2
$\begingroup$

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!

$\endgroup$
  • 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. $\endgroup$ – Keerthi Madapusi Pera Apr 18 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 at 22:31
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.