李超代数在数学、物理等多个领域都有着广泛的应用. 但至今模李超代数(素特征域上的李超代数)的分类工作还没有完成, 其中Cartan型模李超代数成为分类问题的关键, 也取得了很多丰硕的成果^[1-8]. 我们知道, 上同调对于李超代数扩张以及其模扩张的研究具有重大作用, 如^[9-11]. 其中特殊线性李超代数到Cartan型模李超代数W、S和H的低维上同调已经被计算^[12-14]. 因奇Hamilton超代数$ {\mathrm{HO}} $的零阶化分支中包含典型李超代数$ \tilde{P}(2) $, 故在伴随表示意义下, $ {\mathrm{HO}} $可视为$ \tilde{P}(2) $-模. 本文通过对$ {\mathrm{HO}} $进行适当的$ \tilde{P}(2) $-子模分解和相应的权空间计算, 并利用权导子(保持权不变的导子), 计算了$ \tilde{P}(2) $到$ {\mathrm{HO}} $的零维和一维上同调.

令$ \mathbb{N} $与$ \mathbb{N}_{0} $分别表示正整数集与非负整数集, $ \mathbb{F} $表示特征$ p>3 $的域, $ \mathbb{Z}_{2}:=\{\bar{0}, \bar{1}\} $表示整数模2的剩余类环, $ |x| $表示$ \mathbb{Z}_{2} $-阶化向量空间中齐次元素$ x $的$ \mathbb{Z}_{2} $-次数, $ {\rm{zd}}(x) $表示$ \mathbb{Z} $-阶化向量空间中齐次元素$ x $的$ \mathbb{Z} $-次数. 记$ \langle $$ x_{1}, x_{2}, \cdots, x_{k} $$ \rangle $为$ x_{1}, x_{2}, \cdots, x_{k} $在数域$ \mathbb{F} $上线性生成的向量空间. 设$ {\alpha=(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{m})\in\mathbb{N}^{m}_{0}} $, $ {\beta=(\beta_{1}, \beta_{2}, \cdots, \beta_{m})\in\mathbb{N}^{m}_{0}} $, 定义$ \alpha\choose \beta $: =$ \prod^{m}_{i=1}{\alpha_{i}\choose \beta_{i}} $. 令$ \mathcal{U}(x) $是具有生成元集$ \{x^{(\alpha)}=x^{\alpha_{1}}_{1}x^{\alpha_{2}}_{2}\cdots x^{\alpha_{m}}_{m}|\alpha\in\mathbb{N}^{m}_{0}\} $的$ \mathbb{F} $上的除幂代数, 乘法运算为$ {x^{(\alpha)}x^{(\beta)}} $=$ {\alpha+\beta}\choose \beta $$ x^{(\alpha+\beta)}, \; \alpha, \beta\in\mathbb{N}^{m}_{0}. $ $ \Lambda(n) $是由变元$ x_{m+1}, x_{m+2}, \cdots, x_{m+n} $生成$ \mathbb{F} $上的外代数, 基元素用$ x^{u} $表示. 令

$\Lambda(m, n)=\mathcal{U}(m)\otimes\Lambda(n)=\langle{x^{(\alpha)}x^{u}}|\alpha\in\mathbb{N}^{m}_{0}\rangle, $

其中$ x^{(\alpha)}\otimes x^{u} $简记为$ x^{(\alpha)}x^{u} $, 称为除外代数. $ \mathcal{U}(m) $的平凡的$ \mathbb{Z}_{2} $-阶化和$ \Lambda(n) $的自然的$ \mathbb{Z}_{2} $-阶化诱导了$ \Lambda(m, n) $的一个$ \mathbb{Z}_{2} $-阶化:

$\Lambda(m, n)_{\bar{0}}=\mathcal{U}(m)\otimes\Lambda(n)_{\bar{0}}, \; \Lambda(m, n)_{\bar{1}}=\mathcal{U}(m)\otimes\Lambda(n)_{\bar{1}}, $

从而$ \Lambda(m, n) $是一个结合超代数. 令$ {\rm{zd}}(x_{i})=1, \; i=1, 2, \cdots, m+n $, 则$ \Lambda(m, n) $有一个自然的$ \mathbb{Z} $-阶化:

$\Lambda(m, n)=\bigoplus^{m(p-1)+n}_{i=0}\Lambda(m, n)_{i}$.

