TITLE: Bruhat-Tits building of $SL_n(\mathbb{Q})$, hyperbolic isometries and its axis
QUESTION [6 upvotes]: Consider $G=SL_n(\mathbb{Q})$ and $p$ a prime integer. Associated to $G$ and $p$ we have its Bruhat-Tits building $\Delta$.
It is well known that  $\Delta$ can be provided with a canonical $CAT(0)$ metric and an action of $G$ by isometries. Hence every element in $G$ determines an isometry that it is either:
Elliptic: it has a fixed point, or
Hyperbolic: it has an axis, i.e. a geodesic $L$ on which acts as a nontrivial translation.
Associated to $G$ there is a system of apartments $\mathcal{A}=\{g\Sigma : \Sigma \text{ the fundamental apartment}\}$  of $\Delta$ which is not the full system of apartments. 
My question is the following:
Suppose $\alpha\in G$ is a hyperbolic isometry of $\Delta$ and $L$ is an axis of $\alpha$, is there an apartment in $\mathcal{A}$ containing $L$?
There is certainly an apartment in the full system of apartments of $\Delta$ that contains $L$.
Thank you all.

REPLY [2 votes]: If my understanding of the question is correct then the answer is "not necessarily".
I understand that $\Delta$ is the Bruhat-Tits building of $H=\text{SL}_n(\mathbb{Q}_p)$, while $G=\text{SL}_n(\mathbb{Q})<H$ is a countable subgroup. In particular, $\mathcal{A}$ is countable.
Let $A<H$ be the diagonal subgroup and fix a regular element $\alpha\in A$.
The axis of translation of $\alpha$ is contained in the basic apartment $\Sigma$, and only in this apartment, by regularity.
Let $N$ be the normalizer of $A$ in $H$. $A<N$ is of finite index ($n!$), and thus the index of $N<H$ is of continuum cardinality.
Every element $hN\in H/N$ could be identified with the unique apartment $h\Sigma$ containing the axis of translation of the hyperbolic element $\alpha^h$. Not all could be contained in $\mathcal{A}$ by the cardinality constraint.