# 一类五维李代数的交叉模Crossed Modules of a Class of Five-Dimensional Lie Algebras

DOI: 10.12677/PM.2019.93035, PDF, HTML, XML, 下载: 486  浏览: 694  国家自然科学基金支持

Abstract: The set of equivalence classes of crossed modules of four-dimensional unsolvable Lie algebras and the third relative cohomology groups were studied in [1]. Based on this work, the present paper will study the crossed modules of a certain five-dimensional Lie algebra and determine the condition of its equivalent class. Furthermore, it is shown that the third relative cohomology group of the five-dimensional Lie algebra is not trivial.

1. 引言

2. 交叉模的基本概念与性质

$\mu \left(\eta \left(n\right)\cdot m\right)=\left[n,\mu \left(m\right)\right]$,

$\eta \left(\mu \left(m\right)\cdot {m}^{\prime }\right)=\left[m,{m}^{\prime }\right]$ .

$\eta \left(\left[n,{n}^{\prime }\right]\right)\cdot m=\eta \left(n\right)\cdot \left(\eta \left({n}^{\prime }\right)\cdot m\right)-\eta \left({n}^{\prime }\right)\cdot \left(\eta \left(n\right)\cdot m\right)$,

,

$\left[x,y\right]\text{\hspace{0.17em}}.\text{\hspace{0.17em}}r=x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}r-y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}r$ (1)

$x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left[r,{r}^{\prime }\right]=\left[x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}r,{r}^{\prime }\right]+\left[r,x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}^{\prime }\right]$ (2)

$\partial \left(x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}r\right)=\left[x,\partial \left(r\right)\right]$ ; (3)

$\partial \left(r\right)\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}^{\prime }=\left[r,{r}^{\prime }\right]$, (4)

$\text{0}\to V\stackrel{\iota }{\to }R\stackrel{\partial }{\to }L\stackrel{\nu }{\to }P\to 0$ (5)

i) 根据(3)式，显然P是一个李代数的正合列。因此，正合列(5)是一个李代数的正合列。

ii) 根据(4)式可知V属于李代数R的中心，即 $\left[V,R\right]=0$ 。特别地，V是一个Abel李代数。

iii) L在R上的作用包含了一个在V上的P-模结构，i.e. $\stackrel{¯}{x}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}m:=x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}m$，这里 $m\in V$$x\in L$$\stackrel{¯}{x}\in P$$v\left(x\right)=\stackrel{¯}{x}$

i) 下面的图是交换的，

$\begin{array}{ccccccccccc}0& \to & V& \to & R& \stackrel{\partial }{\to }& L& \to & P& \to & 0\\ & & ↓& & ↓& & ↓& & ↓& & \\ 0& \to & V& \to & {R}^{\prime }& \stackrel{{\partial }^{\prime }}{\to }& L& \to & P& \to & 0\end{array}$

${\partial }^{\prime }f=\partial$ ;

${f|}_{V}=i{d}_{V}$ .

ii) 李代数的同态f是相容的，即

$f\left(\eta \left(x\right)\cdot r\right)={\eta }^{\prime }\left(x\right)\cdot f\left(r\right)$ .

$CML\left(P,L,V\right)$ 为所有固定核V和余核P的交叉模等价类的集合。

3. 一类五维李代数的交叉模

$\text{0}\to V\to R\stackrel{\partial }{\to }L\to P\to 0$ .

$x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}=0$ ; $x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}=0$ ; $x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}={r}_{1}$ ;

$y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}=0$ ; $y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}={r}_{2}$ ; $y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}=\text{0}$ ;

$h\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}=0$ ; $h\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}={r}_{\text{1}}$ ; $x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}=-{r}_{\text{2}}$ .

$x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}=0$ ; $x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}=\text{2}{r}_{0}$ ; $x\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}={r}_{1}$ ;

$y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}={r}_{1}$ ; $y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}=\text{2}{r}_{2}$ ; $y\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}=\text{0}$ ;

$h\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{0}=\text{2}{r}_{0}$ ; $h\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{1}}=\text{0}$ ; $h\text{\hspace{0.17em}}.\text{\hspace{0.17em}}{r}_{\text{2}}=-\text{2}{r}_{\text{2}}$ .

${c}_{1}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({r}_{0},{r}_{1},{r}_{2}\right)M$ ;

${c}_{\text{2}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({r}_{0},{r}_{1},{r}_{2}\right)N$ ;

$\partial \left({r}_{0},{r}_{1},{r}_{2}\right)=\left(\text{0},{c}_{1},{c}_{2}\right)A$,

$M=\left(\begin{array}{cc}{m}_{11}& {M}_{1}\\ 0& {M}_{2}\end{array}\right)$, $N=\left(\begin{array}{cc}{n}_{11}& {N}_{1}\\ 0& {N}_{2}\end{array}\right)$, $A=\left(\begin{array}{cc}\text{0}& \text{0}\\ \text{0}& {A}_{\text{1}}\end{array}\right)$,

$\mathrm{det}{A}_{\text{1}}\ne 0$ 时， $P\text{\hspace{0.17em}}.\text{\hspace{0.17em}}R=0$$M=-\frac{{n}_{12}}{{a}_{22}}\left(\begin{array}{ccc}0& {a}_{32}& {a}_{33}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$