令$ {{\mathrm{Y}}_{0}}=\{1, 2, \cdots, m\} $, $ {{\mathrm{Y}}_{1}}=\{m+1, \cdots, m+n\} $, $ {\mathrm{Y}}={\mathrm{Y}}_{0}\bigcup{\mathrm{Y}}_{1} $. 设$ {\mathrm{D}}_{1}, {\mathrm{D}}_{2}, \cdots, {\mathrm{D}}_{s} $是$ \Lambda(m, n) $的线性变换, 并满足

$ {\mathrm{D}}_{i}(x^{(\alpha)}x^{u})=\left\{ \begin{aligned} x^{(\alpha-\varepsilon_{i})}x^{u}, \; \; \; &\forall i\in{\mathrm{Y}}_{0}, \\ x^{(\alpha)}\partial_{i}(x^{u}), \; \; &\forall i\in{\mathrm{Y}}_{1}.\\ \end{aligned} \right. $

其中$ \varepsilon_{i}=(\delta_{i1}, \delta_{i2}, \cdots, \delta_{im}), i\in{\mathrm{Y}}_{0} $; $ \partial_{i} $是$ \Lambda(n) $的特殊导子, $ \forall i\in{\mathrm{Y}}_{1} $. 则$ {\mathrm{D}}_{i}\in{\mathrm{Der}}_{\bar{0}}(\Lambda(m, n)) $, $ \forall i\in{\mathrm{Y}}_{0} $; $ {\mathrm{D}}_{i}\in{\mathrm{Der}}_{\bar{1}}(\Lambda(m, n)) $, $ \forall i\in{\mathrm{Y}}_{1} $.

令$ {\mathrm{W}}(m, n)=\{\sum^{m+n}_{i=1}f_{i}{\mathrm{D}}_{i}|\; f_{i}\in\Lambda(m, n), \; \forall i\in{\mathrm{Y}}\} $, 则$ {\mathrm{W}}(m, n) $是$ {\mathrm{Der}}(\Lambda(m, n)) $的子代数, 称为$ {\textbf{Witt}} $超代数  .

设$ {\mathrm{T_{H}}}:\Lambda(n, n)\rightarrow {\mathrm{W}}(n, n) $是线性映射, 使得$ {\mathrm{T_{H}}}(f)=\sum^{2n}_{i=1}(-1)^{\tau(i)|f|}\partial_{i}(f)\partial_{i'}, $这里

$ \\i'=\left\{ \begin{aligned} i+n, \; \; &i=1, 2, \cdots, n, \\ i-n, \; \; &i=n+1, \cdots, 2n.\\ \end{aligned} \right. $

$ \\\tau(i)=\left\{ \begin{aligned} \bar{0}, \; \; &i=1, 2, \cdots, n, \\ \bar{1}, \; \; &i=n+1, \cdots, 2n.\\ \end{aligned} \right. $

易见$ {\mathrm{zd(T_{H}})}=-2, \; |{\mathrm{T_{H}}}|=\bar{1} $. 令$ {\mathrm{HO}}(n, n):=\{{{\mathrm{T_{H}}}(f)|\; f\in\Lambda(n, n)}\}, $它是$ {\mathrm{W}}(n, n) $的子代数, 称之为$ {\textbf{奇Hamilton}} $超代数  . 它的$ \mathbb{Z}_{2} $-阶化结构为:

${\mathrm{HO}}(n, n)_{\bar{0}}=\langle{\mathrm{T_{H}}}(f)|\; f\in\Lambda(n, n)_{\bar{1}}\rangle, \; \; {\mathrm{HO}}(n, n)_{\bar{1}}=\langle{\mathrm{T_{H}}}(f)|\; f\in\Lambda(n, n)_{\bar{0}}\rangle.$

它的$ \mathbb{Z} $-阶化结构为: $ {\mathrm{HO}}(n, n)_{i}={\mathrm{T_{H}}}(\Lambda(n, n)_{i+2}) $.

由文献^[7]可知, $ \forall f, g\in\Lambda(n, n), \; [{\mathrm{T_{H}}}(f), {\mathrm{T_{H}}}(g)]={\mathrm{T_{H}}}({\mathrm{T_{H}}}(f)(g))={\mathrm{T_{H}}}(h) $, 其中$ h={\mathrm{T_{H}}}(f)(g) $, 为此可在$ \Lambda(n, n) $中定义一个方括号运算$ [f, g]:={\mathrm{T_{H}}}(f)(g), \; f, g\in\Lambda(n, n), $使得$ \Lambda(n, n) $作成李超代数. 因为$ {\mathrm{kerT_{H}}}=\mathbb{F} $, 所以

$\bar{\Lambda}(n, n):=\Lambda(n, n)/\mathbb{F}\simeq{\mathrm{HO}}(n, n)$.

