# Uniform spanning forest on the integer lattice
with drift in one coordinate

By Guillermo Martinez Dibene
Mathematics Department
The University of British Columbia e-mail<EMAIL_ADDRESS>
###### Abstract

In this article we investigate the Uniform Spanning Forest $(\mathsf{USF})$ in
the nearest-neighbour integer lattice
$\mathds{Z}^{d+1}=\mathds{Z}\times\mathds{Z}^{d}$ with an assignment of
conductances that makes the underlying (Network) Random Walk ($\mathsf{NRW}$)
drifted towards the right of the first coordinate. This assignment of
conductances has exponential growth and decay; in particular, the measure of
balls can be made arbitrarily close to zero or arbitrarily large. We establish
upper and lower bounds for its Green’s function. We show that in dimension
$d=1,2$ the $\mathsf{USF}$ consists of a single tree while in $d\geq 3,$ there
are infinitely many trees. We then show, by an intricate study of multiple
$\mathsf{NRW}$s, that in every dimension the trees are one-ended; the
technique for $d=2$ is completely new, while the technique for $d\geq 3$ is a
major makeover of the technique for the proof of the same result for the graph
$\mathds{Z}^{d}.$ We finally establish the probability that two or more
vertices are $\mathsf{USF}$-connected and study the distance between different
trees.

_Keywords._ Uniform Spanning Tree, Uniform Spanning Forest, Biased Random
Walk, Drifted Random Walk, Green’s function, Harmonic Functions, Dirichlet
Functions, Liouville Property, Second Moment Method.

_Acknowledgements._ This research was partially supported by an NSERC (Canada)
grant.

###### Contents

  1. 1 Introduction
    1. 1.1 Basic definitions
    2. 1.2 Wilson’s algorithm
    3. 1.3 Definition of $\Gamma_{d}(\lambda)$ and summary of this paper
      1. 1.3.1 Biased random walk
  2. 2 Green’s function bounds
    1. 2.1 Basic statistics of $\Gamma_{d}(\lambda)$
    2. 2.2 Continuous time network random walk
    3. 2.3 Some estimates of sums
    4. 2.4 Some properties of (network) random walk
    5. 2.5 A hitting time for the continuous time network random walk
    6. 2.6 Statement and proof of the Theorem
  3. 3 The number of trees in $\mathsf{USF}$
    1. 3.1 Coupling of random walks
    2. 3.2 The Liouville property
    3. 3.3 Liouville property and the uniform spanning forest measure
    4. 3.4 The number of components
  4. 4 Crossings of $\mathsf{NRW}$
    1. 4.1 Dimensions $d=1,3,4,\ldots$
    2. 4.2 Two elementary results
    3. 4.3 Setup for $\Gamma_{2}(\lambda)$
      1. 4.3.1 Some preliminary sums of the Green’s function of $\Gamma_{2}(\lambda).$
      2. 4.3.2 Estimates on the Green’s function on some regions of $\Gamma_{2}(\lambda).$
    4. 4.4 Dimension $d=2;$ “logarithmic scale of crossings” of $\Gamma_{2}(\lambda)$
  5. 5 One endedness for $\Gamma_{d}(\lambda)$
    1. 5.1 One endedness for $\Gamma_{1}(\lambda)$
    2. 5.2 One endedness for $\Gamma_{2}(\lambda)$
    3. 5.3 Some preliminary results on Markov processes
      1. 5.3.1 Invariance of the chronology of Wilson’s algorithm rooted at infinity.
      2. 5.3.2 Preliminary estimates in the proof of one-endedness for $\Gamma_{2}(\lambda).$
      3. 5.3.3 The proof of the main theorem.
    4. 5.4 One endedness for $\Gamma_{d}(\lambda),$ with $d\geq 3$
    5. 5.5 On the rays of $\Gamma_{d}(\lambda)$
  6. 6 Probability of same component
    1. 6.1 Probability that two vertices are $\mathsf{USF}$-connected
    2. 6.2 Upper bounds on the probability of $\mathsf{USF}$-connectedness
  7. 7 Separation between components
    1. 7.1 Random relations and stochastic dimension
    2. 7.2 Tail triviality of random relations
    3. 7.3 The $\mathsf{WSF}$ connectedness relation
    4. 7.4 The main theorem

##  1 Introduction

_All graphs are assumed directed, connected, without multiple edges or self-
loops and denumerable. Given an edge $e,$ the direction of $e$ is given by a
starting vertex $e^{-}$ and a ending vertex $e^{+}.$ All conductances are
assumed strictly positive._

This work is based on the PhD thesis of the author; should the reader need
more details, consult [MD20]. A graph without cycles is called a tree; the
tree is spanning if it contains all vertices of the graph. Given a finite
connected graph $\mathrm{G},$ the number of subgraphs is also finite and thus,
there are only finitely many spanning trees in that graph. The uniform
spanning tree is, by definition, the probability measure on the set of
spanning trees of $\mathrm{G}$ satisfying $\mathds{P}(t)=\left(\\#\text{ of
spanning trees}\right)^{-1}.$ This can readily be generalised to the network
case, $\mathds{P}(t)=k\cdot\mathsf{weight}(t),$ where $k$ is a normalising
constant. Any of these measures (or the underlying random object) is called
“uniform spanning tree.”

Suppose now that $(\mathrm{G},\mu)$ is an infinite network. The first paper to
introduce $\mathsf{UST}$ on infinite graphs was Robin Pemantle’s [Pem91]. He
considered $\mathds{Z}^{d}$-balls corresponding to the norm
$\displaystyle\left\lVert x\right\rVert_{\infty}=\max_{1\leq i\leq d}|x_{i}|.$
He showed that the uniform spanning tree measure $\mu_{n}$ on the trees of
$\mathrm{B}_{n}$ converges weakly to some measure $\mu.$ This is the uniform
spanning measure on $\mathds{Z}^{d}.$ We will employ the notation
$\mathsf{USF}$ to denote it.

Later, Itai Benjamini, Russel Lyons, Yuval Peres and Oded Schramm, in [BLPS01]
and using tools unavailable to Pemantle, extended his result from
$\mathds{Z}^{d}$ to general networks. They started by considering an
exhaustion of the graph created by considering increasing families of finite
sets of vertices and then considering the networks induced on these sets. If
$(\mathrm{G}_{n})$ is any such exhaustion, they showed that the uniform
spanning forests measure of $\mathrm{G}_{n}$ defines a sequence of measures
that converge weakly to some measure $\mu^{F}.$ They named $\mu^{F}$ as “free
uniform spanning forest measure.” The second way to construct the uniform
spanning forest measure is by considering constraints in the boundary of
$\mathrm{G}_{n}.$ Indeed, the considered a sequence $(\mathrm{V}_{n})$ of
families of finite sets of vertices and then constructed the network
$\mathrm{G}_{n}^{W},$ which is the network induced on $\mathrm{V}_{n}$ with
the boundary of $\mathrm{V}_{n}$ wired into a vertex. Again, they showed that
the uniform spanning forest measures of $\mathrm{G}_{n}^{W}$ converge to a
measure $\mu^{W}$ and they called this measure the “wired uniform spanning
forest measure.” Several properties of $\mu^{F}$ and $\mu^{W}$ were studied,
in particular, necessary and sufficient conditions were given as to when
$\mu^{F}=\mu^{W}.$ We will employ $\mathsf{FSF}$ and $\mathsf{WSF}$ instead of
$\mu^{F}$ and $\mu^{W},$ respectively.

###  1.1 Basic definitions

Consider any network $(\mathrm{G},\mu).$ For a vertex $v,$ we define the
conductance of $v$ by $\mu(v)=\sum\limits_{e^{-}=v}\mu(v).$ The network random
walk ($\mathsf{NRW}$) in $(\mathrm{G},\mu)$ is the Markov chain with state
space $\mathrm{V}$ (the vertex set of $\mathrm{G}$) and transition probability
$p(u,v)=\mu(u)^{-1}\left(\sum\limits_{e^{-}=u,e^{+}=v}\mu(e)\right).$ If
$(S_{n})_{n\in\mathds{Z}_{+}}$ is the $\mathsf{NRW}$ of $(\mathrm{G},\mu)$, we
will write $\tau_{\mathrm{H}}=\inf\\{n\in\mathds{Z}_{+}\mid
S_{n}\in\mathrm{H}\\}$ and $\tau_{\mathrm{H}}^{+}=\inf\\{n\in\mathds{N}\mid
S_{n}\in\mathrm{H}\\}$ where $\mathrm{H}$ is any set of vertices. For sake of
simplicity, we write $\tau_{a}=\tau_{\\{a\\}},$
$\tau_{a}^{+}=\tau_{\\{a\\}}^{+}.$ $\tau_{\mathrm{H}}$ is the hitting time of
$\mathrm{H}$ and $\tau_{a}$ the visit time of $a;$ if
$\mathrm{H}=\mathrm{K}^{\complement},$ we will say exit time of $\mathrm{K}.$

Suppose $S=(S_{n})_{n\in\mathds{Z}_{+}}$ is the $\mathsf{NRW}$ on
$\mathrm{G}.$ Often we will use
$\mathds{P}{{}^{x}}\left(S_{n}=y\right)=\mathds{P}{}\left(S_{n}=y\mid
S_{0}=x\right)$ and
$\mathds{E}{{}^{x}}\left(f(S_{n})\right)=\mathds{E}{}\left(f(S_{n})\mid
S_{0}=x\right).$ The Green’s function “of the network $\mathrm{G}$” is, by
definition
$G(x,y)=\sum\limits_{n\in\mathds{Z}_{+}}\mathds{P}{{}^{x}}\left(S_{n}=y\right)=\mathds{E}{{}^{x}}\left(\sum\limits_{n=0}^{\infty}\mathds{1}_{\\{S_{n}=y\\}}\right),$
for all $(x,y)\in\mathrm{V}^{2}.$ Similarly, define the Green’s function
restricted to leaving $\mathrm{V}_{0}$ by
$G_{\mathrm{V}_{0}}(x,y)=\sum\limits_{m=0}^{\infty}\mathds{P}{{}^{x}}\left(S_{m}=y,m<\tau_{\mathrm{V}_{0}}^{\complement}\right).$

Suppose that $S_{0}=o$ is some fixed vertex of the network and suppose that
$N=(N_{t})_{t\geq 0}$ is a Poisson process in $\mathds{Z}_{+}$ with unitary
constant intensity which is independent of the process $S.$ We define the
continuous time $\mathsf{NRW}$ to be the process
$\tilde{S}=\left(\tilde{S}_{t}\right)_{t\geq 0}$ given by
$\tilde{S}_{t}=S_{N_{t}}.$ Define the continuous time $\mathsf{NRW}$
transition probability by
$q_{t}^{S}(o,v)=\mathds{P}{{}^{o}}\left(\tilde{S}_{t}=v\right),$ for
$(t,z)\in[0,\infty)\times\mathrm{V}.$

Let $\gamma=(v_{j})_{0\leq j\leq m}$ be a path of vertices. The loop-erasure
of $\gamma,$ denoted as $\mathsf{LE}(\gamma),$ is the path obtained from
$\gamma$ by deleting cycles in chronological order. More specifically, it is
defined as follows. Set $\gamma_{0}=\gamma.$

$(\mathsf{LE})$

Suppose we are at stage $k$ and we receive a path
$\gamma_{k-1}=\left({u}_{1},\ldots,{u}_{m_{k}}\right).$ If no vertex of this
path is repeated, then return $\mathsf{LE}(\gamma)=\gamma_{k-1}.$ Else,
consider $\tau_{k}$ to be the _first index_ $j$ such that the vertex $u_{j}$
appears twice in $\gamma_{k-1}.$ Then, let $\sigma_{k}$ to be the _last index_
$j$ such that $u_{j}=u_{\tau_{k}},$ consider now the path
$\gamma_{k}=\left({u}_{1},\ldots,{u}_{\tau_{k}},{u}_{\sigma_{k}+1},\ldots,{u}_{m_{k}}\right).$
Repeat.

A network $\Gamma$ is transitive if for every pair of vertices $x$ and $y$
there exists a network automorphism $\varphi$ of $\Gamma$ such that
$\varphi(x)=y.$

Let $\mathrm{A}$ be a subset of vertices. The interior of $\mathrm{A}$ is the
set of all vertices of $\mathrm{A}$ whose all neighbours are vertices of
$\mathrm{A}$ as well; we denote it with $\mathring{\mathrm{A}}.$ The closure
of $\mathrm{A},$ denoted $\overline{\mathrm{A}},$ is the set of all vertices
of $\mathrm{A}$ or that are adjacent to $\mathrm{A}.$ The (exterior vertex)
boundary of $\mathrm{A},$ denoted $\partial\mathrm{A},$ is the set of all
vertices _outside_ $\mathrm{A}$ that are adjacent to $\mathrm{A}.$

Let $\mathrm{G}$ be a network with vertex set $\mathrm{V}.$ A harmonic
function on the network $\mathrm{G}$ is any function
$h:\mathrm{V}\to\mathds{R}$ such that $h(x)=\sum\limits_{y\sim x}p(x,y)h(y).$

Let $(\mathrm{G},\mu)$ be a network with vertex set $\mathrm{V}.$ A function
$f:\mathrm{V}\to\mathds{R}$ is said to be a Dirichlet function if
$\mathscr{D}\left(f\right)=\sum\limits_{x,y}\mu(x,y)(f(x)-f(y))^{2}<\infty.$
The finite or infinite number $\mathscr{D}\left(f\right)$ is known as
Dirichlet energy of $f.$

###  1.2 Wilson’s algorithm

David Wilson published his algorithm in [Wil96]. (This algorithm is for finite
networks.)

$(\mathsf{WA})$

Order the vertices and set $t_{0}$ to be the tree consisting solely of the
first vertex. Having defined $t_{k-1},$ start an independent $\mathsf{NRW}$
from the first vertex not in $t_{k-1}$ and run this random walk until it hits
$t_{k-1},$ consider the $\mathsf{LE}$ of this path and call $t_{k}$ the tree
obtained as the union of this $\mathsf{LE}$-path with $t_{k-1}.$ When all
vertices have been searched, return $\mathfrak{T}$ the final tree constructed.

It is known that $\mathfrak{T}$ follows the $\mathsf{UST}$ law of the given
(finite) network. A typical way of using Wilson’s algorithm is via a more
intricate structure; while Wilson introduced this structure, a more detailed
exposition can be found in [Bar16]. We refer to this version of Wilson’s
algorithm as “Wilson’s algorithm with stacks.” See also [LP16], Ch. 4. With
the definitions of these references, we have the following result.

######  Proposition ( 1.1 )

Assume $(\mathrm{G},\mu)$ is a finite network. Fix a root $r\in\mathrm{V}.$
With probability one, there are finitely many cycles that can ever be popped.
Any two ordering of the searches of the vertices in
$\mathrm{V}\setminus\\{r\\}$ will pop all these cycles and the popping
procedure will leave the same visible graph for the two orderings. This
visible graph sitting on top of the stacks is a spanning tree and the
distribution of this random tree is that of the $\mathsf{UST}$ of the given
network.

Observe that Wilson’s algorithm run with stacks on a finite graph produces
always the same tree provided the root is fixed in advance.

We mention now the usual way to construct $\mathsf{WSF}$ on an infinite
network $(\mathrm{G},\mu).$ Consider an exhaustion $\mathrm{V}_{q}$ of the
vertex set and denote by $(\mathrm{G}_{q},\mu)$ the network induced on
$\mathrm{V}_{q}$ with its boundary $\partial\mathrm{V}_{q}$ wired into a
vertex. Run Wilson’s algorithm with stacks taking $\partial\mathrm{V}_{q}$ as
root. This produces a random tree $\mathfrak{T}_{q}.$ Then,
$\mathfrak{T}_{q}\xrightarrow[q\to\infty]{\mathrm{weakly}}\mathfrak{F},$ where
$\mathfrak{F}$ has distribution $\mathsf{WSF},$ see [LP16], sect. 10.1. There
exists an alternative more dynamic (it provides a.s. convergence as opposed to
weakly) method to sample the $\mathsf{WSF}$-distributed random object, which
was first introduced in [BLPS01].

Suppose $\mathrm{G}$ is any network. Denote by $\mathrm{V}$ the set of
vertices of $\mathrm{G}$ and by $\mathrm{E}$ that of edges. Suppose
$\xi:\mathds{N}\to\mathrm{V}$ is a bijection (an “ordering” of the vertex
set). Let $(\Omega,\mathscr{F},\mathds{P})$ be a probability space in which is
possible to define an independent family $(S^{v})_{v\in\mathrm{V}}$ of
$\mathsf{NRW}$s of this network, such that $S^{v}$ is a random walk started at
$v.$ Define inductively the following random spanning subgraphs of
$\mathrm{G}:$

Start.

$L_{1}^{\xi}=\mathfrak{F}_{1}^{\xi}=\mathsf{LE}\left(S^{\xi(1)}_{m}\right)_{m\in\mathds{Z}_{+}}.$

Inductive step.

Having defined $L_{k}^{\xi}$ and $\mathfrak{F}_{k}^{\xi},$ define
$L_{k+1}^{\xi}=\mathsf{LE}\left(S^{\xi(k+1)}_{m}\right)_{m=0,\ldots,\tau_{k}}$
and $\mathfrak{F}_{k+1}=\mathfrak{F}_{k}\cup L_{k+1}^{\xi},$ where
$\tau_{k}=\tau_{\mathfrak{F}_{k}^{\xi}}\left(S^{\xi(k+1)}\right),$ that is,
$\tau_{k}$ is the hitting time of the forest currently present
$\mathfrak{F}_{k}^{\xi}$ by the random walk $S^{\xi(k+1)}.$

Consider finally the random spanning subgraph of $\mathrm{G}:$
$\mathfrak{F}^{\xi}=\bigcup\limits_{k\in\mathds{N}}\mathfrak{F}_{k}^{\xi}.$
What Wilson’s algorithm rooted at infinity [BLPS01, Theorem 5.1] states is
that $\mathfrak{F}^{\xi}\sim\mathsf{WSF}_{\mathrm{G}},$ in other words, the
distribution of $\mathfrak{F}^{\xi}$ is that of the wired spanning forest of
the network $\mathrm{G},$ and this happens independently of the choice of
ordering $\xi.$

###  1.3 Definition of $\Gamma_{d}(\lambda)$ and summary of this paper

In this paper we will study $\Gamma_{d}(\lambda)=(\mathds{Z}^{d+1},\mu),$
where $\mu$ is given by

(1.2)
$\mu((n,x),(n^{\prime},x^{\prime}))=e^{\lambda\max(n,n^{\prime})},\quad\text{
for all }(n,x)\sim(n^{\prime},x^{\prime}).$

It can easily be seen that this assignment of conductances makes the
underlying $\mathsf{NRW}$ to have a uniform drift to the right of the first
axis. We will then investigate the main basic properties of the forest. In the
remainder of this section we provide an overview of the results of this paper.

In sect. 2 we will prove the following theorem

######  Theorem.

(Theorem (2.22)) There exist four constants $c_{i}$ ($i=1,\ldots,4$) such that
for all vertices $z=(n,x)\neq 0$ of the network $\Gamma_{d}(\lambda),$

$\textstyle G(0,z)\leq c_{1}\begin{cases}e^{-c_{2}\left\lVert
z\right\rVert}&\text{ if }\left\lVert x\right\rVert>n,\\\
e^{-c_{2}\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-\frac{d}{2}}&\text{ if }\left\lVert x\right\rVert\leq
n,\end{cases}$

and

$\textstyle G(0,z)\geq c_{3}\begin{cases}e^{-c_{4}\left\lVert
z\right\rVert}&\text{ if }\left\lVert x\right\rVert>n,\\\
e^{-c_{4}\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-\frac{d}{2}}&\text{ if }\left\lVert x\right\rVert\leq
n,\end{cases}$

To prove this bounds we will employ a decomposition of the Green’s function
using the continuous-time $\mathsf{NRW}$ and then use several “well-known”
estimates on the latter.

In sect. 3 we will establish is the following.

######  Theorem.

(Theorem (3.8)) On $\Gamma_{d}(\lambda),$ we have $\mathsf{WSF}=\mathsf{FSF}.$

This is done using the Liouville property and a coupling argument. This
theorem then shows that $\mathsf{USF}$ is well defined on
$\Gamma_{d}(\lambda).$ Next, using Fourier inversion theorem together with
Plancherel’s theorem, we will prove the following result, which we call
“bubble condition.”

######  Theorem.

(Theorem (3.10)) The Green’s function $z\mapsto G(0,z)$ of
$\Gamma_{d}(\lambda)$ belongs to $\mathscr{L}^{2}$ if and only if $d\geq 3.$

Shortly after, as a corollary of the bubble condition, we will also show that
$\mathsf{USF}$ consists of a single tree when $d=1,2,$ and it consists of
infinitely many infinite trees for $d\geq 3.$

######  Theorem.

(Theorem (3.9)) In $d=1,2,$ $\mathsf{WSF}$ of $\Gamma_{d}(\lambda)$ is a.s.
one tree; in $d\geq 3,$ there are a.s. infinitely many trees.

