﻿ 三维实李代数的分类
 上海理工大学学报  2020, Vol. 42 Issue (3): 220-223 PDF

Classification of three-dimensional real Lie algebra
HU Jianhua, LI Feifei, LIU Jing
College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China
Abstract: The classification of three-dimensional real Lie algebra was completed according to the properties of derivative algebras and isomorphic conditions of Lie algebra. When the dimension of derivative algebra is 0 or 1, Lie algebra can be divided into three types, $L\left( {3,0} \right)$ , $L\left( {3, - 1} \right)$ and $L\left( {3,1} \right)$ , according to the properties of Lie bracket operation and the transformation of basis.When the dimension of derivative algebra is 2 or 3, Lie algebras can be divided into five types, $L\left( {3,2,a} \right)$ , $L\left( {3,3} \right)$ , $L\left( {3,4,c} \right)$ , $L\left( {3,5} \right)$ , $L\left( {3,6} \right)$ , according to the properties of eigenvalues of matrix corresponding to internal derivations.
Key words: Lie algebra     derivative algebra     isomorphism

1 基础知识

a. $\left[ {{\lambda _1}{x_1} + {\lambda _2}{x_2},y} \right] = {\lambda _1}\left[ {{x_1},y} \right] + {\lambda _2}\left[ {{x_2},y} \right],\forall {\lambda _1},{\lambda _2} \in F, {x_1},{x_2},y \in L$

b. $\left[ {x,x} \right] = 0,\forall x \in L$

c. $\left[ {x,\left[ {y,z} \right]} \right] + \left[ {y,\left[ {z,x} \right]} \right] + \left[ {z,\left[ {x,y} \right]} \right] = 0,\forall x,y,z \in L$

${\rm{ad}}\;x:L \to L$ 为伴随表示，可诱导出线性表示 ${\rm{ad}}\;x:L' \to L'$ ${\rm{ad}}\;x$ 称为内导子。

 $\begin{split} {\rm{ad}}\;Y =& {\rm{ad}}\;[{Y_1},{Y_2}] = \left[ {{\rm{ad}}\;{Y_1},{\rm{ad}}\;{Y_2}} \right] =\\ & {\rm{ad}}\;{Y_1}\;{\rm{ad}}\;{Y_2} -{\rm{ad}}\;{Y_2}\;{\rm{ad}}\;{Y_1} {\rm{tr}}\left( {{\rm{ad}}\;Y} \right) =\\ & {\rm{tr}}\left( {{\rm{ad}}\;{Y_1}\;{\rm{ad}}\;{Y_2}} \right) - {\rm{tr}}\left( {{\rm{ad}}\;{Y_2}\;{\rm{ad}}\;{Y_1}} \right) = 0 \end{split}$

2 三维实李代数的分类 2.1 导代数维数为0和1

 $\left[ {{X_1},{X_2}} \right] = a{X_1}, \left[ {{X_1},{X_3}} \right] = b{X_1}, \left[ {{X_2},{X_3}} \right] = c{X_1}$

2.2 导代数维数为2

 ${\left( {{\rm{ad}}\;Y} \right)_{\left\{ {Y,Z} \right\}}} = \left( {\begin{array}{*{20}{c}} 0&\alpha \\ 0&\beta \end{array}} \right),\;{\left( {{\rm{ad}}\;Z} \right)_{\left\{ {Y,Z} \right\}}} = \left( {\begin{array}{*{20}{c}} { - \alpha }&0 \\ { - \beta }&0 \end{array}} \right)$

a. ${\rm{ad}}\;X$ 有2个不同的实特征值。

${\lambda _1},{\lambda _2}$ ${\rm{ad}}\;X$ 的2个不同的实特征值，则 ${\rm{ad}}\;X$ 在实数域上可对角化，

