since $\kappa$ is a fixed constant. It remains to bound $\widehat{\mu}$. Recall that $\mathbb{E}(f(X)) \leq \eta \cdot n$ for some constant $\eta$ by Lemma <ref>. Now by (<ref>) and (<ref>) we have $\PP(f(X) \geq \widehat{\mu}-n) = 1-o(1)$, thus

\[\eta \cdot n\geq \mathbb{E}(f(X)) \geq \EE(f(X)\mid f(X) \geq \widehat{\mu}-n )\cdot \PP(f(X) \geq \widehat{\mu}-n) \geq (\widehat{\mu}-n)\cdot (1-o(1)),   \]
and thus
$\widehat{\mu} \leq \eta' \cdot n$ for some constant $\eta'>0$. It follows from (<ref>) that $\widehat{\PP}(f(X) \ge (\eta'+1)n ) = o(1/n^{d+1})$ and thus  ${\PP}(f(X) \ge 3(\eta'+1)n/2 ) = o(1/n^{d})$ by (<ref>), as claimed.

The following lemma will help us deal with the intersection of balls of radius $R$ around vertices when bounding the variance of sums of squares of degrees.

If $p, q\in B_O(R)$ are such that $0\leq\theta_q-\theta_p\leq \pi$ and $r_p+r_q\geq R+\omega(1)$, then

\[
B_p(R)\cap B_q(R) \subseteq B_O(r^*)
\]

where $r^*=r^*(p,q):=R-\min\{r_p,r_q\}-2\ln(\frac{1}{2\pi}(\theta_q-\theta_p))$.
If moreover $r_p, r_q\geq\rho>0$ and $\theta_q-\theta_p\geq 2\pi e^{R-\rho}/(n/\nu)$, then $r^*\leq R$.

Recall that $\theta_R(r,r')$ denotes the largest possible angle between two vertices at distances $r$ and $r'$ from the origin, which are at hyperbolic distance at most $R$ from each other.
Observe that, for $s\in B_p(R)\cap B_q(R)$, it must hold that
$\theta_R(r_p,r_s)+\theta_R(r_q,r_s)\geq \theta_q-\theta_p$
or else $s$ could not be at distance less than $R$ from both $p$ and $q$ simultaneously.
Assuming, without loss of generality, that $r_p\leq r_q$,
by Remark <ref>,
we get $2\theta_R(r_p,r_s)\geq \theta_q-\theta_p$.
Since, by hypothesis and Lemma <ref>, we have $\theta_R(r_p,r_s)\leq 2(1+o(1))e^{\frac12(R-r_p-r_s)}\leq \pi e^{\frac12(R-r_p-r_s)}$, thence
$R-r_p-r_s\geq 2\ln((\theta_q-\theta_p)/(2\pi))$.
The moreover part of the claim follows immediately.

We now establish a key result used when proving concentration of the target time.

Let $\frac12< \alpha <1$ and  $\nu >0$ be fixed. Then, for any $C>0$ and $\rho:=\rho(C)$ as in (<ref>),

\[
\operatorname{Var}\big(\sum_{v\in V(\calC)\setminus B_O(\rho)}d(v)^2\big) = \mathcal{O}\big(n^{1+\frac{1}{2\alpha}} \cdot (\ln n)^{\frac{3\alpha}{1-\alpha}} \big).
\]

Let $X:=\sum_{v\in V(\calC)\setminus B_O(\rho)}d(v)^{2}$ and recall  $\operatorname{Var}(X) = \EE(X^2) - \EE(X)^2$.
Using  the Campbell-Mecke formula <cit.>, we have

\begin{equation}\label{conc:eqn:firstMoment}
\EE(X) = \int_{0}^{2\pi}\int_\rho^R\EE (d(p)^2)\,
\lambda(r_p,\theta_p)\,\mathrm{d}r_p\mathrm{d}\theta_p,
\end{equation}

where $\lambda(r,\theta)$ is from (<ref>). Also, by the Campbell-Mecke formula (again) and then by symmetry,

\begin{align}
\EE(X^2) & = \int_{0}^{2\pi}\int_\rho^R \int_{0}^{2\pi}\int_\rho^R \EE (d(p)^2d(q)^2)\,
\lambda(r_p,\theta_p)\lambda(r_q,\theta_q)\,\mathrm{d}r_q\mathrm{d}\theta_q\,\mathrm{d}r_p\mathrm{d}\theta_p \notag
\\
& = 4\int_{0}^{2\pi}\int_\rho^R \int_{\theta_p}^{\theta_p+\pi}\int_{r_p}^R \EE (d(p)^2d(q)^2)\,
\lambda(r_p,\theta_p)\lambda(r_q,\theta_q)\,\mathrm{d}r_q\mathrm{d}\theta_q\,\mathrm{d}r_p\mathrm{d}\theta_p, \label{eq:fullint}
\end{align}

By partitioning  the range of integration of $\theta_q$ (in the last quadruple integral in (<ref>) above)
between $[\theta_p,\theta_p+\eta)$ and $[\theta_p+\eta,\theta_p+\pi)$, for $\eta:=1/n^{1-\frac{1}{2\alpha}}$
chosen with hindsight,
we obtain two integrals, say $I$ and $J$ respectively, such that $\EE(X^2)=4(I+J)$, and

\begin{equation}\label{eq:varBnd}
\operatorname{Var}(X) = 4(I+J)-\EE(X)^2.
\end{equation}

In what follows, we bound $I$ and $J$, starting with the latter using a very crude bound.
Since, for $a,b\geq 0$, it holds that $ab\leq 2ab\leq a^2+b^2$, we have $d(p)^2d(q)^2\leq d(p)^4+d(q)^4$.
Taking expectation and applying Lemma <ref>, we get
$\EE(d(p)^2d(q)^2) \leq \EE(d(p)^4) + \EE(d(q)^4)$.
Thus, as $\EE(d(p)) = \mathcal{O}(n\cdot e^{-\frac12 r_p})=\calO(e^{\frac12(R-r_p)})$ by Lemma <ref>, we have

\begin{align*}
J	&=\int_{0}^{2\pi}\int_\rho^R \int_{\theta_p}^{\theta_p+\eta}\int_{r_p}^R \EE(d(p)^2d(q)^2)\,
\lambda(r_p,\theta_p)\lambda(r_q,\theta_q)\,\mathrm{d}r_q\mathrm{d}\theta_q\,\mathrm{d}r_p\mathrm{d}\theta_p \\
& =\calO(n\cdot\eta)\int_{0}^{\pi}\int_\rho^R \EE(d(p))^4\,
\lambda(r_p,\theta_p)\,\mathrm{d}r_p\,\mathrm{d}\theta_p
\\ & =\calO(n^2\cdot\eta)\int_{0}^{\pi}\int_\rho^R e^{(2-\alpha)(R-r_p)}\,\mathrm{d}r_p\,\mathrm{d}\theta_p.
\end{align*}

Hence, by our choice of $\rho$ and $\eta$, since $2-\alpha>0$,

\begin{equation}\label{eq:J}
J = \calO\big(n^2\cdot\eta\cdot e^{(2-\alpha)(R-\rho)}\big)
\\
= \calO\big(n^2\cdot\eta\cdot (\ln n)^{\frac{2-\alpha}{1-\alpha}}\big)
= \calO\big(n^{1+\frac{1}{2\alpha}}\cdot (\ln n)^{\frac{2-\alpha}{1-\alpha}}\big).
\end{equation}

In order to bound $I$, let $d_p :=|V(\calG)\cap (B_p(R)\setminus B_O(r^*))|$ and $d'_p :=|V(\calG)\cap B_p(R)\cap B_O(r^*)|$ where $r^*:=r^{*}(p,q)$ is as in Lemma <ref>. Similarly, define $d_q$ and $d'_q$ with respect to $q$ instead of $p$. Then, $d(p)^2d(q)^2  = (d_p + d'_p)^2 \cdot (d_q + d'_q)^2$.

\[
d(p)^2d(q)^2 =\sum_{i_p=0}^{2}\sum_{i_q=0}^2\binom{2}{i_p}\binom{2}{i_q}d_p^{2-i_p}d_q^{2-i_q}{d'_p}^{i_p}{d'_q}^{i_q}.
\]

By Lemma <ref>, since $r_p+r_q\geq 2\rho=2R-\Omega(\ln R)=R+\omega(1)$, we have that $B_p(R)\setminus B_O(r^*)$, $B_q(R)\setminus B_O(r^*)$, $B_O(r^*)$ are pairwise disjoint, implying that $d_p^{i_p},d_q^{i_q},{d'_p}^{2-i_p}{d'_q}^{2-i_q}$ are independent.
So, taking expectation, from the last displayed equation, we get that  $\EE(d(p)^2d(q)^2)$ can be bounded from above by a sum of terms of the form $\EE(d_p^{j_p})\EE(d_q^{j_q})\EE({d'_p}^{j'_p}{d'_q}^{j'_q})$ where $j_p,j_q\in \{0,1,2\}$, $j_p+j'_p=j_q+j'_q=2$. In particular, there are  $c_{j_p,j_q}$'s in $\{1,2,4\}$ such that

\begin{equation}\label{eq:partsofI} I\leq
I_{2,2} + \sum_{(j_p,j_q)\neq (2,2)} c_{j_p,j_q} \cdot I_{j_p,j_q},
\end{equation}

where the  $I_{j_p,j_q}$'s are integrals given by

\begin{equation*}
I_{j_p,j_q}	:=\int_{0}^{2\pi}\int_\rho^R \int_{\theta_p+\eta}^{\theta_p+\pi}\int_{r_p}^R \EE(d_p^{j_p})\EE(d_q^{j_q})\EE({d'_p}^{2-j_p}{d'_q}^{2-j_q})\,
\lambda(r_p,\theta_p)\lambda(r_q,\theta_q)\,\mathrm{d}r_p\mathrm{d}\theta_p\,\mathrm{d}r_q\mathrm{d}\theta_q.
\end{equation*}

Observe now that by (<ref>),

\begin{equation}\label{eq:220}
I_{2,2}\leq\int_{0}^{2\pi}\int_\rho^R \int_{\theta_p}^{\theta_p+\pi}\!\!\int_{r_p}^R \EE(d(p)^2)\EE(d(q)^2)\,
\lambda(r_p,\theta_p)\lambda(r_q,\theta_q)\,\mathrm{d}r_q\,\mathrm{d}\theta_q\,\mathrm{d}r_p\,\mathrm{d}\theta_p = \tfrac14\EE(X)^2,
\end{equation}

so this term will cancel out. We now focus on the case $(j_p,j_q)\neq (2,2)$.

Let $s\in B_O(R)$ then $d_s\leq d(s)$, where the latter is Poisson distributed with expectation bounded from below by a constant by Lemma <ref>. Thus, by Lemmas <ref> and <ref>, for any $j\in[2]$,

\[
\EE(d_s^j) \leq \EE(d(s)^j)=\mathcal{O}(\EE(d(s))^j) = n^j\cdot\mu(B_s(R)\cap B_O(R))^j
= \mathcal{O}\big(n^j\cdot e^{-j\cdot \frac12 r_s}\big)
= \calO\big(e^{\frac{j}{2}(R-r_s)}\big).
\]

Observe that $d'_p,d'_q\leq d':=|V(\calG)\cap B_{O}(r^*)|$,
where $ d'$ is Poisson distributed. We wish to bound the moments of $d'$ however for some values of $r^*$ it may be that $\EE(d') = o(1)$, and so Lemma <ref> will not give a good bound. If $Z$ is Poisson with mean $\lambda>0$ then by [64], for any $k\geq 1$,
\begin{equation}\label{eq:poimoment}\EE(Z^k)= \sum _{i=0}^{k}\lambda ^{i}\cdot \Big\{ { k \atop i} \Big\}, \end{equation}
where $\{ { k \atop i} \}$ denotes the Stirling partition number. Observe that for $k\in [4]$ and $0\leq i\leq k$ we have $0\leq \{ { k \atop i} \}\leq 10$. Thus, by using (<ref>) when $\EE(d') \leq 1/2$ and Lemma <ref> when $\EE(d') > 1/2$ that $\EE(d'^j) = \mathcal{O}(\EE(d')) + \mathcal{O}(\EE(d')^j)$ for any $j\in [4]$. Moreover, provided $\eta\leq\theta_q-\theta_p\leq\pi$, since $\eta=1/n^{1-\frac{1}{2\alpha}}=\omega(\ln(RC/\nu)/n)=\omega(e^{R-\rho}/n)$, we know that $r^*:=r^*(p,q)\leq R$ by Lemma <ref>.
Thus, from Lemma <ref>, we obtain that
So, by the characterization of $r^*(p,q)$ from Lemma <ref>,
=\calO\big(ne^{-\alpha r_p}/(\theta_q-\theta_p)^{2\alpha}\big)$. Thus, for $j\in[4]$,

\[\EE(d'^j) =\calO\big(ne^{-\alpha r_p}/(\theta_q-\theta_p)^{2\alpha}\big) + \calO\big(n^j e^{-\alpha r_p\cdot j}/(\theta_q-\theta_p)^{2\alpha \cdot j}\big) \]

Now, if  $j':=j'_p+j'_q$ then $\EE({d'_p}^{j'_p}{d'_q}^{j'_q})\leq \EE(d'^{j'})$, so it follows that
\begin{align}\label{eq:momentsofdegparts}
\EE(d_p^{j_p})\EE(d_q^{j_q})\EE({d'_p}^{j'_p}{d'_q}^{j'_q})
&= \calO\big(\EE(d(p))^{j_p}\EE(d(q))^{j_q}\big[\EE(d') + \EE(d')^{j'}\big]\big)\notag \\ &= \calO\Bigg(e^{\frac{j_p}{2}(R-r_p)}e^{\frac{j_q}{2}(R-r_q)}\Bigg[ n\Big(\frac{e^{-\frac12 r_p}}{\theta_q-\theta_p}\Big)^{2\alpha}+ n^{j'}\Big(\frac{e^{-\frac12 r_p}}{\theta_q-\theta_p}\Big)^{2\alpha j'}\Bigg]\Bigg) .
\end{align}

Let $j\in [j']$ and assume first that $j_q\neq2$. Then, $\alpha>\frac12\geq\frac{j_q}{2}$, and since $2\alpha j\neq 1$, we have

\begin{align*}
I_{j_p,j_q}^j & =n^{2+j}\int_{0}^{2\pi}\int_{\rho}^R\int_{\theta_p+\eta}^{\theta_p+\pi}\int_{r_p}^R
\Big(\frac{e^{-\frac12 r_p}}{\theta_q-\theta_p}\Big)^{2\alpha j}\,\mathrm{d}r_q\,\mathrm{d}\theta_q\,\mathrm{d}r_p\,\mathrm{d}\theta_p \\
& = \calO(n^{2(1-\alpha)+j_p+j}(1/\eta)^{2\alpha j-1})\int_{0}^{2\pi}\int_{\rho}^R
e^{-(\alpha(j-1)+\frac{j_p}{2})r_p}\,\mathrm{d}r_p\,\mathrm{d}\theta_p \\
& =
\calO\big(n^{2(1-\alpha)+j_p+j}(1/\eta)^{2\alpha j-1}n^{-j_p-2\alpha(j-1)}(e^{R-\rho})^{\alpha(j-1)+\frac{j_p}{2}}\big) \\
& = \calO\big(n^{2-(2\alpha-1)j}(1/\eta)^{2\alpha j-1}(e^{R-\rho})^{\alpha(j-1)+\frac{j_p}{2}}\big).
\end{align*}

