0}\big{|}T_{\dagger}>T-n\big{)},$

and where $(K_{i}(m))_{i\geq 0,m\geq 0}$, resp. $(\widehat{K}_{i}(m))_{i\geq
0,m\geq 0}$ are the Markov chains of clusters starting from the corresponding
initial conditions.

###### Proof.

Consider the following events for the loop soup
$\mathcal{L}^{\theta}_{\cap(0,T-n]}$. Let $E_{1,n,T}$ be the event

$E_{1,n,T}=\Big{\\{}\partial\mathtt{C}_{n}\stackrel{{\scriptstyle\mathcal{L}^{\theta}_{\cap[(c_{1}-1)n,T-n]}}}{{\longleftrightarrow}}\partial\mathtt{C}_{T-n}\Big{\\}}.$

Let $E_{2,n}$ be the event that there is a loop of
$\mathcal{L}^{\theta}_{\cap(0,T-n]}$ contained in the cylinder
$(n,2n)\times\mathbb{S}^{1}$ that surrounds the cylinder. Let $E_{3,n}$ be the
event that there is a loop of $\mathcal{L}^{\theta}_{\cap(0,T-n]}$ contained
in the cylinder $((c_{1}-2)n,(c_{1}-1)n)\times\mathbb{S}^{1}$ that surrounds
the cylinder. See Figure 3.2. Note that the events $E_{1,n,T}$, $E_{2,n}$ and
$E_{3,n}$ are independent since they involve disjoint types of loops.

$n$
---
$2n$
---
$(c_{1}-2)n$
---
$(c_{1}-1)n$
---
$T-n$
---
$\widehat{K}_{0}(0)$
---
$K_{0}(0)$
---
$c_{1}n$
---
Figure 3.2: Schematic representation of the events considered in the proof of
Lemma 3.8. Only the marked compact $K_{0}(0)$ and $\widehat{K}_{0}(0)$ of the
initial conditions $(K_{i}(0))_{i\geq 0}$ and $(\widehat{K}_{i}(0))_{i\geq 0}$
have been depicted (light and dark grey respectively). The existence of the
blue cluster corresponds to the event $E_{1,n,T}$ whereas the existence of the
two red loops corresponds to the events $E_{2,n,T}$ and $E_{3,n,T}$.

Now we consider the Markov chains $(K_{i}(m))_{i\geq 0,m\geq 0}$ and
$(\widehat{K}_{i}(m))_{i\geq 0,m\geq 0}$ on the same probability space,
obtained from the same Brownian loop soup $\mathcal{L}^{\theta}_{0}$. Note
that in this coupling, neither of the two Markov processes is conditioned on
$T_{\dagger}>T-n$. Not yet.

Since $(K_{i}(0))_{i\geq 0}\in\mathtt{F}_{n}$ and $(\widehat{K}_{i}(0))_{i\geq
0}\in\mathtt{F}_{n}$, we have

$E_{1,n,T}\cap E_{2,n}\subset\\{T_{\dagger}((K_{i})_{i\geq
0})>T-n\\}\quad\text{and}\quad E_{1,n,T}\cap
E_{2,n}\subset\\{T_{\dagger}((\widehat{K}_{i})_{i\geq 0})>T-n\\};$

see Figure 3.2. Further, on the event $E_{1,n,T}\cap E_{2,n}\cap E_{3,n}$, we
have $(K_{i}((c_{1}-1)n))_{i\geq 0}=(\widehat{K}_{i}((c_{1}-1)n))_{i\geq 0}$
a.s. Indeed, if two Brownian loops $\wp$ and $\wp^{\prime}$ intersecting
$\partial\mathtt{C}_{(c_{1}-1)n}$ are connected by a finite chain that uses
$(K_{i}(0))_{i\geq 0}$ and $(\widehat{K}_{i}(0))_{i\geq 0}$, then this chain
has to intersect the non-contractible loops in
$((c_{1}-2)n,(c_{1}-1)n)\times\mathbb{S}^{1}$, that exist on the event
$E_{3,n}$, and thus the connection between $\wp$ and $\wp^{\prime}$ can also
be done without using $(K_{i}(0))_{i\geq 0}$ or $(\widehat{K}_{i}(0))_{i\geq
0}$, by using just the Brownian loop; see Figure 3.2.

Thus, we get some overlap in total variation between the non-conditioned laws
$\operatorname{Law}\big{(}(K_{i}((c_{1}-1)n))_{i\geq 0}\big{)}$ and
$\operatorname{Law}\big{(}(\widehat{K}_{i}((c_{1}-1)n))_{i\geq 0}\big{)}$.
Moreover, this overlap happens on the event

$\\{T_{\dagger}((K_{i})_{i\geq
0})>T-n\\}\cap\\{T_{\dagger}((\widehat{K}_{i})_{i\geq 0})>T-n\\}.$

Therefore, we get a non-trivial overlap for the conditioned laws too:

$1-\dfrac{1}{2}d_{\rm
TV}(L_{n,T},\widehat{L}_{n,T})\geq\dfrac{\mathbb{P}(E_{1,n,T}\cap E_{2,n}\cap
E_{3,n})}{\mathbb{P}(T_{\dagger}((K_{i})_{i\geq
0})>T-n)\vee\mathbb{P}(T_{\dagger}((\widehat{K}_{i})_{i\geq 0})>T-n)}.$

