# Why inherit the Tits systems structure by a $B$-adapted homomorphism?

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.

• Should the equation concerning $\phi(C(w))$ mention $\xi$ somewhere?
– HJRW
Jan 12, 2022 at 8:14
• Thank you. It's a careless mistake. Jan 12, 2022 at 9:12
• 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. Jan 12, 2022 at 10:00
• 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 $\hat{B}\cap \phi(N)\triangleleft \phi(N)$. Jan 12, 2022 at 11:55
• TeX note: please use, e.g., $\operatorname{Ker} \phi$ \operatorname{Ker} \phi instead of $\rm Ker \phi$ \rm Ker \phi; particularly note the spacing. Jan 12, 2022 at 14:18