0 引言

可解李代数的结构在李代数的结构研究中起着非常重要的作用^[1-3]，随着李代数理论的不断发展和完善，其理论与方法已逐步渗透到数学物理的许多领域.文献[4]研究了Rota-Bater李代数在理论物理上的应用.文献[5]给出了导代数维数等于一的二维和三维李代数，且权为零的Rota-Baxter算子的具体表达式, 并通过Rota-Baxter算子的可逆性讨论了李代数的幂零性.李代数的Rota-Baxter算子与经典的Yang-Baxter方程的解从另一方面反映出李代数的代数结构^[5-8].近几年Rota-Baxter代数在Yang-Baxter方程、Hopf代数、微分代数、量子域理论等方面也得到很好的应用^[9-12].本文研究复数域上导代数维数等于一的四维可解李代数的权为零的Rota-Baxter算子的结构.

设L是域F上的线性空间，如果在L上存在斜对称的二元线性运算: [, ]:L×L→L，满足对任意x, y, z∈L, [x, [y, z]]=[[x, y], z]+[y, [x, z]]，则称L是域F上的李代数.L的由所有[x, y]生成的子代数，称为李代数L的导代数，记为L₁.对每个自然数s，记L^(s+1)=[L^(s), L^(s)]，如果存在自然数s，使得L^(s)=0，则称L是可解的李代数.显然，如果L是导代数维数等于一的李代数，则L是可解李代数.设L是域F上的李代数，λ∈F，如果线性变换P:L→L满足下列等式，则称P为李代数L的权为λ的Rota-Baxter算子.对任意x, y∈L,

$ \left[ {P\left( x \right),P\left( y \right)} \right] = P\left( {\left[ {P\left( x \right),y} \right] + \left[ {x,P\left( y \right)} \right]} \right) + \lambda P\left[ {x,y} \right]. $

(1)

如果λ=0，则称P为李代数L的权为零的Rota-Baxter算子，以下简称P为李代数L的Rota-Baxter算子.

命题1  线性变换P:L→L为李代数L的权为零的Rota-Baxter算子的充要条件是，P/α为李代数L的权为零Rota-Baxter算子，其中α为域F中任意非零的数.

证明  由等式(1) 知，λ=0时，P为李代数L的权为零的Rota-Baxter算子的充要条件是，[P(x), P(y)]=P([P(x), y]+[x, P(y)])成立.显然，此等式成立的充要条件是，两边都乘以非零数k, 等式仍然成立.

1 主要结论

在本文中，主要讨论复数域C上的李代数.

引理1^[7]  设L是复数域C上导代数维数等于一的李代数，x₁, x₂, x₃, x₄是L的一组基, 则在同构的意义下仅有下面3类：

$ {L_1}:\left[ {{x_1},{x_2}} \right] = {x_3};\;\;\;\;\;{L_2}:\left[ {{x_1},{x_2}} \right] = {x_2};\;\;\;\;{L_3}:\left\{ \begin{array}{l} \left[ {{x_1},{x_2}} \right] = {x_3},\\ \left[ {{x_1},{x_3}} \right] = {x_3}. \end{array} \right. $

为了讨论李代数L的Rota-Baxter算子，首先引入下面一些符号.设P:L→L是权为零的Rota-Baxter算子, P在基x₁, x₂, x₃, x₄表示为

$ P\left( {{x_i}} \right) = \sum\limits_{j = 1}^4 {{a_{ij}}{x_j}} ,\;\;\;1 \le i \le 4, $

(2)

即P在基x₁, x₂, x₃, x₄下的矩阵为$\boldsymbol{A}{\rm{ = }}\left( \begin{gathered} {a_{11}}\;\;{a_{12}}\;\;{a_{13}}\;\;{a_{14}} \hfill \\ {a_{21}}\;\;{a_{22}}\;\;{a_{23}}\;\;{a_{24}} \hfill \\ {a_{31}}\;\;{a_{32}}\;\;{a_{33}}\;\;{a_{34}} \hfill \\ {a_{41}}\;\;{a_{42}}\;\;{a_{43}}\;\;{a_{44}} \hfill \\ \end{gathered} \right) $.