Since $\alpha>\frac12$, $j\leq j'$, and $j_p+j_q+j'=4$, we have $\alpha(j-1)+\frac{j_p}{2}\leq \alpha(j_p+j'-1)
\leq 3\alpha$.
So, by definition of $\rho$, for $j_q\neq 2$ and $j\in[j']$, we obtain

\[
I_{j_p,j_q}^j= \calO\big(n^{2-(2\alpha-1)j}(1/\eta)^{2\alpha j-1}(\ln n)^{\frac{3\alpha}{1-\alpha}}\big).
\]

For $j_q=2$ and $j_p\neq 2$ (thus, $j_p+j'=2$ and $j'\geq 1$), the crude upper bound $I_{j_p,j_q}^j\leq e^{\frac12(R-\rho)}I_{j_p,j_q-1}^j$ and the fact that $\alpha(j'-1)+\frac{j_p+1}{2}
\leq \alpha(j_p+j')=2\alpha$ imply that the last displayed bound for $I_{j_p,j_q}^j$ holds also when $j_q=2$ and $j_p\neq 2$.
Our choice of $\eta=1/n^{1-\frac{1}{2\alpha}}$ yields $n^{2-(2\alpha-1)j}/\eta^{2\alpha j-1}=n^{1+\frac{1}{2\alpha}}$ for any $j\in [j']$. Thus, by (<ref>), for all $(j_p,j_q)\neq (2,2)$, we have

\begin{equation}\label{eq:maxboundI}
I_{j_p,j_q}=\mathcal{O}\left( I_{j_p,j_q}^1\right)+ \mathcal{O}\left( I_{j_p,j_q}^{j'}\right) =  \calO\big(n^{1+\frac{1}{2\alpha}}\cdot(\ln n)^{\frac{3\alpha}{1-\alpha}}\big).
\end{equation}

Combining the bounds (<ref>) through (<ref>), we obtain

\[\operatorname{Var}(X) = \EE(X^2) - \EE(X)^2  =\mathcal{O}(n^{1+\frac{1}{2\alpha}}\cdot (\ln n)^{\frac{3\alpha}{1-\alpha}})+ \mathcal{O}(n^{1+\frac{1}{2\alpha}}\cdot (\ln n)^{\frac{2-\alpha}{1-\alpha}}). \]

Since $\alpha>\frac12$ implies that  $3\alpha> 2-\alpha$,
the sought after result follows.

The result above gave us the variance of sums of squared degrees, to prove concentration for the target time we will instead need to control a sum of degrees raised to some non-integer power less than two. The following lemma lets us transfer the variance bound from Lemma <ref>.
Let $A>0$ be real and $X$ be any non-negative integer valued random variable satisfying $\EE(X^{2A}\ln X)<\infty $. Then, $\operatorname{Var}(X^a)$ is non-decreasing in $0\leq a \leq A$.

Observe that for $a\leq A$ each element of the sequence $(f_n(a))$, where $f_n(a):=\sum_{x=1}^n x^{2a}\cdot \PP(X=x)$, is differentiable and $f_n$ converges to $\EE(X^{2a})<\infty $. Furthermore, the sequence $(\frac{\mathrm{d}}{\mathrm{d}a}f_n(a))_n$ converges uniformly to $\sum_{x \geq 1} (2a\ln x ) x^{2a}\cdot\PP(X=x) = 2a\cdot  \EE(X^{2a}\ln X) <\infty$. Thus, it follows from a standard convergence result <cit.> that
\begin{align*}\frac{\mathrm{d}}{\mathrm{d}a} \operatorname{Var}(X^a) &= \frac{\mathrm{d}}{\mathrm{d}a}  \sum_{x\geq 1}x^{2a}\PP(X=x) -  \frac{\mathrm{d}}{\mathrm{d}a}\Big( \sum_{x\geq 1}x^{a}\PP(X=x)\Big)^2\\
&=  \sum_{x\geq 1}(2a\ln x ) x^{2a}\cdot\PP(X=x) -  2\Big(\sum_{x\geq 1} (a\ln x) x^a \cdot\PP(X=x)\Big)\cdot\Big( \sum_{x\geq 1}x^{a}\PP(X=x)\Big)\\
&= 2a \cdot\big(\EE(X^{2a}\ln X)  - \EE(X^{a}\ln X)\EE(X^{a})\big) \\&\geq 0,
\end{align*}
where the last inequality follows from the fact that $\operatorname{Cov}(f(X),g(X))\geq 0$ for increasing functions $f,g$, see for example [67]. Thus, since $\operatorname{Var}(X^0)= 0$, $\operatorname{Var}(X^a)$ is non-decreasing for $0\leq a\leq A$.

We can now prove the main result of this section,
that is the a.a.s. convergence claims in Theorem <ref> and <ref>, which together with Propositions <ref> and <ref> establish the said theorems.

If $\calC:=\calC_{\alpha,\nu}(n)$, then a.a.s. $t_{\odot}(\calC) = \Theta(n)$, and w.h.p.

\[\mathcal{K}(\calC) = \Theta(n^2), \qquad \text{and} \qquad   \frac{1}{|V(\calC)|^2}\sum_{u,v\in V(\calC)} \Res{u}{v}= \Theta(1).  \]

We begin with $\mathcal{K}(\calC)$. The lower bound follows since the giant component has $\Omega(n)$ vertices of degree $\mathcal{O}(1)$ by Lemmas <ref> and <ref>. If $S:= V(\calC)\setminus B_O(\rho)$ then by Corollary <ref> the following holds w.h.p.,

\[\sum_{u,v\in V(\calC)} \Res{u}{v}\leq   \mathcal{O}(n)\cdot \sum_{w \in S}|V(\calG)\cap\Upsilon(w)| + \mathcal{O}(n^2),\]

so the result follows from Lemma <ref> with $\kappa=1$.

For the average resistance the upper and lower bound follow immediately from the bounds on $\mathcal{K}(\mathcal{C})$ and the fact that the center component has $\Omega(n)$ many vertices w.h.p. by Lemma <ref>.

For the target time $t_{\odot}(\calC)$, the lower bound follows directly from Lemmas <ref> and <ref> as a.a.s $t_{\odot}(\calC)\geq |V(\mathcal{C})|/8 =\Omega(n)$. For the upper bound, let $S:= V(\calC)\setminus B_O(\rho)$ and recall that by Corollary <ref>, w.h.p.,

\begin{equation*}
\leq    \frac{1}{2}\sum_{w\in S}|V(\calG)\cap\Upsilon(w)|\cdot d(w)  + \calO(n), \end{equation*}

As $1/2<\alpha<1$ is fixed, we set $\kappa = 3/4 + \alpha /2$; which satisfies $1<\kappa < 2\alpha<2$, and has Hölder conjugate  $1<\kappa/(\kappa-1) < \infty$. Thus, by Hölder's inequality, w.h.p.
\begin{equation}\label{eq:holder}t_{\odot}(\calC)  \leq   \frac{1}{2}\cdot \Big(\sum_{w\in S}|V(\calG)\cap\Upsilon(w)|^{\frac{\kappa}{\kappa-1}}\Big)^{\frac{\kappa-1}{\kappa}} \cdot\Big(\sum_{w\in S} d(u)^\kappa\Big)^{\frac{1}{\kappa}} + \calO(n).\end{equation}
Observe that the first non-trivial term is $\mathcal{O}(n^{\frac{\kappa-1}{\kappa}})$ w.h.p. by Lemma <ref>.  To control the second, observe that by Lemmas <ref> and <ref>, since $\kappa <2$, we have
\[\operatorname{Var}\Big( \sum_{v\in S}d(v)^\kappa \Big) \leq \operatorname{Var}\Big( \sum_{v\in S}d(v)^2\Big) = \mathcal{O}\Big(n^{1+\frac{1}{2\alpha}} \cdot (\ln n)^{\frac{3\alpha}{1-\alpha}} \Big)=o(n^2).\]
Additionally $\EE(\sum_{v\in V(\mathcal{G})}d(v)^{\kappa}) = \calO(n)$ by Lemma <ref>. Thus, $\sum_{v\in V(\mathcal{G})}d(v)^{\kappa} =\mathcal{O}(n)$ a.a.s. by Chebychev's inequality, and the second non-trivial term in (<ref>) is $\mathcal{O}(n^{\frac{1}{\kappa}})$ giving the result.

§ COVER TIME AND MAXIMUM HITTING TIME

In this section, we first determine the asymptotic behavior
of the cover time of the giant component of the HRG.
To do so, we rely on several intermediate structural results concerning HRGs that were established
in the previous section. However, the arguments developed here are not flow based.
In particular, the ones concerning commute times rely on
extensions of Matthew's Bound, a classical useful method for bounding cover times.
This result is then complemented, by the determination of the maximum hitting time also for the giant.

As in the previous sections, we restrict our discussion exclusively to the parameter range where $\frac12<\alpha<1$.

§.§ Upper Bounds on Hitting and Cover times

To determine the cover time of the center component of the HRG we rely on a classical result concerning random walks in graphs known as Matthew's bound (stated below) and one of its extensions (mentioned later). The bound and its extensions relate the cover time of a graph with its  minimum and maximum hitting times.
For any finite irreducible Markov
chain on $n$ states we have
\[\tcov \leq \thit \cdot \left(1 +\frac{1}{2}+ \cdots + \frac{1}{n} \right).  \]

A well known relation between hitting times and diameter of graphs together with established results
concerning the diameter of the HRG allows us to easily derive upper bounds on
hitting and, via Matthew's bound, also on the cover time of simple random walks on the center component of $\calG_{\alpha,\nu}(n)$.
If $\calC:=\calC_{\alpha,\nu}(n)$, then
a.a.s. and in expectation we have
\[\thit (\calC)=\calO(n\log n) \qquad \text{and } \qquad \tcov(\calC)=\calO(n\log^2 n).\]

Recall that $d_G(u,v)$ denotes the length of the shortest path
between a pair of vertices $u,v\in V(G)$, of a graph $G$. Then, by Lemmas <ref> and <ref>, we have $\Exu{u}{\tau_v}\leq 2|E(G)|d_G(u,v)$. Let $\mathcal{D}_1 := \{\max_{u,v\in V(\calC)}d_{\mathcal{C}}(u,v)\leq \kappa \ln n \}$ for a suitably large constant $\kappa$, then we have $\PP(\mathcal{D}_1^c) =n^{-\kappa'}$ for some fixed $\kappa'>0$ by [61],
as $1/2<\alpha <1$ is fixed. We note that Müller and Staps [61] only state that their $\calO(\log n)$ diameter bound holds a.a.s., however if one checks the proofs in their paper they hold with probability $1-n^{-1000}$, and they additionally use a coupling <cit.> that holds with probability $1-n^{-\kappa'}$, where $\kappa'>0$ depends on $\alpha$. Recall also that $|E(\mathcal{C})|=\calO(n)$ a.a.s. by Lemma <ref>. Thus, we get that $\thit =\calO(n\log n)$ holds a.a.s..  We also define the event $\mathcal{L} := \{\sum_{k=1}^{|V(\mathcal{C})|} k^{-1} \leq 2\ln n \}$ and observe that $\PP(\mathcal{L}^c) \leq \PP(|V(\calC )|> 100 n)\leq e^{-\Omega(n)}  $ by Lemma <ref>. The desired a.a.s. upper bound for the cover time then follows from Matthew's bound (see Theorem <ref>).

We now show that these bounds also hold in expectation. Before we begin, we must define an additional event $\mathcal{D}_2 := \{ \max_{u,v\in V(\calC)}d_{\mathcal{C}}(u,v) \leq \kappa (\ln n)^{\frac{1}{1-2\alpha}} \} $ and note that $\PP(\mathcal{D}_2^c) =\calO(n^{-100})$ by <cit.> (also see the remark on <cit.> after Theorem 3).

We begin with $\EE(t_{\mathsf{hit}})$, conditioning on  the event $\mathcal{D}_1 = \{\max_{u,v\in V(\calC)}d_{\mathcal{C}}(u,v)\leq \kappa \ln n \}$ gives

\begin{equation*}%\label{eq:exhit1}
\EE\left(t_{\mathsf{hit}}\cdot \mathbf{1}_{\mathcal{D}_1}\right)\leq (\kappa\ln n)\cdot 2\EE\left(|E(\mathcal{C})|\right) = \calO(n\log n), \end{equation*}
as $\EE\left(|E(G)|\right) =\calO(n)$ by Lemma <ref>. By the Cauchy–Schwarz inequality we have
\[\EE\left(t_{\mathsf{hit}}\cdot\mathbf{1}_{\mathcal{D}_1^c\cap \mathcal{D}_2}\right) \leq \kappa(\ln n)^{\frac{1}{1-2\alpha}}\cdot 2\EE\left(|E(\mathcal{C})|\mathbf{1}_{\mathcal{D}_1^c} \right) \leq  2\kappa(\ln n)^{\frac{1}{1-2\alpha}}\cdot\sqrt{\EE\left(|E(\calC)|^2\right)\PP(\mathcal{D}_1^c)}, \]

then as   $\EE(|E(\mathcal{G})|^2) = \mathcal{O}(n^2)$ by <cit.>
and by the bound on $\PP(\mathcal{D}_1^c)$ above we have

\begin{equation*}\label{eq:exhit2}\EE\left(t_{\mathsf{hit}}\cdot\mathbf{1}_{\mathcal{D}_1^c\cap \mathcal{D}_2}\right) \leq 2\kappa(\ln n)^{\frac{1}{1-2\alpha}}\cdot\sqrt{\calO(n^2) \cdot n^{-\kappa'}} = o(n).  \end{equation*}

Finally, since $d_{\mathcal{C}}(u,v)\leq |V(\calG)|$, $|E(\calC)|\leq |V(\calC)|^2$, and since $\EE|V(\calC)|^k = \calO(n^k)$ for any fixed $k$ by Lemma <ref>, we have
\begin{equation*}\label{eq:exhit3}\EE\left(t_{\mathsf{hit}}\cdot\mathbf{1}_{\mathcal{D}_1^c\cap \mathcal{D}_2^c }\right) \leq\EE\left(|V(\calC)|^3\cdot \mathbf{1}_{\mathcal{D}_2^c}\right)\leq  \sqrt{\EE\left(|V(\calC)|^6\right)\PP(\mathcal{D}_2^c)} =o(1), \end{equation*}

The result follows from the last 3 displayed equations.

