Let $L$ be a $p$-adic field, let $G$ be a reductive group over $L$ (I'm even okay assuming semisimplicity for now). Let $T$ be a maximal torus of $G$. Let $B$ be the building for $G(L)$. (**Edit 1:** "split" has been removed from "maximal torus"; the building of $U\_3(L/F)$, where $L/F$ is a ramified quadratic extension of $p$-adic fields, provides a counterexample to the question if we require $T$ to be split).


I'm interested in the relationship between, on the one hand: $T(L)$-orbits in $B$ and, on the other hand: fixed sets $B^{T\_0}$ where $T\_0 \subseteq T(L)$ is an open-compact subgroup.


Clearly, since $T\_0$ is commutative, any set $B^{T\_0}$ is closed under multiplication by $T$ and is therefore a union of $T(L)$-orbits. It is, of course, too much to hope that every $T(L)$-orbit is of the form $B^{T\_0}$ for some fixed compact-open subgroup $T\_0$ of $T(L)$, for two reasons: the first is that all elements of a $T(L)$-orbit are of the same type, which is not true of sets of the form $B^{T\_0}$. The second is that for any open-compact subgroup $T\_0$, $B^{T\_0}$ contains the apartment of $T(L)$ (and possibly contains more).


My question is as follows: 



> 
> Let $X$ be a $T(L)$-orbit in $B$, and let $Typ(x) = t$ for every
>  $x\in X$. When are there compact-open subgroups $T\_0 \subseteq T\_0'$ such
>  that $X = (B^{T\_0} - B^{T\_0'})\cap Typ^{-1}(t)$?
> 
> 
> 


and, more, specifically: 



> 
> For a reductive (semisimple) group over a global field $F$, are there
>  only finitely many primes at which the above fails?
> 
> 
> 


Let me give a little bit of clarity by way of example. Let's take $G = SL\_2$, let $T$ be the diagonal torus, and let $A\_0$ be the standard apartment. Then one can check by direct computation that two vertices $v,\, v'$ are in the same $T$-orbit if they are of the same type and if $d(v, \,A\_0) = d(v', \,A\_0)$ (i.e. they are the same distance from the standard apartment).


On the other hand if define the open-compact subgroups 
$$T\_{r} := \bigg\{\begin{pmatrix} t & 0 \\ 0 & t^{-1}\end{pmatrix}: t \in 1 + \varpi^r \mathcal{O}\_L\bigg\}.$$
Then if the residue characteristic of $L$ is at least $3$, the fixed set $B^{T\_r}$ consists of all vertices that are at most distance $r$ from the standard apartment $A\_0$. As such, in the case where the residue characteristic is $> 2$, we can answer in the affirmative; if an orbit $X$ consists of vertices that are distance $r$ from $A\_0$, then we can write $X = B^{T\_r} - B^{T\_{r-1}}$.


On the other hand if the residue characteristic of $L$ is $2$, and $T\_0$ is the maximal compact subgroup of $T\_0$, then by the notation above $T\_0 = T\_1$ so $T\_0$ fixes the standard apartment and all vertices of distance $1$. Therefore, we cannot write the standard apartment as a difference of fixed sets.


This question is extremely open-ended, and I'm really interested in anything that can be said in this area. Even some illumination of examples where $G$ is split-rank $1$ (so $B$ is a tree) beyond $SL\_2$ would be of great interest to me.



When are toral orbits in buildings the difference of fixed-sets?
(Editted: a "weaker" example about $GL\_3$ at the end)
If I didn't make a mistake in my computation, then the second question doesn't hold for $G=Sp\_4$, as it doesn't hold for any $\mathbb{Q}\_p$. Allow me to use $F$ as my p-adic field and $k$ its residue field. Let $V/\_F$ be spanned by $e\_2,e\_1,f\_1,f\_2$ with the symplectic form $(e\_i,f\_j)=\delta\_{ij}$, $(e\_i,e\_j)=(f\_i,f\_j)=0$, so that we identify $G(F)=Sp(V)$.


The hyperspecial vertices on the building correspond to self-dual lattices of $V$. Let $T$ be the diagonal torus which acts on $V$ by $(c\_1,c\_2):e\_i\mapsto c\_ie\_i, f\_i\mapsto c\_i^{-1}f\_i$. Let $\pi\in\mathcal{O}\_F$ be a fixed uniformizer. Consider the lattices


$$\Lambda\_{a,b}=\mathcal{O}\_F\langle e\_2+\pi^{-1} e\_1+\pi^{-2}af\_1+\pi^{-3}bf\_2, e\_1+\pi^{-1}f\_1+\pi^{-2}(a-1)f\_2,f\_1-\pi^{-1}f\_2,f\_2\rangle,$$


where we let $a,b$ runs over a fixed set of representatives of $k$ in $\mathcal{O}\_F$. One checks that all $\Lambda\_{a,b}$ are in different $T(F)$-orbits by essentially showing that they are in different $T(\mathcal{O}\_F)$-orbits.


Next one compute the stabilizer of $\Lambda\_{a,b}$. What's necessary then is to check the following:


**Claim.** (1) $\text{Stab}\_{T(\mathcal{O}\_F)}(\Lambda\_{a,b})\supset T\_3=T(1+\pi^3\mathcal{O}\_F)$.


(2) If $ab+b-a+1\not\in\pi\mathcal{O}\_F$, then $\text{Stab}\_{T(\mathcal{O}\_F)}(\Lambda\_{a,b})\subset T\_2=T(1+\pi^2\mathcal{O}\_F)$.


However, there are only $q+3$ ($q=\#k$) subgroups between $T\_2$ and $T\_3$. In other words, we have $q^2-q+1$ such lattices, but only $q+3$ possible choices for stabilizers for them; an open compact subgroup of $T(F)$ have to correspond to about $O(q)$ orbits.


Note. I did some brute force (using Iwasawa decomp.) and it seems that such an assertion holds for $GL\_3$ in most or all cases of $GL\_3$. However it also seems to me that if one fix a (semisimple) split rank, then it will be difficult for similar assertion to holds for larger $\dim(G)$; in other words, I'd say such an assertion probably only holds for $GL\_n$ (and not for any other classical groups or non-split groups), if it does.


--


Editted: Here is an example about $GL\_3$ where things fail. Let $T$ again be the diagonal torus. Let $a>0$ be any integer. Consider the two lattices in $F^3$


$$\Lambda\_a=\mathcal{O}\_F\langle e\_1+\pi^{-a}e\_2+\pi^{-a}e\_3,e\_2,e\_3\rangle$$
$$\Lambda\_a'=\mathcal{O}\_F\langle e\_1+\pi^{-a}e\_3,e\_2+\pi^{-a}e\_3,e\_3\rangle$$


These two lattices correspond to two hyperspecial vertices which are not in the same $T(\mathcal{O}\_F)$-orbit and thus $T(F)$-orbit. But they have the same stabilizer $Z(F)\cdot T(1+\pi^a\mathcal{O}\_F)$.


This is however a weaker example because in this case a stabilizer correspond to $2$ orbits. It's probably the case that for $GL\_3$ every possible stabilizer corresponds to at most $2$ orbits, but I have no good intuition about why it should be true. (Note that the number of orbits with distant $r$ from the apartment no longer have a uniform bound independent of $p$.)


