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I am looking for definition of polynilpotent Lie algebras. Is there any equivalent concept for that?

Nil
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2 Answers2

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Usually poly-P means that there is a subnormal series (or normal series; here it doesn't matter) in which all subquotients have Property P.

When P is "abelian" or "solvable", clearly poly-P is the same as solvable. Since nilpotent is between solvable and abelian, poly-nilpotent would just be the same as solvable.

So the definition has no interest. Unless one counts the number of steps. Then being $n$-step poly-nilpotent, i.e. polynilpotent with a subnormal series of length $n$ has a reasonable meaning. For instance for $n=2$ it is called meta-nilpotent.

Over a field, every solvable finite-dimensional Lie algebra is nilpotent-by-abelian, hence 2-step polynilpotent.

In infinite dimension one can construct solvable Lie algebra of arbitrary large minimal step of polynilpotency.

YCor
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  • Thank you very much for your answer. How can I define the concept of poly-nilpotent for Leibniz algebras? The definition of nilpotency of Leibniz algebras is similar to Lie algebras. – Nil Apr 24 '19 at 15:03
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    Just by thinking about it. – YCor Apr 24 '19 at 16:44
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If $L$ is a Lie algebra then by iteration the lower central series is defined $L^1=L$; $L^{i+1}=[L; L^{i}]$; $i = 1,2, \dots$. Now $L$ is called nilpotent of class $s$ iff $L^{s+1}=0$ and $L^{s} \neq 0$. All Lie algebras of class $s$ form a variety $N_s$. So, $L$ is called polynilpotent with tuple $(s_q, \dots, s_2, s_1)$ iff there is a chain of ideals $0=L_{q+1} \subset L_{q} \subset \dots L_{2} \subset L_{1}=L $ with $L_{i} / L_{i+1} \in N_{s_{i}}$.

Nil
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    This answer sounds perfectly meaningful and I was puzzled by the downvote. Possibly they are due to the answerer being the same as the OP, which I agree sounds funny. – YCor Apr 23 '19 at 22:02
  • Minor correction: you mean Lie algebras of class $\le s$ form a variety, not class $=s$ (variety should be closed under taking quotients). – YCor Apr 23 '19 at 22:02