Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{p}$ a subalgebra. As we all know, $\mathfrak{p}$ is parabolic if it contains a Borel (thus maximal solvable) subalgebra. In this case, with $\mathfrak{p}^\perp$ the orthocomplement of $\mathfrak{p}$ with respect to the Killing form of $\mathfrak{g}$, $\mathfrak{p}^\perp$ is the nilradical of $\mathfrak{p}$.

There is a handy converse to this statement which goes as follows: a subalgebra $\mathfrak{p}$ of $\mathfrak{g}$ is parabolic if $\mathfrak{p}^\perp$ is a nilpotent (thus central descending series terminates) subalgebra of $\mathfrak{g}$. Note that there is no a priori demand that $\mathfrak{p}^\perp$ is even contained in $\mathfrak{p}$ (though that is, of course. part of the conclusion).

My question: does anyone know a reference for this (not difficult to prove) fact? (I have, in the past, incorrectly attributed it to Grothendieck.)

  • $\begingroup$ I think you need to be more careful about what you mean by nilpotent subalgebra here, since you are implicitly requiring that it consist of "nilpotent" elements in the sense of the abstract Jordan decomposition in $\mathfrak{g}$. A Cartan subalgebra is also nilpotent, for example, but consists of "semisimple" elements. An arbitrary nilpotent subalgebra could involve both types. Unless you assume $\mathfrak{n}$` consists of nilpotent elements, the discussion gets more subtle (and the orthocomplement need not even be a subalgebra of $\mathfrak{g}$) $\endgroup$ Jun 22, 2010 at 12:35
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    $\begingroup$ I think I am being careful, Jim: I only require that $\mathfrak{p}^\perp$ is nilpotent in the usual sense that the central descending series terminates. However, I am definitely requiring that the orthocomplement be a subalgebra, else, as you say, a CSA would provide a counterexample. I will edit my question to make my meaning clearer. $\endgroup$ Jun 22, 2010 at 17:58
  • $\begingroup$ Maybe I understand better: the essential statement not already implied by the Bourbaki theorem is that the orthocomplement of a nilpotent subalgebra containing nonzero semisimple elements is never a Lie subalgebra (since if it were, it would have to be parabolic and thus its orthocomplement in turn would be a nil algebra)? This is somewhat roundabout to state though probably true. I haven't seen it in print, however. $\endgroup$ Jun 22, 2010 at 22:16
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    $\begingroup$ A more "forward" way is to say that if $\mathfrak{n}$ is a nilpotent Lie subalgebra of $\mathfrak{g}$ such that its orthocomplement $\mathfrak{p}:=\mathfrak{n}^\perp$ is a Lie subalgebra then $\mathfrak{p}$ is a parabolic subalgebra (and hence $\mathfrak{n}$ is its nilradical). I can't help thinking that Ozeki and Wakimoto paper, which proves that any polarizable subalgebra is parabolic, is somehow relevant; at least, it gives the right conclusion. $\endgroup$ Jun 23, 2010 at 0:09
  • $\begingroup$ Victor: thank you for drawing the Ozeki-Wakimoto paper to my attention. It is indeed interesting but does not, as far as I can see, prove my statement. In fact, their result seems much deeper and uses 'non-algebraic' considerations: they look at the analytic subgroup corresponding to a w-polarisable subalgebra and see that the resulting coset space is compact whence the subgroup and so, eventually, the subalgebra is parabolic. $\endgroup$ Jun 23, 2010 at 18:29

3 Answers 3


I think that this follows from Bourbaki's Éléments de Mathématique. Groupes et algèbres de Lie, Chapitre VIII, §10, Theorem 1 (see below) applied to the adjoint representation. Alas, I cannot provide the google books link because the book that Google Books claims to be this one, is actually Algèbre commutative, Chapitres 5 à 7! (And the "Feedback" link does not allow me to point this out, since in their arrogance, Google does not even allow for the possibility of such an error!)

Théorème 1. --- Soient $V$ un espace vectoriel de dimension finie, $\mathfrak{g}$ une sous-algèbre de Lie réductive dans $\mathfrak{gl}(V)$, $\mathfrak{q}$ une sous-algèbre de Lie de $\mathfrak{g}$ et $\Phi$ la forme bilinéaire $(x,y) \mapsto \mathrm{Tr}(xy)$ sur $\mathfrak{g} \times \mathfrak{g}$. On suppose que l'orthogonal $\mathfrak{n}$ de $\mathfrak{q}$ par rapport à $\Phi$ est une sous-algèbre de Lie de $\mathfrak{g}$ composée d'endomorphismes nilpotents de $V$. Alors, $\mathfrak{q}$ est une sous-algèbre parabolique de $\mathfrak{g}$.

