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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.

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  • $\begingroup$ It would be helpful if you define the system $\mathcal{A}$. $\endgroup$ – ThiKu Mar 22 '18 at 20:03
  • $\begingroup$ @ThiKu: I added the definition of $\mathcal{A}$. $\endgroup$ – user113290 Mar 22 '18 at 20:08
  • $\begingroup$ @YCor: You are right, I already corrected the definition of hyperbolic isometry. $\endgroup$ – user113290 Mar 22 '18 at 20:43
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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.

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