定理1  设L是引理1中的四维李代数L₁, 设P:L→L是线性变换，P是L的权为零的Rota-Baxter算子的充要条件是P=0或P在基x₁, x₂, x₃, x₄下的矩阵为下列情形之一：

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&1&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&1&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0;\\ {\mathit{\boldsymbol{A}}_3} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&1&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_4} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1 \end{array}} \right);{\mathit{\boldsymbol{A}}_5} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&{{a_{44}}} \end{array}} \right);\\ {\mathit{\boldsymbol{A}}_6} = \left( {\begin{array}{*{20}{c}} 0&0&0&0\\ 0&0&0&1\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_7} = \left( {\begin{array}{*{20}{c}} 0&0&0&1\\ 0&0&0&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right); $

$ {\mathit{\boldsymbol{A}}_8} = \left( {\begin{array}{*{20}{c}} 0&0&0&{{a_{14}}}\\ 0&0&0&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_9} = \left( {\begin{array}{*{20}{c}} 0&0&1&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right);\\ {\mathit{\boldsymbol{A}}_{10}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&1&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{11}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&1&{{a_{34}}}\\ {{a_{41}}}&0&0&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{12}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&1&{{a_{34}}}\\ 0&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{42}} \ne 0;{\mathit{\boldsymbol{A}}_{13}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ {{a_{41}}}&1&0&{{a_{44}}} \end{array}} \right);\\ {\mathit{\boldsymbol{A}}_{14}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 1&0&0&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{15}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right); $

$ {\mathit{\boldsymbol{A}}_{16}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0;{\mathit{\boldsymbol{A}}_{17}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&0&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0,{a_{41}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{18}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0,{a_{42}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{19}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0,{a_{42}} \ne 0;{\mathit{\boldsymbol{A}}_{20}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ {{a_{41}}}&0&0&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{21}} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{42}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{22}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{23}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0,\\ {\mathit{\boldsymbol{A}}_{24}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&0&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0,{a_{41}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{25}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0,{a_{42}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{26}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0,{a_{42}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{27}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ {{a_{41}}}&0&0&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0;{\mathit{\boldsymbol{A}}_{28}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&{{a_{42}}}&0&{{a_{44}}} \end{array}} \right),{a_{42}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{29}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ { - 1/{a_{12}}}&{ - 1}&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&0&0&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0,{a_{34}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{30}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ { - 1/{a_{12}}}&{ - 1}&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0;{\mathit{\boldsymbol{A}}_{31}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ { - 1/{a_{12}}}&{ - 1}&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ {{a_{41}}}&{{a_{12}}}&{{a_{41}}}&{{a_{43}}} \end{array}} \right),{a_{12}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{32}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{33}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{34}} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&{ - 1}&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ {{a_{41}}}&{{a_{12}}}&{{a_{41}}}&{{a_{43}}} \end{array}} \right),{a_{12}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{35}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ { - {a_{33}}/{a_{12}}}&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ 0&0&{{a_{41}}}&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0,{a_{33}} \ne 0;{\mathit{\boldsymbol{A}}_{36}} = \left( {\begin{array}{*{20}{c}} 1&0&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right); $

$ {\mathit{\boldsymbol{A}}_{37}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {\left( {{a_{12}}{a_{22}}} \right)/{a_{12}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{42}}/{a_{12}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0,{a_{22}} \ne - 1;\\ {\mathit{\boldsymbol{A}}_{38}} = \left( {\begin{array}{*{20}{c}} 1&0&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{39}} = \left( {\begin{array}{*{20}{c}} 1&0&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ {\left( { - {a_{21}}{a_{42}}} \right)/{a_{33}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{33}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{40}}\left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ { - a_{22}^2/{a_{12}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{22}}}&0\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{22}} \ne - 1,{a_{22}} \ne 0,{a_{12}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{41}}\left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {\left( {{a_{22}} - {a_{33}} - {a_{22}}{a_{33}}} \right)/{a_{12}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{22}} \ne {a_{33}},{a_{33}} \ne 0,{a_{22}} \ne - 1,{a_{12}} \ne 0. $

证明  设L是李代数L₁，P:L→L是权为零的Rota-Baxter算子，P在基x₁, x₂, x₃, x₄的表示为等式(2).显然，P=0是Rota-Baxter算子.如果P≠0，由权为零的Rota-Baxter算子的定义、等式(1) 及L₁的乘法表可得:a₂₁a₄₂－a₂₂a₄₁=－a₄₁a₃₃；a₁₁a₄₂－a₁₂a₄₁=a₄₂a₃₃；(a₁₁+a₂₂)a₃₃=a₁₁a₂₂－a₁₂a₂₁；(a₁₁+a₂₂)a₃₄=0；a₃₁=a₃₂=0；a₄₂a₃₄=a₄₁a₃₄=0.下面分几种情形进行讨论.

