§ 4. Nilpotent Lie algebras

Recall that henceforth $K$ denotes a commutative field. In the rest of the chapter the Lie algebras are assumed to be finite-dimensional over $K$ .

1. Definition of nilpotent Lie algebras

Definition 1

A Lie algebra $g$ is called nilpotent if there exists a decreasing finite sequence of ideals $(g_i)_{1 \leq i \leq p}$ of $g$ with $g_0 = g, g_p = \{0\}$ , such that* $[g, g_i] \subset g_{i+1}$ for $0 \leq i < p$ .

A commutative Lie algebra is nilpotent.

Proposition 1

Let $g$ be a Lie algebra. The following conditions are equivalent:* (a) $g$ is nilpotent; (b) $\mathscr{C}^k g = \{0\}$ for sufficiently large $k$ ; (c) $\mathscr{C}_k g = g$ for sufficiently large $k$ ; (d) there exists an integer $k$ such that $\operatorname{ad} x_1 \circ \operatorname{ad} x_2 \circ \cdots \circ \operatorname{ad} x_k = 0$ for all elements $x_1, x_2, \ldots, x_k$ in $g$ ; (e) there exists a decreasing sequence of ideals $(g_i)_{0 \leq i \leq n}$ of $g$ with $g_0 = g, g_n = \{0\}$ , such that $[g, g_i] \subset g_{i+1}$ and $\dim g_i / g_{i+1} = 1$ for $0 \leq i < n$ .

If $\mathscr{C}^k g = \{0\}$ (resp. $\mathscr{C}_k g = g$ ), clearly the sequence $\mathscr{C}^1 g, \ldots, \mathscr{C}^k g$ (resp. $\mathscr{C}_k g, \mathscr{C}_{k-1} g, \ldots, \mathscr{C}_0 g$ ) has the properties of Definition 1 and hence $g$ is nilpotent. Conversely, suppose that there exists a sequence $(g_i)_{0 \leq i \leq p}$ with the properties of Definition 1. It is seen by induction on $n$ that $g_i \supset \mathscr{C}^{i+1} g$ and $g_{p-i} \subset \mathscr{C}_i g$ . Hence $\mathscr{C}^{p+1} g = \{0\}$ and $\mathscr{C}_p g = g$ . We have thus proved that conditions (a), (b) and (c) are equivalent. On the other hand, $\mathscr{C}^1 g$ is the set of linear combinations of elements of the form

[x_1, [x_2, \ldots, [x_{i-2}, [x_{i-1}, x_i]] \ldots]]

where $x_1, x_2, \ldots, x_i$ run through $g$ . Hence conditions (b) and (d) are equivalent. Finally, if there exists a sequence $(g_i)_{0 \leq i \leq p}$ of ideals with the properties of

Definition 1, there exists a decreasing sequence $(\mathfrak{h}_i)_{0 \leq i \leq n}$ of vector subspaces of $g$ of dimensions $n, n-1, n-2, \ldots, 0$ and a sequence of indices

i_0 < i_1 < \cdots < i_p

with

g_0 = \mathfrak{h}_{i_0}, g_1 = \mathfrak{h}_{i_1}, \ldots, g_p = \mathfrak{h}_{i_p}

; then as

[g, \mathfrak{h}_{i_k}] \subset \mathfrak{h}_{i_{k+1}}

the

\mathfrak{h}_i

are ideals and

[g, \mathfrak{h}_i] \subset \mathfrak{h}_{i+1}

for all

i

. Hence conditions (a) and (e) are equivalent.

Corollary 1

The centre of a non-zero nilpotent Lie algebra is non-zero.

Corollary 2

The Killing form of a nilpotent Lie algebra is zero.* For all $x$ and $y$ in a nilpotent Lie algebra $\operatorname{ad} x \circ \operatorname{ad} y$ is nilpotent and hence of zero trace.

Proposition 2

Subalgebras, quotient algebras and central extensions of a nilpotent Lie algebra are nilpotent. A finite product of nilpotent Lie algebras is a nilpotent Lie algebra.* Let $g$ be a Lie algebra, $g'$ a subalgebra of $g$ , $\mathfrak{h}$ an ideal of $g$ , $\mathfrak{k} = g/\mathfrak{h}$ and $\phi$ the canonical mapping of $g$ onto $\mathfrak{k}$ . If $g$ is nilpotent, then $\mathscr{C}^k g = \{0\}$ for some integer $k$ , hence $\mathscr{C}^k g' \subset \mathscr{C}^k g = \{0\}$ and $\mathscr{C}^k \mathfrak{k} = \phi(\mathscr{C}^k g) = \{0\}$ and hence $g'$ and $\mathfrak{k}$ are nilpotent. If $\mathfrak{k}$ is nilpotent and $\mathfrak{h}$ is contained in the centre of $g$ , then $\mathscr{C}^k \mathfrak{k} = \{0\}$ for some integer $k$ , hence $\mathscr{C}^k g \subset \mathfrak{h}$ and therefore $\mathscr{C}^{k+1} g \subset [\mathfrak{h}, g] = \{0\}$ , so that $g$ is nilpotent. Finally, the assertion concerning products follows for example from the assertion (a) $\Leftrightarrow$ (d) of Proposition 1.

