本文约定基域$ \mathbb{F} $的特征$ p>2 $. 令$ \mathbb{Z}_{2} = \{\bar{0}, \bar{1}\} $. 域$ \mathbb{F} $上向量空间$ V $, 连同它的值和分解$ V = V_{\bar{0}}\oplus V_{\bar{1}} $称为一个$ \mathbb{Z}_{2} $-阶化空间. 若$ v\in V_{\theta} $, 其中$ \theta\in\mathbb{Z}_{2} $, 则称$ v $为$ \mathbb{Z}_{2} $- 齐次元素, 用$ |v| $表示$ \mathbb{Z}_{2} $-齐次元素的次数. 为方便, 如果符号$ |v| $出现, 那么约定$ v $是$ \mathbb{Z}_{2} $-齐次元素. 若$ V $的一组基为$ v_{1}, \dots, v_{n} $, 则可记为$ V: = \langle v_{1}, \dots, v_{n}\rangle $. 用$ \mathrm{hg}(V) $表示$ \mathbb{Z}_{2} $-阶化空间$ V $的所有$ \mathbb{Z}_{2} $-齐次元素的集合. 设$ L $是李超代数, $ M $为$ L $-模. 称$ \mathbb{Z}_{2} $-齐次线性映射$ \varphi: L\longrightarrow M $为次数为$ |\varphi| $的齐次导子, 如果

$ \begin{eqnarray*} \label{low00} \varphi([x, y]) = (-1)^{|\varphi| |x|}x \varphi(y)-(-1)^{|y| (|\varphi|+|x|)}y \varphi(x), \; \mbox{其中}\; x, y\in L. \end{eqnarray*} $

令$ \mathrm{Der}(L, M) $为$ L $到$ M $的所有$ \mathbb{Z}_{2} $-齐次导子张成的向量空间, 称该空间中的元素为导子. 对于$ \mathbb{Z}_{2} $-齐次元素$ m\in M $, 令

$ {{{\frak{D}}}}_m: L\longrightarrow M, \quad x\longmapsto (-1)^{|x||m|}x m, \; \mbox{其中}\; x\in \mathrm{hg}(L). $

那么$ {{\frak{D}}}_m $是次数为$ |m| $的$ \mathbb{Z}_{2} $-齐次导子. 令$ \mathrm{Ider}(L, M) $为由$ \{{{\frak{D}}}_m\mid m\in \mathrm{hg}(M)\} $张成的向量空间, 称该空间中的元素为内导子. 令$ \frak{h} $为$ L_{\bar{0}} $的Cartan子代数. 设$ L $与$ M $都有关于$ \frak{h} $的权空间分解: $ L = \oplus _{\gamma \in \frak{h}^*}L_{\gamma}, \quad M = \oplus _{\gamma \in \frak{h}^*}M_{\gamma}. $令

$ \begin{align*} \mathrm{Hom}_{\mathbb{F}}(L, M)_{(0)}& = \{\phi\in \mathrm{Hom}_{\mathbb{F}}(L, M)\mid \phi(L_{\alpha})\subset M_{\alpha}, \forall \alpha \in \frak{h}^*\}, \\ \mathrm{Der}(L, M)_{(0)}& = \{\phi\in \mathrm{Der}(L, M)\mid \phi(L_{\alpha})\subset M_{\alpha}, \forall \alpha \in \frak{h}^*\}. \end{align*} $

称$ \mathrm{Hom}_{\mathbb{F}}(L, M)_{(0)} $与$ \mathrm{Der}(L, M)_{(0)} $中元素分别为关于$ \frak{h} $的权映射与权导子. 那么

$ \begin{align} \mathrm{Der}(L, M) = \mathrm{Der}(L, M)_{(0)}+\mathrm{Ider}(L, M), \end{align} $

(1.1)

这个结论可参见文献[1, Lemma 3.2] 或[2, Lemma 2.1]. 值得注意的是, $ L $到$ M $的一阶上同调

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

(1.2)

(1.1) 与(1.2)得到, 计算一阶上同调可以先计算权导子空间. 本文目的在于决定Witt型模李超代数$ W(2) $到其Kac模$ K(\lambda) $的权导子空间, 取得下面主要结论, 其对刻画$ W(2) $的一阶上同调具有重要意义.

定理1  当$ \lambda = 0 $时, $ W(2) $到Kac模$ K(\lambda) $的权导子空间是一维的, 否则权导子空间是零维的.

