相容Malcev代数的表示

Representation of Compatible Malcev Algebra

摘要: 本文首先给出了相容Malcev代数的定义和两个Malcev代数相容的条件，然后给出了相容Malcev代数的表示，并构造了它的一类特殊表示。证明了相容Malcev代数的表示的对偶仍是相容Malcev代数的表示。最后给出了相容pre-Malcev代数的定义和两个pre-Malcev代数相容的条件。

Abstract: In this paper, we first give the definition of compatible Malcev algebra and the conditions that two Malcev algebras are compatible. Then, we give the representation of compatible Malcev algebra and construct a special representation of compatible Malcev algebra. It is proved that the dual mapping of the representation of compatible Malcev algebra is still the representation of compatible Malcev algebra. Finally, we give the definition of compatible pre-Malcev algebra and the conditions that two pre-Malcev algebras are compatible.

1. 引言

Malcev代数和相容的代数结构在数学和物理领域均有重要的应用。Malcev代数是李代数的推广。许多学者对Malcev代数和相容的代数结构展开了研究。例如在文献 [1] 中介绍了特征不是2的任意基域F上维数小于等于6的所有幂零Malcev代数和7维非交换非李幂零Malcev代数的分类。在文献 [2] 中介绍了秩为3的自由Malcev代数可分解为秩为3的自由李代数和由7维单Malcev代数生成的各种秩为3的自由代数的次直和。在文献 [3] 中介绍了pre-Malcev代数的表示和配对。在文献 [4] 中介绍了相容李双代数与相容李代数中经典Yang-Baxter方程的密切关系以及相容pre-李代数与相容李代数中经典Yang-Baxter方程的反对称解的密切关系。在此基础上本文研究了相容Malcev代数的等价条件、表示以及相容pre-Malcev代数的等价条件。

2. 预备知识

定义2.1 ( [5]). 设M是域F上的线性空间，如果M中的二元双线性运算 $\left(x,y\right)\to \left[x,y\right]$ 满足下列条件

$\left[x,x\right]=0$， (2.1)

$J\left(x,y,\left[x,z\right]\right)=\left[J\left(x,y,z\right),x\right]$， $\forall x,y,z\in M$， (2.2)

其中

$J\left(x,y,z\right)=\left[\left[x,y\right],z\right]+\left[\left[y,z\right],x\right]+\left[\left[z,x\right],y\right]$，

则称 $\left(M,[,]\right)$ 为Malcev代数。

注记2.1 ( [5]). 等式(2.2)等价于

$\left[\left[x,z\right],\left[y,t\right]\right]=\left[\left[\left[x,y\right],z\right],t\right]+\left[\left[\left[y,z\right],t\right],x\right]+\left[\left[\left[z,t\right],x\right],y\right]+\left[\left[\left[t,x\right],y\right],z\right]$， (2.3)

$\forall x,y,z,t\in M$。

定义2.2 ( [6]). 设 $\left(M,[,]\right)$ 是Malcev代数，V是线性空间， $\rho :M\to End\left(V\right)$ 为线性映射，若 $\rho$ 满足

$\rho \left(\left[x,y\right]\right)\rho \left(z\right)=\rho \left(\left[\left[x,z\right],y\right]\right)-\rho \left(x\right)\rho \left(\left[z,y\right]\right)+\rho \left(z\right)\rho \left(x\right)\rho \left(y\right)-\rho \left(y\right)\rho \left(z\right)\rho \left(x\right)$， (2.4)

$\forall x,y,z\in M$，则称 $\left(\rho ,V\right)$ 为Malcev代数 $\left(M,[,]\right)$ 的表示。

定理2.1 ( [3]). 设 $\left(M,[,]\right)$ 是Malcev代数，V为线性空间。 $\rho :M\to End\left(V\right)$ 为线性映射，在 $M\oplus V$ 上定义

$\left\{x+u,y+v\right\}=\left[x,y\right]+\rho \left(x\right)v-\rho \left(y\right)u$， $\forall x,y\in M,u,v\in V$，

则 $\left(M\oplus V,\{,\}\right)$ 是Malcev代数当且仅当 $\left(\rho ,V\right)$ 为 $\left(M,[,]\right)$ 的表示。

例2.1 ( [3]). 设 $\left(M,[,]\right)$ 是Malcev代数，定义 $ad:M\to End\left(M\right)$，其中

$adx\left(y\right)=\left[x,y\right]$， $\forall x,y\in M$，

则 $\left(ad,M\right)$ 是 $\left(M,[,]\right)$ 的表示，称为 $\left(M,[,]\right)$ 的伴随表示。

定理2.2 ( [3]). 设 $\left(M,[,]\right)$ 是Malcev代数， $\left(\rho ,V\right)$ 是 $\left(M,[,]\right)$ 的表示，定义 ${\rho }^{\ast }:M\to End\left(V^{\ast }\right)$ 其中

$\langle {\rho }^{\ast }\left(x\right)f,v\rangle =-\langle f,\rho \left(x\right)v\rangle$， $\forall x\in M,v\in V,f\in V^{\ast }$，

则 $\left({\rho }^{\ast },V^{\ast }\right)$ 是 $\left(M,[,]\right)$ 的表示，称为 $\left(\rho ,V\right)$ 的对偶表示。

定义2.3 ( [7]). 设A是线性空间，A中有双线性运算 $\left(x,y\right)\to x\circ y$，如果对于任意的 $x,y,z,t\in A$ 满足

$\begin{array}{l}\left(\left(x\circ y\right)\circ z\right)\circ t-x\circ \left(\left(y\circ z\right)\circ t\right)+\left(z\circ \left(y\circ x\right)\right)\circ t-z\circ \left(y\circ \left(x\circ t\right)\right)+y\circ \left(x\circ \left(z\circ t\right)\right)\\ -\left(x\circ z\right)\circ \left(y\circ t\right)+x\circ \left(\left(z\circ y\right)\circ t\right)-\left(z\circ \left(x\circ y\right)\right)\circ t+\left(z\circ x\right)\circ \left(y\circ t\right)-\left(\left(y\circ x\right)\circ z\right)\circ t=0\end{array}$， (2.5)

则称 $\left(A,\circ \right)$ 为pre-Malcev代数。

3. 相容Malcev代数的表示

定义3.1设 $\left(M,{[,]}_1\right)$ 和 $\left(M,{[,]}_2\right)$ 是Malcev代数，如果对于任意的 $k_1,k_2\in F$，M关于下面定义的代数运算

$\left[x,y\right]=k_1{\left[x,y\right]}_1+k_2{\left[x,y\right]}_2$， $\forall x,y\in M$，

是Malcev代数，则称 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数。

定理3.1设 $\left(M,{[,]}_1\right)$ 和 $\left(M,{[,]}_2\right)$ 是Malcev代数， $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数当且仅当对于任意的 $x,y,z,t\in M$ 满足

