Exercise 11, § 3

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{r}$ its radical.

a) Let $\mathfrak{s}$ be a Levi subalgebra of $\mathfrak{g},\mathfrak{k}$ a Cartan subalgebra of $\mathfrak{s}$. Show that $\mathfrak{k}$ is contained in a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ which is the sum of $\mathfrak{k}$ and a subalgebra of $\mathfrak{r}$. (Use §2, Th. 2, Prop. 10 and Cor. 2 of Th. 1.)

b) Let $\mathfrak{h}'$ be a Cartan subalgebra of $\mathfrak{g}$. Show that there exists a Levi subalgebra $\mathfrak{s}'$ of $\mathfrak{g}$ such that $\mathfrak{h}'$ is the sum of a Cartan subalgebra of $\mathfrak{s}'$ and a subalgebra of $\mathfrak{r}$. (The subalgebras $\mathfrak{s},\mathfrak{k},\mathfrak{h}$ in $a$) can be chosen so that $\mathfrak{h}+\mathfrak{r}=\mathfrak{h}'+\mathfrak{r}$. Put $\mathfrak{a}=\mathfrak{h}+\mathfrak{r}$, which is solvable. By Th. 3, there exists $x\in \mathscr{C}^{\infty}(\mathfrak{a})$ such that $e^{ad_{\mathfrak{a}}x}\mathfrak{h}=\mathfrak{h}'$. Then $e^{ad_{\mathfrak{g}}x}$ is a special automorphism of $\mathfrak{g}$ which transforms $\mathfrak{s}$ into the required Levi subalgebra.)

c) Let $\mathfrak{s}$ be a Levi subalgebra of $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ which is the sum of a Cartan subalgebra $\mathfrak{k}$ of $\mathfrak{s}$ and a subalgebra $\mathfrak{l}$ of $\mathfrak{r}$. Let $\mathfrak{c}$ be the commutant of $\mathfrak{k}$ in $\mathfrak{r}$. Show that $\mathfrak{l}$ is a Cartan subalgebra of $\mathfrak{c}$. (For $x\in \mathfrak{k}$, ad$_{\mathfrak{g}}x$ is semi-simple, but ad$_{\mathfrak{h}}x$ is nilpotent, so $[\mathfrak{k},\mathfrak{h}] = 0$. If $y\in \mathfrak{c}$ is such that $[y,\mathfrak{l}]\subset \mathfrak{l}$, then $[y,\mathfrak{h}]\subset \mathfrak{h}$ so $y\in \mathfrak{h}\cap \mathfrak{r}=\mathfrak{l}.$)

d) Let $\mathfrak{s}$ be a Levi subalgebra of $\mathfrak{g},\mathfrak{k}$ a Cartan subalgebra of $\mathfrak{s},\mathfrak{c}$ the commutant of $\mathfrak{k}$ in $\mathfrak{r},\mathfrak{l}$ a Cartan subalgebra of $\mathfrak{c}$. Then $\mathfrak{h}=\mathfrak{k}+\mathfrak{l}$ is a Cartan subalgebra of $\mathfrak{g}$. (Let $x=y+z(y\in \mathfrak{s}, z\in \mathfrak{r})$ be an element of the normalizer $\mathfrak{u}$ of $\mathfrak{h}$ in $\mathfrak{g}$. Show that $[y,\mathfrak{k}]\subset \mathfrak{k}$, so $y\in \mathfrak{k}$ and $z\in \mathfrak{u}$. Then show that $[z,\mathfrak{k}]\subset \mathfrak{h}\cap \mathfrak{r}\subset \mathfrak{c}$, so $[\mathfrak{k},[\mathfrak{k}, z]] = 0$, and hence that $[\mathfrak{k}, z] = 0$ and $z\in \mathfrak{c}$. Finally, $[z,\mathfrak{l}]\subset \mathfrak{l}$ hence $z\in \mathfrak{l}$ and $x\in \mathfrak{h}.$)

e) Let $\mathfrak{s},\mathfrak{k},\mathfrak{c}$ be as in d), and $\mathfrak{q}= [\mathfrak{g},\mathfrak{r}]$ the nilpotent radical of $\mathfrak{g}$. Let $x\in \mathfrak{q}$ and $u$ the special automorphism $e^{adx}$. If $u(\mathfrak{k})\subset \mathfrak{k}+\mathfrak{c}$, then $x\in \mathfrak{c}$. (Consider the adjoint representation $\rho$ of $\mathfrak{s}$ on $\mathfrak{q}$, and let $\mathfrak{q}_i$ be a complement of $\mathscr{C}^{i+1}\mathfrak{q}$ in $\mathscr{C}^i\mathfrak{q}$ stable under $\rho$; let $\rho_i$ be the subrepresentation of $\rho$ defined by $\mathfrak{q}_i$. Let $\sigma_i=\rho_i|\mathfrak{k}$. Let $\mathfrak{q}'_i$ be the commutant of $\mathfrak{k}$ in $\mathfrak{q}_i$ and $\mathfrak{q}''_i$ a complement of $\mathfrak{q}'_i$ in $\mathfrak{q}_i$ stable under $\sigma_i$. Let $x=x'_1+x''_1+\cdots +x'_n+x''_n$ with $x'_i\in \mathfrak{q}'_i, x''_i\in \mathfrak{q}''_i$. Arguing by contradiction, assume that the $x''_i$ are not all zero and $x''_1=\cdots =x''_{p-1}= 0$, $x''_p\not= 0$, for example. If $h\in \mathfrak{k},u(h) =h+ [x''_p, h] +y$ with $y\in \mathscr{C}^{p+1}\mathfrak{q}$. Since $u(h)\in \mathfrak{k}+\mathfrak{c}$, this gives $[x''_p, h] +y\in \mathfrak{q}'_p+\mathfrak{q}'_{p+1}+\cdots +\mathfrak{q}'_n$, hence $[h, x''_p]\in \mathfrak{q}'_p$, so $[h, x''_p] = 0$. Then $x''_p= 0$, a contradiction.)

f) Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. Then $\mathfrak{h}$ can be expressed uniquely as the sum of $\mathfrak{h}\cap \mathfrak{r}$ and a Cartan subalgebra of a Levi subalgebra of $\mathfrak{g}$. (For the uniqueness, use $e$) and Th. 5 of Chap. I, §6, no. 8.) The Levi subalgebra in question is not unique in general.

g) Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$, and $\mathfrak{t}=\mathfrak{g}^0(\mathfrak{h}\cap \mathfrak{r})$. Then $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{t}$. We have $\mathfrak{g}=\mathfrak{t}+\mathfrak{r}$ (use a Fitting decomposition for the adjoint representation of $\mathfrak{h}\cap \mathfrak{r}$ on $\mathfrak{g}$). The algebra $\mathfrak{t}\cap \mathfrak{r}$ is the radical of $\mathfrak{t}$ and is nilpotent (use Exerc. 5 of §2).

h) Assume that $k$ is algebraically closed. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. There exists a Levi subalgebra $\mathfrak{s}$ of $\mathfrak{g}$ such that, for all $\lambda \in \mathfrak{h}^*$,

(With the notations of $g$), take for $\mathfrak{s}$ a Levi subalgebra of $\mathfrak{t}$ such that $\mathfrak{h}=$ $(\mathfrak{h}\cap \mathfrak{s}) + (\mathfrak{h}\cap \mathfrak{r})$; this exists by $b).)^7$