# Canonicality of group of integers for reductive groups over non-Archimedean local field

$$\DeclareMathOperator\GL{GL}$$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.

• @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 Commented Sep 9, 2023 at 23:35
• 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!) Commented Sep 10, 2023 at 0:01
• @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 Commented Sep 10, 2023 at 9:21
• which David Loeffler is refering to, namely the existence of such " nice enough" open compact $U \subset G(K)$? Commented Sep 10, 2023 at 9:22
• 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 $\mathcal G(\mathcal O_K)$. Commented Sep 10, 2023 at 10:39

## 1 Answer

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).

• 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? Commented Sep 10, 2023 at 9:33
• * integral model Commented Sep 10, 2023 at 9:55
• 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. Commented Sep 10, 2023 at 10:42
• 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.
– anon
Commented Sep 13, 2023 at 23:39
• 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? Commented Oct 1, 2023 at 11:44