本文研究的$ W(2) $属于Cartan型模李超代数的范畴, 具体地说, $ W(2) $是一类Witt型模李超代数. 1986年, 沈光宇老师对这类代数的阶化模做了研究^[4]; 2007年, 刘文德老师和张永正老师对其导子代数做了研究^[5]; 2010年, 舒斌老师和张朝文老师对其限制表示做了研究, 并且定义了Kac模^[6]. 李超代数的上同调是重要的研究课题, 其定义可追溯到D. A. Leites在1975年发表的文章^[7]. 2014年, 孙丽萍老师、刘文德老师和吴勃英老师利用权空间分解的方法, 刻画了在特征大于2的代数闭域上, $ \mathfrak{sl}(m, n) $到Cartan型李超代数$ W $和$ S $的低阶上同调, 并且指出这一计算结果与特征零的情形不同^[8].

本节简要介绍一下本文所用到的定义, 符号以及一些基本的模作用关系等.

定义2.1 ^[3]  设$ \Lambda(n) $是由$ x_{1}, x_{2}, \cdots x_{n} $生成的外代数. 定义

$ W(n): = \mathrm{Der}(\Lambda(n)) = \left\{\sum{f_{i}\partial_{i}}\mid f_{i}\in\Lambda(n)\right\}, $

其中$ \partial_{i} $是$ \Lambda(n) $的导子, 且满足$ \partial_{i}(x_{i}) = \delta_{ij}, \forall i, j\in\{1, 2, \cdots n\} $, 这里$ \delta_{ij} $是Kronecker符号. 约定$ |\partial_{i}| = |x_{i}| = \bar{1}, \forall1\leq i\leq n $, 且

$ [f\partial_{i}, g\partial_{j}]: = f\partial_{i}(g)\partial_{j}-(-1)^{|f\partial_{i}||g\partial_{j}|}g\partial_{j}(f)\partial_{i}, $

其中$ f\partial_{i}, g\partial_{j}\in W(n) $. 那么$ W(n) $构成一个限制李超代数, 称之为秩$ n $的Witt型李超代数.

下文主要研究$ W(2) $及其Kac模的结构. 取$ W(2) $的一组基:

$ \partial_{1}, \quad \partial_{2}, \quad x_{1}\partial_{1}, \quad x_{2}\partial_{2}, \quad x_{1}\partial_{2}, \quad x_{2}\partial_{1}, \quad x_{1}x_{2}\partial_{1}, \quad x_{1}x_{2}\partial_{2}. $

设

$ h_{1}: = x_{1}\partial_{1}-x_{2}\partial_{2}, \quad h_{2}: = x_{1}\partial_{1}+x_{2}\partial_{2}. $

令$ \frak{h} $为$ h_{1}, h_{2} $张成的子空间, 则$ \frak{h} $为$ W(2)_{\bar{0}} $的Cartan子代数. 用$ \eta_{1}, \eta_2 $表示$ h_1, h_2 $的对偶基, 即

$ \eta_{i}(h_{j}) = \delta_{ij}, \forall i, j = 1, 2. $

为方便, 下文用$ \frak{g} $表示$ W(2) $. 令$ x_i $与$ \partial_{i} $的$ \mathbb{Z} $-次数分别为$ 1 $与$ -1 $. 那么$ \frak{g} $成为$ \mathbb{Z} $-阶化李超代数$ \frak{g} = \frak{g}_{-1}\oplus\frak{g}_0\oplus\frak{g}_1. $设$ \Lambda: = \{\lambda = \lambda_{1}\eta_{1}+\lambda_{2}\eta_{2}|\lambda_{1}, \lambda_{2}\in\mathbb{F}_{p}\}. $ $ \frak{g}_0 $(同构于一般线性李代数$ \frak{gl}(2) $)的所有单模的同构类为$ \{L^{0}(\lambda)\mid \lambda\in \Lambda\} $. 令$ v_0 $为$ L^{0}(\lambda) $的属于权$ \lambda $的最高权向量. 设$ e: = x_{1}\partial_{2}, \; f: = x_{2}\partial_{1}. $归纳定义$ v_{i+1} = fv_i, 0\leq i<\lambda_{1} $. 本文约定符号$ 0\leq i<\lambda_{1} $或者$ 0\leq i\leq\lambda_{1} $一旦出现, 则$ \lambda_{1} $表示$ 0\sim p-1 $的自然数(与$ \lambda_1 $模$ p $同余). 那么$ L^{0}(\lambda) $有一组基$ \{v_0, v_1, \ldots, v_{\lambda_1}\} $. 如果$ i\notin \{0, 1, \ldots, \lambda_1\} $, 那么约定$ v_i = 0 $. 下面引理给出了$ L^{0}(\lambda) $的模结构.

