Exercise 28, § 9

(a) Let k be a commutative field of characteristic 0, L an n-dimensional Lie algebra over k, $L_1$ an $(n-1)$-dimensional subalgebra containing no non-zero ideal of L and $a_0$ an element of L not belonging to $L_1$. For $i = 2, 3, \ldots$, we define $L_i$ inductively to be the set of $x \in L_{i-1}$ such that $[x, a_0] \in L_{i-1}$. We write $L_i = L$ for $i \leq 0$. Show that $[L_i, L_j] \subset L_{i+j}$ (by induction on $i + j$) and then that, for $0 \leq i \leq n$, $L_i$ is a subalgebra of L of codimension i (by induction on i).

(b) For $0 < i < n$, choose in $L_i$ an element $a_i$ not belonging to $L_{i+1}$ such that $[a_0, a_i] \equiv ia_{i-1} \pmod{L_i}$. Show, by induction on the ordered pairs $(i, j)$ ordered lexicographically, that, for $0 \leq i < j < n, i + j - 1 < n$ and

(c) Deduce that L is either of dimension 1, or of dimension 2 and non-commutative, or isomorphic to $\mathfrak{sl}(2, k)$.

(d) Let A be a finite-dimensional real Lie algebra. A subalgebra B of A is called extendable if there exists a subalgebra $B_1 \supset B$ such that

† For more details concerning injective linear representations of Lie groups, cf. G. Hochschild, The structure of Lie groups, Holden-Day, 1965, and in particular Chapter XVIII.

dim $B_1 = \dim B + 1$. Let $R'(A)$ denote the intersection of all the non-extendable subalgebras of $A$. Let $R(A)$ denote the largest ideal of $A$ with a composition series as an $A$-module all of whose quotients are of dimension 1.

Show that $R'(A)$ is a characteristic ideal of $A$. (Observe that $R'(A)$ is stable under $\mathrm{Aut}(A)$.)

(e) Show that $R'(A) \supset R(A)$ and that $R'(A/R(A)) = R'(A)/R(A)$.

(f) Show that, if $B$ is a subalgebra of $A$, then $R'(B) \supset R'(A) \cap B$.

(g) If $A$ is solvable, $R'(A) = R(A)$. (It can be assumed that $R(A) = \{0\}$, by (e). Suppose that $R'(A) \neq \{0\}$. By (d), there exists a minimal non-zero ideal $I$ of $A$ contained in $R'(A)$. Using (f) and Chapter I, § 5, Corollary 1 to Theorem 1, show that $\dim I = 1$, whence a contradiction.)

(h) Let $P$ be the radical of $R'(A)$, $B$ a semi-simple subalgebra of $A$ and $C = P + B$ which is a subalgebra of $A$ by (d). Using (f), show that $\mathrm{ad}_P B$ can be expressed in a triangular form with respect to a suitable basis of $P$. Conclude that $[P, B] = \{0\}$.

(k) $R(A)$ is the radical of $R'(A)$. (Use (e), (f), (g), (h) and a Levi decomposition of $A$.) Let $R'(A) = R(A) \oplus T$ be a Levi decomposition of $R'(A)$. By (h), $R'(A) = R(A) \times T$. Then $R(T) = \{0\}$. Applying (c) to the simple factors of $T$, conclude that $T = \{0\}$. It has therefore been shown that $R(A) = R'(A)$.

(l) Let $G$ be a connected real Lie group with Lie algebra $A$. Let $\mathscr{S}(G)$ be the set of subgroups $N$ of $G$ such that there exists a decreasing sequence $(N_m, N_{m-1}, \ldots, N_0)$ of subgroups with the following properties: $N_m = N$, $N_0 = \{e\}$, each $N_i$ is a normal connected Lie subgroup of $G$ and $\dim N_i/N_{i-1} = 1$ for all $i > 0$. Show that the integral subgroup $R(G)$ of $G$ with Lie algebra $R(A)$ is the largest element of $\mathscr{S}(G)$. (Argue by induction on $\dim R(A)$. Let $I$ be a 1-dimensional ideal of $A$ and $N$ the corresponding integral subgroup of $G$. Pass to the quotient by $\overline{N}$. If $N$ is not closed, use Exercise 21 (a) of § 6.)

(m) A Lie subgroup $H$ of $G$ is called extendable if there exists a Lie subgroup $H_1 \supset H$ such that $\dim H_1 = \dim H + 1$. Let $R'(G)$ denote the intersection of all the non-extendable connected Lie subgroups of $G$.

Let $B$ be a non-extendable subalgebra of $A$. Show that the corresponding integral subgroup of $G$ is closed. (Use Proposition 5.) Deduce that

whence $R'(G) \subset R(G)$.

(n) Show that $R(G) = R'(G)$. (Reduce it to the case where $R'(G) = \{e\}$. Then use Exercise 22 of § 6.)†