 ${\left( {{\rm{ad}}\;X} \right)_{\left\{ {Y,Z} \right\}}} \sim \left( {\begin{array}{*{20}{c}} {{\lambda _1}}&0 \\ 0&{{\lambda _2}} \end{array}} \right)$

b. ${\rm{ad}}\;X$ 有1个重数为2的实特征值。

 ${\left( {{\rm{ad}}\;X} \right)_{\left\{ {Y,Z} \right\}}} \sim \left( {\begin{array}{*{20}{c}} \lambda &0 \\ 0&\lambda \end{array}} \right)$

${\rm{ad}}\;X$ 不可对角化，即

 ${\left( {{\rm{ad}}\;X} \right)_{\left\{ {Y,Z} \right\}}} \sim \left( {\begin{array}{*{20}{c}} \lambda &1 \\ 0&\lambda \end{array}} \right)$

c. ${\rm{ad}}\;X$ 有2个复共轭特征值。

 ${\left( {{\rm{ad}}\;X} \right)_{\left\{ {Y,Z} \right\}}} \sim \left( {\begin{array}{*{20}{c}} \lambda &0 \\ 0&{\overline \lambda } \end{array}} \right)$

$\left\{ {{X_1},Y,Z} \right\}$ L的一组基，则 ${\rm{ad}}\;{X_1}:L' \to L'$ 为同构的，若L中存在另一组基 $\left\{ {{X_2},Y,Z} \right\}$ ，令 ${X_2} = k{X_1} + lY + mZ$ ，则有 ${\rm{ad}}\;{X_2} = k{\rm{ad}}\;{X_1} + l{\rm{ad}}\;Y + m{\rm{ad}}\;Z$ ，因此，有 ${\rm{ad}}\;{X_2}\left| {_{L'} = } \right.k{\rm{ad}}\;{X_1}\left| {_{L'}} \right.$ ，它们的特征值仅相差 $k$ 倍，故选取不同扩基时，情况a中的 $a = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$ 和情况c中的 $c = \dfrac{a}{b}$ 不变，情况b中的结论与特征值无关。

2.3 导代数维数为3

${\rm{ad}}\;{ H}$ 的特征值全为零，则 ${\rm{ad}}\;{ H}$ 是幂零的，即存在一组基 $B:{ H},{{ H}'},{{ H}''}$ ，使得

 ${\left( {{\rm{ad}}\;{{H}}} \right)_B} = \left( {\begin{array}{*{20}{c}} 0&a&b \\ 0&0&c \\ 0&0&0 \end{array}} \right),\;a,c \ne 0$

${\rm{ad}}\;{ H} \cdot { H}' = a{ H} \Rightarrow \left( {{\rm{ad}}\;{ H}'} \right){ H} = - a{ H}$ ，所以， ${ H}$ ${\rm{ad}}\;{ H}'$ 的非零特征值 $- a$ 的特征向量。

a. ${\rm{ad}}\; { H}$ 有2个实特征值。

 ${\left( {{\rm{ad}}\;{{H}}} \right)_{\left\{ {{ M},{ N}} \right\}}} = \left( {\begin{array}{*{20}{c}} \lambda &0 \\ 0&{ - \lambda } \end{array}} \right)$