引理2.1 ^[3]

$ \begin{align*} h_{1}v_{i}& = (\lambda_{1}-2i)v_{i}, \quad h_{2}v_{i} = \lambda_{2}v_{i}, \\ ev_{i}& = \begin{cases} 0, & i = 0, \\ i(\lambda_{1}-i+1)v_{i-1}, &1\leq i\leq\lambda_{1}, \end{cases}\\ fv_{i}& = \begin{cases} v_{i+1}, &0\leq i<\lambda_{1}, \\ 0, & i = \lambda_{1}. \end{cases} \end{align*} $

注意到, 若规定$ \frak{g}_1 $平凡作用在$ L^{0}(\lambda) $上, 则$ L^{0}(\lambda) $可视为$ \mathfrak{g}_0\oplus\frak{g}_1 $-模. 定义$ \frak{g} $的Kac模

$ K(\lambda): = u(\mathfrak{g})\otimes_{u(\mathfrak{g}_0\oplus\frak{g}_1)}L^{0}(\lambda), $

其中$ u(L) $表示李超代数$ L $的限制包络代数. 那么作为向量空间, $ K(\lambda)\cong\Lambda(\mathfrak{g}_{-1})\otimes_{\mathbb{F}}L^{0}(\lambda). $显然, $ K(\lambda) $具有基:

$ \{\partial_{1}\otimes v_{i}, \quad \partial_{2}\otimes v_{i}, \quad \partial_{1}\partial_{2}\otimes v_{i}, \quad 1\otimes v_{i} \mid 0\leq i\leq\lambda_{1}\}. $

设$ f\partial_{s}\in \frak{g} $. 根据文献[7], 可得如下等式:

$ \begin{equation} f\partial_{s}(1\otimes v_{i}) = f\partial_{s}\otimes v_{i}, \end{equation} $

(2.1)

$ \begin{equation} f\partial_{s}(\partial_{j}\otimes v_{i}) = [f\partial_{s}, \partial_{j}]+(-1)^{|f\partial_{s}||\partial_{j}|}\partial_{j}\cdot(f\partial_{s}\otimes v_{i}) , (j = 1, 2), \end{equation} $

(2.2)

$ \begin{equation} \begin{split} f\partial_{s}(\partial_{1}\partial_{2}\otimes v_{i})& = [f\partial_{s}, \partial_{1}]\partial_{2}\otimes v_{i}+(-1)^{|f\partial_{s}||\partial_{1}|}\partial_{1}[f\partial_{s}, \partial_{2}]\otimes v_{i}\\ &\quad+(-1)^{|f\partial_{s}||\partial_{1}|+|f\partial_{s}||\partial_{2}|} \partial_{1}\partial_{2}\cdot(f\partial_{s}\otimes v_{i}). \end{split} \end{equation} $

(2.3)

利用$ (2.1)-(2.3) $及引理2.1, 可以得到Kac模$ K(\lambda) $的模作用表.

本节主要是分步骤证明定理1.

通过计算, 容易得到$ \frak{g} $的权空间为:

$ \begin{align*} &\frak{g}_{-\eta_{1}+\eta_{2}} = \langle x_{1}x_{2}\partial_{1}\rangle, \quad \frak{g}_{-\eta_{1}-\eta_{2}} = \langle\partial_{1}\rangle, \\ &\frak{g}_{\eta_{1}+\eta_{2}} = \langle x_{1}x_{2}\partial_{2}\rangle, \quad\;\;\;\frak{g}_{\eta_{1}-\eta_{2}} = \langle\partial_{2}\rangle, \\ &\frak{g}_{0} = \langle x_{1}\partial_{1}, x_{2}\partial_{2}\rangle, \quad\quad \frak{g}_{-2\eta_{1}} = \langle x_{2}\partial_{1}\rangle, \quad \frak{g}_{2\eta_{1}} = \langle x_{1}\partial_{2}\rangle. \end{align*} $

根据表 1, $ K(\lambda) $的权空间为:

