# When are toral orbits in buildings the difference of fixed-sets?

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.

• Maybe I am being stupid - In the example of $SL(2)$ and $p$ odd, it seems to me that for any distance $r$ there are two $T(L)$-orbits? Does type means $G(L)$-orbits on the Bruhat-Tits building? – Cheng-Chiang Tsai Feb 21 '15 at 3:54
• @Cheng-ChiangTsai; yes, you are correct; that is the meaning of type. More specifically, a given chamber (in this case edge) contains one vertex of each type and, more generally, one facet of each type (so an edge contains two vertices, and the edge itself). If $G$ is semisimple, at least, then $G(L)$ acts transitively on facets of the same type. – John Binder Feb 21 '15 at 14:15

(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 $p$.)

• Comment on my note: I was thinking about the case when $G$ is split and $T$ is the split maximal torus. I think in general we at least want $T$ to contain the maximal split torus. If this is the case, I would then guess the possible cases for the assertion are when $G$ is (isogenous to product of) a form of $GL_n$. – Cheng-Chiang Tsai Feb 21 '15 at 20:22
• Thanks for your excellent answer. I'd love to take a look at your computations for $GL_3$ sometime, as these were computations which I was unable to finish and which motivated the question in the first place! – John Binder Feb 25 '15 at 19:39
• I am sorry that I haven't came back to it for long - it's really an ugly and lengthy computation. Iwasawa decomp. (which is quite simple for $GL_3$) tells us we can use "triangular" generators for a lattice as above. I'll edit and write part of it about $GL_3$ now. – Cheng-Chiang Tsai Mar 2 '15 at 2:46
• Or rather, I should say that Iwasawa decomp. makes us choose such lattice with "strictly triangular" generator as a canonical representative (with respect the choice of Borel and max. compact) of the $T(F)$-orbit. By the way, I think I once do some computation of this sort when I tried to computed the dimension of some "generalized" affine Springer fiber. May I ask what was the motivation for this observation and question? Thanks! – Cheng-Chiang Tsai Mar 2 '15 at 3:40
• Sure; the goal was to bound orbital integrals $O_{\gamma}(\Gamma_0(p^r))$, where $\Gamma_0(p^r)$ are the matrices in $GL_3(\mathbb{Z}_p)$ that are upper-triangular mod $p^r$. When we replace $GL_3$ with $GL_2$, it is fairly easy to do this using the Bruhat-Tits building, since you can describe the fixed set of $\gamma$ acting on the tree of $SL_2$ in a very nice way. But for $GL_3$ (and, really, the other examples I've worked with) it is much harder to get a handle on the fixed sets. – John Binder Mar 3 '15 at 4:20