近些年来, hom-结构, 如hom-李代数, hom-代数, hom-李超代数, hom-超代数, hom-双代数, n-元hom-Nambu-李代数以及hom-李2-代数等代数结构及性质被广泛研究, 而且取得了许多重要研究结果. hom-李代数最早是由Hartwig, Larsson以及Silvestrov在文献[1]中研究Witt代数和Virasoro代数的形变时提出来的.文献[2]给出了hom-李代数及其表示的定义, 而且对hom-李代数的表示理论进行了严谨细致的研究.文献[3]在半单李代数上研究了hom-李代数的结构.二次hom-李代数在文献[4]中得到了研究.文献[5]提供了保积hom-李代数可解的充要条件.

而在李理论的研究中, 李环是一个重要且非常有趣的课题.在文献[6]中, 研究了幂零李环的条件及幂零李环的中心序列.对于4-Engel李环的幂零类的研究在文献[7]中得到体现.文献[8, 9]对李环的二阶上同调及Schreier's扩张理论进行了探索.特别的, 在文献[10]中, 作者在李环的一个短正合序列$ A\rightarrow E\xrightarrow{\pi} L $中的$ A $的扩张的自同构群上构造正合序列, 这里$ A $是$ E $的阿贝尔子环.他们通过一系列的研究得出如下结论.

设$ A\rightarrow E\xrightarrow{\pi} L $是阿贝尔扩张, 则存在以下两个正合序列

$ \begin{eqnarray*} 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha), \end{eqnarray*} $

$ \begin{eqnarray*} 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha). \end{eqnarray*} $

设$ A\rightarrow E\rightarrow L $是中心扩张, 则存在以下正合序列

$ \begin{eqnarray*} 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}_{A}(E)\xrightarrow{\tau} \mathrm{Aut}(A)\times \mathrm{Aut}(L)\xrightarrow{\lambda} H^{2}(H, N, \alpha). \end{eqnarray*} $

文献[11, 12]研究了有限维$ p $ -群上的$ p $ -自同构.文献[13, 14]分别在群扩张的自同构中得到了一个正合序列和在群扩张的自同构群上得到Wells正合序列.特别地, 在文献[15]中, 作者在群的一个短正合序列$ 1\rightarrow N\rightarrow G\xrightarrow{\pi} H\rightarrow 1 $中的$ N $的扩张的自同构群上构造正合序列, 这里$ N $是$ G $的阿贝尔子群.他们通过一系列的研究得出如下结论.

若$ 1\rightarrow N\rightarrow G\xrightarrow{\pi} H\rightarrow 1 $是阿贝尔扩张, 则存在以下两个正合序列:

$ \begin{eqnarray*} &&1\rightarrow \mathrm{Aut}^{N, H}(G)\rightarrow \mathrm{Aut}^{H}_{N}(G)\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(H, N), \\ &&1\rightarrow \mathrm{Aut}^{N, H}(G)\rightarrow \mathrm{Aut}^{N}(G)\xrightarrow{\tau_{2}} C_{2}\xrightarrow{\lambda_{2}} H^{2}(H, N). \end{eqnarray*} $

若$ 1\rightarrow N\rightarrow G\xrightarrow{\pi} H\rightarrow 1 $是中心扩张, 则存在以下正合序列:

$ \begin{eqnarray*} 1\rightarrow \mathrm{Aut}^{N, H}(G)\rightarrow \mathrm{Aut}_{N}(G)\xrightarrow{\tau} \mathrm{Aut}(N)\times\mathrm{Aut}(H)\xrightarrow{\lambda} H^{2}(H, N). \end{eqnarray*} $

文献[16]首次研究了hom-李环, 给出了hom-李环的定义并讨论了hom-李环幂零的条件.自然地, 通过对阿贝尔群扩张的自同构和阿贝尔李环扩张的自同构的研究, 希望在阿贝尔hom-李环中也能构造出相应的正合序列, 并对其进行相关应用.本文具体结果如下:

定理1.1   令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张, 则有以下两个正合序列:

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha), \\ &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}\xrightarrow{\lambda_{2}} H^{2}(L, A, \alpha). \end{eqnarray} $

定理1.2   令$ A\rightarrow E\xrightarrow{\pi} L $是一个中心扩张, 则存在下列正合序列:

$ \begin{eqnarray} 1\rightarrow\mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}_{A}(E)\xrightarrow{\tau} \mathrm{Aut}(A)\times \mathrm{Aut}(L)\xrightarrow{\lambda} H^{2}(L, A, \alpha), \end{eqnarray} $

这里$ \lambda(\theta, \phi) = [m_{\theta, \phi}, m'_{\theta, \phi}] $, $ \tau(\gamma) = (\theta, \phi) $.

定理1.3   令$ E $是一个hom-李环, 且$ A $是$ E $的一个阿贝尔理想, 如果序列$ A\rightarrow E\rightarrow L $分裂, 则序列

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}^{\ast}\rightarrow 1, \\ &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota}\mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}^{\ast}\rightarrow 1. \end{eqnarray} $

也可裂.

定理1.4   令$ E $是一个阶为$ p^{n}(p\geq2) $的hom-李$ p $环, $ Z(E) $是循环的且$ \Phi(E)\subseteq Z(E) $.则每一个$ Z(E) $的$ p $自同构都可以被扩张成依赖于$ \mathrm{Aut}^{L}_{A}(E) $的$ E $上的自同构, 这里, $ A = Z(E) $, $ L = E/Z(E) $.

本文内容安排如下.第2节介绍与hom-李环及其正合序列等有关的一些基本概念.第3节, 在$ A\rightarrow E\xrightarrow{\pi} L $是阿贝尔hom-李环扩张的前提下, 讨论了正合序列的实现过程, 并给出了不同情况下的正合序列.在阿贝尔hom-李环扩张$ A\rightarrow E\xrightarrow{\pi} L $是可分裂的情况下, 对不同形式扩张下的正合序列进行了改造, 使得被改造后的正合序列依然可分裂.第4节, 对定理3.10中所得到的正合序列(3.4)进行了应用.

本节给出本文中将涉及的一些相关概念和基本结论.

定义2.1 ^[16]    hom-李环是指一个三元数组$ (L, [\cdot, \cdot], \beta) $, 其中$ L $是一个阿贝尔群, $ [\cdot, \cdot]: L\times L\rightarrow L $是反对称双线性映射, 线性映射$ \beta: L\rightarrow L $满足hom-Jacobi等式

$ \begin{eqnarray*} [\beta(x), [y, z]]+[\beta(y), [z, x]]+[\beta(z), [x, y]] = 0, \quad \forall\, x, \, y, \, z\in L. \end{eqnarray*} $

定义2.2   设$ (L, [\cdot, \cdot], \beta) $为hom-李环, 若线性映射$ \beta: L\rightarrow L $满足

$ \begin{eqnarray*} \beta[x, y] = [\beta(x), \beta(y)], \end{eqnarray*} $

其中$ x, y\in L $.则称$ (L, [\cdot, \cdot], \beta) $为保积hom-李环.

定义2.3   设$ (L, [\cdot, \cdot], \beta) $是一个hom-李环, 并且$ M $为$ L $的子群.若$ \beta(M)\subseteq M, $ $ [M, M]\subseteq M $, 则称$ (M, [\cdot, \cdot], \beta) $是$ (L, [\cdot, \cdot], \beta) $的hom-子环.若$ L $的子群$ M $满足$ \beta(M)\subseteq M, [M, L]\subseteq M $, 称$ M $为$ (L, [\cdot, \cdot], \beta) $的理想.

定义2.4   设$ L $为有限维hom-李环, 若$ L\neq 0 $令$ \Phi(L) $为$ L $的所有极大hom-子环的交.若$ L = 0 $, 令$ \Phi(L) = 0 $, 则称$ \Phi(L) $是$ L $的Frattini hom-子环.

定义2.5   设$ (L, [\cdot, \cdot], \beta) $是一个保积的hom-李环, 令$ \beta^{k} = \beta\circ \beta\circ\cdot\cdot\cdot\circ\beta $, 特别地$ \beta^{0} = id $, $ \beta^{1} = \beta $, 对于任一非负整数$ k $, 一个线性映射$ D: L\rightarrow L $, 如果满足下面两个条件

(1) $ [D, \beta] = 0 $,

(2) $ D([x, y]) = [D(x), \beta^{k}(y)]+[\beta^{k}(x), D(y)], \; \forall\; x, y\in L $.

则称$ D $为hom-李环的$ \beta^{k} $导子.所有的hom-李环的$ \beta^{k} $导子的集合记为$ \mathrm{Der}_{\beta^{k}}(L) $.

定义2.6   定义hom-李环$ (L, [\cdot, \cdot], \beta) $的降中心列为

$ \begin{eqnarray*} L^{1}& = &[L, L], \\ L^{2}& = &[L^{1}, L^{1}], \\ &\vdots&\\ L^{s}& = &[L^{s-1}, L^{s-1}], \\ &\vdots& \end{eqnarray*} $

若存在$ n\in Z^{+} $, 使得$ L^{n} = 0 $, 则称$ (L, [\cdot, \cdot], \beta) $幂零.

定义2.7   令$ L $是一个hom-李环且$ n\in N_{+} $, 定义

$ \begin{eqnarray*} nL = \{nx\mid x\in L\}. \end{eqnarray*} $

则易知$ nL $是$ L $的一个理想.一个幂零hom-李环$ L $的阶为$ p $, 称$ L $为$ p $ -hom-李环.称$ L' = [L, L] $为$ L $的导出子环.