We now bound $\EE(t_{\mathsf{cov}})$. Recall the event $\mathcal{L} = \{\sum_{k=1}^{|V(\mathcal{C})|} k^{-1} \leq 2\ln n \}$ and observe that

\begin{equation*}%\label{eq:excov1}
\EE\left(t_{\mathsf{cov}}\cdot \mathbf{1}_{\mathcal{D}_1\cap \mathcal{L}}\right)\leq (\kappa\ln n)\cdot 2\EE\left(|E(\mathcal{C})|\right)\cdot (2\ln n) = \calO(n\log^2 n).
\end{equation*}

Now, since $ d_{\calC}(u,v) <|V(\calC)|$, $\sum_{k=1}^{|V(\mathcal{C})|} k^{-1}<|V(\calC)| $, and $|E(\calC)|\leq |V(\calC)|^2$, we have

\begin{equation*}%\label{eq:excov2}
\EE\left(t_{\mathsf{cov}}\cdot \mathbf{1}_{\mathcal{L}^c}\right)\leq \EE\left(|V(\mathcal{C})|^4\mathbf{1}_{\mathcal{L}^c}\right) \leq \sqrt{\EE(|V(\mathcal{C})|^8)\PP(\mathcal{L}^c)} = o(1), \end{equation*}

as $\PP(\mathcal{L}^c) = e^{-\Omega(n)} $ and $\EE|V(\calC)|^k = \calO(n^k)$ for any fixed $k$ by Lemma <ref>. Similarly,

\begin{equation*}%\label{eq:excov3}
\EE\left(t_{\mathsf{cov}}\cdot\mathbf{1}_{\mathcal{D}_1^c\cap\mathcal{L}\cap  \mathcal{D}_2}\right) \leq \kappa(\ln n)^{\frac{1}{1-2\alpha}} \cdot 2\sqrt{\EE\left(|E(\calC)|^2\right)\PP(\mathcal{D}_1^c)}\cdot  (2\ln n)  =o(n).   \end{equation*}

Finally, again similarly to before, we have

\begin{equation*}%\label{eq:excov4}
\EE\left(t_{\mathsf{cov}}\cdot\mathbf{1}_{\mathcal{D}_1^c\cap \mathcal{L}\cap \mathcal{D}_2^c }\right) \leq\EE\left(|V(\calC)|^4\cdot \mathbf{1}_{\mathcal{D}_2^c}\right)\leq  \sqrt{\EE\left(|V(\calC)|^8\right)\PP(\mathcal{D}_2^c)} =o(1). \end{equation*}
The bound on $\EE(t_{\mathsf{cov}})$ follows from combining the last 4 displayed equations.

§.§ Lower Bounds on Hitting and Cover times
The lower bounds in the classical version of Matthew's bound <cit.> do not directly (or at least
not obviously) imply a
lower bound for the cover time of $\calC_{\alpha,\nu}(n)$ matching the upper bound obtained in  Proposition <ref>.
Fortunately, several extensions of Matthew's bound have been established. Among these,
we will take advantage of the following result by Kahn, Kim, Lovász and Vu.
If $G=(V,E)$ is a connected graph, then

\begin{equation*}
\geq \frac12 \cdot \max_{U\subseteq V}\big(\kappa_U\cdot\ln |U|\big)
\qquad \text{where $\kappa_{U} := \min_{u,v\in U}[\Exu{u}{\tau_{v}}+ \Exu{v}{\tau_{u}}]$.}
\end{equation*}

The main difficulty in applying the preceding result, in order to lower bound the cover time of the center component of the HRG, is finding an adequate set $U$ of vertices of the center component of $\calG_{\alpha,\nu}(n)$ which is both sufficiently large and such that for every distinct pair of elements $u,v\in U$
the commute time between $u$ and $v$ is large enough, say $|U|=n^{\Omega(1)}$ and $\kappa_U=\Omega(\ln n)$.
In order to lower bound the cover time of the center component of the HRG we show that such a set of vertices of $\calC_{\alpha,\nu}(n)$ is likely to exist.
The set's existence and the fact that it satisfies the necessary conditions
relies on a structural result about induced paths of length $\Theta(\log n)$ in the HRG which
implies that a.a.s. there are many (specifically, $n^{\Omega(1)}$) vertices of the center component of
$\HRG_{\alpha,\nu}(n)$ whose
removal gives rise to a path component of order $\Omega(\log n)$.
The result is implicit in <cit.> and explicitly known (in an equivalent form) to the first author and Mitsche
since their characterization of the size of the second-largest component of the HRG <cit.>.
We state and prove it below, since it is essential for our study of the commute time of the giant component, might also be useful in other settings,
and now seems of interest in its own right. The proof argument is somewhat different
(we believe also simpler) from the one implicit in <cit.> and relies on the tiling structure constructed in the previous section.
Before going into its proof, we give an overview of it.
To start, $B_O(R)$ is partitioned into several regions.
The first region will be a ball centered at the origin with radius
as large as possible but still so it will be unlikely that it contains vertices of $\calG_{\alpha,\nu}(n)$.
This region will be denoted $\calR_O$ and correspond to $B_O(\rho_{O})$ for a $\rho_{O}$ appropriately chosen.
Next, for a positive integer $m$ to be determined, we consider $m$ sectors of central angle $2\pi/m$ and
intersect each one of them with $B_O(R)\setminus\calR_O$ thus
obtaining a partition $\{\calR_1,...,\calR_m\}$ of $B_O(R)\setminus\calR_O$.
We then show that we can pick $m=\Omega(n^{\zeta+\chi})$ for some $0<\zeta+\chi<1-\frac{1}{2\alpha}$ so that in each region $\calR_a$, $a\in [m]$, with a small but non-negligible probability (specifically, with probability $\Omega(n^{-\widetilde{\chi}})$ for a small constant $0<\widetilde{\chi}<\chi$) and independent of what
happens in the other regions except for $\calR_O$, there is a
vertex $v_a$ of the center component of $\HRG_{\alpha,\nu}(n)$ whose removal disconnects $\HRG_{\alpha,\nu}(n)$
into two connected components, one of which is a path $\calP_a$ of length $\Theta(\log n)$.
This suffices for our purpose, since it implies that the expected number of induced
paths of length $\Theta(\log n)$ in $\calC_{\alpha,\nu}(n)$ is $\Omega(n^{\zeta})$, so
conditioned on $\calR_O$ being empty, the independence of events within different regions together with
standard concentration bounds would yield the result we seek to establish.
This completes the high-level overview of our proof argument.
We discuss its details next.

Before stating the result, we introduce one additional term.
We say that a subgraph $P$ of a graph $G$ is a dangling path
if $P$ is a path in $G$ and there is a single edge of $G$ whose removal disconnects the
graph into two connected components, one of which is $P$.
Furthermore, we say that $P$ is a maximal dangling path if it is not a proper subgraph of
some other dangling path of $G$.

For all $0<\zeta<1-\frac{1}{2\alpha}$ there is a positive constant $\Lambda:=\Lambda(\zeta)$
such that with probability at least $1-e^{-\Omega(n^{\zeta})}$ there are $\Omega(n^{\zeta})$
maximal dangling paths of length at least $\Lambda\ln n$ in $\calC_{\alpha,\nu}(n)$.

Henceforth, let $V:=V(\HRG_{\alpha,\nu}(n))$. Set $\rho_{O}:=(1-\frac{1+\eta}{2\alpha})R$ where $0<\eta<2\alpha-1$ is a constant.
Let $\calR_O= B_O(\rho_{O})$ and note that, by Lemma <ref>, we have $\mu(\calR_O)=(1+o(1))e^{-(1+\eta)\frac{R}{2}}$.
Since $R=2\ln(\frac{n}{\nu})$,

\begin{equation}\label{cover:eqn:empty}
\PP(\text{$V\cap\calR_O=\emptyset$}) = e^{-n\mu(\calR_O)} = e^{-\calO(n^{-\eta})} = 1-\calO(n^{-\eta}).
\end{equation}

[scale=1.3, rotate=180]
(O) at (,̧)̧ $\quad _O$;

[yellow!60,fill] (,̧)̧ ++(360/12*1.5:) – ++(360/12*1.5:-̧) arc (360/12*1.5:360/12*3.5:)̧ – ++(360/12*3.5:-+̧) arc(360/12*3.5:360/12*1.5:) – cycle;

[Latex-Latex] (,̧)̧ ++(360/12*1.5:3) arc(360/12*1.5:360/12*2.35:3) node[midway,below,rotate=-30] $_{\theta_R(h_{\ell'}\!{-}\!\delta,\rho_O)}$;
[Latex-Latex] (,̧)̧ ++(360/12*2.35:3) arc(360/12*2.35:360/12*2.65:3) node[midway,below,rotate=-15] $_{k\theta_{\ell'}}$;
[Latex-Latex] (,̧)̧ ++(360/12*2.65:3) arc(360/12*2.65:360/12*3.5:3) node[midway,below] $_{\theta_R(h_{\ell'}\!{-}\!\delta},\rho_O)$;

(,̧)̧ ++(360/12*1.5:) – ++(360/12*1.5:-̧);
(,̧)̧ ++(360/12*3.5:) – ++(360/12*3.5:-̧);
[gray] (,̧)̧ ++(360/12*2.35:) – ++(360/12*2.35:-̧);
[gray] (,̧)̧ ++(360/12*2.65:) – ++(360/12*2.65:-̧);

at (+̧ *̧0.6 * cos(37.5),+̧ *̧0.6 * sin(37.5)) $\calR_{a+1}$;
at (+̧ *̧0.6 * cos(57.5),+̧ *̧0.6 * sin(57.5)) $\calR^{\mathsf{ct}}_{a}$;
at (+̧ *̧0.6 * cos(75),+̧ *̧0.6 * sin(75)) $\calR^{\mathsf{pt}}_{a}$;
at (+̧ *̧0.6 * cos(92.5),+̧ *̧0.6 * sin(92.5)) $\calR^{\mathsf{bf}}_{a}$;
at (+̧ *̧0.6*cos(112.5),+̧ *̧0.6* sin(112.5)) $\calR_{a-1}$;

[pattern=north east lines] (,̧)̧ ++(360/12*2.35:-̧0.2) – ++(360/12*2.35:0.1) arc (360/12*2.35:360/12*2.65:-̧0.15) – ++(360/12*2.65:-0.1) arc(360/12*2.65:360/12*2.35:-̧0.2) node[midway, yshift=4pt] $_{\calB_a}$;

[dotted] (+̧ *̧ cos(22.5),+̧ *̧ sin(22.5)) arc (22.5:130:)̧;
[-Latex,gray] (,̧)̧ – ++(22.5:)̧ node[midway,above,black] $_R$;
[black,thick] (+̧ *̧ cos(25.0),+̧ *̧ sin(25.0)) arc (25.0:128.75:)̧;

[blue, thick] (+̧ (-̧0.4) * cos(27.5),+̧ (-̧0.4) * sin(27.5)) arc (27.5:125:-̧0.4);
[red, thick] (+̧ (-̧0.1) * cos(26.25),+̧ (-̧0.1) * sin(26.25)) arc (26.25:127:-̧0.1);

[dotted] (+̧ (-̧0.4) * cos(25),+̧ (-̧0.4) * sin(25)) arc (25:127.5:-̧0.4);
[dotted] (+̧ (-̧0.1) * cos(23.75),+̧ (-̧0.1) * sin(23.75)) arc (23.75:130:-̧0.1);
[gray,-Latex] (,̧)̧ – ++(120:) node[midway,left,black] $_{\rho_O}$;
[-Latex,gray] (,̧)̧ – ++(126:-̧0.4) node[midway,left,blue] $_\rho$;
[-Latex,gray] (,̧)̧ – ++(130:-̧0.1) node[midway,right,red] $_{\rho'}$;

[dotted] (+̧ * cos(10),+̧ * sin(10)) arc (10:142.5:);
[black] (+̧ * cos(30),+̧ * sin(30)) arc (30:120:);
at (+̧ *̧0.15*cos(60),+̧ *̧0.15* sin(60)) $\calR_{O}$;

(a) Shaded region corresponds to $\calR_a:=(\calR_a^{\mathrm{ct}}\cup\calR_a^{\mathrm{pt}}\cup\calR_a^{\mathrm{bf}})\setminus\calR_O$, and (b) hatched region corresponds to band $\calB_a$.
(Picture not at scale.)

Next, we equipartition $B_O(R)$ into $m$ sectors of central angle $2\pi/m$, say $\Upsilon_1,...,\Upsilon_m$, and focus the ensuing discussion on a specific $\Upsilon_a$, i.e., we fix $a\in [m]$.
The intersection of $\Upsilon_a$ with $B_O(R)\setminus\calR_O$
gives rise to $\calR_a$, i.e., $\calR_a=(\Upsilon_a\cap B_O(R))\setminus\calR_O$.
For the sake of concreteness, let $\phi_a\in [0,2\pi)$ be such that

\[
\Upsilon_a = \{(r,\theta) \mid \phi_a\leq\theta<\phi_a+\tfrac{2\pi}{m}\}.
\]

The sector $\Upsilon_a$ will contain three contiguous sub-sectors. Moving clockwise around the
origin, the order in which these sub-sectors are encountered are; first a buffer sub-sector $\Upsilon^{\mathrm{bf}}_a$,
next a path sub-sector $\Upsilon_a^{\mathrm{pt}}$, and
finally a connection sub-sector $\Upsilon_a^{\mathrm{ct}}$.
Their intersections with $\calR_a$ will be denoted $\calR_a^{\mathrm{bf}}$, $\calR_a^{\mathrm{pt}}$
and $\calR_a^{\mathrm{ct}}$, respectively. (See Figure <ref> (a).)

Our goal is to show that in a narrow band of $\calR_a^{\mathrm{\mathrm{pt}}}$ close
to the boundary of $B_O(R)$,
with non-negligible probability, there is a collection of vertices that induce a path $\calP_a$
of length $\Theta(\log n)$.
We will establish that with constant probability there is a path
(all of whose vertices belong to $\calR_a^{\mathrm{ct}}$) that guarantees that one of the end vertices of $\calP_a$
(and thus also all vertices of $\calP_a$) belongs to the center component of $\HRG_{\alpha,\nu}(n)$.
Still, it could be that internal vertices of the path $\calP_a$ might have other neighbours not in $\calP_a$.
So, our choices of path, and buffer regions will be such that we can guarantee that all potential neighbours
of vertices of $\calP_a$ other than one of its end vertex $v_a$  would belong to either $\calR_O$ or some sub-region of $\calR_a^{\mathrm{pt}}\cup\calR_a^{\mathrm{bf}}$ which, with non-negligible probability,
do not contain vertices of $\calG_{\alpha,\nu}(n)$ (we already this for $\calR_O$).
Thus, we will be
able to conclude that $\calP_a$ is a dangling path of $\calC_{\alpha,\nu}(n)$. Next, we formalize this paragraph's discussion.