In sect. 4 we estimate the frequency at which two independent $\mathsf{NRW}$
of $\Gamma_{d}(\lambda)$ will cross each other. We will prove the following
result

######  Theorem.

(Theorems (4.1) and (4.2), and Corollary (4.15.2)) Let $S$ and $S^{\prime}$ be
two independent network random walks in $\Gamma_{d}(\lambda)$ started at $0.$

  1. (a)

If $d=1,$ then a.s. there exists infinitely many $n$ such that
$S_{n}=S_{n}^{\prime}=0.$

  2. (b)

If $d=2,$ then a.s. there exist infinitely many pairs $(n,n^{\prime})$ such
that $S_{n}=S_{n^{\prime}}^{\prime}.$

  3. (c)

If $d\geq 3,$ there there exists a positive probability that for no pair
$(n,n^{\prime}),$ $S_{n}=S_{n^{\prime}}^{\prime}.$

In fact, in $d=2,$ we will estimate the _probability_ that they will cross is
roughly $\log n$ if they walk $n$ steps (a precise formulation of this is
Theorem (4.15)). The proofs of dimensions $d\neq 2$ are relatively easy. The
proof of dimension $d=2$ is rather involved. We explain the gist of the idea
now. We will consider a paraboloid given by the vertices
$z=(n,x)\in\mathds{Z}_{+}\times\mathds{Z}^{d}$ such that $\left\lVert
x\right\rVert\leq n^{2}$ and inside each of these, we will consider cylinders
(4.7) of appropriate dimensions. We will show that for certain regions (4.10),
the $\mathsf{NRW}$ started in the boundary of these regions will not go back
to the testing cylinder (except with a small probability that can be
controlled) and therefore, the Green’s function virtually does not change
value when we restrict it to leave these regions. We then employ a second-
moment method to estimate the aforementioned probability. The calculations are
quite tight and require a careful set up.

In sect. 5, we prove

######  Theorem.

(Theorems (5.1) and (5.15) and (5.16)) In $\Gamma_{d}(\lambda),$ a.s. every
tree in $\mathsf{USF}$ has one end.

The proofs of dimensions $d=1,$ $d=2$ and $d\geq 3$ need to be done
separately. In $d=1$ we use planar duality and the rather easy-to-establish
fact that $\Gamma_{d}(\lambda)$ is essentially its own dual; the idea is a
makeover of the proof of the same result for the graph $\mathds{Z}^{2}.$ In
$d\geq 3,$ we followed closely the proof of one-endedness for the graphs
$\mathds{Z}^{d}$ given in [LP16]. We made adaptation according to our
conveniences but the underlying idea is that should a tree have two ends,
there will be strong conductivity in the boundary of boxes. The proof of case
$d=2$ is perhaps the most innovative contribution of this paper and was
introduced from scratch. The first property we establish is that Wilson’s
algorithm rooted at infinity has some stochastic-stability (the law remains
unaltered) if we fix the first steps and reorder future vertices (a stronger
result for finite-graphs is proposition (1.1)). This will show that
probabilities of future events depending on the past do not actually depend of
the past. The next step is to show that it suffices to prove that the
component of zero is one-ended. To this aim, we employ Wilson’s algorithm
rooted at infinity and take the first step of the algorithm to be the origin.
Call $L_{0}$ this part of the forest. If the component at zero had at least
two ends, then it would cross the boundary of each cylinder (centred at the
origin) in at least two vertices. The aforementioned invariance of the law
from reordering future searches having fixed the past will allow us to fix
$L_{0}$ and then prove that, in a given a large cylinder
$\mathrm{C}=\left\\{|n|\leq r,\left\lVert x\right\rVert\leq r\right\\},$ most
vertices on $\partial\mathrm{C}$ have infinitesimal probability to connect to
zero, this will reduce the work to study a part of the left base of
$\mathrm{C},$ namely $\mathrm{C}_{r,0}=\left\\{n=-r,\left\lVert
x\right\rVert\leq r^{\frac{1}{2}}\right\\}.$ We will then take a sparse subset
here $(\mathrm{C}_{r,0}^{\prime})$ and construct the branches from its
vertices. The subset $\mathrm{C}_{r,0}^{\prime}$ needs to be sparse enough so
that the probability that the $\mathsf{NRW}$ started from its vertices create
a second end from the origin is small. If now $z$ is any vertex in
$\mathrm{C}_{r,0}\setminus\mathrm{C}_{r,0}^{\prime},$ then we will show that
the $\mathsf{NRW}$ started at $z$ has an overwhelming probability to hit the
branch created by some $z^{\prime}\in\mathrm{C}_{r,0}^{\prime}$ and since this
branch does not create a second end from the origin, $z$ will also not create
a second end. The proof will conclude by gluing together all the estimates.
Admittedly, the proof is elaborate and to ease the reading, we have provided
well-detailed arguments.

In sect. 6 we will show

######  Theorem.

(Theorems (6.2) and (6.5)) There exists a metric
$\eta(z)=\max\left(|n|^{\frac{1}{2}},\left\lVert x\right\rVert\right)$ such

$\mathds{P}{}\left(z\text{ and }z^{\prime}\text{ are in the same
}\mathsf{USF}\text{-component}\right)\asymp\eta(z-z^{\prime})^{-(d-2)}.$

To prove this, we follow similar techniques as those established in [BKPS04].
In a nutshell, we will need to investigate bounds of the convolution
$\displaystyle\sum_{z^{\prime\prime}\in\mathds{Z}^{d+1}}G(z,z^{\prime\prime})G(z^{\prime},z^{\prime\prime}).$

In sect. 7 we will establish the following

######  Theorem.

(Theorem (7.13)) Let $D(z,z^{\prime})$ be the minimal number of
$\mathds{Z}^{d+1}$-edges outside the $\mathsf{USF}$-forest that connects the
tree at $z$ and that at $z^{\prime}.$ Almost surely,
$\displaystyle\max_{z,z^{\prime}}D(z,z^{\prime})=\left\lceil\dfrac{d-2}{4}\right\rceil.$

The proof of this is done by adaptation of most of the methods of [BKPS04] to
the network $\Gamma_{d}(\lambda).$

####  1.3.1 Biased random walk

We make special mention of an article appearing in the arXiv.org in 2018. In
[SSS+18] Zhan Shi, Vladas Sidoravicius, He Song, Longmin Wang, Kainan Xiang
investigated properties of $\mathsf{USF}$ for a random walk they called
“biased random walk.” They studied the network $\mathds{Z}^{d}$ with
assignment of conductances $c_{\lambda}(e)=\lambda^{-|e|},$ where $|e|$ is the
graph-distance from the origin to the edge $e$ and $\lambda\in(0,1)$ is a
fixed parameter. The network considered in this paper and
$\Gamma_{d}(\lambda)$ both have conductances which are far from uniform.
However, the two assignments of conduces give rise to very different random
walks, and very different properties of the uniform spanning forest.

In [SSS+18] the $\mathsf{NRW}$ has a drift away from the origin, while in our
case it has a uniform drift towards the right. Our assignments of conductances
make the network transitive (indeed, translation-invariant), while theirs is
not. They exploited this fact by noticing that the axes play a special rôle.
This also leads to quite different properties of the uniform spanning forest.
For example, the $\mathsf{USF}$ in [SSS+18] has $2^{d}$ trees if $d=2,3,$
while in our case this number is one. Their methods and ours are also
different: they used “spectral radius” and “speed” of the random walk, while
we will not mention these. Another display of the difference in nature of the
results in their paper when compared to this paper is when they count the
number of intersections of random walk paths. They estimated these
intersections by using a local limit theorem. Typically, local limit theorems
have error terms of a polynomial decay in $\left\lVert z\right\rVert.$ We will
show that the Green’s function in $\Gamma_{d}(\lambda)$ has exponential decay
in some directions. Thus, a local limit theorem did not prove useful for us.
Finally, another contrast in the two assignment of conductances. The fact that
$c_{\lambda}(e)$ is bounded from below allowed them to prove one-endedness
using the isoperimetric condition [LP16, Theorem 10.43]. Our conductances
decay exponentially, and so we cannot use the isoperimetric condition for one-
endedness (or any other known technique so far); a new technique was derived
from scratch to prove it.

##  2 Green’s function bounds

###  2.1 Basic statistics of $\Gamma_{d}(\lambda)$

We will often write $z\in\mathds{Z}^{d+1}=\mathds{Z}\times\mathds{Z}^{d}$ as
$z=(n,{x}_{1},\ldots,{x}_{d})=(n,x)$ and we will refer to $n$ as the “drifted
coordinate” or the “zeroth coordinate.” A unit vector in $\mathds{Z}^{d+1}$
has one coordinate equal to 1 or -1 and all the rest equal to zero; for
simplicity, $u$ and $u_{i}$ will denote unit vectors in $\mathds{Z}^{d+1}$ and
$u_{i}$ is such that its $i$th coordinate is non zero ($0\leq i\leq d$). The
functions $\mathrm{pr}_{i}:\mathds{Z}^{d+1}\to\mathds{Z},$ for
$i=0,1,\ldots,d,$ are the projections and are defined in the obvious way. We
will call $\mathds{Z}u_{i}$ the “$i$th factor” as well as “$i$th axis”
($i=0,\ldots,d$). Let $\lambda>0,$ fixed throughout the rest of this document.
Recall $\Gamma_{d}(\lambda)$ is the graph $\mathds{Z}^{d+1}$ with conductances
(1.2). From these conductances, one can get the transition kernel for network
random walk. The probability that the next step should be
$u\in\mathds{Z}^{d+1},$ a unit vector with our conventions, is given by

(2.1) $\textstyle
p(u)=\left\\{\begin{array}[]{lcl}(2d+1+e^{\lambda})^{-1}&\text{ if
}&u=(-1,0)\text{ or }u=u_{i},i=1,\ldots,d,\\\
e^{\lambda}(2d+1+e^{\lambda})^{-1}&\text{ if }&u=(1,0).\end{array}\right.$

The $\frac{1}{2}$-lazy version of the above transition density is
$p^{L}(0)=\frac{1}{2},$ $p^{L}(u)=\frac{1}{2}p(u)$ for unit $u.$

Denote by $\zeta\cdot\varepsilon_{t}$ the measure on $\mathds{Z}^{d}$ such
that it has total mass equal to $\zeta$ at the vertex $t.$ Notice that if
$Y=\left({Y}^{(0)},\ldots,{Y}^{(d)}\right)$ is a random element of
$\mathds{Z}^{d+1}$ with distribution (2.1), then ($i=1,\ldots,d$)

(2.2) $\textstyle
Y^{(0)}\sim(2d+1+e^{\lambda})^{-1}\Big{(}\varepsilon_{-1}+2d\cdot\varepsilon_{0}+e^{\lambda}\cdot\varepsilon_{1}\Big{)},\quad
Y^{(i)}\sim(2d+1+e^{\lambda})^{-1}\Big{(}\varepsilon_{-1}+(2d-1+e^{\lambda})\cdot\varepsilon_{0}+\varepsilon_{1}),$

which immediately allows one to conclude that

(2.3)
$\textstyle\mathds{E}{}\left(Y\right)=(a,0)=\left(\frac{e^{\lambda}-1}{2d+1+e^{\lambda}},0\right),\quad\mathds{C}\mathrm{ov}\left(Y\right)=\left[\begin{matrix}\sigma_{0}^{2}&0\\\
0&\sigma^{2}I_{d}\end{matrix}\right]=\left[\begin{matrix}\frac{e^{\lambda}+1}{2d+1+e^{\lambda}}&0\\\
0&\frac{2}{2d+1+e^{\lambda}}I_{d}\end{matrix}\right],$

where $I_{d}$ is the identity on $\mathds{R}^{d}.$ From the Central Limit
Theorem, it follows immediately that if $(Y_{n})$ is a family of independent
random elements following (2.1),

$\textstyle\frac{1}{\sqrt{n}}\sum\limits_{j=1}^{n}(Y_{j}-(a,0))\xrightarrow[n\to\infty]{\text{weakly}}\mathsf{Norm}\left(0;\left[\begin{matrix}\sigma_{0}^{2}&0\\\
0&\sigma^{2}I_{d}\end{matrix}\right]\right).$

The Fourier transform of $Y$ is

(2.4)
$\textstyle\textstyle\varphi_{Y}(h)=\mathds{E}{}\left(e^{i\left(h\middle|Y\right)}\right)=\frac{e^{\lambda}e^{ih_{0}}+e^{-ih_{0}}}{2d+1+e^{\lambda}}+\frac{2}{2d+1+e^{\lambda}}\sum\limits_{j=1}^{d}\cos
h_{j},$

where $h=({h}_{0},\ldots,{h}_{d})\in[-\pi,\pi]^{d+1}.$ Observe also
$\varphi_{Y}(h)=1$ if and only if $1=e^{ih_{0}}=\cos h_{j}$ for
$j=0,\ldots,d,$ and this is the same as saying that $h_{0}=\ldots=h_{d}=0.$

###  2.2 Continuous time network random walk

Recall the definition of the continuous time network random walk. In
$\Gamma_{d}(\lambda),$ we can write

(2.5)
$\textstyle\tilde{S}_{t}=\left(\tilde{B}_{t},\tilde{X}_{t}\right),\quad(t\geq
0).$

The following properties come from well-known facts regarding Poisson
processes or are otherwise given a reference. Set
$p=\mathds{P}{}\left(Y=u\right)=(2d+1+e^{\lambda})^{-1}.$

  1. (a)

$\tilde{B}$ and $\tilde{X}$ are independent continuous-time Markov processes.

  2. (b)

$\tilde{X}_{t}=X_{M_{t}},$ where $X$ is a standard random walk in
$\mathds{Z}^{d}$ and $M=(M_{t})_{t\geq 0}$ is a Poisson process in
$\mathds{Z}_{+}$ with constant intensity equal to
$2dp=\frac{2d}{2d+1+e^{\lambda}}.$

  3. (c)

$\tilde{B}_{t}=B_{L_{t}},$ where $B$ is a biased random walk in $\mathds{Z}$
with jumps to $\pm 1,$ $L=(L_{t})_{t\geq 0}$ is a Poisson process in
$\mathds{Z}_{+}$ with constant intensity equal to
$\sigma_{0}^{2}=(1+e^{\lambda})p=\frac{1+e^{\lambda}}{2d+1+e^{\lambda}},$ and
the probability of $B$ having a positive jump is
$\frac{e^{\lambda}}{1+e^{\lambda}}.$

  4. (d)

Let $q^{X}$ be the _transition probability_ of $\tilde{X},$ that is,
$q^{X}(t,y)=q_{t}^{X}(y)=\mathds{P}{{}^{0}}\left(\tilde{X}_{t}=y\right).$ Then
[Bar17, Theorem 5.17] provides upper bounds

(2.6) $\textstyle
q^{X}(t,y)\leq\begin{cases}\frac{2}{d}\exp\left(-\frac{\left\lVert
y\right\rVert^{2}}{2dpe^{2}t}\right)&\text{ if }\left\lVert y\right\rVert\leq
2dpet,\\\ \frac{1}{2d}\exp\left(-2dpt-\left\lVert
y\right\rVert\log\left(\frac{\left\lVert
y\right\rVert}{2dpet}\right)\right)&\text{ if }\left\lVert y\right\rVert\geq
2dpet;\end{cases}$

and lower bounds, valid for $\left\lVert y\right\rVert\geq 2dpt>0,$

(2.7) $\textstyle q^{X}(t,y)\geq c\exp\left(-c^{\prime}\left\lVert
y\right\rVert\left[1+\log\left(\frac{\left\lVert
y\right\rVert}{2dpt}\right)\right]\right).$

  1. (e)

By Theorem 6.19 bearing in mind Definition 3.28, Corollary 3.30 and Definition
5.19 of [Bar17] there exists four constants $c_{i}>0$ ($i=1,\ldots,4$) such
that for all $(t,y)$ for which $2dpt>\left\lVert y\right\rVert,$

(2.8) $\textstyle\begin{split}q^{X}_{t}(y)&\leq
c_{1}t^{-\frac{d}{2}}e^{-c_{2}\frac{\left\lVert y\right\rVert^{2}}{t}},\\\
q^{X}_{t}(y)&\geq c_{3}t^{-\frac{d}{2}}e^{-c_{4}\frac{\left\lVert
y\right\rVert^{2}}{t}}.\end{split}$

  1. (f)

There exists a pair of universal constants $c,c^{\prime}>0$ such that for all
$r,T\geq 1,$ $\mathds{P}{}\left(\sup\limits_{0\leq s\leq
T}\left\lVert\tilde{X}_{s}\right\rVert\geq r\right)\leq
ce^{-c^{\prime}\frac{r^{2}}{T}}.$ This follows from Theorem 4.33 of [Bar17]
applied to the standard random walk (so $\beta=2$), you will need to recall
Definitions 4.14 and 4.18, and use Lemma 4.20, which holds with $\alpha=d$ and
$\beta=2$ for the standard random walk on $\mathds{Z}^{d}.$

We conclude this section with the following easy remark.

######  Remark ( 2.9 )

Let $c>0.$

  1. (a)

If $z=(n,x)$ satisfies $|n|\leq c\left\lVert x\right\rVert,$ then $\left\lVert
x\right\rVert\leq\left\lVert z\right\rVert\leq\sqrt{1+c^{2}}\left\lVert
x\right\rVert.$

  2. (b)

If $z=(n,x)$ satisfies $\left\lVert x\right\rVert\leq c|n|,$ then
$|n|\leq\left\lVert z\right\rVert\leq\sqrt{1+c^{2}}|n|.$

  3. (c)

There exists $c^{\prime}>0$ such that for all $z\in\mathds{Z}^{d+1},$
$e^{-c\left\lVert z\right\rVert}\leq c^{\prime}\left\lVert
z\right\rVert^{-\frac{d}{2}}.$

###  2.3 Some estimates of sums

We collect together here some estimates of sums which will be used later. All
proofs can be found in [MD20], sect. 2.3.

######  Proposition ( 2.10 )

Let $L\in\mathds{N}$ and $f:[L,\infty)\to\mathds{R}_{+}^{*}$ be a monotone
function such for a pair $a,b>0,$ we have $af(j+1)\leq f(j)\leq bf(j+1)$ for
integers $j\geq L.$ Let $r>L+1.$ Then, $\sum\limits_{L\leq\left\lVert
x\right\rVert\leq r}f(\left\lVert
x\right\rVert)\asymp\int\limits_{L\leq\left\lVert x\right\rVert\leq r}dx\
f(\left\lVert x\right\rVert)=\sigma_{d}\int\limits_{L}^{r}dt\ f(t)t^{d-1},$
where $\sigma_{d}$ is the “surface area” of the $d$-dimensional sphere and any
implicit constant may depend solely on dimension and on the pair $(a,b).$

######  Proposition ( 2.11 )

Let $a\geq 0,$ $b,\gamma>0$ and $r>2.$

  1. (a)

If $a>0,$ $\sum\limits_{\left\lVert x\right\rVert\leq r}\left\lVert
x\right\rVert^{a}e^{-\gamma\left\lVert
x\right\rVert^{b}}\asymp\min\left(r^{b},\gamma^{-1}\right)^{\frac{a+d}{b}}.$

  2. (b)

If $a=0,$ $\sum\limits_{0<\left\lVert x\right\rVert\leq
r}e^{-\gamma\left\lVert
x\right\rVert^{b}}\asymp\min\left(r^{b},\gamma^{-1}\right)^{\frac{d}{b}};$
thus, if $r^{-b}\leq\gamma\leq 1,$ $\sum\limits_{\left\lVert x\right\rVert\leq
r}e^{-\gamma\left\lVert x\right\rVert^{b}}\asymp\gamma^{-\frac{d}{b}}.$

(Any implicit constant may depend on $a,b$ and dimension but not on $\gamma$
or on $r.$)

######  Proposition ( 2.12 )

Let $a,b>0.$ There exists a pair of constants $c_{1},c_{2}>0$ such that for
all $t\in\mathds{Z}^{d}$ and all $m\in\mathds{N},$

  1. (a)

if $\left\lVert t\right\rVert>bm,$ $e^{-a\left\lVert
t\right\rVert}\leq\sum\limits_{\left\lVert x\right\rVert\leq
bm}e^{-a\left(\frac{\left\lVert x\right\rVert^{2}}{m}+\left\lVert
x-t\right\rVert\right)}\leq c_{1}e^{-c_{2}\left\lVert t\right\rVert};$

  2. (b)

if $\left\lVert t\right\rVert\leq bm,$ $e^{-a\frac{\left\lVert
t\right\rVert^{2}}{m}}\leq\sum\limits_{\left\lVert x\right\rVert\leq
bm}e^{-a\left(\frac{\left\lVert x\right\rVert^{2}}{m}+\left\lVert
x-t\right\rVert\right)}\leq c_{1}e^{-c_{2}\frac{\left\lVert
t\right\rVert^{2}}{m}}.$

######  Proposition ( 2.13 )

For every $\alpha>1$ and all integers $m\geq 2,$
$\frac{1}{\alpha-1}m^{-(\alpha-1)}\leq\sum\limits_{k\geq
m}k^{-\alpha}\leq\frac{2^{\alpha-1}}{\alpha-1}m^{-(\alpha-1)}.$

######  Proposition ( 2.14 )

For every $a\in\mathds{R}$ and $b\geq 1,$ there exists a constant $c=c(a,b)>0$
such that for all $\gamma>1,$ $\sum\limits_{1\leq j\leq
b\gamma}j^{a}e^{-\frac{\gamma}{j}}\leq c\gamma^{a+1}.$

######  Proposition ( 2.15 )

There exists a universal constant $c>0$ such that for all systems
$(a,p,q,r,t)$ with $r>2,$ $a,p,q>0$ and $t\in\mathds{Z}^{d},$

$\textstyle\sum\limits_{\left\lVert x\right\rVert\leq
r}e^{-a\left(\frac{\left\lVert x\right\rVert^{2}}{p}+\frac{\left\lVert
x-t\right\rVert^{2}}{q}\right)}\leq\begin{cases}\displaystyle
c\min\left(r^{2},\frac{p}{a}\right)^{\frac{d}{2}}e^{-\frac{a}{4}\frac{\left\lVert
t\right\rVert^{2}}{q}}&\text{ if }\left\lVert t\right\rVert>2r\\\
\displaystyle
c\left[\min\left(r^{2},\frac{p}{a}\right)^{\frac{d}{2}}e^{-\frac{a}{4}\frac{\left\lVert
t\right\rVert^{2}}{q}}+\min\left(\frac{\left\lVert
t\right\rVert^{2}}{4},\frac{q}{a}\right)^{\frac{d}{2}}e^{-\frac{a}{4}\frac{\left\lVert
t\right\rVert^{2}}{p}}\right]&\text{ if }\left\lVert t\right\rVert\leq
2r.\end{cases}$

###  2.4 Some properties of (network) random walk

These properties are well-known and are included for completeness of the
exposition. We denote by $S=(S_{n})$ the network random walk of a network
$\Gamma$ and $G$ will denote its Green’s function.

