4
$\begingroup$

$\DeclareMathOperator\GL{GL}$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.

$\endgroup$
5
  • $\begingroup$ @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

2
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.