Further,

$\\{T_{\dagger}((K_{i})_{i\geq
0})>T-n\\}\cup\\{T_{\dagger}((\widehat{K}_{i})_{i\geq 0})>T-n\\}\subset
E_{4,n,T},$

where $E_{4,n,T}$ is the event

$E_{4,n,T}=\Big{\\{}\partial\mathtt{C}_{(c_{1}-1)n}\stackrel{{\scriptstyle\mathcal{L}^{\theta}_{\cap[(c_{1}-1)n,T-n]}}}{{\longleftrightarrow}}\partial\mathtt{C}_{T-n}\Big{\\}}.$

Thus,

$\mathbb{P}(T_{\dagger}((K_{i})_{i\geq
0})>T-n)\vee\mathbb{P}(T_{\dagger}((\widehat{K}_{i})_{i\geq
0})>T-n)\leq\mathbb{P}(E_{4,n,T}).$

Thus,

$1-\dfrac{1}{2}d_{\rm
TV}(L_{n,T},\widehat{L}_{n,T})\geq\dfrac{\mathbb{P}(E_{1,n,T}\cap E_{2,n}\cap
E_{3,n})}{\mathbb{P}(E_{4,n,T})}\geq\dfrac{\mathbb{P}(E_{1,n,T})}{\mathbb{P}(E_{4,n,T})}\mathbb{P}(E_{2,n})\mathbb{P}(E_{3,n}).$

By Lemma 3.7,

$\dfrac{\mathbb{P}(E_{1,n,T})}{\mathbb{P}(E_{4,n,T})}\geq\hat{p}_{0}>0,$

where the lower bound is uniform in $n$ and $T$. Further,
$\mathbb{P}(E_{2,n})=\mathbb{P}(E_{3,n})=1-\hat{p}_{2}^{\theta n}$, for some
$\hat{p}_{2}\in(0,1)$. This concludes the proof of the lemma. ∎

Lemma 3.8 is exactly Condition 1 of Theorem 3.3. So it applies to our Markov
chains of clusters in a Brownian loop soup and concludes the proof of Theorem
3.2.

###  3.5 Convergence of the ratio of survival probabilities

We finish this section by proving Theorem 3.1. We first state and prove a
corollary of Theorem 3.2. We use the same notations as the ones introduced
above Theorem 3.2.

######  Corollary 3.9.

For every initial conditions $K_{0}$ and $\hat{K}_{0}$ not identical to
$\partial\mathtt{C}_{0}$,
$\mathbb{P}(T^{\dagger}>T)/\mathbb{P}(\hat{T}^{\dagger}>T)$ converges as
$T\to\infty$.

######  Remark 3.10.

By FKG inequality, it is clear that the ratio
$\mathbb{P}(T^{\dagger}>T)/\mathbb{P}(\hat{T}^{\dagger}>T)$ remains uniformly
bounded away from zero and from infinity. Indeed, recalling the notation
$\rho(K_{0})$ (3.9), let $E$ be the event that there is a loop intersecting
$\hat{K}_{0}$ that disconnects $\partial\mathtt{C}_{\rho(K_{0})}$ and
$\partial\mathtt{C}_{\rho(K_{0})+1}$. Since
$E\cap\\{T^{\dagger}>T\\}\subset\\{\hat{T}^{\dagger}>T\\}$, by FKG inequality
we get that

$\mathbb{P}(T^{\dagger}>T)/\mathbb{P}(\hat{T}^{\dagger}>T)\leq
1/\mathbb{P}\left(E\right).$

The lower bound is similar. On the other hand, showing that this ratio
actually converges is much more involved and relies on the coupling results we
developed.

###### Proof.

Let $T\gg t\gg 1$ be large integers and let $E_{t,T}$ be the event that under
the coupling of Theorem 3.2, the coalescence time $\mathcal{T}^{T}$ is smaller
than $t$. By Theorem 3.2, $\mathbb{P}\left(E_{t,T}\right)\to 1$ as
$T\to\infty$ followed by $t\to\infty$. In this proof, we will denote by
$(\mathcal{F}_{s})_{s\geq 0}$ the filtration
$\mathcal{F}_{s}=\sigma(K_{0}(u),\hat{K}_{0}(u),u\in\\{0,\dots,s\\})$, $s\geq
0$. We have

$\displaystyle\mathbb{P}(T^{\dagger}>T$
$\displaystyle|T^{\dagger}>t)=\mathbb{E}\left[P(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t})\mathbf{1}_{E_{t,T}}\right]+\mathbb{E}\left[\mathbb{P}(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t})\mathbf{1}_{E_{t,T}^{c}}\right]$
$\displaystyle=\mathbb{E}\left[\mathbb{P}(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t,\mathcal{F}_{t})\mathbf{1}_{E_{t,T}}\right]+\mathbb{E}\left[\mathbb{P}(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t})\mathbf{1}_{E_{t,T}^{c}}\right]$
$\displaystyle=\mathbb{P}(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t)+\mathbb{E}\left[\left(\mathbb{P}(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t})-\mathbb{P}(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t,\mathcal{F}_{t})\right)\mathbf{1}_{E_{t,T}^{c}}\right].$

The rest of the proof is dedicated to showing that the second right hand side
term is much smaller than the first one. Let $p,q>1$ be such that $1/p+1/q=1$.
By Hölder’s inequality and the above inequality,
$\left|\mathbb{P}(T^{\dagger}>T|T^{\dagger}>t)-\mathbb{P}(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t)\right|$
is bounded by

$\displaystyle\mathbb{P}\left(E_{t,T}^{c}\right)^{1/p}\mathbb{E}\left[\mathbb{P}\left(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t}\right)^{q}+\mathbb{P}(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t,\mathcal{F}_{t})^{q}\right]^{1/q}.$
(3.15)

