Bourbaki

Proposition 1

Tag 09TO, permanent. Cite it as [Bourbaki, Tag 09TO].

Lie Groups and Lie Algebras, chapter I, § 1. Definition of Lie algebras

*Let g be a Lie algebra. For all x∈gx \in g, ad⁡x\operatorname{ad} x is a derivation. The mapping x↦ad⁡xx \mapsto \operatorname{ad} x is a homomorphism of the Lie algebra g into the Lie algebra d\mathfrak{d} of derivations of g. If D∈dD \in \mathfrak{d} and x∈gx \in g, [D,ad⁡x]=ad⁡(Dx)[D, \operatorname{ad} x] = \operatorname{ad}(Dx).

Identity (4) can be written:

(ad⁡x)⋅[y,z]=[(ad⁡x)⋅y,z]+[y,(ad⁡x)⋅z] (\operatorname{ad} x) \cdot [y, z] = [(\operatorname{ad} x) \cdot y, z] + [y, (\operatorname{ad} x) \cdot z]

or:

(ad⁡[x,y])⋅z=(ad⁡x)⋅((ad⁡y)⋅z)−(ad⁡y)⋅((ad⁡x)⋅z) (\operatorname{ad}[x, y]) \cdot z = (\operatorname{ad} x) \cdot ((\operatorname{ad} y) \cdot z) - (\operatorname{ad} y) \cdot ((\operatorname{ad} x) \cdot z)

whence the first two assertions. On the other hand, if D∈dD \in \mathfrak{d}, x∈gx \in g, y∈gy \in g, then [D,ad⁡x]⋅y=D([x,y])−[x,Dy]=[Dx,y]=(ad⁡Dx)⋅y[D, \operatorname{ad} x] \cdot y = D([x, y]) - [x, Dy] = [Dx, y] = (\operatorname{ad} Dx) \cdot y, whence the last assertion.

The mapping ad⁡x\operatorname{ad} x is also called the inner derivation defined by xx.

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.