b. ${\rm{ad}}\;{ H}$ 有一对共轭特征值。

${\rm{ad}}\;{ H}$ 的共轭特征值为 $\lambda$ $\overline \lambda$ 。由 ${\rm{tr}}\;\left( {{\rm{ad}}\;{ H}} \right) = 0$ 可知 $\lambda + \overline \lambda = 0$ ，故 $\lambda$ 为纯虚数。设 $\lambda = m{\rm{i}},m \ne 0 \in {\bf R}$ 且不妨设 $m > 0$ ，则存在 ${ M} + {\rm{i}}{ N}$ ，使 $\left( {{\rm{ad}}\;{ H}} \right)\left( {{ M} + {\rm{i}}{ N}} \right) = {\rm{i}}m\left( {{ M} + {\rm{i}}{ N}} \right)$ ，即 $\left[ {{ H},{ M} + {\rm{i}}{ N}} \right] = - m{ N} + {\rm{i}}m{ M} = \left[ {{ H},{ M}} \right] + {\rm{i}}\left[ {{ H},{ N}} \right]$ ，因此，有 $\left[ {{ H},{ N}} \right] = m{ M},\left[ {{ H},{ M}} \right] = - m{ N}$ 。又因为， $\left[ {{ M},{ N}} \right] \in \ker \left( {{\rm{ad}}\;{ H}} \right) = {\rm{span}}\left\{ { H} \right\}$ ，设 $\left[ {{{M}},{{N}}} \right] = l{{H}},l \ne 0$ 。令 ${{M}}' = \dfrac{1}{{\sqrt {\left| {ml} \right|} }}{{M}}$ ${{N}}' = - \dfrac{1}{{\sqrt {\left| {ml} \right|} }}{{N}}$ ${ H}' = \dfrac{1}{m}{ H}$ ，则 $\left[ {{{M}}',{{N}}'} \right] = - \dfrac{l}{{\left| l \right|}}{{H}}'$ $\left[ {{ H}',{ M}'} \right] = { N}'$ $\left[ {{ H}',{ N}'} \right] = - { M}'$

$l > 0$ $l < 0$ 时，只是改变了 $\left[ {X',Y'} \right]$ 的符号，但产生了两类不同的李代数，这两类李代数在实数域是不同构的，但在复数域则同构。将b中的基 ${{M}}'$ ${\rm{i}}{{M}}'$ 代替， ${{N}}'$ ${\rm{i}}{{N}}'$ 代替，再进行讨论，即可得出在复数域上这两类李代数同构，属于一类李代数。

a. $L\left( {3,0} \right)$ $\left[ {{e_1},{e_2}} \right] = 0$ $\left[ {{e_1},{e_3}} \right] = 0$ $\left[ {{e_2},{e_3}} \right] = 0$

b. $L\left( {3, - 1} \right)$ $\left[ {{e_1},{e_2}} \right] = {e_1}$ $\left[ {{e_1},{e_3}} \right] = 0$ $\left[ {{e_2},{e_3}} \right] = 0$

c. $L\left( {3,1} \right)$ $\left[ {{e_1},{e_2}} \right] \!=\! {e_3}$ $\left[ {{e_1},{e_3}} \right]\! =\! 0$ $\left[ {{e_2},{e_3}} \right]\! =\! 0$

d. $L\left( {3,2,a} \right)$ $\left[ {{e_1},{e_2}} \right] \!\!=\!\! 0$ $\left[ {{e_1},{e_3}} \right]\!\!=\! \!{e_1}$ $\left[ {{e_2},{e_3}} \right]\!\! =\!\! a{e_2}$ $0 < \left| a \right| \leqslant 1$

e. $L\left( {3,3} \right)$ $\left[ {{e_1},{e_2}} \right] \!\!=\!\! 0$ $\left[ {{e_1},{e_3}} \right] \!\!=\!\! {e_1}$ $\left[ {{e_2},{e_3}} \right] \!=\! {e_1} \!\!+\!\! {e_2}$

f. $L\left( {3,4,c} \right)$ $\left[ {{e_1},{e_2}} \right] \!\!=\!\! 0$ $\left[ {{e_1},{e_3}} \right]\! \!=\!\! c{e_1} \!-\! {e_2}$ $\left[ {{e_2},{e_3}} \right]\! = {e_1} + c{e_2}$ $c \geqslant 0$

g. $L\left( {3,5} \right)$ $\left[ {{e_1},{e_2}} \right] \!\!=\!\! {e_1}$ $\left[ {{e_1},{e_3}} \right] \!\!=\!\! - \!2{e_2}$ $\left[ {{e_2},{e_3}} \right] \!\!=\!\! {e_3}$

h. $L\left( {3,6} \right)$ $\left[ {{e_1},{e_2}} \right]\! =\! {e_3}$ $\left[ {{e_1},{e_3}} \right]\!= \! - {e_2}$ $\left[ {{e_2},{e_3}} \right] \!=\! {e_1}$

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