2
$\begingroup$

$\DeclareMathOperator\SL{SL}$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.

$\endgroup$
3
  • 1
    $\begingroup$ By the way, the more usual term than "affine parabolic" is "parahoric". $\endgroup$
    – LSpice
    Apr 6, 2022 at 16:43
  • 3
    $\begingroup$ 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
    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
    Apr 7, 2022 at 1:42

1 Answer 1

1
$\begingroup$

$\DeclareMathOperator\SL{SL}\newcommand\O{\mathcal O_F}\newcommand\P{\pi\mathcal O_F}\newcommand\Pi{\pi^{-1}\mathcal O_F}$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 $B w_1 B$.

$\endgroup$
2
  • 1
    $\begingroup$ Did you mean to intersect with $SL_3(F)$ rather than $SL_3(\mathcal O_F)$? $\endgroup$
    – Will Sawin
    Apr 6, 2022 at 19:06
  • $\begingroup$ @WillSawin, yes, thanks; edited. $\endgroup$
    – LSpice
    Apr 6, 2022 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.