking b21=a, b23=b, we obtain P28.Case 3.
Assumeb12≠0,  (b12+2b32)=0. Taking b12=1, then b32=-1/2. Equation (16) implies b13=2b11+2b22. Equations (12) and (18) imply b31=b22+b33/2. Equation (17) implies b23=-2b112-2b11b22-2b11b33-2b22b33. Equation (15) implies b21=-b11b22-b11b33-b22b33-b112-b222. Equations (11), (15), and (17) imply b112+b222+b332+2b11b22+2b11b33+2b22b33=0. Then we have (b11+b22+b33)2=0. So b33=-(b11+b22), b21=b11b22, b23=-2b222-2b11b22, b31=-b11/2. We obtain(27)P=b1112b11+2b22b11b22b22-2b222-2b11b22-b112-12-b11+b22. Taking b11=0, b22=a, we obtain P29. Taking b11=a, b22=0, we obtain P30. Taking b11=0, b22=0, we obtain P31. Taking b11=a, b22=b, we obtain P32.


## 3. Solutions of the CYBE ing⋉ad∗g∗

In this section, we will give some solutions of CYBE ing⋉ad∗g∗. Let (g,[,]) be a Lie algebra and β:g→gl(V) a representation of g. On the vector space g⊕V, there is natural Lie algebra structure (denoted by g⋉βV) given by (28)x1+v1,x2+v2=x1,x2+βx1v2-βx2v1,x1,x2∈g,v1,v2∈V. Let β∗:g→gl(V∗) be the dual representation of β. A linear map P:V→g can be identified as an element P~ in g⊗V∗⊂(g⋉β∗V∗)⊗(g⋉β∗V∗) as follows. Let {v1,v2,…,vm} be a basis of V and {v1∗,v2∗,…,vm∗} the dual basis in V∗: that is, vi∗(vj)=δij. Let {e1,e2,…,en} be a basis of g. Set P(vi)=∑i=1naijej,1≤i≤n. Since, as a vector space, Hom(V,g)≅g⊗V∗, then(29)P~=∑i=1nPvi⊗vi∗=∑i=1m∑j=1naijej⊗vi∗⊆g⋉β∗V∗⊗g⋉β∗V∗.Lemma 2 (see [15]).
Letg be a Lie algebra; let (V,β) be a g-module. A linear map P:g→g is a Rota-Baxter operator if and only if r=P-P21 is a skew-symmetric solution of CYBE in g⋉ad∗g∗.Now consider the adjoint representation ofg,(g,ad) which is a g-module. Let e1, e2, e3 be the Cartan-Weyl basis. Using Lemma 2 and relation (29), we can obtain a family of solutions of CYBE in g⋉ad∗g∗ through the Rota-Baxter operators on g given in Theorem 1.Theorem 3.