定义2.8   设$ (L_{1}, [\cdot, \cdot], \beta_{1}) $与$ (L_{2}, [\cdot, \cdot], \beta_{2}) $为hom-李环.若线性映射$ \phi: L_{1}\rightarrow L_{2} $满足

(1) $ \beta_{2}\phi = \phi\beta_{1} $,

(2) $ \phi([x, y]_{1}) = [\phi(x), \phi(y)]_{2}, \; \forall\; x, y\in L_{1} $,

则称$ \phi $为hom-李环同态.当$ \phi $既是单射又是满射时, 则称它为同构映射.

定义2.9   设$ (L, [\cdot, \cdot], \beta) $是一个hom-李环, $ N $是$ L $的理想. $ L/N = \{x+N\mid x\in L\} $在$ L/N $的元素之间定义方括号积$ [\cdot, \cdot] $如下

$ \begin{eqnarray*} [x+N, y+N] = [x, y]+N, \; \forall\; x, y\in L. \end{eqnarray*} $

若$ \overline{\beta} $是$ L/N $上的一个同态且$ \overline{\beta}(\overline{x}) = \beta(x)+N $, 则$ L/N $关于上述方括号积构成一个hom-李环, 称$ L/N $为hom-李环$ L $模掉理想$ N $的商环, 记为$ \overline{L} = L/N $.

定义2.10  若$ E\rightarrow L $是hom-李环的满同态, 所以对$ \forall\; x\in L $, 则存在$ \; tx\in E $, 使得$ tx\rightarrow x $, 称$ tx $为$ x $的一个提升.

定义2.11   设$ L_{1} $, $ L_{2} $, $ \cdots $, $ L_{k} $, $ \cdots $均为域$ \mathbb{F} $上的hom-李环, 又$ f_{i} $是$ L_{i} $到$ L_{i+1} $的同态.如果$ \mathrm{Ker}f_{i+1} = f_{i}(L_{i}) $, i = 1, 2, $ \cdots $, 则称序列

$ \begin{eqnarray*} L_{1}\xrightarrow{f_{1}} L_{2}\xrightarrow{f_{2}}\cdots \rightarrow L_{i} \xrightarrow{f_{i}} L_{i+1}\rightarrow \cdots. \end{eqnarray*} $

为正合序列.

定义2.12   设$ (L, [\cdot, \cdot], \beta) $是一个hom-李环, $ V $是线性空间, 并且$ \beta_{v} $是$ V $上的线性映射. $ (V, \beta_{v}) $就被称为hom- L -模, 如果存在一个映射

$ \begin{eqnarray*} &&L\times V\rightarrow V\\ &&(x, v)\mapsto x\cdot v, \end{eqnarray*} $

并且$ \forall\; v\in V, x, y\in L $, 满足以下条件

(1) $ \beta_{v}(x\cdot v) = \beta(x)\beta_{v}(v) $,

(2) $ [x, y]\beta_{v}(v) = \beta(x)yv-\beta(y)xv $.

定义2.13   设$ a $, $ b $与$ L $均为hom-李环, 若有$ L $的理想$ n $与$ a $同构, 而商环$ L/n $与$ b $同构, 则称$ L $是$ b $通过$ a $的扩张.

定义2.14   设$ L $是一个hom-李环, $ A $是一个平凡的hom-$ L $-模, 令$ Z^{2}(L, A, \alpha) $是所有函数对$ (f, g, \alpha) $的集合.这里$ f, g: L\times L\rightarrow A $, $ \alpha: L\rightarrow \mathrm{Der}_{\beta^{k}}(A) $, 对$ \forall\; l, l_{1}, l_{2}, l_{3}\in L $满足

(1) $ f(l, 0) = 0, f(l_{1}, l_{2}) = f(l_{2}, l_{1}) $,

(2) $ f(l_{1}, l_{2}+l_{3})+f(l_{2}, l_{3}) = f(l_{1}+l_{2}, l_{3})+f(l_{1}, l_{2}) $,

(3) $ g(l, l) = 0 $,

(4) $ g(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(f(l_{1}, l_{2})) = f([l_{1}, l_{3}], [l_{2}, l_{3}])+g(l_{1}, l_{3})+g(l_{2}, l_{3}) $,

$ g(l_{1}, l_{2}+l_{3})-\alpha_{l_{1}}(f(l_{2}, l_{3})) = f([l_{1}, l_{2}], [l_{1}, l_{3}])+g(l_{1}, l_{2})+g(l_{1}, l_{3}) $,

(5) $ g([l_{1}, l_{2}], l_{3})+g([l_{2}, l_{3}], l_{1})+g([l_{3}, l_{1}], l_{2})+f([l_{1}, l_{2}, l_{3}], [l_{2}, l_{3}, l_{1}])+ $ $ f(-[l_{3}, l_{1}, l_{2}], [[l_{3}, l_{1}, l_{2}]) $

$ = \alpha_{l_{3}}(g(l_{1}, l_{2}))+\alpha_{l_{1}}(g(l_{2}, l_{3}))+\alpha_{l_{2}}(g(l_{3}, l_{1})) $.

称$ Z^{2}(L, A, \alpha) $为在$ A $上的$ L $上的上闭链.

现给出两个上闭链$ (f, g, \alpha), \, (f', g', \alpha)\in Z^{2}(L, A, \alpha) $, 令$ (f, g, \alpha)+(f', g', \alpha): = (f+f', g+g', \alpha) $, 这里$ f+f', \, g+g' $在逐点加法(如果$ f, \, g: L\times L\rightarrow A $, 则$ (f, g, \alpha)(l_{1}, l_{2}) = f(l_{1}, l_{2})+g(l_{1}, l_{2}) $)的运算下构成阿贝尔群.

定义2.15   设$ L $是一个hom-李环, $ A $是一个平凡的hom-$ L $-模, 令$ t: L\rightarrow A $是一个满足$ t(0) = 0 $的映射.且令$ B^{2}(L, A, \alpha) $是所有函数对$ (f, g, \alpha) $的集合.这里$ f, \, g: L\times L\rightarrow A $, $ \alpha: L\rightarrow \mathrm{Der}_{\beta^{K}}(A) $, 对$ \; \forall\; l_{1}, \, l_{2}\in L $满足

$ \begin{eqnarray*} f(l_{1}, l_{2}) = t(l_{1})+t(l_{2})-t(l_{1}+l_{2}), \; \; \; g(l_{1}, l_{2}) = \alpha_{l_{1}}(t(l_{2}))-\alpha_{l_{2}}(t(l_{1}))-t([l_{1}, l_{2}]). \end{eqnarray*} $

称函数对$ (f, g, \alpha) $为在$ A $上的$ L $上的边界余映射.

不难证明$ B^{2}(L, A, \alpha) $是$ Z^{2}(L, A, \alpha) $的子集.

定义2.16   称商群$ H^{2}(L, A, \alpha) = Z^{2}(L, A, \alpha)/B^{2}(L, A, \alpha). $为hom-李环$ L $的二阶上同调群.

定义2.17   一个被阿贝尔hom-李环$ A $的被hom-李环$ L $的扩张$ 0\rightarrow A\rightarrow E\xrightarrow{\pi} L\rightarrow 0 $称为是可裂的.如果存在一个hom-李环同态$ \lambda: L\rightarrow E $使得$ \pi\lambda = 1_{L} $.

本节的目的是构造阿贝尔hom-李环的正合序列, 由于hom-李环本身内部存在一个映射和两种运算并且还具有复杂的二阶上同调群, 显然会比群和李环的构造更加复杂.在定理3.10和3.12的基础之上通过阿贝尔李扩张推广了三个正合序列, 并证明了在序列$ A\rightarrow E\xrightarrow{\pi} L $可裂的情况下所推广的这三个正合序列也可裂.

令$ E $和$ L $是保积hom-李环, $ A $是$ E $的理想. $ \beta: L\rightarrow L $限制在$ A $上是$ A $上的线性变换, $ \beta $是同态映射且与$ \gamma, \, \pi $可交换, 这一规定贯穿全文.首先来介绍一些记号.

定义3.1 ^[10]   令

$ \begin{eqnarray*} &&\mathrm{Aut}^{A}(E) = \{\sigma\in \mathrm{Aut}(E)\mid\sigma(x) = x, \forall\; x\in A\}, \\ &&\mathrm{Aut}_{A}(E) = \{\sigma\in \mathrm{Aut}(E)\mid\sigma(A) = A\}, \\ &&\mathrm{Aut}^{A, L}(E) = \{\sigma\in \mathrm{Aut}(E)\mid\sigma(x) = x, \forall\; x\in A, \sigma(x)-x\in A, \forall\; x\in E\}, \\ &&\mathrm{Aut}^{L}_{A}(E) = \{\sigma\in \mathrm{Aut}(E)\mid\sigma(A) = A, \sigma(x)-x \in A, \forall\; x\in E\}. \end{eqnarray*} $

引理3.1   设$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔hom-李环扩张, 其中$ A $是$ E $的阿贝尔子环.则阿贝尔hom-李环$ A $的被hom-李环$ E $的一个扩张确定了$ E $到$ A $的自同构群$ \mathrm{Aut}(A) $的一个hom-李环同态.

