$\varepsilon>0$ such that $(\rho-\delta,\rho+\delta)\subset(k-1,k)$, and such
that

$\displaystyle\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{n^{a}\leq W_{x}\leq\varepsilon
n\\}}>\delta n\Big{)}=o(n^{-k(\beta-2)}).$ (5.27)

By Lemma 5.6 such choices are possible. A lower bound then easily follows
using Lemma 5.8, by observing

$\displaystyle\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq n^{a}\\}}>\rho n\Big{)}$
$\displaystyle\geq\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq\varepsilon n\\}}>\rho
n\Big{)}$ (5.28) $\displaystyle=(F_{\varepsilon}(\rho)+o(1))n^{-k(\beta-2)},$

since $n^{a}=o(n)$. Thus,

$\displaystyle n^{k(\beta-2)}\,{\mathbb{P}}\left(\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq n^{a}\\}}>\rho
n\right)\geq F_{\varepsilon}(\rho)+o(1).$

The desired lower bound in Proposition 5.3 follows by letting
$\varepsilon\searrow 0$ and using monotone convergence. To establish an upper
bound, we write

$\displaystyle\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq n^{a}\\}}>\rho n\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{n^{a}\leq W_{x}\leq\varepsilon
n\\}}>\delta n\Big{)}+\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq\varepsilon
n\\}}>(\rho-\delta)n\Big{)}$ (5.29)
$\displaystyle=o(n^{-k(\beta-2)})+(F_{\varepsilon}(\rho-\delta)+o(1))n^{-k(\beta-2)}\leq(F(\rho-\delta)+o(1))n^{-k(\beta-2)},$

where the equality follows from (5.27) and Lemma 5.8 and the last inequality
holds since $F_{\varepsilon}(\rho-\delta)\leq F(\rho-\delta)$. Therefore,

$\displaystyle n^{k(\beta-2)}\mathbb{P}\Big{(}\sum_{x\in
V_{n}}{\lambda}_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq n^{a}\\}}>\rho n\Big{)}$
$\displaystyle\leq F(\rho-\delta)+o(1).$

The desired upper bound follows by first letting $n\to\infty$, and then
letting $\delta\searrow 0$ and using the continuity of $F$ on the interval
$(k-1,k)$.∎

##  6\. Bulk in edge-length distribution: Proof of Proposition 2.5

###### Proof of Proposition 2.5.

We need to show that

$\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{\\{x,y\\}\in
E_{n}}f(d_{n}(x,y))-\frac{1}{2}{\mathbb{E}}\left[\lambda_{f}(W)\right]\Big{|}>\delta,|E_{n}|>n(\mu+\rho)\bigg{)}=o(n^{-k(\beta-2)}).$

Let $R>0$ be such that $\text{supp}(f)\subseteq(0,R)$. Then

$\displaystyle\frac{1}{n}\sum_{\\{x,y\\}\in E_{n}}f(d_{n}(x,y))$
$\displaystyle=\frac{1}{n}\sum_{\\{x,y\\}\in
E_{n}}f(d_{n}(x,y)){\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}\vee W_{y}\leq n^{a}\\}}$ $\displaystyle\hskip
10.0pt+\frac{1}{n}\sum_{\\{x,y\\}\in
E_{n}}f(d_{n}(x,y)){\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}\vee W_{y}>n^{a}\\}}.$

Since $R<\varepsilon_{n}n^{1/d}$ for $n$ sufficiently large, Proposition 3.1
implies that

$\displaystyle\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{\\{x,y\\}\in
E_{n}}f(d_{n}(x,y)){\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}\vee W_{y}\leq
n^{a}\\}}-\frac{1}{2}{\mathbb{E}}\left[\lambda_{f}(W)\right]\Big{|}>\delta\bigg{)}=o(n^{-k(\beta-2)}).$

Therefore it suffices to show that

$\mathbb{P}\bigg{(}\frac{1}{n}\sum_{\\{x,y\\}\in
E_{n}}f(d_{n}(x,y)){\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}\vee
W_{y}>n^{a}\\}}>\delta\bigg{)}=o(n^{-k(\beta-2)}).$

We bound $|f(d_{n}(x,y))|\leq{\sf F}$, and

$\frac{1}{n}\sum_{\\{x,y\\}\in E_{n}}{\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}\vee W_{y}>n^{a}\\}}\leq\frac{2}{n}\sum_{x,y\in
V_{n}}{\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}>n^{a}\\}},$

so that we are left to show that

$\mathbb{P}\bigg{(}\sum_{x,y\in V_{n}}{\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}>n^{a}\\}}>\frac{\delta}{2{\sf
F}}n\bigg{)}=o(n^{-k(\beta-2)}).$

We first condition on the number of nodes $N$ and then consider the event
$\mathcal{E}_{\varepsilon_{n},M}$, which indicates that the number of points
in each each cube of length $\varepsilon_{n}n^{1/d}$ is inside the interval
$[\varepsilon_{n}^{d}n-M\sqrt{\varepsilon_{n}^{d}n\log(\varepsilon_{n}^{d}n)},\varepsilon_{n}^{d}n+M\sqrt{\varepsilon_{n}^{d}n\log(\varepsilon_{n}^{d}n)}]$.
Recall that for $M$ sufficiently large,
${\mathbb{P}}\left(\mathcal{E}_{\varepsilon_{n},M}\right)=1-o(n^{-k(\beta-2)})$
(see (3.7)). Hence, it is enough to study the sum under the event
$\mathcal{E}_{\varepsilon_{n},M}$. Recall that on this event,

$\sum_{j=1}^{N}{\mathbbm{1}}_{\\{d_{n}(X_{i},X_{j})<R\\}}<2^{d+1}R^{d}n^{-1}N\;\text{
for each $1\leq i\leq N$, }$

see (3.23). Hence, we can bound

$\sum_{i,j=1}^{N}{\mathbbm{1}}_{\\{d_{n}(X_{i},X_{j})\leq
R\\}}{\mathbbm{1}}_{\\{W_{i}>n^{a}\\}}\leq
2^{d+1}R^{d}n^{-1}N\sum_{i=1}^{N}{\mathbbm{1}}_{\\{W_{i}>n^{a}\\}}.$

We now consider

$\mathbb{P}_{N}\Big{(}\sum_{i=1}^{N}{\mathbbm{1}}_{\\{W_{i}>n^{a}\\}}>\frac{\delta}{2^{2+d}R^{d}{\sf
F}}n^{2}N^{-1}\Big{)}.$

Conditioned on $N$, the sum is just a
$\mathrm{Bin}(N,{\mathbb{P}}\left(W>n^{a}\right))$ random variable. Now we
condition on the event that $N\in I_{M}(n)$, which happens with probability
$1-o(n^{-k(\beta-2)})$ for $M$ large enough by (3.3). Then, on this event,
$\frac{\delta}{2^{2+d}R^{d}{\sf
F}}n^{2}N^{-1}>N{\mathbb{P}}\left(W>n^{a}\right)$ for $n$ sufficiently large.
Hence, an application of Hoeffding’s inequality yields that (writing
$Q:=\frac{\delta}{2^{2+d}R^{d}{\sf F}}$)

$\displaystyle\mathbb{P}_{N}\Big{(}\sum_{i=1}^{N}{\mathbbm{1}}_{\\{W_{i}>n^{a}\\}}>\frac{\delta}{2^{2+d}R^{d}{\sf
F}}n^{2}N^{-1}\Big{)}$
$\displaystyle={\mathbb{P}}_{N}\left(\mathrm{Bin}(N,{\mathbb{P}}\left(W>n^{a}\right))>Qn^{2}N^{-1}\right)$
$\displaystyle\leq 2\exp\left(-2Q^{2}n^{4}N^{-3}\right).$

