Decomposition of parabolic subgroup in reductive group

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!

• 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. – Keerthi Madapusi Pera Apr 18 at 1:44
• Could you explain how projectivity of $P_2/P_1$ implies that $P_1$ contains $N_2$? – Mehta May 26 at 22:31

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.