$\begin{array}{l}{\left[{\left[x,z\right]}_1,{\left[y,t\right]}_1\right]}_2+{\left[{\left[x,z\right]}_1,{\left[y,t\right]}_2\right]}_1+{\left[{\left[x,z\right]}_2,{\left[y,t\right]}_1\right]}_1-{\left[{\left[{\left[x,y\right]}_1,z\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_1,x\right]}_2-{\left[{\left[{\left[z,t\right]}_1,x\right]}_1,y\right]}_2-{\left[{\left[{\left[t,x\right]}_1,y\right]}_1,z\right]}_2-{\left[{\left[{\left[x,y\right]}_2,z\right]}_1,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_1,x\right]}_1-{\left[{\left[{\left[z,t\right]}_2,x\right]}_1,y\right]}_1-{\left[{\left[{\left[t,x\right]}_2,y\right]}_1,z\right]}_1-{\left[{\left[{\left[x,y\right]}_1,z\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_2,x\right]}_1-{\left[{\left[{\left[z,t\right]}_1,x\right]}_2,y\right]}_1-{\left[{\left[{\left[t,x\right]}_1,y\right]}_2,z\right]}_1=0\end{array}$ (3.1)

$\begin{array}{l}{\left[{\left[x,z\right]}_2,{\left[y,t\right]}_2\right]}_1+{\left[{\left[x,z\right]}_1,{\left[y,t\right]}_2\right]}_2+{\left[{\left[x,z\right]}_2,{\left[y,t\right]}_1\right]}_2-{\left[{\left[{\left[x,y\right]}_2,z\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_2,x\right]}_1-{\left[{\left[{\left[z,t\right]}_2,x\right]}_2,y\right]}_1-{\left[{\left[{\left[t,x\right]}_2,y\right]}_2,z\right]}_1-{\left[{\left[{\left[x,y\right]}_2,z\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_1,x\right]}_2-{\left[{\left[{\left[z,t\right]}_2,x\right]}_1,y\right]}_2-{\left[{\left[{\left[t,x\right]}_2,y\right]}_1,z\right]}_2-{\left[{\left[{\left[x,y\right]}_1,z\right]}_2,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_2,x\right]}_2-{\left[{\left[{\left[z,t\right]}_1,x\right]}_2,y\right]}_2-{\left[{\left[{\left[t,x\right]}_1,y\right]}_2,z\right]}_2=0\end{array}$ (3.2)

证明 显然 $[,]$ 是双线性运算。由 $\left(M,{[,]}_1\right)$ 和 $\left(M,{[,]}_2\right)$ 是Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(2.1)。因此对于任意的 $k_1,k_2\in F,x\in M$ 直接计算得

$\left[x,x\right]=k_1{\left[x,x\right]}_1+k_2{\left[x,x\right]}_2=0$，

即 $[,]$ 满足等式(2.1)。由 $\left(M,{[,]}_1\right)$ 和 $\left(M,{[,]}_2\right)$ 是Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(2.3)。因此对于任意的 $x,y,z,t\in M$ 直接计算得

$\begin{array}{l}\left[\left[x,z\right],\left[y,t\right]\right]-\left[\left[\left[x,y\right],z\right],t\right]-\left[\left[\left[y,z\right],t\right],x\right]-\left[\left[\left[z,t\right],x\right],y\right]+\left[\left[\left[t,x\right],y\right],z\right]\\ =k_1^2{k}_2({\left[{\left[x,z\right]}_1,{\left[y,t\right]}_1\right]}_2+{\left[{\left[x,z\right]}_1,{\left[y,t\right]}_2\right]}_1+{\left[{\left[x,z\right]}_2,{\left[y,t\right]}_1\right]}_1-{\left[{\left[{\left[x,y\right]}_1,z\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_1,x\right]}_2-{\left[{\left[{\left[z,t\right]}_1,x\right]}_1,y\right]}_2-{\left[{\left[{\left[t,x\right]}_1,y\right]}_1,z\right]}_2-{\left[{\left[{\left[x,y\right]}_2,z\right]}_1,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_1,x\right]}_1-{\left[{\left[{\left[z,t\right]}_2,x\right]}_1,y\right]}_1-{\left[{\left[{\left[t,x\right]}_2,y\right]}_1,z\right]}_1-{\left[{\left[{\left[x,y\right]}_1,z\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_2,x\right]}_1-{\left[{\left[{\left[z,t\right]}_1,x\right]}_2,y\right]}_1-{\left[{\left[{\left[t,x\right]}_1,y\right]}_2,z\right]}_1)\end{array}$

$\begin{array}{l}+k_1{k}_2^2({\left[{\left[x,z\right]}_2,{\left[y,t\right]}_2\right]}_1+{\left[{\left[x,z\right]}_1,{\left[y,t\right]}_2\right]}_2+{\left[{\left[x,z\right]}_2,{\left[y,t\right]}_1\right]}_2-{\left[{\left[{\left[x,y\right]}_2,z\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_2,x\right]}_1-{\left[{\left[{\left[z,t\right]}_2,x\right]}_2,y\right]}_1-{\left[{\left[{\left[t,x\right]}_2,y\right]}_2,z\right]}_1-{\left[{\left[{\left[x,y\right]}_2,z\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_2,t\right]}_1,x\right]}_2-{\left[{\left[{\left[z,t\right]}_2,x\right]}_1,y\right]}_2-{\left[{\left[{\left[t,x\right]}_2,y\right]}_1,z\right]}_2-{\left[{\left[{\left[x,y\right]}_1,z\right]}_2,t\right]}_2\\ -{\left[{\left[{\left[y,z\right]}_1,t\right]}_2,x\right]}_2-{\left[{\left[{\left[z,t\right]}_1,x\right]}_2,y\right]}_2-{\left[{\left[{\left[t,x\right]}_1,y\right]}_2,z\right]}_2)\end{array}$

由 $k_1$ 和 $k_2$ 的任意性可知， $[,]$ 满足等式(2.3)当且仅当 ${[,]}_1,{[,]}_2$ 满足等式(3.1)和(3.2)。因此结论成立。

定义3.2 设 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数， $\left({\rho }_i,V\right)$ 为 $\left(M,{[,]}_i\right)$ $\left(i=1,2\right)$的表示。如果 ${\rho }_1,{\rho }_2$ 满足

$\begin{array}{l}{\rho }_2\left({\left[x,y\right]}_1\right){\rho }_1\left(z\right)+{\rho }_1\left({\left[x,y\right]}_1\right){\rho }_2\left(z\right)+{\rho }_1\left({\left[x,y\right]}_2\right){\rho }_1\left(z\right)-{\rho }_2\left({\left[{\left[x,z\right]}_1,y\right]}_1\right)\\ +{\rho }_2\left(x\right){\rho }_1\left({\left[z,y\right]}_1\right)-{\rho }_2\left(z\right){\rho }_1\left(x\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_1\left(z\right){\rho }_1\left(x\right)-{\rho }_1\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +{\rho }_1\left(x\right){\rho }_1\left({\left[z,y\right]}_2\right)-{\rho }_1\left(z\right){\rho }_1\left(x\right){\rho }_2\left(y\right)+{\rho }_1\left(y\right){\rho }_1\left(z\right){\rho }_2\left(x\right)-{\rho }_1\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +{\rho }_1\left(x\right){\rho }_2\left({\left[z,y\right]}_1\right)-{\rho }_1\left(z\right){\rho }_2\left(x\right){\rho }_1\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(z\right){\rho }_1\left(x\right)=0\end{array}$ (3.3)