Since the last term is bounded by $2\exp\left(-2^{-2}Q^{2}n\right)$ on the
event $N\in I_{M}(n)$, we conclude that, for $M$ sufficiently large,

$\displaystyle\mathbb{P}\Big{(}\sum_{x,y\in
V_{n}}{\mathbbm{1}}_{\\{\\{x,y\\}\in E_{n}\\}}{\mathbbm{1}}_{\\{d_{n}(x,y)\leq
R\\}}{\mathbbm{1}}_{\\{W_{x}n^{a}\\}}>\frac{\delta}{2{\sf F}}n\Big{)}$
$\displaystyle\leq{\mathbb{E}}\left[{\mathbbm{1}}_{\\{N\in
I_{M}(n)\\}}{\mathbb{P}}_{N}\left(\mathrm{Bin}(N,{\mathbb{P}}\left(W>n^{a}\right))>Qn^{2}N^{-1}\right)\right]+o(n^{-k(\beta-2)})$
$\displaystyle\leq
2\exp\left(-2^{-2}Q^{2}n\right)+o(n^{-k(\beta-2)})=o(n^{-k(\beta-2)}),$

which is what we needed to show. ∎

##  7\. Travelling wave in edge-length distribution: Proof of Proposition 2.6

Our proof strategy is to show that the edges of macroscopic length are those
in $E_{\rm high}$ and the form of the condensate emerges from the fact that
$E_{\rm high}$ originates from $k$ uniform vertices that all have weight of
order $n$. The total mass, that is at least $\rho n$, of these corresponds to
edges with length of order $n^{1/d}$. For the proof, we rely on the proofs of
the previous sections.

To make the intuition outlined above precise, define the function
$g:(0,1)^{2}\rightarrow[0,1)$ by

$\displaystyle
g(x,y):=\text{Vol}\left([-\tfrac{1}{2},\tfrac{1}{2}]^{d}\cap\left(\mathcal{B}(0,y)\setminus\mathcal{B}(0,x)\right)\right).$
(7.1)

Fix $0\leq b_{1}<b_{2}\leq\sqrt{d}/2$. Define the vertex set
${\mathcal{U}}_{n}(x)={\mathcal{U}}_{n}^{(b_{1},b_{2})}(x)$ as the
intersection of $V_{n}$ and the annulus
$\mathcal{B}(x,b_{2}n^{1/d})\setminus\mathcal{B}(x,b_{1}n^{1/d})$ centred at
$x\in V_{n}$. Write ${\mathcal{U}}_{n}$ for
${\mathcal{U}}_{n}^{(b_{1},b_{2})}({0})$. Remark that for all $x\in V_{n}$,
for the lattice case, we have that
$|{\mathcal{U}}_{n}(x)|=(1+o(1))g(b_{1},b_{2})n$, while for the Poisson case,
$|{\mathcal{U}}_{n}(x)|$ is Poisson distributed with mean $g(b_{1},b_{2})n$.
Recall that $(W_{\scriptscriptstyle(i)})_{i\in[N]}$ is the order statistic of
the weights $(W_{x})_{x\in V_{n}}$. For $i\in[N]$, recall that
$X_{\scriptscriptstyle(i)}\in V_{n}$ denotes the random vertex with the
$i^{\text{th}}$ largest weight, i.e.,
$W_{X_{\scriptscriptstyle(i)}}=W_{\scriptscriptstyle(i)}$. Define
$\mathbf{X}_{\scriptscriptstyle(i)}:=\\{X_{\scriptscriptstyle(1)},\ldots,X_{\scriptscriptstyle(i)}\\}$,
to be the set of locations of the first $i$ largest weight vertices. Remark
that to prove Proposition 2.6, it suffices to prove the statement for
$f={\mathbbm{1}}_{[b_{1},b_{2})}$, for any $b_{1},b_{2}$ satisfying $0\leq
b_{1}<b_{2}\leq\sqrt{d}/2$. For $w\in(0,\infty)$, define
$\Lambda_{[b_{1},b_{2})}(w)$ as

$\displaystyle\Lambda_{[b_{1},b_{2})}(w):=\int_{[-\frac{1}{2},\frac{1}{2}]^{d}\cap\left(\mathcal{B}(0,b_{2})\setminus\mathcal{B}(0,b_{1})\right)}{{\Lambda}}(w,z)\,dz,$
(7.2)

where
${{\Lambda}}(w,z)={\mathbb{E}}\left[\varphi\left(\frac{\|z\|^{d}}{\mathcal{K}(w,W)}\right)\right]$,
defined in (1.9).

Recall that $\mathbb{P}^{x}$ denotes the probability $\mathbb{P}$ conditioned
on having a point at $x\in\mathbb{T}_{n}^{d}$.

######  Lemma 7.1.

Let $w\in(0,\infty)$ and $x\in\mathbb{T}_{n}^{d}$. For both lattice and
Poisson models,

$\displaystyle\frac{1}{n}\sum_{z\in{\mathcal{U}}_{n}(x)}A_{x,z}\to\Lambda_{[b_{1},b_{2})}(w),$
(7.3)

where the convergence is in probability with respect to $\mathbb{P}^{\,x}$
conditionally on $W_{x}=wn$.

###### Proof of Lemma 7.1.

Fix $w\in(0,\infty)$ throughout the proof. We begin by proving the lemma for
the lattice case. Representing the neighbourhood ${\mathcal{U}}_{n}(x)$ of $x$
by $\mathbb{Z}^{d}\cap\mathbb{Q}_{n}$ with $x$ mapped to the origin, we get
${\mathcal{U}}_{n}=\mathbb{Z}^{d}\cap\mathbb{Q}_{n}\cap\mathcal{B}({0},b_{2}n^{1/d})\setminus\mathcal{B}({0},b_{1}n^{1/d})$.
Conditionally on $W_{{0}}=wn$, we note that
$(A_{0,y})_{y\in{\mathcal{U}}_{n}}$ are independent
Bernoulli-${\mathbb{E}}\left[p_{{0},y}(wn,W_{y})\right]$ distributed random
variables. Using standard concentration arguments (see e.g. [17, Theorem
2.8.]), it suffices to show that
$\frac{1}{n}\mathbb{E}^{0}[\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}\mid{W_{0}=wn}]\to\Lambda_{[b_{1},b_{2})}(w)$
as $n\to\infty$. We write

$\displaystyle\frac{1}{n}\mathbb{E}^{\,{0}}\left[\left.\sum_{y\in
U_{n}}A_{{0},y}\;\right|\;W_{{0}}=wn\right]$
$\displaystyle=\frac{|{\mathcal{U}}_{n}|}{n}\int_{1}^{\infty}f_{W}(t)\frac{1}{|{\mathcal{U}}_{n}|}\sum_{y\in{\mathcal{U}}_{n}}\varphi\left(\frac{\|y\|^{d}}{\kappa(wn,t)}\right)dt.$
(7.4)

By a slight modification of Lemma 5.7, working with ${\mathcal{U}}_{n}$
instead of $V_{n}$, we get that

$\displaystyle\lim_{n\to\infty}\frac{1}{|{\mathcal{U}}_{n}|}\sum_{y\in{\mathcal{U}}_{n}}\varphi\left(\frac{\|y\|^{d}}{\kappa(wn,t)}\right)=\frac{1}{g(b_{1},b_{2})}\int_{[-\frac{1}{2},\frac{1}{2}]^{d}\cap\left(\mathcal{B}(0,b_{2})\setminus\mathcal{B}(0,b_{1})\right)}\varphi\left(\tfrac{\|y\|^{d}}{\mathcal{K}(w,t)}\right)dy,$

since $|{\mathcal{U}}_{n}|=(1+o(1))g(b_{1},b_{2})n$. We can further conclude
by the dominated convergence theorem that

