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Let $k$ be a finite extension of $\mathbb{Q}_p$ and let $G$ be a semisimple Lie group over $k$. We consider the action of $G$ on its Bruhat-Tits building $X$.

Question: If $x,y,x',y'$ are vertices, all of the same type, and $d(x,y)=d(x',y')$, is there a $g\in G$ with $gx=x'$ and $gy=y'$?

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    $\begingroup$ What is the meaning of "the same type"? for instance $PGL_d(Q_p)$ and $SL_d(Q_p)$ acts on the same BT-building but don't have the same vertex orbits. $\endgroup$
    – YCor
    Commented Dec 5, 2018 at 17:13
  • $\begingroup$ Assume for simplicity sake that the building is a simplicial complex. Then it is labellable : if $\Delta$ is the standard simplex with vertex set $\{ 0,1,...,d\}$, $d={\rm dim}(X)$, there is a simplicial map $l$ : $X\rightarrow \Delta$ whose restriction to any chamber is an isomorphism. For a vertex $s$, its label $l(s)$ is sometimes called its type. Is this what you mean? $\endgroup$
    – Paul Broussous
    Commented Dec 5, 2018 at 17:58
  • $\begingroup$ Remark. You can reduce your problem to an appartment. Any pair of vertices are in a commun appartement and $G$ acts transitively on the appartments of $X$, so you may assume that $x$, $y$, $x'$, $y'$ all lie in some fixed apparment, and you can replace $G$ by the affine Weyl group. $\endgroup$
    – Paul Broussous
    Commented Dec 5, 2018 at 18:02
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    $\begingroup$ The question seems to me to be about unravelling the definition of the Bruhat--Tits building associated to $G$. If it is defined as the building associated to the affine $BN$-pair of $G$, then the answer is straightforward by observing that 2-transitivity is the same as Weyl transitivity, i.e. that a chamber stabilizer $B$ acts transitively on each $w$-sphere. This is implied by the Bruhat decomposition $G=BWB$ associated to the $BN$-pair. See Abramenko--Brown's "Buildings" chapter 6.2.1, especially 6.34 and 6.35, for a more detailled discussion. $\endgroup$
    – Guntram
    Commented Dec 5, 2018 at 19:35
  • $\begingroup$ I think @PaulBroussous just about has a counterexample. Let $x = x'$. Then the subset of the affine Weyl group fixing $x$ is just a finite Weyl group. And that definitely does not act transitively on the set of pts in the apartment at a fixed distance from $x$, when that fixed distance is chosen appropriately. E.g., $S_3$ does not act transitively on the set of 12 vertices at distance $\sqrt{7}$, in type $A_2$. $\endgroup$
    – Marty
    Commented Dec 5, 2018 at 22:06

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