Given the group $\mathrm{SL}(2,\mathbb{Q}_p)$ one can describe precisely the number of vertices in the Bruhat-Tits building of distance $k$ from a fixed vertex $v$, where the distance between two vertices $v$ and $v'$ is the minimal number of edges in a path connecting the two. Denoting this quantity by $N_k$, we see in particular that for $\mathrm{SL}(2,\mathbb{Q}_p)$ $$N_k=(q+1)q^{k-1},$$ where $q$ is the size of the residue field.

I am curious if one can come up with an analogous closed formula for $G=\mathrm{SL}(n,\mathbb{Q}_p)$. In the case that $k=1$, this is not difficult and well known by using the lattice interpretation of the building for $\mathrm{SL}$ and counting the number of non-trivial $k$-dimensional subspaces of $\mathbb{F}_p^n$ (one can also use spherical buildings for this and more general groups). For instance, in the case of $n=3$, we have $$N_1=2(q^2+q+1).$$ Continuing the example where $n=3$, to compute $N_2$ is equivalent to finding the number of lattices $\Lambda$ such that if $$\Lambda_0=\mathbb{Z}_p\oplus\mathbb{Z}_p\oplus\mathbb{Z}_p,$$ then $$p^2\Lambda_0\subsetneq \Lambda\subsetneq\Lambda_0,$$ with the additional requirements that $\Lambda$ does not contain $p\Lambda_0$ and that $\Lambda$ is not contained in $p\Lambda_0$ (this ensures that $\Lambda$ is truly distance $2$). However, I have not been able to make the combinatorics of this count work on paper. In general when $n=3$, to compute $N_k$ one would need to count the analogous $\Lambda$s in a chain with $p^2\Lambda_0$ replaced by $p^k\Lambda_0$ subject to the condition that $\Lambda$ does not contain any $p^i\Lambda_0$ and is not contained in any $p^j\Lambda_0$ where $1\le i,j\le k-1$ (to ensure genuine distance $k$).

I am looking to see if a closed formula for $N_k$ is known in the literature or is easy to compute and I am missing the computation. I would be extremely pleased with just the case of $G=\mathrm{SL}(3,\mathbb{Q}_p)$, even if just the case of $n=3$ and $p=2$.

  • $\begingroup$ You will have $n-1$ types of vertices. The $i$th type will correspond to the number of submodules of $(\mathbb{Z}/p^k\mathbb{Z})^n$ isomorphic to $(\mathbb{Z}/p^k\mathbb{Z})^i$. It is not hard to make an explicit computation (which I currently don't have the time to do), but you can surely find it also in Macdonald "Symmetric functions and Hall polynomials". $\endgroup$ – Uri Bader Mar 1 '17 at 18:39
  • $\begingroup$ @UriBader Thank you for the response! If I understand it correctly, I think this gives you some but not all of the vertices, since vertices of the same type as the original vertex may show up. For instance if $k=2$, $n=3$ then the class of the lattice $\mathbb{Z}_p\oplus p\mathbb{Z}_p\oplus p^2\mathbb{Z}_p$ is such a vertex. In the case of $k=2$ these are fairly easy to count (by counting vertices of distance 2 lying in a chamber which intersects another chamber containing the original vertex on a codimension 1 simplex), however I don't quite see the count for these types in general. $\endgroup$ – Guest Mar 2 '17 at 4:25
  • $\begingroup$ Sorry, I guess I was not focused when commenting. Fix $n$ integers $k= \lambda_1\geq \lambda_2 \ldots \lambda_n=0$ and let $\lambda=(\lambda_1,\ldots,\lambda_n)$. A module of type $\lambda$ is one which is isomorphic to $\oplus\mathbb{Z}/p^{\lambda_i}\mathbb{Z}$. You want to count all the submodules of type $\lambda$ in a module of type $(k,k,\ldots,k)$. Now you want to sum up over all such $\lambda$s. In claim 4.5(1) in arxiv.org/pdf/math/0404408.pdf you will find such a computation (with notation introduced before prop 3.2). Probably you can find it already in Macdonald book. $\endgroup$ – Uri Bader Mar 2 '17 at 7:32
  • $\begingroup$ @UriBader Thanks, this is exactly what I was looking! I was able to turn it into a closed formula for $n=3$ (surely for larger $n$ it can't be much harder, just more tedious). I'm happy to accept your comment as an answer if you want to turn it into one. $\endgroup$ – Guest Mar 2 '17 at 18:37
  • $\begingroup$ In my post below we have a summation over various partitions. I didn't make an effort to give a simpler expression. If you happen to get one, please share it with me. $\endgroup$ – Uri Bader Mar 4 '17 at 13:56