Definition 1 and Proposition 2 show that nilpotent Lie algebras are precisely the algebras obtained from commutative Lie algebras by a sequence of central extensions.

Proposition 3

Let $g$ be a nilpotent Lie algebra and $\mathfrak{h}$ a subalgebra of $g$ distinct from $g$ . The normalizer of $\mathfrak{h}$ in $g$ is distinct from $\mathfrak{h}$ .* Let $k$ be the greatest integer such that $\mathscr{C}^k g + \mathfrak{h} \neq \mathfrak{h}$ . Then

[\mathscr{C}^k g + \mathfrak{h}, \mathfrak{h}] \subset \mathscr{C}^{k+1} g + \mathfrak{h} \subset \mathfrak{h}

and hence the normalizer of

\mathfrak{h}

g

contains

\mathscr{C}^k g + \mathfrak{h}

2. Engel's theorem

Lemma 1

Let $V$ be a vector space over $\mathbf{K}$ . If $x$ is a nilpotent endomorphism of $V$ , the mapping $y \mapsto [x, y]$ of $\mathscr{L}(V)$ into $\mathscr{L}(V)$ is nilpotent.* If $f$ denotes this mapping, $f^m(y)$ is a sum of terms of the form $\pm x^i y x^j$ with $i + j = m$ . If $x^k = 0$ , then $f^{2k-1}(y) = 0$ for all $y$ .

Theorem 1 (Engel)

Let $V$ be a vector space over $\mathbf{K}$ and $g$ a finite-dimensional subalgebra of $\mathfrak{gl}(V)$ whose elements are nilpotent endomorphisms of $V$ . If $V \neq \{0\}$ , there exists $u \neq 0$ in $V$ such that $x.u = 0$ for all $x \in g$ .

The proof proceeds by induction on the dimension $n$ of $g$ . The theorem is obvious if $n = 0$ . Suppose that it is true for algebras of dimension $< n$ .

Let $h$ be a Lie subalgebra of $g$ of dimension $m < n$ . If $x \in h$ , and $\mathrm{ad}_g x$ maps $h$ into itself and defines on passing to the quotient an endomorphism $\sigma(x)$ of the space $g/h$ . By Lemma 1 $\mathrm{ad}_g x$ is nilpotent and hence $\sigma(x)$ is nilpotent. By the induction hypothesis there exists a non-zero element of $g/h$ which is annihilated by all the $\sigma(x),\ x \in h$ . It follows that $h$ is an ideal in a certain $(m+1)$ -dimensional subalgebra of $g$ .

We conclude (by iteration starting with $h = \{0\}$ ) that $g$ has an ideal $h$ of dimension $n-1$ . Let $a \in g,\ a \notin h$ . We again use the induction hypothesis: the $u \in V$ such that $x.u = 0$ for all $x \in h$ form a non-zero vector subspace U of V. This subspace is stable under $a$ (§ 3, no. 5, Proposition 5). Since $a$ is a nilpotent endomorphism of V, there exists a non-zero element of U which is annihilated by $a$ and hence by every element of $g$ .

Corollary 1

For a Lie algebra g to be nilpotent, it is necessary and sufficient that, for all $x \in g$ , $\mathrm{ad}\ x$ be nilpotent.

The condition is necessary (Proposition 1). Suppose that its sufficiency has been proved for Lie algebras of dimension $< n$ ( $n \neq 0$ ). Let $g$ be an $n$ -dimensional Lie algebra such that, for all $x \in g$ , $\mathrm{ad}\ x$ is nilpotent. Theorem 1, applied to the set of $\mathrm{ad}\ x\ (x \in g)$ , proves that the centre $c$ of $g$ is non-zero. Then $g$ is a central extension of the Lie algebra $g/c$ , which is nilpotent by our induction hypothesis. The proof is completed by applying Proposition 2.

Corollary 2

Let g be a Lie algebra and h an ideal of g. Suppose that $g/h$ is nilpotent and that, for all $x \in g$ , the restriction of $\mathrm{ad}\ x$ to h is nilpotent. Then g is nilpotent.

