Bourbaki

Proposition 21

Tag 0AJA, permanent. Cite it as [Bourbaki, Tag 0AJA].

Lie Groups and Lie Algebras, chapter III, § 9. Commutators, centralizers and normalizers in a Lie group

Let GG be a simply connected finite-dimensional solvable Lie group over R\mathbf{R} or C\mathbf{C} and MM an integral subgroup of GG. Then MM is a Lie subgroup of GG and is simply connected.

We continue to use the notation n,g,gi,xi,ϕn, g, g_i, x_i, \phi of Proposition 20 but impose on the xix_i the following supplementary condition: let ip>ip−1>⋯>i1i_p > i_{p-1} > \cdots > i_1 be the integers ii such that L(M)∩gi≠L(M)∩gi−1L(M) \cap g_i \neq L(M) \cap g_{i-1}; then we take

xik∈L(M)∩gik x_{i_k} \in L(M) \cap g_{i_k}

\dagger Such a sequence exists by Chapter I, § 5, Proposition 2.

for k=1,2,…,pk = 1, 2, \ldots, p. By induction on nn, it is easily seen that (xip,xip−1,…,xi1)(x_{i_p}, x_{i_{p-1}}, \ldots, x_{i_1}) is a basis of L(M)L(M). Let NN be a simply connected Lie group such that there exists an isomorphism hh of L(N)L(N) onto L(M)L(M). Let

yp=h−1(xip),…,y1=h−1(xi1). y_p = h^{-1}(x_{i_p}), \ldots, y_1 = h^{-1}(x_{i_1}).
By Proposition 20, the mapping
(λ1,λ2,…,λp)↦(exp⁡λ1y1)(exp⁡λ2y2)⋯(exp⁡λpyp) (\lambda_1, \lambda_2, \ldots, \lambda_p) \mapsto (\exp \lambda_1 y_1)(\exp \lambda_2 y_2) \cdots (\exp \lambda_p y_p)
is an isomorphism of the manifold KpK^p onto the manifold NN. There exists a Lie group morphism τ\tau of NN into GG such that h=L(τ)h = L(\tau) and τ(N)=M\tau(N) = M (§ 6, no. 2, Corollary 1 to Proposition 1). Hence MM is the set of elements of GG of the form
τ((exp⁡λ1y1)⋯(exp⁡λpyp))=exp⁡(λ1L(τ)y1)⋯exp⁡(λpL(τ)yp)=exp⁡(λ1xi1)⋯exp⁡(λpxip). \tau((\exp \lambda_1 y_1) \cdots (\exp \lambda_p y_p)) = \exp(\lambda_1 L(\tau)y_1) \cdots \exp(\lambda_p L(\tau)y_p) = \exp(\lambda_1 x_{i_1}) \cdots \exp(\lambda_p x_{i_p}).
Thus M=ϕ(T)M = \phi(T) where TT is a vector subspace of KnK^n.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.