Proposition 9

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

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

Let $G$ be a finite-dimensional real or complex Lie group, $g$ its Lie algebra, $A$ an integral subgroup of $G$ and $a = \mathbf{L}(A)$ . Then $Z_G(A) = Z_G(a)$ , $\mathfrak{z}_g(A) = \mathfrak{z}_g(a)$ and $Z_G(A)$ is a Lie subgroup of $G$ with Lie algebra $\mathfrak{z}_g(a)$ .

Let $x \in G$ . Then

x \in Z_G(A) \iff A \subset Z_G(\{x\}) \\ \iff a \subset \mathbf{L}(Z_G(\{x\})) \quad (\text{§ 6, Corollary 2 to Proposition 3}) \\ \iff a \subset \mathfrak{z}_g(\{x\}) \quad (\text{Proposition 8}) \\ \iff x \in Z_G(a)

and hence

Z_G(A) = Z_G(a)

. Let

u \in g

. Then

u \in \mathfrak{z}_g(A) \iff A \subset Z_G(\{u\}) \\ \iff a \subset \mathbf{L}(Z_G(\{u\})) \quad (\text{§ 6, Corollary 2 to Proposition 3}) \\ \iff a \subset \mathfrak{z}_g(\{u\}) \quad (\text{Proposition 7}) \\ \iff u \in \mathfrak{z}_g(a)

and hence

\mathfrak{z}_g(A) = \mathfrak{z}_g(a)

. The last assertion then follows from Proposition 7 or Proposition 8.

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

Proposition 9

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