Let us denote by
$\tau:=\sup\\{s>0:\exists\wp\in\mathcal{L}_{\cap(0,t]}^{\theta}:\operatorname{Range}(\wp)\cap\\{s\\}\times\mathbb{S}^{1}\neq\varnothing\\}.$
Conditionally on $\tau,\mathcal{F}_{t},T^{\dagger}>t$, in order to have
$T^{\dagger}>T$, there must be a cluster of loops included in
$(t,\infty)\times\mathbb{S}^{1}$ that crosses $[\tau,T]\times\mathbb{S}^{1}$.
By scaling, this probability is equal to
$\mathbb{P}\Big{(}\tau-t\overset{\mathcal{L}_{0}^{\theta}}{\longleftrightarrow}T-t|\tau\Big{)}$.
We obtain that

$\displaystyle\mathbb{E}\left[\mathbb{P}\left(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t}\right)^{q}\right]=\mathbb{P}\left(\tau>T\right)+\int_{t}^{T}\mathbb{P}\left(\tau\in\mathrm{d}s\right)\mathbb{P}\left(s-t\overset{\mathcal{L}_{0}^{\theta}}{\longleftrightarrow}T-t\right)^{q}.$
(3.16)

Let $E_{1},E_{2}$ and $E_{3}$ be the following events:

  * •

$E_{1}$: there is a cluster of $\mathcal{L}_{0}^{\theta}$ crossing
$[t-1,s-t+1]\times\mathbb{S}^{1}$;

  * •

$E_{2}$: there is a non contractible loop inside
$[t-1,t]\times\mathbb{S}^{1}$;

  * •

$E_{3}$: there is a non contractible loop inside
$[s-t,s-t+1]\times\mathbb{S}^{1}$.

We have $E_{1}\cap E_{2}\cap
E_{3}\cap\\{T^{\dagger}>t\\}\cap\\{s-t\overset{\mathcal{L}_{0}^{\theta}}{\longleftrightarrow}T-t\\}\subset\\{T^{\dagger}>T-t\\}$.
The five events on the left hand side are increasing events. By FKG inequality
we deduce that the product of the five individual probabilities is at most
$\mathbb{P}\left(T^{\dagger}>T-t\right)$. By Lemma 2.5,
$\mathbb{P}\left(E_{2}\right)=\mathbb{P}\left(E_{3}\right)\geq c$ and by
Theorem 2.4, $\mathbb{P}\left(E_{1}\right)\geq
c_{\varepsilon}(s/t)^{-1+\theta-\varepsilon}$. This leads to

$\mathbb{P}\Big{(}s-t\overset{\mathcal{L}_{0}^{\theta}}{\longleftrightarrow}T-t\Big{)}\leq
C(s/t)^{1-\theta+\varepsilon}\mathbb{P}\left(T^{\dagger}>T-t\right)/\mathbb{P}\left(T^{\dagger}>t\right).$

Plugging this estimate back in (3.16) and integrating by parts
(differentiating $s^{q(1-\theta+\varepsilon)}$ and integrating
$\mathbb{P}\left(\tau\in\mathrm{d}s\right)$) gives the following upper bound
for the left hand side of (3.16):

$\displaystyle
C(1/t)^{q(1-\theta+\varepsilon)}\mathbb{P}\left(T^{\dagger}>T-t\right)^{q}/\mathbb{P}\left(T^{\dagger}>t\right)^{q}\int_{t}^{T}\mathbb{P}\left(\tau\geq
s\right)s^{q(1-\theta+\varepsilon)-1}\mathrm{d}s.$

By Lemma 2.6, $\mathbb{P}\left(\tau\geq s\right)\leq Ct/s$. With the change of
variable $s=xt$, we have obtained that

$\displaystyle\mathbb{E}\left[\mathbb{P}\left(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t}\right)^{q}\right]\leq
C\mathbb{P}\left(T^{\dagger}>T-t\right)^{q}/\mathbb{P}\left(T^{\dagger}>t\right)^{q}\int_{1}^{\infty}x^{q(1-\theta+\varepsilon)-2}\mathrm{d}x.$

We choose $1<q<1/(1-\theta+\varepsilon)$ to make sure that the integral above
converges. We have overall obtained that
$\mathbb{P}\left(T^{\dagger}>t\right)^{q}\mathbb{E}\left[\mathbb{P}\left(T^{\dagger}>T|T^{\dagger}>t,\mathcal{F}_{t}\right)^{q}\right]\leq
C\mathbb{P}\left(T^{\dagger}>T-t\right)^{q}$. Since
$\mathbb{P}(\hat{T}^{\dagger}>t)/\mathbb{P}\left(T^{\dagger}>t\right)\in[1/C,C]$,
we also obtain that

$\mathbb{P}\left(T^{\dagger}>t\right)^{q}\mathbb{E}\left[\mathbb{P}\left(\hat{T}^{\dagger}>T|\hat{T}^{\dagger}>t,\mathcal{F}_{t}\right)^{q}\right]\leq
C\mathbb{P}\left(T^{\dagger}>T-t\right)^{q}.$

Multiplying (3.15) by
$\mathbb{P}\left(T^{\dagger}>t\right)/\mathbb{P}(\hat{T}^{\dagger}>T)$, we
deduce that

