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If $\cal h$ is a nonzero ideal in a nilpotent Lie algebra $\cal g$. How to prove that $\mathcal h\cap Z(\mathcal g)\not =0$, where $Z(\mathcal g)$ is the center of $\mathcal g$?

Ronald
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1 Answers1

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Let $I$ be a nontrivial ideal of $L$. Then $L$ acts on $I$ by the adjoint action, because $I$ is an ideal. Then by Engel's theorem (or lemma to it), applied to the Lie algebra ${\rm ad}(L)$, there exists a $v\neq 0$ in $I$ with $0={\rm ad}(L).v=[L,v]$, because $L$ is nilpotent. But this just means that $$ v\in I\cap Z(L), $$ so that the intersection is nontrivial. In particular, the center of a nilpotent Lie algebra itself is nontrivial.

Dietrich Burde
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  • This seems to show that the center is non-trivial, rather than it intersecting all non-zero ideals. – Tobias Kildetoft Feb 16 '16 at 19:58
  • Yes, thank you, I was thinking of the nontrivial center, indeed. – Dietrich Burde Feb 16 '16 at 20:06
  • Do you happen to know the answer to his earlier question, about intersecting the last non-zero term of the central series? I think there should be a counterexample (as I mentioned in the comments above), but it is eluding me right now. – Tobias Kildetoft Feb 16 '16 at 20:09
  • Well, if the Lie algebra has a direct abelian factor it cannot be true. – Dietrich Burde Feb 16 '16 at 20:44
  • I have a question regarding the equation $0=L.v=[L,v]$, which holds becasue $L$ is nilpotent. It seems that this holds by the theorem of chapter $3.3$ of Humphreys. However, for that it requires the Lie algebra $L$ to be made up of nilpotent endomorphisms. But for a nilpotent Lie algebra, its elements are ad-nilpotent, and not nilpotent. Thus how can you use this result? – QuantizedObject Jun 02 '25 at 16:36
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    @QuantizedObject But $ad(L)$ is nilpotent and consists of nilpotent endomorphisms. Then apply Theorem $3.3$ to the Lie algebra $ad(L)$. It follows $ad(L).v=0$, which we want. – Dietrich Burde Jun 02 '25 at 18:12