And here's a possible translation:

Theorem 1. --- Let $V$ be a finite-dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra of $\mathfrak{gl}(V)$, $\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x,y) \mapsto \mathrm{Tr}(xy)$ on $\mathfrak{g} \times \mathfrak{g}$. If the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ relative to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of $V$, then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.


As Fran points out in the comments below, my original translation was incorrect and had $\mathfrak{n}$ nilpotent instead of consisting of nilpotent endomorphisms. Happily, for the case of the adjoint representation, one has Engel's theorem, which says that the the two notions agree.

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    $\begingroup$ Almost, but not quite. What Bourbaki has is Grothendieck's result which, for the adjoint representation, requires that each element of $\xi\in\mathfrak{n}$ have $\mathrm{ad}\xi$ nilpotent on $\mathfrak{g}$. This is a stronger condition than simply requiring $\mathfrak{n}$ to be nilpotent---a distinction which got Lost in Translation! $\endgroup$ Jun 22, 2010 at 7:09
  • $\begingroup$ But isn't that just Engel's theorem? i.e., a Lie algebra $\mathfrak{g}$ is nilpotent if and only if the endomorphisms $\mathrm{ad}_x$ are nilpotent for every $x \in \mathfrak{g}$. $\endgroup$ Jun 22, 2010 at 11:19
  • $\begingroup$ I'll edit the translation, though -- since as you point out it is not literal! $\endgroup$ Jun 22, 2010 at 11:20
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    $\begingroup$ Engel's theorem isn't directly applicable here, since you have to look at the adjoint representation of the semisimple Lie algebra and not that of its nilpotent subalgebra. The Bourbaki theorem has to be applied carefully to an abstract Lie algebra and its Killing form. $\endgroup$ Jun 22, 2010 at 11:56
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    $\begingroup$ Disclaimer: I wrote that book when much younger to learn the subject better, but still haven't gotten to "expert" level. Writing a book is definitely the best way for the author to learn a subject in mathematics, though not invariably so for the reader. $\endgroup$ Jun 22, 2010 at 22:22

As alluded to in the comments above, there is a rather old paper by Jacques Dixmier that contains a result in this direction. The reference is

Dixmier, Jacques. Polarisations dans les algèbres de Lie. II. Bull. Soc. Math. France 104 (1976), no. 2, 145--164.

The result is Lemme 1.1, but the proof is attributed to P. Tauvel. The proof proceeds with a judicious application of the invariance of the Killing form and, for the nilpotency of $\mathfrak{p}^{\perp}$, Bourbaki's Groupes et algèbres de Lie, Chapitre I, §5, Lemme 3.

However, there is a caveat: a slightly different notion of co-isotropy is assumed. Dixmier's notion of co-isotropy is to define the orthogonal complement $\mathfrak{p}^f$ with respect to an anti-symmetric bilinear form $B_f$ derived from the Killing form.

Edit: I claimed earlier that Dixmier's notion of co-isotropy implies that $\mathfrak{p}^{\perp}$ is contained in $\mathfrak{p}$. This is not quite correct: you can prove, as is done there, that $\mathfrak{p}^{\perp} = [x,\mathfrak{p}^f]$ is an ideal of $\mathfrak{p}$ without using the assumption of co-isotropy that Dixmier made. The assumption kicks in only for the proof of nilpotency.

  • $\begingroup$ You're welcome! Your proof of the statement is also very interesting. $\endgroup$
    – ulnor
    Mar 23, 2012 at 4:30

If $\mathfrak{p}^\perp$ is a nipotent subalgebra then it must consist of nilpotent elements. Indeed, let $h\in \mathfrak{p}^\perp$ and suppose ${\rm ad}\,h$ is not nilpotent. Let $\mathfrak{g}^0(h)$ be the Fitting null-component of ${\rm ad}\,h$ (it consists of all $x\in\mathfrak{g}$ such that $({\rm ad}\,h)^N(x)=0$ for $N\gg 0$). Since $\mathfrak{p}^\perp$ is nipotent and $[h,\mathfrak{p}]\subseteq \mathfrak{p}^\perp$ we have that $\mathfrak {p}+\mathfrak{p}^\perp\subseteq \mathfrak{g}^0(h)$. If $\mathfrak{g}^0(h)$ is a proper Lie subalgebra of $\mathfrak {g}$, there is a nonzero $a\in \mathfrak{g}^0(h)^\perp$. Then $a\in \mathfrak{p}^\perp\subset \mathfrak{g}^0(h)$ forcing $a\in \mathfrak{g}^0(h)\cap \mathfrak{g}^0(h)^\perp$. However, decomposing $\mathfrak{g}$ into the direct sum of generalised eigenspaces for ${\rm ad}\,h$ it is easy to see that the restriction of the Killing form of $\mathfrak g$ to $\mathfrak{g}^0(h)$ is non-degenerate. So ${\rm ad}\,h$ has to be nilpotent for every $h\in\mathfrak{p}^\perp$.


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