$\displaystyle\frac{1}{n}\mathbb{E}^{\,{0}}\left[\left.\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}\;\right|\;W_{{0}}=wn\right]\to\int_{1}^{\infty}\int_{[-\frac{1}{2},\frac{1}{2}]^{d}\cap\left(\mathcal{B}(0,b_{2})\setminus\mathcal{B}(0,b_{1})\right)}f_{W}(t)\,\varphi\left(\tfrac{\|z\|^{d}}{\mathcal{K}(w,t)}\right)dzdt.$

The proof for the Poisson case is of a similar flavour. First note that given
${0}\in V_{n}$ and the weight $W_{{0}}=wn$, the locations and the weights of
the vertices contributing to $\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}$ are
distributed as a Poisson point process on $\mathbb{R}^{d}\times\mathbb{R}_{+}$
with intensity

$\displaystyle
h(y,t)=f_{W}(t)\,\varphi\left(\frac{\|y\|^{d}}{\kappa(wn,t)}\right){\mathbbm{1}}_{\\{t\geq
1\\}}{\mathbbm{1}}_{\\{y\in\mathbb{T}_{n}^{d}\cap(\mathcal{B}(0,b_{2}n^{1/d})\setminus\mathcal{B}(0,b_{1}n^{1/d}))\\}}.$
(7.5)

This implies that given ${0}\in V_{n}$ and the weight $W_{{0}}=wn$, the sum
$\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}$ is a Poisson random variable with
parameter

$\int_{1}^{\infty}\int_{\mathbb{T}_{n}^{d}\cap(\mathcal{B}(0,b_{2}n^{1/d})\setminus\mathcal{B}(0,b_{1}n^{1/d}))}h(y,t)dydt.$

This is an analogue of Lemma 5.4 and we omit the proof as it is a simple
adaptation of the proof of [15, Lemma 2.1]. Using Poisson concentration
arguments [17, Remark 2.6], it suffices to prove that the mean of the Poisson
random variable (7.5) converges to the right-hand side of (7.3). Just as in
the lattice case, this follows from a modified version of Lemma 5.7, using
that ${\mathbb{E}}\left[|{\mathcal{U}}_{n}|\right]=g(b_{1},b_{2})n$. We omit
the details. ∎

######  Lemma 7.2.

For lattice and Poisson models and $w_{1}\geq\dots\geq w_{k}>0$, conditionally
on
$(W_{\scriptscriptstyle(1)},\dots,W_{\scriptscriptstyle(k)})=(w_{1}n,\dots,w_{k}n)$,

$\displaystyle\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}(w_{i}),$

where $\Lambda_{[b_{1},b_{2})}(w)$ is given by (7.2).

###### Proof.

The following proof holds for both lattice and Poisson models. Let
$\varepsilon>0$. A union bound gives that

$\displaystyle\mathbb{P}\Bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in\,{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon\;\Bigg{|}\;(W_{(j)})_{j\in[k]}=(w_{j}n)_{j\in[k]}\Bigg{)}$
$\displaystyle\leq\sum_{i=1}^{k}\mathbb{P}\Bigg{(}\Big{|}\frac{1}{n}\sum_{y\in\,{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon/k\;\Bigg{|}\;(W_{(j)})_{j\in[k]}=(w_{j}n)_{j\in[k]}\Bigg{)}.$
(7.6)

We will show that, for any $i\in[k]$, the summand of the right-hand side above
is $o(1)$. To be able to apply Lemma 7.1 requires still some work since the
summand above conditions on the $k$ highest weights instead of on
$W_{X_{\scriptscriptstyle(i)}}$. Fix $i\in[k]$ and recall
$\mathbf{X}_{\scriptscriptstyle(i)}:=\\{X_{\scriptscriptstyle(1)},\ldots,X_{\scriptscriptstyle(i)}\\}$.
Using the triangle inequality we can upper bound the summand by

$\displaystyle\mathbb{P}\Big{(}\Big{|}\frac{1}{n}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\end{subarray}\setminus\mathbf{X}_{\scriptscriptstyle(k)}}A_{X_{\scriptscriptstyle(i)},y}-\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon/2k\;\Big{|}\;(W_{\scriptscriptstyle(j)})_{j\in[k]}=(w_{j}n)_{j\in[k]}\Big{)}$
(7.7)
$\displaystyle+\mathbb{P}\Big{(}\Big{|}\frac{1}{n}\sum_{j=1}^{k}A_{X_{\scriptscriptstyle(i)},X_{\scriptscriptstyle(j)}}\Big{|}>\varepsilon/2k\;\Big{|}\;(W_{\scriptscriptstyle(j)})_{j\in[k]}=(w_{j}n)_{j\in[k]}\Big{)}.$
(7.8)

Note that the second summand equals zero for sufficiently large $n$ since we
have the deterministic bound
$\frac{1}{n}\sum_{j=1}^{k}A_{X_{\scriptscriptstyle(i)},X_{\scriptscriptstyle(j)}}\leq
k/n<\varepsilon/2k$. Therefore it remains to show that the first summand (7.7)
tends to zero as $n$ tends to infinity. The fact that the weights
$(W_{x})_{x}$ are independent and that the random variable $A_{x,y}$ only
depends on the locations $x,y$ and weights $W_{x},W_{y}$, gives that, for
$\ell\geq 0$,

$\displaystyle\mathbb{P}\Big{(}\sum_{{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\setminus\mathbf{X}_{\scriptscriptstyle(k)}}}A_{X_{\scriptscriptstyle(i)},y}\geq\ell\;\Big{|}\;(W_{(j)})_{j\in[k]}=(w_{j}n)_{j\in[k]}\Big{)}$
(7.9)
$\displaystyle=\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\setminus\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{X_{\scriptscriptstyle(i)},y}\geq\ell\;\Big{|}\;W_{X_{\scriptscriptstyle(i)}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall
y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\setminus\mathbf{X}_{\scriptscriptstyle(k)}\Big{)}$
$\displaystyle=\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\setminus\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{X_{\scriptscriptstyle(i)},y}\geq\ell\;\Big{|}\;W_{X_{\scriptscriptstyle(i)}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\end{subarray}}A_{X_{\scriptscriptstyle(i)},y}\geq\ell\;\Big{|}\;W_{X_{\scriptscriptstyle(i)}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\Big{)}.$
(7.10)

Similarly, we can lower bound (7.9) by

$\displaystyle\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\end{subarray}}A_{X_{\scriptscriptstyle(i)},y}-k\geq\ell\;\Big{|}\;W_{X_{\scriptscriptstyle(i)}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\Big{)}.$
(7.11)

Thus, for sufficiently large $n$, (7.7) can be bounded from above by

$\displaystyle\mathbb{P}\Big{(}\Big{|}\frac{1}{n}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon^{\prime}\;\Big{|}\;W_{X_{\scriptscriptstyle(i)}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\Big{)},$

for some $\varepsilon^{\prime}>0$ such that
$\varepsilon^{\prime}<\varepsilon/2k-k/n$. Observing that the location of the
vertex $X_{\scriptscriptstyle(i)}$ is just a uniformly chosen (lattice) point
in $\mathbb{T}_{n}^{d}$, the above equals

$\displaystyle\mathbb{P}^{\,{0}}\Big{(}\Big{|}\frac{1}{n}\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}-\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon^{\prime}\;\Big{|}\;W_{{0}}=w_{i}n,\,W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}\Big{)}$
$\displaystyle\leq\frac{\mathbb{P}^{\,{0}}\Big{(}\Big{|}\frac{1}{n}\sum_{y\in{\mathcal{U}}_{n}}A_{{0},y}-\Lambda_{[b_{1},b_{2})}(w_{i})\Big{|}>\varepsilon^{\prime}\;\Big{|}\;W_{{0}}=w_{i}n\Big{)}}{\mathbb{P}^{\,{0}}\Big{(}W_{y}\leq
w_{k}n\,\forall y\in{\mathcal{U}}_{n}\;\Big{|}\;W_{{0}}=w_{i}n\Big{)}}.$
(7.12)