$ \begin{equation} \begin{split} K(\lambda)_{(\lambda_{1}-2i-1)\eta_{1}+(\lambda_{2}-1)\eta_{2}}& = \langle\partial_{1}\otimes v_{i}\rangle, \quad K(\lambda)_{(\lambda_{1}-2i)\eta_{1}+\lambda_{2}\eta_{2}} = \langle 1\otimes v_{i}\rangle, \\ K(\lambda)_{(\lambda_{1}-2i+1)\eta_{1}+(\lambda_{2}-1)\eta_{2}}& = \langle\partial_{2}\otimes v_{i}\rangle, \quad K(\lambda)_{(\lambda_{1}-2i)\eta_{1}+(\lambda_{2}-2)\eta_{2}} = \langle\partial_{1}\partial_{2}\otimes v_{i}\rangle. \end{split} \end{equation} $

(3.1)

进而可得如下引理:

引理3.1  $ \lambda_{2}\ \notin \{0, 1, 2, 3, -1\} $时, $ \frak{g} $到Kac模的权映射必为零映射.

设$ \varphi $是一个非零保权映射, 那么$ \lambda_{2} $只能取$ 0, \; 1, \; 2, \; 3, \; -1 $. 下面的注记给出了$ K(\lambda)_{\mu} $的结构, 其中$ \mu $为$ \frak{g} $的任一权.

注记3.2   当$ 0\leq i\leq\lambda_{1}, \; \lambda_{2}\in\{0, 1, 2, 3, -1\} $时, 有以下结论:

(1) 当$ \lambda_{2} = 0 $时,

$ \begin{align*} &K(\lambda)_{\eta_{1}-\eta_{2}} = \langle\partial_{1}\otimes v_{\frac{\lambda_{1}-2}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}}{2}}\rangle, \\ &K(\lambda)_{-\eta_{1}-\eta_{2}} = \langle\partial_{1}\otimes v_{\frac{\lambda_{1}}{2}}, \partial_{1}\otimes v_{\frac{\lambda_{1}+2}{2}}\rangle, \\ &K(\lambda)_{-\eta_{1}+\eta_{2}} = K(\lambda)_{\eta_{1}+\eta_{2}} = 0, \quad K(\lambda)_{0} = \langle1\otimes v_{\frac{\lambda_{1}}{2}}\rangle, \\ &K(\lambda)_{2\eta_{1}} = \langle1\otimes v_{\frac{\lambda_{1}-2}{2}}\rangle, \quad \quad K(\lambda)_{2\eta_{1}} = \langle1\otimes v_{\frac{\lambda_{1}+2}{2}}\rangle. \end{align*} $

(2) 当$ \lambda_{2} $ = 1时,

$ \begin{align*} &K(\lambda)_{2\eta_{1}} = \langle\partial_{1}\otimes v_{\frac{\lambda_{1}-3}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}-1}{2}}\rangle, \\ &K(\lambda)_{-2\eta_{1}} = \langle\partial_{1}\otimes v_{\frac{\lambda_{1}+1}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}+3}{2}}\rangle, \\ &K(\lambda)_{0} = \langle\partial_{1}\otimes v_{\frac{\lambda_{1}+1}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}-1}{2}}\rangle, \\ &K(\lambda)_{\eta_{1}-\eta_{2}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}-1}{2}}\rangle, \quad\;\; K(\lambda)_{\eta_{1}+\eta_{2}} = \langle1\otimes v_{\frac{\lambda_{1}-2}{2}}\rangle, \\ &K(\lambda)_{-\eta_{1}-\eta_{2}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}+1}{2}}\rangle, \quad K(\lambda)_{\eta_{2}-\eta_{1}} = \langle1\otimes v_{\frac{\lambda_{1}+2}{2}}\rangle. \end{align*} $

(3) 当$ \lambda_{2} = 2 $时,

$ \begin{align*} &K(\lambda)_{\eta_{1}+\eta_{2}} = \langle1\otimes v_{\frac{\lambda_{1}-2}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}}{2}}\rangle, \\ &K(\lambda)_{\eta_{2}-\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}}{2}}, \partial_{2}\otimes v_{\frac{\lambda_{1}+2}{2}}\rangle, \\ &K(\lambda)_{-\eta_{1}-\eta_{2}} = K(\lambda)_{\eta_{1}-\eta_{2}} = 0, \quad K(\lambda)_{0} = \langle1\otimes v_{\frac{\lambda_{1}}{2}}\rangle, \\ &K(\lambda)_{2\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}-2}{2}}\rangle, \quad K(\lambda)_{-2\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}+2}{2}}\rangle. \end{align*} $