证  设正合列$ 0\rightarrow A\rightarrow E\rightarrow L\rightarrow 1 $是阿贝尔hom-李环$ A $的被hom-李环$ E $的一个扩张.则$ E\rightarrow L $是满同态, 进而对$ \; \forall\; x\in L $, 都存在一个提升$ tx\in E $, 即$ tx\rightarrow x $.于是可以规定

$ \begin{eqnarray*} &&\theta_{x}: A\rightarrow A, \\ &&a\mapsto tx+a-tx. \end{eqnarray*} $

又因为$ A $是$ E $的hom-李环, 所以$ \mathrm{Im}(\theta_{x})\subset A $, 当然$ \theta_{x} $是$ A $的一个自同构.从而存在

$ \begin{eqnarray*} &&\theta: L\rightarrow \mathrm{Aut}(A), \\ &&\; \; \; \; x\mapsto \theta_{x}. \end{eqnarray*} $

接下来利用$ \alpha(l) = \alpha_{l} $, 这里$ \alpha_{l}(a) = [\beta^{k}(l), a] $, $ l\in L $, $ a\in A $, 来定义一个hom-李环同态$ \alpha: L\rightarrow \mathrm{Der}_{\beta^{k}}(A) $.所以, 可以考虑第二上同调群$ H^{2}(L, A, \alpha) $.与群和李环的情况类似, 对于任意的阿贝尔扩张$ A\rightarrow E\xrightarrow{\pi} L $, $ (\theta, \phi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L) $被称为是相容的, 如果$ \forall\; l\in L $, 满足$ \theta\alpha_{l}\theta^{-1} = \alpha_{\phi(l)} $.把所有的相容对记为$ C $.类似于阿贝尔群扩张的Wells序列及阿贝尔李环扩张的正合序列, 对每一个$ \gamma\in \mathrm{Aut}_{A}(E) $都包含了一个相容对$ (\theta, \phi) $, 而且映射$ \tau(\gamma) = (\theta, \phi) $是一个同态.令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔hom-李环扩张, 其中$ A $是$ E $的阿贝尔子环.设$ t: L\rightarrow E $是$ E $的一个正规截口.则对任意的$ l\in L $, 必存在一个$ x\in E $使得$ \pi(x) = l $, 而且$ t(l) $由$ A+t(l) = A+x $所确定.这就说明了$ \pi(t(l)) = l $成立.

对$ \forall\; l_{1}, \, l_{2}\in L $, 定义

$ \begin{eqnarray*} &&\mu, \, \, \delta: L\times L\rightarrow A\\ &&\mu(l_{1}, l_{2}) = \beta^{k}(t(l_{1}))+\beta^{k}(t(l_{2})-\beta^{k}(t(l_{1}+l_{2}), \\ &&\delta(l_{1}, l_{2}) = \beta^{k}[t(l_{1}), t(l_{2})]-\beta^{k}(t([l_{1}, l_{2}])). \end{eqnarray*} $

引理3.2   根据以上定义, 则有$ (\mu, \delta, \alpha)\in Z^{2}(L, A, \alpha) $.

证要证$ (\mu, \delta, \alpha)\in Z^{2}(L, A, \alpha) $, 只需要证对$ \forall\; l_{1}, \, l_{2}, \, l_{3}\in L $有以下等式成立即可

$ \begin{eqnarray*} &&\delta(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(\mu(l_{1}, l_{2})) = \delta(l_{1}, l_{3})+\delta(l_{2}, l_{3})+\mu([l_{1}, l_{3}], [l_{2}, l_{3}]), \\ &&\delta(l_{1}, l_{2}+l_{3})+\alpha_{l_{1}}(\mu(l_{2}, l_{3})) = \delta(l_{1}, l_{2})+\delta(l_{1}, l_{3})+\mu([l_{1}, l_{2}], [l_{1}, l_{3}]). \end{eqnarray*} $

由于

$ \begin{eqnarray*} \delta(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(\mu(l_{1}, l_{2})) & = &\beta^{k}([t(l_{1}+l_{2}), t(l_{3})])-\beta^{k}(t[l_{1}+l_{2}, l_{3}])-[\beta^{k}(t(l_{3})), \mu(l_{1}, l_{2})]\\ & = &[\beta^{k}(t(l_{1}))+\beta^{k}(t(l_{2}))-\mu(l_{1}, l_{2}), \beta^{k}(t(l_{3}))]-\beta^{k}(t([l_{1}, l_{3}])\\ &&-\beta^{k}(t([l_{2}, l_{3}])+\mu([l_{1}, l_{3}], [l_{2}, l_{3}])-[\beta^{k}(t(l_{3})), \mu(l_{1}, l_{2})]\\ & = &\delta(l_{1}, l_{3})+\delta(l_{2}, l_{3})+\mu([l_{1}, l_{3}], [l_{2}, l_{3}]), \end{eqnarray*} $

且

$ \begin{eqnarray*} \delta(l_{1}, l_{2}+l_{3})+\alpha_{l_{1}}(\mu(l_{2}, l_{3})) & = &\beta^{k}([t(l_{1})+t(l_{2}+l_{3})])-\beta^{k}(t[l_{1}, l_{2}+l_{3}])+[\beta^{k}(t(l_{1})), \mu(l_{2}, l_{2})]\\ & = &[\beta^{k}(t(l_{1})), \beta^{k}(t(l_{2}))+\beta^{k}(t(l_{3}))-\mu(l_{2}, l_{3})]-\beta^{k}(t([l_{1}, l_{2}])\\ &&-\beta^{k}(t([l_{1}, l_{3}])+\mu([l_{1}, l_{2}], [l_{1}, l_{3}])+[\beta^{k}(t(l_{1})), \mu(l_{2}, l_{3})]\\ & = &\delta(l_{1}, l_{2})+\delta(l_{1}, l_{3})+\mu([l_{1}, l_{2}], [l_{1}, l_{3}]). \end{eqnarray*} $

故有$ (\mu, \delta, \alpha)\in Z^{2}(L, A, \alpha) $.证毕.

引理3.3   令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张, 若$ \sigma\in \mathrm{Aut}_{A}(E) $, 则存在一个三元组$ (\theta, \phi, \chi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L)\times A^{L} $, 使得对任意的$ l, \, l_{1}, \, l_{2}\in L, \, a\in A $.有下列等式成立

(ⅰ) $ \gamma(\beta^{k}(t(l)+a)) = \beta^{k}(t(\phi(l)))+\chi(l)+\theta(a), $

(ⅱ) $ \mu(\phi(l_{1}), \phi(l_{2}))-\theta(\mu(l_{1}, l_{2})) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}) $,

(ⅲ) $ \theta[\beta^{k}(t(l)), a] = [\beta^{k}(t(\phi(l))), \theta(a)] $,

(ⅳ) $ \delta(\phi(l_{1}), \phi(l_{2}))-\theta(\delta(l_{1}, l_{2})) = \chi[l_{1}, l_{2}]-[\chi(l_{1}), \beta^{k}(t(\phi(l_{2})))]-[\beta^{k}(t(\phi(l_{1}))), \chi(l_{2})] $.

证 (ⅰ)   令$ \theta = \gamma\mid_{A} $, 定义$ \phi: L\rightarrow L $使得对任意的$ l\in L $, 满足$ \phi(l) = \pi(\gamma(t(l))) $.显然$ \theta\in \mathrm{Aut}(A) $, $ \phi\in \mathrm{Aut}(L) $.

现在进一步定义$ \chi\in A^{L} $, 由于对任意的$ l\in L $, 都有$ \pi(t(l)) = l $, 故有$ \pi(t(\phi(l))) = \phi(l) $, 所以$ \pi(t(\phi(l)))-\phi(l) = 0 $.由此, 定义$ \chi: L\rightarrow A $满足$ \chi(l) = \beta^{k}(\gamma(t(l)))-\beta^{k}(t(\phi(l))) $.现在令$ l\in L, a\in A $, 则得到$ \gamma(\beta^{k}(t(l)+a)) = \gamma(\beta^{k}(t(l)))+\theta(a) = \chi(l)+\beta^{k}(t(\phi(l)))+\theta(a) $.

(ⅱ)   令$ l_{1}, l_{2}\in L $, 根据定义有,

$ \begin{eqnarray*} \theta(\mu(l_{1}, l_{2}))+\chi(l_{1}+l_{2})+\beta^{k}(t(\phi(l_{1}+l_{2}))) & = &\gamma\mid_{A}(\beta^{k}(t(l_{1}))+\beta^{k}(t(l_{2}))\\ &&-\beta^{k}(t(l_{1}+l_{2})))+\beta^{k}(\gamma(t(l_{1}+l_{2})))\\ & = &\gamma(\beta^{k}(t(l_{1})))+\gamma(\beta^{k}(t(l_{2})))\\ & = &\beta_{k}(t(\phi(l_{1})))+\chi(l_{1})+\beta_{k}(t(\phi(l_{2})))+\chi(l_{2}). \end{eqnarray*} $

因此有,

$ \begin{eqnarray*} \mu(\phi(l_{1}), \phi(l_{1}))-\theta(\mu(l_{1}, l_{2})) & = &\beta^{k}(t(\phi(l_{1})))+\beta^{k}(t(\phi(l_{2})))-\beta^{k}(t(\phi(l_{1})+\phi(l_{2})))-\beta^{k}(t(\phi(l_{1})))\\ &&-\chi(l_{1})-\beta^{k}(t(\phi(l_{2})))-\chi(l_{2})+\chi(l_{1}+l_{2})+\beta^{k}(t(\phi(l_{1}+l_{2})))\\ & = &\chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}). \end{eqnarray*} $

(ⅲ)令$ l\in L, a\in A $, 故

$ \begin{eqnarray*} \theta[\beta^{k}(t(l)), a] & = &\gamma[\beta^{k}(t(l)), a] = [\chi(l), \theta(a)]+[\beta^{k}(t(\phi(l))), \theta(a)] = [\beta^{k}(t(\phi(l))), \theta(a)]. \end{eqnarray*} $