######  Proposition ( 2.16 )

Let $\Gamma$ be any network. For any two vertices $o$ and $v$ of $\Gamma,$
$G(o,v)=\int\limits_{0}^{\infty}dt\ q_{t}^{S}(o,v).$ If $\Gamma$ is transitive
and $v$ is any vertex,
$G(o,v)=G(o,o)\mathds{P}{{}^{o}}\left(\tau_{v}<\infty\right),$ where
$\tau_{v}=\tau_{v}\left(\tilde{S}\right)=\inf\left\\{t\geq 0;\
\tilde{S}_{t}=v\right\\}.$

######  Proposition ( 2.17 )

Consider a random walk $S$ on $\mathds{Z}$ with step-lengths distributed as
$q\cdot\varepsilon_{-1}+r\cdot\varepsilon_{0}+p\cdot\varepsilon_{1}.$ Assume
$1>p+q>p>q>0.$ Consider the events $\mathrm{C}_{u}=\\{\text{for exactly one
}n,S_{n}=u\\}=\bigcup\limits_{n\in\mathds{Z}_{+}}\bigcap\limits_{m\neq
n}\\{S_{n}=u,S_{m}\neq u\\}$ and $\mathrm{E}_{u}=\\{\text{for at least one
}n,S_{n}=u\\}=\bigcup\limits_{n\in\mathds{Z}_{+}}\\{S_{n}=u\\}.$ Then, for
$u\geq 0$ and $v<0,$

  1. (a)

$\mathds{P}{{}^{0}}\left(\mathrm{C}_{u}\right)=\mathds{P}{{}^{0}}\left(\tau_{0}^{+}=\infty\right)=p-q,$

  2. (b)

$\mathds{P}{{}^{0}}\left(\bigcup\limits_{u=0}^{n^{2}}\mathrm{C}_{u}\right)\geq
1-(1-(p-q))^{n}-n\left(\frac{q}{p}\right)^{n},$

  3. (c)

$\mathds{P}{{}^{0}}\left(\mathrm{E}_{v}\right)=\left(\frac{p}{q}\right)^{v}.$

###### Proof.

Items (a) and (c) are well-known. For convenience of the reader, we provide
details of item (b).

Consider $\mathrm{D}_{u}$ the complement of $\mathrm{C}_{u}.$ It suffices to
show
$\mathds{P}{{}^{0}}\left(\bigcap\limits_{u=0}^{n}\mathrm{D}_{un}\right)\leq(1-(p-q))^{n}+n\left(\frac{q}{p}\right)^{n}.$
With respect to $\mathds{P}^{0},$ the events $\mathrm{D}_{u}$ ($u\geq 0$) are
a.s. the events where the state $u$ is visited twice or more. Introduce
$\theta_{t}$ defined by “the time of first return to $t$”:
$\theta_{t}=\inf\\{k>\tau_{t}\mid S_{k}=t\\}\quad(t\in\mathds{Z}_{+}).$ It is
clear that $(\tau_{un})_{u=1,\ldots,n}$ is strictly increasing, that
$\tau_{t}<\theta_{t}$ for all $t\in\mathds{Z}_{+},$ and also that the
complement of the event
$\bigcap\limits_{j=1}^{n}\\{\theta_{(j-1)n}<\tau_{jn}\\}$ is contained in the
event $\bigcup\limits_{j=1}^{n}\\{\tau_{jn}<\theta_{(j-1)n}\\}.$ Therefore,
$\mathds{P}{{}^{0}}\left(\bigcap\limits_{u=0}^{n}\mathrm{D}_{un}\right)\leq\mathds{P}{{}^{0}}\left(\theta_{0}<\tau_{n}<\theta_{n}<\ldots<\tau_{n^{2}}<\theta_{n^{2}}<\infty\right)+\sum\limits_{j=1}^{n}\mathds{P}{{}^{0}}\left(\tau_{jn}<\theta_{(j-1)n}<\infty\right).$
By the strong Markov property,
$\mathds{P}{{}^{0}}\left(\tau_{jn}<\theta_{(j-1)n}<\infty\right)=\mathds{P}{{}^{0}}\left(\mathrm{E}_{-n}\right)=\left(\frac{q}{p}\right)^{n}.$
Set $\mathscr{E}_{n}=\mathscr{F}_{\tau_{n}}^{S}=\sigma(S_{k};0\leq
k\leq\tau_{n})$ and

$\textstyle\gamma_{k}\mathop{=}\limits^{\mathrm{def.}}\mathds{P}{{}^{0}}\left(\theta_{0}<\tau_{n}<\theta_{n}<\ldots<\tau_{kn}<\theta_{kn}<\infty\right).$

Observe that the strong Markov property implies

$\displaystyle\gamma_{n}$
$\displaystyle=\mathds{P}{{}^{0}}\left(\theta_{0}<\tau_{n}<\theta_{n}<\ldots<\tau_{n^{2}}<\theta_{n^{2}}<\infty\right)$
$\displaystyle=\mathds{E}{{}^{0}}\left(\mathds{1}_{\\{\theta_{0}<\tau_{n}<\infty\\}}\mathds{E}{{}^{0}}\left(\mathds{1}_{\\{\theta_{n}<\ldots<\tau_{n^{2}}<\theta_{n^{2}}<\infty\\}}\middle|\
\mathscr{E}_{n}\right)\right)$
$\displaystyle=\mathds{P}{{}^{0}}\left(\theta_{0}<\tau_{n}<\infty\right)\gamma_{n-1}\leq\mathds{P}{{}^{0}}\left(\tau_{0}^{+}<\infty\right)\gamma_{n-1},$

which leads at once to $\gamma_{n}\leq(1-(p-q))^{n}$ since $\gamma_{1}\leq
1-(p-q).$ ∎

###  2.5 A hitting time for the continuous time network random walk

Again, we compile some properties of certain random times associated with the
continuous time random walk (2.5). These properties are arguably well-known
and are included for completeness of the exposition. In what follows, we use
(2.3) $p=(2d+1+e^{\lambda})^{-1},$ $a=(e^{\lambda}-1)p$ and
$\sigma_{0}^{2}=(e^{\lambda}+1)p.$ Introduce

$\textstyle\tau_{n}^{0}=\inf\left\\{t\geq
0;\tilde{B}_{t}=n\right\\}\quad(n\in\mathds{N}).$

Then, $\tau_{n}^{0}$ is an almost surely finite stopping time, it is
independent of the process $\tilde{X}$ and by successive applications of the
strong Markov property at the times $\tau_{j}^{0}$ ($1\leq j\leq n-1$), it is
immediately seen that the law of $\tau_{n}^{0}$ is the $n$th convolution power
of the law of $\tau_{1}^{0}.$ Let $\zeta$ denote the Laplace transform of
$\tau_{1}^{0},$ that is,
$\zeta(t)=\mathds{E}{{}^{0}}\left(e^{t\tau_{1}^{0}}\right).$ The following
proposition follows from elementary results and its proof is therefore
omitted, see [MD20], propositions (2.5.1) and (2.5.2).

######  Proposition ( 2.18 )

  1. (a)

We have $\mathds{E}{{}^{0}}\left(\tau_{1}^{0}\right)=\frac{1}{a}.$

  2. (b)

Then $\zeta(t)<\infty$ for $t<(e^{\frac{\lambda}{2}}-1)^{2}p$ and
$\zeta(t)=\infty$ for $t>(e^{\frac{\lambda}{2}}-1)^{2}p.$ We have the
following bounds

$\textstyle\zeta(t)\leq\begin{cases}\frac{2e^{\lambda}p}{2e^{\lambda}p-t}&\text{
for }-\infty<t\leq 0,\vspace{0.1cm}\\\
\frac{2e^{\lambda}p}{\sigma_{0}^{2}-t}&\text{ for
}0<t<(e^{\frac{\lambda}{2}}-1)^{2}p.\end{cases}$

######  Proposition ( 2.19 )

  1. (a)

For $\alpha\leq\frac{n}{2e^{\lambda}p},$
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\alpha\right)\leq\exp\left(-2e^{\lambda}p\alpha+n\log\left(\frac{2e^{\lambda+1}p\alpha}{n}\right)\right).$

  2. (b)

For $0<\beta<\frac{1}{2e^{\lambda}p}$ there exists a constant $u(\beta)>0$
such that for all $n,$ $\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\beta
n\right)\leq e^{-u(\beta)n}.$

  3. (c)

There exists two constants $T_{0},K_{0}>0$ such that
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\geq\alpha\right)\leq
e^{-T_{0}(\alpha-\frac{n}{a})}$ valid for $\alpha\geq(K_{0}+\frac{1}{a})n.$

###### Proof.

The bound in (c) is a Corollary of Theorem 1 and Lemma 5, in Chapter III of
[Pet75], and observe we can apply these results since the previous proposition
shows the Laplace transform of $\tau_{1}^{0}$ to be absolutely convergent on
an interval around zero. We now prove the bound for
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\alpha\right).$ Markov’s inequality
shows, for $t<0,$
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\alpha\right)=\mathds{P}{{}^{0}}\left(e^{t\tau_{n}^{0}}\geq
e^{t\alpha}\right)\leq e^{-t\alpha}\zeta(t)^{n},$ where $\zeta$ is the Laplace
transform of $\tau_{1}^{0}.$ From the previous proposition, we know
$\zeta(t)\leq\frac{2e^{\lambda}p}{2e^{\lambda}p-t}$ for $t\leq 0,$ and so
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\alpha\right)\leq
e^{-t\alpha+n\log\left(\frac{2e^{\lambda}p}{2e^{\lambda}p-t}\right)}$ for
$t<0.$ Minimising gives the minimiser $t=2e^{\lambda}p-\frac{n}{\alpha},$
which is valid only when $t<0,$ that is, only when
$\alpha<\frac{n}{2e^{\lambda}p},$ the bound follows readily upon evaluation at
the minimiser. When $\alpha=\frac{n}{2e^{\lambda}p},$ we reach the trivial
bound $\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\alpha\right)\leq 1,$ and so it
is also valid in this case. Finally, the substitution $\alpha=\beta n$ implies
$\mathds{P}{{}^{0}}\left(\tau_{n}^{0}\leq\beta n\right)\leq e^{-u(\beta)n},$
where $u(\beta)=2e^{\lambda}p\beta-\log(2e^{\lambda+1}p\beta).$ A standard
minimisation shows $u(\beta)>u\left(\frac{1}{2e^{\lambda}p}\right)=0$ for
$0<\beta<\frac{1}{2e^{\lambda}p}.$ ∎

######  Proposition ( 2.20 )

Recall from (2.16) the definition of $\tau_{z}$ for
$z=(n,x)\in\mathds{Z}^{d+1}.$ There exists $c_{1},c_{2}>0$ such that, for all
$x\in\mathds{Z}^{d},$ $\mathds{P}{{}^{0}}\left(\tau_{(0,x)}<\infty\right)\leq
c_{1}e^{-c_{2}\left\lVert x\right\rVert}.$

###### Proof.

Let $r=r_{x}$ be the smallest integer $k\geq\frac{\left\lVert
x\right\rVert}{2}.$ Then
$\mathds{P}{{}^{0}}\left(\tau_{(0,x)}<\infty\right)\leq\mathds{P}{{}^{0}}\left(\tau_{(0,x)}\leq\tau_{r}^{0}\leq
cr\right)+\mathds{P}{{}^{0}}\left(\tau_{(0,x)}\leq\tau_{r}^{0},\tau_{r}^{0}>cr\right)+\mathds{P}{{}^{0}}\left(\tau_{r}^{0}<\tau_{(0,x)}<\infty\right),$
where $c\geq K_{0}+\frac{1}{a}$ is chosen so that
$cr_{x}\geq(K_{0}+\frac{1}{a})r_{x}$ for all $x\in\mathds{Z}^{d},$ $x\neq 0$
and $K_{0}$ is as in (2.19). Then, recalling
$\tilde{S}=\left(\tilde{B},\tilde{X}\right)$ and item (f) in §2.2,

$\textstyle\mathds{P}{{}^{0}}\left(\tau_{(0,x)}\leq\tau_{r}^{0}\leq
cr\right)\leq\mathds{P}{{}^{0}}\left(\sup\limits_{0\leq s\leq
cr}\left\lVert\tilde{X}_{s}\right\rVert\geq\left\lVert
x\right\rVert\right)\leq c^{\prime}e^{-c^{\prime\prime}\frac{\left\lVert
x\right\rVert^{2}}{c\left\lVert
x\right\rVert}}=c^{\prime}e^{-\frac{c^{\prime\prime}}{c}\left\lVert
x\right\rVert},$

for two constants $c^{\prime},c^{\prime\prime}>0.$ From (2.19),
$\mathds{P}{{}^{0}}\left(\tau_{(0,x)}\leq\tau_{r}^{0},\tau_{r}^{0}>cr\right)\leq
e^{-c^{\prime}r},$ for $c^{\prime}=T_{0}(c-\frac{1}{a}).$ Applying the strong
Markov property at the time $\tau_{r}^{0},$ we reach that

$\displaystyle\mathds{P}{{}^{0}}\left(\tau_{r}^{0}<\tau_{(0,x)}<\infty\right)$
$\displaystyle\leq\mathds{E}{{}^{0}}\left(\mathds{P}{{}^{\tilde{S}_{\tau_{r}^{0}}}}\left(\tau_{(0,x)}<\infty\right)\right)\leq\mathds{P}{{}^{0}}\left(\tilde{B}\text{
ever reaches }-r\right)$ $\displaystyle=\mathds{P}{{}^{0}}\left(B\text{ ever
reaches }-r\right)\leq e^{-c^{\prime}\left\lVert x\right\rVert},$

for some $c^{\prime}>0.$ ∎

######  Proposition ( 2.21 )

Recall from (2.16) the definition of $\tau_{z}$ for $z\in\mathds{Z}^{d+1}.$ If
$z=(n,x),$ we have
$G(0,z)=G(0,0)\sum\limits_{y\in\mathds{Z}^{d}}\mathds{P}{{}^{0}}\left(\tau_{(0,x-y)}<\infty\right)\int\limits_{0}^{\infty}q^{X}_{t}(y)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right).$

###### Proof.

By (2.16), we have
$G(0,z)=G(0,0)\mathds{P}{{}^{0}}\left(\tau_{z}<\infty\right).$ Apply the
strong Markov property at the time $\tau_{n}^{0}$ and use that $\tau_{n}^{0}$
and $\tilde{X}$ are mutually independent,

$\displaystyle\textstyle\mathds{P}{{}^{0}}\left(\tau_{z}<\infty\right)$
$\displaystyle\textstyle=\mathds{E}{{}^{0}}\left(\mathds{P}{{}^{\tilde{S}_{\tau_{n}^{0}}}}\left(\tau_{z}<\infty\right)\right)=\sum\limits_{y\in\mathds{Z}^{d}}\mathds{P}{{}^{(n,y)}}\left(\tau_{(n,x)}<\infty\right)\mathds{P}{{}^{0}}\left(\tilde{S}_{\tau_{n}^{0}}=(n,y)\right)$
$\displaystyle\textstyle=\sum\limits_{y\in\mathds{Z}^{d}}\mathds{P}{{}^{0}}\left(\tau_{(0,x-y)}<\infty\right)\int\limits_{0}^{\infty}\mathds{P}{{}^{0}}\left(\tilde{X}_{t}=y\right)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right).$

Since $q^{X}_{t}(y)=\mathds{P}{{}^{0}}\left(\tilde{X}_{t}=y\right),$ we reach
the desired conclusion. ∎

###  2.6 Statement and proof of the Theorem

Consider the Green’s function $G$ of $\Gamma_{d}(\lambda).$ By transience, the
probability $p_{0}=\mathds{P}{{}^{0}}\left(\tau_{0}<\infty\right)$ that the
network random walk started at zero returns to zero is (strictly) less than
one. Thus, the number of visits to zero, started at zero, has finite
expectation since this number follows a geometric law with probability of
success $p_{0},$ in other words, $G(0,0)$ is some finite number, depending
only on $\lambda$ (as such, regarded as universal). The next theorem deals
with the bounds $G(0,z)$ for $z\neq 0.$

######  Theorem ( 2.22 )

There exist four constants $c_{i}$ ($i=1,\ldots,4$) such that for all vertices
$z=(n,x)\neq 0$ of the network $\Gamma_{d}(\lambda),$

$\textstyle G(0,z)\leq c_{1}\begin{cases}e^{-c_{2}\left\lVert
z\right\rVert}&\text{ if }\left\lVert x\right\rVert>n,\\\
e^{-c_{2}\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-\frac{d}{2}}&\text{ if }\left\lVert x\right\rVert\leq
n,\end{cases}$

and

$\textstyle G(0,z)\geq c_{3}\begin{cases}e^{-c_{4}\left\lVert
z\right\rVert}&\text{ if }\left\lVert x\right\rVert>n,\\\
e^{-c_{4}\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-\frac{d}{2}}&\text{ if }\left\lVert x\right\rVert\leq
n,\end{cases}$

###### Proof.

The proof is rather lengthy and we divide it into several steps.

Proof of the upper bounds.

We begin with proving the upper bound when
$z\in\mathds{Z}_{-}\times\mathds{Z}^{d}.$ We start with the following simple
observation: bearing in mind Remark (2.9), if
$z=(n,x)\in\mathds{Z}_{-}\times\mathds{Z}^{d},$ the upper bound will be
established if we prove $G(0,z)\leq ce^{-c^{\prime}\max(|n|,\left\lVert
x\right\rVert)}$ for a pair of constants $c,c^{\prime}>0$ independent of $z.$

We analyse separately the cases $\left\lVert x\right\rVert<|n|$ and
$\left\lVert x\right\rVert\geq|n|.$

Assume first $z\in\mathds{Z}_{-}\times\mathds{Z}^{d}$ is such that
$\left\lVert x\right\rVert<|n|,$ notice $n\leq 0.$ We claim there exists a
pair of constants $c,c^{\prime}>0$ (independent of $z=(n,x)$) such that
$G(0,z)\leq ce^{-c^{\prime}|n|}.$ To see this, recall (2.16) and and (2.17),
so that

$\textstyle G(0,z)=G(0,0)\mathds{P}{{}^{0}}\left(\tau_{z}<\infty\right)\leq
G(0,0)\mathds{P}{{}^{0}}\left(B\text{ ever reaches level
}n\right)=G(0,0)e^{c^{\prime}n},$

where $\tau_{z}=\tau_{z}\left(\tilde{S}\right)$ is the visit time of the point
$z$ by the continuous time random walk $\tilde{S}$ (2.5), proving the claim.

Assume now $z\in\mathds{Z}_{-}\times\mathds{Z}^{d}$ satisfies $\left\lVert
x\right\rVert\geq|n|,$ notice $n\leq 0.$ We have that
$G(0,z)=\frac{\mu(z)}{\mu(0)}G(0,-z),$ and since $\mu(z)\asymp e^{\lambda n},$
we have the following preliminary result.

######  Lemma ( 2.22.1 )

If the Green’s function bounds hold on the region of
$z\in\mathds{Z}_{+}\times\mathds{Z}^{d}$ such that $0\leq n\leq\left\lVert
x\right\rVert,$ then they also hold on the region of points
$z\in\mathds{Z}_{-}\times\mathds{Z}^{d}$ such that $|n|\leq\left\lVert
x\right\rVert.$

Observe that (2.20) shows that the Green’s function bounds hold on the set
$\\{0\\}\times\mathds{Z}^{d}.$ Thus, with the previous lemma in mind, the case
$z\in\mathds{Z}_{-}\times\mathds{Z}^{d}$ has been established.

Let us focus now on the case
$z=(n,x)\in\mathds{Z}_{+}^{*}\times\mathds{Z}^{d}.$ By virtue of (2.20) and
(2.21), we have

($\ast$) $\textstyle G(0,z)\leq
c\sum\limits_{y\in\mathds{Z}^{d}}e^{-c^{\prime}\left\lVert
x-y\right\rVert}\int\limits_{0}^{\infty}q_{t}^{X}(y)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right),$

where $\tau_{n}^{0}$ is the visit time of $n$ by the walk $\tilde{B}$ and
$q_{t}^{X}$ is the continuous time transition probability of $\tilde{X}.$

######  Lemma ( 2.22.1 )

There exist two constants $c_{1},c_{2}>0,$ independent of $z,$ such that

$\textstyle\int\limits_{0}^{\infty}q^{X}_{t}(y)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right)\leq\begin{cases}c_{1}e^{-c_{2}n}+n^{-\frac{d}{2}}e^{-c_{2}\frac{\left\lVert
y\right\rVert^{2}}{n}}&\text{ if }\left\lVert
y\right\rVert\leq\frac{d}{e^{\lambda}+1}n\vspace{0.15cm}\\\
c_{1}e^{-c_{2}\left\lVert y\right\rVert}&\text{ if }\left\lVert
y\right\rVert>\frac{d}{e^{\lambda}+1}n.\end{cases}$

###### Proof of lemma..

Recall $p=(2d+1+e^{\lambda})^{-1}.$ We have
$\int\limits_{0}^{\infty}q^{X}_{t}(y)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right)=\int\limits_{0}^{\infty}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right).$ We will tackle the two cases.

