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Edit: both questions are resolved in comments. Let $F$ be a local field and $O$ its integral points. Let $G$ be a split reductive group over $O$. The Bruhat decomposition states that there is a decomposition $$G(F)=\bigsqcup_{w \in W}B(F)wB(F)$$ where $B$ is borel subgroup of $G$. It is not true that, $$ G(O)=\bigsqcup_{w \in W}B(O)wB(O)$$ I am interested in the cell $N_{-}(F) B(F) \hookrightarrow G(F)$, where $N_-$ is the opposite unipotent of $B$. Is it true that $$ G(O)\cap N_{-}(F)B(F)= N_-(O)B(O)?$$

As LSpice, explained this question seems to answer the question as false.

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  • $\begingroup$ Despite my hasty (now-deleted) answer saying otherwise, doesn't the question you linked answer both parts negatively after all? For example, the only expression of $\begin{pmatrix} 1&\\p&1\end{pmatrix} \in G(\mathbb Z_p)$ as an element of $U(\mathbb Q_p)\begin{pmatrix}&-1\\1\end{pmatrix}B(\mathbb Q_p)$ is $\begin{pmatrix}1&p^{-1}\\&1\end{pmatrix}\begin{pmatrix}&-1\\1\end{pmatrix}\begin{pmatrix}p&1\\&p^{-1} \end{pmatrix}$. $\endgroup$
    – LSpice
    Commented Jul 30 at 23:45
  • $\begingroup$ yes, it is a typo. Thanks, that answers 1, but does it answer 2? $\endgroup$
    – W. Zhan
    Commented Jul 30 at 23:49
  • $\begingroup$ (I removed my reference to the typo. Sorry for my own inadvertent conflict wth your notation, using $U$ rather than $N_+$ for the unipotent radical.) Re, yes: the unique expression of $\begin{pmatrix}&1\\-1\end{pmatrix}\begin{pmatrix}1\\p&1\end{pmatrix} \in G(\mathbb Z_p)$ as an element of $N_-(\mathbb Q_p)B(\mathbb Q_p)$ is $\begin{pmatrix}1\\-p^{-1}&1\end{pmatrix}\begin{pmatrix}p&1\\&p^{-1}\end{pmatrix}$. $\endgroup$
    – LSpice
    Commented Jul 30 at 23:52
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    $\begingroup$ Interesting counterexample — do you happen to know if it's true that $G(\mathbb Z_p) \cap N_-(\mathbb Q_p) N_+(\mathbb Q_p) = N_-(\mathbb Z_p) N_+(\mathbb Z_p)$ in general? Like if you cut the torus out of the picture? This at least appears to be true for $\mathrm{GL}_n$ (I think?). $\endgroup$
    – Ashwin Iyengar
    Commented Jul 31 at 0:30
  • $\begingroup$ @AshwinIyengar, re, I believe that that is true, but the proof that I have in mind would probably take more space than a comment box (reduce to $\operatorname{GL}_n$ easily, then some tiresome bookkeeping). If you would like to ask it as a separate question, then I can try to answer it. $\endgroup$
    – LSpice
    Commented Jul 31 at 0:45

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