(ⅳ)令$ l_{1}, l_{2}\in L $, 根据定义有,

$ \begin{eqnarray*} &&\theta(\delta(l_{1}, l_{2}))+\chi[l_{1}, l_{2}]+\beta^{k}(t(\phi[l_{1}, l_{2}]))\\ & = &\theta(\beta^{k}([t(l_{1}), t(l_{2})])-\beta^{k}(t[l_{1}, l_{2}))+\chi[l_{1}, l_{2}]+\beta^{k}(t(\phi[l_{1}, l_{2}]))\\ & = &[\beta^{k}(\gamma(t(l_{1}))), \beta^{k}(\gamma(t(l_{1})))]-\gamma(\beta^{k}(t[l_{1}, l_{2})) +\chi[l_{1}, l_{2}]+\beta^{k}(t([\phi(l_{1}), \phi(l_{2})])\\ & = &[\chi(l_{1}), \beta^{k}(t(\phi(l_{2})))]+[\beta^{k}(t(\phi(l_{1}))), \chi(l_{2})] +[\beta^{k}(t(\phi(l_{1}))), \beta^{k}(t(\phi(l_{2})))]. \end{eqnarray*} $

故有,

$ \begin{eqnarray*} \delta(\phi(l_{1}), \phi(l_{2}))-\theta(\delta(l_{1}, l_{2})) & = &\beta^{k}[t(\phi(l_{1})), t(\phi(l_{2}))]-\beta^{k}(t[\phi(l_{1}), \phi(l_{2})])-[\chi(l_{1}), \beta^{k}(t(\phi(l_{2})))]\\ &&-[\beta^{k}(t(\phi(l_{1}))), \chi(l_{2})]-[\beta^{k}(t(\phi(l_{1}))), \beta^{k}(t(\phi(l_{2})))]\\ &&+\chi[l_{1}, l_{2}]+\beta^{k}(t(\phi[l_{1}, l_{2}]))\\ & = &\chi[l_{1}, l_{2}]-[\chi(l_{1}), \beta^{k}(t(\phi(l_{2})))]-[\beta^{k}(t(\phi(l_{1}))), \chi(l_{2})]. \end{eqnarray*} $

证毕.

引理3.4   令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张, 若三元组$ (\theta, \phi, \chi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L)\times A^{L} $满足引理3.3中的(ii)–(iv), 则由引理3.3的(i)定义的映射$ \gamma: E\rightarrow E $是属于$ \mathrm{Aut}_{A}(E) $中的在$ E $上的自同构.

证  首先注意到对任意的$ x\in E $, 其都可以被表示为$ x = \beta^{k}(t(l))+a $的形式.令

$ \begin{eqnarray*} x_{1} = \beta^{k}(t(l_{1}))+a_{1}, \; \; \; x_{2} = \beta^{k}(t(l_{2}))+a_{2}, \end{eqnarray*} $

其中$ l_{1}, l_{2}\in L $, $ a_{1}, a_{2}\in A $.故有,

$ \begin{eqnarray*} \gamma(x_{1}+x_{2}) & = &\gamma(\beta^{k}(t(l_{1}))+a_{1}+\beta^{k}(t(l_{2}))+a_{2})\\ & = &\gamma(\beta_{k}(t(l_{1}+l_{2}))+\mu(l_{1}, l_{2})+a_{1}+a_{2})\\ & = &\beta^{k}(t(\phi(l_{1})))+\beta^{k}(t(\phi(l_{2})))+\chi(l_{1})+\chi(l_{2})+\theta(a_{1})+\theta(a_{2}), \\ & = &\gamma(x_{1})+\gamma(x_{2}), \\ \gamma[x_{1}, x_{2}] & = &\gamma[\beta^{k}(t(l_{1}))+a_{1}, \beta^{k}(t(l_{2}))+a_{2}]\\ & = &\gamma(\delta(l_{1}, l_{2})+\beta^{k}(t[l_{1}, l_{2}]))+\theta[\beta^{k}(t(l_{1})), a_{2}]+\theta[a_{1}, \beta^{k}(t(l_{2}))]\\ & = &\beta^{k}(t(\phi[l_{1}, l_{2}]))+\theta(\delta(l_{1}, l_{2}))+\chi[l_{1}, l_{2}]\\ &&+[\beta^{k}(t(\phi(l_{1}))), \theta(a_{2})]+[\theta(a_{1}), \beta^{k}(t(\phi(l_{2})))]\\ & = &[\gamma(x_{1}), \gamma(x_{2})]. \end{eqnarray*} $

现在令$ x\in \mathrm{ker}(\gamma) $, 这里$ x = \beta^{k}(t(l))+a $, 其中$ l\in L, a\in A $.这意表明$ \beta^{k}(t(\phi(l)))\in A $或者说$ \phi(l) = 0 $, 因此$ l = 0 $.由此可知$ \theta(a) = 0 $, 所以$ x = 0 $.故该映射是单射.下证$ \gamma $是满射.令$ u\in E $, 则可以写成$ u = \beta^{k}(t(l')+a') $, 其中$ l'\in L, a'\in A $.取$ l\in L, a\in A $满足: $ \phi(l) = l' $且$ \theta(a) = a'-\chi(l) $.故有,

$ \begin{eqnarray*} \gamma(\beta^{k}(t(l))+a) = \beta^{k}(t(\phi(l)))+\chi(l)+\theta(a) = \beta^{k}(t(l'))+a' = u. \end{eqnarray*} $

故该映射是满射.证毕.

当$ A\subseteq Z(E) $时, 说$ A\rightarrow E\xrightarrow{\pi} L $是中心扩张.若$ A\rightarrow E\xrightarrow{\pi} L $是一个中心扩张, 则$ L $在$ A $上的作用是平凡的, 且引理3.3, 3.4会有如下改变:

引理3.5   令$ A\rightarrow E\xrightarrow{\pi} L $是中心扩张, 若$ \gamma\in \mathrm{Aut}_{A}(E) $, 则存在一个三元组$ (\theta, \phi, \chi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L)\times A^{L} $对$ \forall\; l, \, l_{1}, \, l_{2}\in L, \, a\in A $满足

(ⅰ) $ \gamma(\beta^{k}(t(l))+a) = \beta^{k}(t(\phi(l)))+\chi(l)+\theta(a) $,

(ⅱ) $ \mu(\phi(l_{1}), \phi(l_{2}))-\theta(\mu(l_{1}, l_{2})) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}) $,

(ⅲ) $ \delta(\phi(l_{1}), \phi(l_{2}))-\theta(\delta(l_{1}, l_{2})) = \chi[l_{1}, l_{2}] $.

引理3.6   令$ A\rightarrow E\xrightarrow{\pi} L $是中心扩张, 若三元组$ (\theta, \phi, \chi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L)\times A^{L} $满足引理3.5中的(ⅱ)和(ⅲ), 则由引理3.5的(i)定义的映射$ \gamma: E\rightarrow E $是属于$ \mathrm{Aut}_{A}(E) $中的在$ E $上的自同构.

任给一个阿贝尔扩张$ A\rightarrow E\xrightarrow{\pi} L $, 若对$ \forall\; l\in L $都有$ \theta\alpha_{l}\theta^{-1} = \alpha_{\phi(l)} $, 则称$ (\theta, \phi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L) $是相容的.

引理3.7   由所有的相容对组成的集合是$ \mathrm{Aut}(A)\times \mathrm{Aut}(L) $的子群.将所有相容对组成的子群记为$ C $.并由此定义$ C_{1} $和$ C_{2} $如下

$ \begin{eqnarray*} C_{1} = \{\theta\in \mathrm{Aut}(A)\mid(\theta, 1)\in C\}, \; \; \; C_{2} = \{\phi\in \mathrm{Aut}(L)\mid(1, \phi)\in C\}. \end{eqnarray*} $

记

(ⅰ) $ \theta\in C_{1}\Leftrightarrow \theta_{\alpha_{l}} = \alpha_{l}\theta, \; \forall\; l\in L\Leftrightarrow \theta[\beta^{k}(t(l)), a] = [\beta^{k}(t(l)), \theta(a)], \; \forall\; l\in L, \, a\in A, $

(ⅱ) $ \phi\in C_{2}\Leftrightarrow \alpha_{l} = \alpha_{\phi(l)}, \; \forall\; l\in L\Leftrightarrow [\beta^{k}(t(l)), a] = [\beta^{k}(t(\phi(l))), a], \; \forall\; l\in L, \, a\in A. $

令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张, 且令$ \theta\in C_{1} $, $ \phi\in C_{2} $.定义$ k_{\theta}, k'_{\theta}, k_{\phi}, k'_{\phi}: L\times L\rightarrow A $如下

$ \begin{eqnarray*} &&k_{\theta}(l_{1}, l_{2}) = \mu(l_{1}, l_{2})-\theta(\mu(l_{1}, l_{2})), \; \; \; k'_{\theta}(l_{1}, l_{2}) = \delta(l_{1}, l_{2})-\theta(\delta(l_{1}, l_{2})), \\ &&k_{\phi}(l_{1}, l_{2}) = \mu(\phi(l_{1}), \phi(l_{2}))-\mu(l_{1}, l_{2}), \; \; \; k'_{\phi}(l_{1}, l_{2}) = \delta(\phi(l_{1}), \phi(l_{2})-\delta(l_{1}, l_{2}). \end{eqnarray*} $

引理3.8   根据以上定义, $ (k_{\theta}, k'_{\theta}) $和$ (k_{\phi}, k'_{\phi}) $是$ Z^{2}(L, A, \alpha) $中的元素.