$\begin{array}{l}{\rho }_1\left({\left[x,y\right]}_2\right){\rho }_2\left(z\right)+{\rho }_2\left({\left[x,y\right]}_1\right){\rho }_2\left(z\right)+{\rho }_2\left({\left[x,y\right]}_2\right){\rho }_1\left(z\right)-{\rho }_1\left({\left[{\left[x,z\right]}_2,y\right]}_2\right)\\ +{\rho }_1\left(x\right){\rho }_2\left({\left[z,y\right]}_2\right)-{\rho }_1\left(z\right){\rho }_2\left(x\right){\rho }_2\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(z\right){\rho }_2\left(x\right)-{\rho }_2\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +{\rho }_2\left(x\right){\rho }_1\left({\left[z,y\right]}_2\right)-{\rho }_2\left(z\right){\rho }_1\left(x\right){\rho }_2\left(y\right)+{\rho }_2\left(y\right){\rho }_1\left(z\right){\rho }_2\left(x\right)-{\rho }_2\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +{\rho }_2\left(x\right){\rho }_2\left({\left[z,y\right]}_1\right)-{\rho }_2\left(z\right){\rho }_2\left(x\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_2\left(z\right){\rho }_1\left(x\right)=0\end{array}$ (3.4)

$\forall x,y,z\in M$，则称 $\left({\rho }_1,{\rho }_2,V\right)$ 为 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

定理3.2 设 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数，V为线性空间。 ${\rho }_i:M\to End\left(V\right)$ 为线性映射，在 $M\oplus V$ 上定义

${\left\{x+u,y+v\right\}}_i={\left[x,y\right]}_i+{\rho }_i\left(x\right)v-{\rho }_i\left(y\right)u$， $\forall x,y\in M,u,v\in V,i=1,2$，

则 $\left(M\oplus V,{\{,\}}_1,{\{,\}}_2\right)$ 是相容Malcev代数当且仅当 $\left({\rho }_1,{\rho }_2,V\right)$ 为 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

证明 由定理2.1可知 $\left(M\oplus V,{\{,\}}_i\right)$ 是Malcev代数当且仅当 $\left({\rho }_i,V\right)$ 为 $\left(M,{[,]}_i\right)$ 的表示。由 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(3.1)，因此对于任意的 $x,y,z,t\in M,u,v,w,s\in V$ 直接计算得

$\begin{array}{l}{\left\{{\left\{x+u,z+w\right\}}_1,{\left\{y+v,t+s\right\}}_1\right\}}_2+{\left\{{\left\{x+u,z+w\right\}}_1,{\left\{y+v,t+s\right\}}_2\right\}}_1+{\left\{{\left\{x+u,z+w\right\}}_2,{\left\{y+v,t+s\right\}}_1\right\}}_1\\ -{\left\{{\left\{{\left\{x+u,y+v\right\}}_1,z+w\right\}}_1,t+s\right\}}_2-{\left\{{\left\{{\left\{y+v,z+w\right\}}_1,t+s\right\}}_1,x+u\right\}}_2-{\left\{{\left\{{\left\{z+w,t+s\right\}}_1,x+u\right\}}_1,y+v\right\}}_2\\ -{\left\{{\left\{{\left\{t+s,x+u\right\}}_1,y+v\right\}}_1,z+w\right\}}_2-{\left\{{\left\{{\left\{x+u,y+v\right\}}_2,z+w\right\}}_1,t+s\right\}}_1-{\left\{{\left\{{\left\{y+v,z+w\right\}}_2,t+s\right\}}_1,x+u\right\}}_1\\ -{\left\{{\left\{{\left\{z+w,t+s\right\}}_2,x+u\right\}}_1,y+v\right\}}_1-{\left\{{\left\{{\left\{t+s,x+u\right\}}_2,y+v\right\}}_1,z+w\right\}}_1-{\left\{{\left\{{\left\{x+u,y+v\right\}}_1,z+w\right\}}_2,t+s\right\}}_1\\ -{\left\{{\left\{{\left\{y+v,z+w\right\}}_1,t+s\right\}}_2,x+u\right\}}_1-{\left\{{\left\{{\left\{z+w,t+s\right\}}_1,x+u\right\}}_2,y+v\right\}}_1-{\left\{{\left\{{\left\{t+s,x+u\right\}}_1,y+v\right\}}_2,z+w\right\}}_1\end{array}$

$\begin{array}{l}=({\rho }_2\left({\left[y,t\right]}_1\right){\rho }_1\left(z\right)+{\rho }_1\left({\left[y,t\right]}_1\right){\rho }_2\left(z\right)+{\rho }_1\left({\left[y,t\right]}_2\right){\rho }_1\left(z\right)-{\rho }_2\left({\left[{\left[y,z\right]}_1,t\right]}_1\right)+{\rho }_2\left(y\right){\rho }_1\left({\left[z,t\right]}_1\right)\\ -{\rho }_2\left(z\right){\rho }_1\left(y\right){\rho }_1\left(t\right)+{\rho }_2\left(t\right){\rho }_1\left(z\right){\rho }_1\left(y\right)-{\rho }_1\left({\left[{\left[y,z\right]}_2,t\right]}_1\right)+{\rho }_1\left(y\right){\rho }_1\left({\left[z,t\right]}_2\right)-{\rho }_1\left(z\right){\rho }_1\left(y\right){\rho }_2\left(t\right)\\ +{\rho }_1\left(t\right){\rho }_1\left(z\right){\rho }_2\left(y\right)-{\rho }_1\left({\left[{\left[y,z\right]}_1,t\right]}_2\right)+{\rho }_1\left(y\right){\rho }_2\left({\left[z,t\right]}_1\right)-{\rho }_1\left(z\right){\rho }_2\left(y\right){\rho }_1\left(t\right)+{\rho }_1\left(t\right){\rho }_2\left(z\right){\rho }_1\left(y\right))u\\ +({\rho }_2\left({\left[z,x\right]}_1\right){\rho }_1\left(t\right)+{\rho }_1\left({\left[z,x\right]}_1\right){\rho }_2\left(t\right)+{\rho }_1\left({\left[z,x\right]}_2\right){\rho }_1\left(t\right)-{\rho }_2\left({\left[{\left[z,t\right]}_1,x\right]}_1\right)+{\rho }_2\left(z\right){\rho }_1\left({\left[t,x\right]}_1\right)\end{array}$

$\begin{array}{l}-{\rho }_2\left(t\right){\rho }_1\left(z\right){\rho }_1\left(x\right)+{\rho }_2\left(x\right){\rho }_1\left(t\right){\rho }_1\left(z\right)-{\rho }_1\left({\left[{\left[z,t\right]}_2,x\right]}_1\right)+{\rho }_1\left(z\right){\rho }_1\left({\left[t,x\right]}_2\right)-{\rho }_1\left(t\right){\rho }_1\left(z\right){\rho }_2\left(x\right)\\ +{\rho }_1\left(x\right){\rho }_1\left(t\right){\rho }_2\left(z\right)-{\rho }_1\left({\left[{\left[z,t\right]}_1,x\right]}_2\right)+{\rho }_1\left(z\right){\rho }_2\left({\left[t,x\right]}_1\right)-{\rho }_1\left(t\right){\rho }_2\left(z\right){\rho }_1\left(x\right)+{\rho }_1\left(x\right){\rho }_2\left(t\right){\rho }_1\left(z\right))v\\ +({\rho }_2\left({\left[t,y\right]}_1\right){\rho }_1\left(x\right)+{\rho }_1\left({\left[t,y\right]}_1\right){\rho }_2\left(x\right)+{\rho }_1\left({\left[t,y\right]}_2\right){\rho }_1\left(x\right)-{\rho }_2\left({\left[{\left[t,x\right]}_1,y\right]}_1\right)+{\rho }_2\left(t\right){\rho }_1\left({\left[x,y\right]}_1\right)\\ -{\rho }_2\left(x\right){\rho }_1\left(t\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_1\left(t\right)-{\rho }_1\left({\left[{\left[t,x\right]}_2,y\right]}_1\right)+{\rho }_1\left(t\right){\rho }_1\left({\left[x,y\right]}_2\right)-{\rho }_1\left(x\right){\rho }_1\left(t\right){\rho }_2(\; y\; )\end{array}$