The following tensors are solutions of the classical Yang-Baxter equation ing⋉ad∗g∗, where a, b, and c are nonzero complex numbers(30)r1=ae1+e2⊗e3∗-e3∗⊗ae1+e2,r2=ae2-a2e3⊗e2∗+e2-ae3⊗e3∗-e2∗⊗ae2-a2e3-e3∗⊗e2-ae3,r3=e2⊗e3∗-e3∗⊗e2,r4=abe1+ae2-a2e3⊗e2∗+be1+e2-ae3⊗e3∗-e2∗⊗abe1+ae2-a2e3-e3∗⊗be1+e2-ae3,r5=ae1+e3⊗e2∗-e2∗⊗ae1+e3,r6=e1⊗e2∗-e2∗⊗e1,r7=e3⊗e2∗-e2∗⊗e3′,r8=0,r9=ae1+be3⊗e2∗+e1⊗e3∗-e2∗⊗ae1+be3-e3∗⊗e1,r10=ae1⊗e2∗+e1⊗e3∗-e2∗⊗ae1-e3∗⊗e1,r11=ae3⊗e2∗+e1⊗e3∗-e2∗⊗ae3-e3∗⊗e1,r12=e1⊗e3∗-e3∗⊗e1,r13=ae1+e2+be3⊗e2∗-e2∗⊗ae1+e2+be3,r14=ae1+e2⊗e2∗-e2∗⊗ae1+e2,r15=e2+ae3⊗e2∗-e2∗⊗e2+ae3,r16=e2⊗e2∗-e2∗⊗e2,r17=e3⊗e1∗+ae1+be3⊗e2∗-e3⊗e3∗-e1∗⊗e3-e2∗⊗ae1+be3+e3∗⊗e3,r18=ae1+e3⊗e1∗+be3⊗e2∗+a2-ae1+a-1e3⊗e3∗-e1∗⊗ae1+e3-e2∗⊗be3-e3∗⊗a2-ae1+a-1e3,a≠12,r19=ae1+e3⊗e1∗+be1⊗e2∗+a2-ae1+a-1e3⊗e3∗-e1∗⊗ae1+e3-e2∗⊗be1-e3∗⊗a2-ae1+a-1e3,a≠12,r20=e3⊗e1∗+ae3⊗e2∗-e3⊗e3∗-e1∗⊗e3-e2∗⊗ae3+e3∗⊗e3,r21=e3⊗e1∗+ae1⊗e2∗-e3⊗e3∗-e1∗⊗e3-e2∗⊗ae1+e3∗⊗e3,r22=ae1+e3⊗e1∗+a2-ae1+a-1e3⊗e3∗-e1∗⊗ae1+e3-e3∗⊗a2-ae1+a-1e3a≠12,r23=e3⊗e1∗-e3⊗e3∗-e1∗⊗e3+e3∗⊗e3,r24=ae1+e3⊗e1∗+be1+ce3⊗e2∗+a2-ae1+a-1e3⊗e3∗-e1∗⊗ae1+e3-e2∗⊗be1+ce3-e3∗⊗a2-ae1+a-1e3,a≠12,r25=12e1+e3⊗e1∗+ae3⊗e2∗-14e1+12e3⊗e3∗-e1∗⊗12e1+e3-e2∗⊗ae3+e3∗⊗14e1+12e3,r26=12e1+e3⊗e1∗+ae1⊗e2∗-14e1+12e3⊗e3∗-e1∗⊗12e1+e3-e2∗⊗ae1+e3∗⊗14e1+12e3,r27=12e1+e3⊗e1∗-14e1+12e3⊗e3∗-e1∗⊗12e1+e3+e3∗⊗14e1+12e3,r28=12e1+e3⊗e1∗+ae1+be3⊗e2∗-14e1+12e3⊗e3∗-e1∗⊗12e1+e3-e2∗⊗ae1+be3+e3∗⊗14e1+12e3,r29=e2+2ae3⊗e1∗+ae2+2a2e3⊗e2∗-12e2+ae3⊗e3∗-e1∗⊗e2+2ae3-e2∗⊗ae2+2a2e3+e3∗⊗12e2+ae3,r30=ae1+e2+2ae3⊗e1∗-a2e1+12e2+ae3⊗e3∗-e1∗⊗ae1+e2+2ae3+e3∗⊗a2e1+12e2+ae3,r31=e2⊗e1∗-12e2⊗e3∗-e1∗⊗e2+e3∗⊗12e2,r32=ae1+e2+2a+be3⊗e1∗+abe1+be2-2b2+abe3⊗e2∗-a2e1+12e2+a+be3⊗e3∗-e1∗⊗ae1+e2+2a+be3-e2∗⊗abe1+be2-2b2+abe3⊗e2∗+e3∗⊗a2e1+12e2+a+be3.One can check that all of the tensors above are solutions of the classical Yang-Baxter equation ing⋉ad∗g∗.


