Here is the context in which I heard it. If $G$ is a reductive group over a non-archimedean local field $F$, then let $K$ be the hyperspecial maximal compact subgroup $G(\mathcal{O})$. For example, if $G=SO(n)$, and assuming that $G$ is split, then there is a natural choice of $SO(n)(\mathcal{O})$. What exactly does the last sentence mean? (I know that a hyperspecial maximal compact subgroup exists if and only if $G$ is quasisplit and splits over an unramified extension -- this is stated in Tits' Corvallis article.) The only context I am aware of is in the theory of Lie groups.

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    $\begingroup$ math.stackexchange.com/questions/2077999/what-is-a-split-son $\endgroup$ – Carlo Beenakker Jan 3 '17 at 17:48
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    $\begingroup$ Wikipedia to the rescue: en.wikipedia.org/wiki/… $\endgroup$ – Andrei Smolensky Jan 3 '17 at 18:25
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    $\begingroup$ @AndreiSmolensky The problem is that I'm trying to learn what a split $SO(n)$ is, over a non-archimedean local field, not over the reals or a finite field. $\endgroup$ – Not a grad student Jan 3 '17 at 18:39
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    $\begingroup$ Well, it's the same thing: you take a vector space of dimension $n$, equip it with a quadratic form $\langle 1,\ldots,1,-1,\ldots,-1\rangle$ and consider the group of all isometries ot this form. The keywords are "a group of points of a split semi-simple algebraic group". $\endgroup$ – Andrei Smolensky Jan 3 '17 at 19:25
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    $\begingroup$ As indicated, the context here is the Borel-Tits theory of reductive (e.g., simple) algebraic groups defined over an arbitrary field $k$, not the more limited context of Lie groups. Here "split' just means that the group has a maximal torus defined and split over $k$: isomorphic over $k$ to a product of copies of the multiplicative group. $\endgroup$ – Jim Humphreys Jan 3 '17 at 19:31

In fact, to properly understand what is $SO(n)(\mathcal{O})$, you need to know what are split groups over a ring. A summary of this story goes as follows:

Given a ring $R$, and given a finite Dynkin diagram (https://en.wikipedia.org/wiki/Dynkin_diagram), you have a unique simply connected (resp. adjoint) absolutely simple split algebraic group over $R$. For example, for the $A_n$ diagram, the corresponding algebraic group is $SL_{n+1}$ (resp. $PGL_{n+1}\simeq SL_{n+1}/\mu_{n+1}$ ) considered as an algebraic group over $R$.

But you also have some groups "in between" the simply connected one and the adjoint one. In the $A_n$ case, this corresponds to the fact that you could quotient by proper subgroups of $\mu_{n+1}$.

Finally, your group $K$ in the $A_n$ case would just be $SL_{n+1}(\mathcal{O})$, or if working with the adjoint guy, it would be $PGL_{n+1}(\mathcal{O}) = GL_{n+1}(\mathcal{O})/\mathcal{O}^{\times} $. Warning here : computing rational points of quotients of algebraic groups is a complicated story. But it's ok for this particular quotient using that $\mathcal{O}$ is a PID.

Now, $SO(2n+1)(\mathcal{O})$ (resp. $SO(2n)(\mathcal{O})$) is just the group of rational points of a well-chosen quotient of THE simply connected absolutely simple split algebraic group over $\mathcal{O}$ of type $B_n$ (resp. $D_n$).

To make this more concrete and link that back to the comments, let us assume that $n$ is odd, or that $2$ is invertible in $\mathcal{O}$. Then actually $SO(n)(\mathcal{O})\simeq \lbrace g\in SL_n(\mathcal{O})~\vert~ q(g.x) = q(x)\rbrace $, where $q$ is the quadratic form $(x_{-k},...,x_k)\mapsto \sum x_{-i}x_i$ (we can have $2k+1$ or $2k$ variables, depending on whether or not we wish to include the index $0$).

If you want to know what $SO(n)$ is when $n$ is even and $2$ is not invertible in the base ring, you can for example look in "The Book of Involutions" by Knus, Merkurjev, Rost and Tignol (their definition of the split special orthogonal group works over an arbitrary ring). There is also the Appendix C of "Reductive Group Schemes" by Conrad, for that matter.

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    $\begingroup$ It may be worth noting that the definition of "split" over a general ring entails some subtleties that one doesn't notice over a field (such as in relation to "root spaces" being free of rank 1 rather than merely line bundles), but fortunately these finer issues don't arise over local rings such as $\mathcal{O}$ (so not strictly necessary for present purposes). This comes up when one wants to state (and prove) versions of the Existence and Isomorphism Theorem for "split" reductive groups over rings (e.g., labeling isomorphism classes by root data). $\endgroup$ – nfdc23 Jan 5 '17 at 21:57

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