Note that

$\displaystyle\mathbb{P}^{\,{0}}\Big{(}W_{y}\leq w_{k}n\,\forall
y\in{\mathcal{U}}_{n}\;\Big{|}\;W_{{0}}=w_{i}n\Big{)}$
$\displaystyle={\mathbb{P}}^{\,{0}}\Big{(}W_{y}\leq w_{k}n\,\forall
y\in{\mathcal{U}}_{n}\Big{)}={\mathbb{E}}^{\,{0}}\left[{\mathbb{P}}\left(W\leq
w_{k}n\right)^{|{\mathcal{U}}_{n}|}\right],$

and recall that $|{\mathcal{U}}_{n}|=(1+o(1))g(b_{1},b_{2})n$ for lattice
models and $|{\mathcal{U}}_{n}|$ is ${\rm Poi}(g(b_{1},b_{2})n)$-distributed
for Poisson models. Using the distribution of $W$ given by (1.1), we have
$\mathbb{E}^{0}[{\mathbb{P}}\left(W\leq
w_{k}n\right)^{|{\mathcal{U}}_{n}|}]\rightarrow 0$ for $n\rightarrow\infty$.
We can conclude that (7.7) tends to zero by applying Lemma 7.1 to the first
factor of (7.12). ∎

######  Corollary 7.3.

Conditionally on the event $|E_{n}|>n(\rho+\mu)$,

$\displaystyle\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in\,{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\left(W_{\scriptscriptstyle(i)}/n\right)\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}0,$

where $\Lambda_{[b_{1},b_{2})}(w)$ is given by (7.2) for $w\in(0,\infty)$.

###### Proof.

By Theorem 1.1, it suffices to show that, for all $\varepsilon>0$

$\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\left(W_{\scriptscriptstyle(i)}/n\right)\Big{|}>\varepsilon,\,|E_{n}|>n(\rho+\mu)\bigg{)}=o(n^{-k(\beta-2)}).$

Let $b>0$ and recall $\mathcal{A}_{k}(b)$ from (5.25). We begin by noting that

$\displaystyle\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\left(W_{\scriptscriptstyle(i)}/n\right)\Big{|}>\varepsilon,\,|E_{n}|>n(\rho+\mu),\mathcal{A}_{k}(b)\bigg{)}$
$\displaystyle\leq\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\left(W_{\scriptscriptstyle(i)}/n\right)\Big{|}>\varepsilon,\,\mathcal{A}_{k}(b)\bigg{)}$
$\displaystyle\leq\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\left(W_{\scriptscriptstyle(i)}/n\right)\Big{|}>\varepsilon,\,W_{\scriptscriptstyle(i)}>bn\,\forall
i\in[k]\bigg{)}.$

By Lemma 7.2, the last line is
$o({\mathbb{P}}\left(W_{\scriptscriptstyle(k)}>bn\right))$. Using the density
of $W$ and the fact that $(W_{x})_{x\in V_{n}}$ are i.i.d., we get that
${\mathbb{P}}\left(W_{\scriptscriptstyle(k)}>bn\right)=O(n^{-k(\beta-2)})$.
Next, remark that

$\displaystyle\mathbb{P}\bigg{(}\Big{|}\frac{1}{n}\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\Lambda_{[b_{1},b_{2})}\Big{(}W_{\scriptscriptstyle(i)}/n\Big{)}\Big{|}>\varepsilon,\,|E_{n}|>n(\rho+\mu),\mathcal{A}_{k}(b)^{c}\bigg{)}$
$\displaystyle\leq{\mathbb{P}}\left(|E_{n}|>n(\rho+\mu),\mathcal{A}_{k}(b)^{c}\right).$

The last line equals $o(n^{-k(\beta-2)})$ by the bound (2.5) combined with
Propositions 2.1, 2.2 and inequalities (5.11) and (5.26). ∎

Finally, we have all the tools to prove Proposition 2.6.

###### Proof of Proposition 2.6.

It is enough to prove the statement for $f={\mathbbm{1}}_{(b_{1},b_{2})}$,
where $0<b_{1}<b_{2}<\sqrt{d}/2$. For this choice of $f$, it holds that

$\displaystyle\int
f\Big{(}\frac{x}{n^{1/d}}\Big{)}\,d\mu_{n}=\frac{1}{n}\sum_{{}_{\\{x,y\\}\in
E_{n}}}{\mathbbm{1}}_{\\{d_{n}(x,y)/n^{1/d}\in(b_{1},b_{2})\\}}.$

By Corollary 7.3 it therefore suffices to prove that, conditionally on
$|E_{n}|\geq n(\rho+\mu)$,

$\displaystyle\frac{1}{n}\bigg{|}\sum_{{}_{\\{x,y\\}\in
E_{n}}}{\mathbbm{1}}_{\\{d_{n}(x,y)/n^{1/d}\in(b_{1},b_{2})\\}}-\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}\bigg{|}\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}0.$
(7.13)

Remark that

$\displaystyle\sum_{{}_{\\{x,y\\}\in
E_{n}}}{\mathbbm{1}}_{\\{d_{n}(x,y)/n^{1/d}\in(b_{1},b_{2})\\}}$
$\displaystyle=\frac{1}{2}\sum_{x\in
V_{n}}\sum_{y\in{\mathcal{U}}_{n}(x)}A_{x,y}.$

We can rewrite the double sum as

$\displaystyle\sum_{x\in
V_{n}}\sum_{y\in{\mathcal{U}}_{n}(x)}A_{x,y}=\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}+\sum_{{x\in
V_{n}}\atop{x\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}\sum_{y\in{\mathcal{U}}_{n}(x)}A_{x,y}\phantom{thisistobetteralign}$

$\displaystyle\phantom{X}=\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}+\sum_{i=1}^{k}\sum_{{x\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}\atop{x\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}A_{X_{\scriptscriptstyle(i)},x}+\sum_{{x\in
V_{n}}\atop{x\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}\sum_{{y\in{\mathcal{U}}_{n}(x)}\atop{y\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}A_{x,y}$
(7.14)
$\displaystyle\phantom{X}=2\sum_{i=1}^{k}\sum_{y\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})}A_{X_{\scriptscriptstyle(i)},y}-\sum_{i=1}^{k}\sum_{j=1}^{k}A_{X_{\scriptscriptstyle(i)},X_{(j)}}{\mathbbm{1}}_{\\{X_{(j)}\in{\mathcal{U}}_{n}({X_{\scriptscriptstyle(i)}})\\}}+\sum_{{x\in
V_{n}}\atop{x\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}\sum_{{y\in{\mathcal{U}}_{n}(x)}\atop{y\not\in\mathbf{X}_{\scriptscriptstyle(k)}}}A_{x,y}.$

Further, bounding $A_{x,y}$ from above by one, we get that
$\frac{1}{n}\sum_{i=1}^{k}\sum_{j=1}^{k}A_{X_{\scriptscriptstyle(i)},X_{\scriptscriptstyle(j)}}\leq
k^{2}/n$, which vanishes as $n$ increases. Putting everything together, in
order to prove (7.13) we must show that, conditionally on $|E_{n}|\geq
n(\rho+\mu)$,

$\displaystyle\frac{1}{n}\sum_{\begin{subarray}{c}x\in
V_{n}\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}(x)\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{x,y}\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}0.$
(7.15)

First, remark that $\sum_{\begin{subarray}{c}x\in
V_{n}\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}(x)\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{x,y}{\mathbbm{1}}_{\\{W_{x}\vee
W_{y}\leq n^{a}\\}}\leq|E_{\rm long}|$, since $b_{1}>\varepsilon_{n}$ for
sufficiently large $n$. Thus, Proposition 2.2 and Theorem 1.1 give that, for
all $\delta>0$,

