Suppose $M$ is a smooth manifold with Riemannian metric $g$. Recently I have dealt with some problem which lead me to the following question:
Can a Lie algebra of Killing vector fields on $M$ has a nilpotent Lie subalgebra which is not abelian?
I have thought about it a while and no idea has come to my mind. Googling didn't give me an answear either so I would be gratefoul for help/references.
If $G$ is a (nilpotent) Lie group with a left invariant metric, its Lie algebra is a (nilpotent) Lie algebra of killing vector fields. So the answer is obviously yes for non compact manifolds.
Now, if $M$ is compact, its isometry group is a compact lie group, and admits therefore a bi-invariant Riemanian metric. Therefore its Lie algebra admits an $ad$-invariant scalar product. But a non abelian nilpotent Lie algebra cannot admits such a scalar product, has the orthogonal of its center would be a Lie subalgebra.