$\displaystyle\left|\frac{\mathbb{P}\left(T^{\dagger}>T\right)}{\mathbb{P}(\hat{T}^{\dagger}>T)}-\frac{\mathbb{P}\left(T^{\dagger}>t\right)}{\mathbb{P}(\hat{T}^{\dagger}>t)}\right|\leq
C\mathbb{P}\left(E_{t,T}^{c}\right)^{1/p}\frac{\mathbb{P}\left(T^{\dagger}>T-t\right)}{\mathbb{P}(\hat{T}^{\dagger}>T)}.$

Using FKG inequality and the fact that the probability of crossing
$[T-t,T]\times\mathbb{S}^{1}$ tends to 1 as $T\to\infty$ ($t$ fixed), it can
be checked that
$\limsup_{T\to\infty}\mathbb{P}\left(T^{\dagger}>T-t\right)/\mathbb{P}(\hat{T}^{\dagger}>T)$
is bounded uniformly in $t$; see Remark 3.10 for a very similar statement. By
then sending $t\to\infty$, we obtain that

$\limsup_{t\to\infty}\limsup_{T\to\infty}\left|\frac{\mathbb{P}\left(T^{\dagger}>T\right)}{\mathbb{P}(\hat{T}^{\dagger}>T)}-\frac{\mathbb{P}\left(T^{\dagger}>t\right)}{\mathbb{P}(\hat{T}^{\dagger}>t)}\right|=0.$

This shows that
$\mathbb{P}\left(T^{\dagger}>T\right)/\mathbb{P}(\hat{T}^{\dagger}>T)$
converges as $T\to\infty$. ∎

We are now ready to prove Theorem 3.1.

###### Proof of Theorem 3.1.

The proof of the convergence of the first ratio in (3.1) follows quickly from
Corollary 3.9 and from conformal invariance of the loop soup. Indeed, let
$f:D\to\mathbb{D}$ be a conformal map with $f(x)=0$. By conformal invariance
of the loop soup, the first ratio in (3.1) is equal to

$\mathbb{P}\Big{(}f(K)\overset{\mathcal{L}_{\mathbb{D}}^{\theta}}{\longleftrightarrow}f(D(x,r))\Big{)}\Big{/}\mathbb{P}\Big{(}f(\hat{K})\overset{\mathcal{L}_{\mathbb{D}}^{\theta}}{\longleftrightarrow}f(D(x,r))\Big{)}.$

There exists a constant $C>1$ such that if $r$ is small enough
$C^{-1}r\mathbb{D}\subset f(D(x,r))\subset Cr\mathbb{D}$. Because the
probability of connecting $K$ to $C^{-1}r\mathbb{D}$ is asymptotically
equivalent to that of connecting $K$ to $Cr\mathbb{D}$, we deduce that the
first ratio in (3.1) is equal to

$(1+o(1))\mathbb{P}\Big{(}f(K)\overset{\mathcal{L}_{\mathbb{D}}^{\theta}}{\longleftrightarrow}r\mathbb{D}\Big{)}\Big{/}\mathbb{P}\Big{(}f(\hat{K})\overset{\mathcal{L}_{\mathbb{D}}^{\theta}}{\longleftrightarrow}r\mathbb{D}\Big{)}.$

By Corollary 3.9, the above ratio converges as $r\to 0$ ($r=e^{-T}$ in the
notations of Corollary 3.9).

We now move to the proof of the convergence of the second ratio in (3.1) and
the identification of the limit. Let us denote by $\ell$ the limit of the
first ratio in (3.1). Let $\delta>0$ and let
$\varepsilon\in(0,\mathrm{d}(z,\partial D))$ be small enough so that for all
$r\in(0,\varepsilon)$,

$\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(z,r)\Big{)}\Big{/}\mathbb{P}\Big{(}\hat{K}\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(z,r)\Big{)}\leq(1+\delta)\ell.$

As in (2.18), we can replace $\Xi_{a}^{z}$ by a disc centred at $z$ with
random radius $\left\|\Xi_{a}^{z}-z\right\|_{\infty}$:

$\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}z\Big{)}=(1+o(1))\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(z,\left\|\Xi_{a}^{z}-z\right\|_{\infty})\Big{)}.$

By Lemma 2.9, the probability that
$\left\|\Xi_{a}^{z}-z\right\|_{\infty}>\varepsilon$ is at most
$Ca=C\gamma^{2}$ which is much smaller than the probability we are looking at
(it behaves like $\gamma^{2(1-\theta)+o(1)}$, see Theorem 1.2). By Lemma 2.9,
we get that

$\displaystyle\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}z\Big{)}=(1+o(1))\int_{0}^{\varepsilon}\mathrm{d}r\frac{ar^{a-1}}{\operatorname{CR}(z,D)^{a}}\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(z,r)\Big{)}$
$\displaystyle\leq(1+o(1))(1+\delta)\ell\int_{0}^{\varepsilon}\mathrm{d}r\frac{ar^{a-1}}{\operatorname{CR}(z,D)^{a}}\mathbb{P}\Big{(}\hat{K}\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(z,r)\Big{)}\leq(1+o(1))(1+\delta)\ell\mathbb{P}\Big{(}\hat{K}\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}z\Big{)}$

where in the last inequality we simply reversed the above procedure. This
proves that

$\limsup_{a\to
0}\mathbb{P}\Big{(}K\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}z\Big{)}\Big{/}\mathbb{P}\Big{(}\hat{K}\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}z\Big{)}\leq\ell.$

The lower bound is proved similarly. This concludes the proof. ∎

##  4 Construction of $h_{\theta}$: proof of Theorems 1.2, 1.11 and 1.12

We start by stating the intermediate results needed in order to prove Theorems
1.2, 1.11 and 1.12, and then use them to prove these theorems. The rest of the
section then consists of the proofs of these intermediate results.