Let $x \in g$ . As $g/h$ is nilpotent, there exists an integer $k$ such that $(\mathrm{ad}\ x)^k(g) \subset h$ . By hypothesis there exists an integer $k'$ such that $(\mathrm{ad}\ x)^{k'}(h) = \{0\}$ . Hence $(\mathrm{ad}\ x)^{k+k'}(g) = \{0\}$ . Corollary 2 is hence a consequence of Corollary 1.

Corollary 3

Let V be a vector space and g a finite-dimensional subalgebra of $\mathfrak{gl}(V)$ whose elements are nilpotent endomorphisms of V. Then g is a nilpotent Lie algebra.

This follows immediately from Lemma 1 and Corollary 1.

Example

The algebra $\mathfrak{n}(n, K)$ (§ 1, no. 2, Example 2 (3)) is nilpotent.

3. The largest nilpotency ideal of a representation

Lemma 2

Let g be a Lie algebra, a an ideal of g and M a simple g-module. If, for all $x \in a,\ x_M$ is nilpotent, then $x_M = 0$ for all $x \in a$ .

Let N be the subspace of M consisting of the $m \in M$ such that $x_M.m = 0$ for all $x \in a$ . By Theorem 1, $N \neq \{0\}$ . On the other hand, for all $y \in g$ , N is stable under $y_M$ (§ 3, no. 5, Proposition 5). Hence $N = M$ , which proves the lemma.

Lemma 3

Let g be a Lie algebra, a an ideal of g, M a g-module of finite dimension over

K and $(M_i)_{0 \leq i \leq n}$ a Jordan-Hölder series of the $g$ -module $M$ . The following conditions are equivalent:

(a) for all $x \in a$ , $x_M$ is nilpotent; (b) for all $x \in a$ , $x_M$ is in the Jacobson radical of the associative algebra $A$ generated by 1 and the $y_M$ where $y \in g$ ; (c) for all $x \in a$ ,

x_M(M_0) \subset M_1, x_M(M_1) \subset M_2, \ldots, x_M(M_{n-1}) \subset M_n.

If these conditions are fulfilled, $a$ is orthogonal to $g$ with respect to the bilinear form associated with the $g$ -module $M$ .

(b) $\Rightarrow$ (a): as $A$ is finite-dimensional over $K$ , the Jacobson radical of $A$ is a nilpotent ideal (Algebra, Chapter VIII, § 6, no. 4, Theorem 3) and hence every element of this radical is nilpotent.

(a) $\Rightarrow$ (c): each $Q_i = M_i/M_{i+1}$ ( $0 \leq i < n$ ) is a simple $g$ -module. For all $x \in a$ , the endomorphism $x_{Q_i}$ (which is derived from $x_M$ by restricting to $M_i$ and passing to the quotient) is nilpotent if condition (a) holds and hence zero by Lemma 2; in other words, $x_M(M_i) \subset M_{i+1}$ .

(c) $\Rightarrow$ (b): suppose condition (c) holds; let $x \in a$ and $z \in A$ . Then $z(M_i) \subset M_i$ ( $0 \leq i < n$ ) and hence $(zx_M)^n(M) = \{0\}$ ; thus $Ax_M$ is a left nilideal of $A$ and hence is contained in the Jacobson radical of $A$ (Algebra, Chapter VIII, § 6, no. 3, Corollary 3 to Theorem 1).

Finally, suppose conditions (a), (b) and (c) hold. Let $x \in a$ and $y \in g$ . We have just seen that $y_M x_M$ is nilpotent and hence $\operatorname{Tr}(y_M x_M) = 0$ , which proves the last assertion of the lemma.

Proposition 4

Let $g$ be a Lie algebra, $M$ a $g$ -module of finite dimension over $K$ and $A$ the associative algebra generated by 1 and the set of $x_M$ ( $x \in g$ ).

(a) The ideals $a$ of $g$ such that $x_M$ is nilpotent for all $x \in a$ are all contained in one of them, $n$ .

(b) The ideal $n$ is the set of $x \in g$ such that $x_M$ belongs to the Jacobson radical of $A$ .

(c) Let $(M_i)_{0 \leq i \leq n}$ be a Jordan-Hölder series of the $g$ -module $M$ ; then $n$ is also the set of $x \in g$ such that $(x)_{M_i/M_{i+1}} = 0$ for all $i$ .

(d) $n$ is orthogonal to $g$ with respect to the bilinear form associated with $\varphi$ .

The set of $x \in g$ such that $x_M$ belongs to the Jacobson radical of $A$ is obviously an ideal of $g$ . The proposition then follows immediately from Lemma 3.