Henceforth, consider the tiling $\calF(c)$ of Section <ref> where $\eps:=\frac12\ln 2$ and $c:=c(\eps)$ is like in Lemma <ref>.
Also, let $\rho$ and $\rho'$ be defined as usual, i.e., as in (<ref>) and (<ref>), respectively,
and fix $\ell$ and $\ell'$ as always, i.e., let them be the largest integers such that
$h_{\ell}\leq\rho$ and $h_{\ell'}\leq\rho'$, respectively.

We let $\Upsilon_a^{\mathrm{bf}}$ be the sector of $\HH^2$ which contains points with angular coordinates
at least $\phi_a$ and strictly less than $\phi_a+\theta_{R}(h_{\ell'}{-}\delta,\rho_O)$
where (with hindsight) we set $\delta:=\ln\frac{9}{8}$.
Thus, $\Upsilon_a^{\mathrm{bf}}$ and $\Upsilon_a$ share one of their boundaries.
Let $\Upsilon_a^{\mathrm{pt}}$ be the sector of $\HH^2$ of central angle $3k\theta_{\ell'}$, where $k$ will be
fixed later, and contiguous to $\Upsilon_a^{\mathrm{bf}}$ when moving clockwise around the origin,
i.e., $\Upsilon_a^{\mathrm{pt}}$ contains all points of $\HH^2$ whose angular coordinates are at least
$\phi_a+\theta_R(h_{\ell'}{-}\delta,\rho_O)$ and strictly less than $\phi_a+\theta_R(h_{\ell'}{-}\delta,\rho_O)+3k\theta_{\ell'}$.
Also, let $\calR^{\mathrm{bf}}_a:=\Upsilon_a^{\mathrm{bf}}\cap (B_O(R)\setminus\calR_O)$ and $\calR^{\mathrm{pt}}_a:=\Upsilon_a^{\mathrm{pt}}\cap (B_O(R)\setminus\calR_O)$.
We consider a band $\calB_a$ of points of $\Upsilon^{\mathrm{pt}}_a$ with radial distance to the
origin between $h_{\ell'}-\delta$ and $h_{\ell'}$, specifically $\calB_a:=\Upsilon_a^{\mathrm{pt}}\cap (B_{O}(h_{\ell'})\setminus B_{O}(h_{\ell'}{-}\delta))$. See Figure <ref> (b).

[scale=0.85, rotate=175,every node/.style=inner sep=0,outer sep=0]

(13.5,31) rectangle (31.7,41.1);
(O) at (,̧)̧ $O$;

[gray] (O) – ++(-:);
[gray] (O) – ++(+27*:);

[yellow!40,fill] (O) ++(-:+̧) arc (-:+27*:+̧) – ++(+27*:) arc (+27*:-:+̧2*) – cycle;
[yellow!40,pattern=north west lines] (O) ++(-:+̧) arc (-:+27*:+̧) – ++(+27*:) arc (+27*:-:+̧2*) – cycle;

ıin 0,1,...,12,19,20,...,27
(O) ++(+*ı:)̧ – ++(+*ı:2*) ;

in 0,1,2

[fill,magenta!50] (O) ++(+*:-̧2**) arc (+*:+*:-̧2**) – ++(+*:2*) arc (+*:+*:-̧2**+2*) – cycle;
ıin ,,...,
(O) ++(+*ı:-̧2**) – ++(+*ı:2*);

in 0,2,4, 6, 12
[dashed] (O) ++(-32.5:-̧*) arc (-32.5:+27.5:-̧*);

[dotted] (O) ++(-22.5:) arc (-22.5:+27.5:);
(O) ++(-20.0:) arc (-20.0:+25.0:);

[dotted,red] (O) ++(-22.5:+̧2*) arc (-22.5:+27.5:+̧2*);
[red,thick] (O) ++(-20.0:+̧2*) arc (-20.0:+25.0:+̧2*);

[dotted,blue] (O) ++(-22.5:-̧14*) arc (-22.5:+27.5:-̧14*);
[blue,thick] (O) ++(-20.0:-̧14*) arc (-20.0:+25.0:-̧14*);

[Latex-Latex] (O) ++(-:-̧30*) arc(-:+27*:-̧30*) node[midway,above] $k\theta_{\ell'}$;

[thick,dotted] (O) ++(+15*:-̧10*) – ++(+19*:2*);
[thick,dotted] (O) ++(-14*:-̧10*) – ++(-14*:2*);

(O) ++(-14*:-̧22*) node $\Upsilon_{a}^{\mathrm{ct}}$;
(O) ++(+15*:-̧22*) node $\Upsilon_a^{\mathrm{pt}}$;
(O) ++(+35*:-̧22*) node $\Upsilon_{a}^{\mathrm{bf}}$;

[blue] (O) ++(-28*:-̧14*) node ${\rho}$;
[red] (O) ++(-27.5*:+̧2*) node ${\rho'}$;
(O) ++(-27.5*:+̧4*) node ${R}$;
(O) ++(+33.5*:-̧13*) node[rotate=12] level $\ell\!\!+\!\!1$;
(O) ++(+32.5*:+̧1*) node[rotate=12] level $\ell'$;

å/in 26.5/0.3, 22.4/0.33, 11.7/0.35, 7.2/0.31, 3.7/0.365, -0.7/0.31, -2.7/0.1, -4.7/0.4, -6.7/0.4, -8.7/0.2, -10.7/0.3, -12.7/0.1, -11.3/-0.2, -14.8/-0.4, -16.4/-0.3, -18.4/-0.3, -20.4/-0.1, -22.3/-0.25, -22.5/-0.9, -24.5/-0.7, -26.4/-0.6, -28.6/-0.55, -30.8/-0.65, -32.6/-0.75
[fill] (O) ++(+å*:+̧) circle (0.02);

[fill] (O) ++(+26.5*:+̧0.3) node (AA) ;
[fill] (O) ++(+22.4*:+̧0.33) node (BB) ;
[fill] (O) ++(+11.7*:+̧0.35) node (CC) ;
[fill] (O) ++(+7.2*:+̧0.31) node (DD) ;
[fill] (O) ++(+3.7*:+̧0.365) node (EE) ;
[fill] (O) ++(-0.7*:+̧0.31) node (A) ;
[fill] (O) ++(-2.7*:+̧0.1) node (B) ;
[fill] (O) ++(-4.7*:+̧0.4) node (C) ;
[fill] (O) ++(-6.7*:+̧0.4) node (D) ;
[fill] (O) ++(-8.7*:+̧0.2) node (E) ;
[fill] (O) ++(-10.7*:+̧0.3) node (F) ;
[fill] (O) ++(-12.7*:+̧0.1) node (G) ;
[fill] (O) ++(-11.3*:-̧0.2) node (H) ;
[fill] (O) ++(-14.8*:-̧0.4) node (I) ;
[fill] (O) ++(-16.4*:-̧0.3) node (J) ;
[fill] (O) ++(-18.4*:-̧0.3) node (K) ;
[fill] (O) ++(-20.5*:-̧0.1) node (L) ;
[fill] (O) ++(-22.3*:-̧0.25) node (M) ;
[fill] (O) ++(-22.4*:-̧0.9) node (N) ;
[fill] (O) ++(-24.5*:-̧0.7) node (OO) ;
[fill] (O) ++(-26.4*:-̧0.6) node (P) ;
[fill] (O) ++(-28.6*:-̧0.55) node (Q) ;
[fill] (O) ++(-30.8*:-̧0.65) node (R) ;
[fill] (O) ++(-32.6*:-̧0.75) node (S) ;

(AA) to [in=135,out=45] (BB) to [in=-135, out=-45] (CC) to [in=135, out=45] (DD) to [in=-135, out=-45] (EE) – (A) – (B) – (C) – (D) – (E) – (F) – (G) – (H) – (I) – (J) – (K) – (L) – (M) – (N) – (OO) – (P) – (Q) – (R) – (S);

(a) Hatched region corresponds to band $\calB_a$. Each division of the hatched region corresponds to a $\calB'_j$. (b)
The curve segment corresponds to path $\calP_a$.
(c) Non-hatched shaded region correspond to half-tile $H_s^{(i)}$ for $s\in\{\ell,...,\ell'\}$ and $i\in [6]$.
The piecewise linear segment corresponds to a way in which the path $\calP_a$ connects to the rest of the center component.
(Picture not at scale. For simplicity, we have illustrated the case where $h_{\ell}=\rho$ and $h_{\ell'}=\rho'$.)

Partitioning $\Upsilon^{\mathrm{pt}}_a$ into sectors of central angle $\frac14\theta_{\ell'}$ we end up with $12k$
sectors whose intersections with $\calB_a$ partitions the latter into $\calB'_{1},...,\calB'_{12k}$ (these parts depend on the index $a$, but we ommitt the index from our notation in order to avoid over cluttering).
Without loss of generality, the indexing of the $\calB'_{j}$'s reflects the order in which each piece is encountered when moving anti-clockwise around the origin. Thus, $\calB'_{12k}$'s boundary intersects the one of $\Upsilon^{\mathrm{bf}}_a$
while $\calB'_{1}$'s boundary intersects the one of $\Upsilon^{\mathrm{ct}}_a$. See Figure <ref> (a).

We claim that our choice of central angle for $\Upsilon^{\mathrm{bf}}_a$ ensures that for every point
$p\in\calB_{a}$ the elements of $B_{p}(R)\setminus\calR_O$ with angular coordinate at most $\theta_{p}$
are completely contained in $\Upsilon^{\mathrm{pt}}_a\cup\Upsilon^{\mathrm{bf}}_a$.
Indeed, all such elements span an angle at the origin with $p$ which is at most $\theta_R(r_p,\rho_O)$.
By Remark <ref>, the latter angle is at most $\theta_R(h_{\ell'}-\delta,\rho_O)$ (because $p\not\in B_{0}(h_{\ell'}-\delta)$). Our last claim follows.

Now, suppose the following event  occurs: there is exactly one vertex,
say $v_{j}$, that belongs to $\calB'_{j}$ if $j-1$ is a multiple of $3$ and
there is no vertex in $\calB'_{j}$ for all other $j$'s. We claim that the $v'_{j}$'s induce a path $\calP_a$ on $4k$ vertices. Indeed, note that the angle at the origin between $v'_{j}$ and $v'_{j+1}$ is at most $\theta_{\ell'}$.
Since both $v'_{j}$ and $v'_{j+1}$ belong to $B_O(h_{\ell'})\setminus B_O(h_{\ell'}-\delta)$,
by definition of $\theta_R(\cdot,\cdot)$, both vertices are within distance at most $R$ of each other, so there is an edge between them in $\calG_{\alpha,\nu}(n)$. On the other hand, the angle at the origin spanned by $v'_{j-1}$ and $v'_{j+1}$ is at least  $\frac{5}{4}\theta_{\ell'}=\frac54(1+o(1))e^{-\delta}\theta_{R}(h_{\ell'}{-}\delta,h_{\ell'}{-}\delta)$, so by our choice of $\delta$, the said angle is at least $\frac{10}{9}(1+o(1))\theta_{\ell'}$, hence $v'_{j-1}$ and $v'_{j+1}$ are
at distance at least $R$ from each other and no edge between them exists in $\calG_{\alpha,\nu}(n)$. This completes the proof of the claim stated at the start of this paragraph.
We argue next that the probability of the claim's event is $\Omega(n^{-\chi})$ provided that $C'$ is chosen a
large enough constant.
Thus, by Lemma <ref>
and Lemma <ref>, the expected number of vertices in any single one of the $\calB'_{j}$'s is

\[
\EE|V\cap\calB'_{j}| = n\frac{\theta_{\ell'}}{4}\mu(B_{O}(h_{\ell'})\setminus B_{O}(h_{\ell'}{-}\delta))
= \Theta(n\theta_{\ell'}e^{-\alpha(R- h_{\ell'})})
= \Theta(\nu e^{(1-\alpha)(R-h_{\ell'})}).
\]

By our choice of $h_{\ell'}$, we have that $R-h_{\ell'}=\frac{1}{1-\alpha}\ln(\frac{2C'}{\nu})+\Theta(1)$ and conclude that $\lambda:=\EE|V\cap\calB'_{j}|=\Theta(C')$.
The probability that any single one of the $\calB'_{j}$'s does not contain a vertex of $\HRG_{\alpha,\nu}(n)$
is $e^{-\lambda}$, and the probability that it contains exactly one vertex is $\lambda e^{-\lambda}$.
Since the $\calB'_{j}$'s are disjoint, the number of vertices of $\calG_{\alpha,\nu}(n)$ contained within
each one is independent. Thus, we can choose $C'$ large enough so

\begin{equation}\label{cover:eqn:path}
\PP(\text{$V\cap\calB_a$ induces a path in $\calG_{\alpha,\nu}(n)$})\geq (\lambda e^{-3\lambda})^{4k}=e^{-\calO(C'k)}.
\end{equation}

So far we have exposed the band $\calB_a$ and shown that with non-negligible probability it contains an induced path of $\HRG_{\alpha,\nu}(n)$. Any neighbour that a vertex of $\calP_a$ could have, except maybe the vertices $\{v'_{1},v'_{2},v'_{3}\}$ of $\calP_a$, would have to belong to $\calR_O$ or $\calR_a^{\mathrm{pt}}\setminus\calB_a$ or $(B_{v'_{4}}(R)\cup B_{v'_{4k}}(R)\setminus\calR_O)\setminus\calB_a$. We have already seen that the first region is unlikely to contain vertices. In contrast, we shall see that for the latter two regions we can only guarantee that with non-negligible probability they do not contain vertices. The measure of the second region is upper bounded by the one of the sector $\Upsilon_a^{\mathrm{pt}}$ (because $\calR^{\mathrm{pt}}_a\subseteq\Upsilon_a^{\mathrm{pt}}$), which by Lemma <ref> and Lemma <ref>, equals $3k\theta_{\ell'}=\Theta(k\frac{\nu}{n}(\frac{2C'}{\nu})^{\frac{1}{1-\alpha}})=\Theta(\frac{k}{n}(C')^{\frac{1}{1-\alpha}})$. Using Lemma <ref> we get that the measure of the third region is at most  $\mu(B_{v'_4}(R))+\mu(B_{v'_{4k}})=\calO(e^{-\frac12 h_{\ell'}})=\calO(\frac{1}{n}(C')^{\frac{1}{1-\alpha}})$. Thus,

\begin{equation}\label{cover:eqn:exposed}
\PP(V\cap (\calR_a^{\mathrm{pt}}\setminus\calB_a)\cup ((B_{v'_4}(R)\cup B_{v'_{4k}}(R) \setminus \calR_O)\setminus\calB_a)=\emptyset)
\end{equation}

Now, let $H$ be the half-tile to which $v'_{1}$ belongs. Since $v'_{j}\in B_O(h_{\ell'})\setminus B_O(h_{\ell'}{-}\delta)$, the half-tile $H$ must
be at level $\ell'$. Next, we recursively specify a sequence of half-tiles of $\calF(c)$. First, we
set $s=\ell'$ and let $H^{(1)}_{s}$ be the half-tile that is contiguous to $H$ obtained by rotating the latter clockwise by the adequate angle (i.e., by $\frac12\theta_{\ell'}$).  Giving a half-tile $H^{(i)}_{s}$ at level $s$ of $\calF(c)$, $i\in [5]$, let $H^{(i+1)}_{s}$ be the half-tile, also at level $s$, that is contiguous to $H^{(i)}_{s}$ and is again obtained by rotating the latter clockwise by the adequate angle (i.e., by $\frac12\theta_s$). Next, let $H^{(1)}_{s-1}$ be the parent half-tile of $H^{(6)}_{s}$. Note that $H^{(1)}_{s-1}$ is at level $s-1$. The process is repeated with $s-1$ instead of $s$ until reaching level $\ell$ where as usual $\ell$ is the largest integer such that $h_{\ell}\leq\rho$ for $\rho$ defined in (<ref>) (i.e., $\rho:=R-\frac{1}{1-\alpha}\ln(\frac{CR}{\nu})$). See Figure <ref> (c).