We start with a lemma that shows that computing the second moment of
$h_{\gamma}$ boils down to computing the crossing probabilities between $x$
and $y$ in a soup whose loops come from 1) an unconditioned Brownian loop soup
$\mathcal{L}_{D}^{\theta}$, 2) a thick loop $\Xi_{a}^{x}$ at $x$ and 3) a
thick loop $\Xi_{a}^{y}$ at $y$.

######  Lemma 4.1.

Let $\varepsilon>0$ and $F:D\times D\times\mathfrak{L}\to\mathbb{R}$ be a
bounded measurable admissible function such that for all $x,y\in D$,
$F(x,y,\cdot)$ does not depend on loops with diameter smaller than
$\varepsilon$. For all $\gamma,\gamma^{\prime}\in(0,\sqrt{2})$,

$\displaystyle\mathbb{E}\left[\int_{D\times
D}F(x,y,\mathcal{L}_{D}^{\theta})h_{\gamma}(x)h_{\gamma^{\prime}}(y)\mathrm{d}x\mathrm{d}y\right]=C(\gamma,\gamma^{\prime},\varepsilon)\frac{\gamma^{2}{\gamma^{\prime}}^{2}}{Z_{\gamma}Z_{\gamma^{\prime}}}$
(4.1) $\displaystyle+\frac{1}{Z_{\gamma}Z_{\gamma^{\prime}}}\int_{D\times
D}\mathbb{E}\Big{[}F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}}{\longleftrightarrow}y\Big{\\}}\Big{]}\mathrm{d}x\mathrm{d}y$

where the three collections of loops $\mathcal{L}_{D}^{\theta}$,
$\\{\Xi_{a_{i}}^{x}\\}_{i\geq 1}$ and $\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}$ are independent and distributed as in Lemma 2.3. The constant
$C(\gamma,\gamma^{\prime},\varepsilon)$ is bounded uniformly w.r.t.
$\gamma,\gamma^{\prime}$ in compact subsets of $[0,\sqrt{2})$ by a quantity
which may depend on $\varepsilon$.

The next two results are then dedicated to estimating the crossing probability
appearing in Lemma 4.1. The first result has a very similar flavour as [JLQ23,
Proposition 4.1]. On the other hand, the second result is novel and crucially
relies on Theorem 3.1.

######  Lemma 4.2.

For any $\eta>0$, there exists $C>0$ such that for all $\gamma\in(0,1]$ and
$x,y\in D$,

$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y\right)\leq
C|\log|x-y||^{2(1-\theta)+\eta}.$ (4.2)

######  Proposition 4.3.

For any distinct points $x,y\in D$, the limit

$C_{\theta}(x,y):=\lim_{\gamma\to
0}\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y\Big{)}$
(4.3)

exists and

$C_{\theta}(x,y)\leq\liminf_{\gamma_{1},\gamma_{2}\to
0}\frac{1}{Z_{\gamma_{1}}Z_{\gamma_{2}}}\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a_{1}}^{x}\cup\Xi_{a_{2}}^{y}}{\longleftrightarrow}y\Big{)}.$
(4.4)

Moreover,

$C_{\theta}(x,y)=\left(\log\frac{1}{|x-y|}\right)^{2(1-\theta)+o(1)}\quad\quad\text{as}\quad
x-y\to 0.$ (4.5)

More generally, let $\varepsilon>0$ and $F:\mathfrak{L}\to\mathbb{R}$ be a
bounded measurable function that does not depend on loops with diameter
smaller than $\varepsilon>0$. Then the following two limits exist and are
identical

$\lim_{r\to
0}\frac{1}{Z_{r}^{2}}\mathbb{E}\left[F(\mathcal{L}_{D}^{\theta})\mathbf{1}\Big{\\{}D(x,r)\overset{\mathcal{L}_{D}^{\theta}}{\longleftrightarrow}D(y,r)\Big{\\}}\right]=\lim_{a\to
0}\frac{1}{Z_{\gamma}^{2}}\mathbb{E}\left[F(\mathcal{L}_{D}^{\theta})\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y\Big{\\}}\right].$
(4.6)

######  Notation 4.4.

If we want to keep track of the domain $D$, we will denote by
$C_{\theta,D}(x,y)$ the limit of (4.3).

We can now show Theorems 1.2, 1.11 and 1.12.

###### Proof of Theorem 1.11.

It follows directly from (4.6). Indeed, let $x,y\in D$ be two distinct points.
By a density-type argument, to prove Theorem 1.11, it is enough to show that
for any $\varepsilon>0$ and any bounded measurable function
$F:\mathfrak{L}\to\mathbb{R}$ which does not depend on loops with diameter
smaller than $\varepsilon$, $\mathbb{E}_{x\leftrightarrow
y,r}\left[F(\mathcal{L}_{\mathbb{D}}^{\theta})\right]$ converges as $r\to 0$.
This is a mere restatement of the convergence (4.6). ∎

###### Proof of Theorem 1.2.

Let $f:D\to\mathbb{R}$ be a bounded measurable function. By decomposing $f$
into its positive and negative parts, we can assume without loss of generality
that $f$ is nonnegative. We want to show that
$((h_{\gamma},f),\gamma\in(0,\sqrt{2})$ is Cauchy in $L^{2}$, i.e. that
$\mathbb{E}\left[|(h_{\gamma_{1}},f)-(h_{\gamma_{2}},f)|^{2}\right]\to 0$ as
$\gamma_{1},\gamma_{2}\to 0$. By Lemma 4.1, this second moment is equal to