1) 当a₁₁+a₂₂=0时，得到a₁₁a₂₂－a₁₂a₂₁=0, 如果a₁₁=0，则a₂₂=0，所以，a₁₂a₂₁=0.如果a₃₄≠0，得到a₄₂=a₄₁=0，再由命题1，可得矩阵A₁, A₂, A₃.如果a₃₄=0，得到a₁₂a₂₁=0，a₄₁a₄₂a₃₃=0，分别讨论各种情形可得到矩阵A₄, …, A₂₈.如果a₁₁≠0，由命题1，可以徦设a₁₁=1，则a₂₂=－1，a₁₂a₂₁=－1.如果a₃₄≠0，得到a₄₂=a₄₁=0，得到A₂₉.如果a₃₄=0, 得到矩阵A₃₀, A₃₁.

2) 当a₁₁+a₂₂≠0时，则a₃₁=a₃₂=a₃₄=0.如果a₁₁=0，则a₂₂≠0，不妨设a₂₂=1.如果a₃₃=0，得到a₁₂a₂₁=0, a₄₁=a₂₁a₄₂, 得到矩阵A₃₂, A₃₃, A₃₄.如果a₃₃≠0，得到a₁₂a₂₁=a₃₃且a₄₁=a₄₂=0, 得到矩阵A₃₅.如果a₁₁≠0，不妨设a₁₁=1, 则a₂₂≠－1，在a₃₃=0的情形下，得到a₁₂a₂₁=a₂₂, 可得到矩阵A₃₆, A₃₇.在a₃₃≠0的情形下，得到a₁₂a₂₁=－a₃₃(1+a₂₂)+a₂₂.如果a₁₂=0，得到a₂₂=0, a₄₂=a₄₁=0，或是a₄₁=－a₂₁a₄₂/a₃₃, a₄₂≠0, a₃₃≠0, 分别得到矩阵A₃₈, A₃₉.如果a₁₂≠0, 得到等价方程:a₂₁a₄₂－a₂₂a₄₁=－a₄₁a₃₃; a₄₂－a₁₂a₄₁=a₄₂a₃₃; (1+a₂₂)a₃₃=a₂₂－a₁₂a₂₁.如果a₂₂=a₃₃，则a₄₁=a₄₂=0，得到矩阵A₄₀.如果a₂₂≠a₃₃，由上述方程组进行讨论可知a₄₁=a₄₂=0，得到矩阵A₄₁.证毕.

定理2  设L是引理1中的四维李代数L₂, 设P:L→L是线性变换，P是L的权为零的Rota-Baxter算子的充要条件是, P在基x₁, x₂, x₃, x₄下的矩阵为下列情形之一:

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 1&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ 0&{{a_{32}}}&{{a_{33}}}&0\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);\\ {\mathit{\boldsymbol{A}}_3} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&0\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{23}} \ne 0\;aaa\;{a_{24}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_4} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ 1&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_5} = \left( {\begin{array}{*{20}{c}} 0&0&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ 1&0&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{42}} \ne 0,\\ {\mathit{\boldsymbol{A}}_6} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ 0&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{32}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_7} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_8} = \left( {\begin{array}{*{20}{c}} 0&1&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{42}} \ne 0;\\ {\mathit{\boldsymbol{A}}_9} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {1/{a_{12}}}&{ - 1}&{{a_{23}}}&{{a_{24}}}\\ 0&0&{{a_{33}}}&{{a_{34}}}\\ 0&0&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{12}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{10}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&0&0&0\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right). $

证明  由权为零的Rota-Baxter算子的定义及引理1可知，当L是李代数L₂时，设P:L→L是权为零的Rota-Baxter算子.由等式(1)、(2) 及L₁的乘法表得到下列关系同时成立:

