Bourbaki

Exercise 6, § 4

Lie Groups and Lie Algebras, I, 92

Let gg be a nilpotent Lie algebra and aa a commutative ideal of gg. The following conditions are equivalent: (a) aa is a maximal commutative ideal of gg; (b) aa is a maximal commutative subalgebra of gg; (c) aa is equal to its centralizer a′a' in gg. (The implications not (a) ⇒\Rightarrow not (b) ⇒\Rightarrow not (c) are obvious. If a′≠aa' \neq a, there exists by Proposition 1 an ideal a′′a'' of gg such that a⊂a′′⊂a′a \subset a'' \subset a' and dim⁡a′′/a=1\dim a''/a = 1. Then a′′=a+Kxa'' = a + Kx, hence [a′′,a′′]⊂[a,a]+[x,a]={0}[a'', a''] \subset [a, a] + [x, a] = \{0\} and hence (a) is false.)

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