Let $\calE_a$ be the event such that for all $s\in\{\ell,...,\ell'\}$ and $i\in [6]$ the half-tile $H^{(i)}_s$ contains at least one vertex of $\HRG_{\alpha,\nu}(n)$. By Remark <ref>, if $\calE_a$ occurs, then $v_a:=v'_{1}$ is connected via a path to a vertex, say $u_a\in V\cap H_{\ell}^{(1)}$, and since Lemma <ref> implies that, a.a.s., all vertices in $B_O(\rho)$ belong to $\calC_{\alpha,\nu}(n)$, so does $v_a=v'_{1}$ and hence also the vertices of $\calP_a$. Before calculating the probability that $\calE_a$ occurs, we need to make sure that it does not involve regions already exposed. For this, it will suffice
to show that the $H_s^{(i)}$'s do not intersect $B_{v'_4}(R)$.
By construction the angle spanned at the origin by $v'_4$ and a point in $\bigcup_{i\in [6]} H^{(i)}_{\ell'}$ is
at least $\frac{9}{4}\theta_{\ell'}= \frac{9}{8}\theta_{\ell'-1}=\frac94(1+o(1))e^{\frac12(R-2h_{\ell'-1})}$ where for the last equality we have used Lemma <ref>. Using said lemma again,
we get that $\frac98\theta_{\ell'}=\frac98(1+o(1)) e^{\frac12(h_{\ell'}-\delta-h_{\ell'-1})}\theta_R(h_{\ell'-1},h_{\ell'}-\delta)\geq \theta_R(h_{\ell'-1},h_{\ell'}-\delta)$ (the last inequality is obtained recalling that $h_{\ell'}\geq h_{\ell'-1}+\ln 2-\epsilon$, and because of our choice of $\epsilon$ and $\delta$).
Thus, by definition of $\theta_R(\cdot,\cdot)$, the intersection  $B_{v'_4}(R)\cap\bigcup_{i\in [6]} H^{(i)}_{\ell'}$
is empty.
For values of $s<\ell'$, we have that the angle spanned by any point in $\bigcup_{i\in [6]} H^{(i)}_{s}$
and $v'_{4}$ is at least

\begin{align*}
\frac{9}{4}\theta_{\ell'}
& > 2\theta_{\ell'}\Big(1+\sum_{j=0}^{\ell'-s-1}2^j\Big)=\theta_{\ell'}2^{\ell'-s+1}\geq\theta_{s-1}.\end{align*}

By definition of $\theta_{s-1}$, Remark <ref>, and since $s< \ell'$, it follows that  $\theta_{s-1}=\theta_R(h_{s-1},h_{s-1})\geq\theta_R(h_{s-1},h_{\ell'-1})$,
so we conclude that the intersection
$B_{v'_4}\cap\bigcup_{i\in [6]} H^{(i)}_s$ is empty as we had claimed.
We now determine the probability of the event $\calE_a$.
Because the $H_s^{(i)}$'s are disjoint, independence and standard arguments imply that

\begin{align*}
\PP(\calE_a) & = \prod_{s=\ell}^{\ell'}\prod_{i\in [6]}\big(1-\exp\big({-}\EE|V\cap H_{s}^{(i)}|\big)\big).
\end{align*}

Since the expected number of vertices in a tile is twice that of a half-tile of the same level, by  Claim <ref>, we have
$\EE|V\cap H_{s}^{(i)}|\geq \frac{1}{4}\nu e^{(1-\alpha)(R-h_s)}$. Thus, since $h_{\ell'}\leq\rho'$, by definition of
$\rho'$, we get $\EE|V\cap H_s^{(i)}|\geq \frac14\nu e^{(1-\alpha)(R-h_{\ell'})}\geq \frac12 C'$.
So, recalling that $\ln(1-x)\geq -\frac{x}{1-x_0}$ for $0\leq x\leq x_0<1$ and provided $C'\geq\ln 4$,

\[
\ln\PP(\calE_a) \geq -2\sum_{s=\ell}^{\ell'}\exp\big({-}\tfrac{1}{4}\nu e^{(1-\alpha)(R-h_s)}\big)=
\]

By Claim <ref>, $h_{\ell'}-h_s\geq (\ell'-s)\ln 2-\epsilon$, so taking $\gamma:=\frac{1}{2}C' e^{-(1-\alpha)\epsilon}$ and $\beta:=(1-\alpha)\ln 2\geq \frac12(1-\alpha)$,

\[
\sum_{s=0}^{\ell-\ell'}\exp\big(-\gamma e^{\beta s}\big) \leq e^{-\gamma}+\tfrac{1}{\beta\gamma}e^{-\gamma}
\leq \tfrac{2}{1-\alpha}\big(1+\tfrac{1}{\gamma}\big)e^{-\gamma}
\]

Taking $C'$ a large constant we can make $\gamma$ large and $\xi$ as small as needed, say small
enough so that
$\PP(\calE_a)\geq e^{-2\xi}\geq\frac12$ (any constant instead of $\frac12$ will suffice).
Summarizing, for $C'$ large enough,

\begin{equation}\label{cover:eqn:connect}
\PP(\text{vertex $v_a=v'_1$ belongs to $\calC_{\alpha,\nu}(n)$})\geq \frac12.
\end{equation}

On the other hand, we have that all $H_s^{(i)}$'s are contained within a sector contiguous to $\Upsilon_a^{\mathrm{pt}}$ of angle

\[
6\sum_{j=\ell}^{\ell'}\frac{\theta_{j}}{2} = 3\theta_{\ell'}\sum_{j=0}^{\ell'-\ell}2^j
\leq 3\theta_{\ell'}2^{\ell'-\ell+1}=6\theta_\ell.
\]

Let $\calR_a^{\mathrm{ct}}:=\Upsilon^{\mathrm{ct}}_a\setminus\calR_O$ where $\Upsilon^{\mathrm{ct}}_a$ is a sector clockwise contiguous to $\Upsilon^{\mathrm{pt}}_a$ which is
guaranteed to contain all $H_s^{(i)}$'s for every
$s\in\{\ell,...,\ell'\}$ and $i\in [6]$, and moreover contains $B_{v'_4}(R)\setminus\calR_O$.
For this to hold, it suffices to let $\Upsilon_a^{\mathrm{ct}}$ be
the sector clockwise contiguous to $\Upsilon^{\mathrm{pt}}_a$
with central angle $\max\{6\theta_{\ell},\theta_R(h_{\ell'}-\delta,\rho_O)\}$.

Next, we will fix the $\eta$ in the definition of $\rho_O$
so that $6\theta_{\ell}\leq\theta_R(h_{\ell'}-\delta,\rho_O)$ and set
the central angle of the sector
$\Upsilon_a$ to $3k\theta_{\ell'}+2\theta_{\ell}(h_\ell'-\delta,\rho_O)$ so $\Upsilon_a^{\mathrm{bf}}\cup\Upsilon_a^{\mathrm{pt}}\cup\Upsilon_a^{\mathrm{ct}}\subseteq\Upsilon_a$. Recalling that $\rho_O=(1{-}\frac{1+\eta}{2\alpha})R$, for any $\zeta,\chi>0$ such that $0<\zeta+\chi<1-\frac{1}{2\alpha}$, we can take $\eta>0$ small enough so that $\theta_R(h_{\ell'}-\delta,\rho_O)=\calO(n^{-(\zeta+\chi)})$.
Also, for some small constant $\widetilde{\chi}>0$ which we will fix later, if we set $k=\widetilde{\chi}\ln n$ we have
$k\theta_{\ell'}=\calO(n^{-1}\log n)$.
Moreover, note that $\theta_{\ell}=\calO(n^{-1}(\log n)^{\frac{1}{1-\alpha}})$.
Thus, we need to make sure that the central angle of $\Upsilon_a$
is at least  $C''n^{-(\zeta+\chi)}$
for some appropriately chosen large constant $C''$.
Hence, we may choose $m=\lfloor \frac{2\pi}{C''}n^{\zeta+\chi}\rfloor$, partition $B_O(R)$ into
$m$ sectors $\Upsilon_1,...,\Upsilon_m$ so within $\Upsilon_a$ we can accommodate
It follows that $\calR_a:=(\Upsilon_a\cap B_{O}(R))\setminus\calR_O$ contains $\calR_a^{\mathrm{ct}}\cup\calR_a^{\mathrm{pt}}\cup\calR_a^{\mathrm{bf}}$.

To conclude, let
$X_a$ be the indicator of the following event: there is a vertex in the subgraph of $\calC_{\alpha,\nu}(n)$ induced by $V\cap\calR_a$ whose removal gives rise to a path component on $4k-3=\Omega(\log n)$ vertices.
Let $X_a$ be the indicator of the latter event, and let $X:=\sum_{a\in [m]}X_a$.
From (<ref>)-(<ref>) we conclude that $\EE X_a=e^{-\calO(k)}$ and so we can choose $\widetilde{\chi}$ sufficiently
small so that

\[
\EE X=m\cdot e^{-\calO(k)}=e^{(\zeta+\chi)\ln n}\cdot e^{-\calO(\widetilde{\chi}\ln n)}=\Omega(n^{\zeta}).
\]

By Chernoff bounds (see <cit.>), $\PP(X\leq\frac12\EE X) =\exp(-\Omega(n^{\zeta}))$.

We now have all the necessary ingredients to derive the lower bounds matching the bounds in Proposition <ref>. Together these two propositions prove Theorem <ref>.

If $\calC:=\calC_{\alpha,\nu}(n)$, then a.a.s. $t_{\mathsf{hit}}(\calC)=\Omega(n\ln n)$ and $t_{\mathsf{cov}}(\calC)=\Omega(n\ln^2 n)$.

The lower bound on $t_{\mathsf{hit}}(\calG)$ is a direct consequence of the claimed lower bound on $t_{\mathsf{cov}}(\calG)$
and Matthew's bound, so we concentrate on proving the latter one.

Let $\zeta$ and
$\Lambda:=\Lambda(\zeta)$ be as in
the statement of Lemma <ref>. Let $U\subseteq V(\calC)$ be the collection of vertices $u$
of $\calC$ which have degree $2$ in $\calC$ and are end vertices of a maximal dangling path $P_u$ of $\calC$ of length at least
Now, let $u,v\in U$, $u\neq v$,
and contract into a single vertex all vertices
of $\calC$ except for those that
belong to $P_{u}$ or $P_{v}$.
One obtains a path of length at least $2\Lambda\ln(n)$.
Rayleigh's Monotonicity Law (RML) (Theorem <ref>)
implies that the contraction of edges does
not increase the effective resistance between any of the remaining pair of vertices. Since the effective resistance between the endpoints of a path with $\ell$ edges is $\ell$ (this follows directly from our working definition of effective resistance – see (<ref>)), by RML we get that $\Res{u}{v}\geq 2\Lambda\ln(n)$ for all distinct $u,v\in U$.
Since $\Exu{u}{\tau_{v}}+ \Exu{v}{\tau_{u}}=2|E(\calC)|\cdot \Res{u}{v}$ by Lemma <ref>,

\[
\kappa_{U} := \min_{u,v\in U}[\Exu{u}{\tau_{v}}+ \Exu{v}{\tau_{u}}]\geq 2\Lambda|E(\calC)|\cdot\ln(n).
\]

By Theorem <ref>, we know that a.a.s. $\calC$ is of order
$\Theta(n)$, and since $\calC$ is connected, we also have that a.a.s. $|E(\calC)|=\Omega(n)$.
Thus, we conclude that $\kappa_{U}=\Omega(n\ln n)$ a.a.s., so the sought after lower bound follows directly
from Lemma <ref> and Theorem <ref>.

We also remark that Proposition <ref> shows that the maximum effective resistance between any two vertices in the center component is a.a.s. $\Theta(\log n)$.

§ COMMUTE TIMES

In previous sections we analyzed parameters of simple random walks on the giant of the HRG, with a focus on parameters which depend on the graph as a whole and not the start and end vertices of walks. In this section we give  (under mild assumptions) tight estimates (up to polylogarithmic factors) for the commute time between a pair $u, v$ of vertices added to $\HRG_{\alpha,\nu}(n)$. The estimates depend on the radius of the added vertices.

We address the upper and lower bounds in two separate sections below. First, we establish the upper bound via the commute time identity and bounding effective resistances from above via the energy dissipated by a specially crafted network flow.
Then, we consider the lower bound, establishing it by exhibiting small edge-cuts whose removal leaves $u$ and $v$ in distinct connected components.
Put together, these two bounds yield Theorem <ref>.

§.§ Upper bound

Let $\HRG:=\HRG_{\alpha,\nu}(n)$ and consider $\rho$ as given by (<ref>).
Fix two points in $B_O(\rho)$ and
add to $\HRG$ vertices $s$ and $t$ at the previously chosen points (one vertex per point).
By Lemma <ref> and Proposition <ref>, a.a.s., the effective resistance between $s$ and $t$ is $\calO(1)$. However, the proof of Proposition <ref> shows something stronger.
Specifically, that for $s,t\in V(\HRG)\cap (B_{O}(\rho)\setminus B_{O}(\frac{R}{2}))$ it holds that
On the other hand, it is well known (and follows easily from Lemma <ref>) that the degree $d(w)$ of  a vertex $w\in V(\HRG)\cap B_O(\rho)$ is, w.h.p., $\Theta(e^{\frac12(R-r_w)})$.
Summarizing, we have that if $s,t\in V(\HRG)\cap (B_{O}(\rho)\setminus B_{O}(\frac{R}{2}))$, then a.a.s.,[If $w\in\{s,t\}$ is in $V(\calG)\cap B_O(\frac{R}{2})$, the proof of Proposition <ref> establishes the claim but with $d(w)$ replaced by $\max\{d(w),\sqrt{n}\}$. However, this is an artifact of the simplified way we defined the flow $f_{s,t}$ in Section <ref> which sufficed for proving the aforementioned result.]

\[
\Res{s}{t} =
\calO(d(s)^{-2(1-\alpha)}+d(t)^{-2(1-\alpha)}).
\]

The goal of this section is to strengthen this bound.

