Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$. Let $B$ be the corresponding reduced building. Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G\_{x,r}$. Our hypotheses give $G\_x=G\_{x,0}$.


Define 
$\Omega\_{A,r}$ as the set
$$ \{ y \in A : \forall \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r \}.$$



> 
> Is it true that $T(R)G\_{x,r} = \cap\_{y\in \Omega\_{A,r}} G\_y$ ?
> 
> 
> 


(With $r>0$ the product gives a group which is contained in the given intersection. Equality seems to be about whether all hyperplanes $H\_{\alpha,r}$ meet $\Omega\_{A,r}$.)


Let $A(x)$ be the set of all apartments of $B$ containing $x$. Let $Z$ be the (finite) center of $G$.



> 
> Is it true that $ZG\_{x,r} = \cap\_{A \in A(x)}\cap\_{y\in \Omega\_{A,r}}G\_y$ ?  
> 
>  Equiv by (1): Is $ZG\_{x,r} =\cap\_{g \in G\_x} (gT(R)g^{-1})G\_{x,r}$? 
> 
> 
> 


This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$). Pointers to any literature much appreciated.



When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)?

> 
> Regretfully the answer to the first question is no.
> 
> 
> 


For example consider a group of type $G\_2$. Normalize your roots so the short roots have length 1, for example, and let $x=0$. Write $\Phi = \Phi^l \cup \Phi^s$ as a union of long and short roots, respectively, and define the corresponding sets $\Omega^l\_{A,r}$ and $\Omega^s\_{A,r}$. Write $B\_r$ for the disk of radius $r$. 


Then $\Omega^l\_{A,r} \subseteq B\_{2r/3}$ whereas $B\_r \subseteq \Omega^s\_{A,r}$. Consequently the parenthetical comment fails. 


There was also a typo in the parenthetical comment, so let's be explicit. (Given a root $\alpha$, write $U\_\alpha$ for the corresponding root subgroup and normalize so $U\_\alpha(R)= U\_\alpha \cap G\_0$.) 


To see that $T(R)G\_{x,r} \subseteq \cap\_{y\in \Omega\_{A,r}} G\_y$, it suffices to verify this on a set of generators. We have $T(R) \subseteq G\_y$ for all $y\in A$. Then the rest of $G\_{x,r}$ is generated by the groups $U\_\alpha(P^{\lceil r-\alpha(x)\rceil})$.  

For each $y \in \Omega\_{A,r}$ we have $\alpha(x)-\alpha(y)\leq r$ so $-\alpha(y) \leq r-\alpha(x)$ which implies $U\_\alpha(P^{\lceil r-\alpha(x)\rceil}) \subseteq U\_\alpha(P^{\lceil -\alpha(y)\rceil})$. Thus $T(R)G\_{x,r}$ is contained in each of these $G\_y$.


Conversely, in the case of $G\_2$ described above, the inequalities imply that for sufficiently large $r$, there is no $y \in \Omega\_{A,r}$ for which the equality $\lceil -\alpha(y) \rceil = \lceil r-\alpha(x) \rceil$ ($=\lceil r \rceil$) holds; therefore the intersection will be strictly larger than $T(r)G\_{x,r}$.



> 
> It follows that the answer to the first version of the second question is also no. The second version is true for $SL(2)$, for example, so perhaps there is an argument via Chevalley generators and relations. 
> 
> 
> 


