I am interested to identify (ideally classify) nilpotent Lie algebras that occur as nilradicals of parabolic subalgebras in (say) reductive Lie algebras. For example, all Heisenberg Lie algebras appear as such, the same holds for free 2-step nilpotent Lie algebras. But what about general free n-step nilpotent Lie algebras?

Let $\mathfrak{m}_n$ denote the free 2-step nilpotent Lie algebra with $n$ generators over $\mathbb{C}$. Consider the simple Lie algebra $\mathfrak{g} = \mathfrak{so}(2n+1)$ and the set of positive roots $$\Delta_+ = \{e_i - e_j, e_i + e_j \mid 1 \leq i \lt j \leq n\} \cup \{e_i \mid 1 \leq i\leq n\}.$$ So we have $$\mathfrak{g} = \mathfrak{n}_- \oplus \mathfrak{gl}(n) \oplus \mathfrak{n}_+$$ whith a root space decomposition of $$\mathfrak{n}_+ = \bigoplus_{i} \mathfrak{g}_{e_i} \oplus \bigoplus_{i \lt j} \mathfrak{g}_{e_i + e_j}$$
We see that $\mathfrak{m}_n \simeq \mathfrak{n}_+ $ appears as a nilradical of the parabolic $\mathfrak{p} = \mathfrak{gl}(n) \oplus \mathfrak{n}_+ $. For a free nilpotent Lie algebra of higher index (say 3), we'd need to find a (perhaps generalized) root system that "matches" the corresponding Hall basis. The reductive part $\mathfrak{gl}(n)$ would stay the same as long as we fix $n$ as the number of generators.

The dim$(\mathfrak{g}^{k}) \gt d^2$ argument in Yves' answer below is indeed strong. Still, I was wondering if one can start to look at this from the other way around. I.e. starting with any finite-dimensional nilpotent Lie algenbra $\mathfrak{n}$ over $\mathbb{C}$, then looking at the derivation algebra Der$(\mathfrak{n})$ of $\mathfrak{n}$, deducting the reductive part $\mathfrak{l}$ of Der$(\mathfrak{n})$ (let's assume that $\mathfrak{n}$ is not characteristically nilpotent) so $\mathfrak{l}$ acts on $\mathfrak{n}$ in a natuarl way. We may then try to treat $\mathfrak{n}$ similar as it was a nilpotent radical of something like $\mathfrak{l} \oplus \mathfrak{n}$ and hence constucting a Lie algebra like $$\mathfrak{g} := \mathfrak{n}_- \oplus \mathfrak{l} \oplus \mathfrak{n}$$ where $\mathfrak{n}_-$ is another copy of $\mathfrak{n}$ analogous to a root space decomposition. This works particularly fine with the free nilpotent Lie algebras $\mathfrak{f}(k,d)$ where in this case we have $\mathfrak{l} = \mathfrak{gl}_d$. I was thinking if it would be possible to define something like roots and a Weyl group to the above constructed $\mathfrak{g}$. Perhaps in the sense of G. Favre, Système de poids sur une algèbre de Lie nilpotente, Manuscripta Math. 9, 1973. However, I'd love to see this in the context of the bigger $\mathfrak{g}$ using the "big" Weyl group $W(\mathfrak{g})$ as well as the "small" Weyl group $W(\mathfrak{l})$. So I was just wondering if at all it makes sense to glue together this $\mathfrak{g}$ as described above. It will in general surely not be semi-simple anymore as we have found out, but perhaps something "similar".

  • $\begingroup$ Does your question concern only free nilpotent Lie algebras or all nilpotent Lie algebras, free nilpotent being only a special case? $\endgroup$ – YCor Mar 3 '14 at 15:20
  • $\begingroup$ I am looking at all nilpotent Lie algebras and mentioned the free nilpotent ones as an example. We could even look at the analogous question for Kac-Moody algebras, but reductive (or semi-simple) would be ok for now. $\endgroup$ – Bizfold Mar 3 '14 at 15:29
  • $\begingroup$ how to you get all free 2-step nilpotent as nilradical of parabolics? $\endgroup$ – YCor Mar 3 '14 at 21:54
  • $\begingroup$ In case of n generators, take so(2n+1) and a well chosen parabolic such that the Levi factor is gl(n). $\endgroup$ – Bizfold Mar 3 '14 at 22:17
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    $\begingroup$ @Bizfold: do I miss something, or isn't it true that any parabolic subalgebra is the product of parabolic subalgebras of the simple factors and of the center? If so it means that the possible nilradicals are direct products of nilradicals of the factors (and hence a non-abelian free nilpotent can only occur in a simple Lie algebra). $\endgroup$ – YCor Mar 4 '14 at 9:16

Among free nilpotent Lie algebras the possible nilradicals of parabolics are exactly (up to 1 case):

  • the abelian ones
  • the 2-nilpotent ones (as you mentioned)
  • the free 3-nilpotent on 2 generators (5-dimensional).