$ \left\{ \begin{array}{l} \left( {{a_{11}} + {a_{22}}} \right){a_{22}} = {a_{11}}{a_{22}} - {a_{12}}{a_{21}},\\ \left( {{a_{11}} + {a_{22}}} \right){a_{21}} = 0,\\ \left( {{a_{11}} + {a_{22}}} \right){a_{23}} = 0,\\ \left( {{a_{11}} + {a_{22}}} \right){a_{24}} = 0, \end{array} \right.\;\;\;\left\{ \begin{array}{l} {a_{32}}{a_{22}} = {a_{11}}{a_{32}} - {a_{12}}{a_{31}},\\ {a_{32}}{a_{23}} = 0,\\ {a_{32}}{a_{21}} = 0,\\ {a_{32}}{a_{24}} = 0, \end{array} \right.\;\;\;\\ \left\{ \begin{array}{l} {a_{42}}{a_{22}} = {a_{11}}{a_{42}} - {a_{12}}{a_{41}},\\ {a_{42}}{a_{23}} = 0,\\ {a_{42}}{a_{21}} = 0,\\ {a_{42}}{a_{24}} = 0, \end{array} \right. $

$ \;\;\;\;\;\left\{ \begin{array}{l} - {a_{31}}{a_{22}} = {a_{21}}{a_{32}} - {a_{22}}{a_{31}},\\ - {a_{31}}{a_{21}} = 0,\\ - {a_{31}}{a_{23}} = 0,\\ - {a_{31}}{a_{24}} = 0, \end{array} \right.\;\;\;\;\;\;\left\{ \begin{array}{l} - {a_{41}}{a_{22}} = {a_{21}}{a_{42}} - {a_{22}}{a_{41}},\\ - {a_{21}}{a_{41}} = 0,\\ - {a_{41}}{a_{23}} = 0,\\ - {a_{41}}{a_{24}} = 0, \end{array} \right.\;\;\;\;{a_{31}}{a_{42}} - {a_{32}}{a_{41}} = 0. $

与定理1的证明方法类似，分几种情形进行讨论.

1) a₁₁+a₂₂=0时，得到a₁₁a₂₂－a₁₂a₂₁=0.如果a₂₂=0，则a₁₁=0，a₁₂a₂₁=0, a₁₂a₃₁=a₃₂a₂₃=a₃₂a₂₄=0, a₁₂a₄₁=a₂₃a₄₁=a₄₁a₂₄=a₄₂a₂₃=a₄₂a₂₄=a₂₃a₃₁=a₃₁a₂₄=0, 分别讨论各种情况可得到矩阵A₁, …, A₈.如果a₂₂≠0，由命题1，不妨设a₂₂=1, 则a₁₁=－1.上述讨论可知a₁₂a₂₁=1.再由上面的方程组可得到矩阵A₉.

2) 当a₁₁+a₂₂≠0时，a₂₁=a₂₃=a₂₄=0, 得到(a₁₁+a₂₂)a₂₂=a₁₁a₂₂，所以a₂₂=0, 得到矩阵A₁₀.证毕.

定理3  设L是引理1中的四维李代数L₃, P:L→L是线性变换，P是L的权为零的Rota-Baxter算子的充要条件是，P在基x₁, x₂, x₃, x₄下的矩阵为下列情形之一：

$ {\mathit{\boldsymbol{A}}_1} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&0&1&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_2} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&0&1&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0;\\ {\mathit{\boldsymbol{A}}_3} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&0&1&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_4} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&0&1&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0,{a_{41}} \ne 0;{\mathit{\boldsymbol{A}}_5} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&1&0&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);\\ {\mathit{\boldsymbol{A}}_6} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&1&0&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_7} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&1&0&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0;\\ {\mathit{\boldsymbol{A}}_8} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&1&0&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0,{a_{41}} \ne 0;{\mathit{\boldsymbol{A}}_9} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right), $

$ {a_{23}} \ne 0, - 1;{\mathit{\boldsymbol{A}}_{10}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ 0&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{41}} \ne 0,{a_{23}} \ne 0, - 1;\\ {\mathit{\boldsymbol{A}}_{11}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0,{a_{23}} \ne 0, - 1; $