$o(1)+\int_{D\times
D}f(x)f(y)\Bigg{(}\frac{\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a_{1}}^{x}\cup\Xi_{a_{1}}^{y}}{\longleftrightarrow}y\Big{)}}{Z_{\gamma_{1}}^{2}}+\frac{\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a_{2}}^{x}\cup\Xi_{a_{2}}^{y}}{\longleftrightarrow}y\Big{)}}{Z_{\gamma_{2}}^{2}}-2\frac{\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a_{1}}^{x}\cup\Xi_{a_{2}}^{y}}{\longleftrightarrow}y\Big{)}}{Z_{\gamma_{1}}Z_{\gamma_{2}}}\Bigg{)}.$

Proposition 4.3 together with dominated convergence theorem and Lemma 4.2 (and
Fatou’s lemma for the crossed term involving both $a_{1}$ and $a_{2}$) then
shows that $\limsup_{\gamma_{1},\gamma_{2}\to
0}\mathbb{E}\left[|(h_{\gamma_{1}},f)-(h_{\gamma_{2}},f)|^{2}\right]\leq 0$.
This concludes the $L^{2}$ convergence of $(h_{\gamma},f)$.

We now explain how one can lift this convergence to a convergence in the
Sobolev space $H^{-\varepsilon}(\mathbb{C})$, where $\varepsilon\in(0,1)$ is
arbitrary small. Recall that $H^{-\varepsilon}(\mathbb{C})$ is the set of
tempered distribution $f$ such that

$\left\|f\right\|_{H^{-\varepsilon}(\mathbb{C})}^{2}:=\int_{\mathbb{C}}(1+|\xi|^{2})^{-\varepsilon}|\hat{f}(\xi)|^{2}\mathrm{d}\xi<\infty.$
(4.7)

Let $\gamma_{1},\gamma_{2}\in(0,\sqrt{2})$. Denoting
$C_{\gamma_{1},\gamma_{2}}(x,y)=\mathbb{E}\left[(h_{\gamma_{1}}(x)-h_{\gamma_{2}}(x))(h_{\gamma_{1}}(y)-h_{\gamma_{2}}(y))\right]$,
we have

$\displaystyle\mathbb{E}\left\|h_{\gamma_{1}}-h_{\gamma_{2}}\right\|_{H^{-\varepsilon}(\mathbb{C})}^{2}$
$\displaystyle=\int_{\mathbb{R}^{2}}\frac{\mathbb{E}\left|\hat{h}_{\gamma_{1}}(\xi)-\hat{h}_{\gamma_{2}}(\xi)\right|^{2}}{(1+\left|\xi\right|^{2})^{\varepsilon}}\mathrm{d}\xi\leq\int_{\mathbb{R}^{2}}\frac{\mathbb{E}\left|\hat{h}_{\gamma_{1}}(\xi)-\hat{h}_{\gamma_{2}}(\xi)\right|^{2}}{\left|\xi\right|^{2\varepsilon}}\mathrm{d}\xi$
(4.8) $\displaystyle=\int_{D\times
D}\mathrm{d}x\mathrm{d}y~{}C_{\gamma_{1},\gamma_{2}}(x,y)\int\mathrm{d}\xi~{}\frac{e^{2\pi
i\xi\cdot(x-y)}}{\left|\xi\right|^{2\varepsilon}}=C_{\varepsilon}\int_{D\times
D}\mathrm{d}x\mathrm{d}y~{}\frac{C_{\gamma_{1},\gamma_{2}}(x,y)}{|x-y|^{2-2\varepsilon}}$

where we used the fact that the Fourier transform of $|\cdot|^{-2\varepsilon}$
is $C_{\varepsilon}|\cdot|^{-2+2\varepsilon}$ for some constant
$C_{\varepsilon}$ depending on $\varepsilon$. We can then conclude as before
using Proposition 4.3 that $h_{\gamma}$ is Cauchy in
$L^{2}(H^{-\varepsilon}(\mathbb{C}),\mathbb{P})$.

Finally, the proof that the normalising constant $Z_{\gamma}$ (1.3) satisfies
$Z_{\gamma}=\gamma^{2(1-\theta)+o(1)}$ as $\gamma\to 0$ can be found in
Section 2.4. ∎

###### Proof of Theorem 1.12.

By a simple density-type argument, it is enough to check the identity
statement in Theorem 1.12 for a function $F:\mathfrak{L}\to\mathbb{R}$
depending only on loops with diameter larger than a given threshold. For such
functions, it then follows from Lemma 4.1, Theorem 1.11 and (4.6). ∎

###  4.1 Proof of Lemma 4.1

###### Proof of Lemma 4.1.

By definition of $h_{\gamma}$, the left hand side of (4.1) is equal to

$\displaystyle\frac{1}{Z_{\gamma}Z_{\gamma^{\prime}}}\int_{D\times
D}\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta})(\mathcal{M}_{a}^{+}(\mathrm{d}x)\mathcal{M}_{a^{\prime}}^{+}(\mathrm{d}y)-\mathcal{M}_{a}^{+}(\mathrm{d}x)\mathrm{d}y-\mathrm{d}x\mathcal{M}_{a^{\prime}}^{+}(\mathrm{d}y)+\mathrm{d}x\mathrm{d}y)\right].$

