# Ideals of a Nilpotent Complex Lie Algebra

Let $\frak{g}$ be a finite-dimensional nilpotent complex Lie algebra. What is known about the dimensions of the ideals of $\frak{g}$? For instance, $\frak{g}$ admits an ideal of codimension 1, if $\frak{g}$ is non-trivial. Also, what (if any) results are known concerning the existence of ideals of certain dimensions and the index of $\frak{g}$?

Thanks

-
Probably you are combining too many questions, some too broad for a realistic answer. Since the number of nonisomorphic nilpotent Lie algebras grows very rapidly, the ideal structures seem unapproachable in general. Similarly, the study of "index" is usually geared more restrictively toward nilpotent Lie algebras which arise in the study of semisimple Lie algebras, etc. (I believe for instance that there was a paper by A.N. Panov in that setting.) A more focused question is needed. Searching MathSciNet, if that's accessible, could be useful. –  Jim Humphreys Jul 26 '12 at 20:48
nilpotent $\mathfrak{g}$ admits ideals of all possible dimensions, this is a straightforward exercise (hint: mod out by 1-dimensional central subalgebra). –  Yves Cornulier Jul 27 '12 at 3:30
Since the dimensions of ideals of a nilpotent Lie algebra $\mathfrak{g}$ are not too interesting in general, one might ask for the maximal dimension of ideals with special properties, such as abelian ideals. Let $\mathfrak{g}$ be nilpotent, non-abelian of dimension $n$. If $\beta(\mathfrak{g})$ denotes the maximal dimension of an abelian ideal in $\mathfrak{g}$, then it is known that $$\frac{\sqrt{8n+1}-1}{2}\le \beta(\mathfrak{g})\le n-1.$$ Any abelian subalgebra (ideal) of maximal dimension contains the center.
The index $i(\mathfrak{g})$ of $\mathfrak{g}$, which is mentioned in the question, is closely related to $\beta(\mathfrak{g})$, because we have $$\beta(\mathfrak{g})\le \frac{n+i(\mathfrak{g})}{2}.$$