# Global bound for number of vertices in Bruhat-Tits building

Let $G$ denote a semisimple linear algebraic group over $\mathbb Q$ and let $r$ be its absolute rank. For any prime $p$ let $v_p$ be a vertex of the Bruhat-Tits building of $G({\mathbb Q}_p)$ and let $D(p)$ denote the number of vertices neighboring $v_p$. Can we say that there exists a constant $C>0$ such that $$D(p)\le C p^r$$ holds for all $p$? If so, what is the minimal value for $C$?

• The question you formulate may be reasonable, but it would help a lot to indicate what your primary sources are, along with some motivation and/or evidence. A natural example to start with is $G= \mathrm{SL}_2$, where the Bruhat-Tits building is a tree. What value of $C$ is optimal there? Are there other examples where you can answer the question directly? – Jim Humphreys Jan 9 '16 at 21:31
• Notice that it may happen that $C=0$. Indeed take for $G$ a simple group of absolute rank $r>0$ and assume it is anisotropic over ${\mathbb Q}_p$. Then the building is reduced to a single vertex $v$ and $D(p)=0$. I.e. Take $G({\mathbb Q}_p )= D^\times /{\mathbb Q}_p^\times$, where $D$ is a central division algebra of degree $d$ over ${\mathbb Q}_p$ (in that case $r=d^2 -1$). – Paul Broussous Jan 12 '16 at 15:50

To make things simpler assume that $G$ is simple so that its building is a genuine simplicial complex. Let $X$ be the Bruhat-Tits building of $G({\mathbb Q}_p )$ and fix a vertex $v$ of $X$. Using Bruhat-Tits theory, one attaches to $v$ a connected reductive group $G_v$ over the residue field ${\mathbb F}_p$. For instance when $G={\rm SL}(n)$, $G_v ={\rm SL}(n,{\mathbb F}_p )$ for any $v$. Then it is a classical fact that the neighbourg vertices of $v$ are in $1-1$ correspondence with the maximal parabolic subgroups of $G_v$. Your are thus reduced to a problem of combinatorics of flag varieties over finite fields.