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Let $\mathfrak{g}$ be a real semisimple Lie algebra with complexification $\mathfrak{g}_\mathbb{C}$. Recall that a parabolic subalgebra in $\mathfrak{g}_\mathbb{C}$ is one which contains a Borel subalgebra, and a parabolic subalgebra in $\mathfrak{g}$ is one which complexifies to a parabolic one in $\mathfrak{g}_\mathbb{C}$.

Question: Given a Borel subalgebra $\mathfrak{b} \subset \mathfrak{g}_\mathbb{C}$, is it contained in the complexification $\mathfrak{q}_\mathbb{C}$ of some minimal parabolic subalgebra $\mathfrak{q} \subset \mathfrak{g}$?


Problem VII.13 in Knapp's "Lie Groups beyond an introduction" says this is true, under the additional condition that $\mathfrak{b}$ is built from the complexification of a maximally noncompact Cartan subalgebra of $\mathfrak{g}$, in the usual way via a root space decomposition and choice of positive roots. I am wondering to what extent we can drop this requirement -- or if this is a strong requirement at all?

Note that Knapp defines minimal parabolic subalgebras differently than I do here, but I believe their notion should be equivalent (this is related to the first part of my question here on math.stackexchange, for which I believe I have found an argument now, but I yet need to formalize it).

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  • $\begingroup$ Doesn't the minimality condition (combined with your definition of parabolic subgroup) imply that $\mathfrak{b}$ should equal $\mathfrak{q}_\mathfrak{C}$ (if and when such $\mathfrak{q}$ exists) rather than being merely contained in it? Am I missing something? $\endgroup$
    – Vincent
    Jun 29, 2021 at 13:04
  • $\begingroup$ About maximally non-compact: the Cartan subalgebra from which $\mathfrak{b}$ is built in the Knapp-world is supposed to be the complexification of the Cartan subalgebra $\mathfrak{t} \oplus \mathfrak{a}$ of $\mathfrak{g}$, with $\mathfrak{t}$ a Cartan subalgebra of $\mathfrak{m}$ and with $\mathfrak{m}$ and $\mathfrak{a}$ as in your MSE question. Now intuitively it seems to me that the fact that $\mathfrak{t} \oplus \mathfrak{a}$ is maximally non-compact is the direct result of $\mathfrak{a}$ being maximal itself (subject to some conditions) as part of its definition. So it seems unavoidable. $\endgroup$
    – Vincent
    Jun 29, 2021 at 13:17
  • $\begingroup$ Also in my first comment I meant $\mathfrak{q}_\mathbb{C}$ where I wrote $\mathfrak{q}_\mathfrak{C}$ but it is too late to edit now. $\endgroup$
    – Vincent
    Jun 29, 2021 at 13:18
  • $\begingroup$ Very interesting question by the way $\endgroup$
    – Vincent
    Jun 29, 2021 at 13:19
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    $\begingroup$ Thank you for the plentiful comments! I don't think minimality implies $\mathfrak{b} = \mathfrak{q}_\mathbb{C}$, the problem being that $\mathfrak{b}$ cannot necessarily be expressed as the complexification of a subalgebra in the real form $\mathfrak{g}$. Indeed, this seems to only happen in real Lie algebras $\mathfrak{g}$ which Knapp calls quasisplit -- if I understand it correctly, Knapp even defines the quasisplit property by the existence of a Borel subalgebra in $\mathfrak{g}_\mathbb{C}$ which is itself a complexification of a real subalgebra, see Chapter VI, Problems 28 - 35. $\endgroup$ Jun 29, 2021 at 13:41

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NB: for brevity I'm going to refer to Borel subalgebras and parabolic subalgebra as Borels and parabolics, respectively

Even for a split (or quasi-split) Lie algebra not all complex Borels are the complexification of real Borels. Otherwise, for example, the real and complex flag manifolds would be identical. Naturally, a Borel cannot contain another one since they are supposed to be maximal solvable/minimal parabolic. So for these cases the condition would appear to be necessary as well as sufficient.

What is true is that a real parabolic $ \mathfrak{p} \leq \mathfrak{g} $ complexifies to a complex parabolic $ \mathfrak{p}^{\mathbb{C}} \leq \mathfrak{g}^{\mathbb{C}} $. Such a complex parabolic contains a family of Borels (of $\mathfrak{g}^{\mathbb{C}}$) and any one of these determines $ \mathfrak{p}^{\mathbb{C}} $ and thus $\mathfrak{p}$ uniquely. To be more precise parabolics come in a number of conjugacy classes in correspondence with subsets of the Dynkin diagram and a given Borel subalgebra defines a unique parabolic in each conjugacy class that contains it. In particular we get a unique one in the class which contains the complexifications of the minimal parabolics in our real form. Because we chose it to be a complexification we can recover its real form but that wouldn't work otherwise. The uniqueness we mentioned above tells us that if a Borel $\mathfrak{b}$ is contained in some complex parabolic $\mathfrak{q}$ which is conjugate to $\mathfrak{p}^{\mathbb{C}}$ but is not a complexified parabolic itself then $\mathfrak{b}$ is not contained in any complexified minimal parabolic.

So now we have to ask if all Borel subalgebras contained within a given $ \mathfrak{p}^{\mathbb{C}} $ are of the form described by Knapp. I couldn't work that out off the top of my head but certainly there is some necessary condition.

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