(4) 当$ \lambda_{2} = 3 $时,

$ \begin{align*} &K(\lambda)_{\eta_{1}+\eta_{2}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}-1}{2}}\rangle, \quad K(\lambda)_{\eta_{2}-\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{\frac{\lambda_{1}+1}{2}}\rangle, \\ &K(\lambda)_{\eta_{1}-\eta_{2}} = K(\lambda)_{-\eta_{1}-\eta_{2}} = K(\lambda)_{ -2\eta_{1}} = K(\lambda)_{2\eta_{1}} = K(\lambda)_{0} = 0. \end{align*} $

(5) 当$ \lambda_{2} = -1 $时,

$ \begin{align*} &K(\lambda)_{\eta_{1}+\eta_{2}} = \langle1\otimes v_{\frac{\lambda_{1}-1}{2}}\rangle, \quad K(\lambda)_{-\eta_{1}+\eta_{2}} = \langle1\otimes v_{\frac{\lambda_{1}+1}{2}}\rangle, \\ &K(\lambda)_{-\eta_{1}-\eta_{2}} = K(\lambda)_{\eta_{1}-\eta_{2}} = K(\lambda)_{-2\eta_{1}} = K(\lambda)_{2\eta_{1}} = K(\lambda)_{0} = 0. \end{align*} $

记$ K = \{-3, -2, -1, 0, 1, 2, 3\} $. 当$ \lambda_{2}\in\{0, 1, 2, 3, -1\} $时, 若$ K(\lambda)_{\mu}\neq0 $且$ \mu $为$ \frak{g} $的权, 则(3.1)中出现的$ i $都是$ \frac{\lambda_{1}+k}{2} $的形式, 其中$ k\in K $. 对于$ \forall k\in K $, 当$ \lambda_{1}\geq3 $时, 作为自然数$ \frac{\lambda_{1}+k}{2}\in\mathbb{F}_{p} $; 而$ 0\leq\lambda_{1}\leq2 $时, 作为自然数, $ \exists\frac{\lambda_{1}+k}{2}\leq0 $, 所以调整$ p $, 使得$ \frac{\lambda_{1}+k+p}{2}\geq0 $. 下文将分为$ 0\leq\lambda_{1}\leq2 $和$ \lambda_{1}\geq3 $两种情况讨论.

引理3.3

(1) 当$ \lambda_{1} = 0, \lambda_{2} = 0 $时, $ \frak{g} $到Kac模的权导子空间是一维的.

(2) 当$ \lambda_{1} = 0, \lambda_{2}\in\{1, 2, 3, -1\} $时, $ \frak{g} $到Kac模的权导子空间是零空间.

(3) 当$ \lambda_{1} = 0, \lambda_{2}\in\{0, 1, 2, 3, -1\} $时, $ \frak{g} $到Kac模的权导子空间是零空间.

证  由于特征$ p $不同时, $ \frac{\lambda_{1}+k}{2} $作为自然数未必满足$ 1\leq i\leq\lambda_{1} $, 因此证明过程中我们根据特征$ p $分情况讨论:

● 当$ \lambda_{1} = 0 $时, 考虑$ p = 3 $和$ p\geq5 $两种情况.

● 当$ \lambda_{1} = 1 $时, 考虑$ p = 3 $和$ p\geq5 $两种情况.

● 当$ \lambda_{1} = 2 $时, 考虑$ p = 3, p = 5 $和$ p\geq7 $三种情况.

下面只证明$ \lambda_{1} = 0, \lambda_{2} = 0 $的情形, 其他情形可做类似证明.

(1) 当$ p = 3 $时, 作为自然数, 显然有

$ \tfrac{\lambda_{1}-2}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}}{2} = \lambda_{1} = 0, \quad \tfrac{\lambda_{1}+2}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}-1}{2}>\lambda_{1}, $

$ \tfrac{\lambda_{1}+1}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}-3}{2} = \lambda_{1} = 0, \quad \tfrac{\lambda_{1}+3}{2} = \lambda_{1} = 0. $

由注记3.2可知,

$ K(\lambda)_{-\eta_{1}-\eta_{2}} = \langle\partial_{1}\otimes v_{0}\rangle, \quad K(\lambda)_{\eta_{1}-\eta_{2}} = \langle\partial_{2}\otimes v_{0}\rangle, \quad K(\lambda)_{\mu} = 0, $

其中$ \mu = -\eta_{1}+\eta_{2}, \eta_{1}+\eta_{2}, -2\eta_{1}, 2\eta_{1}, 0. $设权映射$ \varphi $满足:

