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This is a follow-up to Does the bruhat decomposition induces decomposition on integral points (on an open cell)?

Given a split connected reductive group $G$ over a $p$-adic local field $F$ with ring of integers $\mathcal O$, suppose $N_+$ is the unipotent radical of a Borel and $N_-$ its opposite. Is it true that $$G(\mathcal O) \cap N_-(F)N_+(F) = N_-(\mathcal O)N_+(\mathcal O)?$$ I can prove this for $G=\mathrm {GL}_n$ basically by a sort of induction on the rows and columns in the matrix, so I'd be happy with a reduction to this case. But I'm also curious if there are ways to prove this that don't involve reducing to $\mathrm{GL}_n$.

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    $\begingroup$ By the way, connectedness is irrelevant; $N_-$ and $N_+$ are both contained in $G^\circ$. $\endgroup$ Commented Jul 31, 2024 at 1:11

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I am sure that there is some elegant way to proceed without reduction to $\operatorname{GL}_n$, but the reduction is so straightforward that it's a shame not to do it. It's also not as silly a proof strategy as it might seem; this is the strategy used in the ‘dynamic’ study of subgroups associated to cocharacters in Chapter 2 of Conrad, Gabber, and Prasad - Pseudo-reductive groups (for example, see Proposition 2.1.8).

All schemes over $\mathscr O$ (which could be $\mathbb Z_p$ if you'd like to stick with the ambient field $\mathbb Q_p$, or could be the ring of integers in any locally compact, non-Archimedean field in general), to avoid having to speak of $\mathscr O$-models for schemes over a field that I never need to reference explictly.

The important thing is that your statement becomes weaker if you replace your Borel by a larger parabolic, so you might as well make the claim for arbitrary parabolics. Now choose a faithful representation $G \to \operatorname{GL}_n$. Let $T$ be a split maximal torus in the Borel (or just parabolic) subgroup $B$, let $\lambda$ be a cocharacter of $T$ so that $B$ is the parabolic $P_G(\lambda)$ of $G$ associated to $\lambda$, and—here's the key—replace $G$ by $\operatorname{GL}_n$, and $B$ by $P_{\operatorname{GL}_n}(\lambda)$, to make a stronger claim. This parabolic subgroup probably won't be a Borel subgroup even if $B$ was, but, as we observed, that's OK.

Now, as you observed, one can prove this stronger claim by induction—the way that I'd think to do it is by downward induction on $\lvert i + j\rvert$, where we consider the $(i, j)$ entries of the two triangular matrices.

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