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Oct 1, 2023 at 16:57 comment added David Loeffler A hyperspecial maximal compact subgroup is, in particular, a maximal compact subgroup. So I don't understand your question.
Oct 1, 2023 at 11:44 comment added user267839 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?
Sep 13, 2023 at 23:39 comment added anon 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.
Sep 10, 2023 at 12:05 vote accept user267839
Sep 10, 2023 at 10:42 comment added LSpice 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.
Sep 10, 2023 at 9:55 comment added user267839 * integral model
Sep 10, 2023 at 9:33 comment added user267839 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?
Sep 10, 2023 at 7:43 history answered David Loeffler CC BY-SA 4.0