Proposition 17

Let $g$ be a finite-dimensional nilpotent Lie algebra over $K$ and $(g_0, g_1, \ldots, g_n)$ a decreasing sequence of ideals of $g$ such that $g_0 = g, g_n = \{0\}$ and $[g, g_i] \subset g_{i+1}$ for $0 \leq i < n$. Let $a_1, a_2, \ldots, a_p$ be vector subspaces of $g$ such that each $g_i$ is the direct sum of its intersections with the $a_j$. Let $g$ be given the Hausdorff law of composition $H$. Let $\phi$ be the mapping

of $a_1 \times a_2 \times \cdots \times a_p$ into $g$.

(i) $\phi$ is a bijection of $a_1 \times a_2 \times \cdots \times a_p$ onto $g$; (ii) $\phi$ and $\phi^{-1}$ are polynomial mappings; (iii) the mapping $(x, y) \mapsto \phi^{-1}(\phi(x) \cdot \phi(y)^{-1})$ of $(a_1 \times a_2 \times \cdots \times a_p)^2$ into $a_1 \times a_2 \times \cdots \times a_p$ is polynomial.

The proposition is obvious for $\dim g = 0$. Suppose that $\dim g > 0$ and that the proposition has been established for dimensions $< \dim g$. It can be assumed that $g_{n-1} \neq \{0\}$ and $g_{n-1}$ is then a non-zero central ideal of $g$. There exists an index $j$ such that $h = g_{n-1} \cap a_j \neq \{0\}$. Let $g' = g/h$, $\theta$ be the canonical morphism of $g$ onto $g'$, $g'_i = \theta(g_i)$ and $a'_i = \theta(a_i)$. Then $(g'_0, g'_1, \ldots, g'_n)$ is a decreasing sequence of ideals of $g'$ such that $g'_0 = g', g'_n = \{0\}$,

for $0 \leq i < n$ and each $g'_i$ is the direct sum of its intersections with the $a'_j$. Let $\phi'$ be the mapping

of $a'_1 \times a'_2 \times \cdots \times a'_p$ into $g'$. By the induction hypothesis, $\phi'$ is bijective and $\phi', {\phi'}^{-1}$ are polynomial mappings.

Let $x \in g$. We write

(1)

Then (2) Let $h_1$ be a vector subspace supplementary to $h$ in $g$, the sum of the $a_k$ with $k \neq j$ and a supplement of $h$ in $a_j$. There exists a bijection $\eta$ of $g'$ onto $h_1$ such that $\theta \circ \eta = \mathrm{Id}_{g'}$. For $x \in g$, we write (3) (4)

By (2) and (3), $\theta(\zeta(x)) = \theta(x)$ and hence $y(x) \in \mathfrak{h}$. Finally we write

As $y(x)$ is central in $\mathfrak{g}$, Hence $\phi \circ \psi = \mathrm{Id}_g$. Now let $(x_1, x_2, \ldots, x_p) \in a_1 \times a_2 \times \cdots \times a_p$ and write $x = \phi(x_1, x_2, \ldots, x_p) = x_1 \mathbin{\mathbf{H}} x_2 \mathbin{\mathbf{H}} \cdots \mathbin{\mathbf{H}} x_p$. Then hence $x_i'(x) = \theta(x_i)$ for $1 \leq i \leq p$ and therefore Then by (5) Hence $\psi \circ \phi = \mathrm{Id}_{a_1 \times \ldots \times a_p}$. This proves (i). As the Hausdorff law is polynomial, $\phi$ is polynomial. By the induction hypothesis, ${\phi'}^{-1}$ is polynomial; by formula (1), the functions $x_j'$ are polynomial, hence $\zeta$ is polynomial (formula (3)), $y$ is polynomial (formula (4)) and $\psi$ is polynomial (formula (5)). This proves (ii). Assertion (iii) follows from (i) and (ii) and the fact that the Hausdorff law is polynomial.

Example of a nilpotent Lie group. Let $G$ be the lower strict triangular subgroup of $\mathbf{GL}(n, K)$. It is a Lie subgroup of $\mathbf{GL}(n, K)$ and $L(G) \subset \mathfrak{gl}(n, K)$ is the Lie algebra of lower triangular matrices with zero diagonal (§ 3, no. 10, Proposition 36). By Chapter II, § 4, no. 6, Remark, $G$ is nilpotent. Suppose henceforth that $K = \mathbf{R}$ or $\mathbf{C}$. As $G$ is homeomorphic to $K^{n(n-1)/2}$, $G$ is simply connected. The exponential mapping of $L(G)$ into $G$ is just the mapping

(§ 6, no. 4, Example). By Proposition 13, the exponential is an isomorphism of the manifold $L(G)$ onto the manifold $G$. Proposition 17 of § 6, no. 9 gives the inverse bijection $\log$. We give $K^n$ a norm. By Spectral Theories, Chapter I, § 4, no. 9, for $g \in G$ and $\|g - 1\| < 1$,

that is

But the two sides of (6) are analytic functions of $g$ for $g \in G$ and are therefore equal for all $g \in G$.