$\begin{array}{l}+{\rho }_1\left(y\right){\rho }_1\left(x\right){\rho }_2\left(t\right)-{\rho }_1\left({\left[{\left[t,x\right]}_1,y\right]}_2\right)+{\rho }_1\left(t\right){\rho }_2\left({\left[x,y\right]}_1\right)-{\rho }_1\left(x\right){\rho }_2\left(t\right){\rho }_1\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_1\left(t\right))w\\ +({\rho }_2\left({\left[x,z\right]}_1\right){\rho }_1\left(y\right)+{\rho }_1\left({\left[x,z\right]}_1\right){\rho }_2\left(y\right)+{\rho }_1\left({\left[x,z\right]}_2\right){\rho }_1\left(y\right)-{\rho }_2\left({\left[{\left[x,y\right]}_1,z\right]}_1\right)+{\rho }_2\left(x\right){\rho }_1\left({\left[y,z\right]}_1\right)\\ -{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_1\left(z\right)+{\rho }_2\left(z\right){\rho }_1\left(y\right){\rho }_1\left(x\right)-{\rho }_1\left({\left[{\left[x,y\right]}_2,z\right]}_1\right)+{\rho }_1\left(x\right){\rho }_1\left({\left[y,z\right]}_2\right)-{\rho }_1\left(y\right){\rho }_1\left(x\right){\rho }_2\left(z\right)\\ +{\rho }_1\left(z\right){\rho }_1\left(y\right){\rho }_2\left(x\right)-{\rho }_1\left({\left[{\left[x,y\right]}_1,z\right]}_2\right)+{\rho }_1\left(x\right){\rho }_2\left({\left[y,z\right]}_1\right)-{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_1\left(z\right)+{\rho }_1\left(z\right){\rho }_2\left(y\right){\rho }_1\left(x\right))s\end{array}$

由 $u,v,w,s$ 的任意性可知， ${\{,\}}_1,{\{,\}}_2$ 满足等式(3.1)当且仅当 ${\rho }_1,{\rho }_2$ 满足等式(3.3)。由 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(3.2)，因此直接计算得

$\begin{array}{l}{\left\{{\left\{x+u,z+w\right\}}_2,{\left\{y+v,t+s\right\}}_2\right\}}_1+{\left\{{\left\{x+u,z+w\right\}}_1,{\left\{y+v,t+s\right\}}_2\right\}}_2+{\left\{{\left\{x+u,z+w\right\}}_2,{\left\{y+v,t+s\right\}}_1\right\}}_2\\ -{\left\{{\left\{{\left\{x+u,y+v\right\}}_2,z+w\right\}}_2,t+s\right\}}_1-{\left\{{\left\{{\left\{y+v,z+w\right\}}_2,t+s\right\}}_2,x+u\right\}}_1-{\left\{{\left\{{\left\{z+w,t+s\right\}}_2,x+u\right\}}_2,y+v\right\}}_1\\ -{\left\{{\left\{{\left\{t+s,x+u\right\}}_2,y+v\right\}}_2,z+w\right\}}_1-{\left\{{\left\{{\left\{x+u,y+v\right\}}_2,z+w\right\}}_1,t+s\right\}}_2-{\left\{{\left\{{\left\{y+v,z+w\right\}}_2,t+s\right\}}_1,x+u\right\}}_2\\ -{\left\{{\left\{{\left\{z+w,t+s\right\}}_2,x+u\right\}}_1,y+v\right\}}_2-{\left\{{\left\{{\left\{t+s,x+u\right\}}_2,y+v\right\}}_1,z+w\right\}}_2-{\left\{{\left\{{\left\{x+u,y+v\right\}}_1,z+w\right\}}_2,t+s\right\}}_2\\ -{\left\{{\left\{{\left\{y+v,z+w\right\}}_1,t+s\right\}}_2,x+u\right\}}_2-{\left\{{\left\{{\left\{z+w,t+s\right\}}_1,x+u\right\}}_2,y+v\right\}}_2-{\left\{{\left\{{\left\{t+s,x+u\right\}}_1,y+v\right\}}_2,z+w\right\}}_2\end{array}$

$\begin{array}{l}=({\rho }_1\left({\left[y,t\right]}_2\right){\rho }_2\left(z\right)+{\rho }_2\left({\left[y,t\right]}_1\right){\rho }_2\left(z\right)+{\rho }_2\left({\left[y,t\right]}_2\right){\rho }_1\left(z\right)-{\rho }_1\left({\left[{\left[y,z\right]}_2,t\right]}_2\right)+{\rho }_1\left(y\right){\rho }_2\left({\left[z,t\right]}_2\right)\\ -{\rho }_1\left(z\right){\rho }_2\left(y\right){\rho }_2\left(t\right)+{\rho }_1\left(t\right){\rho }_2\left(z\right){\rho }_2\left(y\right)-{\rho }_2\left({\left[{\left[y,z\right]}_2,t\right]}_1\right)+{\rho }_2\left(y\right){\rho }_1\left({\left[z,t\right]}_2\right)-{\rho }_2\left(z\right){\rho }_1\left(y\right){\rho }_2\left(t\right)\\ +{\rho }_2\left(t\right){\rho }_1\left(z\right){\rho }_2\left(y\right)-{\rho }_2\left({\left[{\left[y,z\right]}_1,t\right]}_2\right)+{\rho }_2\left(y\right){\rho }_2\left({\left[z,t\right]}_1\right)-{\rho }_2\left(z\right){\rho }_2\left(y\right){\rho }_1\left(t\right)+{\rho }_2\left(t\right){\rho }_2\left(z\right){\rho }_1\left(y\right))u\\ +({\rho }_1\left({\left[z,x\right]}_2\right){\rho }_2\left(t\right)+{\rho }_2\left({\left[z,x\right]}_1\right){\rho }_2\left(t\right)+{\rho }_2\left({\left[z,x\right]}_2\right){\rho }_1\left(t\right)-{\rho }_1\left({\left[{\left[z,t\right]}_2,x\right]}_2\right)+{\rho }_1\left(z\right){\rho }_2\left({\left[t,x\right]}_2\right)\end{array}$

