Let $ G $ be a simple linear algebraic group. Let $ G\_\mathbb{R} $ be the real points of $ G $. Let $ G\_\mathbb{Z} $ be the integer points of $ G $. Is $ G\_\mathbb{Z} $ a maximal closed subgroup? In other words, are the only closed subgroups $ H $ of $ G\_\mathbb{R} $ such that
$$
G\_\mathbb{Z} \subset H \subset G\_\mathbb{R}
$$
just $ G\_\mathbb{R} $ and $ G\_{\mathbb{Z}} $?


The answer is yes for for $\operatorname{SO}\_3$. $ \operatorname{SO}\_3(\mathbb{Z}) $ is the 24 element octohedral group (isomorphic to the symmetric group $ S\_4 $) and this group is a maximal finite subgroup of rotations and indeed it turns out that adding any other rotation and taking the closure will generate all of $ \operatorname{SO}\_3(\mathbb{R})$.


The group of integer points is also maximal in $ \operatorname{SL}\_n(\mathbb R) $. See the [answer](https://math.stackexchange.com/a/3390666) by YCor given to [Is $\operatorname{SL}\_2(\mathbb Z)$ a maximal discrete subgroup in $\operatorname{SL}\_2(\mathbb R)$?](https://math.stackexchange.com/questions/3388661/is-sl-2-mathbb-z-a-maximal-discrete-subgroup-in-sl-2-mathbb-r).


Although this only states that $ \operatorname{SL}\_n(\mathbb{Z}) $ is maximal among discrete subgroups I'm guessing that it is in fact maximal among closed subgroups as well in this case.



Is the group of integer points of a simple real linear algebraic group a maximal closed subgroup?
Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}\_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL\_n(\mathbb Z)$ and hence under all of $\operatorname{SL}\_n(\mathbb R)$ (since $\operatorname{SL}\_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}\_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}\_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}\_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}\_n(\mathbb R)$.


If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.


In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K\_p$ is a maximal compact open subgroup of $G(\mathbb Q\_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K\_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.