$\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}x\in
V_{n}\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}(x)\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{x,y}{\mathbbm{1}}_{\\{W_{x}\vee
W_{y}\leq n^{a}\\}}>\delta n\;\Big{|}\;|E_{n}|\geq n(\rho+\mu)\Big{)}=o(1).$

Further, by a slightly modified version of the upper bound (5.11) together
with Lemma 5.6,

$\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}x\in
V_{n}\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}(x)\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{x,y}{\mathbbm{1}}_{\\{n^{a}\leq
W_{x}\leq\varepsilon n\\}}>\delta n\;\Big{|}\;|E_{n}|\geq
n(\rho+\mu)\Big{)}=o(1).$

Lastly,

$\displaystyle\mathbb{P}\Big{(}\sum_{\begin{subarray}{c}x\in
V_{n}\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}\sum_{\begin{subarray}{c}y\in{\mathcal{U}}_{n}(x)\setminus\,\mathbf{X}_{\scriptscriptstyle(k)}\end{subarray}}A_{x,y}{\mathbbm{1}}_{\\{W_{x}\geq\varepsilon
n\\}}>\delta n\;\Big{|}\;|E_{n}|>n(\rho+\mu)\Big{)}$
$\displaystyle\leq\mathbb{P}\left(\left.W_{\scriptscriptstyle(k+1)}\geq\varepsilon
n\;\right|\;|E_{n}|\geq n(\rho+\mu)\right)=o(1),$

where the equality follows from the proof of Lemma 5.8. By (7), combining the
above three bounds concludes the proof of (7.15). ∎

##  8\. Limit for the weight order statistics: Proof of Proposition 2.7

###### Proof of Proposition 2.7.

Fix positive reals $a_{1}\geq\dots\geq a_{k}>0$. Similar to the definition
(5.25), define the event

$\displaystyle\mathcal{H}_{k}=\mathcal{H}_{k}(a_{1},\ldots,a_{k}):=\\{W_{\scriptscriptstyle(i)}>a_{i}n\;\;\text{for
all}\;\;1\leq i\leq k\\}.$ (8.1)

To prove Proposition 2.7 for lattice and Poisson models, we will show that

$\displaystyle{\mathbb{P}}\left(\mathcal{H}_{k}\cap\\{|E_{n}|\geq(\mu+\rho)n\\}\right)$
$\displaystyle=(1+o(1))n^{-k(\beta-2)}(\beta-1)^{k}\int_{a_{1}}^{\infty}\dots\int_{a_{k}}^{\infty}{\mathbbm{1}}_{\\{y_{1}>\dots>y_{k}\\}}{\mathbbm{1}}_{\\{\sum_{i=1}^{k}{\Lambda}(y_{i})>\rho\\}}\prod_{i=1}^{k}y_{i}^{-\beta}dy_{i}.$
(8.2)

By the proof of Theorem 1.1, it suffices to show that
${\mathbb{P}}\left(\mathcal{H}_{k}\cap\\{|E_{\rm high}|>n\rho\\}\right)$
equals the right-hand side of (8.2) in order to conclude (8.2). Following the
proof of (5.11) and using Lemma 5.1, we can deduce that, for small $\delta>0$
and sufficiently large $n$,

$\displaystyle\mathbb{P}(\mathcal{H}_{k}\,\cap\,$ $\displaystyle\\{|E_{\rm
high}|>\rho n\\})$
$\displaystyle\leq{\mathbb{P}}\left(\mathcal{H}_{k}\cap\left\\{\sum_{x\in
V_{n}}\lambda_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq
n^{a}\\}}>(\rho/(1+\delta))n\right\\}\right)+o(n^{-k(\beta-2)}),$

and

$\displaystyle\mathbb{P}(\mathcal{H}_{k}\,\cap\,$ $\displaystyle\\{|E_{\rm
high}|>\rho n\\})$
$\displaystyle\geq{\mathbb{P}}\left(\mathcal{H}_{k}\cap\left\\{\sum_{x\in
V_{n}}\lambda_{n}(W_{x}){\mathbbm{1}}_{\\{W_{x}\geq
n^{a}\\}}>(\rho/(1-\delta))n\right\\}\right)+o(n^{-k(\beta-2)}).$

Combining (5.28) and (5.29) together with slight variants of Lemma 5.6 and the
proof of Lemma 5.8, we can see that (8.2) follows if we can prove that
$\mathbb{P}(\mathcal{H}_{k}\cap\\{\sum_{i=1}^{k}\lambda_{n}(W_{\scriptscriptstyle(i)})>\rho
n\\})$ equals the right-hand side of (8.2). We begin to prove this for lattice
models. Note that

$\displaystyle{\mathbb{P}}\left(\mathcal{H}_{k}\cap\Big{\\{}\sum_{i=1}^{k}\lambda_{n}(W_{\scriptscriptstyle(i)})>\rho
n\Big{\\}}\right)$
$\displaystyle=(N)_{k}(\beta-1)^{k}\int_{a_{1}n}^{\infty}\dots\int_{a_{k}n}^{\infty}{\mathbbm{1}}_{\\{\sum_{i=1}^{k}\lambda_{n}(x_{i})>\rho
n\\}}{\mathbbm{1}}_{\\{x_{1}>\dots>x_{k}\\}}\prod_{i=1}^{k}x_{i}^{-\beta}dx_{i},$

where $(N)_{k}:=N(N-1)\dots(N-k+1)$ is the falling factorial. By the change of
variables $x_{i}/n=y_{i}$ for all $i\in[k]$, the above equals up to a
$(1+o(1))$ factor,

$\displaystyle
n^{-k(\beta-2)}(\beta-1)^{k}\int_{a_{1}}^{\infty}\dots\int_{a_{k}}^{\infty}{\mathbbm{1}}_{\\{\sum_{i=1}^{k}\frac{\lambda_{n}(ny_{i})}{n}>\rho\\}}{\mathbbm{1}}_{\\{y_{1}>\dots>y_{k}\\}}\prod_{i=1}^{k}y_{i}^{-\beta}dy_{i},$

since $N=n$ in the lattice case. Using Lemma 5.7, we conclude that

$\displaystyle{\mathbb{P}}\left(\mathcal{H}_{k}\cap\Big{\\{}\sum_{i=1}^{k}\lambda_{n}(W_{\scriptscriptstyle(i)})>\rho
n\Big{\\}}\right)$
$\displaystyle=(1+o(1))n^{-k(\beta-2)}(\beta-1)^{k}\int_{a_{1}}^{\infty}\dots\int_{a_{k}}^{\infty}{\mathbbm{1}}_{\\{y_{1}>\dots>y_{k}\\}}{\mathbbm{1}}_{\\{\sum_{i=1}^{k}{\Lambda}(y_{i})>\rho\\}}\prod_{i=1}^{k}y_{i}^{-\beta}dy_{i},$

and so Proposition 2.7 follows. The proof for Poisson models follow by
conditioning on the number of points $N$ and using that $N$ is a
$\text{Poi}(n)$-distributed random variable. More precisely, recall $I_{M}(n)$
given by (3.1) for $M>0$, and note that for a given inverse power of $n$, we
can choose a sufficiently large $M$ such that ${\mathbb{P}}\left({N}\notin
I_{M}(n)\right)$ tends to zero faster than this inverse power as $n\to\infty$.
Thus,

$\displaystyle{\mathbb{P}}\left(\mathcal{H}_{k}\cap\big{\\{}\sum_{i=1}^{k}\lambda_{n}(W_{\scriptscriptstyle(i)})>\rho
n\big{\\}}\right)$ $\displaystyle={\mathbb{E}}\left[{\mathbbm{1}}_{\\{{N}\in
I_{M}(n)\\}}\mathbb{P}\left(\left.\mathcal{H}_{k}\cap\Big{\\{}\sum_{i=1}^{k}\lambda_{n}(W_{\scriptscriptstyle(i)})>\rho
n\Big{\\}}\;\right|\;{N}\right)\right]+o(n^{-k(\beta-2)}).$