下面简要回顾李超代数$ \tilde{P}(2) $. 在一般线性李超代数$ {\mathrm{gl}}(m, n) $中,

$ \\ \tilde{P}(m):=\left\{ \begin{gathered} \begin{pmatrix}{\mathrm{A}}&{\mathrm{B}}\\{\mathrm{C}}&{\mathrm{-A}}^{T}\end{pmatrix} \end{gathered} \in {\mathrm{gl}}(m, m)|\; {\mathrm{B}}={\mathrm{B}}^{T}, \; {\mathrm{C}}={\mathrm{-C}}^{T}\right\}. $

易见$ \tilde{P}(2) $的一组基为

$E_{33}-E_{11}, \; E_{44}-E_{22}, \; E_{13}, \; E_{24}, \; E_{23}+E_{14}, \; E_{43}-E_{12}, \; E_{34}-E_{21}, \; E_{41}-E_{32}.$

我们可以定义$ \tilde{P}(2) $到$ {\mathrm{HO}}(2, 2)_{0} $的线性映射$ \varphi $:

$ \begin{eqnarray*} &&E_{23}+E_{14}\rightarrow {\mathrm{T_{H}}}(x_{1}x_{2});\; \; E_{41}-E_{23}\rightarrow {\mathrm{T_{H}}}(x_{3}x_{4});\\ &&E_{ij}\rightarrow {\mathrm{T_{H}}}(x_{i}x_{i}), \; i=1, 2, \; j=i+2;\\ &&E_{ij}-E_{kl}\rightarrow {\mathrm{T_{H}}}(x_{k}x_{i}), \; i, j=3, 4, \; k=j-2, \; l=i-2. \end{eqnarray*} $

显然, $ \varphi $是$ \tilde{P}(2) $到$ {\mathrm{HO}}(2, 2)_{0} $的同构映射, 故$ \tilde{P}(2) $$ \simeq{\mathrm{HO}}(2, 2)_{0} $. 又由于$ {\mathrm{HO}}(2, 2)_{0}\simeq\Lambda(2, 2)_{2}, $其中$ \Lambda(2, 2)_{2} $的一组基为

$\jmath=\{x_{1}x_{1}, x_{1}x_{2}, x_{1}x_{3}, x_{1}x_{4}, x_{2}x_{2}, x_{2}x_{3}, x_{2}x_{4}, x_{3}x_{4}\}.$

故计算$ \tilde{P}(2) $到$ {\mathrm{HO}}(n, n) $的低维上同调可转化为计算$ \Lambda(2, 2)_{2} $到$ \bar{\Lambda}(n, n) $的低维上同调. 下文简记同构的李超代数$ \tilde{P}(2) $, $ {\mathrm{HO}}(2, 2)_{0} $或$ \Lambda(2, 2)_{2} $为$ {\mathrm{g}} $, $ {\mathrm{HO}}(n, n) $和$ \bar{\Lambda}(n, n) $ $ (n\geq2) $分别为$ {\mathrm{HO}} $和$ \bar{\Lambda} $.

若$ M $为李超代数$ {\mathrm{g}} $-模, 记$ kM:=M\oplus\cdots\oplus M $ ($ k $-个), 当$ T $为平凡$ {\mathrm{g}} $-模, 且$ {\mathrm{dim}}T=k $, 那么$ T\otimes M\simeq kM $.

由上同调的性质, 我们计算$ {\mathrm{g}} $到$ {\mathrm{HO}} $的上同调可以计算$ {\mathrm{g}} $到$ {\mathrm{HO}} $子模的上同调. 为此, 我们将$ {\mathrm{g}} $-模$ {\mathrm{HO}} $进行适当的分解. 根据前文的内容, 我们只需对$ \bar{\Lambda}(2, 2)_{2} $-模$ \bar{\Lambda} $进行子模分解.

设$ \gamma=\{1, 2, n+1, n+2\}, \; S={\mathrm{Y}}\backslash\gamma $. 则$ \bar{\Lambda} $首先有下面的子模分解:

$ \bar{\Lambda}=M\oplus T.$

其中,

$ M\simeq\langle x^{(\alpha)}x^{u}|\; \partial_{i}(x^{(\alpha)}x^{u})\neq0, \; \exists i\in\gamma\rangle, $

$ T\simeq\langle x^{(\alpha)}x^{u}|\; \partial_{i}(x^{(\alpha)}x^{u})=0, \; \forall i\in\gamma\rangle/\mathbb{F}.$

