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

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

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.