$ \partial_{1}\mapsto a_{1}\partial_{1}\otimes v_{0}, \quad \partial_{2}\mapsto a_{2}\partial_{2}\otimes v_{0}, \quad x_{1}\partial_{1}\mapsto 0, $

$ x_{2}\partial_{2}\mapsto 0, \quad x_{1}\partial_{2}\mapsto 0, \quad x_{2}\partial_{1}\mapsto 0, \quad x_{1}x_{2}\partial_{1}\mapsto 0, \quad x_{1}x_{2}\partial_{2}\mapsto 0, $

则$ |\varphi| = \bar{0} $. 进而$ \varphi $是权导子当且仅当下列方程有非零解:

$ \begin{equation*} \left\{ \begin{array}{l} \varphi([\partial_{1}, \partial_{2}]) = \partial_{1}\varphi(\partial_{2})+\partial_{2}\varphi(\partial_{1}), \\ \varphi([\partial_{2}, x_{2}\partial_{2}]) = \partial_{2}\varphi(x_{2}\partial_{2})-x_{2}\partial_{2}\varphi(\partial_{2}). \end{array} \right. \end{equation*} $

即:

$ \begin{equation*} \left\{ \begin{array}{l} a_{2}\partial_{1}\partial_{2}\otimes v_{0}-a_{1}\partial_{1}\partial_{2}\otimes v_{0} = 0, \\ a_{2}\partial_{2}\otimes v_{0} = a_{1}\partial_{2}\otimes v_{0}.\\ \end{array} \right. \end{equation*} $

整理上述方程组可得:

$ a_{1} = a_{2}. $

(2) 当$ p\geqslant5 $时, 作为自然数, 显然有

$ \tfrac{\lambda_{1}-2}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}}{2}>\lambda_{1} = 0, \quad \tfrac{\lambda_{1}+2}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}-1}{2}>\lambda_{1}, $

$ \tfrac{\lambda_{1}+1}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}-3}{2}>\lambda_{1}, \quad \tfrac{\lambda_{1}+3}{2}>\lambda_{1}. $

由注记3.2可知,

$ \begin{align*} &K(\lambda)_{-\eta_{1}-\eta_{2}} = \langle\partial_{1}\otimes v_{0}\rangle, \quad K(\lambda)_{0} = \langle1\otimes v_{0}\rangle, \\ &K(\lambda)_{\eta_{1}-\eta_{2}} = \langle\partial_{2}\otimes v_{0}\rangle, \quad K(\lambda)_{\mu} = 0, \end{align*} $

其中$ \mu = -\eta_{1}+\eta_{2}, \eta_{1}+\eta_{2}, -2\eta_{1}, 2\eta_{1}. $设权映射$ \varphi $满足:

$ \partial_{1}\mapsto a_{1}\partial_{1}\otimes v_{0}, \quad \partial_{2}\mapsto a_{2}\partial_{2}\otimes v_{0}, \quad x_{1}\partial_{1}\mapsto a_{3}1\otimes v_{0}, \quad x_{2}\partial_{2}\mapsto a_{4}1\otimes v_{0}, $

$ x_{1}\partial_{2}\mapsto 0, \quad x_{2}\partial_{1}\mapsto 0, \quad x_{1}x_{2}\partial_{1}\mapsto 0, \quad x_{1}x_{2}\partial_{2}\mapsto 0, $

则$ |\varphi| = \bar{0} $. 于是$ \varphi $是权导子当且仅当下列方程有非零解:

$ \begin{equation*} \left\{ \begin{array}{l} \varphi([\partial_{1}, \partial_{2}]) = \partial_{1}\varphi(\partial_{2})+\partial_{2}\varphi(\partial_{1}), \\ \varphi([\partial_{1}, x_{2}\partial_{2}]) = \partial_{1}\varphi(x_{2}\partial_{2})-x_{2}\partial_{2}\varphi(\partial_{1}), \\ \varphi([\partial_{2}, x_{2}\partial_{2}]) = \partial_{2}\varphi(x_{2}\partial_{2})-x_{2}\partial_{2}\varphi(\partial_{2}). \end{array} \right. \end{equation*} $

即:

$ \begin{equation*} \left\{ \begin{array}{l} a_{2}\partial_{1}\partial_{2}\otimes v_{0}-a_{1}\partial_{1}\partial_{2}\otimes v_{0} = 0, \\ a_{3}\partial_{1}\otimes v_{0} = 0, \\ a_{4}\partial_{2}\otimes v_{0} = 0. \end{array} \right. \end{equation*} $

整理上述方程组可得:

$ \begin{equation*} \left\{ \begin{array}{l} a_{1}-a_{2} = 0, \\ a_{3} = 0, \\ a_{4} = 0. \end{array} \right. \end{equation*} $

综上所述, $ \lambda_{1} = 0, \; \lambda_{2} = 0 $时, 即$ \lambda = 0 $时, 权导子空间是一维的.