作为$ \bar{\Lambda} $的$ {\mathrm{g}} $-子模, 易见$ M\simeq {M_{1}\otimes T} $, 其中

$ M_{1}\simeq\langle x^{\alpha_{1}}_{1}x^{\alpha_{2}}_{2}x^{\beta_{1}}_{n+1}x^{\beta_{2}}_{n+2}|\; 0\leq\alpha_{1}, \alpha_{2}\leq p-1, \beta_{1}, \beta_{2}=0\; {\mbox{或}}\; 1\rangle/\mathbb{F}\simeq\bar{\Lambda}(2, 2).$

这样, 作为$ {\mathrm{g}} $-模有$ \bar{\Lambda}\simeq\bar{\Lambda}(2, 2)\otimes T\oplus T $, 易见$ T $为平凡$ {\mathrm{g}} $-模, 且

$ {\mathrm{dim}}T=2^{n-2}p^{n-2}-1.$

综上, 作为$ {\mathrm{g}} $-模有

$ \begin{align} \bar{\Lambda}(n, n)\simeq{\mathrm{HO}}(n, n)\simeq(2^{n-2}p^{n-2}-1)\bar{\Lambda}(2, 2)\oplus T. \end{align} $

(1)

这样, 计算$ {\mathrm{g}} $到$ {\mathrm{HO}} $的低维上同调只需计算$ {\mathrm{g}} $到$ {\mathrm{g}} $-子模$ \bar{\Lambda}(2, 2) $与$ T $的上同调.

下面给出本文计算一维上同调所需的概念和引理. 设$ L $为李超代数, $ M $为$ L $-模. 一个$ L $到$ M $的$ \mathbb{Z}_{2} $-齐次线性映射$ \varphi $叫做导子, 如果满足:

$ \begin{equation} \varphi([x, y])=(-1)^{|\varphi||x|}x\cdot\varphi(y)-(-1)^{|y|(|\varphi|+|x|)}y\cdot\varphi(x), \; \forall x, y\in L. \end{equation} $

(2)

若存在$ m\in M $使得

$ \begin{align} \varphi(x)=(-1)^{|x||m|}x\cdot m, \; \forall x\in L, \end{align} $

则$ \varphi $称为内导子, 否则称为外导子. 记$ {\mathrm{Der}}(L, M) $和$ {\mathrm{Ider}}(L, M) $分别为$ L $到模$ M $的导子空间和内导子空间. 李超代数$ L $到模$ M $的零维上同调与一维上同调分别为

$ \begin{align} H ^{0}(L, M):=\{{m\in M|L\cdot m=0}\}, \end{align} $

$H^{1}(L, M):={\mathrm{Der}}(L, M)/{\mathrm{Ider}}(L, M).$

定义3.1  设$ H $是$ L $的Cartan子代数, $ L $与$ L $-模$ M $关于$ H $的权空间分解为$ L=\oplus_{\alpha\in H^{\ast}}L_{\alpha} $和$ M=\oplus_{\alpha\in H^{\ast}}M_{\alpha} $. 一个$ L $到$ M $的导子$ \varphi $称为关于$ H $的权导子, 若$ \varphi(L_{\alpha})\subseteq M_{\alpha}, \; \forall\alpha\in H^{\ast} $.

引理3.2 ^{[15, 16]}  任何一个李超代数$ L $到$ L $-模$ M $的导子都是一个权导子与内导子之和.

由引理3.2和(1) 式, 计算$H$ ^{1}({\mathrm{g}}, {\mathrm{HO}}) $只需分别计算$ {\mathrm{g}} $到$ T $和$ \bar{\Lambda}(2, 2) $的权导子, 为此我们首先需要确定$ {\mathrm{g}} $的一个Cartan子代数$ h=\langle x_{1}x_{3}, x_{2}x_{4}\rangle $, 相应的权分别为$ (0, 0), \; (-1, 1) $, $ (-1, -1), \; (1, -1), \; (1, 1), \; (-2, 0), \; (0, -2) $.

根据定义3.1和引理3.2, 计算$H$ ^{1}({\mathrm{g}}, {\mathrm{HO}}) $只需考虑$ \bar{\Lambda}(2, 2) $中与$ {\mathrm{g}} $有相同权的权向量. 由以下公式

$ \begin{align} [x_{1}x_{3}, x^{k_{1}}_{1}x^{k_{2}}_{2}x^{\beta_{3}}_{3}x^{\beta_{4}}_{4}] &=(\beta_{3}-k_{1})x^{k_{1}}_{1}x^{k_{2}}_{2}x^{\beta_{3}}_{3}x^{\beta_{4}}_{4}, \end{align} $

$ \begin{align} [x_{2}x_{4}, x^{k_{1}}_{1}x^{k_{2}}_{2}x^{\beta_{3}}_{3}x^{\beta_{4}}_{4}] &=(\beta_{4}-k_{2})x^{k_{1}}_{1}x^{k_{2}}_{2}x^{\beta_{3}}_{3}x^{\beta_{4}}_{4}. \end{align} $

可做如下图表:

引理3.3   李超代数$ {\mathrm{g}} $到$ \bar{\Lambda}(2, 2) $的非零权导子都是外导子.