Case 1: $\left\lVert y\right\rVert\leq\frac{d}{e^{\lambda}+1}n.$ We split the
integral as follows

$\textstyle\int\limits_{0}^{\infty}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)=\left[\int\limits_{0}^{\frac{\left\lVert
y\right\rVert}{e}}+\int\limits_{\frac{\left\lVert
y\right\rVert}{e}}^{\frac{d}{e^{\lambda}+1}n}+\int\limits_{\frac{d}{e^{\lambda}+1}n}^{2dp(K_{0}+\frac{1}{a})n}+\int\limits_{2dp(K_{0}+\frac{1}{a})n}^{\infty}\right]q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right),$

where $K_{0}$ is as in (2.19). In what follows, we will use the bounds for
$\mathds{P}{}\left(\tau_{n}^{0}\leq\alpha\right)$ and
$\mathds{P}{}\left(\tau_{n}^{0}\leq\beta n\right)$ of this proposition without
further notice. The first two integrals will be bounded using the bounds
(2.6). In the region $0\leq t\leq\frac{\left\lVert y\right\rVert}{e},$ we have
$q^{X}_{\frac{t}{2dp}}(y)\leq\frac{1}{2d}e^{-t-\left\lVert
y\right\rVert\log\left(\frac{\left\lVert y\right\rVert}{et}\right)}$ and the
function $t\mapsto-t-\left\lVert y\right\rVert\log(\frac{\left\lVert
y\right\rVert}{et})$ is easily seen to be increasing on $(0,\left\lVert
y\right\rVert),$ hence

$\displaystyle\textstyle\int\limits_{0}^{\frac{\left\lVert
y\right\rVert}{e}}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)$
$\displaystyle\textstyle\leq\frac{1}{2d}e^{-\frac{\left\lVert
y\right\rVert}{e}}\mathds{P}{}\left(\tau_{n}^{0}\leq\frac{\left\lVert
y\right\rVert}{2dpe}\right)\leq\frac{1}{2d}e^{-\frac{\left\lVert
y\right\rVert}{e}}\mathds{P}{}\left(\tau_{n}^{0}\leq\frac{n}{2e^{\lambda+1}p}\right)\leq
e^{-\frac{\left\lVert y\right\rVert+n}{e}}\leq e^{-\frac{n}{e}}.$

Similarly, in the region $\frac{\left\lVert y\right\rVert}{e}\leq t<\infty,$
we have the bound
$q^{X}_{\frac{t}{2dp}}(y)\leq\frac{2}{d}e^{-\frac{\left\lVert
y\right\rVert^{2}}{e^{2}t}},$ and the term on the right is an increasing
function of $t>0.$ Hence,

$\displaystyle\textstyle\int\limits_{\frac{\left\lVert
y\right\rVert}{e}}^{\frac{d}{e^{\lambda}+1}n}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)$
$\displaystyle\textstyle\leq\frac{2}{d}e^{-\left(\frac{e^{\lambda-2}}{d}\right)\frac{\left\lVert
y\right\rVert^{2}}{n}}\mathds{P}{}\left(\tau_{n}^{0}\leq\frac{n}{2p(e^{\lambda}+1)}\right)\leq
e^{-\left(\frac{e^{\lambda-2}}{d}\right)\frac{\left\lVert
y\right\rVert^{2}}{n}-cn}\leq e^{-cn}$

for some constant $c=u\left(\frac{1}{2p(e^{\lambda}+1)}\right)>0.$ We use now
the bounds (2.8). Consider the interval $\frac{d}{e^{\lambda}+1}n\leq t\leq
2dp\left(K_{0}+\frac{1}{a}\right)n,$ we have a Gaussian decay
$q^{X}_{\frac{t}{2dp}}(y)\leq c_{1}t^{-\frac{d}{2}}e^{-c_{2}\frac{\left\lVert
y\right\rVert^{2}}{t}}.$ On this interval, $t\asymp n$ and so, there is a pair
of constants $c,c^{\prime}>0$ such that
$\int\limits_{\frac{d}{e^{\lambda}+1}n}^{2dp(K_{0}+\frac{1}{a})n}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)\leq cn^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert
y\right\rVert^{2}}{n}}.$ Next, for $t\geq 2dp\left(K_{0}+\frac{1}{a}\right)n,$
we use the factor $t^{-\frac{d}{2}}$ of the Gaussian estimate for
$q^{X}_{\frac{t}{2dp}}(y)$ to conclude there exists $c>0$ such that
$q^{X}_{\frac{t}{2dp}}(y)\leq c_{1}t^{-\frac{d}{2}}\leq cn^{-\frac{d}{2}}.$
Thus,

$\textstyle\int\limits_{2dp(K_{0}+\frac{1}{a})n}^{\infty}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in
dt\right)=cn^{-\frac{d}{2}}\mathds{P}{}\left(\tau_{n}^{0}\geq\left(K_{0}+\frac{1}{a}\right)n\right)=cn^{-\frac{d}{2}}e^{-T_{0}K_{0}n}\leq
e^{-T_{0}K_{0}n},$

again by (2.19). Putting all the bounds found, this concludes the case
$\left\lVert y\right\rVert\leq\frac{d}{e^{\lambda}+1}n.$

Case 2: $\left\lVert y\right\rVert>\frac{d}{e^{\lambda}+1}n.$ Here we divide

$\textstyle\int\limits_{0}^{\infty}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)=\left[\int\limits_{0}^{\frac{\left\lVert
y\right\rVert}{e}}+\int\limits_{\frac{\left\lVert
y\right\rVert}{e}}^{2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert}+\int\limits_{2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert}^{\infty}\right]q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right).$

Recall the function $t\mapsto-t-\left\lVert
y\right\rVert\log\left(\frac{\left\lVert y\right\rVert}{et}\right)$ is
increasing on $(0,\left\lVert y\right\rVert).$ As in Case 1, we can bound

$\textstyle q^{X}_{\frac{t}{2dp}}(y)\leq\begin{cases}e^{-\frac{\left\lVert
y\right\rVert}{e}}&\text{ if }t\leq\frac{\left\lVert
y\right\rVert}{e},\vspace{0.05cm}\\\ \frac{2}{d}e^{-c\left\lVert
y\right\rVert}&\text{ if }\frac{\left\lVert y\right\rVert}{e}\leq t\leq
2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert y\right\rVert,\vspace{0.05cm}\\\
c_{1}&\text{ if }t\geq 2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert,\end{cases}$

where $c_{1}$ is the constant appearing on the Gaussian bound of $q_{t}^{X}$
and $c=\big{(}2\sigma_{0}^{2}e^{2}(K_{0}+\frac{1}{a})\big{)}^{-1}.$ We may
assume, should the need arise, $K_{0}$ is so large so that
$2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\geq 1.$ It is obvious then that the first
two integrals decay exponentially in $\left\lVert y\right\rVert,$ as for the
third one, we employ (2.19) to obtain

$\textstyle\int\limits_{2\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert}^{\infty}q^{X}_{\frac{t}{2dp}}(y)\mathds{P}{}\left(2dp\
\tau_{n}^{0}\in dt\right)\leq
c_{1}\mathds{P}{}\left(\tau_{n}^{0}\geq\frac{\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert}{dp}\right),$

the condition $\frac{d}{e^{\lambda}+1}n<\left\lVert y\right\rVert$ guarantees
$\mathds{P}{}\left(\tau_{n}^{0}\geq\frac{\sigma_{0}^{2}(K_{0}+\frac{1}{a})\left\lVert
y\right\rVert}{dp}\right)\leq e^{-T_{0}K_{0}\frac{e^{\lambda}+1}{d}\left\lVert
y\right\rVert}.$ This completes Case 2 and the lemma. ∎

With the lemma proven, let us return to the proof of the upper bounds for the
Green’s function. The reader is reminded that there only remains to prove the
bounds for $z=(n,x)$ in $\mathds{Z}_{+}^{*}\times\mathds{Z}^{d}.$ The previous
lemma and ($\ast$) imply that there exist two constants $c,c^{\prime}>0$ such
that

$\textstyle G(0,z)\leq c\left(\sum\limits_{\left\lVert y\right\rVert\leq
bn}e^{-c^{\prime}(\left\lVert x-y\right\rVert+n)}+\sum\limits_{\left\lVert
y\right\rVert\leq bn}n^{-\frac{d}{2}}e^{-c^{\prime}\left(\left\lVert
x-y\right\rVert+\frac{\left\lVert
y\right\rVert^{2}}{n}\right)}+\sum\limits_{\left\lVert
y\right\rVert>bn}e^{-c^{\prime}(\left\lVert x-y\right\rVert+\left\lVert
y\right\rVert)}\right),$

where $b=\frac{d}{e^{\lambda}+1}.$

Bearing in mind (2.9), we only need to prove there exists a pair of constants
$c,c^{\prime}>0$ such that $G(0,z)\leq ce^{-c^{\prime}\left\lVert
x\right\rVert}$ for all $\left\lVert x\right\rVert>bn$ and that there exists a
pair of constants $c,c^{\prime}>0$ such that $G(0,z)\leq
cn^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert x\right\rVert^{2}}{n}}$ for
all $\left\lVert x\right\rVert\leq bn.$ We consider first a sum of the form
$\sum\limits_{\left\lVert y\right\rVert\leq bn}e^{-c(\left\lVert
x-y\right\rVert+\alpha n)}.$ Set
$c^{\prime}=\min\left(c,\frac{c\alpha}{2b}\right),$ then

$\displaystyle\textstyle\sum\limits_{\left\lVert y\right\rVert\leq
bn}e^{-c(\left\lVert x-y\right\rVert+\alpha n)}$
$\displaystyle\textstyle\leq\sum\limits_{\left\lVert y\right\rVert\leq
bn}e^{-c^{\prime}\left\lVert x-y\right\rVert-c\alpha n}\leq
e^{-c^{\prime}\left\lVert x\right\rVert}\sum\limits_{\left\lVert
y\right\rVert\leq bn}e^{\frac{c\alpha}{2b}\left\lVert y\right\rVert-c\alpha
n}$ $\displaystyle\textstyle\leq e^{-c^{\prime}\left\lVert
x\right\rVert}Ln^{d}e^{-\frac{c\alpha}{2}n}\leq
c^{\prime\prime}e^{-c^{\prime}\left\lVert x\right\rVert-\frac{c\alpha}{4}n},$

in which $L$ is a constant, depending only in dimension, satisfying that
$\mathrm{card}\left(\mathrm{B}_{\mathds{Z}^{d}}\left(0;r\right)\right)\leq
Lr^{d}$ for all $r\geq 1$ and $\displaystyle
c^{\prime\prime}=\sup_{n\in\mathds{N}}Ln^{d}e^{-\frac{c\alpha}{4}n}<\infty.$
Thus, we have proven the following lemma.

######  Lemma ( 2.22.2 )

Let $\alpha,\beta,\gamma>0.$ There exist constants $c,c^{\prime}>0$ such that
$\sum\limits_{\left\lVert y\right\rVert\leq\beta n}e^{-\gamma(\left\lVert
x-y\right\rVert+\alpha n)}\leq ce^{-c^{\prime}\left\lVert(n,x)\right\rVert},$
for all $(n,x)\in\mathds{Z}\times\mathds{Z}^{d}.$

Next, we consider the sum $\sum\limits_{\left\lVert
y\right\rVert>bn}e^{-c^{\prime}(\left\lVert x-y\right\rVert+\left\lVert
y\right\rVert)}.$ In the case that $\left\lVert x\right\rVert\leq bn,$ then we
may bound $e^{-c^{\prime}(\left\lVert x-y\right\rVert+\left\lVert
y\right\rVert)}\leq e^{-c^{\prime}bn}e^{-c^{\prime}\left\lVert
x-y\right\rVert},$ thus $\sum\limits_{\left\lVert
y\right\rVert>bn}e^{-c^{\prime}(\left\lVert x-y\right\rVert+\left\lVert
y\right\rVert)}\leq
e^{-c^{\prime}bn}\sum\limits_{y\in\mathds{Z}^{d}}e^{-c^{\prime}\left\lVert
y\right\rVert}.$ In the case that $\left\lVert x\right\rVert>b|n|,$ we may
bound $\left\lVert x-y\right\rVert\geq\left\lVert x\right\rVert-\left\lVert
y\right\rVert,$ thus $e^{-c^{\prime}(\left\lVert x-y\right\rVert+\left\lVert
y\right\rVert)}\leq e^{-\frac{c^{\prime}}{2}\left\lVert
x-y\right\rVert-c^{\prime}\left\lVert y\right\rVert}\leq
e^{-\frac{c^{\prime}}{2}\left\lVert
x\right\rVert-\frac{c^{\prime}}{2}\left\lVert y\right\rVert}.$ Therefore
$\sum\limits_{\left\lVert y\right\rVert>bn}e^{-c^{\prime}(\left\lVert
x-y\right\rVert+\left\lVert y\right\rVert)}\leq
e^{-\frac{c^{\prime}}{2}\left\lVert
x\right\rVert}\sum\limits_{y\in\mathds{Z}^{d}}e^{-\frac{c^{\prime}}{2}\left\lVert
y\right\rVert}.$ We have established the following lemma.

######  Lemma ( 2.22.3 )

Let $\alpha,\beta>0.$ There exists a pair of constants $c,c^{\prime}>0$ such
that $\sum\limits_{\left\lVert y\right\rVert>\beta n}e^{-\alpha(\left\lVert
x-y\right\rVert+\left\lVert y\right\rVert)}\leq
ce^{-c^{\prime}\left\lVert(n,x)\right\rVert}$ for every
$(n,x)\in\mathds{Z}\times\mathds{Z}^{d}.$

By virtue of the previous lemmas, it remains to show that

$\textstyle\sum\limits_{\left\lVert y\right\rVert\leq
bn}n^{-\frac{d}{2}}e^{-c^{\prime}\left(\left\lVert
x-y\right\rVert+\frac{\left\lVert
y\right\rVert^{2}}{n}\right)}\leq\begin{cases}ce^{-c^{\prime}\left\lVert
x\right\rVert}&\text{ if }\left\lVert x\right\rVert>bn,\\\
ce^{-c^{\prime}\frac{\left\lVert x\right\rVert^{2}}{n}}n^{-\frac{d}{2}}&\text{
if }\left\lVert x\right\rVert\leq bn.\end{cases}$

Assume first that $\left\lVert x\right\rVert>bn>0.$ The estimate (2.12) shows
at once $\sum\limits_{\left\lVert y\right\rVert\leq bn}e^{-c\left(\left\lVert
x-y\right\rVert+\frac{\left\lVert y\right\rVert^{2}}{n}\right)}\leq
ce^{-c^{\prime}\left\lVert x\right\rVert}$ for a pair of constants
$c,c^{\prime}>0.$ This concludes the case $\left\lVert x\right\rVert>bn>0.$
Assume now that $\left\lVert x\right\rVert\leq bn.$ By (2.12), we also have
$\sum\limits_{\left\lVert y\right\rVert\leq
bn}n^{-\frac{d}{2}}e^{-c^{\prime}\left(\left\lVert
x-y\right\rVert+\frac{\left\lVert y\right\rVert^{2}}{n}\right)}\leq
cn^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert x\right\rVert^{2}}{n}}$ for
(possibly) another pair $c,c^{\prime}>0.$ This concludes the case $\left\lVert
x\right\rVert\leq bn,$ and thus, the proof of the upper bounds by virtue of
Lemma (2.22.1).

Proof of lower bounds.

These are much simpler. We have $G(0,z)=\frac{\mu(z)}{\mu(0)}G(0,-z).$ If
$z=-(n,x)$ with $n>0,$ then $\mu(z)\asymp e^{-\lambda n}$ and so $G(0,z)\geq
ce^{-\lambda n}G(0,-z)\geq ce^{-\lambda\left\lVert z\right\rVert}G(0,-z).$
Therefore, if we prove lower bounds for $n>0,$ we will reach the lower bounds
for $n<0.$ For the case $n=0$ observe the following: by (2.16),

$\textstyle
G(0,(0,x))=G(0,0)\mathds{P}{{}^{0}}\left(\tau_{(0,x)}<\infty\right)\geq
G(0,0)\mathds{P}{{}^{0}}\left(\tau_{(0,x)}<\infty,\tilde{S}_{\sigma_{1}}=(1,0)\right),$

where $\sigma_{1}$ is the time of the first jump of the Poisson process $N$ in
the definition of $\tilde{S}.$ Now,

$\textstyle\mathds{P}{{}^{0}}\left(\tau_{(0,x)}<\infty,\tilde{S}_{N_{\sigma_{1}}}=(1,0)\right)=e^{\lambda}p\int\limits_{0}^{\infty}dt\
e^{-t}\mathds{P}{{}^{(1,0)}}\left(\tau_{(0,x)}<\infty\right)=e^{\lambda}p\mathds{P}{{}^{0}}\left(\tau_{(-1,x)}<\infty\right).$

Thus, if the correct bounds hold for $n<0,$ then $G(0,(0,x))\geq
e^{\lambda}pG(0,(-1,x))\geq
ce^{-c^{\prime}\left\lVert(1,x)\right\rVert}\left\lVert(1,x)\right\rVert^{-\frac{d}{2}},$
and it is obvious that
$\left\lVert(1,x)\right\rVert\asymp\left\lVert(0,x)\right\rVert$ (2.9).
Whence, the proof of the lower bounds reduces to the case $n>0.$

When $n>0,$ we can apply (2.21). Now,

$\textstyle\int\limits_{0}^{\infty}q_{t}^{X}(y)\mathds{P}{}\left(\tau_{n}^{0}\in
dt\right)\geq\int\limits_{\frac{2d}{e^{\lambda}}n}^{4(K_{0}+\frac{1}{a})dp\
n}q_{\frac{t}{2dp}}^{X}(y)\mathds{P}{}\left(2dp\ \tau_{n}^{0}\in dt\right).$

Observe that the lower bounds of (2.7) and (2.8) are of the same type (with
possibly different constants) whenever $\left\lVert y\right\rVert\asymp t.$ On
the interval of integration, $t\asymp n,$ therefore

$\textstyle
q_{\frac{t}{2dp}}^{X}(y)\geq\begin{cases}cn^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert
y\right\rVert^{2}}{n}}&\text{ if }n\geq\left\lVert y\right\rVert,\\\
c\left\lVert y\right\rVert^{-\frac{d}{2}}e^{-c^{\prime}\left\lVert
y\right\rVert}&\text{ if }n<\left\lVert y\right\rVert.\end{cases}$

Also, (2.19) gives $\mathds{P}{}\left(\frac{2d}{e^{\lambda}}n\leq 2dp\
\tau_{n}^{0}\leq 4(K_{0}+\frac{1}{a})dp\ n\right)\geq c,$ for a constant $c>0$
independent of $n.$ Therefore,

$\textstyle\frac{G(0,z)}{G(0,0)}\geq c\left(\sum\limits_{n\geq\left\lVert
y\right\rVert}\mathds{P}{{}^{0}}\left(\tau_{(0,x-y)}<\infty\right)n^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert
y\right\rVert^{2}}{n}}+\sum\limits_{n<\left\lVert
y\right\rVert}\mathds{P}{{}^{0}}\left(\tau_{(0,x-y)}<\infty\right)\left\lVert
y\right\rVert^{-\frac{d}{2}}e^{-c^{\prime}\left\lVert y\right\rVert}\right).$

We discard all the terms except $y=x,$ and reach

$\textstyle
G(0,z)\geq\begin{cases}cG(0,0)n^{-\frac{d}{2}}e^{-c^{\prime}\frac{\left\lVert
x\right\rVert^{2}}{n}}&\text{ if }n\geq\left\lVert x\right\rVert\\\
cG(0,0)\left\lVert x\right\rVert^{-\frac{d}{2}}e^{-c^{\prime}\left\lVert
x\right\rVert}&\text{ if }n<\left\lVert x\right\rVert.\end{cases}$

This terminates the proof of Theorem (2.22). ∎

##  3 The number of trees in $\mathsf{USF}$

In this section we will establish a number of basic properties of the uniform
spanning forest. $\Gamma_{d}(\lambda)$ has the Liouville property, this
implies both measures $\mathsf{WSF}$ and $\mathsf{FSF}$ coincide in this
network. Finally, establish when the samples are trees or forests.

###  3.1 Coupling of random walks

Consider two vertices $x$ and $y$ in the network $(\mathrm{G},\mu).$ We will
say that the network random walk started at $x$ can be classically coupled
with the network random walk started at $y$ if the following condition holds:

($\mathsf{CC}$)

There exists a probability space $(\Omega,\mathscr{F},\mathds{P}),$ a
filtration $(\mathscr{F}_{n})_{n\in\mathds{Z}_{+}}$ in this probability space,
two Markov processes with respect to this filtration and defined on this
probability space, denoted by $(S_{n},\mathscr{F}_{n})$ and
$\left(S_{n}^{\prime},\mathscr{F}_{n}\right),$ both of them having for
transition probability that defined by the conductances of the network,
$S_{0}=x,$ $S_{0}^{\prime}=y$ and there exists a stopping time $\tau$ relative
to the filtration $(\mathscr{F}_{n})_{n\in\mathds{Z}_{+}},$ such that
$\tau<\infty$ $\mathds{P}$-a.e. and for those $\omega\in\Omega$ for which
$\tau(\omega)<\infty,$
$S_{\tau(\omega)+n}(\omega)=S_{\tau(\omega)+n}^{\prime}(\omega)$ for any
$n\in\mathds{N}.$

When this happens, we will call $\tau$ the coupling time and the condition
$\tau<\infty$ $\mathds{P}$-a.e. is expressed by saying that “$\tau$ is
successful.” Obviously, for a classical coupling to exist, it is necessary
that $x$ and $y$ should be connected in $(\mathrm{G},\mu),$ but this is far
from sufficient.

######  Remark ( 3.1 )

Observe that by definition, a classical coupling is a coupling
$T=\left(S,S^{\prime}\right)$ of the network random walk $S$ started at $x$
and the network random walk $S^{\prime}$ started at $y$ satisfying strong
conditions and it does not necessarily exists; for instance, the standard
lattice $\mathds{Z}^{1}$ with $x=0$ and $y=1,$ any two standard random walks
will always be at distance at least one and so, no classical coupling can
exist (the “parity problem”).