The latter appears as nilradical of a 9-dimensional parabolic subalgebra of the (14-dimensional) exceptional simple Lie algebra of type $G_2$.

It remains to exclude the other ones. One way to do so (which also works for many other nilpotent Lie algebras) is to use the following fact: if $P$ is a parabolic subgroup in a semisimple complex algebraic group with unipotent radical $P_u$, and $A$ is a normal abelian subgroup of $P$ contained in $P_u$, then $P$ has finitely many orbits on $A$. This result is due to G. Röhrle, On normal abelian subgroups in parabolic groups. Ann. Fourier 48(5) (1998) 1455-1482; still I'll only use it in the case of a subgroup central in $P_u$, which Röhrle claims as a standard well-known case (as well as the characteristic zero case, which he attributes to earlier work).

It remains to deduce the above list. Let $\mathfrak{g}$ be a nilpotent Lie algebra of nilpotency length $k$ and whose abelianization $\mathfrak{a}$ has dimension $d$. Let $\mathfrak{g}^k$ be the last term in its lower central series. Then the automorphism group of $\mathfrak{g}$ acts on $\mathfrak{g}^k$, in such a way that those automorphisms acting trivially on $\mathfrak{a}$ also act trivially on $\mathfrak{g}^k$. Hence the action of $\mathrm{Aut}(\mathfrak{g})$ on $\mathfrak{g}^k$ factors through a subgroup of $\mathrm{GL}_d$. Therefore, if $\dim(\mathfrak{g}^k)>d^2$ then the action of $\mathrm{Aut}(\mathfrak{g})$ on $\mathfrak{g}^k$ has infinitely many orbits.

It remains to compute this in the case of the free $k$-nilpotent Lie algebra $\mathfrak{f}(k,d)$ on $d$ generators. For $k=3$, we have $\dim(\mathfrak{f}(3,d)^3)=(d^3-d)/3$, which is $>d^2$ as soon as $d\ge 4$ (for $(k,d)=(3,3)$ see below). For $k=4$, we have $\dim(\mathfrak{f}(4,d)^4)=(d^4-d^2)/4$, which is $>d^2$ as soon as $d\ge 3$ (for $(k,d)=(4,2)$ see below). For $k=5$, we have $\dim(\mathfrak{f}(5,d)^5)=(d^5-d)/5$ which is $>d^2$ for all $d\ge 2$, since $\dim(\mathfrak{f}(k,d)^k)$ increases, for $d$ fixed, when $k\ge 2$, this rules out $k\ge 5$ as well.

It remains $(k,d)=(3,3)$ or $(4,2)$. In the case $(4,2)$, the dimension of $\dim(\mathfrak{f}(4,2)^4)$ is 3, so the Levi part of $P$ should be at least 3 (while it's at most 4) since it embeds into $\mathrm{GL}_2$). The dimension of $\dim(\mathfrak{f}(4,2))$ is $2+1+2+3=8$, so the dimension of the Lie algebra should be $2\times 8+3/\!\!/4=19/\!\!/20$, but there is no simple complex Lie algebra of this dimension (and non-simple semisimple are excluded, in view of the comments). Hence $(4,2)$ is discarded. In the case $(3,3)$, the dimension of $\dim(\mathfrak{f}(3,3)^3)$ is 8, so the Levi part of $P$ should be at least 8 (while it's at most 9 since it embeds into $\mathrm{GL}_3$). The dimension of $\dim(\mathfrak{f}(3,3))$ is $3+3+8=14$. Hence the dimension of the Lie algebra should be $2\times 14+8/\!\!/9=36/\!\!/37$ (// means or). By the classification of simple complex Lie algebras, there are 2 possibilities then, namely $\mathrm{SO}_9$ and $\mathrm{Sp}_8$, both 36-dimensional. But here it means that the Levi subalgebra is $\mathrm{SL}_3$, and it follows (by a simple argument) that $P$ is unimodular, which is a contradiction.

