Let $G$ be a split reductive group over a perfect field $k$ (not necessarily algebraically closed) with split maximal torus $T$ and Borel $B \supset T$.

Then there is(/should be) an inclusion-preserving bijection from the set of parabolic subgroups $P$ of $G$ containing $B$ to the set of subsets of the set of simple roots $\Delta(B)$ associated to $B$.

Does someone know a (good) reference for this?

I know that this question is more suitable for MathStackExchange but I got the impression, that there is not much activity at the moment.

  • 3
    $\begingroup$ The fact that MSE is not very active doesn't make an MSE question more appropriate for MO, though. $\endgroup$
    – LSpice
    Jul 19 at 18:08

1 Answer 1


Gunter Malle and Donna Testerman, Linear algebraic groups and finite groups of Lie type. Cambridge Studies in Advanced Mathematics, 133. Cambridge University Press, Cambridge, 2011.

See Proposition 12.2.

  • $\begingroup$ Mmmh but they consider linear algebraic groups over an algebraically closed field. $\endgroup$
    – KKD
    Jul 20 at 10:42
  • $\begingroup$ This plays no role. Your torus $T$ is split. You construct the parabolic subgroups (containing $B$) over an algebraic closure, and they are defined over the base field. $\endgroup$ Jul 20 at 14:51
  • $\begingroup$ Is this explained somewhere? I need to quote this for a thesis and unsure if I can take this for granted. $\endgroup$
    – KKD
    Jul 20 at 15:47
  • 2
    $\begingroup$ Any parabolic subgroup defined over $k$ is defined over an algebraic closure $\bar k$ of $k$ and hence it comes from a subset $I$ of the set of simple roots $S$. Conversely, Malle and Testerman construct a parabolic subgroup $P_I$ from a subset $I$ of the set of simple roots $S$ without any additional choices. The Galois group $\Gamma={\rm Gal}(\bar k/k)$ acts trivially on $S$ and hence preserves $I$ and $P_I$. Since your field $k$ is perfect, the algebraic subgroup $P_I$ of $G$ is defined over $k$. $\endgroup$ Jul 21 at 17:55

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.