######  Proposition ( 3.2 )

Denote by $u_{i}$ ($1\leq i\leq d$) the canonical unit vectors of
$\mathds{Z}^{d}.$ Let $\rho$ be a probability measure on $\mathds{Z}^{d}$ such
that

  1. (a)

the relation $\rho(x)>0$ implies $x$ is an integer multiple of some $u_{i};$

  2. (b)

$\rho(\mathds{Z}^{*}u_{i})>0$ for each $i=1,\ldots,d.$

Introduce two probability measures, first
$\varphi(i)=\rho(\mathds{Z}^{*}u_{i})+\frac{\rho(0)}{d}$ ($1\leq i\leq d$) and
then $\nu_{i}(k)=\frac{\rho(ku_{i})}{\varphi(i)}$ for $k\in\mathds{Z}^{*}$ and
$\nu_{i}(0)=\frac{\rho(0)}{d\varphi(i)}.$ Suppose $F_{j}\sim\varphi$ and
$X_{j}^{(i)}\sim\nu_{i}$ are independent random elements ($1\leq i\leq d$ and
$j\in\mathds{Z}_{+}$). Set
$Y_{j}=X_{j}^{(F_{j})}=\sum\limits_{i=1}^{d}X_{j}^{(i)}\mathds{1}_{\\{F_{j}=i\\}}.$
Then, for any $S_{0}$ independent of the families $X_{j}^{(i)}$ and $F_{j},$
$S_{n}=S_{0}+Y_{1}+\ldots+Y_{n}$ is a random walk on $\mathds{Z}^{d}$ started
at $S_{0}$ with transition probability $\rho:$
$\mathds{P}{}\left(Y_{j}=x\right)=\rho(x)$ ($x\in\mathds{Z}^{d}$).

######  Remark ( 3.3 )

When $\rho$ of (3.2) is given by the $\frac{1}{2}$-lazy version of (2.1), the
appropriate distributions are
$\varphi(0)=(1+e^{\lambda})(2(2d+1+e^{\lambda}))^{-1}+(2(d+1))^{-1},$ and for
$i=1,\ldots,d,$ $\varphi(i)=(2d+1+e^{\lambda})^{-1}+(2(d+1))^{-1}.$ While the
$\nu_{i}$ are given as $\nu_{0}(-1)=(2(2d+1+e^{\lambda})\varphi(0))^{-1},$
$\nu_{0}(0)=(2(d+1)\varphi(0))^{-1},$
$\nu_{0}(1)=e^{\lambda}(2(2d+1+e^{\lambda})\varphi(0))^{-1}$ and, for
$i=1,\ldots,d,$ $\nu_{i}(1)=\nu_{i}(-1)=(2(2d+1+e^{\lambda})\varphi(i))^{-1},$
$\nu_{i}(0)=(2(d+1)\varphi(i))^{-1}.$

The following theorem is arguably well-known while a source where it is proven
is elusive to find. The reader can find a complete proof in [MD20], theorem
(3.2.3). The proof-strategy follows by classically coupling each coordinate at
a time.

######  Theorem ( 3.4 )

Let $\rho$ be a probability measure on $\mathds{Z}^{d}$ such that

  1. (a)

the relation $\rho(x)>0$ implies $x=ku_{i}$ for some $k\in\mathds{Z}$ and some
canonical unitary vector $u_{i};$

  2. (b)

$\rho(\mathds{Z}^{\ast}u_{i})>0$ for each $i=1,\ldots,d;$

  3. (c)

for every $i=1,\ldots,d,$ the probability measures $\nu_{i},$ defined in
(3.2), and $\bar{\nu_{i}}$ defined by $\bar{\nu}_{i}:k\mapsto\nu_{i}(-k)$
($k\in\mathds{Z}$) are such that their convolution $\nu_{i}\ast\bar{\nu}_{i}$
defines a transition probability kernel for a Markov chain on $\mathds{Z}$
that is irreducible and recurrent.

Then, for any pair of vertices $x$ and $y$ in $\mathds{Z}^{d},$ there exists a
classical coupling of random walks started at $x$ and $y,$ respectively.

###  3.2 The Liouville property

We will say that a network satisfies

($\mathsf{LP}$)

The Liouville property if every _bounded_ harmonic function on the network is
constant.

($\mathsf{SLP}$)

The strong Liouville property if every _positive_ harmonic function on the
network is constant.

($\mathsf{DLP}$)

The Dirichlet Liouville property if every _Dirichlet_ harmonic function is
constant.

######  Remark ( 3.5 )

  1. (a)

For any network, we have the following implications from the above properties

$\textstyle(\mathsf{SLP})\implies(\mathsf{LP})\implies(\mathsf{DLP}).$

The first of these implications is trivial for if $h$ is a bounded harmonic
function, then $h+c$ is a positive harmonic function for large enough constant
$c,$ then condition ($\mathsf{SLP}$) implies $h+c$ is constant, and so is $h.$
The second of this implications is harder and it fully proved in [MD20],
appendix D (the author followed closely the steps of Soardi [Soa94]).

  2. (b)

It is a well-known fact that the graph $\mathds{Z}^{d}$ satisfies the strong
Liouville property. One proof is given in [Bar17]. Here, M. Barlow followed an
already established strategy: showing that $\mathds{Z}^{d}$ satisfies the so-
called elliptic Harnack inequality ($\mathsf{EHI}$) and also showing that for
any network $(\mathsf{EHI})\implies(\mathsf{SLP}).$

  3. (c)

The network $\Gamma_{d}(\lambda)$ does not satisfy ($\mathsf{SLP}$). Consider
the function $h(z)=e^{-\lambda n}$ (with $z=(n,x)$ as usual). Then, with
$p=(2d+1+e^{\lambda})^{-1},$

$\textstyle\sum\limits_{z^{\prime}\sim
z}p(z,z^{\prime})h(z^{\prime})=pe^{-\lambda(n-1)}+2dpe^{-\lambda
n}+e^{\lambda}pe^{-\lambda(n+1)}=e^{-\lambda n}=h(z),$

so that $h$ is harmonic, and proving the claim.

  4. (d)

If on a given network and for any pair of vertices, two lazy network random
walks, one started at each vertex, can be classically coupled, then this
network satisfies the Liouville property. In other words,

$\textstyle(\mathsf{CC})\implies(\mathsf{LP}).$

First notice that a function $h:\mathrm{V}\to\mathds{R}$ is harmonic relative
to the $\mathsf{NRW}$ if and only if it is harmonic relative to the
$\frac{1}{2}$-lazy $\mathsf{NRW}.$ Next, consider a bounded harmonic function
$h,$ any two vertices $x,y$ and the two $\frac{1}{2}$-lazy $\mathsf{NRW}$s $S$
and $S^{\prime},$ started at $x$ and $y,$ respectively, with successful
coupling time $\tau.$ Since $h(S)$ and $h(S^{\prime})$ are bounded
martingales, we obtain
$h(x)=\mathds{E}{}\left(h(S_{\tau})\right)=\mathds{E}{}\left(h(S_{\tau}^{\prime})\right)=h(y).$

######  Theorem ( 3.6 )

The network $\Gamma_{d}(\lambda)$ has the Liouville property.

###### Proof.

By virtue of the previous proposition, it suffices to prove that any two lazy
network random walks (started at any two vertices) of $\Gamma_{d}(\lambda)$
can be classically coupled. In this case, the hypothesis of theorem (3.4) are
satisfied: the measure $\nu_{i}\ast\bar{\nu}_{i},$ where $\nu_{i}$ is as in
(3.3) is clearly symmetric and bounded, it is aperiodic by lazyness, thus
$\nu_{i}\ast\bar{\nu}_{i}$ is also recurrent and irreducible. ∎

###  3.3 Liouville property and the uniform spanning forest measure

The Liouville property is relevant to the context of uniform spanning forests
due to the following theorem.

######  Theorem ( 3.7 )

Let $\Gamma$ be any network. A sufficient condition for the measures
$\mathsf{WSF}$ and $\mathsf{FSF},$ of the wired and free uniform spanning
forest on $\mathrm{G},$ to be the same, is that this network should satisfy
the Liouville property.

For the proof, see [BLPS01, Theorem 7.3]. As a corollary of the foregoing.

######  Theorem ( 3.8 )

$\mathsf{WSF}=\mathsf{FSF}$ on $\Gamma_{d}(\lambda).$

By virtue of the equality between $\mathsf{WSF}$ and $\mathsf{FSF}$ on
$\Gamma_{d}(\lambda),$ we will denote this measure as $\mathsf{USF}$ from now
onwards.

###  3.4 The number of components

To count the number of components, the reader should recall the definition of
transitive network. We know $\Gamma_{d}(\lambda)$ is transitive by considering
translations.

We need the following results:

[LP16], Theorem 10.24.

In any transient transitive network $(\mathrm{G},\mu),$ with vertex set
$\mathrm{V}$ and Green’s function $G,$ the number of intersections of any two
independent network random walks is almost surely infinity if
$\sum\limits_{z\in\mathrm{V}}G(0,z)^{2}=\infty$ and is almost surely finite
otherwise.

[BLPS01], Theorem 9.2.

On any network, for the number of components in $\mathsf{WSF}$ to be one, it
is necessary and sufficient that, two independent network random walk started
at any two vertices, should intersect (a.s.).

[BLPS01], Theorem 9.4.

If there exists a vertex $o$ such that, for almost every realisation, the
number of intersections of two independent copies of the network random walk
started at $o$ is finite, then the number of components in $\mathsf{WSF}$ is
infinite.

######  Theorem ( 3.9 )

Consider the network $\Gamma_{d}(\lambda).$ The $\mathsf{USF}$ here (3.8)
consists of a single tree when $d=1,2$ and it has an infinite number of
components when $d\geq 3.$ From now onwards, we use $\mathsf{UST}$ in lieu of
$\mathsf{USF}$ for $d=1,2.$

###### Proof.

By the theorems stated above, it all reduces to show that
$\sum\limits_{z\in\mathds{Z}^{d+1}}G(0,z)^{2}$ diverges for $d=1,2$ and that
it converges otherwise. In the sake of clarity, we separate the calculations
in the next proposition, with them the theorem is proved. ∎

As mentioned in the proof above, the next theorem states when the Green’s
function belongs to $\mathscr{L}^{2}.$

######  Theorem ( 3.10 )

(The “bubble condition.”)

  1. (a)

If $d=1,2,$ $\sum\limits_{z\in\mathds{Z}^{d+1}}G(0,z)^{2}=\infty.$

  2. (b)

If $d\geq 3,$ $\sum\limits_{z\in\mathds{Z}^{d+1}}G(0,z)^{2}<\infty.$

###### Proof.

Recall the Fourier transform of the step-lengths of the network random (2.4).
Now, if $(S_{n})$ is the network random walk started at zero and $\varphi_{n}$
is its Fourier transform of $S_{n},$ then $\varphi_{n}(h)=\varphi_{Y}(h)^{n}.$
Hence, for all $h\neq 0,$
$\widehat{G}(h)=\sum\limits_{n=0}^{\infty}\varphi_{n}(h)=(1-\varphi_{Y}(h))^{-1},$
where $\widehat{G}$ denotes the Fourier transform of the Green’s function $G.$
By Parseval’s relation,

(3.11)
$\textstyle\sum\limits_{z\in\mathds{Z}^{d+1}}G(0,z)^{2}=\frac{1}{(2\pi)^{d+1}}\int\limits_{[-\pi,\pi]^{d+1}}dh\
\left|1-\varphi_{Y}(h)\right|^{-2},$

where the two integrals will converge to the same value, or they will both
diverge.

Using second order expansion in (2.4) and with $a$ as in (2.3) and
$p=(2d+1+e^{\lambda})^{-1},$ we reach

$\textstyle\varphi_{Y}(h)=1+iah_{0}-\frac{a}{2}h_{0}^{2}-p\sum\limits_{j=0}^{d}h_{j}^{2}+\mathrm{o}\left(h_{0}^{2}\right)+\mathrm{O}\left(\left\lVert
h\right\rVert^{4}\right),\quad h\to 0.$

Denote $\alpha_{j}=p$ for $j=1,\ldots,d$ and $\alpha_{0}=p+\frac{a}{2},$
$\varphi_{Y}(h)=1+iah_{0}-\sum\limits_{j=0}^{d}\alpha_{j}h_{j}^{2}+\mathrm{o}\left(h_{0}^{2}\right)+\mathrm{O}\left(\left\lVert
h\right\rVert^{4}\right).$ Then,

($\ast$) $\displaystyle\textstyle\left|1-\varphi_{Y}(h)\right|^{2}$
$\displaystyle\textstyle=a^{2}h_{0}^{2}+iah_{0}\sum\limits_{j=0}^{d}\alpha_{j}h_{j}^{2}+\mathrm{o}\left(h_{0}^{2}\right)+\mathrm{O}\left(h_{0}\left\lVert
h\right\rVert^{4}\right)$ $\displaystyle\textstyle-
iah_{0}\sum\limits_{j=0}^{d}\alpha_{j}h_{j}^{2}+\left(\sum\limits_{j=0}^{d}\alpha_{j}h_{j}^{2}\right)^{2}+\mathrm{O}\left(\left\lVert
h\right\rVert^{6}\right)$
$\displaystyle\textstyle=a^{2}h_{0}^{2}+\left(\sum\limits_{j=0}^{d}\alpha_{j}h_{j}^{2}\right)^{2}+\mathrm{o}\left(h_{0}^{2}\right)+\mathrm{O}\left(h_{0}\left\lVert
h\right\rVert^{4}\right)+\mathrm{o}\left(\left\lVert
h\right\rVert^{4}\right).$

Divide the right-hand side of (3.11) as the sum of
$I_{\varepsilon}+J_{\varepsilon},$ where $I_{\varepsilon}$ has domain of
integration
$\mathrm{K}_{\varepsilon}=[-\varepsilon,\varepsilon]\times\mathrm{B}_{\mathds{R}^{d}}\left(0;\varepsilon\right)$
and $J_{\varepsilon}$ has domain of integration
$[-\pi,\pi]^{d+1}\setminus\mathrm{K}_{\varepsilon}.$ It follows at once
$J_{\varepsilon}<\infty.$ We will show the existence of a positive
$\varepsilon$ such that $I_{\varepsilon}=\infty$ for $d=1,2$ and
$I_{\varepsilon}<\infty$ for $d\geq 3.$

Assume first $d=1,2.$ Then by ($\ast$) for $\varepsilon$ small enough,
$|1-\varphi_{Y}(h)|^{2}\leq c^{-1}(h_{0}^{2}+\left\lVert
h^{\prime}\right\rVert^{4})$ for some $c>0.$ Then,

$\textstyle I_{\varepsilon}\geq
c\int\limits_{\mathrm{B}_{\mathds{R}^{d}}\left(0;\varepsilon\right)}dh^{\prime}\
\int\limits_{-\varepsilon}^{\varepsilon}dh_{0}\ \frac{1}{h_{0}^{2}+\left\lVert
h^{\prime}\right\rVert^{4}}=2c\int\limits_{\mathrm{B}_{\mathds{R}^{d}}\left(0;\varepsilon\right)}dh^{\prime}\
\int\limits_{0}^{\varepsilon}dh_{0}\ \frac{1}{h_{0}^{2}+\left\lVert
h^{\prime}\right\rVert^{4}}.$

Changing measures to surface measure on the spheres of $\mathds{R}^{d},$ there
is a constant $c>0$ such that

$\textstyle I_{\varepsilon}\geq c\int\limits_{0}^{\varepsilon}d\rho\
\rho^{d-1}\int\limits_{0}^{\varepsilon}dh_{0}\
\frac{1}{h_{0}^{2}+\rho^{4}}=c\int\limits_{0}^{\varepsilon}d\rho\
\rho^{d-3}\arctan\left(\frac{\varepsilon}{\rho^{2}}\right)=\infty.$

Similarly from ($\ast$), we can assume $\varepsilon>0$ is small enough so that
for some constant $c>0,$ $\left|1-\varphi_{Y}(h)\right|^{2}\geq
c^{-1}(h_{0}^{2}+\left\lVert h^{\prime}\right\rVert^{4}).$ As before, we
change to surface measure of spheres and use that $\rho^{d-1}\leq\rho^{2}$
since $\varepsilon$ can be assumed smaller than unity and $d\geq 3$, then

$\textstyle I_{\varepsilon}\leq
c\int\limits_{\mathrm{K}_{\varepsilon}}\frac{1}{h_{0}^{2}+\left\lVert
h^{\prime}\right\rVert^{4}}d(h_{0},h^{\prime})\leq
c^{\prime}\int\limits_{0}^{\varepsilon}d\rho\
\rho^{2}\cdot\frac{1}{\rho^{2}}\arctan\left(\frac{h_{0}}{\rho^{2}}\right)\Bigg{|}_{h_{0}=0}^{h_{0}=\varepsilon}<\infty.$

Therefore, we reach that the sum (3.11) is finite for all $d\geq 3.$ ∎

##  4 Crossings of $\mathsf{NRW}$

###  4.1 Dimensions $d=1,3,4,\ldots$

######  Theorem ( 4.1 )

Let $d\geq 3.$ Consider two independent network random walks $S$ and
$S^{\prime}$ on $\Gamma_{d}(\lambda).$ Suppose they have arbitrary (random)
initial points $S_{0}$ and $S_{0}^{\prime},$ respectively. Then, almost
surely, the paths of $S$ and $S^{\prime}$ will cross finitely many times.

###### Proof.

By the bubble condition (3.10),
$\sum\limits_{z\in\mathds{Z}^{d+1}}G(0,z)^{2}<\infty$ in this range of
dimensions. The theorem then follows from [LP16, Theorem 10.24]. ∎

The previous theorem is not true in dimension $d=1,2.$ However, there is an
important distinction between the cases $d=1$ and $d=2.$ The gist of the
distinction lies in how frequently the two random paths cross. In this section
we consider the case $d=1.$

######  Theorem ( 4.2 )

Let $d=1.$ Consider two independent network random walks $S$ and $S^{\prime}$
on $\Gamma_{1}(\lambda)$ with arbitrary starting points, $S_{0}$ and
$S_{0}^{\prime}.$ Then, $S-S^{\prime}$ is a recurrent random walk; in
particular, for almost every realisation, there will be infinitely many $n$
for which $S_{n}-S_{n}^{\prime}=S_{0}-S_{0}^{\prime}.$ Thus, if both $S$ and
$S^{\prime}$ start at the same point, then $S$ and $S^{\prime}$ will collide
infinitely often.

###### Proof.

By means of a translation by $-(S_{0}-S_{0}^{\prime}),$ we can assume
$S_{0}-S_{0}^{\prime}=0.$ The difference $S-S^{\prime}$ is a random walk,
started at zero and has step-length distribution

$\displaystyle\mu$
$\displaystyle=p^{2}e^{\lambda}\Big{(}\varepsilon_{(-2,0)}+\varepsilon_{(2,0)}\Big{)}+p^{2}\Big{(}\varepsilon_{(0,2)}+\varepsilon_{(0,-2)}\Big{)}$
$\displaystyle+p^{2}(e^{\lambda}+1)\Big{(}\varepsilon_{(-1,-1)}+\varepsilon_{(1,1)}+\varepsilon_{(-1,1)}+\varepsilon_{(1,-1)}\Big{)}+p^{2}(e^{2\lambda}+3)\varepsilon_{(0,0)},$

where $\varepsilon_{(h,k)}$ denotes the measures of unitary total mass at
$(h,k)\in\mathds{Z}^{2}$ and $p=\frac{1}{3+e^{\lambda}}.$ Observe that
$S-S^{\prime}$ is a symmetric, aperiodic and irreducible random walk with
bounded step lengths on the lattice generated by $u_{1}=(1,1)$ and
$u_{2}=(-1,1).$ By [LL10, Theorem 4.1.1] $S-S^{\prime}$ is recurrent. ∎

The remainder of this chapter is devoted into _estimating how frequently_
there will be a crossing between two network random walk paths on
$\Gamma_{2}(\lambda).$ Whence, in the upcoming sections we will develop the
necessary tools to tackle dimension $d=2.$

###  4.2 Two elementary results

For the sake of reference, we write the following. The “second moment
inequality” states that for a non-negative random variable

(4.3)
$\textstyle\mathds{P}{}\left(Z>\frac{1}{2}\mathds{E}{}\left(Z\right)\right)\geq\frac{\mathds{E}{}\left(Z\right)^{2}}{4\mathds{E}{}\left(Z^{2}\right)}.$

Consider now an increasing sequence of $\sigma$-fields
$(\mathscr{E}_{n})_{n\in\mathds{Z}_{+}},$ with $\mathscr{E}_{0}$ the
$\sigma$-field generated by $\varnothing,$ on some probability space
$(\mathrm{E},\mathscr{E},\mathds{P}).$ Suppose that $\mathrm{A}_{n}$ is an
event measurable up to time $n,$ that is $\mathrm{A}_{n}\in\mathscr{E}_{n}.$
Set $q_{n}=\mathds{P}{}\left(\mathrm{A}_{n}\mid\mathscr{E}_{n-1}\right)$ (so
that $q_{1}=\mathds{P}{}\left(\mathrm{A}_{1}\right)$). We have the following
result.

######  Proposition ( 4.4 )

The event $\mathrm{N}=\\{\mathrm{A}_{n}\mathrm{\
i.o.}\\}\triangle\left\\{\sum\limits_{n=1}^{\infty}q_{n}=\infty\right\\}$ is
$\mathds{P}$-null (“Levy’s extension of Borel-Cantelli lemma”).

The proof of this result is a standard application of martingale theory to
$X_{n}=\sum\limits_{j=1}^{n}(\mathds{1}_{\mathrm{A}_{j}}-q_{j}).$ See section
7.8 of [Ash72] (apply theorem 7.8.4).