证  设$ \varphi $是李超代数$ {\mathrm{g}} $到$ \bar{\Lambda}(2, 2) $的权导子, 根据表 1可见, $ {\mathrm{zd}}(\varphi)=0, 2, 4, p-2, p, 2p-4 $. 下面我们按$ \varphi $的$ \mathbb{Z} $-次数分情况进行讨论. 约定在下文中, 在$ \varphi $的映射没有出现的$ \jmath $中元素的像为0.

情况1   $ {\mathrm{zd}}(\varphi)=0 $时. 根据表 1可设

$ \begin{align} \varphi(x_{1}x_{3})&=a_{13}x_{1}x_{3}+b_{13}x_{2}x_{4}, \; \; \; \; \; \; \; \; \varphi(x_{2}x_{4})=a_{24}x_{1}x_{3}+b_{24}x_{2}x_{4}, \end{align} $

$ \begin{align} \varphi(x_{i}x_{j})&=a_{ij}x_{i}x_{j}, \; \; x_{i}x_{j}\in{\jmath\backslash\{x_{1}x_{3}, \; x_{2}x_{4}\}}, \; \; \; \; \; \; \; a_{ij}, b_{ij}\in\mathbb{F}. \end{align} $

由(2) 式可知

$ \begin{align} -a_{14}x_{1}x_{4}&=\varphi(-x_{1}x_{4})=\varphi([x_{1}x_{3}, x_{1}x_{4}]) \end{align} $

$ \begin{align} &=(x_{3}\partial_{3}-x_{1}\partial_{1})a_{14}x_{1}x_{4}+(x_{4}\partial_{3}-x_{1}\partial_{2})(a_{13}x_{1}x_{3}+b_{13}x_{2}x_{4}) \end{align} $

$ \begin{align} &=(a_{13}-b_{13}-a_{14})x_{1}x_{4}, \end{align} $

比较系数可得$ a_{13}-b_{13}=0 $. 又由于

$ \begin{align} a_{34}x_{3}x_{4}&=\varphi(x_{3}x_{4})=\varphi([x_{1}x_{3}, x_{3}x_{4}]) \end{align} $

$ \begin{align} &=(x_{3}\partial_{3}-x_{1}\partial_{1})a_{34}x_{3}x_{4}-(x_{4}\partial_{1}-x_{3}\partial_{2})(a_{13}x_{1}x_{3}+b_{13}x_{2}x_{4}) \end{align} $

$ \begin{align} &=(a_{34}+a_{14}+b_{14})x_{3}x_{4}, \end{align} $

比较系数可得$ a_{13}+b_{13}=0 $. 那么

$ \begin{equation} a_{13}=b_{13}=0, \; \; {\mbox{即}}\; \varphi(x_{1}x_{3})=0. \end{equation} $

(3)

同理可知

$ \begin{equation} a_{24}=b_{24}=0, \; \; {\mbox{即}}\; \varphi(x_{2}x_{4})=0. \end{equation} $

(4)

再由(2) 得以下两式

$ \begin{align} -a_{11}x_{1}x_{1}=\varphi(-x_{1}x_{1})=\varphi([x_{1}x_{4}, x_{1}x_{2}])=-(a_{11}+a_{14})x_{1}x_{1}, \end{align} $

与

$ \begin{align} 0=\varphi(x_{2}x_{4}-x_{1}x_{3})=\varphi([x_{1}x_{4}, x_{2}x_{3}])=(a_{23}-a_{14})x_{2}x_{4}+(a_{14}-a_{23})x_{1}x_{3}, \end{align} $

可得

$ \begin{equation} a_{23}=a_{14}=0, \; \; {\mbox{即}}\; \varphi(x_{2}x_{3})=\varphi(x_{1}x_{4})=0. \end{equation} $

(5)