证  要证$ (k_{\theta}, k'_{\theta})\in Z^{2}(L, A, \alpha) $, 只需要证对$ \forall\; l_{1}, l_{2}, l_{3}\in L $, 有以下等式成立即可

$ \begin{eqnarray*} &&k'_{\theta}(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(k_{\theta}(l_{1}, l_{2})) = k_{\theta}([l_{1}, l_{3}], [l_{2}, l_{3}])+k'_{\theta}(l_{1}, l_{3})+k'_{\theta}(l_{2}, l_{3}), \\ &&k'_{\theta}(l_{1}+l_{2}, l_{3})-\alpha_{l_{1}}(k_{\theta}(l_{2}, l_{3})) = k'_{\theta}(l_{1}, l_{2})+k'_{\theta}(l_{1}, l_{3})+k_{\theta}([l_{1}, l_{2}], [l_{1}, l_{3}]). \end{eqnarray*} $

由引理3.2知以下等式成立

$ \begin{eqnarray*} &&\delta(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(\mu(l_{1}, l_{2})) = \delta(l_{1}, l_{3})+\delta(l_{2}, l_{3})+\mu([l_{1}, l_{3}], [l_{2}, l_{3}]), \\ &&\delta(l_{1}, l_{2}+l_{3})+\alpha_{l_{1}}(\mu(l_{2}, l_{3})) = \delta(l_{1}, l_{2})+\delta(l_{1}, l_{3})+\mu([l_{1}, l_{2}], [l_{1}, l_{3}]). \end{eqnarray*} $

故有,

$ \begin{eqnarray*} && k'_{\theta}(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(k_{\theta}(l_{1}, l_{2}))\\ & = &\delta(l_{1}+l_{2}, l_{3})-\theta(\delta(l_{1}+l_{2}, l_{3}))-\alpha_{l_{3}}(\mu(l_{1}, l_{2}))+\theta(\alpha_{l_{3}}(\mu(l_{1}, l_{2})))\\ & = &\delta(l_{1}, l_{3})+\delta(l_{2}, l_{3})+\mu([l_{1}, l_{3}], [l_{2}, l_{3}])-\theta(\delta(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(\mu(l_{1}, l_{2})))\\ & = &k'_{\theta}(l_{1}, l_{3})+k'_{\theta}(l_{2}, l_{3})+k_{\theta}([l_{1}, l_{3}], [l_{2}, l_{3}]), \\ && k'_{\theta}(l_{1}, l_{2}+l_{3})-\alpha_{l_{1}}(k_{\theta}(l_{2}, l_{3}))\\ & = &\delta(l_{1}, l_{2}+l_{3})-\theta(\delta(l_{1}, l_{2}+l_{3}))+\alpha_{l_{1}}(\mu(l_{2}, l_{3}))-\theta(\alpha_{l_{1}}(\mu(l_{2}, l_{3})))\\ & = &\delta(l_{1}, l_{2})+\delta(l_{1}, l_{3})+\mu([l_{1}, l_{2}], [l_{1}, l_{3}])-\theta(\delta(l_{1}+l_{2}, l_{3})+\alpha_{l_{1}}(\mu(l_{2}, l_{3})))\\ & = &k'_{\theta}(l_{1}, l_{3})+k'_{\theta}(l_{2}, l_{3})+k_{\theta}([l_{1}, l_{3}], [l_{2}, l_{3}]). \end{eqnarray*} $

由以上可知$ (k_{\theta}, k'_{\theta})\in Z^{2}(L, A, \alpha) $.同理可证$ (k_{\phi}, k'_{\phi})\in Z^{2}(L, A, \alpha) $.证毕.

根据上面的假设和标记, 定义$ \lambda_{i}: C_{i}\rightarrow H^{2}(L, A, \alpha)(i = 1, 2) $如下

$ \begin{eqnarray*} \lambda_{1}(\theta) = [(k_{\theta}, k'_{\theta})], \; \; \; \lambda_{2}(\phi) = [(k_{\phi}, k'_{\phi})]. \end{eqnarray*} $

引理3.9   映射$ \lambda_{1} $和$ \lambda_{2} $是定义明确的.

证   首先证明$ \lambda_{1} $是定义明确的.令$ t, t': L\rightarrow E $是两个正常映射, 则有$ t(l)-t'(l)\in A $, $ \forall\; l\in L $.因此对每一个$ l\in L $都会唯一的存在$ \lambda(l)\in A $使得: $ \beta^{k}(t(l)) = \beta^{k}(t'(l))+\lambda(l) $.这就由等式: $ \lambda(l) = \beta^{k}(t(l))+\beta^{k}(t'(l)) $, 定义了一个映射$ \lambda: L\rightarrow A $.又由于

$ \begin{eqnarray*} \mu(l_{1}, l_{2}) = \beta^{k}(t(l_{1}))+\beta^{k}(t(l_{2}))-\beta^{k}(t(l_{1}+l_{2})), \; \; \; \delta(l_{1}, l_{2}) = \beta^{k}[t(l_{1}), t(l_{2})]-\beta^{k}(t[l_{1}, l_{2}]), \end{eqnarray*} $

其中$ l_{1}, l_{2}\in L $.同样地, 考虑$ t' $, 得到两个映射$ \mu', \delta': L\times L\rightarrow A $.使得其满足

$ \begin{eqnarray*} \mu'(l_{1}, l_{2}) = \beta^{k}(t'(l_{1}))+\beta^{k}(t'(l_{2}))-\beta^{k}(t'(l_{1}+l_{2})), \; \; \; \delta'(l_{1}, l_{2}) = \beta^{k}[t'(l_{1}), t'(l_{2})]-\beta^{k}(t'[l_{1}, l_{2}]), \end{eqnarray*} $

其中$ l_{1}, l_{2}\in L $, 可以看出

$ \begin{eqnarray} \mu(l_{1}, l_{2})-\mu'(l_{1}, l_{2}) = \lambda(l_{1})+\lambda(l_{2})-\lambda(l_{1}+l_{2}). \end{eqnarray} $

(3.1)

又由于

$ \begin{eqnarray*} &&\delta'(l_{1}, l_{2})+\alpha_{l_{1}}(\lambda(l_{2}))-\alpha_{l_{2}}(\lambda(l_{1}))-\lambda[l_{1}, l_{2}]\\ & = &\beta^{k}([t'(l_{1}), t'(l_{2})])-\beta^{k}(t'[l_{1}, l_{2}])+[\beta^{k}(t(l_{1})), \lambda(l_{2})] -\lambda[l_{1}, l_{2}]-[\beta^{k}(t(l_{2})), \lambda(l_{1})]\\ & = &[\beta^{k}(t(l_{1})), \beta^{k}(t(l_{2}))]-\beta^{k}(t'[l_{1}, l_{2}])-\lambda[l_{1}, l_{2}]\\ & = &[\beta^{k}(t(l_{1})), \beta^{k}(t(l_{2}))]-\beta^{k}(t[l_{1}, l_{2}])+\lambda[l_{1}, l_{2}]-\lambda[l_{1}, l_{2}]\\ & = &\delta(l_{1}, l_{2}), \end{eqnarray*} $

故有,

$ \begin{eqnarray} \delta(l_{1}, l_{2}) = \delta'(l_{2})+\alpha_{l_{1}}(\lambda(l_{2}))-\alpha_{l_{2}}(\lambda(l_{1}))-\lambda([l_{1}, l_{2}]). \end{eqnarray} $

(3.2)

现在令$ \theta\in C_{1} $且$ \phi\in C_{2} $, 定义映射$ m_{\mu}, m'_{\mu}: L\times L\rightarrow A $使得其满足

$ \begin{eqnarray*} m_{\theta}(l_{1}, l_{2}) = \mu'(l_{1}, l_{2})-\theta(\mu'(l_{1}, l_{2})), \; \; \; m'_{\theta}(l_{1}, l_{2}) = \delta'(l_{1}, l_{2})-\theta(\delta'(l_{1}, l_{2})). \end{eqnarray*} $

则由(3.1)式可知下面等式成立

$ \begin{eqnarray*} k_{\theta}(l_{1}, l_{2})-m_{\theta}(l_{1}, l_{2}) = \lambda(l_{1})+\lambda(l_{2})-\lambda(l_{1}+l_{2})-\theta(\lambda(l_{1})+\lambda(l_{2})-\lambda(l_{1}+l_{2})). \end{eqnarray*} $

定义映射$ \lambda': L\rightarrow A $使得其满足对$ \forall\; l\in L $都有, $ \lambda'(l) = \lambda(l)-(\theta\lambda)(l). $自然地有,

$ \begin{eqnarray} (k_{\theta}-m_{\theta})(l_{1}, l_{2}) = \lambda'(l_{1})+\lambda'(l_{2})-\lambda'(l_{1}+l_{2}). \end{eqnarray} $

(3.3)

由(3.2)及$ \theta\in C_{1} $, 可以得知,

$ \begin{eqnarray*} (k'_{\theta}-m'_{\theta})(l_{1}, l_{2}) & = &k'_{\theta}(l_{1}, l_{2})-m'_{\theta}(l_{1}, l_{2})\\ & = &\delta(l_{1}, l_{2})-\delta'(l_{1}, l_{2})-\theta(\delta(l_{1}, l_{2})+\delta'(l_{1}, l_{2}))\\ & = &\alpha_{l_{1}}(\lambda(l_{2}))-\alpha_{l_{2}}(\lambda(l_{1}))-\lambda([l_{1}, l_{2}])-\theta(\alpha_{l_{1}}(\lambda(l_{2}))-\alpha_{l_{2}}(\lambda(l_{1}))-\lambda([l_{1}, l_{2}])\\ & = &\alpha_{l_{1}}(\lambda'(l_{2}))-\alpha_{l_{2}}(\lambda'(l_{1}))-\lambda'([l_{1}, l_{2}]). \end{eqnarray*} $

通过上式及(3.3)式, 可以证得$ (k_{\theta}-m_{\theta}, k'_{\theta}-m'_{\theta})\in B^{2}(L, A, \alpha) $, 或者说可以得到,

$ \begin{eqnarray*} [(k_{\theta}, k'_{\theta})] = [(m_{\theta}, m'_{\theta})]. \end{eqnarray*} $

这就说明了$ \lambda_{1} $的选取不依赖于所取得空间.同理可证$ \lambda_{2} $.证毕.