###  4.3 Setup for $\Gamma_{2}(\lambda)$

As before, we will write $z=(n,x);$ any affix, such as a sub or superscript,
will be given to the three symbols: if we write $z^{\prime}$ at some point, it
will be assumed that $z^{\prime}=(n^{\prime},x^{\prime}),$ similarly, if we
talk about $n_{1}$ and $x_{1},$ we will be assuming $z_{1}=(n_{1},x_{1}),$
etc.

Divide the space $\mathds{Z}\times\mathds{Z}^{2}$ into the two sets
$\mathrm{S}_{1}=\\{z;n\leq\left\lVert x\right\rVert\\}$ and
$\mathrm{S}_{2}=\\{z;n>\left\lVert x\right\rVert\\}.$ Notice
$0\in\mathrm{S}_{1}$ and $\mathrm{S}_{1}\cap\mathrm{S}_{2}=\varnothing.$ To
simplify notation, we define (for any $y\in\mathds{Z}^{r},r\geq 1$)

(4.5) $\textstyle\left\langle y\right\rangle=\max(\left\lVert
y\right\rVert,1),$

and notice $\left\langle y\right\rangle=\left\lVert y\right\rVert$ unless
$y=0;$ also $\left\langle y\right\rangle\asymp\left\lVert
y\right\rVert+1\asymp\sqrt{\left\lVert y\right\rVert^{2}+1}.$

Consider the following tessellation ($p,q\in\mathds{N}$ and $i=1,2$) of the
“positive half-space” $\mathds{Z}_{+}\times\mathds{Z}^{2}$:

(4.6) $\textstyle\begin{split}\mathrm{R}_{1}(p,q)&=\\{z:9^{p}<n\leq 2\cdot
9^{p},3^{q-1}<\left\lVert x\right\rVert\leq 3^{q}\\},\\\
\mathrm{R}_{2}(p,q)&=\\{z:2\cdot 9^{p}<n\leq 9^{p+1},3^{q-1}<\left\lVert
x\right\rVert\leq 3^{q}\\}\\\ \mathrm{R}_{1}(p,0)&=\\{z:9^{p}<n\leq 2\cdot
9^{p},\left\lVert x\right\rVert\leq 1\\},\\\ \mathrm{R}_{2}(p,0)&=\\{z:2\cdot
9^{p}<n\leq 2\cdot 9^{p+1},\left\lVert x\right\rVert\leq 1\\},\\\
\mathrm{R}_{1}(0,q)&=\\{z:0\leq n\leq 2,3^{q-1}<\left\lVert x\right\rVert\leq
3^{q}\\}\\\ \mathrm{R}_{2}(0,q)&=\\{z:3\leq n\leq 9,3^{q-1}<\left\lVert
x\right\rVert\leq 3^{q}\\}\\\ \mathrm{R}_{1}(0,0)&=\\{z:0\leq n\leq
2,\left\lVert x\right\rVert\leq 1\\}\\\ \mathrm{R}_{2}(0,0)&=\\{z:3\leq n\leq
9,\left\lVert x\right\rVert\leq 1\\}\end{split}$

For $p\in\mathds{N},$ we shall consider the following region

(4.7)
$\textstyle\mathrm{D}_{p}=\bigcup\limits_{q=0}^{p}\mathrm{R}_{1}(p,q)=\\{z:9^{p}<n\leq
2\cdot 9^{p},\left\lVert x\right\rVert\leq 3^{p}\\}.$

The following easy remark contains all the technical estimates that will be
employed later in this section. Consider the Green’s function bounds and fix
the constants $c_{i}$ ($1\leq i\leq 4$) of (2.22).

######  Remark ( 4.8 )

For $i=1,2$ and $(p,q)\in\mathds{Z}_{+}^{2},$ we have the following.

  1. (a)

$\mathrm{card}\left(\mathrm{R}_{i}(p,q)\right)\asymp 9^{p+q}.$

  2. (b)

$\mathrm{R}_{i_{1}}(p_{1},q_{1})$ and $\mathrm{R}_{i_{2}}(p_{2},q_{2})$ are
disjoint unless $(i_{1},p_{1},q_{1})=(i_{2},p_{2},q_{2}).$

  3. (c)

$\mathrm{card}\left(\mathrm{D}_{p}\right)\asymp
9^{p}\sum\limits_{q=0}^{p}9^{q}\asymp 9^{2p}.$

  4. (d)

We have $\mathrm{R}_{i}(p,q)\subset\mathrm{S}_{1}$ if $2p+3\leq q$ and
$\mathrm{R}_{i}(p,q)\subset\mathrm{S}_{2}$ if $q\leq 2p.$

  5. (e)

If $z\in\mathds{Z}_{+}\times\mathds{Z}^{2}$ is in $\mathrm{S}_{1}$ and $c$ is
a positive constant, then $0\leq n\leq\left\lVert x\right\rVert,$ this implies
$\left\lVert x\right\rVert\leq\left\lVert z\right\rVert\leq\sqrt{2}\left\lVert
x\right\rVert,$ and then $\frac{1}{\sqrt{2}}e^{-\sqrt{2}c\left\lVert
x\right\rVert}\left\langle x\right\rangle^{-1}\leq e^{-c\left\lVert
z\right\rVert}\left\langle z\right\rangle^{-1}\leq e^{-c\left\lVert
x\right\rVert}\left\langle x\right\rangle^{-1}.$ Therefore, if
$3^{q-1}\leq\left\lVert x\right\rVert\leq 3^{q},$
$\frac{1}{\sqrt{2}}e^{-\sqrt{2}c3^{q}}3^{-q}\leq e^{-c\left\lVert
z\right\rVert}\left\langle z\right\rangle^{-1}\leq
3e^{-\frac{c}{3}3^{q}}3^{-q}.$

  6. (f)

If $z\in\mathds{Z}_{+}\times\mathds{Z}^{2}$ is in $\mathrm{S}_{2}$ and $c$ is
a positive constant, then $\left\lVert x\right\rVert<n,$ this implies
$0<n\leq\left\lVert z\right\rVert\leq\sqrt{2}n,$ and then
$\frac{1}{\sqrt{2}}e^{-c\frac{\left\lVert x\right\rVert^{2}}{n}}n^{-1}\leq
e^{-c\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-1}\leq e^{-c\frac{\left\lVert x\right\rVert^{2}}{n}}n^{-1}.$
Therefore, if $9^{p}\leq n\leq 9^{p+1}$ and $3^{q-1}\leq\left\lVert
x\right\rVert\leq 3^{q}$ $\frac{1}{9\sqrt{2}}e^{-c9^{q-p}}9^{-p}\leq
e^{-c\frac{\left\lVert x\right\rVert^{2}}{n}}\left\lVert
z\right\rVert^{-1}\leq e^{-\frac{c}{81}9^{q-p}}9^{-p}.$

  7. (g)

Define the function
$\varphi_{\mathrm{U}}:\mathds{Z}\times\mathds{Z}^{2}\to\mathds{R}$ given by
$\varphi_{\mathrm{U}}(z)=c_{1}e^{-c_{2}\left\lVert z\right\rVert}\left\langle
z\right\rangle^{-1}$ if $z$ lies in region $\mathrm{S}_{1}$ and
$\varphi_{\mathrm{U}}(z)=c_{1}e^{-c_{2}\frac{\left\lVert
x\right\rVert^{2}}{n}}\left\lVert z\right\rVert^{-1}$ for $z$ within
$\mathrm{S}_{2}.$ Then $G(0,z)\leq\varphi_{\mathrm{U}}(z).$ If
$z\in\mathrm{R}_{1}(p,q)\cup\mathrm{R}_{2}(p,q),$

$\varphi_{\mathrm{U}}(z)\leq\begin{cases}3c_{1}e^{-\frac{c_{2}}{3}3^{q}}3^{-q}&\text{
if }2p+3\leq q,\vspace{0.2cm}\\\
c_{1}e^{-\frac{c_{2}}{81}9^{q-p}}9^{-p}&\text{ if }q\leq 2p+2.\end{cases}$

  8. (h)

Define $\varphi_{\mathrm{L}}(z)=c_{3}e^{-c_{4}\left\lVert
z\right\rVert}\left\langle z\right\rangle^{-1}$ if $z\in\mathrm{S}_{1}$ and
$\varphi_{\mathrm{L}}(z)=c_{3}e^{-c_{4}\frac{\left\lVert
x\right\rVert^{2}}{n}}\left\lVert z\right\rVert^{-1}$ if $z\in\mathrm{S}_{2}.$
Then, $G(0,z)\geq\varphi_{\mathrm{L}}(z).$ If $z\in\mathrm{R}_{i}(p,q)$
($i=1,2$),

$\varphi_{\mathrm{L}}(z)\geq\begin{cases}\frac{c_{3}}{\sqrt{2}}e^{-\sqrt{2}c_{4}3^{q}}3^{-q}&\text{
if }2p+3\leq q,\vspace{0.25cm}\\\
\frac{c_{3}}{9\sqrt{2}}e^{-c_{4}9^{q-p}}9^{-p}&\text{ if }q\leq
2p+2.\end{cases}$

  9. (i)

For $z=(n,x)\in\mathds{Z}\times\mathds{Z}^{2}$ define
$\dot{z}=(-n,x)\in\mathds{Z}\times\mathds{Z}^{2}$ (the “reflection” of $z$
through the “plane” $\\{0\\}\times\mathds{Z}^{2}$). Then, $z\mapsto\dot{z}$ is
a bijection of $\mathds{Z}_{\pm}\times\mathds{Z}^{2}$ onto
$\mathds{Z}_{\mp}\times\mathds{Z}^{2},$ it is the identity when restricted to
$\\{0\\}\times\mathds{Z}^{2}.$ For every
$z\in\mathds{Z}_{-}\times\mathds{Z}^{2},$
$\varphi_{\mathrm{U}}(z)\leq\varphi_{\mathrm{U}}\left(\dot{z}\right).$

  10. (j)

There are constants $d_{1}=3c_{1},$ $d_{2}=\frac{c_{2}}{81},$
$d_{3}=\frac{c_{3}}{18}$ and $d_{4}=2c_{4},$ such that the relation
$z\in\mathrm{R}_{i}(p,q)$ ($p,q\in\mathds{Z}_{+},$ $i=1,2$) imply

$G(0,z)\leq\varphi_{\mathrm{U}}(z)\leq\begin{cases}d_{1}e^{-d_{2}3^{q}}3^{-q}&\text{
if }2p+1\leq q,\\\ d_{1}e^{-d_{2}9^{q-p}}9^{-p}&\text{ if }q\leq
2p.\end{cases}$

and

$G(0,z)\geq\varphi_{\mathrm{L}}(z)\geq\begin{cases}d_{3}e^{-d_{4}3^{q}}3^{-q}&\text{
if }2p+1\leq q,\\\ d_{3}e^{-d_{4}9^{q-p}}9^{-p}&\text{ if }q\leq
2p.\end{cases}$

  11. (k)

$G(0,z)\asymp 9^{-p}$ with any constant being uniform for $z\in\mathrm{D}_{p}$
and $p\in\mathds{N}.$

####  4.3.1 Some preliminary sums of the Green’s function of
$\Gamma_{2}(\lambda).$

In what follows, the estimates in Remark (4.8) will be used, often without
referencing to this remark.

######  Proposition ( 4.9 )

Let $S$ and $S^{\prime}$ be two independent network random walks started at
$0$ in $\Gamma_{2}(\lambda).$ Then, with any implicit constants being
independent of $p\in\mathds{N},$

  1. (a)

$\sum\limits_{z\in\mathrm{D}_{p}}G(0,z)^{2}\asymp 1;$ and

  2. (b)

$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)G(0,-z)\asymp 1.$

  3. (c)

Also, there exists a constant $c>0$ such that
$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)^{2}\leq cp,$ for all
$p\in\mathds{N}.$

###### Proof.

We know $G(0,z)^{2}\asymp 9^{-2p}$ and
$\mathrm{card}\left(\mathrm{D}_{p}\right)\asymp 9^{2p}$ for
$z\in\mathrm{D}_{p}$ and the implicit constants independent of $p;$ this
proves the first assertion $\sum\limits_{z\in\mathrm{D}_{p}}G(0,z)^{2}\asymp
1.$ There remains to show the other two estimates. First, it is clear that
$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)G(0,-z)\geq G(0,0)^{2}.$
Hence, there remains to prove the existence of constants $c>0$ and
$c^{\prime}>0$ such that
$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)G(0,-z)\leq c$ and
$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)^{2}\leq c^{\prime}p,$
for all $p\in\mathds{N}.$ Set $\mathrm{Q}^{(p)}=\mathrm{D}_{p}-\mathrm{D}_{p}$
and
$\mathrm{Q}_{+}^{(p)}=\mathrm{Q}^{(p)}\cap(\mathds{Z}_{+}\times\mathds{Z}^{2}).$
Consider the function $\varphi_{\mathrm{U}}$ introduced in (4.8), we regard
the constants $c_{k}$ and $d_{k}$ ($k=1,\ldots,4$) of (4.8) to be fixed during
the argument. Then $\sum\limits_{z\in\mathrm{Q}^{(p)}}G(0,z)G(0,-z)\leq
2\sum\limits_{z\in\mathrm{Q}_{+}^{(p)}}G(0,z)G(0,-z)\leq
2\sum\limits_{z\in\mathrm{Q}_{+}^{(p)}}\varphi_{\mathrm{U}}(z)\varphi_{\mathrm{U}}(-z),$
and
$\sum\limits_{z\in\mathrm{Q}^{(p)}}G(0,z)^{2}\leq\sum\limits_{z\in\mathrm{Q}^{(p)}}\varphi_{\mathrm{U}}(z)^{2}\leq
2\sum\limits_{z\in\mathrm{Q}_{+}^{(p)}}\varphi_{\mathrm{U}}(z)^{2}$ where the
second inequality follows from considering the reflection $z\mapsto\dot{z}$
(4.8)(i). To simplify notation, introduce $\mathrm{R}(s,t)$ to be the union
$\mathrm{R}_{1}(s,t)\cup\mathrm{R}_{2}(s,t),$ these sets being defined in
(4.6). We know that there exists a constant $L>0$ such that for all
$s,t\in\mathds{Z}_{+},$ $\mathrm{card}\left(\mathrm{R}(s,t)\right)\leq
L9^{s+t}.$ Then, one can write
$\mathrm{Q}_{+}^{(p)}\subset\bigcup\limits_{(s,t)}\mathrm{R}(s,t),$ where the
indices run over all choices of $0\leq s,t\leq p+1.$ Now, the sets
$\mathrm{R}(s,t)$ are pairwise disjoint and, therefore, we split
$\sum\limits_{z\in\mathrm{Q}_{+}^{(p)}}\varphi_{\mathrm{U}}(z)\varphi_{\mathrm{U}}(-z)\leq\sum\limits_{(s,t)}\sum\limits_{z\in\mathrm{R}(s,t)}\varphi_{\mathrm{U}}(z)\varphi_{\mathrm{U}}(-z)=P_{1}+P_{2},$
and, in a similar fashion,
$\sum\limits_{z\in\mathrm{Q}_{+}^{(p)}}\varphi_{\mathrm{U}}(z)^{2}\leq
T_{1}+T_{2},$ where $P_{1}$ and $T_{1}$ are the sums corresponding to $t<s$
and $P_{2}$ and $T_{2},$ the sums for $t\geq s.$ We shall show that $P_{1}$
and $P_{2}$ are bounded by constants independent of $p,$ and $T_{1}\leq cp$
and $T_{2}\leq c^{\prime}p$ for a pair of constants $c,c^{\prime}>0$
independent of $p;$ having this, we will conclude
$\sum\limits_{z\in\mathrm{D}_{p}-\mathrm{D}_{p}}G(0,z)G(0,-z)\leq c$ and
$\sum\limits_{z\in\mathrm{Q}^{(p)}}G(0,z)^{2}\leq c^{\prime}p$ for some
constants $c>0$ and $c^{\prime}>0,$ as desired.

We now prove $P_{1}$ and $P_{2}$ are bounded by constants independent of $p.$
Observe from (4.8)(g) that for all $z\in\mathrm{R}(s,t),$ either
$\varphi_{\mathrm{U}}(z)\leq d_{1}e^{-d_{2}3^{t}}3^{-t}$ or
$\varphi_{\mathrm{U}}(z)\leq d_{1}e^{-d_{2}9^{t-s}}9^{-s}.$ Also, for
$z\in\mathrm{Q}_{+}^{(p)}$ we have
$\varphi_{\mathrm{U}}(-z)=c_{1}e^{-c_{2}\left\lVert z\right\rVert}\left\lVert
z\right\rVert^{-1}.$ Therefore (use (4.8)(a)),

$\displaystyle\textstyle P_{2}$
$\displaystyle\textstyle\leq\sum\limits_{s=0}^{p+1}\sum\limits_{t=s}^{p+1}\sum\limits_{z\in\mathrm{R}(s,t)}\varphi_{\mathrm{U}}(z)\varphi_{\mathrm{U}}(-z)\leq
3c_{1}d_{1}L\sum\limits_{s=0}^{p+1}\sum\limits_{t=s}^{p+1}\big{(}e^{-d_{2}3^{t}}9^{s}+e^{-d_{2}9^{t-s}}3^{t}\big{)}e^{-\frac{c_{2}}{3}3^{t}}$
$\displaystyle\textstyle\leq
3c_{1}d_{1}L\sum\limits_{l=0}^{\infty}e^{-\frac{c_{2}}{3}3^{l}}\left(\left[\sum\limits_{k=0}^{\infty}e^{-d_{2}3^{k}}\right]9^{l}+\left[\sum\limits_{k=0}^{\infty}e^{-d_{2}9^{k}}3^{k}\right]3^{l}\right)<\infty.$

The sum $P_{1}$ is easier,

$\displaystyle\textstyle P_{1}$
$\displaystyle\textstyle\leq\sum\limits_{s=0}^{p+1}\sum\limits_{t=0}^{s-1}\sum\limits_{z\in\mathrm{R}(s,t)}\varphi_{\mathrm{U}}(z)\varphi_{\mathrm{U}}(-z)\leq
c_{1}d_{1}L\sum\limits_{s=0}^{p+1}\sum\limits_{t=0}^{s-1}9^{s+t}e^{-d_{2}9^{t-s}}9^{-s}e^{-c_{2}9^{s}}9^{-s}$
$\displaystyle\textstyle\leq
c_{1}d_{1}L\sum\limits_{s=0}^{p+1}e^{-c_{2}9^{s}}9^{-s}\sum\limits_{t=0}^{s-1}9^{t}=\frac{c_{1}d_{1}L}{8}\sum\limits_{k=0}^{\infty}e^{-c_{2}9^{k}}<\infty.$

We continue with the proof of the bounds of $T_{1}$ and $T_{2},$ beginning
with $T_{2}\leq cp,$ for some constant $c>0.$ By definition of $T_{2},$
$T_{2}=\sum\limits_{s=0}^{p+1}\sum\limits_{t=s}^{p+1}\sum\limits_{z\in\mathrm{R}(s,t)}\varphi_{\mathrm{U}}(z)^{2}.$
As before, we have $\varphi_{\mathrm{U}}(z)^{2}\leq
d_{1}^{2}\Big{(}e^{-2d_{2}3^{t}}9^{-t}+e^{-2d_{2}9^{t-s}}9^{-2s}\Big{)}.$
Therefore, with $C=d_{1}^{2}L,$

$\displaystyle\textstyle T_{2}$ $\displaystyle\textstyle\leq
C\sum\limits_{s=0}^{p+1}\sum\limits_{t=s}^{p+1}\left(e^{-2d_{2}3^{t}}9^{-t}+e^{-2d_{2}9^{t-s}}9^{-2s}\right)9^{s+t}=C\sum\limits_{s=0}^{p+1}\sum\limits_{t=s}^{p+1}\left(e^{-2d_{2}3^{t}}9^{s}+e^{-2d_{2}9^{t-s}}9^{t-s}\right)$
$\displaystyle\textstyle\leq
C\sum\limits_{k=0}^{\infty}e^{-d_{2}3^{k}}9^{k}\sum\limits_{k=0}^{\infty}e^{-d_{2}3^{k}}+C(p+2)\sum\limits_{k=0}^{\infty}e^{-2d_{2}9^{k}}9^{k}\leq
C^{\prime}p,$

where
$C^{\prime}=C\sum\limits_{k=0}^{\infty}e^{-d_{2}3^{k}}9^{k}\sum\limits_{k=0}^{\infty}e^{-d_{2}3^{k}}+3C\sum\limits_{k=0}^{\infty}e^{-2d_{2}9^{k}}9^{k}<\infty$
is independent of $p.$ Similarly, notice one can express
$T_{1}=\sum\limits_{s=0}^{p+1}\sum\limits_{t=0}^{s-1}\sum\limits_{z\in\mathrm{R}(s,t)}\varphi_{\mathrm{U}}(z)^{2},$
bearing this in mind and that $\varphi_{\mathrm{U}}(z)\leq d_{1}9^{-s}$ for
$0\leq t\leq s$ it follows immediately that, for some positive constants $c$
and $c^{\prime},$ $T_{1}\leq
c\sum\limits_{s=0}^{p+1}\sum\limits_{t=0}^{s}9^{s+t}9^{-2s}\leq c^{\prime}p,$
as desired. ∎

####  4.3.2 Estimates on the Green’s function on some regions of
$\Gamma_{2}(\lambda).$

Denote with $S$ and $S^{\prime}$ two independent network random walks of
$\Gamma_{2}(\lambda).$

Consider the following region ($p\in\mathds{N}$)

(4.10) $\textstyle\mathrm{U}_{p}=\left\\{(n,x)\in\mathds{Z}^{3}:|n|\leq 4\cdot
9^{pk_{0}},\left\lVert x\right\rVert\leq 3^{(p+1)k_{0}}\right\\},$

where $k_{0}$ is a positive integer to be determined later on. Define the
“separating cylinders” (creating a barrier between $\mathrm{D}_{pk_{0}}$ and
$\mathrm{D}_{(p+1)k_{0}}$ for $p\in\mathds{N}$) to be the (vertex) boundary
$\mathrm{F}_{p}=\partial\mathrm{U}_{p},$ and set $\mathrm{F}_{0}=\\{0\\}.$
Having these sets, define ($p\in\mathds{Z}_{+}$)

(4.11) $\textstyle T_{p}=\tau_{\mathrm{F}_{p}}(S)$

the hitting time of $\mathrm{F}_{p}$ by $S.$ If $z\in\mathds{Z}^{3},$ let

(4.12) $\textstyle p_{z}=\min\\{p\mid z\in\mathrm{U}_{p}\\}.$

Having defined $p_{z},$ observe that when $S_{0}=o,$ then a.s. the sequence of
random variables $(T_{p})_{p\geq p_{o}}$ is increasing and each one of them is
finite (by transience).

