Let $\phi$ be the continuous mapping $(x, y) \mapsto x^{-1}y^{-1}xy$ of $G \times G$ into G. Then $\phi(A \times B) \subset (A, B)$, hence $\phi(\overline{A} \times \overline{B}) \subset (\overline{A}, \overline{B})$ and hence

the opposite inclusion is obvious and therefore $(\overline{A}, \overline{B}) = (\overline{A}, \overline{B})$. Clearly $\overline{D^0A} = \overline{D^0A}$; assuming the equality $\overline{D^iA} = \overline{D^iA}$, it follows that

and hence $\overline{D^iA} = \overline{D^iA}$ for all $i$. The proof of the formula $\overline{C^iA} = \overline{C^iA}$ is analogous.

§ 9.2

This follows immediately from Proposition 1.

For fixed y in B, the set M_y of (x, y) with x ∈ A is connected (for the mapping x ↦ (x, y) of A into G is continuous). e ∈ M_y, and hence the union R of the M_y with y ∈ B is connected. But (A, B) is the subgroup of G generated by R, whence the proposition.

Let a ∈ h_1, b ∈ h_2. There exist an open neighbourhood I of 0 in K and analytic mappings f_1, f_2 of I into G such that

We write

We identify an open neighbourhood of e in G with an open subset of K' by means of a chart which maps e to 0. Then L(G) is identified with K'. By § 5, no. 2, Proposition 1, the expansion of f(\lambda, \mu) as an integral series about the origin is

where a_{ij} ∈ K' (the terms in \lambda^i and \mu^j in the expansion of f(\lambda, \mu) are zero because f(\lambda, 0) = f(0, \mu) = 0). We fix \mu in I. Letting \lambda tend to 0, we see that Since this is true for all \mu ∈ I, it follows that [a, b] ∈ h.

Suppose that $[L(A), L(B)] \subset L(C)$. The sum $L(A) + L(B) + L(C)$ is a Lie subalgebra of $L(G)$. By considering an integral subgroup of $G$ with Lie algebra $L(A) + L(B) + L(C)$, the problem is reduced to the case where

and $G$ is connected. Then $L(C)$ is an ideal of $L(G)$. Suppose first that $G$ is simply connected. Then $C$ is a normal Lie subgroup of $G$ (§ 6, no. 6, Proposition 14). Let $\phi$ be the canonical morphism of $G$ onto $G/C$. Then

and hence $\phi(A)$ and $\phi(B)$ commute by the Hausdorff formula; therefore $(A, B) \subset C$. In the general case, let $G'$ be the universal covering of $G$ and $A', B', C'$ the integral subgroups of $G'$ such that $L(A') = L(A), L(B') = L(B), L(C') = L(C)$. Then $(A', B') \subset C'$ and $A, B, C$ are the canonical images of $A', B', C'$ in $G$, whence $(A, B) \subset C$. On the other hand, $(A, B)$ is the underlying set of an integral subgroup of $G$ (§ 6, no. 2, Corollary to Proposition 4) and its Lie algebra contains $[L(A), L(B)]$ (Proposition 3). If

then $(A, B) \supset C$, whence $(A, B) = C$.

The first assertion follows from Proposition 4 by induction on $i$. The second follows from the first and § 6, no. 6, Proposition 14.

Let $a = L(A)$. Let $G_1$ be the set of $g \in G$ such that

Then $G_1$ is a closed subgroup of $G$. If $y \in a$, then $\exp y \in G_1$, by § 6, no. 4, Corollary 3 (ii) to Proposition 10. Hence $G_1$ contains $A$ and therefore $\overline{A}$. Thus, for $g \in \overline{A}$, $L(\mathrm{Int}\, g)$ leaves $a$ stable and therefore $\mathrm{Int}\, g$ leaves $A$ stable; more precisely, $L(\mathrm{Int}\, g)$ defines the identity automorphism of $a/\mathscr{D}a$ and hence $\mathrm{Int}\, g$ defines the identity automorphism of $A/DA$. This proves that $(\overline{A}, A) \subset DA$. With the real Lie group structure on $G$, $\overline{A}$ is a Lie subgroup (§ 8, no. 2, Theorem 2); let $b$ be its Lie algebra. Let $G_2$ be the set of $g \in G$ such that

By the above, $G_2 \supseteq A$ and hence $G_2 \supseteq \overline{A}$. Therefore, for $g \in \overline{A}$, Int $g$ leaves DA stable and defines the identity automorphism of $\overline{A}/DA$. Hence $DA \supseteq D\overline{A}$.