类似的, 可以得到$ \lambda_{1} = 0, 1, 2 $时, 每一组满足导子定义的方程组. 经过计算可知, 只有$ \lambda_{1} = 0 $且$ \lambda_{2} = 0 $时才有非零解, 其余情况只有零解.

回忆$ K(\lambda) $的基, 下面刻画一下$ v_{i} $的情形.

引理3.4   对$ \forall k\in K $, 当$ \lambda_{1}+k $为偶数时, 则$ v_{\frac{\lambda_{1}+k}{2}}\neq0 $. 当$ \lambda_{1}+k $为奇数时, 若$ 3\leq\lambda_{1}<p-3 $, 则$ v_{\frac{\lambda_{1}+k}{2}} = 0 $.

证   因为$ \lambda_{1}+k $不能被2整除时, 不一定可以作为$ i $的取值, 故按照$ \lambda_{1}+k $为奇数和偶数两种情况讨论.

当$ \lambda_{1}+k $为偶数时: $ \tfrac{\lambda_{1}+k}{2}\in\{0, 1, \cdots, \tfrac{p+1}{2}, \tfrac{p+2}{2}\}\subset\{0, 1, \cdots, p-1\} $, 此时可以直接按自然数次序与$ \lambda_{1} $比较大小, 又因为$ k\leq\lambda_{1} $, 从而有$ \tfrac{\lambda_{1}+k}{2}\leq\lambda_{1} $. 于是当$ \lambda_{1}+k $为偶数时, $ v_{\tfrac{\lambda_{1}+k}{2}}\neq0 $.

当$ \lambda_{1}+k $为奇数且$ 3\leq\lambda_{1}<p-3 $时: $ \tfrac{\lambda_{1}+k+p}{2}\geq\lambda_{1} $. 如若不然, $ \lambda_{1}\geq p+k $, 这与$ \lambda_{1} $是介于$ 0\sim p-1 $的自然数矛盾，因此$ v_{\tfrac{\lambda_{1}+k}{2}} = 0. $

引理3.5   当$ \lambda_{1}\geq3, \; \lambda_{2}\in\{0, 1, 2, 3, -1\} $时, $ \frak{g} $到Kac模的权导子空间是零空间.

证  只需进行如下分类讨论:

(1) 当$ \lambda_{1} = p-1 $时, 若$ \lambda_{1}+k $为偶数, 则$ v_{\frac{\lambda_{1}+k}{2}}\neq0 $. 若$ \lambda_{1}+k $为奇数, 作为自然数有

$ \tfrac{\lambda_{1}+3+p}{2} = 1<\lambda_{1}, \quad \tfrac{\lambda_{1}+1+p}{2} = 0<\lambda_{1}, \quad \tfrac{\lambda_{1}-1+p}{2} = \lambda_{1}, \quad \tfrac{\lambda_{1}-3+p}{2} = p-2<\lambda_{1}. $

易得$ v_{\tfrac{\lambda_{1}+k}{2}}\neq0 $.

(2) 当$ \lambda_{1} = p-2 $时, 若$ \lambda_{1}+k $为偶数, 则$ v_{\frac{\lambda_{1}+k}{2}}\neq0 $. 若$ \lambda_{1}+k $为奇数, 作为自然数有

$ \tfrac{\lambda_{1}+2+p}{2} = 0<\lambda_{1}, \quad \tfrac{\lambda_{1}+p}{2} = p-1>\lambda_{1}, \quad \tfrac{\lambda_{1}-2+p}{2} = p-2 = \lambda_{1}. $

易得$ v_{\tfrac{\lambda_{1}}{2}} = 0, \; v_{\tfrac{\lambda_{1}+2}{2}}\neq0, \; v_{\tfrac{\lambda_{1}-2}{2}}\neq0 $.