$ {\mathit{\boldsymbol{A}}_{12}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&1&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0,{a_{41}} \ne 0,{a_{23}} \ne 0, - 1;\\ {\mathit{\boldsymbol{A}}_{13}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ 0&0&0&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{14}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ 0&0&0&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right), $

$ {a_{41}} \ne 0;{\mathit{\boldsymbol{A}}_{15}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&0&0&{{a_{24}}}\\ 0&0&0&0\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{16}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&0&0&{{a_{24}}}\\ 0&0&0&0\\ {{a_{41}}}&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0,{a_{41}} \ne 0; $

$ {\mathit{\boldsymbol{A}}_{17}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {{a_{23}}/\left( {{a_{12}} + {a_{23}}} \right)}&0&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {{a_{23}}/\left( {{a_{42}} + {a_{43}}} \right)}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),{a_{23}} \ne - 1,{a_{12}} \ne - {a_{13}},{a_{42}} \ne - {a_{43}};\\ {\mathit{\boldsymbol{A}}_{18}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {{a_{22}}/\left( {{a_{12}} + {a_{13}}} \right)}&{{a_{22}}}&0&{{a_{24}}}\\ 0&0&0&0\\ {{a_{22}}/\left( {{a_{42}} + {a_{43}}} \right)}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right), $

$ {a_{22}} \ne - 1,{a_{12}} \ne - {a_{13}},{a_{42}} \ne - {a_{43}}; $

$ {\mathit{\boldsymbol{A}}_{19}} = \left( {\begin{array}{*{20}{c}} 1&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {\left( {{a_{22}} + {a_{23}}} \right)/\left( {{a_{12}} + {a_{13}}} \right)}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\ 0&0&0&0\\ {\left( {{a_{22}} + {a_{23}}} \right)/\left( {{a_{42}} + {a_{43}}} \right)}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),\\ {a_{22}} + {a_{23}} \ne - 1,{a_{12}} \ne - {a_{13}},{a_{42}} \ne - {a_{43}};{\mathit{\boldsymbol{A}}_{20}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 1&{{a_{22}}}&{ - {a_{22}}}&{{a_{24}}}\\ 1&{{a_{32}}}&{ - {a_{32}}}&{{a_{34}}}\\ 0&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right); $

$ {\mathit{\boldsymbol{A}}_{21}} = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 1&{{a_{22}}}&{ - \left( {{a_{11}} + {a_{22}}} \right)}&{{a_{24}}}\\ 1&{{a_{32}}}&{ - \left( {{a_{11}} + {a_{22}}} \right)}&{{a_{34}}}\\ 0&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{11}} \ne 0;\\ {\mathit{\boldsymbol{A}}_{22}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&{{a_{22}}}&{ - {a_{22}}}&{{a_{24}}}\\ 0&1&{ - {a_{32}}}&{{a_{34}}}\\ 0&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{23}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ 0&{{a_{22}}}&{ - {a_{22}}}&{{a_{24}}}\\ 0&0&0&{{a_{34}}}\\ 0&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right); $

$ {\mathit{\boldsymbol{A}}_{24}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ 0&{{a_{22}}}&{ - {a_{22}}}&{{a_{24}}}\\ 0&0&0&0\\ 1&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right);{\mathit{\boldsymbol{A}}_{25}} = \left( {\begin{array}{*{20}{c}} 0&{{a_{12}}}&{ - {a_{12}}}&{{a_{14}}}\\ {{a_{21}}}&{{a_{22}}}&{ - {a_{22}}}&{{a_{24}}}\\ 0&0&0&0\\ 1&{{a_{42}}}&{ - {a_{42}}}&{{a_{44}}} \end{array}} \right),{a_{21}} \ne 0. $

证明  设P:L→L是权为零的Rota-Baxter算子，由权为零的Rota-Baxter算子的定义及引理1可知，当L是李代数L₃时，由等式(1)、(2) 及L₁的乘法表得到下列关于系数的方程组：

