Bourbaki

Corollary 1

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

Lie Groups and Lie Algebras, chapter I, § 6. Semi-simple Lie algebras

Let gg be a Lie algebra, β\beta its Killing form and aa a semi-simple sub-algebra of gg. The orthogonal hh of aa with respect to β\beta is a supplementary subspace of aa in gg and [a,h]⊂h[a, h] \subset h. If aa is an ideal of gg, so is hh, which is then the centralizer of aa in gg.

Let β′\beta' be the restriction of β\beta to aa: it is the bilinear form associated with the representation x↦ad⁡gxx \mapsto \operatorname{ad}_g x of aa on the space gg. This representation is faithful and hence β′\beta' is non-degenerate (Proposition 1). Hence hh is supplementary to aa in gg. On the other hand, if x,yx, y are in aa and z∈hz \in h, then

β(x,[y,z])=β([x,y],z)=0, \beta(x, [y, z]) = \beta([x, y], z) = 0,

since [x,y]∈a[x, y] \in a, and hence [y,z]∈h[y, z] \in h, which proves that [a,h]⊂h[a, h] \subset h. If aa is an ideal of gg, we know that hh is an ideal of gg (§3§ 3, Proposition 7) and gg is identified with a×ha \times h. As the centre of aa is zero, the centralizer of aa in gg is hh.

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.