We start by giving a high level overview of the argument used to prove the upper bound, this relies on building a unit $(s,t)$-flow $f_{s,t}$ which yields an upper bound for $\Res{s}{t}$. Observe that if only $m$ of the edges incident to either $s$ or $t$ carry non-zero flow, then it is easy to see that $\Res{s}{t}=\Omega(1/m)$, or see the Nash-Williams inequality <cit.>. On the other hand, the further away from the boundary (closer to the origin of $B_O(R)$) the larger (in expectation) number of neighbors both $s$ and $t$ have (in fact, in the HRG, a constant fraction of the neighbors of a vertex are within constant distance from the boundary of $B_O(R)$ and this number decreases exponentially fast as a function of the distance to the boundary). This suggests that a `good' flow should start by pushing flow towards the neighbors of $s$ close to the boundary. However, the closer to the boundary the neighbors are, the smaller the angle of the sector that contains them. In order for the flow to dissipate little energy it can not traverse exceedingly large paths. However, in the HRG, geodesic paths tend to approach the origin (more so the larger the angle formed at the origin between the endvertices of the path). Unfortunately, the paths starting at vertices within a sector of small central angle merge as they approach the origin. So, there is a trade-off in designing low energy unit $(s,t)$-flows; route flow using as many neighbors of vertices $s$ and $t$ versus  distribute flow in a sector of sufficiently large central angle. Our construction relies on using neighbors of $w\in\{s,t\}$ whose radial distance is roughly $R-(2\alpha-1)(R-r_w)$. Flow is sent from neighbors of $s$ towards the origin until it all reaches a single half-tile (we do the same for $t$ but in reverse direction). Finally, we route the flow between the relevant pair of half-tiles using the adaptation of the flow of Section <ref> discussed in Remark <ref>.

Formally, let $\calF(c)$ be the tiling determined by $c>0$ (see Section <ref>).
Assume $w\in\{s,t\}$ belongs to a tile of level $\ell_w$ of the tiling $\calF(c)$ where $(1-\frac{1}{2\alpha})R\leq r_w<\rho$.
As usual, fix $\ell$ with respect to $\rho$ as in (<ref>). In particular, recall that
Furthermore, let

\begin{equation}\label{resist:eqn:deflPrime}
\ell'_w:=\min\big\{
\ell+1,
\min\{ \ell'\in\NN \mid R-h_{\ell'} \leq (2\alpha-1)(R-r_w)\big\}.
\end{equation}

Since $h_i\to\infty$ as $i\to\infty$, it follows that $\ell'_w$ is well-defined.
Moreover, we claim that $r_w\leq h_{\ell'_w}$. Indeed,
the claim is obvious if $\ell'_w=\ell+1$ since $r_w<\rho< h_{\ell+1}$.
If $\ell'_w\neq\ell+1$, then
$R-h_{\ell'_w}\leq (2\alpha-1)(R-r_w)< R-r_w$, so again the claim holds.
A direct consequence of our previous observation is that $\ell_w\leq\ell'_w\leq\ell+1$.
Next, define

\begin{equation}\label{resist:eqn:defk}
\max\{ k\in\NN \mid k\leq\ell'_w \text{ and } \theta_R(h_{\ell'_w- k},h_{\ell'_w-k})\leq \theta_R(r_w,h_{\ell'_w})\}.
\end{equation}

Note that $k_w$ is well-defined since
$r_w\leq h_{\ell'_w}$
and so, by Remark <ref>, we know that,
Also, observe that $k_w\leq\ell'_w-\ell_w$, since otherwise
contradicting that $r_w\geq h_{\ell_w-1}$, $h_{\ell'_w}\ge h_{\ell_w-1}$.
This and Remark <ref> imply that $\theta_R(h_{\ell_w-1},h_{\ell_w-1})>\theta_R(r_w,h_{\ell'_w})$.

For future reference, we establish the following fact relating $k_w$, $\ell'_w$ and $r_w$.

If $(1-\frac{1}{2\alpha})R< r_w<\rho$,  then

First, we observe that $h_{\ell'_w}\geq\frac{R}{2\alpha}$.
Indeed, if $\ell'_w=\ell+1$, then $h_{\ell'_w}=\rho$ and the claim follows by our choice of $\rho$.
By hypothesis on $r_w$, if $\ell'_w\neq \ell+1$, then
$R-h_{\ell'_w}\leq (2\alpha-1)(R-r_w)\leq (1-\frac{1}{2\alpha})R$, and the claim immediately follows.

Next, we argue that $k_w\neq\ell'_w$.
Otherwise, by definition of $k_\omega$, we would have $\pi=\theta(h_0,h_0)\leq\theta_R(r_w,h_{\ell'_w})$, implying that $r_w+h_{\ell'_w}\leq R$.
However, by hypothesis and the observation from the previous paragraph, $r_w+h_{\ell'_w}>R$, a contradiction.

Now, by definition of $h_i$, we know that $\theta_R(h_i,h_i)=\frac12\theta_R(h_{i-1},h_{i-1})$ for all $i>1$.
By choice of $\ell'_w$ and $k_w$
we have $\theta_R(h_{\ell'_w-k_w},h_{\ell'_w-k_w})\leq\theta_R(r_w,h_{\ell'_w})
Lemma <ref> and Claim <ref> yield
that $\theta_R(h_{\ell'_w-k_w},h_{\ell'_w-k_w})=\Theta(\theta_R(h_{\ell'_w-k_w-1},h_{\ell'_w-k_w-1}))$
and we get $\theta_R(h_{\ell'_w-k_w},h_{\ell'_w-k_w})=\Theta(\theta_R(r_w,h_{\ell'_w}))$.
Summarizing, $2^{k_w}\theta_R(h_{\ell'_w},h_{\ell'_w})=\Theta(\theta_R(r_w,h_{\ell'_w}))$.
The result follows applying Lemma <ref> twice.

Now, let $T_w$ be the tile at level $\ell'_w-k_w$ that intersects the ray starting at the origin $O$ and passing through the point where $w$ is located. Do not confuse $T_w$ with the tile containing $w$ (which in previous sections was denoted $T(w)$), they could be the same tile, but typically are not.
Also, let $H_w$ be the halft-tile of $T_w$ intersecting the ray starting at the origin $O$ and passing through $w$.
Our next result shows that all level $\ell'_w$ descendants of $T_w$ are contained in $B_w(R)$.

If $(1-
\frac{1}{2\alpha})R\leq r_w<\rho$,  then every tile in $\pi^{-k_w}(\{T_w\})$ is a level $\ell'_w$ tile contained in $B_w(R)$.

Consider an arbitrary tile $T\in\pi^{-k_w}(\{T_w\})$.
Since $T$ is a $k_w\textsuperscript{th}$ generation descendant of the level $\ell'_w-k_w$ tile $T_w$, it follows that $T$ is at level $(\ell'_w-k_w)+k_w=\ell'_w$. This proves the first part of the lemma. To establish the second part,
let $\Upsilon_{T_w}$ be the smallest sector containing $T_w$. Thus, the central angle of $\Upsilon_{T_w}$ is $\theta_{\ell'_w-k_w}\leq\theta_R(h_{\ell'_w-k_w},h_{\ell'_w-k_w})$.
Consider the sector $\Upsilon_w$ whose central angle is double that of $\Upsilon_{T_w}$, but whose bisector is the ray starting from the origin $O$ and passing through the point where vertex $w$ was added.
Note that $\Upsilon_{T_w}\subseteq\Upsilon_w$ and thus $T\subseteq\Upsilon_w\cap B_{O}(h_{\ell'_w})$.
Hence, every point $q\in T$ is within an angle at most
$\theta_R(h_{\ell'_w-k_w},h_{\ell'_w-k_w})\leq\theta_{R}(r_w,h_{\ell'_w})$ of the bisector of $\Upsilon_w$.
By Remark <ref>, since $r_q<h_{\ell'_w}$ (because $q$ belongs to the level $\ell'_w$ tile $T$), we know that $\theta_R(r_w,h_{\ell'_w})\leq\theta_R(r_w,r_q)$ implying that $q\in B_w(R)$ as claimed.

We now construct a $(\{w\},V\cap H_w)$-flow $f_{w,V\cap H_w}$ in $\calG=(V,E)$.
The idea of the construction is to distribute one unit of flow from $w$ among all vertices located in a half-tile
$H$ which is both a descendant of $H_w$
and belongs
$\pi^{-k_w}(\{T_w\})$ so that each such half-tile $H$
receives $1/2^{k_w}$ units of flow, and so the flow ends up evenly spread among the vertices
in $V\cap H$ (note however that since distinct half-tiles might contain distinct number of vertices,  vertices belonging to distinct half-tiles might receive different amounts of flow).
Subsequently, we initially set $s:=0$ and for every pair of twin half-tiles descendants of $H_w$ whose union is a tile $T\in \pi^{-(k_w-s)}(\{T_w\})$ we re-distribute (re-balance) flow among all vertices in $V\cap T$ so each one ends up with exactly the same amount of flow.
We then spread all the flow at a vertex in $V\cap T$  evenly among its neighbours in the parent half-tile of $T$ (by construction of the tiling, every vertex in the parent half-tile of $T$ is a neighbour of vertices in $V\cap T$).
Provided $s<k_w$, the  process is repeated,  but replacing $s$ by $s+1$ until $s$ becomes $k_w$.
Formally, for every half-tile $H$ which is both a descendant of $H_w$ and belongs to a tile in $\pi^{-k_w}(\{T_w\})$,

\[
f_{w,H_w}(\vec{wv}) := \frac{1/2^{k_w}}{|V\cap H|}\qquad
\text{for all $v\in V\cap H$.}
\]

Moreover, for every tile $T$ which is a descendant of $T_w$ whose level is $\ell'_w-k_w+s$ where $0\leq s< k_w$,
if $H$ is a half-tile of $T$ such that $|V\cap H|<|V\cap (T\setminus H)|$, then

\[
f_{w,H_w}(\vec{uv}) :=
\Big(\frac{1/2^{s}}{|V\cap H|}-\frac{1/2^{s}}{|V\cap(T\setminus H)|}\Big)\cdot\frac{1}{|V\cap T|}
\qquad \text{if $u\in V\cap H$ and $v\in V\cap (T\setminus H)$}.
\]

Finally, if $T$ is a descendant of tile $T_w$ whose level is $\ell'_w-k_w+s$ where $1\leq s\leq k_w$ and $H$ is the parent half-tile of $T$, then

\[
f_{w,H_w}(\vec{vw}) := \frac{1/2^{s-1}}{|V\cap T|}\cdot\frac{1}{|V\cap H|}
\qquad
\text{for all $v\in V\cap T$ and $w\in V\cap H$.}
\]

The mapping $f_{w,H_w}$ is extended to all of $\vec{E}(\calG)$ so it satisfies the antisymmetric property of flows, i.e. $f_{w,H_w}(\vec{uv})=-f_{w,H_w}(\vec{vu})$, and so it is null for all remaining oriented edges $\vec{uv}\in \vec{E}(\calG)$ where $f_{w,H_w}(\vec{uv})$ remains undefined.

The following result establishes several of the properties that hold for $f_{w,H_w}$.

Let $\epsilon, c:=c(\epsilon)$ be as in Claim <ref> and consider the tilling $\calF(c)$.
If none of the tiles of $\calF(c)$ intersecting $B_O(\rho)$ is faulty and a vertex $w$ is added to $\HRG:=\HRG_{\alpha,\nu}(n)$ at a point located within $B_O(\rho)$, then $f_{w,H_w}$ is an $\calF(c)$-compatible balanced unit $(\{w\},V\cap H_w)$-flow in $\HRG$.

\[
\calE(f_{w,H_w}) = \calO\big(\min\big\{e^{-2\alpha(1-\alpha)(R-r_w)},\tfrac{1}{\ln n}\,e^{-(1-\alpha)(R-r_w)}\big\}\big).
\]

The proof that $f_{w,H_w}$ is an $\calF(c)$-compatible balanced unit flow is similar to the proof of Lemma <ref> where the same property was shown but for another flow and hence we omit it.
To show the bound on the energy dissipated by the flow, denote by $\pi^{-p}(\{H_w\})$ the collection  of $p^\textsuperscript{th}$ generation half-tiles descendants of $H_w$.
Also, abbreviate half-tile parent as h-t.prnt.
Say $\{H,H'\}$ is a half-tile partition of tile $T$ (abbreviated h-t.part. of $T$) if $H, H'$ are twin half-tiles of $T$.
By definition of $f_{w,H_w}$,

\begin{align}\label{resist:eqn:energyUBnd}
\calE(f_{w,H_w})
& = \sum_{\text{$H\in\pi^{-k_w}(\{H_w\})$}}\frac{(1/2^{k_w})^2}{|V\cap H|} + \sum_{p=0}^{k_w-1}\, \sum_{\text{$H\in \pi^{-p}(\{H_w\})$}\atop \text{h-t.prnt.~of $T$}}\frac{(1/2^{p})^2}{|V\cap T|}\cdot\frac{1}{|V\cap H|}\\[3pt]
\nonumber   & \qquad + \sum_{p=0}^{k_w-1} \,
\sum_{\text{$\{H,H'\}$ h-t.part.~of $T$}\atop\text{s.t.~$H, H'\in\pi^{-{p+1}}(\{H_w\})$}}
\Big(\frac{1/2^{p}}{|V\cap H|}-\frac{1/2^{p}}{|V\cap H')|}\Big)^2\cdot\frac{|V\cap H|\cdot|V\cap H'|}{|V\cap T|^2}.
\end{align}

In order to bound each of the summations above recall that
since $w\in B_O(\rho)$, it holds that $\ell'_w\leq\ell+1$.

To upper bound the first summation in (<ref>), observe that the $2^{k_w}$ half-tiles in $\pi^{-k_w}(\{H_w\})$ intersect $B_O(\rho)$ so, by hypothesis, they  are sparse.
Using Claim <ref>, it follows that the first summation is $\calO(2^{-k_w}e^{-(1-\alpha)(R-h_{\ell'_w})})$.