Suppose that $[L(A), L(B)] \subset L(C)$. As in the proof of Proposition 4, the problem reduces to the case where $L(C)$ is an ideal of $L(G)$. Then, by replacing G by an open subgroup, it reduces to the case where C is normal in G ($§ 7$, no. 1, Proposition 2). Let $\phi$ be the canonical morphism of G onto $G/C$. Then

By the Hausdorff formula, there exist open subgroups A', B' of A, B such that $\phi(A')$ and $\phi(B')$ commute, whence $(A', B') \subset C$. Suppose further that By Proposition 3, the Lie subalgebra tangent to $(A', B')$ at $e$ contains $L(C)$. Hence $(A', B')$ contains a Lie subgroup germ of G with Lie algebra $L(C)$. Therefore $(A', B')$ is an open subgroup of C.

(a) By Proposition 3 applied inductively, for every open subgroup $G_1$ of G and for all $i$, $D^iG_1$ contains a Lie subgroup germ of G with Lie algebra $\mathscr{D}^ig$.

(b) Let $G'$ be an open subgroup of G such that, for $i \leq n$, $D^iG'$ is a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$. By Proposition 6, there exist open subgroups $H_1, H_2$ of $D^nG'$, such that $(H_1, H_2)$ is a Lie subgroup with Lie algebra $\mathscr{D}^{n+1}g$. Let $G''$ be an open subgroup of $G'$ small enough for $D^nG'' \subset H_1 \cap H_2$. Then $D^{n+1}G'' \subset (H_1, H_2)$. The relations

prove, using (a), that $D^iG''$ is, for $i \leq n + 1$, a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$.

(c) There exists an integer $p$ such that $\mathscr{D}^pg = \mathscr{D}^{p+1}g = \cdots$. By the above, there exists an open subgroup $G_0$ of G such that $D^iG_0$ is, for $i \leq p$, a Lie subgroup of G with Lie algebra $\mathscr{D}^ig$. But, by (a), the same assertion remains true for $i > p$ since $D^pG_0 \supset D^iG_0$ for $i > p$.

(d) The argument is similar for the $C^i$.

This follows from § 3, Proposition 44 and Corollary 2 to Proposition 39.

Suppose that $A$ consists of a single point $a$. Then $Z_G(A)$ is the set of fixed points of $\mathrm{Int}\, a$; hence $Z_G(A)$ is a Lie subgroup of $G$ and $\mathbf{L}(Z_G(A))$ is the set of fixed points of $\mathrm{Ad}\, a$, that is $\mathfrak{z}_g(A)$ (§ 3, no. 8, Corollary 1 to Proposition 29). The general case follows by means of § 6, no. 2, Corollary 3 to Proposition 1.

Let $x \in G$. Then

and hence $Z_G(A) = Z_G(a)$. Let $u \in g$. Then and hence $\mathfrak{z}_g(A) = \mathfrak{z}_g(a)$. The last assertion then follows from Proposition 7 or Proposition 8.

Suppose that $L(G)$ is nilpotent. If $K = \mathbf{R}$ or $\mathbf{C}$, the identity component $G_0$ of G is nilpotent by the Corollary to Proposition 4, no. 2, and $G_0$ is open in G. If K is ultrametric, the Corollary to Proposition 6, no. 2, proves that there exist an open subgroup $G_1$ of G, an integer $i > 0$ and a neighbourhood V of e in G such that $C^iG_1 \cap V = \{e\}$. Then, if $G_0$ is a sufficiently small subgroup of $G_1$, $C^iG_0 \subset V$, hence $C^iG_0 = \{e\}$ and $G_0$ is nilpotent.