## 4. Induced Left-Symmetric Algebras from Rota-Baxter Operators of Weight 0 ong

A left-symmetric algebra structure ong is a bilinear product ·:g⊗g→g satisfying the condition(31)x·y·z-x·y·z=y·x·z-y·x·zfor all x,y,z∈g. There are many examples of Lie algebras which do not admit a left-symmetric product. For example, it is easy to see that there are no left-symmetric algebras with semisimple Lie algebra. Equation (31) implies that the commutators [x,y]=x·y-y·x satisfy the Jacobi identity; that is to say each left-symmetric product has an associated commutation Lie algebra, which is called the subadjacent Lie algebra. If R is a Rota-Baxter operator on a left-symmetric algebra, then R is a solution of CYBE on its subadjacent Lie algebra [17]. Clearly, each associative algebra product is a left-symmetric product. Given a Lie algebra g, it is a fundamental problem to decide whether g admits a left-symmetric product and to give a classification of such products [18]. As an application of Yang-Baxter operators, we can use them to construct left-symmetric algebras with respect to g6.Lemma 4 (see [13]).
Letg be a Lie algebra; P is called a solution of the classical Yang-Baxter equation. Define a new operation on g by (32)x∗y=Px,y,∀x,y∈g. Then (g,∗) is a left-symmetric algebra.According to Theorem3 and Lemma 4, we can get some left-symmetric algebras of g.Theorem 5.
Some left-symmetric algebras ofg (of weight zero) are determined:(1)
e3∗e1=-e1, e3∗e2=ae1, e3∗e3=e1+e3;(2)
e2∗e1=-ae1, e2∗e2=a2(e1+e3), e2∗e3=a(e1+e3), e3∗e1=-e1, e3∗e2=a(e1+e3), e3∗e3=e1+e3;(3)
e3∗e1=-e1, e3∗e3=e1+e3;(4)
e2∗e1=-ae1, e2∗e2=(a2+ab)e1+a2e3, e2∗e3=a(e1+e3), e3∗e1=-e1, e3∗e2=(a+b)e1+ae3, e3∗e3=e1+e3;(5)
e2∗e2=(a-1)e1-e3;(6)
e2∗e2=e1;(7)
e2∗e2=-(e1+e3);(8)
e2∗e2=(a-b)e1-be3, e3∗e2=e1;(9)
e2∗e2=ae1, e3∗e2=e1;(10)
e2∗e2=-a(e1+e3), e3∗e2=e1;(11)
e3∗e2=e1;(12)
e2∗e1=-e1, e2∗e2=(a-b)e1-be3, e2∗e3=e1+e3;(13)
e2∗e1=-e1, e2∗e2=ae1, e2∗e3=e1+e3;(14)
e2∗e1=-e1, e2∗e2=-a(e1+e3), e2∗e3=e1+e3;(15)
e2∗e1=-e1, e2∗e3=e1+e3;(16)
e1∗e2=-(e1+e3), e2∗e2=(a-b)e1-be3, e3∗e2=e1+e3;(17)
e1∗e2=(a-1)e1-e3, e2∗e2=-b(e1+e3), e3∗e2=(a-1)2e1-(a-1)e3;(18)
e1∗e2=(a-1)e1-e3, e2∗e2=be1, e3∗e2=(a-1)2e1-(a-1)e3;(19)
e1∗e2=-(e1+e3), e2∗e2=-a(e1+e3), e3∗e2=e1+e3;(20)
e1∗e2=e1+e3, e2∗e2=ae1, e3∗e2=e1+e3;(21)
e1∗e2=(a-1)e1-e3, e3∗e2=(a-1)2e1-(a-1)e3;(22)
e1∗e2=-(e1+e3), e3∗e2=e1+e3;(23)
e1∗e2=(a-1)e1-e3, e2∗e2=(b-c)e1-ce3, e3∗e2=(a-1)2e1-(a-1)e3;(24)
e1∗e2=-1/2e1-e3, e2∗e2=-a(e1+e3), e3∗e2=1/4e1+1/2e3;(25)
e1∗e2=-1/2e1-e3, e2∗e2=ae1, e3∗e2=1/4e1+1/2e3;(26)
e1∗e2=-1/2e1-e3, e3∗e2=1/4e1+1/2e3;(27)
e1∗e2=-1/2e1-e3, e2∗e2=(a-b)e1-be3, e3∗e2=1/4e1+1/2e3;(28)
e1∗e1=-e1, e1∗e2=-2a(e1+e3), e1∗e3=e1+e3, e2∗e1=-ae1, e2∗e2=-2a2(e1+e3), e2∗e3=a(e1+e3), e3∗e1=1/2e1, e3∗e2=a(e1+e3), e3∗e3=-1/2(e1+e3);(29)
e1∗e1=-e1, e1∗e2=-ae1-2ae3, e1∗e3=e1+e3, e3∗e1=1/2e1, e3∗e2=a/2e1+ae3, e3∗e3=-1/2(e1+e3);(30)
e1∗e3=e1+e3, e3∗e3=-1/2(e1+e3);(31)
e1∗e1=-e1, e1∗e2=-a+2be1-2a+be3, e1∗e3=e1+e3, e2∗e1=-be1, e2∗e2=(2b2+3ab)e1+(2b2+2ab)e3, e2∗e3=b(e1+e3), e3∗e1=1/2e1, e3∗e2=(a/2+b)e1+(a+b)e3, e3∗e3=-1/2(e1+e3).



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