We consider the Bruhat-Tits building of $\text{SL}_n(\mathbb{Q}_p)$. As usual, we identify the vertices with lattices, that is subgroups of the of $\mathbb{Q}^n$ commensurated with the standard lattice $\Lambda_0=\mathbb{Z}_p^n$, defined up to homothety, namely multiplication by $p^k$ for some $k\in \mathbb{Z}$. Getting rid of the homothety relation, we can represent each lattice uniquely as a subgroup $\Lambda$ of $\Lambda_0$ which is not contained in $p\Lambda_0$. Such a $\Lambda$ will contain $p^k\Lambda_0$ for some natural $k$, and the minimal such $k$ is said to be the distance from $\Lambda$ to $\Lambda_0$.

From now on by vertex I mean a representation of homothety class of lattices given as above. Thus the set all vertices of distance $k$ from $\Lambda_0$ is given by $$ \{\Lambda\mid p^k\Lambda_0<\Lambda<\Lambda_0,~\Lambda\notin p\Lambda_0 \mbox{ and the rank of } p^k\Lambda_0/\Lambda\mbox{ is less than }n\}.$$ Dividing by $p^k\Lambda_0$, the set above is in bijection with the set of all $p$-abelian subgroups $$ \{A \mid A < (\mathbb{Z}/p^k\mathbb{Z})^n,\mbox{ the ranks of } A\mbox{ and } (\mathbb{Z}/p^k\mathbb{Z})^n/A \mbox{ is less than }n\}.$$

Recall that every for every $p$-abelian group $A$ there exists a unique vector of natural numbers $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_m)$ with $\lambda_1\geq \lambda_2\geq \ldots\geq \lambda_m> 0$ such that $$ A\simeq \bigoplus_{i=1}^m \mathbb{Z}/p^{\lambda_i}\mathbb{Z}.$$ In that case we say that $A$ is of (isomorphism) type $\lambda$. If for all $i$ $\lambda_i=k$ we say $\lambda=k^m$.

Given types $\nu$ and $\lambda$ we denote by ${\nu\choose \lambda}_p$ the number of submodules of type $\lambda$ of a given module of type $\nu$. We conclude that the number of vertices of distance $k$ is given by $\sum {k^n\choose \lambda}_p$ where the sum runs over all $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_m)$ with $\lambda_1=k$ and $m<n$.

We are left with a concrete computation. Fixing $\nu$ and $\lambda$, ${\nu\choose \lambda}_p$ is a polynomial in the variable $p$ which is computed in claim 4.5(1) in https://arxiv.org/pdf/math/0404408.pdf: $$ {\nu\choose \lambda}_p=p^{\langle \nu'-\lambda',\lambda'\rangle}\prod_i\begin{bmatrix} \nu'_i-\lambda'_{i+1} \\ \nu'_i-\lambda'_i\end{bmatrix}_p$$ where $\nu',\lambda'$ are the transpose diagrams of $\nu,\lambda$, $\langle \nu,\lambda\rangle=\sum \nu_i\lambda_i$ and we use the notations: $$ [i]_1=1-p^{-i},~[m]_p!=\prod_{i=1}^m [i]_p,~\begin{bmatrix} n \\ m\end{bmatrix}_p =\frac{[n]_p}{[m]_p[n-m]_p}.$$


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