Definition 2

The ideal $n$ of Proposition 4 is called the largest nilpotency ideal for the $g$ -module $M$ or the largest nilpotency ideal of the corresponding representation.

Clearly $n$ contains the kernel of this representation. It equals it when $M$ is semi-simple (Proposition 4 (c)), but not in general. It should be noted that an element of $x$ of $g$ such that $x_M$ is nilpotent does not necessarily belong to $n$ .

We also note that a particular case of Lemma 3 immediately gives the following result:

Proposition 5

Let $V$ be a vector space of finite dimension $n$ over $K$ and $g$ a Lie sub-algebra of $gl(V)$ whose elements are nilpotent endomorphisms of $V$ . Then there exists a decreasing sequence of vector subspaces $V_0, V_1, \ldots, V_n$ of $V$ , of dimensions $n, n-1, \ldots, 0$ , such that $x(V_i) \subset V_{i+1}$ for all $x \in g$ and $i = 0, 1, \ldots, n-1$ .

4. The largest nilpotent ideal of a Lie algebra

Let $g$ be a Lie algebra and $a$ an ideal of $g$ . For $a$ to be nilpotent, it is necessary and sufficient that, for all $x \in a$ , $\operatorname{ad}_g x$ be nilpotent; the condition is obviously sufficient and is necessary, for, if $a$ is nilpotent and $x \in a$ , $\operatorname{ad}_a x$ is nilpotent and $\operatorname{ad}_g x$ maps $g$ into $a$ , hence $\operatorname{ad}_g x$ is nilpotent. Then Proposition 4 applied to the adjoint representation of $g$ gives the following result:

Proposition 6

Let $g$ be a Lie algebra and $E$ the associative subalgebra of $\mathscr{L}(g)$ generated by 1 and the $\operatorname{ad}_g x$ ( $x \in g$ ). Let $R$ be the Jacobson radical of $E$ . (a) The set $n$ of $y \in g$ such that $\operatorname{ad}_g y \in R$ is the largest nilpotent ideal of $g$ . (b) It is orthogonal to $g$ under the Killing form.

It should be noted that $g/n$ can have non-zero nilpotent ideals.

5. Extension of the base field

Let $g$ be a Lie $K$ -algebra, $K_1$ an extension of $K$ and $g' = g_{(K_1)}$ . As $\mathscr{C}^k g' = (\mathscr{C}^k g)_{(K_1)}$ , $g$ is nilpotent if and only if $g'$ is nilpotent.

Let $M$ be a $g$ -module of finite dimension over $K$ , $n$ the largest nilpotency, ideal for $M$ and $M' = M_{(K_1)}$ . Let $(M_i)_{0 \leq i \leq n}$ be a Jordan-Hölder series of the $g$ -module $M$ . Then $x_M(M_i) \subset M_{i+1}$ for all $i$ and all $x \in n$ , hence

x'_{M'}((M_i)_{(K_1)}) \subset (M_{i+1})_{(K_1)}

for all $i$ and all $x' \in n_{(K_1)}$ ; hence $x'_{M'}$ is nilpotent for $x' \in n_{(K_1)}$ so that $n_{(K_1)}$ is contained in the largest nilpotency ideal $n'$ for $M'$ . We shall now see that, if $K_1$ is separable over $K$ , then $n' = n_{(K_1)}$ . Let $E$ be the associative $K$ -algebra generated by 1 and the $x_M$ ( $x \in g$ ), $E'$ the associative $K$ -algebra generated by 1 and the $x_{M'}$ ( $x' \in g'$ ) and $R$ and $R'$ the Jacobson radicals of $E$ and $E'$ . The algebra $E'$ is canonically identified with $E_{(K_1)}$ . Then $R' = R_{(K_1)}$ (Algebra, Chapter VIII, § 7, no. 2, Corollary 2 (c) to Proposition 3). Then let $y' \in n'$ and write

y' = \sum_{i=1}^n \lambda_i y_i,

where the $y_i$ are in $g$ and the $\lambda_i \in K_1$ are linearly independent over $K$ . Then $y'_{M'} = \sum_{i=1}^n \lambda_i (y_i)_{M'}$ and $y'_{M'} \in R' = R_{(K_1)}$ . Hence $(y_i)_M \in R$ and therefore $y_i \in n$ for all $i$ . It follows that $y' \in n_{(K_1)}$ , whence $n' \subset n_{(K_1)}$ .

In particular, if $K_1$ is separable over $K$ , the largest nilpotent ideal of $g_{(K_1)}$ is derived from that of $g$ by extending the base field from $K$ to $K_1$ .

Exercises

The conventions of § 4 remain valid unless otherwise mentioned.

See the exercises for § 4.