注记:映射$ \lambda_{1}, \lambda_{2} $不是同态映射, 但这里$ \mathrm{Ker}(\lambda_{i})\, \, (1\leq i\leq 2) $与通常其所表示的意义一致.

令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张.由引理3.3, 每个自同构$ \gamma\in \mathrm{Aut}_{A}(E) $包含了一个自同构$ \theta\in \mathrm{Aut}(A) $和另一个一个自同构$ \phi\in \mathrm{Aut}(L) $, 其中$ \theta = \gamma\mid_A $满足$ \phi(l) = \pi(\gamma(t(l))) $, $ \forall\; l\in L $.现在可以通过令$ \tau(\gamma) = (\theta, \phi) $, 定义一个同态$ \tau: \mathrm{Aut}_{A}(E)\rightarrow \mathrm{Aut}(A)\times \mathrm{Aut}(L) $.同样可以通过令$ \tau_{1}(\gamma) = \theta, \, \tau_{2}(\gamma) = \phi $, 定义$ \tau_{1}: \mathrm{Aut}^{L}_{A}(E) \rightarrow C_{1} $和$ \tau_{2}: \mathrm{Aut}^{A}(E)\rightarrow C_{2} $.显然$ \tau_{1} $和$ \tau_{2} $分别是$ \tau $在$ \mathrm{Aut}^{L}_{A}(E) $和$ \mathrm{Aut}^{A}(E) $上的限制.

定理3.10   令$ A\rightarrow E\xrightarrow{\pi} L $是一个阿贝尔扩张, 则有以下两个正合序列:

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha), \end{eqnarray} $

(3.4)

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}\xrightarrow{\lambda_{2}} H^{2}(L, A, \alpha). \end{eqnarray} $

(3.5)

证   先证(3.4), 由于$ 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E) $和$ \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1} $是显然的.只需证$ \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha) $即可.令$ \gamma\in \mathrm{Aut}^{L}_{A}(E) $, 有, $ (\lambda_{1}\tau_{1})(\gamma) = \lambda_{1}(\theta) = [(k_{\theta}, k'_{\theta})] $, 这里$ \theta = \gamma\mid_A $.因为$ \gamma\in \mathrm{Aut}^{L}_{A}(E) $, 有, $ \phi(l) = (\pi\gamma t)(l) = l $.又由引理3.3可知,

$ \begin{eqnarray*} k_{\theta}(l_{1}, l_{2}) & = &\mu(l_{1}, l_{2})-\theta(\mu(l_{1}, l_{2})) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}), \\ k'_{\theta}(l_{1}, l_{2}) & = &\delta(l_{1}, l_{2})-\theta(\delta(l_{1}, l_{2}))\\ & = &\chi[l_{1}, l_{2}]+[\beta^{k}(t(\phi(l_{2}))), \chi(l_{1})]-[\beta^{k}(t(\phi(l_{1}))), \chi(l_{2})]\\ & = &\chi[l_{1}, l_{2}]+\alpha_{l_{2}}(\chi(l_{1}))-\alpha_{l_{1}}(\chi(l_{2})). \end{eqnarray*} $

这说明了$ (k_{\theta}, k'_{\theta})\in B^{2}(L, A, \alpha) $, 因此$ \mathrm{Im}(\tau_{1})\subseteq \mathrm{Ker}(\lambda_{1}) $.

下面证明$ \mathrm{Ker}(\lambda_{1})\subseteq \mathrm{Im}(\tau_{1}) $.设$ \theta\in C_{1} $且满足$ \lambda_{1}(\theta) = 1. $则有$ (k_{\theta}, k'_{\theta})\in B^{2}(L, A, \alpha) $, 因此存在一个映射$ \chi: L\rightarrow A $, 其中$ \chi(0) = 0 $, 满足等式

$ \begin{eqnarray*} &&k_{\theta}(l_{1}, l_{2}) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}), \\ &&k'_{\theta}(l_{1}, l_{2}) = \chi[l_{1}, l_{2}]+\alpha_{l_{2}}(\chi(l_{1}))-\alpha_{l_{1}}(\chi(l_{2})). \end{eqnarray*} $

根据引理3.3, 令$ \phi = \mathrm{id}\mid_L $, 得到自同构$ \gamma\in \mathrm{Aut}_{A}(E) $满足

$ \begin{eqnarray*} \gamma(\beta^{k}(t(l))+a) = \beta^{k}(t(l))+\chi(l)+\theta(a), \end{eqnarray*} $

其中$ l\in L, a\in A $.因此有,

$ \begin{eqnarray*} \gamma(\beta^{k}(t(l)))-\beta^{k}(t(l))-a = \beta^{k}(t(l))+\chi(l)+\theta(a)-\beta^{k}(t(l))-a = \chi(l)+\theta(a)-a. \end{eqnarray*} $

故$ \gamma(\beta^{k}(t(l)))-(\beta^{k}(t(l))+a)\in A $.这就证明了$ \gamma\in \mathrm{Aut}^{L}_{A}(E) $.进一步有, $ \gamma(a) = \theta(a), $其中$ a\in A $.所以$ \tau_{1}(\gamma) = \theta $.故$ \mathrm{Ker}(\lambda_{1})\subseteq \mathrm{Im}(\tau_{1}) $.因此$ \mathrm{Aut}^{L}_{A}(E))\xrightarrow{\tau_{1}} C_{1}\xrightarrow{\lambda_{1}} H^{2}(L, A, \alpha) $是正合列.

下证(3.5)由于$ 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E) $和$ \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2} $是显然的, 只需证$ \mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}\xrightarrow{\lambda_{2}} H^{2}(L, A, \alpha) $即可.令$ \gamma\in \mathrm{Aut}^{A}(E) $, 有,

$ \begin{eqnarray*} (\lambda_{2}\tau_{2})(\gamma) = \lambda_{2}(\phi) = [(k_{\phi}, k'_{\phi}]. \end{eqnarray*} $

这里$ \phi(l) = (\pi\gamma t)(l) $, $ l\in L $.由于$ \phi = \tau_{2}(\gamma) $, $ \alpha_{l} = \alpha_{\phi(l)} $, 由引理3.3的(ⅱ), (ⅳ)及$ \theta = \gamma\mid_A = id_{A}, $可知,

$ \begin{eqnarray*} &&k_{\phi}(l_{1}, l_{2}) = \mu(\phi(l_{1}), \phi(l_{1}))-\mu(l_{1}, l_{2}) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}), \\ &&k'_{\phi}(l_{1}, l_{2}) = \delta(\phi(l_{1}), \phi(l_{2})-\delta(l_{1}, l_{2}) = \chi([l_{1}, l_{2}])+\alpha_{l_{2}}(\chi(l_{1}))-\alpha_{l_{1}}(\chi(l_{2})). \end{eqnarray*} $

因为$ \gamma\in \mathrm{Aut}^{A}(E) $, 因此$ [(k_{\phi}, k'_{\phi})] = 1 $.故$ \mathrm{Im}(\tau_{2})\subseteq \mathrm{Ker}(\lambda_{2}) $.

下面证明$ \mathrm{Ker}(\lambda_{2})\subseteq \mathrm{Im}(\tau_{2}) $.设$ \phi\in \mathrm{Ker}(\lambda_{2}) $, 则存在一个映射$ \chi: L\rightarrow A $满足$ \lambda(0) = 0. $使得,

$ \begin{eqnarray*} k_{\phi}(l_{1}, l_{2}) = \chi(l_{1}, l_{2})-\chi(l_{1})-\chi(l_{2}), \; \; \; k'_{\phi}(l_{1}, l_{2}) = \chi[l_{1}, l_{2}]+\alpha_{l_{2}}(\chi(l_{1}))-\alpha_{l_{1}}(\chi(l_{2})). \end{eqnarray*} $

由引理3.3, 令$ \theta = id\mid_A $, 则$ \forall\; l\in L, \, \forall\; a\in A $, 有一个自同构$ \gamma\in \mathrm{Aut}_{A}(E) $满足

$ \begin{eqnarray*} \gamma(\beta^{k}(t(l))+a) = \beta^{k}(t(\phi(l)))+\chi(l)+\theta(a). \end{eqnarray*} $

因此有$ \gamma(a) = \theta(a) = a $, $ \gamma\in \mathrm{Aut}^{A}(E) $, 且显然有$ \tau_{2}(\gamma) = \phi $.故$ \mathrm{Ker}(\lambda_{2})\subseteq \mathrm{Im}(\tau_{2}) $.证毕.

引理3.11   令$ A\rightarrow E\xrightarrow{\pi} L $是阿贝尔扩张且$ \theta\in C_{1} $, $ \phi\in C_{2} $, 定义

$ \begin{eqnarray*} &&m_{\theta, \phi}(l_{1}, l_{2}) = k_{\theta}(l_{1}, l_{2})+k_{\phi}(l_{1}, l_{2}) = \mu(\phi(l_{1}), \phi(l_{1}))-\theta(\mu(l_{1}, l_{2})), \\ &&m'_{\theta, \phi}(l_{1}, l_{2}) = k'_{\theta}(l_{1}, l_{2})+k'_{\phi}(l_{1}, l_{2}) = \delta(\phi(l_{1}), \phi(l_{1}))-\theta(\delta(l_{1}, l_{2})). \end{eqnarray*} $

则有$ (m_{\theta, \phi}, m'_{\theta, \phi})\in Z^{2}(L, A, \alpha) $.

