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Let $\mathfrak{g}$ be a Lie algebra with $\mathfrak{a}$ a maximal commutative subalgebra. We then have the root decomposition $$\mathfrak{g}=\mathfrak{a} \oplus \bigoplus _{\alpha \in \Psi} \mathfrak{g}_\alpha$$ for a set of simple roots $\Psi \subset \mathfrak{a}^*$. A maximal solvable subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ is called Borel subalgebra. (n.b. Borel algebras are always conjugated to algebras of the form $\mathfrak{a} \oplus \bigoplus _{\alpha \in \Psi^+} \mathfrak{g}_\alpha$ for a set of positive roots $\Psi^+ \subset \Psi$)

A subalgebra $\mathfrak{p} \subset \mathfrak{g}$ is called parabolic subalgebra if it contains a Borel subalgebra. I have found two different definitions for minimal parabolic subalgebra.

Onishchik, Vinberg: Lie Groups and Lie Algebras III, p. 191: $\mathfrak{q}(\emptyset)=\mathfrak{g}_0 \oplus \bigoplus _{\alpha \in \Psi^+} \mathfrak{g}_\alpha$.

Knapp: Lie Groups Beyond an Introduction, p. 270: $\mathfrak{q}=\mathfrak{b}$.

Clearly, $\mathfrak{q}=\mathfrak{b}$ is the smallest subalgebra containing $\mathfrak{b}$. Why would one want to call the bigger algebra $\mathfrak{q}(\emptyset)$ a minimal parabolic Lie algebra?

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  • $\begingroup$ For the same reason we usually don't consider the Lie algebra itself to be a maximal parabolic. Both extreme cases are in some sense "trivial", and hence we prefer to leave them out when we say "minimal" and "maximal". $\endgroup$
    – Tobias Kildetoft
    Commented Mar 6, 2018 at 16:23

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The notion of "minimal parabolic" applies, meaningfully, more generally than does the notion of "Borel subgroup/subalgebra".

It depends whether you're working over $\mathbb R$ or over $\mathbb C$, since over $\mathbb R$ a Levi-Malcev component of a minimal parabolic may be quite large. For example, in (the Lie algebra of) $O(n,1)$, there is a unique conjugacy class of parabolic subgroup (apart from the group itself), and they all have Levi-Malcev component isomorphic to $O(n)\times O(1)$. In this case, one might hesitate to call this minimal parabolic a Borel subgroup/subalgebra, because of the discrepancy between real and complex root spaces.

But for "split" or "quasi-split" groups/algebras, the minimal parabolics are indeed Borel.

Over $\mathbb C$, they are always the same.

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  • $\begingroup$ I am not sure I would call the Borel minimal parabolic in most cases, simply because then I would have no good term for the second smallest parabolics, which are often an important case. $\endgroup$
    – Tobias Kildetoft
    Commented Mar 6, 2018 at 19:13
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    $\begingroup$ @TobiasKildetoft, I can understand the naming issue, but/and I think the more serious issue is when a minimal-among-parabolics is not Borel, due to some non-splitness, no? $\endgroup$
    – paul garrett
    Commented Mar 6, 2018 at 19:26
  • $\begingroup$ That might indeed be a more important issue (but not one I really know anything about). Certainly, this seems like a natural thing for Knapp to cover given the title of the book in question. But now that I look at it again, it seems that in the complex case the definitions given by both authors agree (both giving the Borel). $\endgroup$
    – Tobias Kildetoft
    Commented Mar 6, 2018 at 19:31
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    $\begingroup$ It's not that $M=O(n)\times O(1)$ is the whole minimal parabolic, but rather that the minimal parabolic is a semi-direct product of $M$ with the unipotent radical of the minimal parabolic $P$. In this situation, $P$ (or its Lie algebra) is not solvable, because of the extreme non-split-ness of the group (over $\mathbb R$). The "solvability" criterion only works over alg closed fields (in char $0$), or for "split" or "quasi-split" groups. Note that $O(p,q)$ is not quasi-split unless $|p-q|\le 2$ (or $\le 1$, depending a technical quibble) and $U(p,q)$ is not quasi-split unless $|p-q|\le 1$. $\endgroup$
    – paul garrett
    Commented Mar 6, 2018 at 20:01
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    $\begingroup$ If by ($\mathbb R$-) "split" one means that there is a maximal torus which is $\mathbb R$-split, then this should follow more-or-less from definitions, using semi-simplicity or reductiveness. This allows $U(n,n)$. "Quasi-split" allows $U(n,n+1)$ also. $\endgroup$
    – paul garrett
    Commented Mar 7, 2018 at 13:19

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