最后, 根据以下三个等式

$ \begin{eqnarray*} &&-2a_{12}x_{1}x_{2}=\varphi(x_{1}x_{2})=\varphi([x_{1}x_{4}, x_{2}x_{2}])=(-a_{22})x_{1}x_{2}, \\ &&0=\varphi(x_{2}x_{4}-x_{1}x_{3})=\varphi([x_{1}x_{2}, x_{3}x_{4}])=(a_{34}-a_{12})x_{2}x_{4}+(a_{12}-a_{34})x_{1}x_{3}, \\ &&0=\varphi(2x_{1}x_{4})=\varphi([x_{3}x_{4}, x_{1}x_{1}])=(a_{11}-2a_{34})x_{1}x_{4}, \end{eqnarray*} $

可得

$ \begin{equation} a_{11}=a_{22}=2a_{12}=2a_{34}. \end{equation} $

(6)

综合(3)-(6) 式, 可知$ \varphi=a\varphi_{1}, \; a\in\mathbb{F} $, 其中$ \varphi_{1} $为$ {\text {g}} $到$ \bar{\Lambda}(2, 2) $的线性映射, 使得

$ \begin{equation} \varphi_{1}:\; x_{1}x_{2}\rightarrow x_{1}x_{2}, \; x_{3}x_{4}\rightarrow x_{3}x_{4}, \; x_{1}x_{1}\rightarrow 2x_{1}x_{1}, \; x_{2}x_{2}\rightarrow 2x_{2}x_{2}. \end{equation} $

(7)

由(2) 验证可知, $ \varphi_{1} $是$ {\text {g}} $到$ \bar{\Lambda}(2, 2) $的权导子.

情况2   $ {\mathrm{zd}}(\varphi)=2, 4, p, p-2 $时. 可根据表 1设出不同次数的权导子, 如$ {\mathrm{zd}}(\varphi)=2 $时, 可设

$ \begin{align} \varphi:\; & x_{1}x_{3}\rightarrow a_{13}x_{1}x_{2}x_{3}x_{4}, \; \; \; \; \; x_{2}x_{4}\rightarrow a_{24}x_{1}x_{2}x_{3}x_{4}, \end{align} $

$ \begin{align} & x_{1}x_{4}\rightarrow a_{14}x^{2}_{1}x_{3}x_{4}, \; \; \; \; \; x_{2}x_{3}\rightarrow a_{23}x^{2}_{1}x_{3}x_{4}, \end{align} $

$ \begin{align} & x_{1}x_{2}\rightarrow a_{12}x^{2}_{1}x_{2}x_{3}+b_{12}x_{1}x^{2}_{2}x_{4}, \end{align} $

$ \begin{align} & x_{2}x_{2}\rightarrow a_{22}x_{1}x^{2}_{2}x_{3}+b_{22}x^{3}_{2}x_{4}, \end{align} $

$ \begin{align} & x_{1}x_{1}\rightarrow a_{11}x^{3}_{1}x_{3}+b_{11}x^{2}_{1}x_{2}x_{4}, \; \; \; \; \; \; \; a_{ij}, b_{ij}\in\mathbb{F}. \end{align} $

再根据(2), 与情况1同样的方法, 计算得系数全为0, 即$ \varphi=0 $.

同理, 可以证明$ {\mathrm{zd}}(\varphi)=4, p, p-2 $时, $ \varphi $为0.

情况3   $ {\mathrm{zd}}(\varphi)=2p-4 $时. 由表 1, 可设$ \varphi(x_{3}x_{4})=bx^{p-1}_{1}x^{p-1}_{2}, \; \; b\in\mathbb{F}. $则$ \varphi=b\varphi_{2} $, 其中,

$ \begin{equation} \varphi_{2}:x_{3}x_{4}\rightarrow x^{p-1}_{1}x^{p-1}_{2}. \end{equation} $

(8)

易见, $ \varphi_{2} $是$ {\text {g}} $到$ \bar{\Lambda}(2, 2) $的权导子.

综上3种情况, 有$ \varphi=a\varphi_{1}+b\varphi_{2}, \; a, b\in\mathbb{F} $. 下证当$ \varphi\neq0 $时为外导子. 假设$ \varphi $是由$ m\in\bar{\Lambda}(2, 2) $决定的内导子, 由于$ \varphi $是权导子, 故$ m $为零权向量. 由表 1, 可设$ m=l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}, \; l_{i}\in\mathbb{F} $. 一方面, 由已知

$ \begin{align} \varphi(x_{3}x_{4})&=(a\varphi_{1}+b\varphi_{2})(x_{3}x_{4})=a\varphi_{1}(x_{3}x_{4})+b\varphi_{2}(x_{3}x_{4})=ax_{3}x_{4}+bx^{p-1}_{1}x^{p-2}_{2}, \end{align} $