Next, by definition $\mathrm{F}_{p}\cap\mathrm{U}_{p}=\varnothing.$ It will be
useful to divide the set $\mathrm{F}_{p}$ into several regions. Geometrically,
the set $\mathrm{F}_{p}$ can be thought as a cylinder and we need to consider
the body and the two bases separately. Analytically, if
$(n,x)\in\mathrm{F}_{p}$ then $|n|>4\cdot 9^{pk_{0}}$ or else $\left\lVert
x\right\rVert>3^{(p+1)k_{0}}.$ We can specialise further for either
$|n|=4\cdot 9^{pk_{0}}+1$ and $\left\lVert x\right\rVert\leq 3^{(p+1)k_{0}}$
or else, $|n|\leq 4\cdot 9^{pk_{0}}$ and $3^{(p+1)k_{0}}\leq\left\lVert
x\right\rVert\leq 3^{(p+1)k_{0}}+1.$ Thus, the exponential scales being
considered permit us to assume that $\left\lVert z\right\rVert,$ for
$z\in\mathrm{F}_{p},$ satisfies :

  1. (a)

$|n|=4\cdot 9^{pk_{0}}$ and $\left\lVert x\right\rVert\leq 3^{(p+1)k_{0}},$ or

  2. (b)

$|n|<4\cdot 9^{pk_{0}}$ and $\left\lVert x\right\rVert=3^{(p+1)k_{0}}.$

With this assumption, we introduce the following sets. The “left-hand base of
the cylinder $\mathrm{F}_{p}$” is the set of $z\in\mathrm{F}_{p}$ such that
$n=-4\cdot 9^{pk_{0}};$ the “right-hand base of the cylinder $\mathrm{F}_{p}$”
is the set of $z\in\mathrm{F}_{p}$ such that $n=4\cdot 9^{pk_{0}}.$ The points
in $\mathrm{F}_{p}$ that belong to neither the left nor the right-hand base of
the cylinder will be referred to as the “body of the cylinder
$\mathrm{F}_{p},$” in other words, the body of the cylinder is the set of
points $z\in\mathrm{F}_{p}$ such that $|n|<4\cdot 9^{pk_{0}}$ and $\left\lVert
x\right\rVert=3^{(p+1)k_{0}}.$

Define $(T_{p}^{\prime})$ as above changing the process $S$ for the process
$S^{\prime}.$ Observe that $(T_{p})$ and $(T_{p}^{\prime})$ are independent
families.

Having the times $(T_{p})$ and $(T_{p}^{\prime})$ we now consider “sections”
of the network random walks and the number of intersections in
$\mathrm{D}_{pk_{0}}$ of two such sections. Thus, define ($p\in\mathds{N}$)

(4.13) $\textstyle
M_{p}=\sum\limits_{z\in\mathrm{D}_{pk_{0}}}\sum\limits_{m=T_{p-1}}^{T_{p}}\sum\limits_{m^{\prime}=T_{p-1}^{\prime}}^{T_{p}^{\prime}}\mathds{1}_{\\{S_{m}=z\\}}\mathds{1}_{\\{S_{m^{\prime}}^{\prime}=z\\}}.$

Observe that $M_{p}$ counts the number of intersections of the two random
walks inside the region $\mathrm{D}_{pk_{0}}$ starting in the separating
cylinder $\mathrm{F}_{p-1}.$ We remark that, if $S_{0}=o,$ then $S$ cannot be
inside $\mathrm{D}_{pk_{0}}$ before the time $T_{p-1}$ provided $p-1\geq
p_{o};$ similarly, $S^{\prime}$ cannot be inside $\mathrm{D}_{pk_{o}}$ before
time $T_{p-1}^{\prime}$ provided $S_{0}^{\prime}=o^{\prime}$ and $p-1\geq
p_{o^{\prime}}.$

We will study the Green’s function restricted to the region $\mathrm{U}_{p}:$
$G_{\mathrm{U}_{p}}(0,z)=\mathds{E}{{}^{0}}\left(\sum\limits_{m=0}^{T_{p}-1}\mathds{1}_{\\{S_{m}=z\\}}\right),$
for $z\in\mathds{Z}^{3}.$ For consistency we denote $\mathrm{U}_{0}=\\{0\\}.$
For the sake of simplicity, we will write $G_{p}=G_{\mathrm{U}_{p}}.$

######  Proposition ( 4.14 )

For every $p\in\mathds{N},$
$\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}\subset\mathrm{U}_{p}.$ Moreover, for any
$\varepsilon>0$ there exists an index $k_{0}=k_{0}(\varepsilon)$ such that for
some $p_{0}\in\mathds{N}$ the relations $p\geq p_{0}$ and
$z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ imply
$G(0,z)\leq(1+\varepsilon)G_{p}(0,z).$

###### Proof.

The proof contains several steps. We begin by noticing that for
$z_{1}\in\mathrm{D}_{pk_{0}}$ and $z_{2}\in\mathrm{U}_{p-1}$ one has by (4.7)
and (4.10) $9^{pk_{0}}<n_{1}\leq 2\cdot 9^{pk_{0}}$ and $\left\lVert
x_{1}\right\rVert\leq 3^{pk_{0}},$ and $-4\cdot 9^{(p-1)k_{0}}\leq n_{2}\leq
4\cdot 9^{(p-1)k_{0}}$ and $\left\lVert x_{2}\right\rVert\leq 3^{pk_{0}},$
which implies

######  ( 4.14.1 )

Any $(n,x)\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ satisfies $\frac{1}{2}\cdot
9^{pk_{0}}\leq n\leq\frac{5}{2}9^{pk_{0}},\left\lVert x\right\rVert\leq 2\cdot
3^{pk_{0}}.$

In particular, $z_{1}-z_{2}$ belongs to the region $\mathrm{U}_{p}$ so this
establishes the first assertion in (4.14).

We now prove the second assertion of (4.14). First,
$G(0,z)=G_{p}(0,z)+\mathds{E}{{}^{0}}\left(G(S_{T_{p}},z)\right).$ Observe
that $S_{T_{p}}$ is in $\mathrm{F}_{p}$ and, therefore, we need to estimate
$G(z_{3},z)$ for $z_{3}\in\mathrm{F}_{p}.$ This will be done in the upcoming
three lemmas.

We will pause momentarily the proof of the proposition in order to establish
these lemmas and resume it once they have been proven.

######  Lemma ( 4.14.2 )

  1. (a)

For every $\varepsilon>0,$ there exists a $p_{0}=p_{0}(\varepsilon)$ such that
for all $p\geq p_{0},$ all $k_{0}\in\mathds{N},$ and all
$z_{1}\in\mathrm{D}_{pk_{0}},z_{2}\in\mathrm{U}_{p-1}$ and
$z_{3}\in\mathrm{F}_{p}$ for which $n_{1}-n_{2}-n_{3}\leq 2\cdot
3^{(p+1)k_{0}}$ and $z_{3}$ does not lie in the left-hand base of
$\mathrm{F}_{p},$ we have $G(z_{3},z_{1}-z_{2})\leq\varepsilon
G(0,z_{1}-z_{2}).$

  2. (b)

There exists a constant $c>0$ such that for $p\in\mathds{N}$ and all
$z_{1}\in\mathrm{D}_{pk_{0}},z_{2}\in\mathrm{U}_{p-1}$ and
$z_{3}\in\mathrm{F}_{p}$ for which $n_{1}-n_{2}-n_{3}\geq 2\cdot
3^{(p+1)k_{0}}$ and $z_{3}$ does not lie in the left-hand base of
$\mathrm{F}_{p}$ then $G(z_{3},z_{1}-z_{2})\leq c9^{-k_{0}}G(0,z_{1}-z_{2}).$

###### Proof of lemma..

Suppose the constants $c_{k}$ ($1\leq k\leq 4$) of (2.22) are given. By
(4.14.1) and the estimates on the Green’s function, we have
$G(0,z_{1}-z_{2})\asymp 9^{-pk_{0}}.$

We prove part (a) first. Assume $z_{1},z_{2},z_{3}$ are as stated. Consider
first the case $n_{3}<4\cdot 9^{pk_{0}},$ that is to say, consider first when
$z_{3}$ is not in the right-hand base of $\mathrm{F}_{p}.$ Then $\left\lVert
x_{3}\right\rVert=3^{(p+1)k_{0}},$ which in turn yields $\frac{1}{3}\cdot
3^{(p+1)k_{0}}\leq\left\lVert x_{1}-x_{2}-x_{3}\right\rVert\leq 2\cdot
3^{(p+1)k_{0}}.$ Suppose first that $n_{1}-n_{2}-n_{3}\leq\frac{1}{3}\cdot
3^{(p+1)k_{0}},$ then $z_{1}-z_{2}-z_{3}$ is in $\mathrm{S}_{1},$ implying by
(2.22)

$\displaystyle G(z_{3},z_{1}-z_{2})$ $\displaystyle\leq
c_{1}e^{-c_{2}\left\lVert z_{1}-z_{2}-z_{3}\right\rVert}\left\lVert
z_{1}-z_{2}-z_{3}\right\rVert^{-1}\leq
3c_{1}e^{-\frac{c_{2}}{3}3^{(p+1)k_{0}}}3^{-(p+1)k_{0}}$
$\displaystyle=\left[3c_{1}e^{-\frac{c_{2}}{3}3^{(p+1)k_{0}}}3^{(p-1)k_{0}}\right]9^{-pk_{0}}\leq\left[3c_{1}e^{-\frac{c_{2}}{3}3^{pk_{0}}}3^{pk_{0}}\right]9^{-pk_{0}}.$

Define $u_{1}(t)=3c_{1}e^{-\frac{c_{2}}{3}3^{t}}3^{t},$ so that
$G(z_{3},z_{1}-z_{2})\leq u_{1}(pk_{0})9^{-pk_{0}},$ for $z_{1},$ $z_{2}$ and
$z_{3}$ as stated. Next, suppose that $\frac{1}{3}\cdot 3^{(p+1)k_{0}}\leq
n_{1}-n_{2}-n_{3}\leq 2\cdot 3^{(p+1)k_{0}}.$ Then,

$\displaystyle G(z_{3},z_{1}-z_{2})$ $\displaystyle\leq
c_{1}\left(e^{-c_{2}\left\lVert
z_{1}-z_{2}-z_{3}\right\rVert}+e^{-c_{2}\frac{\left\lVert
x_{1}-x_{2}-x_{3}\right\rVert^{2}}{n_{1}-n_{2}-n_{3}}}\right)\left\lVert
z_{1}-z_{2}-z_{3}\right\rVert^{-1}$ $\displaystyle\leq
3c_{1}\left(e^{-\frac{c_{2}}{3}3^{(p+1)k_{0}}}+e^{-\frac{c_{2}}{18}3^{(p+1)k_{0}}}\right)3^{-(p+1)k_{0}}$
$\displaystyle=\left[3c_{1}\left(e^{-\frac{c_{2}}{3}3^{(p+1)k_{0}}}+e^{-\frac{c_{2}}{18}3^{(p+1)k_{0}}}\right)3^{(p-1)k_{0}}\right]9^{-pk_{0}}$
$\displaystyle\leq\left[3c_{1}\left(e^{-\frac{c_{2}}{3}3^{pk_{0}}}+e^{-\frac{c_{2}}{18}3^{pk_{0}}}\right)3^{pk_{0}}\right]9^{-pk_{0}}=u_{2}(pk_{0})9^{-pk_{0}},$

where $u_{2}(t)$ is defined to be
$3c_{1}\left(e^{-\frac{c_{2}}{3}3^{t}}+e^{-\frac{c_{2}}{18}3^{t}}\right)3^{t}.$
Consider now the case where $z_{3}$ is in the right-hand base of the cylinder,
that is to say, assume $n_{3}=4\cdot 9^{pk_{0}}$ and $\left\lVert
x_{3}\right\rVert\leq 3^{(p+1)k_{0}}$ but in this case, $-\frac{7}{2}\cdot
9^{pk_{0}}\leq n_{1}-n_{2}-n_{3}\leq-\frac{3}{2}\cdot 9^{pk_{0}}$ and again,
by (2.22), $G(z_{3},z_{1}-z_{2})\leq
c_{1}e^{-c_{2}|n_{1}-n_{2}-n_{3}|}|n_{1}-n_{2}-n_{3}|^{-1}\leq\frac{2c_{1}}{3}e^{-\frac{2c_{2}}{3}9^{pk_{0}}}9^{-pk_{0}}.$
Here we set $u_{3}(t)=\frac{2c_{1}}{3}e^{-\frac{2c_{2}}{3}9^{t}},$ so that in
this case $G(z_{3},z_{1}-z_{2})\leq u_{3}(pk_{0})9^{-pk_{0}}.$ Notice that in
all three cases above, we bounded $G(z_{3},z_{1}-z_{2})\leq
u_{i}(pk_{0})9^{-pk_{0}},$ and it is clear that $u_{i}(t)\to 0$ as
$t\to\infty$ in each case. Then, there exists a $t_{0}$ such that
$\sup\limits_{t\geq t_{0}}\max\limits_{i=1,2,3}u_{i}(t)\leq\varepsilon.$ We
can define $p_{0}$ to be the least integer greater than or equal to $t_{0};$
this $p_{0}$ works for all $k_{0}$ since $pk_{0}\geq p.$ This finishes part
(a).

We now prove part (b). Suppose that $z_{1},z_{2},z_{3}$ are as stated in (b).
This implies the relation $n_{1}-n_{2}-n_{3}\geq 2\cdot 3^{(p+1)k_{0}}.$ By
(4.14.1), we have $n_{1}-n_{2}-n_{3}\leq 7\cdot 9^{pk_{0}}$ and, therefore
$n_{1}-n_{2}-n_{3}$ lies in an interval of the form
$\mathrm{J}_{\alpha}=\left[(\alpha-1)3^{(p+1)k_{0}},\alpha
3^{(p+1)k_{0}}\right]$ where $\alpha=3,\ldots,7\cdot 3^{(p-1)k_{0}}.$
Regardless the value of $\alpha,$ the relation
$n_{1}-n_{2}-n_{3}\in\mathrm{J}_{\alpha}$ implies
$z_{1}-z_{2}-z_{3}\in\mathrm{S}_{2}$ and these relations imply,

$G(z_{3},z_{1}-z_{2})\leq c_{1}e^{-c_{2}\frac{\left\lVert
x_{1}-x_{2}-x_{3}\right\rVert^{2}}{n_{1}-n_{2}-n_{3}}}(n_{1}-n_{2}-n_{3})^{-1}\leq
2c_{1}e^{-\frac{c_{2}}{9}\alpha^{-1}3^{(p+1)k_{0}}}\alpha^{-1}3^{-(p+1)k_{0}}.$

Observe that for any constant $c>0,$ the function
$u_{c}(t)=t^{-1}e^{-ct^{-1}}$ for $t>0$ has an absolute and global maximum at
$t=c.$ Its maximum equals $u_{c}(c)=e^{-1}c^{-1}.$ We deduce
$G(z_{3},z_{1}-z_{2})\leq
2c_{1}e^{-1}\frac{9}{c_{2}}9^{-(p+1)k_{0}}=c^{\prime}9^{-k_{0}}9^{-pk_{0}},$
the desired conclusion since $G(0,z_{1}-z_{2})\asymp 9^{-pk_{0}},$ with any
implicit constant independent of $p$ and of $k_{0}.$ ∎

######  Lemma ( 4.14.3 )

For any $\varepsilon>0,$ there exist indices
$k_{0}=k_{0}(\varepsilon)\in\mathds{N}$ and
$p_{0}=p_{0}(\varepsilon)\in\mathds{N}$ such that the relations $p\geq p_{0}$
and $z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ imply
$\sup\limits_{\begin{subarray}{c}z_{3}\in\mathrm{F}_{p}\\\ n_{3}>-4\cdot
9^{pk_{0}}\end{subarray}}G(z_{3},z)\leq\varepsilon G(0,z).$

###### Proof of lemma..

By (4.14.2), given $\varepsilon>0$ there is a $p_{0}\in\mathds{N}$ such that
if $p\geq p_{0},$ $k_{0}\in\mathds{N}$ and for all
$z_{1}\in\mathrm{D}_{pk_{0}},$ $z_{2}\in\mathrm{U}_{p-1}$ and
$z_{3}\in\mathrm{F}_{p}$ with $n_{3}>-4\cdot 9^{p},$ we either have
$G(z_{3},z_{1}-z_{2})\leq\varepsilon G(0,z_{1}-z_{2})$ or
$G(z_{3},z_{1}-z_{2})\leq c9^{-k_{0}}G(0,z_{1}-z_{2}),$ where the constant $c$
is universal. Choose $k_{0}\in\mathds{N}$ such that
$c9^{-k_{0}}\leq\varepsilon.$ Hence, $G(z_{3},z_{1}-z_{2})\leq\varepsilon
G(0,z_{1}-z_{2})$ uniformly in $z_{3}\in\mathrm{F}_{p}$ with $n_{3}>-4\cdot
9^{p}$ for all $p\geq p_{0},$ which is the conclusion to be reached. ∎

######  Lemma ( 4.14.4 )

For any $\varepsilon>0,$ there exist indices
$k_{0}=k_{0}(\varepsilon)\in\mathds{N}$ and
$p_{0}=p_{0}(\varepsilon)\in\mathds{N}$ such that the relation $p\geq p_{0}$
implies $\mathds{P}{{}^{0}}\left(S_{T_{p}}\in\\{z\in\mathrm{F}_{p}:n=-4\cdot
9^{p}\\}\right)\leq\varepsilon.$ Also, there is a universal constant $c>0$
such that for all $p\in\mathds{N}$ and all
$z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1},$
$\sup\limits_{\begin{subarray}{c}z_{3}\in\mathrm{F}_{p}\\\ n_{3}=-4\cdot
9^{p}\end{subarray}}G(z_{3},z)\leq cG(0,z).$

###### Proof of lemma..

Set $\mathrm{A}_{p}=\\{z\in\mathrm{F}_{p}:n=-4\cdot 9^{pk_{0}}\\}.$ Consider
the network random walk $S$ in $\Gamma_{2}(\lambda)$ started at zero and
define $B=\mathrm{pr}_{0}(S).$ By (2.17),
$\mathds{P}{}\left(S_{T_{p}}\in\mathrm{A}_{p}\right)\leq\mathds{P}{}\left(B\text{
ever reaches }-4\cdot 9^{pk_{0}}\right)=e^{-4\lambda 9^{pk_{0}}}.$ This proved
the first claim. To prove the second claim, use (4.14.1), if
$z_{3}\in\mathrm{A}_{p}$ and $z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ then
$G(z_{3},z)\asymp 9^{-pk_{0}}$ with any implicit constant being universal in
$p,k_{0},z,z_{3}.$ Similarly, $G(0,z)\asymp 9^{-pk_{0}}$ for
$z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ with any implicit constant
universal. The second claim follows readily. ∎

We are ready to continue with the proof of Proposition (4.14). Let us write
$\mathrm{A}_{p}=\\{z\in\mathrm{F}_{p}:n_{3}=-4\cdot 9^{p}\\}$ for the left-
hand base of $\mathrm{F}_{p}$ and
$\mathrm{B}_{p}=\mathrm{F}_{p}\setminus\mathrm{A}_{p}$ for the remainder of
the separating cylinder. For the time being, consider $\varepsilon>0$ to be
any positive number. By Lemmas (4.14.2) and (4.14.4), there are indices
$k_{0}$ and $p_{0},$ only depending on $\varepsilon,$ and a universal constant
$c>0$ (not depending on any index) satisfying for all $p\geq p_{0},$
$z_{1}\in\mathrm{D}_{pk_{0}},$ $z_{2}\in\mathrm{U}_{p-1}$
$\sup\limits_{z_{3}\in\mathrm{B}_{p}}G(z_{3},z_{1}-z_{2})\leq\varepsilon
G(0,z_{1}-z_{2}),$
$\mathds{P}{{}^{0}}\left(S_{T_{p}}\in\mathrm{A}_{p}\right)\leq\varepsilon$ and
for all $z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1},$
$\sup\limits_{z_{3}\in\mathrm{A}_{p}}G(z_{3},z)\leq cG(0,z).$ Write now, for
$z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1},$

$\displaystyle G(0,z)$
$\displaystyle=G_{p}(0,z)+\mathds{E}{{}^{0}}\left(G(S_{T_{p}},z)\mathds{1}_{\mathrm{A}_{p}}(S_{T_{p}})\right)+\mathds{E}{{}^{0}}\left(G(S_{T_{p}},z)\mathds{1}_{\mathrm{B}_{p}}(S_{T_{p}})\right)$
$\displaystyle\leq
G_{p}(0,z)+cG(0,z)\mathds{P}{{}^{0}}\left(S_{T_{p}}\in\mathrm{A}_{p}\right)+\varepsilon
G(0,z)\mathds{P}{{}^{0}}\left(S_{T_{p}}\in\mathrm{B}_{p}\right)$
$\displaystyle\leq G_{p}(0,z)+(c+1)\varepsilon G(0,z).$

Hence, for any $\varepsilon>0,$ there are indices $k_{0}$ and $p_{0},$
depending only on $\varepsilon,$ so that the relations $p\geq p_{0}$ and
$z\in\mathrm{D}_{pk_{0}}-\mathrm{U}_{p-1}$ imply $G(0,z)\leq
G_{p}(0,z)+\frac{\varepsilon}{1+\varepsilon}G(0,z),$ which yields
$G(0,z)\leq(1+\varepsilon)G_{p}(0,z).$ ∎

###  4.4 Dimension $d=2;$ “logarithmic scale of crossings” of
$\Gamma_{2}(\lambda)$

We will establish a _lower bound_ on the probability of how frequently two
random walk paths in $\Gamma_{2}(\lambda)$ will cross each other.

