# Fixed points in the Bruhat-Tits building

Let $G$ be a connected reductive group over a complete discrete valuation field with perfect residue field (or just a non-arch local field). Let $\mathcal{B}$ be its reduced Bruhat-Tits building, and $d(\cdot,\cdot)$ the distance function on $\mathcal{B}$. I am wondering if the following statement is true:

There exists a constant $C$ such that for any compact element $x\in G$ (i.e. $\{x^n|n\in\mathbb{Z}\}$ is contained in a compact subgroup), and any $p_1, p_2\in\mathcal{B}$ such that $x.p_1=p_2$, there exists $p_0\in\mathcal{B}$ with $$x.p_0=p_0,\;d(p_0,p_1)\le C\cdot d(p_1,p_2).$$

What I actually want is have an explicit $C$ in terms of $G$. For example, I'll guess that $C$ can be bounded by a constant multiple of the rank of $G$. E.g. when the rank is $1$ the assertion is true for $C=\frac{1}{2}$.

The motivation is for the following question, which in terms is for the interest of orbital integrals: Can we cover the compact'' part of a $G(\mathcal{O})$-double cosets by a bounded estimable number of conjugates of $G(\mathcal{O})$? Thanks!

## 1 Answer

This is not really an answer but this may help. Write ${\mathcal B}^x$ for the set of points fixed by $x$ in $\mathcal B$. This is a closed convex subset of $x$ and it is stable by the action of $x$. It is classical that any point $p$ of $\mathcal B$ has a unique projection to ${\mathcal B}^x$. So in the inequality that one wants to prove one may take for $p_0$ the projection of $p_1$ onto ${\mathcal B}^x$. Note that by equivariance, it is also the projection of $p_2$ onto ${\mathcal B}^x$. I do not know how to prove the existence of $C$, but on the other hand the following inequality always holds: $$d(p_0 ,p_1 )\geqslant \frac{1}{2}. d(p_1 ,p_2 )$$ (with equality in rank $1$). Indeed write the CAT($0$) inequality for the triangle $(p_0 ,p_1 ,p_2 )$ : $$d(p_0 ,p_1 )^2 +d(p_0 ,p_2 )^2 \geqslant 2d(p_0 ,m)^2 +\frac{1}{2}d(p_1 ,p_2 )^2$$ where $m$ is the middle point of $[p_1 ,p_2]$. Since the action of $x$ is isometric, one has $d(p_0 ,p_1 ) =d(p_0 ,p_2 )$. Hence we obtain $$2d(p_0 ,p_1 )^2 \geqslant \frac{1}{2}d(p_1 ,p_2 )^2$$ and my claim follows.

As a consequence, if your $C$ exists, it must be $\geqslant 1/2$.