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Let $G$ be a linear algebraic group define over $\overline{\mathbb{F}_p}$, consider it as a subgroup of $\operatorname{GL}(n)$. Let $F_p$ be the standard Frobenius. Let $B$ and $Q$ be an $F_p$-stable Borel subgroup and $F_p$-stable parabolic subgroup respectively with $B\subset Q$. Let $\bigcup_{w\in W^{Q}} BwQ/Q$ be the Bruhat decomposition of $G/Q$. Where $W^{Q}$ is the Weyl group of $Q$.

My question: Is the Schubert cell $BwQ/Q$ $F_p$-stable? For the case $G=\operatorname{SL}(n)$?

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    $\begingroup$ Do you mean by $W^Q$ the group $N_{G^{F_p}}(M)/M^{F_p}$, for $M$ a Levi component of $Q$? If so, then this is trivial, so you must not; but then I don't know what you do mean. (Also, please use "$F_p$-stable" $F_p$-stable, not "$F_p-$stable" $F_p-$stable. I have edited accordingly.) $\endgroup$
    – LSpice
    Commented Jun 6, 2022 at 20:51
  • $\begingroup$ @LSpice :Dear LSpice, mathoverflow.net/q/195456/147080 I just use the notation of this question😂 $\endgroup$
    – fool rabbit
    Commented Jun 7, 2022 at 4:18
  • $\begingroup$ That link also unfortunately doesn't define it. A common definition is $W^Q = N_{G^{F_p}}(M)/M^{F_p}$ (and then the decomposition holds as desired), in which case the result is obvious; so, if you want to use some other definition, then you'll need to say what it is. $\endgroup$
    – LSpice
    Commented Jun 7, 2022 at 13:09
  • $\begingroup$ Thanks. But if it defined over $\mathbb{C}$, then how to define $W^Q$? $\endgroup$
    – fool rabbit
    Commented Jun 7, 2022 at 16:12
  • $\begingroup$ I think I may realise what I wasn't seeing. You are assuming that $B$ and $Q$, but not $G$, are defined over $\mathbb F_p$? (I'm not sure what you mean by asking about $\mathbb C$, since we're over $\overline{\mathbb F_p}$ here; maybe you mean to replace $(\overline{\mathbb F_p}, F_p)$ by $(\mathbb C, \text{conjugation})$?) $\endgroup$
    – LSpice
    Commented Jun 7, 2022 at 16:17

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