Consider next the second summation in (<ref>).
Observe that there are $2^p$ distinct tiles $T$ whose half-tile parent $H$ belongs to $\pi^{-p}(\{H_w\})$  and that all of them are at level $\ell'_w-k_w+p+1$, hence again each of them intersects $B_O(\rho)$.
Thus, neither $T$ is faulty nor $H$ is sparse (by hypothesis).
Using again Claim <ref>
and the fact that $h_{\ell'_w-k_w+p}=h_{\ell'_w}-(k_w+p)\ln 2+\Theta(1)$ (see Claim <ref>), it follows that the
second summation in (<ref>) is

\begin{align*}
\Theta\Big(\sum_{p=1}^{k_w}2^{-p}e^{-2(1-\alpha)(R-h_{\ell'_w-k_w+p})}\Big)
& = \Theta\Big((2^{k_w}e^{R-h_{\ell'_w}})^{-2(1-\alpha)}\sum_{p=1}^{k_w}2^{-2(\alpha-\frac12)p}\Big) \\
& = \calO\big((2^{k_w}e^{R-h_{\ell'_w}})^{-2(1-\alpha)}\big).
\end{align*}

Finally, consider the third summation in (<ref>).
As for the previous two summations, all the tiles and half-tiles involved intersect $B_O(\rho)$ so they are non-faulty (by hypothesis).
Applying Claim <ref> as in the previous cases, we get that this last summation is of order at most  the second one.

\[
\calE(f_{w,H_w}) = \calO\big(e^{\alpha(R-h_{\ell'_w})}/(2^{k_w}e^{R-h_{\ell'_w}})
\]

Fact <ref> and the previous displayed equation imply that

\[
\calE(f_{w,H_w})=\calO\big(e^{(\alpha-\frac12)(R-h_{\ell'_w})}/e^{\frac12(R-r_w)}+
\]

To conclude, note that by the definition of $\ell'_w$
(see (<ref>)) and our estimates for the $h_i$'s (see Claim <ref>),
either $R-h_{\ell'_w}=(2\alpha-1)(R-r_w)+\Theta(1)$
or $h_{\ell'_w}=\rho+\Theta(1)$.
Replacing in the last displayed equation, for the former case, we get
$\calE(f_{w,H_w})=\calO(e^{-2\alpha(1-\alpha)(R-r_w)})$ and for the latter
$\calE(f_{w,H_w})=\calO(\tfrac{1}{\ln n}\,e^{-(1-\alpha)(R-r_w)}\big)$.

The following result gives the sought after upper bound.

Let $\epsilon, c:=c(\epsilon)$ be as in Claim <ref>. Then, for every $d>0$ there is a sufficiently large $C>0$ such that for a pair of vertices $s,t$ added to $\HRG:=\HRG_{\alpha,\nu}(n)$ at points located within $B_O(\rho)\setminus B_O((1-\frac{1}{2\alpha})R)$ where $\rho:=\rho(C)$ is as in (<ref>),
with probability $1-o(1/n^d)$,
both $R-r_s$ and $R-r_t$ are at least
$(R-\rho)/(2\alpha-1)$, then

\[
\Res{s}{t}
\]

If $w\in\{s,t\}$ is such that
$w\in B_O(\rho)$ and $(2\alpha-1)(R-r_w)<R-\rho$,
then the claim holds but with $1/d(w)^{4\alpha(1-\alpha)}$ above replaced by $\frac{1}{\ln n} \cdot 1/d(w)^{2\alpha}$.

By definition of effective resistance, it suffices to show that with probability $1-o(1/n^d)$ there is an $f_{s,t}$ unit $\calF(c)$-compatible $(s,t)$-flow in $\HRG$ such that $\calE(f_{s,t})=\calO(d(s)^{-4\alpha(1-\alpha)}+d(t)^{-4\alpha(1-\alpha)})$.

By Lemma <ref>, there is a sufficiently large constant $C>0$ such that for $\rho:=\rho(C)$ as defined in (<ref>), with probability $1-o(1/n^d)$, every tile in $\calF(c)$ intersecting $B_O(\rho)$ is non-faulty.
By hypothesis, if $w\in\{s,t\}$, then $w\in B_O(\rho)$.
Thus, there is a $f_{w,H_w}$ flow in $\calG$ satisfying
the properties guaranteed by Lemma <ref>.

By Remark <ref> (following Proposition <ref>) and since $H_w$ intersects $B_O(\rho)$ and is at level $\ell'_w-k_w$,  there is an $\calF(c)$-compatible balanced unit $(H_s,H_t)$-flow $f_{H_s,H_t}$ in $\calG$ such that

\[
\calE(f_{H_s,H_t}) = \calO(e^{-2(1-\alpha)(R-h_{\ell'_s-k_s})}+e^{-2(1-\alpha)(R-h_{\ell'_t-k_t})}).
\]

Assuming $k_w=\ell'_w$,
since $R-h_0=\frac{R}{2}>\alpha(R-r_w)$ if and only if $r_w>(1-\frac{1}{2\alpha})R$, by hypothesis,

\[
< e^{-2\alpha(1-\alpha)(R-r_w)}.
\]

Assuming $k_w<\ell'_w$, then
by Claim <ref>, we have $h_{\ell'_w-k_w}=h_{\ell'_w}-k_w\ln 2+\Theta(1)$
and thus,
using Fact <ref>,

\[
= \Theta\big((e^{R-r_w}e^{R-h_{\ell'_w}})^{-(1-\alpha)}\big).
\]

Recalling that $R-h_{\ell'_w}=(2\alpha-1)(R-r_w)+\Theta(1)$ or
In the former case (the other case is handled similarly and left to the reader) we get that

\[
= \Theta\big(e^{-2\alpha(1-\alpha)(R-r_w)}\big).
\]

\begin{equation}\label{resist:eqn:energyUBnd2}
\calE(f_{H_s,H_t})=\calO(e^{-2\alpha(1-\alpha)(R-r_s)}+e^{-2\alpha(1-\alpha)(R-r_t)}).
\end{equation}

To conclude, let $f'_{H_t,t}$ be the flow in $\calG$ obtained from $f_{t,H_t}$ by reversing the flows along all edges of $\calG$, that is, $f'_{H_t,t}(\vec{uv})=-f_{t,H_t}(\vec{uv})$ for all $\vec{uv}\in\vec{E}(\calG)$.
It is easy to check that $f'_{H_t,t}$ is indeed a $(V\cap H_t,\{t\})$-flow, actually an $\calF(c)$-compatible balanced unit flow.
By Claim <ref>, we get that
$f_{s,t}:=f_{s,H_s}+f_{H_s,H_w}+f'_{H_t,t}$ is
a unit $(s,t)$-flow in $\calG$ such that

\[
\calE(f_{s,t}) = \calO\big(\calE(f_{s,H_s})+\calE(f_{H_s,H_t})+\calE(f'_{H_t,t})\big)
= \calO(e^{-2\alpha(1-\alpha)(R-r_s)}+e^{-2\alpha(1-\alpha)(R-r_t)}).
\]

where the last equality is a consequence of (<ref>)
and Lemma <ref>.
The desired conclusion follows from the well known fact (which is an immediate consequence of Lemma <ref>) that the degree $d(w)$ of every vertex $w\in V(\HRG)\cap B_O(\rho)$ is, w.h.p., $\Theta(e^{\frac12(R-r_w)})$.

The preceding result together with the commute time identity (see Lemma <ref>) and
the fact that the number of edges of the center component of $\calG_{\alpha,\nu}(n)$ is $\Theta(n)$ (see Lemma <ref>) yield the  upper bound of Theorem <ref>.

§.§ Lower bound

As in the previous section, let $\HRG:=\HRG_{\alpha,\nu}(n)$, consider $s, t\in V(\calG)\cap B_O(R)$ and let $\rho$ be as in (<ref>). The goal of this section is to obtain a lower bound for effective resistances matching, within polylogarithmic factors, the upper bound of Proposition <ref>.

One classical approach to derive a lower bound on $\Res{s}{t}$ is
to observe that the cardinality of a subset $F$ of the set of
edges of $\HRG$ whose
removal leaves $s$ and $t$ in separate connected components gives
a lower bound for $\Res{s}{t}$.
Indeed, the energy dissipated by any unit $(s,t)$-flow must be at least $1/|F|$, so by definition of effective resistance $\Res{s}{t}\geq 1/|F|$, see the Nash-Williams inequality <cit.>.
Clearly, the smaller the size of $F$ the better the lower bound.
To achieve this section's goal, we thus focus on finding small $(s,t)$-edge-cuts in $\HRG$.
We start by identifying small edge-cuts that would isolate a vertex $v$ added to $\HRG$ at a point $p\in B_O(R)$ from some vertex in the center component of $\HRG$.

Throughout our ensuing discussion we assume the radial coordinate of $p$ is $r_p=r$.
Also, we let $\omega:=\omega(n)$ be a positive function going to infinity arbitrarily slowly and
define $\phi_r:=2\pi\nu e^{-\omega}/(n\mu(B_O(r)))$.

If $(1-\frac{1}{2\alpha})R\leq r<R$, then

Direct consequence of our choice of $\phi_r$, the hypothesis on $r$ and Lemma <ref>.

Let $\Upsilon_p$ be the sector of central angle $\phi_r$ on whose bisector $p$ lies.
Let $V:=V(\calG)$ and $v$ be a vertex added at $p$.
We are interested in determining a good upper bound (that holds a.a.s.) for
the edge-cut in $\calG$ induced by $(\{v\}\cup V)\cap\Upsilon_p$, that is,
if for a graph $G$ and a set of its nodes $S\subseteq V(G)$ we define
$\delta_G(S):=\{uv\in E(G) \mid u\in S, v\in V(G)\setminus S\}$, then we seek a
good upper bound for $|\delta_{\calG}((\{v\}\cup V)\cap\Upsilon_p)|$.
The following result reduces (under some mild conditions) the task of bounding
the edge-cut induced by $(\{v\}\cup V)\cap\Upsilon_p$ to the one of bounding
the edge-cut induced by $V\cap\Upsilon_p$.

If $p\in B_O(R)$ is such that $(1-\frac{1}{2\alpha})R<r_p\leq R-\omega/(1-\alpha)(\alpha-\frac12)$,
then a.a.s.,

\[
|\delta_{\calG}((\{v\}\cup V)\cap\Upsilon_p)|
= |\delta_{\calG}(V\cap\Upsilon_p)|+\calO(e^{2\alpha(1-\alpha)(R-r)}).
\]

Let $r:= r_p$.
By Lemma <ref> and since $R=2\ln(n/\nu)$,

\[
\theta_R(R,r)
= 2(1+o(1))e^{-\frac12 r}
= 2(\nu/n)(1+o(1))e^{\frac12(R-r)}.
\]

By hypothesis on $r$ we have $\frac12(R-r)\leq \alpha(R-r)-\omega/(1-\alpha)$ which together with Lemma <ref> implies
that $\theta_R(R,r)= o(\phi_r)$.
On the other hand, by definition of $\theta_R(\cdot,\cdot)$, we
have $\theta_R(R-r,r)=\pi\geq\phi_r$.
Thus, by monotonicity and continuity of the mapping $x\mapsto\theta_R(x,r)$
(see Remark <ref>),
there is a unique $R-r\le x\le R$ such that $\theta_R(x,r)=\phi_r$.

We claim that any
vertex $u\in V\setminus B_O(x)$ at distance at most $R$ from $v$
must necessarily be in $\Upsilon_p$.  Indeed,
since $\theta_R(r,\cdot)$ is non-decreasing
(again by Remark <ref>) we have
and thus $u$ is in
$\Upsilon_p$ (actually, in its closure).
Hence, if we denote by $E_{\calG}(S,T):=\{uv\in E(\calG) \mid u\in S, v\in T\}$ the set of edges of $\calG$ with one
end-vertex in $S$ and the other in $T$,   then we have

\[
|\delta_{\calG}((\{v\}\cup V)\cap\Upsilon_p)|
\leq |\delta_{\calG}(V\cap\Upsilon_p)| + |E_{\calG}(\{v\},V\cap B_O(x))|.
\]

Since $x+r\ge R$, by Lemma <ref>,
we have $n\theta_R(r,x)=\Theta(e^{\frac12(R-x)+\frac12(R-r)})$.
On the other hand, by Claim <ref>,
So $\theta_R(r,x)=\phi_r$ implies that
By Lemma <ref>,
since $|E_{\calG}(\{v\},V\cap B_O(x))|$
equals the number of vertices in $B_O(x)$ at distance at most $R$
from the location of $v$, we deduce that the expectation of
$|E_{\calG}(\{v\},V\cap B_O(x))|$ is

\[
n\mu(B_p(R)\cap B_O(x))= \Theta(e^{\frac12(R-r)-(\alpha-\frac12)(R-x)})
\]

By hypothesis on $r$ we have $2\alpha(1-\alpha)(R-r)\geq 2\alpha\omega/(\alpha-\frac12)\geq 2\alpha\omega
$ so the probability that
$|E_{\calG}(\{v\},V\cap B_O(x))|$ exceeds its expectation by a factor
of at least $\delta:=e^{\frac32}$ is at most $e^{-\delta\omega/2}=o(1)$.

Motivated by the previous result, we focus on bounding $|\delta_{\calG}(V\cap\Upsilon_p)|$ from above.

Let $\calA_a$ be the annulus of points of $B_O(R)$
at radial distance between $a$ and $a+1$ of the origin,
that is, $\calA_a:=B_O(R)\cap (B_O(a+1)\setminus B_O(a))$.
Also, let $\calO_{a,b}$ be the collection of points in $\calA_a$ outside
$\Upsilon_p$ (that is, in $B_O(R)\setminus\Upsilon_p$)
that are at distance at most $R$
from a point belonging to $\calA_b\cap B_O(R)\cap\Upsilon_p$.
Analogously, let $\calI_{a,b}$ be the collection of points in $\calA_b$
inside $B_O(R)\cap\Upsilon_p$ that are at distance at most $R$
from any point belonging to $\calA_a\cap B_O(R)\setminus\Upsilon_p$.

\[
\delta_{\calG}(V\cap\Upsilon_p)
= \bigcup_{a=0}^{\lfloor R\rfloor}\bigcup^{\lfloor R\rfloor}_{b=0}
\]

Let $b_{min}=\lfloor r\rfloor$.
Observe that the expected number of vertices of $\calG$
that are located
in $\Upsilon_p\cap B_O(b_{min})$ is

\[
n\mu(\Upsilon_p\cap B_O(b_{min}))
\]

where the second equality is due to Lemma <ref> and
the third equality is by Claim <ref> and since
$b_{min}\leq r$.
As the expected number of points in $\Upsilon_p\cap B_O(b_{min})$ is $o(1)$, Markov's inequality gives us that a.a.s. $V\cap\Upsilon_p\cap B_O(b_{min})=\emptyset$.
Similarly, for $a_{min}=\lceil (1-\frac{1}{2\alpha})R-\omega\rceil$
it also holds that a.a.s. $V\cap B_O(a_{min})=\emptyset$.
From our ongoing discussion, we get that a.a.s.,

\[
\delta_{\calG}(V\cap\Upsilon_p) = \bigcup_{a=a_{min}}^{\lfloor R\rfloor}\bigcup^{\lfloor R\rfloor}_{b=b_{min}} E_{\calG}(V\cap\calO_{a,b},V\cap\calI_{a,b}),
\]

so if we let $O_{a,b}=|V\cap\calO_{a,b}|$ and $I_{a,b}=|V\cap\calI_{a,b}|$, then a.a.s.,

\begin{equation}\label{ap:eqn:sum}
|\delta_{\calG}(V\cap\Upsilon_p)| \leq \sum_{a=a_{min}}^{\lfloor R\rfloor}\sum_{b=b_{min}}^{\lfloor R\rfloor} O_{a,b}\cdot I_{a,b}.
\end{equation}

To upper bound the righ-hand side expression above we will need to estimate the expected value of $O_{a,b}$ and $I_{a,b}$, or equivalently, compute $\mu(\calO_{a,b})$
and $\mu(\calI_{a,b})$. We claim that if $a+b\geq R$, then

\[
\mu(\calO_{a,b})=\Theta\big(\theta_R(a,b)\cdot \mu(B_O(a))\big)
\qquad\text{and}\qquad
\mu(\calI_{a,b})=\Theta\big(\min\{\phi_r,2\theta_R(a,b)\}\cdot \mu(B_O(b))\big).
\]

Indeed, observe that $\calO_{a,b}$ is contained in the union of two sectors (each one of central angle $\theta_R(a,b)$). By  Lemma <ref> we have $\mu(\calA_{a})=\Theta(B_{O}(a))$.
The claimed formula for $\mu(\calO_{a,b})$ follows.
A similar argument holds for $\mu(\calI_{a,b})$ but taking into account that $\calI_{a,b}$ is contained in the sector $\Upsilon_p$ of central angle $\phi_r$.