$\begin{array}{l}-{\rho }_1\left(t\right){\rho }_2\left(z\right){\rho }_2\left(x\right)+{\rho }_1\left(x\right){\rho }_2\left(t\right){\rho }_2\left(z\right)-{\rho }_2\left({\left[{\left[z,t\right]}_2,x\right]}_1\right)+{\rho }_2\left(z\right){\rho }_1\left({\left[t,x\right]}_2\right)-{\rho }_2\left(t\right){\rho }_1\left(z\right){\rho }_2\left(x\right)\\ +{\rho }_2\left(x\right){\rho }_1\left(t\right){\rho }_2\left(z\right)-{\rho }_2\left({\left[{\left[z,t\right]}_1,x\right]}_2\right)+{\rho }_2\left(z\right){\rho }_2\left({\left[t,x\right]}_1\right)-{\rho }_2\left(t\right){\rho }_2\left(z\right){\rho }_1\left(x\right)+{\rho }_2\left(x\right){\rho }_2\left(t\right){\rho }_1\left(z\right))v\\ +({\rho }_1\left({\left[t,y\right]}_2\right){\rho }_2\left(x\right)+{\rho }_2\left({\left[t,y\right]}_1\right){\rho }_2\left(x\right)+{\rho }_2\left({\left[t,y\right]}_2\right){\rho }_1\left(x\right)-{\rho }_1\left({\left[{\left[t,x\right]}_2,y\right]}_2\right)+{\rho }_1\left(t\right){\rho }_2\left({\left[x,y\right]}_2\right)\\ -{\rho }_1\left(x\right){\rho }_2\left(t\right){\rho }_2\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_2\left(t\right)-{\rho }_2\left({\left[{\left[t,x\right]}_2,y\right]}_1\right)+{\rho }_2\left(t\right){\rho }_1\left({\left[x,y\right]}_2\right)-{\rho }_2\left(x\right){\rho }_1\left(t\right){\rho }_2(\; y\; )\end{array}$

$\begin{array}{l}+{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_2\left(t\right)-{\rho }_2\left({\left[{\left[t,x\right]}_1,y\right]}_2\right)+{\rho }_2\left(t\right){\rho }_2\left({\left[x,y\right]}_1\right)-{\rho }_2\left(x\right){\rho }_2\left(t\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_2\left(x\right){\rho }_1\left(t\right))w\\ +({\rho }_1\left({\left[x,z\right]}_2\right){\rho }_2\left(y\right)+{\rho }_2\left({\left[x,z\right]}_1\right){\rho }_2\left(y\right)+{\rho }_2\left({\left[x,z\right]}_2\right){\rho }_1\left(y\right)-{\rho }_1\left({\left[{\left[x,y\right]}_2,z\right]}_2\right)+{\rho }_1\left(x\right){\rho }_2\left({\left[y,z\right]}_2\right)\\ -{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_2\left(z\right)+{\rho }_1\left(z\right){\rho }_2\left(y\right){\rho }_2\left(x\right)-{\rho }_2\left({\left[{\left[x,y\right]}_2,z\right]}_1\right)+{\rho }_2\left(x\right){\rho }_1\left({\left[y,z\right]}_2\right)-{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_2\left(z\right)\\ +{\rho }_2\left(z\right){\rho }_1\left(y\right){\rho }_2\left(x\right)-{\rho }_2\left({\left[{\left[x,y\right]}_1,z\right]}_2\right)+{\rho }_2\left(x\right){\rho }_2\left({\left[y,z\right]}_1\right)-{\rho }_2\left(y\right){\rho }_2\left(x\right){\rho }_1\left(z\right)+{\rho }_2\left(z\right){\rho }_2\left(y\right){\rho }_1\left(x\right))s\end{array}$

由 $u,v,w,s$ 的任意性可知， ${\{,\}}_1,{\{,\}}_2$ 满足等式(3.2)当且仅当 ${\rho }_1,{\rho }_2$ 满足等式(3.4)。因此结论成立。

例3.1 设 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数，则 $\left(a{d}_1,a{d}_2,M\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示，称为 $\left(M,{[,]}_1,{[,]}_2\right)$ 的伴随表示。

证明 由例2.1可知 $\left(a{d}_i,M\right)$ 是 $\left(M,{[,]}_i\right)\left(i=1,2\right)$ 的表示。由 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(3.1)。因此对于任意的 $x,y,z,t\in M$ 直接计算得

$\begin{array}{l}(a{d}_2\left({\left[x,y\right]}_1\right)a{d}_1\left(z\right)+a{d}_1\left({\left[x,y\right]}_1\right)a{d}_2\left(z\right)+a{d}_1\left({\left[x,y\right]}_2\right)a{d}_1\left(z\right)-a{d}_2\left({\left[{\left[x,z\right]}_1,y\right]}_1\right)\\ +a{d}_2\left(x\right)a{d}_1\left({\left[z,y\right]}_1\right)-a{d}_2\left(z\right)a{d}_1\left(x\right)a{d}_1\left(y\right)+a{d}_2\left(y\right)a{d}_1\left(z\right)a{d}_1\left(x\right)-a{d}_1\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +a{d}_1\left(x\right)a{d}_1\left({\left[z,y\right]}_2\right)-a{d}_1\left(z\right)a{d}_1\left(x\right)a{d}_2\left(y\right)+a{d}_1\left(y\right)a{d}_1\left(z\right)a{d}_2\left(x\right)-a{d}_1\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +a{d}_1\left(x\right)a{d}_2\left({\left[z,y\right]}_1\right)-a{d}_1\left(z\right)a{d}_2\left(x\right)a{d}_1\left(y\right)+a{d}_1\left(y\right)a{d}_2\left(z\right)a{d}_1\left(x\right))t\end{array}$

$\begin{array}{l}={\left[{\left[x,y\right]}_1,{\left[z,t\right]}_1\right]}_2+{\left[{\left[x,y\right]}_1,{\left[z,t\right]}_2\right]}_1+{\left[{\left[x,y\right]}_2,{\left[z,t\right]}_1\right]}_1-{\left[{\left[{\left[x,z\right]}_1,y\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[z,y\right]}_1,t\right]}_1,x\right]}_2-{\left[{\left[{\left[y,t\right]}_1,x\right]}_1,z\right]}_2-{\left[{\left[{\left[t,x\right]}_1,z\right]}_1,y\right]}_2-{\left[{\left[{\left[x,z\right]}_2,y\right]}_1,t\right]}_1\\ -{\left[{\left[{\left[z,y\right]}_2,t\right]}_1,x\right]}_1-{\left[{\left[{\left[y,t\right]}_2,x\right]}_1,z\right]}_1-{\left[{\left[{\left[t,x\right]}_2,z\right]}_1,y\right]}_1-{\left[{\left[{\left[x,z\right]}_1,y\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[z,y\right]}_1,t\right]}_2,x\right]}_1-{\left[{\left[{\left[y,t\right]}_1,x\right]}_2,z\right]}_1-{\left[{\left[{\left[t,x\right]}_1,z\right]}_2,y\right]}_1=0\end{array}$

即 $a{d}_1,a{d}_2$ 满足等式(3.3)。由 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数可知， ${[,]}_1,{[,]}_2$ 满足等式(3.2)。因此直接计算得

$\begin{array}{l}(a{d}_1\left({\left[x,y\right]}_2\right)a{d}_2\left(z\right)+a{d}_2\left({\left[x,y\right]}_1\right)a{d}_2\left(z\right)+a{d}_2\left({\left[x,y\right]}_2\right)a{d}_1\left(z\right)-a{d}_1\left({\left[{\left[x,z\right]}_2,y\right]}_2\right)\\ +a{d}_1\left(x\right)a{d}_2\left({\left[z,y\right]}_2\right)-a{d}_1\left(z\right)a{d}_2\left(x\right)a{d}_2\left(y\right)+a{d}_1\left(y\right)a{d}_2\left(z\right)a{d}_2\left(x\right)-a{d}_2\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +a{d}_2\left(x\right)a{d}_1\left({\left[z,y\right]}_2\right)-a{d}_2\left(z\right)a{d}_1\left(x\right)a{d}_2\left(y\right)+a{d}_2\left(y\right)a{d}_1\left(z\right)a{d}_2\left(x\right)-a{d}_2\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +a{d}_2\left(x\right)a{d}_2\left({\left[z,y\right]}_1\right)-a{d}_2\left(z\right)a{d}_2\left(x\right)a{d}_1\left(y\right)+a{d}_2\left(y\right)a{d}_2\left(z\right)a{d}_1\left(x\right))t\end{array}$

