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$\DeclareMathOperator\GL{GL}$Let $G$ be a semisimple (but I think there is no obstruction to assume it to be reductive) algebraic group over a non-Archimedean local field $K$ and $\mathcal{O}_K$ be its ring of integers.

In context of Satake theory one considers as important object for further constructions the group $G(\mathcal{O}_K)$ of integer points of $G$.

Question: Why is this group well defined at all? Although it seems rather natural to construct, namely we can embed $G$ in some $\GL_n$ by definition of algebraic group and then set $G(\mathcal{O}_K):=G(K) \cap \GL_n(\mathcal{O}_K)$, latter exists obviously "canonically". But here is of course an explicit choice of $\GL_n$ involved where $G$ in going to be embedded, so $G(\mathcal{O}_K)$ seems to be dependent on an explicit embedding $ G \subset \GL_n$, isn' t it?

Or is it nevertheless $G(\mathcal{O}_K)$ "canonical" by some additional argument?

#EDIT: As @David Loeffler' s answer states, the claim without certain additional assumptions on $G$ is wrong, so I would like to additionally assume as @LSpice suggested that $G$ is moreover split, conjecturing that this may provide a sufficient assumption for existence of the group I' m looking for.

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  • $\begingroup$ @DaveBenson: In the question I missed to mention that $G$ is considered as a variety over field $K$, sorry for confusion. So in language schemes it can be considered as a representable functor from finitely generated $K$- algebras to groups. So a priori it's not clear "what is" $G(\mathcal{O}_K)$, since $O_K$ is not a $K$ - algebra. Naively it only make sense when we fix/consider the closed embedding $ G \subset GL_n$ as sketched above. But my concern is if this $G(\mathcal{O}_K)$ really depends on choosen embedding $ G \subset GL_n$ or exist independently of such embedding abstractly $\endgroup$
    – user267839
    Commented Sep 9, 2023 at 23:35
  • $\begingroup$ Is your group split? If so, then you can take the split reductive $\mathcal O_K$-group $\mathcal G$ of the same type, and identify $\mathcal G_K$ with $G$. This identification is not canonical, but the resulting image of $\mathcal G(\mathcal O_K)$ in $G(K)$ is. (This probably works just as well if $G$ is quasisplit, although I'd want to be a bit careful about Galois groups, and every reductive group over $K$ is quasisplit over an unramified extension of $K$ … but somewhere in there I lose the thread of well definedness!) $\endgroup$
    – LSpice
    Commented Sep 10, 2023 at 0:01
  • $\begingroup$ @LSpice: yes thanks, in light of David Loeffler' s answer it seems that in order to assure the existence of such group one should pose for $G$ some additinal assumptions. Beeing 'split' seems to be very natural, I should add it. But could you give a reference or a brief sketch how this $O_K$-group scheme $\mathcal{G}$ is constructed? Is there a "standard" method? Do I understand it correctly, that this splitting assumption is one of possible assumptions which would garantee availability of the result from Bruhat- Tits theory to $\endgroup$
    – user267839
    Commented Sep 10, 2023 at 9:21
  • $\begingroup$ which David Loeffler is refering to, namely the existence of such " nice enough" open compact $U \subset G(K)$? $\endgroup$
    – user267839
    Commented Sep 10, 2023 at 9:22
  • $\begingroup$ Re, as @DavidLoedffler showed, my reference to well definedness of the image was wrong; the identification of $\mathcal G_K$ with $G$ allows inner automorphisms coming from $G(K)$, not just $\mathcal G(\mathcal O_K)$. $\endgroup$
    – LSpice
    Commented Sep 10, 2023 at 10:39

1 Answer 1

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No, there is not a well-defined subgroup "$G(\mathcal{O}_K)$" for a semisimple algebraic group over $K$; if you define it using embeddings into $GL_n$ then the subgroup you get will depend on the embedding.

To see this, consider the two different embeddings $\iota, \iota'$ of $SL_2(K)$ into $GL_2(K)$ where $\iota$ is the obvious inclusion, and $\iota'(\begin{pmatrix} a & b \\ c & d \end{pmatrix}) = \begin{pmatrix} a & \pi b \\ \pi^{-1} c & d \end{pmatrix}$ for $\pi$ a uniformizer of $K$. Then the two subgroups $\iota^{-1}(GL_2(\mathcal{O}_K))$ and $(\iota')^{-1}(GL_2(\mathcal{O}_K))$ are not the same (and not even conjugate in $SL_2(K)$).

One of the major results in Bruhat-Tits theory is to show that if you start with a sufficiently nice open compact subgroup $U \subseteq G(K)$, then you can construct an integral model of $G$ (i.e. a group scheme $\mathcal{G} / \mathcal{O}$ with a specified isomorphism $\mathcal{G} \times_{\mathcal{O}_K} K \cong G$) for which the image of $\mathcal{G}(\mathcal{O}_K)$ is $U$; and there is an explicit characterisation of when $\mathcal{G}$ will have reductive special fibre (this happens iff $U$ is a hyperspecial maximal compact).

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  • $\begingroup$ are there some let me call them "standard" assumptions on $G$, which garantee the existence of such "nice" open compact $U \subset G(K)$? eg "splittness" of $G$ as LSpice suggested? Do you know some classical reference where the construction of such integral is discussed? $\endgroup$
    – user267839
    Commented Sep 10, 2023 at 9:33
  • $\begingroup$ * integral model $\endgroup$
    – user267839
    Commented Sep 10, 2023 at 9:55
  • $\begingroup$ Splitting over an unramified extension shows the extensive of a hyperspecial. You can refer to the original Bruhat--Tits, or better to Tits's exposition, or even better to Rabinoff's or Yu's modern expositions. On my phone, so can't get links easily now. $\endgroup$
    – LSpice
    Commented Sep 10, 2023 at 10:42
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    $\begingroup$ Or better, the new book by Kaletha and Prasad: MR4520154 - Bruhat-Tits theory—a new approach Kaletha, Tasho; Prasad, Gopal New Math. Monogr., 44 Cambridge University Press, Cambridge, 2023, xxx+718 pp. $\endgroup$
    – anon
    Commented Sep 13, 2023 at 23:39
  • $\begingroup$ another point: do I understand it correctly that in case there exist such hyperspecial maximal compact $U$ (...and so $\mathcal{G}(\mathcal{O}_K)$ has isomorphic image to $U$ as the statement claims) that $U$ "plays" in non-Archimedean world the role of a maximal compact group $K$ wrt representation theory of $G$, even thought $U$ is not neccessary a maximal compact subgroup of $G$ as it would be the case for $K$ in classical Lie group setting, right? $\endgroup$
    – user267839
    Commented Oct 1, 2023 at 11:44

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