证   由引理定义可知, 要证$ (m_{\theta, \phi}, m'_{\theta, \phi})\in Z^{2}(L, A, \alpha) $, 只需证明以下两式成立即可:

$ \begin{eqnarray*} &&1.\, \, m'_{\theta, \phi}(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(m_{\theta, \phi}(l_{1}, l_{2})) = m_{\theta, \phi}([l_{1}, l_{3}], [l_{2}, l_{3}])+m'_{\theta, \phi}(l_{1}, l_{3})+m'_{\theta, \phi}(l_{2}, l_{3}), \\ &&2.\, \, m'_{\theta, \phi}(l_{1}, l_{2}+l_{3})+\alpha_{l_{1}}(m_{\theta, \phi}(l_{2}, l_{3})) = m_{\theta, \phi}([l_{1}, l_{2}], [l_{1}, l_{3}])+m'_{\theta, \phi}(l_{1}, l_{2})+m'_{\theta, \phi}(l_{1}, l_{3}). \end{eqnarray*} $

由

$ \begin{eqnarray*} &&\delta(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(\mu(l_{1}, l_{2})) = \mu([l_{1}, l_{3}], [l_{2}, l_{3}])+\delta(l_{1}, l_{3})+\delta(l_{2}, l_{3}), \\ &&\delta(l_{1}, l_{2}+l_{3})-\alpha_{l_{1}}(\mu(l_{2}, l_{3})) = \mu([l_{1}, l_{2}], [l_{1}, l_{3}])+\delta(l_{1}, l_{2})+\delta(l_{1}, l_{3}), \end{eqnarray*} $

及等式

$ \begin{eqnarray*} &&m_{\theta, \phi}(l_{1}, l_{2}) = k_{\theta}(l_{1}, l_{2})+k_{\phi}(l_{1}, l_{2}) = \mu(\phi(l_{1}), \phi(l_{2}))-\theta(\mu(l_{1}, l_{2})), \\ &&m'_{\theta, \phi}(l_{1}, l_{2}) = k'_{\theta}(l_{1}, l_{2})+k'_{\phi}(l_{1}, l_{2}) = \delta(\phi(l_{1}), \phi(l_{2}))-\theta(\delta(l_{1}, l_{2})), \end{eqnarray*} $

可知

$ \begin{eqnarray*} m'_{\theta, \phi}(l_{1}+l_{2}, l_{3})-\alpha_{l_{3}}(m_{\theta, \phi}(l_{1}, l_{2})) & = &\delta(\phi(l_{1}+l_{2}), \phi(l_{3}))-\theta(\delta(l_{1}+l_{2}, l_{3}))\\ &&-\alpha_{l_{3}}(\mu(\phi(l_{1}, \phi(l_{2})))-\theta(\mu(l_{1}, l_{2}))\\ & = &m_{\theta, \phi}([l_{1}, l_{3}], [l_{2}, l_{3}])+m'_{\theta, \phi}(l_{1}, l_{3})+m'_{\theta, \phi}(l_{2}, l_{3}). \end{eqnarray*} $

同理可证

$ \begin{eqnarray*} m'_{\theta, \phi}(l_{1}, l_{2}+l_{3})+\alpha_{l_{1}}(m_{\theta, \phi}(l_{2}, l_{3})) = m_{\theta, \phi}([l_{1}, l_{2}], [l_{1}, l_{3}])+m'_{\theta, \phi}(l_{1}, l_{2})+m'_{\theta, \phi}(l_{1}, l_{3}). \end{eqnarray*} $

故$ (m_{\theta, \phi}, m'_{\theta, \phi})\in Z^{2}(L, A, \alpha) $.证毕.

令$ A\rightarrow E\xrightarrow{\pi} L $是一个中心扩张, 且$ L $平凡地作用在$ A $上.由于$ A\subseteq Z(E) $, 且$ \alpha_{l}(a) = [\beta^{k}(t(l)), a] = 0 $, 故有$ {A} = C_{1} $及$ \mathrm{Aut}(L) = C_{2} $.

定理3.12   令$ A\rightarrow E\xrightarrow{\pi} L $是一个中心扩张, 则存在下列正合序列:

$ \begin{eqnarray} 1\rightarrow\mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}_{A}(E)\xrightarrow{\tau} \mathrm{Aut}(A)\times \mathrm{Aut}(L)\xrightarrow{\lambda} H^{2}(L, A, \alpha), \end{eqnarray} $

(3.6)

这里$ \lambda(\theta, \phi) = [m_{\theta, \phi}, m'_{\theta, \phi}] $, $ \tau(\gamma) = (\theta, \phi) $.

证  在$ \mathrm{Aut}^{A, L}(E) $和$ \mathrm{Aut}_{A}(E) $上序列显然是正合的.在证明此定理之前, 可以证明$ \lambda $是定义明确的.若$ (\theta, \phi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L) $是由$ \gamma\in \mathrm{Aut}_{A}(E) $所诱导的, 则通过引理3.5的(ⅱ)和(ⅲ)存在一个映射$ \chi: L\rightarrow A $满足$ \lambda(0) = 0 $使得

$ \begin{eqnarray*} m_{\theta, \phi}(l_{1}, l_{2}) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}), \; \; \; m'_{\theta, \phi}(l_{1}, l_{2}) = \chi[l_{1}, l_{2}]. \end{eqnarray*} $

故$ (m_{\theta, \phi}, m'_{\theta, \phi})\in B^{2}(L, A, \alpha) $且$ \lambda(\theta, \phi) = 1 $.

相反的, 如果$ (\theta, \phi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L) $满足$ [(m_{\theta, \phi}, m'_{\theta, \phi})] = 1 $, 则存在一个映射$ \chi: L\rightarrow A $满足$ \lambda(0) = 0 $使得

$ m_{\theta, \phi}(l_{1}, l_{2}) = \chi(l_{1}+l_{2})-\chi(l_{1})-\chi(l_{2}), \; \; \; m'_{\theta, \phi}(l_{1}, l_{2}) = \chi[l_{1}, l_{2}] $.

现由引理3.6知存在一个自同构$ \gamma\in \mathrm{Aut}_{A}(E) $使得$ \tau(\gamma) = (\theta, \phi) $.证毕.

令$ A\rightarrow E\xrightarrow{\pi} L $是阿贝尔李扩张, 记$ C_{1}^{\ast} = \{\theta\in C_{1}\mid \lambda_{1}(\theta) = 1\} $和$ C_{2}^{\ast} = \{\phi\in C_{2}\mid \lambda_{2}(\phi) = 1\} $.由定理3.10得到下列两个正合序列:

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}^{L}_{A}(E)\xrightarrow{\tau_{1}} C_{1}^{\ast}\rightarrow 1, \end{eqnarray} $

(3.7)

$ \begin{eqnarray} &&1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota}\mathrm{Aut}^{A}(E)\xrightarrow{\tau_{2}} C_{2}^{\ast}\rightarrow 1. \end{eqnarray} $

(3.8)

接下来, 将证明如果原正合序列可裂, 则这两个正和序列也是可裂的.

定理3.13   令$ E $是一个hom-李环, 且$ A $是$ E $的一个阿贝尔理想, 如果序列$ A\rightarrow E\rightarrow L $分裂, 则序列(3.7), (3.8)也可裂.

证  在$ A\rightarrow E\xrightarrow{\pi} L $中存在一个hom-李环同态$ t:\, L\rightarrow E $满足$ \pi\circ t = id_{L} $, 而且对$ E $中的每个元素都可以唯一的用$ \beta^{k}(t(l))+a $表示出来, 其中$ l\in L; a\in A $.首先证明(3.7)可裂.

定义一个映射$ \psi_{1}: C_{1}^{\ast}\rightarrow \mathrm{Aut}^{L}_{A}(E) $使得其满足: $ \psi_{1}(\theta) = \gamma_{1} $, 这里$ \gamma_{1}: E\rightarrow E $, 满足运算

$ \begin{eqnarray*} \gamma_{1}(\beta^{k}(t(l))+a) = \beta^{k}(t(l))+\theta(a). \end{eqnarray*} $

对任意的$ x_{1} = \beta^{k}(t(l_{1}))+a_{1} $, $ x_{2} = \beta^{k}(t(l_{2}))+a_{2} $, 由于$ t $是同构映射, 则有

$ \begin{eqnarray*} \gamma_{1}(x_{1}+x_{2}) = \gamma_{1}(x_{1})+\gamma_{1}(x_{2}). \end{eqnarray*} $

又由于$ \theta\in C_{1} $, 有,

$ \begin{eqnarray*} \gamma_{1}[x_{1}, x_{2}] & = &\gamma_{1}[\beta^{k}(t(l_{1}))+a_{1}, \beta^{k}(t(l_{2}))+a_{2}]\\ & = &\gamma_{1}[\beta^{k}(t(l_{1})), \beta^{k}(t(l_{2}))]+\gamma_{1}[\beta^{k}(t(l_{1})), a_{2}]+\gamma_{1}[a_{1}, \beta^{k}(t(l_{2}))]+\gamma_{1}[a_{1}, a_{2}]\\ & = &[\gamma_{1}(x_{1}), \gamma_{1}(x_{2})]. \end{eqnarray*} $

又由于$ \gamma_{1}\beta = \beta\gamma_{1} $, 这就证明了$ \gamma_{1} $是hom-李环同态.显然可知: $ \gamma_{1}\in\mathrm{Aut}^{L}_{A}(E) $, $ \gamma_{1} $是群同态且$ \tau_{1}\circ \psi_{1} = id_{C_{1}^{\ast}} $.所以序列(3.7)可裂.