$ \begin{align} \varphi(x_{1}x_{1})&=(a\varphi_{1}+b\varphi_{2})(x_{1}x_{1})=a\varphi_{1}(x_{1}x_{1})+b\varphi_{2}(x_{1}x_{1})=2ax_{1}x_{1}. \end{align} $

另一方由内导子定义有

$ \begin{align} \varphi(x_{3}x_{4})&=[x_{3}x_{4}, l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}] \end{align} $

$ \begin{align} &=(x_{4}\partial_{1}-x_{3}\partial_{2})(l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}) \end{align} $

$ \begin{align} &=-(l_{2}+l_{3})x_{3}x_{4}, \end{align} $

$ \begin{align} \varphi(x_{1}x_{1})&=[x_{1}x_{1}, l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}] \end{align} $

$ \begin{align} &=2x_{1}\partial_{3}(l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}) \end{align} $

$ \begin{align} &=2l_{1}x_{1}x_{2}x_{3}x_{4}+2l_{2}x_{1}x_{1}. \end{align} $

可得

$ \begin{align} l_{1}=0, \; l_{2}+l_{3}=-a. \end{align} $

(9)

同理, 一方面

$ \begin{align} \varphi(x_{1}x_{2})&=a\varphi_{1}(x_{1}x_{2})+b\varphi_{2}(x_{1}x_{2})=ax_{1}x_{2}. \end{align} $

另一方面

$ \begin{align} \varphi(x_{1}x_{2})&=[x_{1}x_{2}, l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}] \end{align} $

$ \begin{align} &=(x_{2}\partial_{3}+x_{1}\partial_{4})(l_{1}x_{1}x_{2}x_{3}x_{4}+l_{2}x_{1}x_{3}+l_{3}x_{2}x_{4}) \end{align} $

$ \begin{align} &=2l_{1}x_{1}x^{2}_{2}x_{4}+2l_{1}x^{2}_{1}x_{2}x_{3}+(l_{2}+l_{3})x_{1}x_{2}. \end{align} $

可得

$ \begin{align} l_{2}+l_{3}=a. \end{align} $

(10)

综合(9)与(10) 得, $ l_{1}=l_{2}=l_{3}=0 $, 即$ m=0 $, 从而$ \varphi=0 $, 与已知矛盾. 引理3.3得证.

引理4.1   $ H^{0}({\text {g}}, {\mathrm{HO}})=T $, 且$ {\mathrm{dim}}H^{0}({\text {g}}, {\mathrm{HO}})=2^{n-2}p^{n-2}-1. $

证  由零维上同调定义以及$ (1) $式知$ T\subseteq H^{0}({\text {g}}, {\mathrm{HO}}) $, 下面只需证$ H^{0}({\text {g}}, \bar\Lambda{(2, 2)})=0 $. 设$ l\in H^{0}({\text {g}}, \bar\Lambda{(2, 2)}) $, 由于

$H^{0}({\text {g}}, \bar\Lambda{(2, 2)})\subseteq\; H^{0}(h, \bar\Lambda{(2, 2)})\subseteq \bar\Lambda{(2, 2)_{(0)}}$

(其中$ \bar{\Lambda}(2, 2)_{(0)} $表示$ \bar{\Lambda}(2, 2) $中的零权空间), 可设$ l=f_{1}x_{1}x_{3}+f_{2}x_{2}x_{4}+f_{3}x_{1}x_{2}x_{3}x_{4} $, 其中$ f_{i}\in\mathbb{F} $, 则

$ \begin{align} [x_{3}x_{4}, l]&=[x_{3}x_{4}, f_{1}x_{1}x_{3}+f_{2}x_{2}x_{4}+f_{3}x_{1}x_{2}x_{3}x_{4}] \end{align} $

$ \begin{align} &=(x_{4}\partial_{1}-x_{3}\partial_{2})(f_{1}x_{1}x_{3}+f_{2}x_{2}x_{4}+f_{3}x_{1}x_{2}x_{3}x_{4}) \end{align} $

$ \begin{align} &=-(f_{1}+f_{2})x_{3}x_{4}=0, \end{align} $

即$ f_{1}+f_{2}=0 $. 又

$ \begin{align} [x_{1}x_{4}, l]&=[x_{3}x_{4}, f_{1}x_{1}x_{3}+f_{2}x_{2}x_{4}+f_{3}x_{1}x_{2}x_{3}x_{4}] \end{align} $

$ \begin{align} &=(x_{4}\partial_{3}-x_{1}\partial_{2})(f_{1}x_{1}x_{3}+f_{2}x_{2}x_{4}+f_{3}x_{1}x_{2}x_{3}x_{4}) \end{align} $

$ \begin{align} &=(f_{1}-f_{2})x_{1}x_{4}-2f_{3}x^{2}_{1}x_{3}x_{4}=0, \end{align} $

得$ f_{1}=f_{2}, \; f_{3}=0 $, 即$ l=0 $, $ H^{0}({\text {g}}, \bar\Lambda{(2, 2)})=0 $. 故$ H^{0}({\text {g}}, {\mathrm{HO}})=T $, 其维数为$ 2^{n-2}p^{n-2}-1, $得证.

