Bourbaki

Proposition 7

Tag 09W8, permanent. Cite it as [Bourbaki, Tag 09W8].

Lie Groups and Lie Algebras, chapter I, § 3. Representations

Let gg be a Lie algebra, β\beta an invariant symmetric bilinear form on gg and aa an ideal of gg.

(a) The orthogonal a′a' of aa with respect to β\beta is an ideal of gg. (b) If aa is characteristic and β\beta is completely invariant, a′a' is characteristic. (c) If β\beta is non-degenerate, a∩a′a \cap a' is commutative.

Let DD be a derivation of gg. Suppose that aa is stable under DD and that β(Dx,y)+β(x,Dy)=0\beta(Dx, y) + \beta(x, Dy) = 0 for x,yx, y in gg. Then z∈a′z \in a' implies Dz∈a′Dz \in a', since, for all t∈at \in a, Dt∈aDt \in a and hence β(Dz,t)=−β(z,Dt)=0\beta(Dz, t) = -\beta(z, Dt) = 0. Thus a′a' is stable under DD. This establishes (a) and (b).

Now let bb be an ideal of gg and suppose that the restriction of β\beta to bb is zero. For x,yx, y in bb and z∈gz \in g, β([x,y],z)=β(x,[y,z])=0\beta([x, y], z) = \beta(x, [y, z]) = 0, for [y,z]∈b[y, z] \in b. Thus [b,b][b, b] is orthogonal to gg. If β\beta is non-degenerate, bb is therefore commutative. This result applied to a∩a′a \cap a' proves (c).

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

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