$\begin{array}{l}={\left[{\left[x,y\right]}_2,{\left[z,t\right]}_2\right]}_1+{\left[{\left[x,y\right]}_1,{\left[z,t\right]}_2\right]}_2+{\left[{\left[x,y\right]}_2,{\left[z,t\right]}_1\right]}_2-{\left[{\left[{\left[x,z\right]}_2,y\right]}_2,t\right]}_1\\ -{\left[{\left[{\left[z,y\right]}_2,t\right]}_2,x\right]}_1-{\left[{\left[{\left[y,t\right]}_2,x\right]}_2,z\right]}_1-{\left[{\left[{\left[t,x\right]}_2,z\right]}_2,y\right]}_1-{\left[{\left[{\left[x,z\right]}_2,y\right]}_1,t\right]}_2\\ -{\left[{\left[{\left[z,y\right]}_2,t\right]}_1,x\right]}_2-{\left[{\left[{\left[y,t\right]}_2,x\right]}_1,z\right]}_2-{\left[{\left[{\left[t,x\right]}_2,z\right]}_1,y\right]}_2-{\left[{\left[{\left[x,z\right]}_1,y\right]}_2,t\right]}_2\\ -{\left[{\left[{\left[z,y\right]}_1,t\right]}_2,x\right]}_2-{\left[{\left[{\left[y,t\right]}_1,x\right]}_2,z\right]}_2-{\left[{\left[{\left[t,x\right]}_1,z\right]}_2,y\right]}_2=0\end{array}$

即 $a{d}_1,a{d}_2$ 满足等式(3.4)。综上可知 $\left(a{d}_1,a{d}_2,M\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

定理3.3 设 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数， $\left({\rho }_1,{\rho }_2,V\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示，则 $\left({\rho }_1^{\ast },{\rho }_2^{\ast },V^{\ast }\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示，称为 $\left({\rho }_1,{\rho }_2,V\right)$ 的对偶表示。

证明 由定理2.2可知 $\left({\rho }_i^{\ast },V^{\ast }\right)$ 是Malcev代数 $\left(M,{[,]}_i\right)\left(i=1,2\right)$ 的表示。由 $\left({\rho }_1,{\rho }_2,V\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示可知， ${\rho }_1,{\rho }_2$ 满足等式(3.3)。因此对于任意的 $x,y,z\in M,v\in V,f\in V^{\ast }$ 直接计算得

$\begin{array}{l}\langle ({\rho }_2^{\ast }\left({\left[x,y\right]}_1\right){\rho }_1^{\ast }\left(z\right)+{\rho }_1^{\ast }\left({\left[x,y\right]}_1\right){\rho }_2^{\ast }\left(z\right)+{\rho }_1^{\ast }\left({\left[x,y\right]}_2\right){\rho }_1^{\ast }\left(z\right)-{\rho }_2^{\ast }\left({\left[{\left[x,z\right]}_1,y\right]}_1\right)\\ +{\rho }_2^{\ast }\left(x\right){\rho }_1^{\ast }\left({\left[z,y\right]}_1\right)-{\rho }_2^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right){\rho }_1^{\ast }\left(y\right)+{\rho }_2^{\ast }\left(y\right){\rho }_1^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right)-{\rho }_1^{\ast }\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +{\rho }_1^{\ast }\left(x\right){\rho }_1^{\ast }\left({\left[z,y\right]}_2\right)-{\rho }_1^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right){\rho }_2^{\ast }\left(y\right)+{\rho }_1^{\ast }\left(y\right){\rho }_1^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right)-{\rho }_1^{\ast }\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +{\rho }_1^{\ast }\left(x\right){\rho }_2^{\ast }\left({\left[z,y\right]}_1\right)-{\rho }_1^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right){\rho }_1^{\ast }\left(y\right)+{\rho }_1^{\ast }\left(y\right){\rho }_2^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right))f,v\rangle \end{array}$

$\begin{array}{l}=\langle f,({\rho }_2\left({\left[z,y\right]}_1\right){\rho }_1\left(x\right)+{\rho }_1\left({\left[z,y\right]}_1\right){\rho }_2\left(x\right)+{\rho }_1\left({\left[z,y\right]}_2\right){\rho }_1\left(x\right)-{\rho }_2\left({\left[{\left[z,x\right]}_1,y\right]}_1\right)\\ +{\rho }_2\left(z\right){\rho }_1\left({\left[x,y\right]}_1\right)-{\rho }_2\left(x\right){\rho }_1\left(z\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_1\left(z\right)-{\rho }_1\left({\left[{\left[z,x\right]}_2,y\right]}_1\right)\\ +{\rho }_1\left(z\right){\rho }_1\left({\left[x,y\right]}_2\right)-{\rho }_1\left(x\right){\rho }_1\left(z\right){\rho }_2\left(y\right)+{\rho }_1\left(y\right){\rho }_1\left(x\right){\rho }_2\left(z\right)-{\rho }_1\left({\left[{\left[z,x\right]}_1,y\right]}_2\right)\\ +{\rho }_1\left(z\right){\rho }_2\left({\left[x,y\right]}_1\right)-{\rho }_1\left(x\right){\rho }_2\left(z\right){\rho }_1\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_1\left(z\right))v\rangle =0\end{array}$

即 ${\rho }_1^{\ast },{\rho }_2^{\ast }$ 满足等式(3.3)。由 $\left({\rho }_1,{\rho }_2,V\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示可知， ${\rho }_1,{\rho }_2$ 满足等式(3.4)。因此直接计算得

$\begin{array}{l}\langle ({\rho }_1^{\ast }\left({\left[x,y\right]}_2\right){\rho }_2^{\ast }\left(z\right)+{\rho }_2^{\ast }\left({\left[x,y\right]}_1\right){\rho }_2^{\ast }\left(z\right)+{\rho }_2^{\ast }\left({\left[x,y\right]}_2\right){\rho }_1^{\ast }\left(z\right)-{\rho }_1^{\ast }\left({\left[{\left[x,z\right]}_2,y\right]}_2\right)\\ +{\rho }_1^{\ast }\left(x\right){\rho }_2^{\ast }\left({\left[z,y\right]}_2\right)-{\rho }_1^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right){\rho }_2^{\ast }\left(y\right)+{\rho }_1^{\ast }\left(y\right){\rho }_2^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right)-{\rho }_2^{\ast }\left({\left[{\left[x,z\right]}_2,y\right]}_1\right)\\ +{\rho }_2^{\ast }\left(x\right){\rho }_1^{\ast }\left({\left[z,y\right]}_2\right)-{\rho }_2^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right){\rho }_2^{\ast }\left(y\right)+{\rho }_2^{\ast }\left(y\right){\rho }_1^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right)-{\rho }_2^{\ast }\left({\left[{\left[x,z\right]}_1,y\right]}_2\right)\\ +{\rho }_2^{\ast }\left(x\right){\rho }_2^{\ast }\left({\left[z,y\right]}_1\right)-{\rho }_2^{\ast }\left(z\right){\rho }_2^{\ast }\left(x\right){\rho }_1^{\ast }\left(y\right)+{\rho }_2^{\ast }\left(y\right){\rho }_2^{\ast }\left(z\right){\rho }_1^{\ast }\left(x\right))f,v\rangle \end{array}$

