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$\DeclareMathOperator\SL{SL}$Let $F$ be a local field and $\mathcal{O}_{F}$ its valuation ring. Let $\pi\in \mathcal{O}_{F}$ be a uniformizer and $\mathfrak{p}=\pi\mathcal{O}_{F}$. Let $G$ be a split semisimple algebraic group over $F$. I think the case $G=\SL_{3}$ as an example. Let us consider a generator system $S=\{s_{1},s_{2},w_{1}\}$ of the affine Weyl group of $\SL_{3}(F)$, where $$ s_{1}= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right),\quad s_{2}= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array} \right),\quad w_{1}= \left( \begin{array}{ccc} 0 & 0 & -\pi^{-1} \\ 0 & 1 & 0 \\ \pi & 0 & 0 \end{array} \right). $$ It is well-known that holds the affine Bruhat decomposition $\SL_{3}(F)=B\langle S\rangle B$ and $\SL_{3}(\mathcal{O}_{F})=B\langle s_{1},s_{2}\rangle B$ where $B$ is the standard Iwahori subgroup of $\SL_{3}(F)$.

Question: What is the remaining component $B\langle w_{1}\rangle B=B\cup Bw_{1}B$? I want to know the explicit form of this subgroup.

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    $\begingroup$ By the way, the more usual term than "affine parabolic" is "parahoric". $\endgroup$
    – LSpice
    Commented Apr 6, 2022 at 16:43
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    $\begingroup$ What do you mean by "remaining component"? We don't have $\langle s_1, s_2 , w_1 \rangle = \langle s_1, s_2 \rangle \cup \langle w_1 \rangle$. $\endgroup$
    – Will Sawin
    Commented Apr 6, 2022 at 19:05
  • $\begingroup$ Sorry. The term of "remaining component" had no special meaning. I understand that $\langle S\rangle \neq \langle s_{1},s_{2} \rangle\cup \langle w_{1}\rangle$. $\endgroup$
    – M masa
    Commented Apr 7, 2022 at 1:42

1 Answer 1

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$\DeclareMathOperator\SL{SL}\newcommand\O{\mathcal O_F}\newcommand\P{\pi\mathcal O_F}\newcommand\Pi{\pi^{-1}\mathcal O_F}$I assume that the standard Iwahori is the group of matrices in $\SL_3(\O)$ that are upper triangular modulo $\pi$. Then $B \cup Bw_1 B$ is $K \mathrel{:=} \begin{pmatrix} \O & \O & \Pi \\ \P & \O & \O \\ \P & \P & \O \end{pmatrix} \cap \SL_3(F)$.

Indeed, it is clear that this is a subgroup, that it contains (hence is stable under left- and right-multiplication by) $B$, and that it contains $w_1$. Therefore, it contains $B \cup Bw_1 B$.

On the other hand, suppose $\gamma = \begin{pmatrix} a & b & \pi^{-1} c \\ \pi d & e & f \\ \pi g & \pi h & i \end{pmatrix}$ belongs to $K$ (so that $a, \dotsc, i$ belong to $\O$). If $c$ belongs to $\P$, then $\gamma$ belongs to $B$. Otherwise, $b \mathrel{:=} \begin{pmatrix} 1 \\ & 1 \\ \pi i/c && 1 \end{pmatrix}$ and $(b w_1)^{-1}\gamma$ both belong to $B$, so $\gamma$ belongs to $B w_1 B$.

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    $\begingroup$ Did you mean to intersect with $SL_3(F)$ rather than $SL_3(\mathcal O_F)$? $\endgroup$
    – Will Sawin
    Commented Apr 6, 2022 at 19:06
  • $\begingroup$ @WillSawin, yes, thanks; edited. $\endgroup$
    – LSpice
    Commented Apr 6, 2022 at 19:22

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