Let $G$ be a reductive group over $\mathbb{Q}$ and let $P_0$ minimal parabolic subgroup.

If $P_1=M_{1}N_{1} \subset P_2=M_2N_2$ are standard parabolic subgroups of $G$, then can we decompose $P_1=(P_1 \cap M_2 )N_2$?

It seems it does hold but I don't know why it holds. Any comments are welcome!

  • 1
    $\begingroup$ By 'standard' here, I'm assuming you mean parabolics that contain the specified minimal parabolic. Not that it really matters. Your question is basically if $P_1\subset P_2$ are parabolic subgroups, then we have the decomposition you wrote down. To see this, just note that both sides are subgroups of $G$. Moreover, the right hand side is contained in $P_1$: This amounts to the fact that $P_1$ contains $N_2$, which is immediate from the projectivity of $P_2/P_1$. The rest is now easy. $\endgroup$ Apr 18, 2019 at 1:44
  • $\begingroup$ Could you explain how projectivity of $P_2/P_1$ implies that $P_1$ contains $N_2$? $\endgroup$
    – Mehta
    May 26, 2019 at 22:31

1 Answer 1


Over an algebraically closed field (of any characteristic), it is fairly obvious using the omitted definition of "standard parabolic" that the assertion here for a pair of included standard parabolics is true: one just wants to trace the pairs of positive and negative roots involved. However, the question is too loosely formulated to be clear. For example, the minimal $k$-parabolic $P_0$ is introduced but then ignored for $k=\mathbb{Q}$. Here $k$ may be arbitrary, but it needs to be clarified what "standard" means in this setting. The 1965 foundational paper by Borel and Tits is now somewhat old-fashioned in language, but the BN-pair setting for parabolics is probably helpful here.

  • $\begingroup$ Sorry. I found your reply late. Thank you very much! $\endgroup$
    – Monty
    May 24, 2019 at 15:55

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