$\begin{array}{l}=\langle f,({\rho }_1\left({\left[z,y\right]}_2\right){\rho }_2\left(x\right)+{\rho }_2\left({\left[z,y\right]}_1\right){\rho }_2\left(x\right)+{\rho }_2\left({\left[z,y\right]}_2\right){\rho }_1\left(x\right)-{\rho }_1\left({\left[{\left[z,x\right]}_2,y\right]}_2\right)\\ +{\rho }_1\left(z\right){\rho }_2\left({\left[x,y\right]}_2\right)-{\rho }_1\left(x\right){\rho }_2\left(z\right){\rho }_2\left(y\right)+{\rho }_1\left(y\right){\rho }_2\left(x\right){\rho }_2\left(z\right)-{\rho }_2\left({\left[{\left[z,x\right]}_2,y\right]}_1\right)\\ +{\rho }_2\left(z\right){\rho }_1\left({\left[x,y\right]}_2\right)-{\rho }_2\left(x\right){\rho }_1\left(z\right){\rho }_2\left(y\right)+{\rho }_2\left(y\right){\rho }_1\left(x\right){\rho }_2\left(z\right)-{\rho }_2\left({\left[{\left[z,x\right]}_1,y\right]}_2\right)\\ +{\rho }_2\left(z\right){\rho }_2\left({\left[x,y\right]}_1\right)-{\rho }_2\left(x\right){\rho }_2\left(z\right){\rho }_1\left(y\right)+{\rho }_2\left(y\right){\rho }_2\left(x\right){\rho }_1\left(z\right))v\rangle =0\end{array}$

即 ${\rho }_1^{\ast },{\rho }_2^{\ast }$ 满足等式(3.4)。综上可知 $\left({\rho }_1^{\ast },{\rho }_2^{\ast },V^{\ast }\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

例 3.2 设 $\left(M,{[,]}_1,{[,]}_2\right)$ 是相容Malcev代数，则 $\left(a{d}_1^{\ast },a{d}_2^{\ast },M^{\ast }\right)$ 是 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

证明 由例3.1可知 $\left(a{d}_1,a{d}_2,M\right)$ 是相容Malcev代数 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示，再由定理3.3可知 $\left(a{d}_1^{\ast },a{d}_2^{\ast },M^{\ast }\right)$ 是相容Malcev代数 $\left(M,{[,]}_1,{[,]}_2\right)$ 的表示。

4. pre-Malcev代数相容的条件

定义4.1 设 $\left(A,{\circ }_1\right)$ 和 $\left(A,{\circ }_2\right)$ 是pre-Malcev代数，如果对于任意的 $k_1,k_2\in F$，A关于下面定义的代数运算

$x\circ y=k_{1}x{\circ }_{1}y+k_{2}x{\circ }_{2}y$， $\forall x,y\in A$，

是pre-Malcev代数，则称 $\left(A,{\circ }_1,{\circ }_2\right)$ 是相容pre-Malcev代数。

定理4.1设 $\left(A,{\circ }_1\right)$ 和 $\left(A,{\circ }_2\right)$ 是pre-Malcev代数， $\left(A,{\circ }_1,{\circ }_2\right)$ 是相容pre-Malcev代数当且仅当对于任意的 $x,y,z,t\in A$ 满足

$\begin{array}{l}\left(\left(x{\circ }_{1}y\right){\circ }_{1}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{1}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{1}x\right)\right){\circ }_{2}t-z{\circ }_2\left(y{\circ }_1\left(x{\circ }_{1}t\right)\right)\\ +y{\circ }_2\left(x{\circ }_1\left(z{\circ }_{1}t\right)\right)+\left(\left(x{\circ }_{2}y\right){\circ }_{1}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{2}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{2}x\right)\right){\circ }_{1}t\\ -z{\circ }_1\left(y{\circ }_1\left(x{\circ }_{2}t\right)\right)+y{\circ }_1\left(x{\circ }_1\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{1}y\right){\circ }_{2}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{1}z\right){\circ }_{2}t\right)\\ +\left(z{\circ }_2\left(y{\circ }_{1}x\right)\right){\circ }_{1}t-z{\circ }_1\left(y{\circ }_2\left(x{\circ }_{1}t\right)\right)+y{\circ }_1\left(x{\circ }_2\left(z{\circ }_{1}t\right)\right)-\left(x{\circ }_{1}z\right){\circ }_2\left(y{\circ }_{1}t\right)\end{array}$

$\begin{array}{l}+x{\circ }_2\left(\left(z{\circ }_{1}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{1}y\right)\right){\circ }_{2}t+\left(z{\circ }_{1}x\right){\circ }_2\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{1}z\right){\circ }_{2}t\\ -\left(x{\circ }_{1}z\right){\circ }_1\left(y{\circ }_{2}t\right)+x{\circ }_1\left(\left(z{\circ }_{2}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{2}y\right)\right){\circ }_{1}t+\left(z{\circ }_{1}x\right){\circ }_1\left(y{\circ }_{2}t\right)\\ -\left(\left(y{\circ }_{2}x\right){\circ }_{1}z\right){\circ }_{1}t-\left(x{\circ }_{2}z\right){\circ }_1\left(y{\circ }_{1}t\right)+x{\circ }_1\left(\left(z{\circ }_{1}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{1}y\right)\right){\circ }_{1}t\\ +\left(z{\circ }_{2}x\right){\circ }_1\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{2}z\right){\circ }_{1}t=0\end{array}$ (4.1)

$\begin{array}{l}\left(\left(x{\circ }_{2}y\right){\circ }_{2}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{2}z\right){\circ }_{2}t\right)+\left(z{\circ }_2\left(y{\circ }_{2}x\right)\right){\circ }_{1}t-z{\circ }_1\left(y{\circ }_2\left(x{\circ }_{2}t\right)\right)\\ +y{\circ }_1\left(x{\circ }_2\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{2}y\right){\circ }_{1}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{2}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{2}x\right)\right){\circ }_{2}t\\ -z{\circ }_2\left(y{\circ }_1\left(x{\circ }_{2}t\right)\right)+y{\circ }_2\left(x{\circ }_1\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{1}y\right){\circ }_{2}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{1}z\right){\circ }_{2}t\right)\\ +\left(z{\circ }_2\left(y{\circ }_{1}x\right)\right){\circ }_{2}t-z{\circ }_2\left(y{\circ }_2\left(x{\circ }_{1}t\right)\right)+y{\circ }_2\left(x{\circ }_2\left(z{\circ }_{1}t\right)\right)-\left(x{\circ }_{2}z\right){\circ }_1\left(y{\circ }_{2}t\right)\end{array}$