Invoking Proposition (4.14), for $\varepsilon=1$ there exists two indices
$k_{0}$ and $p_{0},$ depending only on $\varepsilon,$ such that the
conclusions of this proposition hold. For the remainder of this chapter, we
assume both $k_{0}$ and $p_{0}$ are fixed and given as stated.

######  Theorem ( 4.15 )

Consider two independent network random walks $S$ and $S^{\prime}$ in
$\Gamma_{2}(\lambda),$ with respective starting points $o$ and $o^{\prime}.$
Define the random variables $T_{p}$ and $T_{p}^{\prime}$ in (4.11) using $S$
and $S^{\prime},$ accordingly; now introduce $M_{p}$ as in (4.13). Consider
now the $\sigma$-fields
$\mathscr{E}_{p}=\mathscr{F}^{S}_{T_{p}}\vee\mathscr{F}^{S^{\prime}}_{T_{p}^{\prime}}.$
Define the event $\mathrm{A}_{p}=\\{M_{p}>0\\}$ and the random variable
$q_{p+1}=\mathds{P}{}\left(\mathrm{A}_{p+1}\mid\mathscr{E}_{p}\right).$ Then,
$\mathrm{A}_{p}\in\mathscr{E}_{p}$ and there exists a positive integer
$P=P(o,o^{\prime})$ and a universal constant $c>0$ (depending neither on $o$
nor on $o^{\prime}$) such that, almost surely, $q_{p}>\frac{c}{p}$ for all
$p\geq P.$

###### Proof.

It is clear that $\mathrm{A}_{p}\in\mathscr{E}_{p}.$ Consider
$p_{1}=\max(p_{o},p_{o^{\prime}})$ (4.12), that is, $p_{1}$ is the first index
such that $o$ and $o^{\prime}$ belong to $\mathrm{U}_{p}.$ Suppose $p$ is any
integer $\geq\max(p_{0},p_{1}).$ By (5.3), we have
$q_{p+1}=\mathds{P}{}\left(\mathrm{A}_{p+1}\middle|S_{T_{p}},S_{T_{p}^{\prime}}^{\prime}\right).$
It is now straightforward to check that

($\ast$) $\textstyle
q_{p+1}=\sum\limits_{(z_{1},z_{2})\in\mathrm{F}_{p}^{2}}\mathds{P}{}\left(\mathrm{A}_{p+1}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)\mathds{1}_{\left\\{S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right\\}}.$

We will use (4.3). By the strong Markov property for a pair of independent
Markov processes (5.3) gives (inside the expectation on the right of the first
equality we use abuse of notation and assume $(S_{m})$ and
$(S_{m^{\prime}}^{\prime})$ are independent network random walks started at
zero)

$\displaystyle\textstyle\mathds{E}{}\left(M_{p+1}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)$
$\displaystyle\textstyle=\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}\mathds{E}{}\left(\sum\limits_{m=0}^{T_{p+1}}\sum\limits_{m^{\prime}=0}^{T_{p+1}^{\prime}}\mathds{1}_{\\{S_{m}+z_{1}=z_{3}\\}}\mathds{1}_{\left\\{S_{m^{\prime}}^{\prime}+z_{2}=z_{3}\right\\}}\right)$
$\displaystyle\textstyle=\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}G_{p+1}(0,z_{3}-z_{1})G_{p+1}(0,z_{3}-z_{2})$
$\displaystyle\textstyle\asymp\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}G(0,z_{3}-z_{1})G(0,z_{3}-z_{2}),\quad\text{by
}(\ref{Proposition: Asymptotic bound between the Green function and the
restricted Green function on d equals 2})$
$\displaystyle\textstyle\asymp\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}9^{-2(p+1)k_{0}}\asymp
1,\quad\text{ by }(\ref{Remark: Principal calculations for the number of
crossings in dimension d equals 2})(k).$

We shall now estimate the second moment of $M_{p}$ with respect to the
conditional measure. We have

$\textstyle
M_{p+1}^{2}=\sum\limits_{(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}}\sum\limits_{m_{1},m_{2}=T_{p}}^{T_{p+1}}\sum\limits_{m_{1}^{\prime},m_{2}^{\prime}=T_{p}^{\prime}}^{T_{p+1}^{\prime}}\mathds{1}_{\\{S_{m_{1}}=z_{3},S_{m_{2}}=z_{4}\\}}\mathds{1}_{\left\\{S_{m_{1}^{\prime}}^{\prime}=z_{3},S_{m_{2}^{\prime}}^{\prime}=z_{4}\right\\}}.$

Now apply the conditional expectation, by the strong Markov property (5.3) and
independence between $S$ and $S^{\prime}$ (we again employ abuse of notation
and assume $S$ and $S^{\prime}$ are two independent network random walks
started at zero after having used the strong Markov property for two
independent Markov processes)

$\displaystyle\textstyle\mathds{E}{}\left(M_{p+1}^{2}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)$
$\displaystyle\textstyle=\sum\limits_{(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}}\mathds{E}{}\left(\sum\limits_{m_{1},m_{2}=0}^{T_{p+1}}\mathds{1}_{\\{S_{m_{1}}=z_{3}-z_{1},S_{m_{2}}=z_{4}-z_{1}\\}}\right)$
$\displaystyle\textstyle\times\mathds{E}{}\left(\sum\limits_{m_{1}^{\prime},m_{2}^{\prime}=0}^{T_{p+1}^{\prime}}\mathds{1}_{\left\\{S_{m_{1}^{\prime}}^{\prime}=z_{3}-z_{2},S_{m_{2}^{\prime}}^{\prime}=z_{4}-z_{2}\right\\}}\right)$
$\displaystyle\textstyle\leq B_{1}+B_{2}+B_{3}+B_{4},$

in which

$\begin{split}B_{1}&\textstyle=\sum\limits_{\begin{subarray}{c}m_{1}\leq
m_{2}\\\ m_{1}^{\prime}\leq m_{2}^{\prime}\\\
(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}\end{subarray}}\mathds{P}{}\left(S_{m_{1}}=z_{3}-z_{1},S_{m_{2}}=z_{4}-z_{1}\right)\mathds{P}{}\left(S_{m_{1}^{\prime}}^{\prime}=z_{3}-z_{2},S_{m_{2}^{\prime}}^{\prime}=z_{4}-z_{2}\right),\\\
B_{2}&\textstyle=\sum\limits_{\begin{subarray}{c}m_{1}\leq m_{2}\\\
m_{1}^{\prime}\geq m_{2}^{\prime}\\\
(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}\end{subarray}}\mathds{P}{}\left(S_{m_{1}}=z_{3}-z_{1},S_{m_{2}}=z_{4}-z_{1}\right)\mathds{P}{}\left(S_{m_{1}^{\prime}}^{\prime}=z_{3}-z_{2},S_{m_{2}^{\prime}}^{\prime}=z_{4}-z_{2}\right),\\\
B_{3}&\textstyle=\sum\limits_{\begin{subarray}{c}m_{1}\geq m_{2}\\\
m_{1}^{\prime}\leq m_{2}^{\prime}\\\
(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}\end{subarray}}\mathds{P}{}\left(S_{m_{1}}=z_{3}-z_{1},S_{m_{2}}=z_{4}-z_{1}\right)\mathds{P}{}\left(S_{m_{1}^{\prime}}^{\prime}=z_{3}-z_{2},S_{m_{2}^{\prime}}^{\prime}=z_{4}-z_{2}\right),\end{split}$

and

$\textstyle B_{4}=\sum\limits_{\begin{subarray}{c}m_{1}\geq m_{2}\\\
m_{1}^{\prime}\geq m_{2}^{\prime}\\\
(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}\end{subarray}}\mathds{P}{}\left(S_{m_{1}}=z_{3}-z_{1},S_{m_{2}}=z_{4}-z_{1}\right)\mathds{P}{}\left(S_{m_{1}^{\prime}}^{\prime}=z_{3}-z_{2},S_{m_{2}^{\prime}}^{\prime}=z_{4}-z_{2}\right).$

We handle each sum separately.

  1. (a)

By independent increments and writing
$p^{k}(z,z^{\prime})=\mathds{P}{}\left(S_{k}=z^{\prime}\mid S_{0}=z\right),$

$\displaystyle B_{1}$
$\displaystyle\textstyle=\sum\limits_{\begin{subarray}{c}m_{1}\leq m_{2}\\\
m_{1}^{\prime}\leq m_{2}^{\prime}\\\
(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}\end{subarray}}p^{m_{1}}(0,z_{3}-z_{1})p^{m_{2}-m_{1}}(0,z_{4}-z_{3})p^{m_{1}^{\prime}}(0,z_{3}-z_{2})p^{m_{2}^{\prime}-m_{1}^{\prime}}(0,z_{4}-z_{3})$
$\displaystyle\textstyle=\sum\limits_{(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}}G(0,z_{3}-z_{1})G(0,z_{4}-z_{3})^{2}G(0,z_{3}-z_{2}),$

the last equality being an immediate consequence of Lebesgue-Tonelli’s theorem
and the definition of the Green’s function of $\Gamma_{2}(\lambda).$ By virtue
of (4.14.1) and (4.9) we can proceed further,

$\displaystyle B_{1}$
$\displaystyle\textstyle\asymp\sum\limits_{(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}}9^{-2(p+1)k_{0}}G(0,z_{4}-z_{3})^{2}\leq
ck_{0}(p+1)\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}9^{-2(p+1)k_{0}}\asymp
p.$

  2. (b)

Here we get

$\displaystyle B_{2}$
$\displaystyle\textstyle=\sum\limits_{(z_{3},z_{4})\in\mathrm{D}_{(p+1)k_{0}}^{2}}G(0,z_{3}-z_{1})G(0,z_{4}-z_{3})G(0,z_{4}-z_{2})G(0,z_{3}-z_{4})$
$\displaystyle\textstyle\asymp\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}\sum\limits_{z_{4}\in\mathrm{D}_{(p+1)k_{0}}}9^{-2(p+1)k_{0}}G(0,z_{4}-z_{3})G(0,z_{3}-z_{4})$
$\displaystyle\textstyle\asymp\sum\limits_{z_{3}\in\mathrm{D}_{(p+1)k_{0}}}9^{-2(p+1)k_{0}}\asymp
1,$

where the second $\asymp$ is a consequence of (4.9).

  3. (c)

Similarly to $B_{2},$ we get $B_{3}\asymp 1.$

  4. (d)

As with $B_{1},$ $B_{4}\leq cp,$ for some universal constant $c>0.$

Therefore, we have shown that there exist two positive constants $c$ and
$c^{\prime},$ such that, whenever $p\geq\max(p_{0},p_{1})$ and
$(z_{1},z_{2})\in\mathrm{F}_{p}^{2},$
$\mathds{E}{}\left(M_{p+1}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)\geq
c$ and
$\mathds{E}{}\left(M_{p+1}^{2}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)\leq
c^{\prime}p.$ By (4.3), we finally reach that there exists a constant $c>0$
and a $P=\max(p_{0},p_{1})$ such that, for all $p\geq P$ and all
$(z_{1},z_{2})\in\mathrm{F}_{p}^{2},$
$\mathds{P}{}\left(\mathrm{A}_{p+1}\middle|S_{T_{p}}=z_{1},S_{T_{p}^{\prime}}^{\prime}=z_{2}\right)\geq\frac{c}{p},$
comparing with the expression for $q_{p+1}$ in ($\ast$), we have substantiated
the claims. ∎

######  Corollary ( 4.15.1 )

Let $S$ and $S^{\prime}$ be two independent network random walks in
$\Gamma_{2}(\lambda),$ with starting points $S_{0}=o$ and
$S_{0}^{\prime}=o^{\prime}.$ There is a universal constant $c>0$ and an index
$P=P(o,o^{\prime}),$ such that for all $p\geq P,$ the probability that the two
random walk paths $(S_{m})_{m=T_{p-1},\ldots,T_{p}}$ and
$(S_{m^{\prime}}^{\prime})_{m^{\prime}=T_{p-1}^{\prime},\ldots,T_{p}^{\prime}}$
cross each other inside $\mathrm{D}_{pk_{0}}$ is at least $\frac{c}{p}.$

The following corollary follows using (4.4) and (4.15).

######  Corollary ( 4.15.2 )

Any two independent network random walks on $\Gamma_{2}(\lambda)$ will have
infinitely many intersections.

######  Remark ( 4.16 )

There is a much simpler way to obtain Corollary (4.15.2). Denote $S=(B,X),$
that is, $S_{n}=(B_{n},X_{n})$ for $n\in\mathds{Z}_{+};$ here $B$ is the
biased random walk on $\mathds{Z}$ and $X$ a random walk on $\mathds{Z}^{2}.$
Define $\zeta_{k}=\tau_{\\{k\\}\times\mathds{Z}^{2}}(S)$ for
$k\in\mathds{Z}_{+},$ which is the time at which the random walk $B$ visits
$k.$ With these hitting times, consider $Y_{n}=X_{\zeta_{n}}.$ Similarly,
define $Y_{n}^{\prime}$ using $S^{\prime}=(B^{\prime},X^{\prime}),$ which is
an independent copy of $S.$ In a similar manner as theorem (4.2), it is seen
that $Y_{n}=Y_{n}^{\prime}$ for infinitely many indices $n.$ Having this, it
is now clear that $S_{n}=S_{n^{\prime}}^{\prime}$ for infinitely many pairs
$(n,n^{\prime}).$ Of course, this result is much coarser than (4.15.1).

##  5 One endedness for $\Gamma_{d}(\lambda)$

Let $\mathrm{G}$ be a directed infinite tree. A ray is an infinite path that
does not repeat vertices; any two rays of $\mathrm{G}$ will be disjoint or
will eventually merge, this creates equivalence classes on the set of rays of
$\mathrm{G}.$ Any of these classes is called an end.

###  5.1 One endedness for $\Gamma_{1}(\lambda)$

We refer the reader to [LP16], see Section 6.5 , Section 9.2 and Section 10.3
for the definitions and notations of planar graph and duality of graphs.

######  Theorem ( 5.1 )

$\mathsf{UST}$ in $\Gamma_{1}(\lambda)$ (3.9) has, almost surely, one end.

###### Proof.

It is easy to see that the dual network $\Gamma_{1}(\lambda)^{\dagger}$ of
$\Gamma_{1}(\lambda)$ is the network image of the latter under the reflection
$(n,x)\mapsto(-n,x).$ In particular, the wired and free uniform spanning
forest of $\Gamma_{1}(\lambda)^{\dagger}$ coincide and the number of
$\mathsf{USF}$ trees in $\Gamma_{1}(\lambda)^{\dagger}$ is the same as that of
$\Gamma_{1}(\lambda);$ that is to say, for almost every realisation, just one
tree, by (3.9). Bearing this in mind and the duality between $\mathsf{UST}$ in
$\Gamma_{1}(\lambda)$ with $\mathsf{UST}$ in $\Gamma_{1}(\lambda)^{\dagger},$
we see at once that the tree in $\Gamma_{1}(\lambda)$ is almost surely one
ended; for if it were two ended, it would split the plane into two
disconnected regions making it impossible for the dual tree to be a tree. ∎

######  Remark ( 5.2 )

The core idea of the proof of previous theorem is implicit in [LP16, Theorem
10.36].

###  5.2 One endedness for $\Gamma_{2}(\lambda)$

We will prove the following theorem: $\mathsf{UST}$ in $\Gamma_{2}(\lambda)$
has, almost surely, one end.

The proof is rather involved and to facilitate the reading of it we are going
to develop it in several steps and draw the conclusion at the end.

###  5.3 Some preliminary results on Markov processes

Consider a discrete time Markov process
$(S_{n},\mathscr{F}_{n})_{n\in\mathds{Z}_{+}},$ with values on a metrisable
separable space $\mathrm{E},$ endowed with Borels sets $\mathscr{E}$ as
$\sigma$-algebra. Let $\eta$ be an a.s. finite stopping time for this Markov
process. We may construct the $\sigma$-field
$\mathscr{F}_{\eta}=\mathscr{F}_{\eta}^{S}$ of all events
$\mathrm{G}\in\mathscr{E}$ such that $\mathrm{G}\cap\\{\eta\leq
n\\}\in\mathscr{F}_{n}$ (“stopped filtration up to time $\eta$”). In a similar
way, we may construct $S_{\eta}:\omega\mapsto S_{\eta(\omega)}(\omega).$ If no
filtration is specified, it is assumed the canonical filtration is used
$\mathscr{F}_{n}^{S}=\sigma(S_{k};0\leq k\leq n).$ Consider now the product
space $\mathrm{E}^{\mathds{Z}_{+}},$ which is also a metrisable separable
space, and its Borel $\sigma$-field coincides with
$\displaystyle\bigotimes_{n\in\mathds{Z}_{+}}\mathscr{E}.$ For any bounded
measurable function $\varphi:\mathrm{E}^{\mathds{Z}_{+}}\to\mathds{R},$ there
exists a version of the conditional expectation
$\mathds{E}{}\left(\varphi\left(S_{j}\right)_{j\in\mathds{Z}_{+}}\middle|S_{0}=x\right);$
denote this function as
$\mathds{E}{{}^{x}}\left(\varphi\left(S_{j}\right)_{j\in\mathds{Z}_{+}}\right).$
The strong Markov property states that
$\mathds{E}{}\left(\varphi\left((S_{j})_{j\geq\eta}\right)\middle|\mathscr{F}_{\eta}\right)=\mathds{E}{{}^{S_{\eta}}}\left(\varphi\left((S_{j})_{j\in\mathds{Z}_{+}}\right)\right).$

The following results follow from well-known techniques (monotone-class,
Dynkin’s theorem, etc.). The reader may consult [MD20], propositions (1.12.3)
and (1.12.4).

######  Proposition ( 5.3 )

Suppose $S=(S_{n})_{n\in\mathds{Z}_{+}}$ and
$S^{\prime}=(S_{n}^{\prime})_{n\in\mathds{Z}_{+}}$ are two independent Markov
processes, defined on some probability space
$\left(\Omega,\mathscr{F},\mathds{P}\right)$ with values on some metrisable,
separable space $\mathrm{E},$ endowed with Borel $\sigma$-field $\mathscr{E}.$
Let
$\psi:\mathrm{E}^{\mathds{Z}_{+}}\times\mathrm{E}^{\mathds{Z}_{+}}\to\mathds{R}$
be a bounded measurable function, relative to the Borel sets of both
$\mathds{R}$ and
$\mathrm{E}^{\mathds{Z}_{+}}\times\mathrm{E}^{\mathds{Z}_{+}}.$ Define
$v_{\psi}(x,x^{\prime})=\mathds{E}{}\left(\psi\left(S_{j},S_{j^{\prime}}^{\prime}\right)_{(j,j^{\prime})\in\mathds{Z}_{+}^{2}}\middle|S_{0}=x,S_{0}^{\prime}=x^{\prime}\right),$
for $(x,x^{\prime})\in\mathrm{E}^{2},$ that is $v_{\psi}$ is a version of
$\psi\left(S_{j},S_{j^{\prime}}^{\prime}\right)_{(j,j^{\prime})\in\mathds{Z}_{+}^{2}}$
given $(S_{0},S_{0}^{\prime}).$ Suppose $\eta$ and $\eta^{\prime}$ are two
stopping times, $\eta$ relative to $S$ and $\eta^{\prime},$ to $S^{\prime}.$
With $\psi$ as before,
$\mathds{E}{}\left(\psi(S_{j},S_{j^{\prime}}^{\prime})_{j\geq\eta,j^{\prime}\geq\eta^{\prime}}\middle|\mathscr{F}_{\eta}^{S}\vee\mathscr{F}_{\eta^{\prime}}^{S^{\prime}}\right)=v_{\psi}(S_{\eta},S_{\eta}^{\prime}).$

######  Proposition ( 5.4 )

Let $\mathrm{E}=\mathds{R}^{d}$ (or any separable, metrisable, Fréchet space).
Suppose $(S_{n})_{n\in\mathds{Z}_{+}}$ and
$(S^{\prime}_{n})_{n\in\mathds{Z}_{+}}$ are two $\mathrm{E}$-valued stochastic
processes with stationary independent increments. Suppose that $\eta$ and
$\eta^{\prime}$ are stopping times relative to $S$ and $S^{\prime},$
respectively. Then,
$\mathscr{F}_{\eta}^{S}\vee\mathscr{F}_{\eta^{\prime}}^{S^{\prime}}$ is
independent of
$\mathscr{G}_{\eta}^{S}\vee\mathscr{G}_{\eta^{\prime}}^{S^{\prime}},$ with
$\mathscr{G}_{\eta}^{S}=\sigma(S_{\eta+j}-S_{\eta};j\in\mathds{Z}_{+}),$ and a
corresponding definition for $\mathscr{G}_{\eta^{\prime}}^{S^{\prime}}$ with
$(S^{\prime},\eta^{\prime})$ replacing $(S,\eta).$ In particular, if
$\EuScript{A}$ is an event depending on the paths $(S_{j})_{0\leq j\leq\eta}$
and $(S_{j^{\prime}}^{\prime})_{0\leq j^{\prime}\leq\eta^{\prime}}$ and
$\EuScript{B},$ depending on the paths