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I am reading Humphreys' book "Introduction to Lie algebras and representation theory."

A Lie algebra $\mathfrak{g}$ is called nilpotent if $[\mathfrak{g},[\mathfrak{g},\cdots,[\mathfrak{g},\mathfrak{g}]]]]]=\{0\}$ for some number of repetitions. What are some enlightening examples of nilpotent Lie algebras? Are there any examples of nilpotent Lie algebras for which Engel's theorem is useful in establishing nilpotency?

I am aware that the Lie algebra of strictly upper triangular matrices is nilpotent, as seems very straightforward to verify (Humphreys cites this as a consequence of Engel's theorem but this seems like an overcomplication).

Wikipedia lists two extra examples:

A "Cartan subalgebra" is always nilpotent, but this appears to be part of the definition, so it seems like examples can only be as interesting as good examples of Lie algebras which arise as Cartan subalgebras. Are there any?

And (according to wikipedia) if a Lie algebra has an automorphism (I assume an automorphism of the Lie algebra and not of the vector space) of prime period and no nontrivial fixed points, then the Lie algebra is nilpotent. I have no idea how natural this is and I have no idea what examples of this would look like.

I would prefer examples which are real or complex Lie algebras but I would be interested in any answer.

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    $\begingroup$ It's a bit broad. First, every nilpotent Lie algebra is Cartan subalgebra of itself. So considering Cartan algebras does not really help in "producing" Lie algebras; it is rather interesting to understand the structure of larger Lie algebras (in the most fundamental example, namely semisimple, these are abelian, so not really interesting per sé). $\endgroup$
    – YCor
    Jun 9, 2019 at 10:52
  • $\begingroup$ If a Lie algebra has a grading in positive integers, then it is nilpotent. The converse is not true (but is true in dimension $\le 6$, for any field except maybe characteristic 2,3). To get an idea of the classification in dimension $\le 6$, check de Graaf's paper projecteuclid.org/euclid.em/1120145567. $\endgroup$
    – YCor
    Jun 9, 2019 at 10:53
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    $\begingroup$ Given any filtration (where $0=V_0$ and $V_n=V$) $$ V_0\subset V_1\subset\cdots\subset V_n $$ of a vector space, one may form the lie algebra $\mathfrak{g}$ of all transformations $X$ such that $XV_i\subseteq V_{i-1}$ for each $i$. $\endgroup$
    – anon
    Jun 9, 2019 at 19:54
  • $\begingroup$ thanks YCor. arctic tern, isn't that example (non-canonically) identical to the example of strictly upper triangular matrices? $\endgroup$
    – Quarto Bendir
    Jun 11, 2019 at 19:37
  • $\begingroup$ Almost, by filtration I did not necessarily mean a complete flag but potentially a partial flag. This is essentially the block strictly upper triangular matrices (for some choice of how to split up into blocks, possibly of different sizes). $\endgroup$
    – anon
    Jul 13, 2019 at 1:17

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"What are some enlightening examples of nilpotent Lie algebras? " The answer depends on what you will find enlightening, and in which context. There are several examples arising from geometry. To give an example, there is a paper by Yves Benoist "Une nilvariété non affine" which corresponds to a "very interesting" $11$-dimensional nilpotent Lie algebra having no faithful representation of degree $12$. See also the paper Affine structures on nilmanifolds in this context. Furthermore, interesting nilpotent Lie algebras also arise from group theory, e.g., from $p$-groups. And Zelmanov received a fields medal involving a Lie algebra nilpotency theorem (for the solution of the Restricted Burnside Problem), see for example here.

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