现在证明(3.8)可裂.定义一个映射$ \psi_{2}: C_{2}\ast\rightarrow \mathrm{Aut}^{A}(E) $使得其满足$ \psi_{2}(\phi) = \gamma_{2} $, 这里$ \gamma_{2}: E\rightarrow E $, 满足运算

$ \begin{eqnarray*} \gamma_{2}(\beta^{k}(t(l))+a) = \beta^{k}(t(\phi(l)))+a. \end{eqnarray*} $

对任意的$ x_{1} = \beta^{k}(t(l_{1}))+a_{1} $, $ x_{2} = \beta^{k}(t(l_{2}))+a_{2}\in E $, 由于$ t $是同构映射, 有,

$ \begin{eqnarray*} \gamma_{1}(x_{1}+x_{2}) = \gamma_{1}(x_{1})+\gamma_{1}(x_{2}). \end{eqnarray*} $

又由于$ \phi\in C_{2} $, 有$ \alpha_{l} = \alpha_{\phi(l)}, \; \forall\; l\in L $.故

$ \begin{eqnarray*} \gamma_{2}[x_{1}, x_{2}] & = &\gamma_{2}[\beta^{k}(t(l_{1}))+a_{1}, \beta^{k}(t(l_{2}))+a_{2}] = \beta^{k}(t(\phi[l_{1}, l_{2}]))+\alpha_{l_{1}}(a_{2})-\alpha_{l_{2}}(a_{1})\\ & = &\beta^{k}(t(\phi[l_{1}, l_{2}]))+\alpha_{\phi(l_{1})}(a_{2})-\alpha_{\phi(l_{2})}(a_{1}) = [\gamma_{2}(x_{1}), \gamma_{2}(x_{2})]. \end{eqnarray*} $

又由于$ \gamma_{2}\beta = \beta\gamma_{2} $, 这就证明了$ \gamma_{2} $是hom-李环的自同构, 显然可知: $ \gamma_{2}\in \mathrm{Aut}^{A}(E), $ $ \gamma_{2} $是群自同构且$ \tau_{2}\circ \psi_{2} = id_{C_{2}^{\ast}} $.所以序列(3.8)可裂.证毕.

令$ A\rightarrow E\xrightarrow{\pi} L $是中心李扩张, 记$ C^{\ast} = \{(\theta, \phi)\in \mathrm{Aut}(A)\times \mathrm{Aut}(L)\mid \lambda(\theta, \phi) = 1\} $.由定理3.12得到下列正合序列

$ \begin{eqnarray} 1\rightarrow \mathrm{Aut}^{A, L}(E)\xrightarrow{\iota} \mathrm{Aut}_{A}(E)\xrightarrow{\tau} C^{\ast}\rightarrow 1. \end{eqnarray} $

(3.9)

接下来, 将证明如果原始的正合序列可裂, 则这个正和序列也是可裂的.

定理3.14   令$ E $是一个hom-李环, 且$ A $是$ E $的一个阿贝尔理想, 进一步, 若$ A\subseteq Z(E) $, 如果序列$ A\rightarrow E\rightarrow L $分裂, 则序列(3.9)也可裂.

证   在$ A\rightarrow E\xrightarrow{\pi} L $中存在一个hom-李环同态$ t: L\rightarrow E $, 满足$ \pi\circ t = id_{l} $, 而且对$ E $中的每个元素都可以唯一的用$ \beta^{k}(t(l))+a $表示出来, 其中$ l\in L, \, a\in A $.现在证明(3.9)可裂.

由$ \psi(\theta, \phi) = \gamma $定义一个映射$ \psi: C^{\ast}\rightarrow \mathrm{Aut}_{A}(E) $, 这里$ \gamma: E\rightarrow E $, 满足运算

$ \begin{eqnarray*} \gamma(\beta^{k}(t(l))+a) = \beta^{k}(t(\phi(l)))+\theta(a). \end{eqnarray*} $

对任意的$ x_{1} = \beta^{k}(t(l_{1}))+a_{1} $, $ x_{2} = \beta^{k}(t(l_{2}))+a_{2}\in E $, 有$ \gamma(x_{1}+x_{2}) = \gamma(x_{1})+\gamma(x_{2}). $由于$ t $是自同态映射, 且$ A\subseteq Z(E) $.可知,

$ \begin{eqnarray*} \gamma[x_{1}, x_{2}] = \gamma[\beta^{k}(t(l_{1}))+a_{1}, \beta^{k}(t(l_{2}))+a_{2}] = \gamma(\beta^{k}(t[l_{1}, l_{2}])) = t(\phi[l_{1}, l_{2}]) = [\gamma(x_{1}), \gamma(x_{2})]. \end{eqnarray*} $

又由于$ \gamma\beta = \beta\gamma $, 故$ \gamma $是自同态.不难得知$ \gamma\in \mathrm{Aut}_{A}(E) $, $ \psi $是群同态, 且$ \tau\circ \psi = id_{C^{\ast}} $.所以序列(3.9)可裂.证毕.

本节, 在定理3.10 (3.4)的基础上, 将$ Z(E) $的自同构推广到$ E $的自同构.以下引理显然成立.

引理4.1   令$ E $是一个hom-李环.

(1) 若$ E $是阿贝尔hom-李环, 则$ E $的任何加法子群也都是$ E $的子环.

(2) 若$ E $是不含有非平凡子环的平凡hom-李环, 则$ E $是素数阶循环的.

引理4.2  令$ M $是hom-李环$ E $的子环, 取$ \aleph_{E}(M) = \{x\in E\mid [x, M]\subseteq M\} $, 则以下结论成立.

(ⅰ) $ \aleph_{E}(M) $是hom-李环$ E $的子环, 且$ M\subseteq \aleph_{E}(M) $.

(ⅱ) $ M $是$ E $的一个理想当且仅当$ \aleph_{E}(M) = E $.

(ⅲ) $ [K, M]\subseteq M $当且仅当$ K\subseteq \aleph_{E}(M) $, 这里$ K $是$ E $的子环.

(ⅳ)若$ E $是幂零的且$ M\subset E $, 则$ M\subset \aleph_{E}(M) $.

(ⅴ)若$ E $是幂零的, 且$ M $是$ E $的最大子环, 则$ M $是$ E $理想且$ E/M $是素数阶.

引理4.3   令$ E $是hom-李环, 以下结论成立.

(ⅰ)若$ E $是有限维的, 则$ \Phi(E) $恰好由$ E $的所有非生成元组成.

(ⅱ) $ E/I $是阿贝尔的当且仅当$ E'\subseteq I $.

(ⅲ)若$ E $是幂零的, 则$ E'\subseteq \Phi(E) $.

(ⅳ)若$ E $是幂零的, 则$ \Phi(E) $是$ E $的理想.

(ⅴ)若$ E $是幂零的且非平凡, 则$ Z(E)\neq 0 $.

定理4.4   令$ E $是一个阶为$ p^{n}(p\geq2) $的hom-李P环, $ Z(E) $是循环的且$ \Phi(E)\subseteq Z(E) $.则每一个$ Z(E) $的p自同构都可以被扩张成依赖于$ \mathrm{Aut}^{L}_{A}(E) $的$ E $上的自同构, 这里, $ A = Z(E) $, $ L = E/Z(E) $.

证   令$ A = \langle x\rangle $, $ |A| = p^{\omega} $, 且$ \pi: E\rightarrow L $是自然同态.所以得到一个中心扩张$ A\rightarrow E\rightarrow L $.容易看出每一个$ A $上的每一个循环群都保留着李乘.现在由第二章的注记得到$ C_{1} = \mathrm{Aut}(A)\cong Z_{P^{\omega-1}(P-1)}. $注意到$ \theta = \gamma\mid_ A $.因为$ \gamma\in \mathrm{Aut}^{L}_{A}(E) $, 有$ \phi(l) = (\pi\gamma t)(l) = l. $所以扩张是中心扩张.定义$ \theta: A\rightarrow A $满足$ \theta(x) = mx, $这里$ m = 1+jp $且$ 1\leq m<p^{\omega} $.显然$ \theta $是$ A $上的一个$ p $ -自同构, 而且有$ p^{\omega-1} $这样的$ p $ -自同构.对$ \forall\; l_{1}, l_{2}\in L $, 有

$ \begin{eqnarray*} k_{\theta}(l_{1}, l_{2}) & = &\mu(l_{1}, l_{2})-\theta(\mu(l_{1}, l_{2}))\\ & = &(1-m)(\beta^{k}(t(l_{1}))+\beta^{k}(t(l_{2}))-\beta^{k}(t(l_{1}+l_{2}))). \end{eqnarray*} $

由$ pt(l)\in pE\subseteq \Phi(E)\subseteq A $且$ p $整除$ 1-m $, 可以定义映射$ \chi: L\rightarrow A $使得其满足$ \chi(l) = (1-m)t(l). $因此,

$ \begin{eqnarray} k_{\theta}(l_{1}, l_{2}) = \chi(l_{1})+\chi(l_{2})-\chi(l_{1}+l_{2}). \end{eqnarray} $

(4.1)

同样的, 对$ \forall\; a, b\in E $, 有$ p[a, b] = 0 $.故$ pE\subseteq \Phi(E)\subseteq A $.因此,

$ \begin{eqnarray*} k'_{\theta}(l_{1}, l_{2}) = \delta(l_{1}, l_{2})-\theta(\delta(l_{1}, l_{2})) = (1-m)(\beta^{k}([t(l_{1}), t(l_{2})])-\beta^{k}(t[l_{1}, l_{2}])) = -\chi[l_{1}, l_{2}]. \end{eqnarray*} $

现在由(4.1)及上式可以推出, $ (k_{\theta}, k'_{\theta})\in B^{2}(L, A, \alpha) $.因此由定理3.10的(3.4), 得到$ \lambda_{1}(\theta) = 1 $, 且$ \theta\in \mathrm{Ker}(\lambda_{1}) = \mathrm{Im}(\tau_{1}) $.所以, $ p^{\omega-1}\subseteq|\mathrm{Im}(\tau_{1})| $.故每一个$ A $上的$ p $ -自同构都属于$ \mathrm{Im}(\tau_{1}) $.证毕.