By (2.12),

$\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta})\mathcal{M}_{a}^{+}(\mathrm{d}x)\right]=\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1})\right]\mathrm{d}x$

where the loops $\\{\Xi_{a_{i}}^{x}\\}_{i\geq 1}$ are as defined above Lemma
2.3. A similar result holds at $y$. To handle the term
$\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta})(\mathcal{M}_{a}^{+}(\mathrm{d}x)\mathcal{M}_{a^{\prime}}^{+}(\mathrm{d}y)\right]$,
we first control the contribution of soups containing at least one loop which
visits both $x$ and $y$. By triangle inequality and then by (2.13) and (2.14)
applied to $F\equiv 1$, we have

$\displaystyle\left|\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta})\mathcal{M}_{a}^{+}(\mathrm{d}x)\mathcal{M}_{a^{\prime}}^{+}(\mathrm{d}y)\mathbf{1}_{\left\\{\exists\wp\in\mathcal{L}_{D}^{\theta}:x,y\in\wp\right\\}}\right]\right|\leq\left\|F\right\|_{\infty}\mathbb{E}\left[\mathcal{M}_{a}(\mathrm{d}x)\mathcal{M}_{a^{\prime}}(\mathrm{d}y)\mathbf{1}_{\left\\{\exists\wp\in\mathcal{L}_{D}^{\theta}:x,y\in\wp\right\\}}\right]$
$\displaystyle=4\left\|F\right\|_{\infty}\left(\left(2\pi\sqrt{aa^{\prime}}G_{D}(x,y)\right)^{1-\theta}\Gamma(\theta)I_{\theta-1}\left(4\pi\sqrt{aa^{\prime}}G_{D}(x,y)\right)-1\right)\mathrm{d}x\mathrm{d}y=O(\gamma^{2}{\gamma^{\prime}}^{2}).$
(4.9)

On the other hand, by the Girsanov-type transform (2.14), the contribution
with no such loop is equal to

$\displaystyle\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta})\mathcal{M}_{a}^{+}(\mathrm{d}x)\mathcal{M}_{a}^{+}(\mathrm{d}y)\mathbf{1}_{\left\\{\nexists\wp\in\mathcal{L}_{D}^{\theta}:x,y\in\wp\right\\}}\right]$
$\displaystyle=4\mathbb{E}\Big{[}F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})\mathbf{1}\\{\mathfrak{C}^{+}(\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})\\}\Big{]}\mathrm{d}x\mathrm{d}y.$

We then notice that, conditionally on
$\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq 1}$, the probability that both $x$
and $y$ belong to positive clusters of
$\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq 1}$ is equal to

$\displaystyle\frac{1}{4}\Big{(}1-\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}}{\longleftrightarrow}y\Big{\\}}\Big{)}+\frac{1}{2}\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}}{\longleftrightarrow}y\Big{\\}}$ $\displaystyle=$
$\displaystyle\frac{1}{4}+\frac{1}{4}\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}}{\longleftrightarrow}y\Big{\\}}.$

Putting everything together, we have obtained that the left hand side of (4.1)
is equal to

$\displaystyle\frac{1}{Z_{\gamma}Z_{\gamma^{\prime}}}\int_{D\times
D}\mathbb{E}\left[F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})\mathbf{1}\Big{\\{}x\overset{\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1}}{\longleftrightarrow}y\Big{\\}}\right]\mathrm{d}x\mathrm{d}y$
$\displaystyle+O(1)\frac{\gamma^{2}{\gamma^{\prime}}^{2}}{Z_{\gamma}Z_{\gamma^{\prime}}}+\frac{1}{Z_{\gamma}Z_{\gamma^{\prime}}}\int_{D\times
D}\mathbb{E}[F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})-F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq 1})$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})+F(x,y,\mathcal{L}_{D}^{\theta})]\mathrm{d}x\mathrm{d}y.$

To bound the last integral above, we need to recall that $F$ does not depend
on loops with diameter smaller than a given threshold $\varepsilon$. Hence, if
$\wedge_{i\geq 1}\Xi_{a_{i}}^{x}$ (concatenation of $\Xi_{a_{i}}^{x},i\geq 1$)
or $\wedge_{i\geq 1}\Xi_{a_{i}^{\prime}}^{y}$ has diameter smaller than
$\varepsilon$, then

$F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})-F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}}^{x}\\}_{i\geq
1})-F(x,y,\mathcal{L}_{D}^{\theta}\cup\\{\Xi_{a_{i}^{\prime}}^{y}\\}_{i\geq
1})+F(x,y,\mathcal{L}_{D}^{\theta})=0.$

The probability that both $\wedge_{i\geq 1}\Xi_{a_{i}}^{x}$ and $\wedge_{i\geq
1}\Xi_{a_{i}^{\prime}}^{y}$ have diameter at least $\varepsilon$ is of order
$\gamma^{2}{\gamma^{\prime}}^{2}$ (Lemma 2.9) concluding the proof. ∎

###  4.2 Proof of Lemma 4.2 and Proposition 4.3

We start by proving Lemma 4.2.