The remainder of the proof follows straightforwardly and is omitted. ∎

##  9\. The empirical degree distribution: Proof of Proposition 2.9

We begin the proof of Proposition 2.9(i) by following that of Proposition 2.6
in Section 7. For lattice and Poisson models, recall that Lemma 7.1 holds when
we take the sum over all of $V_{n}$, that is for $b_{1}=0$ and
$b_{2}=\sqrt{d}/2$. This gives, for $w\geq 0$, $x\in\mathbb{T}_{n}^{d}$ and
conditionally on $W_{x}=wn$, that

$\displaystyle\frac{1}{n}\sum_{y\in V_{n}}A_{x,y}\to\Lambda(w),$ (9.1)

where the convergence is in probability with respect to $\mathbb{P}^{x}$. By
adapting the proof of Lemma 7.2, starting from (7.6) and summing again over
all of $V_{n}$, we can show that, conditionally on the event
$(W_{\scriptscriptstyle(1)},\ldots,W_{\scriptscriptstyle(k)})=(y_{1}n,\ldots,y_{k}n)$
for $y_{1}\geq\ldots\geq y_{k}>0$,

$\displaystyle\frac{1}{n}\bigg{(}\sum_{y\in
V_{n}}A_{X_{\scriptscriptstyle(1)},y},\ldots,\sum_{y\in
V_{n}}A_{X_{\scriptscriptstyle(k)},y}\bigg{)}\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}\left(\Lambda(y_{1}),\ldots,\Lambda(y_{k})\right),$
(9.2)

for lattice and Poisson models, proving Proposition 2.9(i).

Next, to show Proposition 2.9(ii), we use a second-moment method. Let
$O_{1},O_{2}$ be two independent uniformly chosen random vertices in $V_{n}$.
For $u_{1},\dots,u_{k}\in\mathbb{T}_{1}^{d}$ we define

$C_{n}:=\\{(W_{\scriptscriptstyle(i)})_{i\in[k]}=(y_{i}n)_{i\in[k]},\,(X_{\scriptscriptstyle(i)})_{i\in[k]}=(n^{1/d}u_{i})_{i\in[k]}\\},$

where, in the lattice case, we project $n^{1/d}u_{i}$ to the nearest lattice
point. Recall (1.20). We can write

$\displaystyle\mathbb{E}[(\pi_{a,b}^{\scriptscriptstyle(n)})^{2}\,|\,C_{n}]=\mathbb{P}\big{(}D_{O_{1}}^{\mathbf{b}}=a,\,D_{O_{1}}^{\mathbf{h}}=b,\,D_{O_{2}}^{\mathbf{b}}=a,\,D_{O_{2}}^{\mathbf{h}}=b\,\big{|}\,C_{n}\big{)}$
$\displaystyle=\mathbb{P}(D_{O_{1}}^{\mathbf{b}}=a,\,D_{O_{1}}^{\mathbf{h}}=b,\,D_{O_{2}}^{\mathbf{b}}=a,\,D_{O_{2}}^{\mathbf{h}}=b;O_{1},O_{2}\notin\\{X_{\scriptscriptstyle(1)},\dots,X_{\scriptscriptstyle(k)}\\}\,|\,C_{n})+o(1).$
(9.3)

We now first focus on the lattice case. For distinct points $o_{1},o_{2}\in
V_{n}$, we define

$f(o_{1},o_{2}):=\mathbb{P}\big{(}D_{o_{1}}^{\mathbf{b}}=a,\,D_{o_{1}}^{\mathbf{h}}=b,\,D_{o_{2}}^{\mathbf{b}}=a,\,D_{o_{2}}^{\mathbf{h}}=b,\,\\{o_{1},o_{2}\\}\notin
E_{n}\,\big{|}\,C_{n}\big{)}.$

Then (9.3) can be written as

$\displaystyle\mathbb{E}\Big{[}f(O_{1},O_{2}){\mathbbm{1}}_{\\{O_{1},O_{2}\notin\\{X_{\scriptscriptstyle(1)},\dots,X_{\scriptscriptstyle(k)}\\}\\}}\Big{|}\,C_{n}\Big{]}+o(1),$

since on the event that $O_{1}$ and $O_{2}$ are not high-weight vertices, with
high probability they do not share an edge. To see this, note that the weights
of the vertices in $V_{n}\setminus\mathbf{X}_{(k)}$ are independent copies of
a variable $\hat{W}_{n}$ having the conditional law of the weights $W$ in
(1.1) conditionally on $\\{W<y_{k}n\\}$, and their locations are distributed
as the locations of $V_{n}$ with $k$ uniformly random vertices removed. In
particular, the probability that $O_{1}$ and $O_{2}$ share an edge on the
event that they are not in $\mathbf{X}_{(k)}$, equals up to an asymptotically
negligible error, the probability that two random vertices share an edge in
the graph $G_{n}$ where each vertex in $V_{n}$ now has an independent copy of
the weight $\hat{W}_{n}$, instead of $W$. The latter probability goes to zero
using Assumption A and a straightforward calculation.

Next, for convenience, let us introduce the notation $D_{x\rightarrow
A}=D_{x\rightarrow A}(G_{n})$ for $x\in V_{n}$ and $A\subseteq V_{n}$ to
denote the number of edges connecting the vertex $x$ to the set $A$ in the
graph $G_{n}$. Recall that ${\mathbf{X}_{\scriptscriptstyle(k)}}$ stands for
the set $\\{X_{\scriptscriptstyle(1)},\dots,X_{\scriptscriptstyle(k)}\\}$.
Then

$\displaystyle f(o_{1},o_{2})=$
$\displaystyle\mathbb{P}\big{(}D_{o_{1}\rightarrow
V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{o_{2}\\})}=a,\,D_{o_{1}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=b,\,D_{o_{2}\rightarrow
V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{o_{1}\\})}=a,$
$\displaystyle\hskip
10.0pt\,D_{o_{2}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=b,\\{o_{1},o_{2}\\}\notin
E_{n}\mid C_{n}\big{)}.$ (9.4)

By conditioning on the weights $W_{o_{1}}$ and $W_{o_{2}}$, and using the
conditional independence of the events,

$\displaystyle f(o_{1},o_{2})$
$\displaystyle=\mathbb{E}\Big{[}\mathbb{P}\left(\left.D_{o_{1}\rightarrow
V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{o_{2}\\})}=a\;\right|\;W_{o_{1}},\,C_{n}\right)\mathbb{P}\left(\left.D_{o_{1}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=b\;\right|\;W_{o_{1}},\,C_{n}\right)$
$\displaystyle\qquad\cdot\mathbb{P}\left(\left.D_{o_{2}\rightarrow
V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{o_{1}\\})}=a\;\right|\;W_{o_{2}},\,C_{n}\right)\mathbb{P}\left(\left.D_{o_{2}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=b\;\right|\;W_{o_{2}},\,C_{n}\right)$
$\displaystyle\qquad\cdot\mathbb{P}\left(\left.\\{o_{1},o_{2}\\}\notin
E_{n}\;\right|\;W_{o_{1}},W_{o_{2}},\,C_{n}\right)\Big{|}\,C_{n}\Big{]}.$
(9.5)

Recall that $D_{\scriptscriptstyle\infty}(w)$ denotes the degree of the root
of the local limit of the unconditional graph given that the weight of the
root is $w$. We claim that, for fixed $w_{1}\geq 1$ and with

