习题 14, § 4

令$g$为李代数，$\mathfrak{b}$为$g$的次不变量子代数，使得$g$中$\mathfrak{b}$的中心化器$\mathfrak{z}(\mathfrak{b})$为零。

(a) The centralizer $\mathfrak{z}(\mathscr{C}^\infty \mathfrak{b})$ of $\mathscr{C}^\infty \mathfrak{b}$ in $g$ is contained in $\mathscr{C}^\infty \mathfrak{b}$. (If $\mathfrak{z}(\mathscr{C}^\infty \mathfrak{b}) \not\subset \mathscr{C}^\infty \mathfrak{b}$, $\mathfrak{z}(\mathscr{C}^\infty \mathfrak{b}) \not\subset \mathfrak{b}$ by Exercise 13; $\mathscr{C}^\infty \mathfrak{b}$ is an ideal of $g$ (§ 1, Exercise 14); let $c = \mathfrak{b} + \mathfrak{z}(\mathscr{C}^\infty \mathfrak{b})$; $c$ is a subalgebra, $\mathfrak{b}$ is subinvariant in $c$; $\mathfrak{b}$ is an ideal of $\mathfrak{b}_1 \subset c$, where $\mathfrak{b} \neq \mathfrak{b}_1$; let $y \in \mathfrak{b}_1, y \notin \mathfrak{b}$; $y = x + x'$, where $x \in \mathfrak{z}(\mathscr{C}^\infty \mathfrak{b})$, $x' \in \mathfrak{b}$; then $x \in \mathfrak{z}(\mathscr{C}^\infty \mathfrak{b}) \cap \mathfrak{b}_1, x \notin \mathfrak{b}$; $\mathfrak{d} = Kx + \mathfrak{b}$ is a subalgebra; $\mathscr{C}^\infty \mathfrak{b} = \mathscr{C}^\infty \mathfrak{b}$ because $c \cap \mathfrak{z}(\mathscr{C}^\infty \mathfrak{b}) \subset \mathscr{C}^\infty \mathfrak{b}$, whence $\mathscr{C}^k \mathfrak{b} \subset \mathscr{C}^k \mathfrak{b}$ for all $k$; $\mathfrak{z}(\mathscr{C}^\infty \mathfrak{b}) \cap \mathfrak{b} \not\subset \mathscr{C}^\infty \mathfrak{b}$ and hence the centre of $\mathfrak{d}$ is non-zero by Exercise 13; a fortiori, $\mathfrak{z}(\mathfrak{b}) \neq \{0\}$, which is absurd.)

(b) 由(a)推导出，如果$\mathfrak{D}$表示$\mathscr{C}^\infty \mathfrak{b}$的推导的李代数和$\mathscr{C}^\infty \mathfrak{b}$的$a$中心，则$\dim g \leq \dim \mathfrak{D} + \dim a$（让$g$通过伴随矩阵对$\mathscr{C}^\infty \mathfrak{b}$进行运算） 表示）。