Let $g$ be a nilpotent Lie algebra. The Hausdorff series $H(X, Y)$ corresponding to $g$ has only a finite number of non-zero terms and we know (Chapter II, § 6, no. 5, Remark 3) that the law of composition $(x, y) \mapsto H(x, y)$ defines a group structure on $g$. Suppose further that $g$ is complete and normable. Clearly the law H is a continuous-polynomial (Differentiable and Analytic Manifolds, R, Appendix). Hence $g$, together with the law H, is a Lie group G, said to be associated with $g$. By § 4, no. 2, Lemmas 2 and 3, $L(G) = g$. The identity mapping $\phi$ of $g$ into $G$ is an exponential mapping of $G$ such that $\phi(\lambda x)\phi(\lambda'x) = \phi((\lambda + \lambda')x)$ for all $x \in g, \lambda \in K, \lambda' \in K$. Every Lie subalgebra $\mathfrak{h}$ of $g$ admitting a topological supplement is a Lie subgroup $H$ of $G$ and $L(H) = \mathfrak{h}$.

(i) $\exp_G$ is an isomorphism of the Lie group associated with $L(G)$ onto $G$.

(ii) Every integral subgroup of $G$ is a simply connected Lie subgroup of $G$.

Let $g = L(G)$, which is nilpotent (Proposition 12). As two simply connected Lie groups over $\mathbf{R}$ or $\mathbf{C}$ which have the same Lie algebra are isomorphic ($§ 6$, no. 3, Theorem 3 (ii)), it suffices to prove the proposition when $G$ is the group associated with $g$. Then (i) and (ii) follow from what was said before the proposition.

(i) If $G$ is nilpotent, $\exp_G$ is étale and surjective.

(ii) If $K = \mathbf{C}$ and $\exp_G$ is étale, then $G$ is nilpotent.

Let $G'$ be the universal covering space of $G$. Let $\phi$ be the canonical morphism of $G'$ onto $G$. Then $\exp_G = \phi \circ \exp_{G'}$ ($§ 6$, no. 4, Proposition 10) and hence (i) follows from Proposition 13 (i).

If $K = \mathbf{C}$ and $\exp$ is étale, then, for all $x \in L(G)$, and $x$ has no eigenvalue belonging to $2i\pi(\mathbf{Z} - \{0\})$ ($§ 6$, no. 4, Corollary to Proposition 12). Applying this to $\lambda x$, where $\lambda$ varies through $\mathbf{C}$, it follows that all the eigenvalues of $\mathrm{ad}\, x$ are zero and hence that $\mathrm{ad}\, x$ is nilpotent. Therefore $L(G)$ is nilpotent (Chapter I, $§ 4$, Corollary 1 to Theorem 1) and hence $G$ is nilpotent (Proposition 12).

By Proposition 9 of no. 3 it suffices to prove that $Z_G(A)$ is connected. Let $g \in Z_G(A)$. There exists $x \in L(G)$ such that $g = \exp x$ (Proposition 14). Then $\mathrm{Ad}\, g|L(A) = 1$ (no. 3, Proposition 9), hence $\mathrm{Ad}\, g^n|L(A) = 1$ for all $n \in \mathbf{Z}$ and hence $\exp(\mathrm{ad}\, nx)|L(A) = 1$ for all $n \in \mathbf{Z}$. As the mapping

of $K$ into $\mathscr{L}(L(A), L(G))$ is polynomial, $\exp(\mathrm{ad}\, \lambda x)|L(A) = 1$ for all $\lambda \in K$, that is $\exp(\lambda x) \in Z_G(A)$ for all $\lambda \in K$.

$N_G(A) \neq A$ (Algebra, Chapter I, § 6, Corollary 1 to Proposition 8). By Proposition 11 of no. 4, we need only prove that $N_G(A)$ is connected. Let $g \in N_G(A)$. There exists $x \in L(G)$ such that $g = \exp x$ (Proposition 14). Let $E$ be the vector subspace of $\mathscr{L}(L(G))$ consisting of the $u \in \mathscr{L}(L(G))$ such that $u(L(A)) \subset L(A)$. Then $\mathrm{Ad}\, g^n \in E$ and hence $\exp(\mathrm{ad}\, nx) \in E$ for all $n \in \mathbf{Z}$. Hence $\exp(\mathrm{ad}\, \lambda x) \in E$ for all $\lambda \in K$, that is $\exp(\lambda x) \in N_G(A)$ for all $\lambda \in K$.

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 $g \in G$.