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Let $G$ be a (connected) reductive linear algebraic group over $\overline{\mathbb{Q}}_p$. By definition, this means that $G$ is a closed subgroup of some $\mathrm{GL}_n$. We can always find a reductive $\mathbb{Z}_p$-group $\mathbb{G}$, which after base-change to $\overline{\mathbb{Q}}_p$ recovers $G$. I.e., $\mathbb{G}_{\overline{\mathbb{Q}}_p} = G$.

Let $K/\mathbb{Q}_p$ be an extension (inside $\overline{\mathbb{Q}}_p$) with ring of integers $R$. We have an inclusion $$ \mathbb{G}(R) \subset \mathbb{G}(\overline{\mathbb{Q}}_p) = G(\overline{\mathbb{Q}}_p)\subset \mathrm{GL}_n(\overline{\mathbb{Q}}_p). $$ Is there any relationship between $\mathbb{G}(R)$ and the intersection $\mathbb{G}(\overline{\mathbb{Q}}_p)\:\cap\:\mathrm{GL}_{n}(R)$ inside $\mathrm{GL}_n(\overline{\mathbb{Q}}_p)$? In particular, is the latter contained in the former? If something like this isn't true in general, is it true for a particular choice of $\mathbb{Z}_p$-model $\mathbb{G}$? Is there anything I can do to achieve a similar inclusion? Any relevant reference is greatly appreciated.

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  • $\begingroup$ Let $V$ be an $n$-dimensional vector space over $\mathbb{Q}_p$. Assume that that the closed immersion $G \to \operatorname{GL}(V)_{\overline{\mathbb{Q}}_p}$ comes from a map $i:\mathbb{G}_{\mathbb{Q}_p} \to \operatorname{GL}(V)$. Then there should be a $\mathbb{Z}_p$-lattice $\Lambda \subset V$ such that $i$ extends (uniquely) to a closed immersion $\mathbb{G} \to \operatorname{GL}(\Lambda)$ over $\mathbb{Z}_p$; this is certainly not true for any lattice. $\endgroup$
    – Pol van Hoften
    Commented Apr 4 at 6:17
  • $\begingroup$ @PolvanHoften I'm not sure I understand sorry. I'm not very comfortable with groups over rings. Why can we assume the first map comes from some $i$? And why does should $i$ extend in the way you describe? Are you saying that this should be possible for some choices of extensions $K$ with ring of integers $R$? $\endgroup$
    – Otto
    Commented Apr 5 at 16:36
  • $\begingroup$ I was trying to say that we likely need to assume that our map is defined over $\mathbb{Q}_p$ in order for your question to have a reasonable answer. Given this, there should be some choice of lattice $\Lambda$ such that $\mathbb{G}(R)$ is equal to $\operatorname{GL}(\Lambda)(R) \cap \mathbb{G}(K)$ for all finite extensions $K$ with ring of integers $R$. For $K=\mathbb{Q}_p$ atleast, the existence of the lattice follows from via a standard argument from the compactness of $\mathbb{G}(\mathbb{Z}_p)$ (any compact group acting on a finite dimensional $\mathbb{Q}_p$-vector space fixes a lattice). $\endgroup$
    – Pol van Hoften
    Commented Apr 5 at 20:30
  • $\begingroup$ On the other hand, if you fix the lattice $\Lambda$, then in general the groups $\mathbb{G}(\mathbb{Z}_p)$ and $\mathbb{G}(\mathbb{Q}_p) \cap \operatorname{GL}(\Lambda)(\mathbb{Z}_p)$ have nothing to do with each other. For instance, there does not have to be a containment in either direction. $\endgroup$
    – Pol van Hoften
    Commented Apr 5 at 20:34

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