$\mathrm{det}{A}_{\text{1}}=0$ 时， $P\text{\hspace{0.17em}}.\text{\hspace{0.17em}}R=0$$M=-\frac{{a}_{32}}{{a}_{22}}N$$N=\left(\begin{array}{ccc}{n}_{11}& {n}_{12}& {n}_{13}\\ 0& -\frac{{a}_{23}}{{a}_{22}}{n}_{32}& -\frac{{a}_{23}}{{a}_{22}}{n}_{33}\\ 0& {n}_{32}& {n}_{33}\end{array}\right)$

$\text{0}=\partial \left({c}_{1}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)\right)=\partial \left({r}_{0},{r}_{1},{r}_{2}\right)M=\left(\text{0},{c}_{1},{c}_{2}\right)AM$,

$\text{0}=\partial \left({c}_{\text{2}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)\right)=\partial \left({r}_{0},{r}_{1},{r}_{2}\right)N=\left(\text{0},{c}_{1},{c}_{2}\right)AN$,

$\left(\text{0,0,}\left[{r}_{1},{r}_{2}\right]\right)=\partial \left({r}_{1}\right)\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({a}_{2\text{2}}{c}_{1}+{a}_{3\text{2}}{c}_{2}\right)\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({r}_{0},{r}_{1},{r}_{2}\right)\left({a}_{22}M+{a}_{32}N\right)$,

$\left(0,-\left[{r}_{1},{r}_{2}\right],0\right)=\partial \left({r}_{\text{2}}\right)\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({a}_{2\text{3}}{c}_{1}+{a}_{3\text{3}}{c}_{2}\right)\cdot \left({r}_{0},{r}_{1},{r}_{2}\right)=\left({r}_{0},{r}_{1},{r}_{2}\right)\left({a}_{2\text{3}}M+{a}_{3\text{3}}N\right)$,

$0=\left[{c}_{1},{c}_{2}\right]\left({r}_{0},{r}_{1},{r}_{2}\right)=\left({r}_{0},{r}_{1},{r}_{2}\right)\left(MN-NM\right)$,

$\mathrm{det}{A}_{\text{1}}\ne 0$，由分块矩阵的运算，则 ${M}_{\text{2}}=\text{0}$${N}_{\text{2}}=\text{0}$ 。经计算可得，

$M=-\frac{{n}_{12}}{{a}_{22}}\left(\begin{array}{ccc}0& {a}_{32}& {a}_{33}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$, $N=-\frac{{n}_{12}}{{a}_{22}}\left(\begin{array}{ccc}0& {a}_{\text{2}2}& {a}_{\text{2}3}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$,

$\mathrm{det}{A}_{\text{1}}=0$，经计算可得，

, $M=-\frac{{a}_{32}}{{a}_{22}}\left(\begin{array}{ccc}{n}_{11}& {n}_{12}& {n}_{13}\\ 0& -\frac{{a}_{23}}{{a}_{22}}{n}_{32}& -\frac{{a}_{23}}{{a}_{22}}{n}_{33}\\ 0& {n}_{32}& {n}_{33}\end{array}\right)=-\frac{{a}_{32}}{{a}_{22}}N$,

${c}_{1}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{{r}^{\prime }}_{1},{{r}^{\prime }}_{2}\right)=\left({r}_{0},{{r}^{\prime }}_{1},{{r}^{\prime }}_{2}\right){M}^{\prime }$ ;

${c}_{\text{2}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left({r}_{0},{{r}^{\prime }}_{1},{{r}^{\prime }}_{2}\right)=\left({r}_{0},{{r}^{\prime }}_{1},{{r}^{\prime }}_{2}\right){N}^{\prime }$ ;

${\partial }^{\prime }\left({r}_{0},{{r}^{\prime }}_{1},{{r}^{\prime }}_{2}\right)=\left(\text{0},{c}_{1},{c}_{2}\right){A}^{\prime }$,

${M}^{\prime }=\left(\begin{array}{cc}{{m}^{\prime }}_{11}& {{M}^{\prime }}_{1}\\ 0& {{M}^{\prime }}_{2}\end{array}\right)$, ${N}^{\prime }=\left(\begin{array}{cc}{{n}^{\prime }}_{11}& {{N}^{\prime }}_{1}\\ 0& {{N}^{\prime }}_{2}\end{array}\right)$, ${A}^{\prime }=\left(\begin{array}{cc}\text{0}& \text{0}\\ \text{0}& {{A}^{\prime }}_{\text{1}}\end{array}\right)$,

$B=\left(\begin{array}{cc}\text{1}& {B}_{1}\\ \text{0}& {B}_{2}\end{array}\right)$,

$A={A}^{\prime }B$,

${N}^{\prime }B=BN$,

${M}^{\prime }B=BM$,

$f\left(\left[{r}_{1},{r}_{2}\right]\right)=\left[f\left({r}_{1}\right),f\left({r}_{2}\right)\right]$,

$\text{0}\to C\left(P,V\right)\stackrel{{\nu }^{*}}{\to }C\left(L,V\right)\stackrel{{k}^{*}}{\to }C\left(P,L;V\right)\to 0$ .

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