$ \left\{ \begin{array}{l} \left( {{a_{11}} + {a_{22}} + {a_{23}}} \right){a_{33}} = {a_{11}}\left( {{a_{22}} + {a_{23}}} \right) - {a_{21}}\left( {{a_{12}} + {a_{13}}} \right),\\ \left( {{a_{11}} + {a_{22}} + {a_{23}}} \right){a_{31}} = 0,\\ \left( {{a_{11}} + {a_{22}} + {a_{23}}} \right){a_{32}} = 0,\\ \left( {{a_{11}} + {a_{22}} + {a_{23}}} \right){a_{34}} = 0, \end{array} \right.\\ \left\{ \begin{array}{l} \left( {{a_{11}} + {a_{32}} + {a_{23}}} \right){a_{33}} = {a_{11}}{a_{32}} - {a_{12}}{a_{31}} + {a_{11}}{a_{33}} - {a_{13}}{a_{31}},\\ \left( {{a_{11}} + {a_{32}} + {a_{23}}} \right){a_{31}} = 0,\\ \left( {{a_{11}} + {a_{32}} + {a_{23}}} \right){a_{32}} = 0,\\ \left( {{a_{11}} + {a_{32}} + {a_{23}}} \right){a_{34}} = 0, \end{array} \right. $

$ \left\{ \begin{array}{l} \left( {{a_{42}} + {a_{43}}} \right){a_{33}} = {a_{11}}{a_{42}} - {a_{12}}{a_{41}} + {a_{11}}{a_{43}} - {a_{13}}{a_{41}},\\ \left( {{a_{42}} + {a_{43}}} \right){a_{31}} = 0,\\ \left( {{a_{42}} + {a_{43}}} \right){a_{32}} = 0,\\ \left( {{a_{42}} + {a_{43}}} \right){a_{34}} = 0, \end{array} \right.\\ \left\{ \begin{array}{l} \left( {{a_{21}} - {a_{31}}} \right){a_{33}} = {a_{21}}{a_{32}} - {a_{22}}{a_{31}} + {a_{21}}{a_{33}} - {a_{23}}{a_{31}},\\ \left( {{a_{21}} - {a_{31}}} \right){a_{31}} = 0,\\ \left( {{a_{21}} - {a_{31}}} \right){a_{32}} = 0,\\ \left( {{a_{21}} - {a_{31}}} \right){a_{34}} = 0, \end{array} \right. $

$ \left\{ \begin{array}{l} - {a_{41}}{a_{33}} = {a_{21}}{a_{42}} - {a_{22}}{a_{41}} + {a_{21}}{a_{43}} - {a_{23}}{a_{41}},\\ - {a_{41}}{a_{31}} = 0,\\ - {a_{41}}{a_{32}} = 0,\\ - {a_{41}}{a_{34}} = 0, \end{array} \right.\\ \left\{ \begin{array}{l} - {a_{41}}{a_{33}} = {a_{31}}{a_{42}} - {a_{32}}{a_{41}} + {a_{31}}{a_{43}} - {a_{33}}{a_{41}},\\ - {a_{41}}{a_{31}} = 0,\\ - {a_{41}}{a_{32}} = 0,\\ - {a_{41}}{a_{34}} = 0. \end{array} \right. $

分几种情形进行讨论.

1) a₁₁+a₂₂+a₂₃≠0时，得到a₃₁=a₃₂=a₃₄=a₃₃=0.且由a₁₁:(a₁₂+a₁₃)=a₂₁:(a₂₂+a₂₃)=a₄₁:(a₄₂+a₄₃), 得到矩阵A₁, …, A₁₉.

2) a₁₁+a₂₂+a₂₃=0时，若a₁₁+a₃₂+a₃₃≠0，仍可得矩阵A₁, …, A₁₉.如果a₁₁+a₃₂+a₃₃=0，a₃₁≠0，得到a₄₁=0, a₂₁=a₃₁，a₄₂+a₄₃=0，且有a₁₁a₃₂－a₁₂a₃₁+a₁₁a₃₃－a₁₃a₃₁=a₁₁a₂₂－a₁₂a₂₁+a₁₁a₂₃－a₁₃a₂₁=0.可得矩阵A₂₀, A₂₁.当a₃₁=0时，得到a₁₁=0，所以，a₂₂+a₂₃=0, a₃₂+a₃₃=0.在a₃₂≠0时，由命题1，得到a₄₁=a₂₁=0，a₄₂=－a₄₃, 得到矩阵A₂₂.在a₃₂=0时，得到a₃₃=0, 得到矩阵A₂₃, A₂₄, A₂₅.证毕.