$\begin{array}{l}+x{\circ }_1\left(\left(z{\circ }_{2}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{2}y\right)\right){\circ }_{1}t+\left(z{\circ }_{2}x\right){\circ }_1\left(y{\circ }_{2}t\right)-\left(\left(y{\circ }_{2}x\right){\circ }_{2}z\right){\circ }_{1}t\\ -\left(x{\circ }_{1}z\right){\circ }_2\left(y{\circ }_{2}t\right)+x{\circ }_2\left(\left(z{\circ }_{2}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{2}y\right)\right){\circ }_{2}t+\left(z{\circ }_{1}x\right){\circ }_2\left(y{\circ }_{2}t\right)\\ -\left(\left(y{\circ }_{2}x\right){\circ }_{1}z\right){\circ }_{2}t-\left(x{\circ }_{2}z\right){\circ }_2\left(y{\circ }_{1}t\right)+x{\circ }_2\left(\left(z{\circ }_{1}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{1}y\right)\right){\circ }_{2}t\\ +\left(z{\circ }_{2}x\right){\circ }_2\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{2}z\right){\circ }_{2}t=0\end{array}$ (4.2)

证明 显然 $\circ$ 是双线性运算。由 $\left(A,{\circ }_1\right)$ 和 $\left(A,{\circ }_2\right)$ 是pre-Malcev代数可知， ${\circ }_1,{\circ }_2$ 满足等式(2.5)。因此对于任意的 $k_1,k_2\in F,x,y,z,t\in A$ 直接计算得

$\begin{array}{l}\left(\left(x\circ y\right)\circ z\right)\circ t-x\circ \left(\left(y\circ z\right)\circ t\right)+\left(z\circ \left(y\circ x\right)\right)\circ t-z\circ \left(y\circ \left(x\circ t\right)\right)+y\circ \left(x\circ \left(z\circ t\right)\right)\\ -\left(x\circ z\right)\circ \left(y\circ t\right)+x\circ \left(\left(z\circ y\right)\circ t\right)-\left(z\circ \left(x\circ y\right)\right)\circ t+\left(z\circ x\right)\circ \left(y\circ t\right)-\left(\left(y\circ x\right)\circ z\right)\circ t\\ =k_1^2{k}_2[\left(\left(x{\circ }_{1}y\right){\circ }_{1}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{1}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{1}x\right)\right){\circ }_{2}t-z{\circ }_2\left(y{\circ }_1\left(x{\circ }_{1}t\right)\right)\\ +y{\circ }_2\left(x{\circ }_1\left(z{\circ }_{1}t\right)\right)+\left(\left(x{\circ }_{2}y\right){\circ }_{1}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{2}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{2}x\right)\right){\circ }_{1}t\\ -z{\circ }_1\left(y{\circ }_1\left(x{\circ }_{2}t\right)\right)+y{\circ }_1\left(x{\circ }_1\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{1}y\right){\circ }_{2}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{1}z\right){\circ }_{2}t\right)\end{array}$

$\begin{array}{l}+\left(z{\circ }_2\left(y{\circ }_{1}x\right)\right){\circ }_{1}t-z{\circ }_1\left(y{\circ }_2\left(x{\circ }_{1}t\right)\right)+y{\circ }_1\left(x{\circ }_2\left(z{\circ }_{1}t\right)\right)-\left(x{\circ }_{1}z\right){\circ }_2\left(y{\circ }_{1}t\right)\\ +x{\circ }_2\left(\left(z{\circ }_{1}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{1}y\right)\right){\circ }_{2}t+\left(z{\circ }_{1}x\right){\circ }_2\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{1}z\right){\circ }_{2}t\\ -\left(x{\circ }_{1}z\right){\circ }_1\left(y{\circ }_{2}t\right)+x{\circ }_1\left(\left(z{\circ }_{2}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{2}y\right)\right){\circ }_{1}t+\left(z{\circ }_{1}x\right){\circ }_1\left(y{\circ }_{2}t\right)\\ -\left(\left(y{\circ }_{2}x\right){\circ }_{1}z\right){\circ }_{1}t-\left(x{\circ }_{2}z\right){\circ }_1\left(y{\circ }_{1}t\right)+x{\circ }_1\left(\left(z{\circ }_{1}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{1}y\right)\right){\circ }_{1}t\\ +\left(z{\circ }_{2}x\right){\circ }_1\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{2}z\right){\circ }_{1}t]\end{array}$

$\begin{array}{l}+k_1{k}_2^2[\left(\left(x{\circ }_{2}y\right){\circ }_{2}z\right){\circ }_{1}t-x{\circ }_1\left(\left(y{\circ }_{2}z\right){\circ }_{2}t\right)+\left(z{\circ }_2\left(y{\circ }_{2}x\right)\right){\circ }_{1}t-z{\circ }_1\left(y{\circ }_2\left(x{\circ }_{2}t\right)\right)\\ +y{\circ }_1\left(x{\circ }_2\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{2}y\right){\circ }_{1}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{2}z\right){\circ }_{1}t\right)+\left(z{\circ }_1\left(y{\circ }_{2}x\right)\right){\circ }_{2}t\\ -z{\circ }_2\left(y{\circ }_1\left(x{\circ }_{2}t\right)\right)+y{\circ }_2\left(x{\circ }_1\left(z{\circ }_{2}t\right)\right)+\left(\left(x{\circ }_{1}y\right){\circ }_{2}z\right){\circ }_{2}t-x{\circ }_2\left(\left(y{\circ }_{1}z\right){\circ }_{2}t\right)\\ +\left(z{\circ }_2\left(y{\circ }_{1}x\right)\right){\circ }_{2}t-z{\circ }_2\left(y{\circ }_2\left(x{\circ }_{1}t\right)\right)+y{\circ }_2\left(x{\circ }_2\left(z{\circ }_{1}t\right)\right)-\left(x{\circ }_{2}z\right){\circ }_1\left(y{\circ }_{2}t\right)\end{array}$

$\begin{array}{l}+x{\circ }_1\left(\left(z{\circ }_{2}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{2}y\right)\right){\circ }_{1}t+\left(z{\circ }_{2}x\right){\circ }_1\left(y{\circ }_{2}t\right)-\left(\left(y{\circ }_{2}x\right){\circ }_{2}z\right){\circ }_{1}t\\ -\left(x{\circ }_{1}z\right){\circ }_2\left(y{\circ }_{2}t\right)+x{\circ }_2\left(\left(z{\circ }_{2}y\right){\circ }_{1}t\right)-\left(z{\circ }_1\left(x{\circ }_{2}y\right)\right){\circ }_{2}t+\left(z{\circ }_{1}x\right){\circ }_2\left(y{\circ }_{2}t\right)\\ -\left(\left(y{\circ }_{2}x\right){\circ }_{1}z\right){\circ }_{2}t-\left(x{\circ }_{2}z\right){\circ }_2\left(y{\circ }_{1}t\right)+x{\circ }_2\left(\left(z{\circ }_{1}y\right){\circ }_{2}t\right)-\left(z{\circ }_2\left(x{\circ }_{1}y\right)\right){\circ }_{2}t\\ +\left(z{\circ }_{2}x\right){\circ }_2\left(y{\circ }_{1}t\right)-\left(\left(y{\circ }_{1}x\right){\circ }_{2}z\right){\circ }_{2}t]\end{array}$

由 $k_1$ 和 $k_2$ 的任意性可知， $\circ$ 满足等式(2.5)当且仅当 ${\circ }_1,{\circ }_2$ 满足等式(4.1)和(4.2)。因此结论成立。

基金项目

辽宁师范大学教改项目LS202002。