I am willing to fill in the gaps I found in the introduction of the following two papers :
- W. Woess, Lamplighters, Diestel-Leader graphs, random walks, and harmonic functions, (link).
- A. Terras and D. Wallace, Selberg's trace formula on the k-regular tree and applications (link).
$\newcommand{\bbN}{\mathbb{N}}$ By $T_p$ I mean the regular $(p+1)$-tree (the tree where each vertex has degree $p+1$). Define a ray of $T_p$ as a sequence $(e_n)_{n\in\bbN}$ of vertices such that $$d(e_i,e_j)=|i-j|$$
for all $i,j\in\bbN$. Define an end of $T_p$ to be an equivalence class of rays, and denote the set of such ends as $\partial T_p$, where two rays $(e_n)_{n\in\bbN}$ and $(f_n)_{n\in\bbN}$ are equivalent if the sets $\{e_n,\;n\in\bbN\}$ and $\{f_n,\;n\in\bbN\}$ have infinite intersection.
Now, fix an end $\omega\in\partial T_p$. The situation can be described picturally :
The line in bold represents a ray leading to an end $\xi\neq\omega$.
I have proven that for any vertex $x\in T_p$, there is a unique ray $\overline{xw}$ starting at $x$ and leading to $\omega$. Now, define the confluent of two vertices $x,y\in T_p$ to be the unique $z=x\curlywedge y$ such that $$\overline{x\omega}\cap\overline{y\omega}=\overline{z\omega}.$$
Finally, by also fixing a vertex $o\in T_p$, one can define a Busemann function $\mathfrak{h}$ on $T_p$ by : $$\mathfrak{h}(x)=d(x,x\curlywedge o)-d(o,x\curlywedge o),$$
with $d$ the graph metric on $T_p$. The horocycle of level $k$ is $$H_k=\{x\in T_p\;/\;\mathfrak{h}(x)=k\}.$$
Picturally, we have :
It is clear that horocycles partition $T_p$. What I am concerned about are the other claims that are made in the papers :
- Horocycles are infinite.
- Each $x\in H_k$ has exactly one neighboor in $H_{k-1}$ and $p$ neighboors in $H_{k+1}$.
If those two properties are very intuitive and visual once we picture the situation, I have no idea how to prove them formally.
How would one proceed to prove these claims ?
So far, for 2., by writing $y_0,...,y_p$ the $p+1$ neighboors of $x\in H_k$, I tried computing some relations between $y_i\curlywedge o$ and $x\curlywedge o$, but without success. Considering a path $\gamma:e_0=x\to e_1\to...\to e_{n-1}\to e_n=x\curlywedge o$ lead me to two cases : either $y_i=e_1$, or $y_i\notin\gamma$, but in both cases, I wasn't able to prove anything useful.
For 1., if I knew that some $H_k$ was infinite, then by 2., I would also get that all the $H_{k-\ell}$, $\ell\geqslant0$ are infinite, by induction, by considering the function $H_k\to H_{k-1}$ sending a vertex $x\in H_k$ to its only neighboor in $H_{k-1}$ (since pre-images of vertices $y\in H_{k-1}$ have at most $p+1$ elements, because we are in $T_p$). But this has two issues : we don't know for sure how to prove that one horocycle is infinite (otherwise, by mimicing the argument, 1. would be solved already), and this would only tell us that somewhat "half" of the horocycles would be infinite...
To get arbitrarily many elements of $H_k$, first take $n$ steps from $o$ along $\overline{o\omega}$ (reaching a vertex we'll call $y$). Then, take $n+k$ more steps that don't follow $\overline{o\omega}$ and don't backtrack. (This requires $n \ge -k$, but otherwise works for arbitrary $n$.) This gives us a vertex $x$ and I claim $x \in H_k$.
To see this, note that following the last $n+k$ steps backward from $x$ to $y$, then following the remainder of $\overline{o\omega}$, gives a ray leading to $\omega$; you've proven that it must be unique, so it is $\overline{x \omega}$. Similarly, starting at $y$ and following the remainder of $\overline{o\omega}$ (without the first $n$ edges) must give $\overline{y\omega}$. This tells us $\overline{x \omega} \cap \overline{o \omega} = \overline{y \omega}$, so $y = x \curlywedge o$, and therefore $$\mathfrak{h}(x) = d(x,y) - d(o,y) = (n+k) - n = k.$$
To prove 2, let's first be more specific. If $x \in H_k$, let $x_0$ be $x$'s neighbor along $\overline{x \omega}$, and $x_1, \dots, x_p$ be $x$'s other neighbors. Then $x_0 \in H_{k-1}$ and $x_1, \dots, x_p \in H_{k+1}$.
Taking $\overline{x\omega}$ and deleting $x$ from the beginning gives a ray starting at $x_0$ and leading to $\omega$; this must be $\overline{x_0 \omega}$. Similarly, for $1 \le i \le p$, taking $\overline{x\omega}$ and prepending $x_i$ must give $\overline{x_i \omega}$.
In most cases, we have $x \curlywedge o = x_i \curlywedge o$ for $0 \le i \le p$, and we get from $x_i$ to $x \curlywedge o$ by walking along $\overline{x_i \omega}$. This tells us that $d(x_0, x \curlywedge o) = d(x, x\curlywedge o) - 1$ and $d(x_i, x \curlywedge o) = d(x, x\curlywedge o) + 1$ for $1 \le i \le p$. From here, we get $x_0 \in H_{k-1}$ and $x_1, \dots, x_p \in H_{k+1}$.
This doesn't quite happen if $x$ lies on $\overline{o\omega}$, in which case two of $x$'s neighbors are also on $\overline{o\omega}$, and you'll have to handle those separately.