(3) 当$ \lambda_{1} = p-3 $时, 若$ \lambda_{1}+k $为偶数, 则$ v_{\frac{\lambda_{1}+k}{2}}\neq0 $. 若$ \lambda_{1}+k $为奇数, 作为自然数有

$ \tfrac{\lambda_{1}+3+p}{2} = 0<\lambda_{1}, \quad \tfrac{\lambda_{1}+1+p}{2} = p-1>\lambda_{1}, \quad\tfrac{\lambda_{1}-1+p}{2} = p-2>\lambda_{1}, \quad \tfrac{\lambda_{1}-3+p}{2} = p-3 = \lambda_{1}. $

易得$ v_{\tfrac{\lambda_{1}+1}{2}} = 0, \; v_{\tfrac{\lambda_{1}-1}{2}} = 0, \; v_{\tfrac{\lambda_{1}+3}{2}}\neq0, \; v_{\tfrac{\lambda_{1}-3}{2}}\neq0 $.

(4) 当$ 3\leq\lambda_{1}<p-3 $时, 若$ \lambda_{1}+k $为奇数, 由引理3.4可知, $ \frak{g} $在$ \varphi $下的象只有零空间, 若$ \lambda_{1}+k $为偶数, 与$ \lambda_{1} = p-1 $的情形下计算过程相似.

下面证明$ \lambda_{1} = p-2, \lambda_{2} = 2 $的情形. 由注记3.2可知,

$ K(\lambda)_{\eta_{1}+\eta_{2}} = \langle1\otimes v_{\lambda_{1}}\rangle, \quad K(\lambda)_{-\eta_{1}+\eta_{2}} = \langle\partial_{2}\otimes v_{0}\rangle, \quad $

$ K(\lambda)_{2\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{\lambda_{1}}\rangle, \quad K(\lambda)_{-2\eta_{1}} = \langle\partial_{1}\partial_{2}\otimes v_{0}\rangle, \quad K(\lambda)_{\mu} = 0, $

其中$ \mu = -\eta_{1}-\eta_{2}, \eta_{1}-\eta_{2}, 0. $设权映射$ \varphi $满足:

$ x_{1}x_{2}\partial_{2}\mapsto a_{1}1\otimes v_{\lambda_{1}}, \quad x_{1}x_{2}\partial_{1}\mapsto a_{2}\partial_{2}\otimes v_{0}, \quad x_{1}\partial_{2}\mapsto a_{3}\partial_{1}\partial_{2}\otimes v_{\lambda_{1}}, \quad $

$ x_{2}\partial_{1}\mapsto a_{4}\partial_{1}\partial_{2}\otimes v_{0}, \quad x_{1}\partial_{1}\mapsto0, \quad x_{2}\partial_{2}\mapsto0, \quad \partial_{1}\mapsto0, \quad \partial_{2}\mapsto0, $

则$ |\varphi| = \bar{0} $. 那么$ \varphi $是权导子当且仅当下列方程有非零解:

$ \begin{equation*} \left\{ \begin{array}{l} \varphi([x_{1}\partial_{2}, x_{1}x_{2}\partial_{1}]) = x_{1}\partial_{2}\varphi(x_{1}x_{2}\partial_{1})-x_{1}x_{2}\partial_{1}\varphi(x_{1}\partial_{2}), \\ \varphi([x_{2}\partial_{1}, x_{1}x_{2}\partial_{1}) = x_{2}\partial_{1}\varphi(x_{1}x_{2}\partial_{1})-x_{1}x_{2}\partial_{1}\varphi(x_{2}\partial_{1}). \end{array} \right. \end{equation*} $

即:

$ \begin{equation*} \left\{ \begin{array}{l} -a_{1}1\otimes v_{\lambda_{1}} = 0-\tfrac{\lambda_{1}}{2}a_{3}\partial_{1}\otimes v_{0}, \\ (\tfrac{p-2}{2}a_{4}-a_{2})\partial_{1}\otimes v_{\lambda_{1}}-(a_{2}+a_{4})\partial_{1}\partial_{2}\otimes v_{0} = 0. \end{array} \right. \end{equation*} $

整理上述方程组可得:

$ \begin{equation*} \left\{ \begin{array}{l} a_{1} = 0, \\ a_{3} = 0, \\ a_{4}+a_{2} = 0, \\ \tfrac{p-2}{2}a_{4}-a_{2} = 0. \end{array} \right. \end{equation*} $

解得:

$ a_{1} = a_{2} = a_{3} = a_{4} = 0. $

经过上述计算, 上述所有情况中, 满足导子定义的每个方程组都只有零解, 结论得证.

由引理3.1, 3.3以及3.5易得定理1.