To bound the summation in (<ref>), we will use the following technical claim which allows us to ignore summation terms corresponding to pairs $(a,b)$ satisfying an specific constraint.

Let $\omega'=\omega'(n)$ be a positive function that goes to infinity with $n$ and let $\varphi:=\varphi(n)=2\ln(\omega' R)$
be such that $\varphi\ge 2\alpha\omega$.
With probability $\calO(1/\omega')$, there is pair $(c,d)$ such that
$a_{min}\leq c\leq R$, $b_{min}\leq d\leq R$,
and $R-d\leq (2\alpha-1)(R-c)-\varphi$, for which either $I_{c,d}\neq 0$
or $O_{d,c}\neq 0$.

Since $c\geq a_{min}\geq (1-\frac{1}{2\alpha})R-\omega$ and by hypothesis
$2(1-\alpha)R\leq c+d-2\alpha c-\varphi\leq c+d-(2\alpha-1)R$,
we have that $c+d\ge R$.

\[
= \Theta(e^{\frac12(R-d)-(\alpha-\frac12)(R-c)}).
%= \calO(e^{-\frac12\varphi}).
\]

Since $\EE(O_{d,c})=n\mu(\calO_{d,c})$
(respectively, $\EE(\calI_{c,d})=n\mu(\calI_{c,d})$), by Markov's inequality, the probability that $O_{d,c}\neq 0$ (respectively, $I_{c,d}\neq 0$) is $\calO(e^{\frac12(R-d)-(\alpha-\frac12)(R-c)})$.
By a union bound,
summing over $R-(2\alpha-1)(R-c)+\varphi\leq d\leq R$ and then over the
at most $\lceil R\rceil=\calO(\ln n)$ possible values taken by $c$ we
get that with probability $\calO(e^{-\frac12\varphi}\cdot\ln(n))$ there
exists a pair $(c,d)$ such that $O_{d,c}\neq 0$ (respectively, $I_{c,d}\neq 0$).
The claim follows by our choice of $\varphi$.

Finally, we can bound the size of the edge-cut induced
by $(\{v\}\cup V(\HRG))\cap\Upsilon_p$.

Suppose $(1-\frac{1}{2\alpha})R<r<R-\omega/(1-\alpha)(\alpha-\frac12)$.
Then, a.a.s. it holds that

\[
|\delta_{\calG}((\{v\}\cup V)\cap\Upsilon_p)|=\calO(e^{2\alpha(1-\alpha)(R-r)}\cdot
(\ln n)^{3}\cdot(\ln\ln n)^2).
\]

By Lemma <ref>, it suffices to upper bound
Let $\varphi$ and $\omega'$ be as in Claim <ref>.
Henceforth, let $\Omega$ be the set of pairs $(a,b)$ such that
$a_{min}\le a\leq \lfloor R\rfloor$, $b_{min}\le b\le \lfloor R\rfloor$,
$R-b\ge (2\alpha-1)(R-a)-\varphi$, and
$R-a\ge (2\alpha-1)(R-b)-\varphi$.

We first bound the sum of $O_{a,b}\cdot I_{a,b}$
over the pairs $(a,b)\in\Omega$ such that $2\theta_R(a,b)\leq\phi_r$.
By Claim <ref> we know that $\phi_r=o(1)$,
so by Lemma <ref>,
if $2\theta_R(a,b)\leq \phi_r$, then $a+b\ge R$ and
Moreover, $R-d\geq (2\alpha-1)(R-c)-\varphi$ when $(c,d)\in\Omega$, so

\begin{align*}
& =  \Theta(e^{-(\alpha-\frac12)(R-a)+\frac12(R-b)})
=  \Omega(e^{-\frac12\varphi}),
\\
n\mu(\calI_{a,b}) & = \Theta(e^{-(\alpha-\frac12)(R-b)+\frac12(R-a)})
=  \Omega(e^{-\frac12\varphi}).
\end{align*}

By Chernoff's bound (Lemma <ref>) and a union bound over the at most
$\lceil R\rceil^2$ pairs $(a,b)\in\Omega$ such that $2\theta_R(a,b)\le\phi_r$,
the probability that some such $O_{a,b}$ (respectively, $I_{a,b}$)
exceeds its expectation by a factor $e^{\frac12\varphi}\cdot 2\ln(\omega' R^2)$ is at most $\calO(1/\omega')$.
Hence, a.a.s.,

\begin{align*}
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)\leq\phi_r}
\!\!\! O_{a,b}\cdot I_{a,b}
& \leq
4e^{\varphi}(\ln(\omega' R^2))^2
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)\leq\phi_r}
\!\!\! n^2\mu(\calO_{a,b})\mu(\calI_{a,b})
\\ & = e^{\varphi}(\ln(\omega' R^2))^2
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)\leq\phi_r}
\!\!\! \calO(e^{(1-\alpha)(2R-a-b)}).
\end{align*}

A direct consequence of Claim <ref> is that

\[
2\theta_R(a,b)\leq \phi_r \text{ if and only if }
R-\tfrac12(a+b)\le \alpha(R-r)-\omega+\Theta(1).
\]

Hence, summing over the pairs $(a,b)\in\Omega$
such that $a+b=s$ and then over the valid
values of $s\geq 2R-2\alpha(R-r)+2\omega+\Theta(1)$
(observing that there are at most
$\lfloor R\rfloor=\calO(\ln n)$ pairs
$(a,b)\in\Omega$ for which $a+b=s$ for a given $s$), we get

\[
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)\leq\phi_r}
O_{a,b}\cdot I_{a,b}
\calO(e^{2\alpha(1-\alpha)(R-r)-2(1-\alpha)\omega}\cdot\ln n\cdot e^{\varphi}(\ln(\omega' R^2))^2).
\]

Next, we bound the sum of $O_{a,b}\cdot I_{a,b}$ over the pairs $(a,b)\in\Omega$
such that $2\theta_R(a,b)>\phi_r$.
Observe that,

\begin{align*}
n\mu(\calO_{a,b}) &
= \begin{cases}
\Theta(e^{\frac{R}{2}-\alpha(R-a)}), & \text{if $a+b\le R$,} \\
\Theta(e^{-(\alpha-\frac12)(R-a)+\frac12(R-b)}), & \text{if $a+b> R$.}
\end{cases}
\end{align*}

Since $a\geq a_{min}$ we have
$\frac{R}{2}-\alpha(R-a)\geq \frac{R}{2}-\alpha(R-a_{min})$.
Moreover, if $a+b\leq R$, then
\frac{R}{2}-\alpha(R-a_{min})$.
Since $a_{min}\geq (1-\frac{1}{2\alpha})R-\omega$, we conclude that
$n\mu(\calO_{a,b})=\Omega(e^{-\alpha\omega})$ independent of whether
$a+b$ is smaller or larger than $R$.

Since by assumption $2\theta_R(a,b)>\phi_r$, from
Claim <ref>, we get

\[
n\mu(\calI_{a,b}) = \Theta(n\phi_r\mu(B_O(b+1)))
= \Theta(e^{\alpha(R-r)-\alpha(R-b)-\omega})=\Omega(e^{-\omega}).
\]

So, by Chernoff's bound and a union bound over all the at most
$\lceil R\rceil^2$ pairs $(a,b)\in\Omega$ such that
the probability that some such $O_{a,b}$ (respectively, $I_{a,b}$)
exceeds its expectation by a factor $e^{\alpha\omega}2\ln(\omega'R^2)$
$e^{\omega}\cdot 2\ln(\omega'R^2)$) is at most $\calO(1/\omega')$.
Hence, a.a.s.,

\begin{align}
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)>\phi_r}
O_{a,b}\cdot I_{a,b}
& \leq
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)>\phi_r, a+b\le R}
\calO\Big(e^{\frac{R}{2}+\alpha(R-r)-\alpha(2R-a-b)-\omega}\Big)
\notag \\ & \qquad
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)>\phi_r, a+b> R}
\calO\Big(e^{\alpha(R-r)-(\alpha-\frac12)(2R-a-b)-\omega}\Big). \label{eq:lastboy}
\end{align}

Recall that $2\theta_R(a,b)>\phi_r$ is equivalent to
$a+b\leq 2R-2\alpha(R-r)+2\omega+\Theta(1)$.
Thus, summing over the pairs $(a,b)\in\Omega$ such
that $a+b=s$ and then over the valid
values of $s\leq 2R-2\alpha(R-r)+2\omega+\Theta(1)$ (observing that there are at most $\lfloor R\rfloor=\calO(\ln n)$
$(a,b)\in\Omega$ for which $a+b=s$ for a given $s$), the second sum
in the right-hand side of (<ref>) above is

\[
\calO\Big(e^{\alpha(R-r)-2(\alpha-\frac12)\alpha(R-r) - 2(1-\alpha)\omega}\cdot\ln n\Big)
= \calO\Big(e^{2\alpha(1-\alpha)(R-r)-2(1-\alpha)\omega}\cdot\ln n\Big)
\]

Similarly, since $r\geq (1-\frac{1}{2\alpha})R$ implies that
$\alpha(R-r)-(\alpha-\frac12)R\leq 2\alpha(1-\alpha)(R-r)$,
summing over the pairs $(a,b)\in\Omega$ such that $a+b=s$ and
then over the valid values of $s\leq R$, the first summation in the
right-hand side of (<ref>) above is

\[
\calO\Big(e^{\alpha(R-r)-(\alpha-\frac12)R-\omega}\cdot\ln n\Big)
\calO\Big(e^{2\alpha(1-\alpha)(R-r)-\omega}\cdot\ln n\Big).
\]

Our ongoing discussion yields (using that $\frac12<\alpha<1$ implies
that $2(1-\alpha)<1$ and $3\alpha-1<2\alpha$)

\[
\sum_{(a,b)\in\Omega \text{ s.t. }\atop 2\theta_R(a,b)>\phi_r}
O_{a,b}\cdot I_{a,b}
\calO(e^{2\alpha(1-\alpha)(R-r)}\ln n\cdot e^{2\alpha\omega}(\ln(\omega' R^2))^2).
\]

Since $\varphi\geq 2\alpha\omega$, we conclude
that a.a.s. $|\delta(V\cap\Upsilon_p)|\leq \calO(e^{2\alpha(1-\alpha)(R-r)}\cdot\ln n\cdot e^{\varphi}(\ln(\omega' R^2))^2)$.
Taking $\omega'=\omega(1)\cap\calO(R)$ gives the sough after bound.

Let $p$ and $q$ be points at radial coordinates $r_p$ and $r_q$, respectively, where $(1-\frac{1}{2\alpha})R < r_p,r_q<R-\omega/(1-\alpha)(\alpha-\frac12)$.
If $s$ and $t$ are vertices added to $\HRG:=\HRG_{\alpha,\nu}(n)$ at points $p$ and $q$ respectively, and the angle formed by $p$ and $q$ at the origin is at least $\min\{\phi_{r_p},\phi_{r_q}\}$, then a.a.s.,

\[
\Res{s}{t} = \Omega\big((e^{-2\alpha(1-\alpha)(R-r_p)}+e^{-2\alpha(1-\alpha)(R-r_q)})/(\ln n\ln\ln n)^{3}\big).
\]

Clearly, the removal of the edges $\delta_{\HRG}(V(\HRG)\cap\Upsilon_p)$
leaves $s$ and $t$ in distinct components.
By Lemma <ref> and since the inverse of the cardinality of an edge-cut between $s$ and $t$ is a lower bound on $\Res{s}{t}$ we obtain the claimed result.

Recall that (by Lemma <ref>) for $\rho$ as in (<ref>) the degree of a vertex $w\in V(\HRG)\cap B_O(\rho)$ is, a.a.s., $\Theta(e^{\frac12(R-r_w)})$.
Thus, the preceding result together with the commute time identity (see Lemma <ref>) and
the fact that a.a.s. the number of edges of the center component of $\calG_{\alpha,\nu}(n)$ is $\Theta(n)$ (see Lemma <ref>) yield the  lower bound of Theorem <ref>.

§ CONCLUDING REMARKS

In this paper we determined the order of the maximum hitting time, cover time, target time and effective resistance between two uniform vertices, the results hold in expectation and a.a.s. (w.r.t. to the HRG). Our main finding to take away is that (in expectation) there are order $\log n$ gaps between each of the quantities. This indicates that most vertices in the giant are well-connected to the center (vertices at distance at most $R/2$ of the origin) of the HRG, but a significant proportion is not.
We also gave sharp estimates for commute times for pairs of vertices $s,t$ added to the HRG for almost all the relevant positions at which $s, t$ could be placed at, provided the angular coordinates of the vertices satisfy some (mild) condition. We speculate that there could be settings in which the commute times between vertices whose angular distances are very close might be relevant. Our estimates do not hold in this case and might be worth deriving.

We restricted our study to the giant component of the graph in the regime $1/2<\alpha <1$. Even though this is arguably the most interesting regime it would still be interesting to determine the aforementioned quantities on the other smaller components or when $\alpha \notin (1/2,1)$. Another problem we leave open is to discover the leading constants hidden behind our asymptotic notation. If the expression for the clustering coefficient of the HRG [31] is anything to go by these constants may have very rich and complex expressions as functions of $\alpha$ and $\nu$.

An interesting problem is to determine the order of meeting time, that is, the expected time it takes two (lazy) random walks to occupy the same vertex when started from the worst case start vertices [42]. Finally, the mixing time of a (lazy) random walk on the giant HRG is known up to polylogarithmic factors by [46]. Closing this gap is of great importance, but it may well be quite challenging.

§ ACKNOWLEDGEMENTS
M.K. gratefully acknowledges the support of grant GrHyDy ANR-20-CE40-0002 and BASAL funds for centers of excellence from ANID-Chile grants ACE210010 and FB210005. M.S. gratefully acknowledges the support of the DIAMANT PhD travel grant for a 2-week research stay in Santiago de Chile. J.S. was supported by EPSRC project EP/T004878/1: Multilayer Algorithmics to Leverage Graph Structure and the ERC Starting Grant 679660 (DYNAMIC MARCH).

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