$h_{w_{1}}(o_{1},o_{2}):=\mathbb{P}\left(\left.D_{o_{1}\rightarrow
V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{o_{2}\\})}=a\;\right|\;W_{o_{1}}=w_{1},C_{n}\right),$

we have

$\displaystyle
h_{w_{1}}(O_{1},O_{2})\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}{\mathbb{P}}\left(D_{\scriptscriptstyle\infty}(w_{1})=a\right).$
(9.6)

To prove (9.6), as observed earlier, we recall that the weights of vertices in
$V_{n}\setminus\mathbf{X}_{\scriptscriptstyle(k)}$ are independent copies of a
variable $\hat{W}_{n}$ having the conditional law of the weights $W$ in (1.1)
conditionally on $\\{W<y_{k}n\\}$. We now show how to use this observation to
couple the graph $G_{n}$, conditionally on $C_{n}$, with an unconditional
geometric random graph $G_{n}^{\prime}$ where the weights are i.i.d. copies of
$\hat{W}_{n}$. Sample independent copies of $\hat{W}$ for every vertex in
$V_{n}$ and, given the weights, edges between pairs of distinct vertices using
the connection function $\varphi$. The graph $G_{n}^{\prime}$ is formed using
these edges. For the conditional graph $G_{n}$, we modify the weights of the
vertices $X_{\scriptscriptstyle(i)}=n^{1/d}u_{i}$ to be $y_{i}n$, for all
$i\in[k]$, and resample (only) the edges adjacent to
$\mathbf{X}_{\scriptscriptstyle(k)}$ using these weights.

Note that the graph $G_{n}^{\prime}$ has the same local limit as the original
unconditioned graph (with weights of law $W$) since
$\hat{W}\stackrel{{\\!{}_{\mathrm{d}}}}{{\rightarrow}}W$ as $n\to\infty$, and
using the fact that the local limit is preserved under removal of a finite set
of vertices. Therefore the degree of $o_{1}$ in $G_{n}^{\prime}$, given
$W_{o_{1}}=w_{1}$, converges in distribution to
$D_{\scriptscriptstyle\infty}(w_{1})$. We also observe that with high
probability the vertex $O_{1}$ has distance of order $n^{1/d}$ from the
locations of the vertices in
$\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{O_{2}\\}$. Hence, with high
probability, $O_{1}$ does not share an edge with the vertices at
$\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{O_{2}\\}$ and the degree of $O_{1}$
in $G_{n}^{\prime}$ is the same as its degree in the subgraph of
$G_{n}^{\prime}$, induced by the vertices
$V_{n}\setminus(\mathbf{X}_{\scriptscriptstyle(k)}\cup\\{O_{2}\\})$. Since in
our coupling, these induced subgraphs of $G_{n}^{\prime}$ and $G_{n}$ match,
we obtain that in fact
$h_{w_{1}}(O_{1},O_{2})\stackrel{{\\!{}_{\mathbb{P}}}}{{\rightarrow}}{\mathbb{P}}\left(D_{\scriptscriptstyle\infty}(w_{1})=a\right)$.
This proves (9.6).

Define
$g_{w_{1}}(o_{1},o_{2}):=\mathbb{P}(D_{o_{1}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=b\mid
W_{o_{1}}=w_{1},C_{n}).$ We analyse the asymptotic behaviour of
$g_{w_{1}}(O_{1},O_{2})$. In the conditioned graph $G_{n}$, we can write
$D_{o_{1}\rightarrow\mathbf{X}_{\scriptscriptstyle(k)}}=\sum_{i=1}^{k}{\tilde{B}^{\scriptscriptstyle(i)}_{o_{1},w_{1}}},$
where the random variables $\tilde{B}^{\scriptscriptstyle(i)}_{o_{1},w_{1}}$
are independent and Bernoulli distributed with parameter
$\varphi({d_{n}(o_{1},n^{1/d}u_{i})^{d}}/{\kappa(w_{1},ny_{i})})$. Using that
$\varphi$ is continuous almost everywhere, we get

$\displaystyle\varphi\bigg{(}\frac{d_{n}(O_{1},n^{1/d}u_{i})^{d}}{\kappa(w_{1},ny_{(i)})}\bigg{)}\stackrel{{\\!{}_{\mathrm{d}}}}{{\rightarrow}}\varphi\left(\frac{d_{1}(U_{1},u_{i})^{d}}{\mathcal{K}(y_{i},w_{1})}\right),$
(9.7)

where $U_{1}$ is uniformly distributed on $\mathbb{T}_{1}^{d}$. The random
variables $\tilde{B}^{\scriptscriptstyle(i)}_{O_{1},w_{1}}$ therefore converge
in distribution to conditionally independent Bernoulli random variables
$B^{\scriptscriptstyle(i)}_{U_{1},w_{1}}$ with parameter
$\varphi({d_{1}(U_{1},u_{i})^{d}}/{\mathcal{K}(y_{i},w_{1})})$ given $U_{1}$.
Therefore,

$g_{w_{1}}(O_{1},O_{2})\stackrel{{\\!{}_{\mathrm{d}}}}{{\rightarrow}}\mathbb{P}\Big{(}\sum_{i=1}^{k}B^{(i)}_{U_{1},w_{1}}=b~{}\Big{|}~{}U_{1}\Big{)}.$

Note that $(W_{O_{1}},O_{1})$ and $(W_{O_{2}},O_{2})$ converge in distribution
to two independent copies $(W_{1},U_{1})$ and $(W_{2},U_{2})$ of the pair
$(W,U)$ with a Pareto-distributed random variable $W$ and an independent
uniform variable $U\in\mathbb{T}_{1}^{d}$. As $g,h$ are continuous and
bounded,

$\displaystyle\mathbb{E}\Big{[}f(O_{1},O_{2})$
$\displaystyle{\mathbbm{1}}_{\\{O_{1},O_{2}\notin\\{X_{(1)},\dots,X_{(k)}\\}}\\}~{}\Big{|}~{}C_{n}\Big{]}$
$\displaystyle=\mathbb{E}\big{[}h_{W_{O_{1}}}(O_{1},O_{2})g_{W_{O_{1}}}(O_{1},O_{2})h_{W_{O_{2}}}(O_{2},O_{1})g_{W_{O_{2}}}(O_{2},O_{1})\mid
C_{n}\big{]}$

converges to

$\displaystyle\mathbb{E}\bigg{[}\mathbb{P}(D_{\scriptscriptstyle\infty}(W_{1})=a|W_{1})\mathbb{P}(D_{\scriptscriptstyle\infty}(W_{2})=a\mid
W_{2})$
$\displaystyle\phantom{xxxxxx}\cdot\mathbb{P}\Big{(}\sum_{i=1}^{k}B^{(i)}_{U_{1},W_{1}}=b~{}\Big{|}~{}U_{1},W_{1}\Big{)}\mathbb{P}\Big{(}\sum_{i=1}^{k}B^{(i)}_{U_{2},W_{2}}=b~{}\Big{|}~{}U_{2},W_{2}\Big{)}\bigg{]}$
$\displaystyle=\mathbb{E}\bigg{[}\mathbb{P}(D_{\scriptscriptstyle\infty}(W)=a\mid
W)\mathbb{P}\Big{(}\sum_{i=1}^{k}B^{(i)}_{U,W}=b~{}\Big{|}~{}W\Big{)}\bigg{]}^{2}.$

Combining the above with (9.3) and (9.5) gives

$\displaystyle\mathbb{E}[(\pi_{a,b}^{{\scriptscriptstyle(n)}})^{2}\mid
C_{n}]=(1+o(1))(\pi_{a,b}((y_{i},u_{i})_{i\in[k]}))^{2}.$

The above computations also imply that

$\displaystyle\mathbb{E}[\pi_{a,b}^{{}_{(n)}}\mid
C_{n}]=(1+o(1))\pi_{a,b}((y_{i},u_{i})_{i\in[k]}).$

Hence the variance of $\pi_{a,b}^{{}_{(n)}}$ vanishes asymptotically and an
application of Chebyshev’s inequality finishes the proof of Proposition
2.9(ii) for the lattice case. The Poisson case is analogous using that
removing finitely many randomly chosen vertices from the Poisson point process
leaves its law intact.