注  在一维上同调中元素仍用导子表示.

命题4.2   令$ \hbar=\begin{pmatrix}E &0\\0 &-E \end{pmatrix} $, 其中$ E $是二阶单位阵, 则$ H^{1}({\text {g}}, T)={\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T) $, 且

${\mathrm{dim}}H^{1}({\text {g}}, T)=2^{n-2}p^{n-2}-1.$

证  由于$ \tilde{P}(2)=P(2)\oplus\mathbb{F}\hbar $, 其中

$ \\ P(2):=\left\{ \begin{gathered} \begin{pmatrix}{\mathrm{A}}&{\mathrm{B}}\\{\mathrm{C}}&{\mathrm{-A}}^{T}\end{pmatrix} \end{gathered} \in \tilde{P}(2)|\; {\mathrm{tr(A)}}=0\right\} $

是单的线性李超代数. 此时$ T $为$ P(2) $与$ \mathbb{F}\hbar $的平凡模. 由导子及单李超代数的性质知$ {\mathrm{Der}}(P(2), T)=0 $, 且$ {\mathrm{Der}}(\mathbb{F}\hbar, T)={\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T) $. 所以,

$ \begin{align} {\mathrm{Der}}({\text {g}}, T)={\mathrm{Der}}(\mathbb{F}\hbar, T)={\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T). \end{align} $

又由任何$ \mathbb{F}\hbar $到$ T $的非零导子都是外导子, 根据引理3.2知$ H^{1}({\text {g}}, T)={\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T) $. 最后由

$ \begin{align} {\mathrm{dim}}{\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T)={\mathrm{dim}}T=2^{n-2}p^{n-2}-1 \end{align} $

知命题成立.

命题4.3   设$ \varphi_{1} $, $ \varphi_{2} $如(7), (8) 所定义, 则

$H^{1}({\text {g}}, \bar{\Lambda}(2, 2))=\langle\varphi_{1}, \varphi_{2}\rangle, \; {\mathrm{dim}}H^{1}({\text {g}}, \bar{\Lambda}(2, 2))=2$.

证  根据引理3.3的证明, 有$ H^{1}({\text {g}}, \bar{\Lambda}(2, 2))\subseteq\langle\varphi_{1}, \varphi_{2}\rangle $, 因为$ \varphi_{1}-\varphi_{2} $是权导子, 由引理3.3知, $ \varphi_{1}-\varphi_{2} $是外导子, 故$ \varphi_{1} $与$ \; \varphi_{2} $模内导子空间$ {\mathrm{Ider}}({\text {g}}, \bar{\Lambda}(2, 2)) $是线性无关的, 又$ \varphi_{1} $与$ \varphi_{2} $在$ \; \mathbb{F} $上也是线性无关的, 所以$ \langle\varphi_{1}, \varphi_{2}\rangle\subseteq H^{1}({\text {g}}, \bar{\Lambda}(2, 2)) $, 即, $ H^{1}({\text {g}}, \bar{\Lambda}(2, 2))=\langle\varphi_{1}, \varphi_{2}\rangle $, 得证.

命题4.4   $ H^{1}({\text {g}}, {\mathrm{HO}})=(2^{n-2}p^{n-2}-1)\langle\varphi_{1}, \varphi_{2}\rangle\oplus{\mathrm{Hom}}_{\mathbb{F}}(\mathbb{F}\hbar, T) $, 特别地,

${\mathrm{dim}}H^{1}({\text {g}}, {\mathrm{HO}})=3(2^{n-2}p^{n-2}-1).$

证  由$ {\mathrm{HO}}\simeq(2^{n-2}p^{n-2}-1)\bar{\Lambda}(2, 2)\oplus T $知, $ H^{1}({\text {g}}, {\mathrm{HO}})=(2^{n-2}p^{n-2}-1)H^{1}({\text {g}}, \bar{\Lambda}(2, 2))\oplus H^{1}({\text {g}}, T) $, 根据命题4.2和4.3知结论成立.