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Let $(G,B,N,S)$ be a Tits system and $\phi\colon G\longrightarrow \hat{G}$ a $B$-adapted in the sense of the paper Groupes réductifs sur un corps local: I of Bruhat–Tits. They said that $\phi$ is a $B$-adapted homomorphism if $\phi$ satisfies the following conditions;

  1. $\operatorname{Ker}\phi\subset B$, and
  2. for any $ g\in \hat{G}$, there exists $h\in G$ such that $g\phi(B)g^{-1}=\phi(hBh^{-1})$.

Then any parabolic subgroup $P$ of $G$ and $g\in \hat{G}$, the subgroup ${}^{g}P=\phi^{-1}(g\phi(P)g^{-1})$ is also a parabolic subgroup of $G$. So $\hat{G}$ acts on the set of parabolic subgroups of $G$. Write $\operatorname{Stab}(P)=\{g\in \hat{G}\mathrel{\vert} {}^{g}P=P\}$.

They provides a homomorphism $\xi\colon \hat{G}\longrightarrow \operatorname{Aut}(W,S)$ such that $$ \phi(C(\xi(g)\cdot w))=\phi(h)^{-1}g\phi(C(w))g^{-1}\phi(h) $$ for all $h\in G$ (satisfies $g\phi(B)g^{-1}=\phi(hBh^{-1})$) and $w\in W$, where $C(w)$ is a Bruhat cell $BwB$). Write $\hat{G}_{0}=\operatorname{Ker} \xi$.

I have two questions;

  1. why is the restriction $\xi\rvert_{\operatorname{Stab}(B)}\colon \operatorname{Stab}(B)\longrightarrow \operatorname{Im} \xi$ surjective?
  2. why is $(\hat{B},\phi(N))$ a $BN$-pair of $\hat{G}_{0}$? Here $\hat{B}=\hat{G}_{0}\cap \operatorname{Stab}(B)$.

These questions are written in Bruhat–Tits's "Groupes réductifs sur un corps local: I" without proof.　I thought about this all day today, but I couldn't prove it.

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  • $\begingroup$ Should the equation concerning $\phi(C(w))$ mention $\xi$ somewhere? $\endgroup$
    – HJRW
    Commented Jan 12, 2022 at 8:14
  • $\begingroup$ Thank you. It's a careless mistake. $\endgroup$
    – M masa
    Commented Jan 12, 2022 at 9:12
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    $\begingroup$ I can prove 1. For any $g\in \hat{G}$, there exists $h\in G$ such that $\phi(h^{-1})g\in {\rm Stab} B$. Since $\phi(G)\subset \hat{G}_{0}$, we have $\hat{G}=\hat{G}_{0}\cdot {\rm Stab} B$. Thus the restriction of $\xi$ to ${\rm Stab} B$ is surjective. $\endgroup$
    – M masa
    Commented Jan 12, 2022 at 10:00
  • $\begingroup$ Now I proved a part of 2. The intersection $\phi(N)\cap \hat{B}$ is normal in $\phi(N)$. For any $x,y\in N$ such that $\phi(x)\in \hat{B}$, we have $\phi(xBx^{-1})=\phi(B)$ from the definition ${\rm Stab} B$. So we have $\phi(xb_{1}x^{-1}b_{2})=0$ for any $b_{1}\in B$ and some $b_{2}\in B$. Since ${\rm Ker}\phi\subset B$, $xb_{1}x^{-1}b_{2}\in B$. As $N_{G}(B)=B$, this means that $x\in B\cap N (\triangleleft N)$. Then we have $\phi(yxy^{-1})\in \hat{B}\cap \phi(N)$. Thus $\hat{B}\cap \phi(N)\triangleleft \phi(N)$. $\endgroup$
    – M masa
    Commented Jan 12, 2022 at 11:55
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    $\begingroup$ TeX note: please use, e.g., $\operatorname{Ker} \phi$ \operatorname{Ker} \phi instead of $\rm Ker \phi$ \rm Ker \phi; particularly note the spacing. $\endgroup$
    – LSpice
    Commented Jan 12, 2022 at 14:18

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