We next prove Proposition 2.9(iii), that is,
$\mathbb{E}[D_{X_{\scriptscriptstyle(j)}}(G_{n})|C_{n}]=o(n)$ for each $j>k$.
By the coupling of the conditional graph $G_{n}$ and the unconditional graph
$G_{n}^{\prime}$, for $j>k$,

$D_{X_{\scriptscriptstyle(j)}}(G_{n})\leq D_{X_{\scriptscriptstyle(j)}\to
V_{n}\setminus\mathbf{X}_{\scriptscriptstyle(k)}}(G_{n})+k=D_{X_{\scriptscriptstyle(j)}\to
V_{n}\setminus\mathbf{X}_{\scriptscriptstyle(k)}}(G_{n}^{\prime})+k\leq
D_{X_{\scriptscriptstyle(j)}}(G_{n}^{\prime})+k.$

Let $Y$ be the maximum of $n$ i.i.d. random variables with law (1.1). Then,
for $\varepsilon>0$ there is $c>0$ such that
${\mathbb{P}}\left(Y>cn^{1/(\beta-1)}\right)<\varepsilon$. In $G_{n}^{\prime}$
we have $W_{\scriptscriptstyle X_{\scriptscriptstyle(j)}}\preceq_{st}Y$ and
thus

$\displaystyle\frac{1}{n}{\mathbb{E}}\left[D_{X_{\scriptscriptstyle(j)}}(G_{n}^{\prime})\right]$
$\displaystyle\leq{\mathbb{E}}\left[\varphi\left(\frac{d_{n}(n^{1/d}u_{j},U)^{d}}{\kappa(Y,W)}\right)\right]\leq{\mathbb{E}}\left[\varphi\left(\frac{\|U\|^{d}}{\kappa(cn^{1/(\beta-1)},W)}\right)\right]+\varepsilon,$

where $W$ has law (1.1) and $U$ is a random uniform point in $V_{n}$, and
where the inequality holds since $\kappa$ is increasing and $\varphi$
decreasing.By Lemma 3.4 we can bound the above by a constant multiple of
$n^{-1}n^{1/(\beta-1)}$. This vanishes as $n$ tends to infinity, concluding
the proof of Proposition 2.9(iii).

### Concluding remarks on local limits.

Note that the proof of Proposition 2.9 shows that the subgraph of the
conditional graph $G_{n}$ given $|E_{n}|\geq n(\mu+\rho)$ induced by removing
the vertices with the $k$ highest weights has a local limit which equals the
unconditional local limit $G_{\infty}$ of $G_{n}$. The local limit of the
conditional graph $G_{n}$ given $|E_{n}|\geq n(\mu+\rho)$ itself, however,
does not exist. This is because with asymptotically positive probability a
randomly sampled vertex connects to one of the $k$ high-weight vertices, and
as these vertices have infinite degree in the limit, the two-neighbourhood of
a randomly sampled vertex becomes infinite with positive probability. However,
one can still give a local description of the conditional graph, which we now
describe informally.

Sample a random vertex $o_{n}$ in the conditional graph. With high
probability, it is not one of the $k$ high-weight vertices. Now start
exploring the neighbourhood of this vertex in some order, e.g., the depth-
first order. At every step of the exploration, when we discover an incidence
to one of the $k$ high-weight vertices, we stop exploring from that vertex,
and go back to the last visited vertex to restart exploring from there. In
particular, this method of exploration avoids discovering the ‘forward’
neighbours of a high-weight vertex, of which there are linearly many, as we
have seen. This ‘restricted’ exploration can be formalized easily, which we
omit to keep the discussion informal. Now, say we explore the conditional
graph in this fashion, and look at the set of explored vertices that are at
graph distance $r$ from $o_{n}$, and call this explored graph $B_{n}(r)$.
Then, $B_{n}(r)$, rooted at $o_{n}$, for any fixed $r$, converges in the local
(Benjamini-Schramm) topology (see [13, Chapter 2]) to the graph $B_{r}$, whose
construction we describe next.

We briefly describe the local limit $G_{\infty}$ of the unconditional graph
$G_{n}$. The vertex set of $G_{\infty}$ is the corresponding infinite vertex
set - the infinite integer lattice $\mathbb{Z}^{d}$ in the lattice case and
the Palm version $\Gamma\cup\\{0\\}$ of a standard Poisson process $\Gamma$ on
$\mathbb{R}^{d}$ in the Poisson case. Then the weight for each vertex is
sampled as an independent random variable with density (1.1), and
conditionally on the weights and locations, edges are placed independently
between vertex pairs with probability (1.2). The graph $G_{\infty}$ is rooted
at $0$.

Now we describe the graph $B_{r}$. Let
$B_{\infty}(r)=(V_{\infty}(r),E_{\infty}(r))$ be the graph neighbourhood of
radius $r$ about $0$ in $G_{\infty}$. Let the weights of the vertices in
$V_{\infty}(r)$ be $(W(v))_{v\in V_{\infty}(r)}$. Note that the $W(v)$ are not
i.i.d., since the weights are tilted due to the fact that they are in the
$r$-neighbourhood of $0$, and the precise law of $(W(v))_{v\in V_{\infty}(r)}$
is rather involved. The construction of the graph $B_{r}$ is as follows.

  * $\rhd$

Append $k$ extra vertices, say $v_{1},\dots,v_{k}$, to $V_{\infty}(r)$ and
call the new vertex set $V_{r}$.

  * $\rhd$

The vertices $(v_{i})_{i\in[k]}$ have marks $(U_{i},Y_{i})$, where the first
coordinates are i.i.d. uniform points of $[-\frac{1}{2},\frac{1}{2}]^{d}$
independent of $(Y_{i})_{i\in[k]}$, and the second coordinates have joint
distribution (1.16).

  * $\rhd$

Given the variables sampled above, we place an edge between each vertex $v$ of
$V_{\infty}(r)$ and each $v_{i},$ $i\in[k],$ independently, with probability
$\varphi(\|U_{i}\|^{d}/\mathcal{K}(Y_{i},W(v))).$ Denote the new edge set by
$E_{r}$.

  * $\rhd$

The graph $B_{r}$ has vertex set $V_{r}$ and edge set $E_{r}$.

This description gives a way to compute the typical distance (graph distance
between two randomly sampled vertices) in the conditional graph $G_{n}$:
Independently sample two random vertices $o_{1}$ and $o_{2}$ in $G_{n}$. Start
exploring in the fashion described above from both $o_{1}$ and $o_{2}$, and
let $d_{1}$ (resp., $d_{2}$) be the distance from $o_{1}$ (resp., $o_{2}$) up
to which one has to explore to see a high-weight vertex. Denote by $H_{i}$ the
set of high-weight vertices exactly at graph distance $d_{i}$ from $o_{i}$,
for $i=1,2$. Recalling from Remark 2.8 that the high-weight vertices form a
clique, we note that the graph distance from $o_{1}$ to $o_{2}$ in the
conditional graph is $d_{1}+d_{2}+{\mathbbm{1}}_{\\{H_{1}\cap
H_{2}=\emptyset\\}}$ with high probability. This quantity as a sequence in $n$
is in fact a tight sequence of random variables - starting from a random
vertex in the conditional graph, one finds a high-weight vertex within tight
graph distance from it.

#### Acknowledgements

CK and PM are supported by DFG project 444092244 “Condensation in random
geometric graphs” within the priority programme SPP 2265. The work of RvdH is
supported in part by the Netherlands Organisation for Scientific Research
(NWO) through Gravitation-grant NETWORKS-024.002.003.

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