  • $\begingroup$ Motivated by your very helpful answer, i have edited/extended my question above. $\endgroup$ – Bizfold Mar 24 '14 at 19:50

The subclass of nilpotent Lie algebras formed by arbitrary ideals of parabolic subalgebras consisting of nilpotent elements in reductive Lie algebras has been classifed in the article Yu.B. Khakimdzhanov, "Standard subalgebras of reductive Lie algebras" Vestn. Moskov. Univ. Mat. Mekh. : 6 (1974) pp. 49–55 (In Russian) (English abstract). I have not seen the paper, but it seems to me that the nilpotent Lie algebras arising this way are somewhat special. For example, they are graded by positive integers. This excludes already all characteristically nilpotent Lie algebras, i.e., those nilpotent Lie algebras having only nilpotent derivations.

Edit: For the question on free nilpotent Lie algebras: Tamaru has proved in 2007 that the nilradical of any parabolic subalgebra of a (real) semisimple Lie algebra is a so-called Einstein nilradical (A nilpotent Lie algebra which can be a nilradical of a standard Einstein metric solvable Lie algebra is called an Einstein nilradical. ). However, a free $p$-step nilpotent Lie algebra on $m$ generators can only be an Einstein nilradical if (a) $p=1,2$, or (b) $p=3$, $m=2,3,4,5$, (c) $p=4$ and $m=2$ or (d) $p=5$ and $m=2$. This is due to Y.Nikolayevsky, see http://arxiv.org/pdf/math/0612117v1.pdf. So this already excludes most free nilpotent Lie algebras, and I suspect that one can exclude further cases by investigating Khakimdzhanov's classification.

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    $\begingroup$ It is immediate (without any classification) that if the nilradical of a (finite-dimensional complex) Lie algebra is characteristically nilpotent then the Lie algebra is the direct product of a semisimple Lie algebra with its nilradical. $\endgroup$ – YCor Mar 3 '14 at 16:43
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    $\begingroup$ These nilradical are indeed always graded by positive integers, say $\mathfrak{n} = \mathfrak{n}_1 + \cdots +\mathfrak{n}_k$. In fact, there is a so called grading element in the center of the Levi subalgebra $\mathfrak{l}$ of $\mathfrak{p} = \mathfrak{l}\oplus\mathfrak{n}$ whose eigenspaces give the grading. Moreover $\mathfrak{n}$ is generated by the first grading component $\mathfrak{n}_1$. $\endgroup$ – Vít Tuček Mar 3 '14 at 17:51
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    $\begingroup$ By the way, the existence of a grading by positive integers is a much weaker condition than the existence of a grading by positive integers generated in degree 1 (aka Carnot grading). For instance, all complex nilpotent Lie algebra up to dimension 6 satisfy the first condition but many in dimension 5 and 6 fail to satisfy the second one. $\endgroup$ – YCor Mar 3 '14 at 18:10
  • $\begingroup$ Thanks for the comments so far. All free nilpotent Lie algebras are naturally graded and generated by the 1st degree. However, it seems to be counter-intuitive that they "fit" as a nilradical into a parabolic of a reductive Lie algebra. $\endgroup$ – Bizfold Mar 3 '14 at 21:34
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    $\begingroup$ I have checked Khakimdzhanov's paper. It only provides a description of nilradicals of parabolic subalgebras in terms of root systems, which can be found in many other places, but does not provide any explicit list. $\endgroup$ – YCor Mar 11 '14 at 19:31

This is not an answer, just several facts that can help single out parabolic nilradicals which are a bit long for comments.

As was already mentioned by Yves de Cornulier, nilradicals of parabolic subalgebras have a Carnot grading, i.e. a grading in positive integers such that the algebra is generated in degree one.

I have just come across Root Systems for Levi Factors and Borel–de Siebenthal Theory by Kostant, where he proves:

  1. The nilradicals have the same (up to reindexing) lower and upper central series.
  2. There is a refinement of the Carnot grading to a grading by $(\mathrm{cent}\;\mathfrak{l})$-root spaces (which are $\mathfrak{l}$-irreducible) such that the algebra is generated by root spaces corresponding to simple $(\mathrm{cent}\;\mathfrak{l})$-roots.
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    $\begingroup$ Kostant claims the first fact for semisimple Lie algebras, but it sounds false for $\mathfrak{sl}_i\times\mathfrak{sl}_j$ when $i<j$ and minimal parabolic subalgebras. For instance for $(i,j)=(2,3)$, the dimensions of terms is the lower central series, resp. upper central series, are: $(4,1,0)$, resp. $(0,2,4)$. $\endgroup$ – YCor Jun 5 '15 at 12:27

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