###### Proof of Lemma 4.2.

Let $\eta>0$ be small and $x,y\in D$. If $|x-y|\leq e^{-a^{-1+\eta}}$, we
simply bound the probability in (4.2) by 1. Using that
$Z_{\gamma}=\gamma^{2(1-\theta)+o(1)}=a^{1-\theta+o(1)}$, we obtain that the
left hand side of (4.2) is at most

$a^{-2(1-\theta)+o(1)}\leq C|\log|x-y||^{2(1-\theta)/(1-2\eta)}.$

We therefore only need to consider the case where $x$ and $y$ are not too
close to each other, i.e. $|x-y|\geq e^{-a^{-1+\eta}}$. For $z\in\\{x,y\\}$,
we will denote by $D_{z}=D(z,|x-y|/2)$. By Lemma 2.9, the minimal disc centred
at $x$ containing $\Xi_{a}^{x}$ has typically a radius of order $e^{-O(1)/a}$.
It is therefore very unlikely that $\Xi_{a}^{x}$ exits $D_{x}$. We are first
going to show that this is still the case conditionally on the existence of a
cluster joining $x$ and $y$:

$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,\exists
z\in\\{x,y\\}:\Xi_{a}^{z}\not\subset D_{z}\right)\to
0\quad\quad\text{as~{}}\gamma\to 0$ (4.10)

and also that the left hand side can be bounded by
$C|\log|x-y||^{\frac{2(1-\theta)}{1-2\eta}}$, uniformly in $\gamma$. Thanks to
a union bound, we can focus on the event that $\Xi_{a}^{x}\not\subset D_{x}$.
With some abuse of notation, we will view $\Xi_{a}^{x}$ both as the collection
of excursions in a Poisson point process with intensity $a\mu_{D}^{x,x}$ and
as the loop formed by the concatenation of all these excursions. Let
$E_{x}^{1}$ (resp. $E_{x}^{2}$) be the event that there is a unique excursion
(resp. at least two distinct excursions) of $\Xi_{a}^{x}$ that exits $D_{x}$.
By Lemma 2.9, $\mu_{D}^{x,x}(\\{\wp\not\subset
D_{x}\\})=\log\frac{2\operatorname{CR}(x,D)}{|x-y|}.$ So

$\mathbb{P}\left(E_{x}^{2}\right)\leq\mathbb{E}\Big{[}\sum_{\wp\neq\wp^{\prime}\in\Xi_{a}^{x}}\mathbf{1}_{\left\\{\wp\not\subset
D_{x},\wp^{\prime}\not\subset
D_{x}\right\\}}\Big{]}=a^{2}\mu_{D}^{z,z}(\\{\wp\not\subset D_{x}\\})^{2}\leq
Ca^{2}|\log|x-y||^{2}.$

In that case, we simply bound

$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{x}^{2}\right)\leq
a^{-2(1-\theta)+o(1)}\mathbb{P}\left(E_{x}^{2}\right)\leq
Ca^{2\theta+o(1)}|\log|x-y||^{2}\leq
C|\log|x-y||^{\frac{2(1-\theta)}{1-2\eta}}.$

The probability of the event $E_{x}^{1}$ is at most $Ca|\log|x-y||$. On that
event, let $R_{x}$ be the minimal distance between $y$ and the unique
excursion of $\Xi_{a}^{x}$ that exists $D_{x}$. Let also $R_{y}$ be the
maximal $r>0$ such that there exists a cluster of loops of
$\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{y}$ that connects $y$ to $\partial
D(y,r)$. On the event $E_{1}^{x}$, in order to have a cluster joining $x$ and
$y$, we need to have $R_{x}\leq R_{y}$. Since $R_{y}$ and $R_{x}$ are
independent, we get that

$\displaystyle\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{x}^{1}\right)$
$\displaystyle\leq\frac{\mathbb{P}\left(E_{x}^{1}\right)}{Z_{\gamma}^{2}}\mathbb{P}\left(R_{x}\leq
R_{y}|E_{x}^{1}\right)$ (4.11)
$\displaystyle\leq\frac{Ca|\log|x-y||}{Z_{\gamma}^{2}}\int_{0}^{|x-y|}\mathbb{P}\left(R_{y}\in\mathrm{d}r\right)\mathbb{P}\left(R_{x}\leq
r|E_{x}^{1}\right).$

For all $r\in(0,|x-y|)$, the probability that $R_{x}$ is smaller than $r$ can
be compared with the probability that a Brownian path starting on the circle
$\partial D_{x}$ hits $D(y,r)$ before exiting the domain $D$. Hence for all
$r\in(0,|x-y|)$,

$\mathbb{P}\left(R_{x}\leq|x-y|\Big{|}E_{x}^{1}\right)\leq
C\frac{|\log|x-y||}{|\log r|}.$

Injecting this estimate in (4.11) and then integrating by parts leads to

$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{x}^{1}\right)\leq\frac{Ca|\log|x-y||^{2}}{Z_{\gamma}^{2}}\int_{0}^{|x-y|}\frac{\mathbb{P}\left(R_{y}\geq
r\right)}{r|\log r|^{2}}\mathrm{d}r.$

Now, by Lemma 2.11 (and conformal invariance to get back to the unit disc),

$\frac{1}{Z_{\gamma}}\mathbb{P}\left(R_{y}\geq r\right)\leq C|\log
r|^{1-\theta+\eta},\quad\quad r\in(0,|x-y|).$

We thus have

$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{x}^{1}\right)\leq\frac{Ca|\log|x-y||^{2}}{Z_{\gamma}}\int_{0}^{|x-y|}\frac{\mathrm{d}r}{r|\log
r|^{1+\theta-\eta}}\leq Ca^{\theta+o(1)}|\log|x-y||^{2-\theta+\eta}.$

In the last inequality we used that $a/Z_{\gamma}=a^{\theta+o(1)}$. This
concludes the proof of (4.10). This also shows that the left hand side of
(4.10) is bounded by $C|\log|x-y||^{\frac{2(1-\theta)}{1-2\eta}}$ because
$|x-y|\geq e^{-a^{-1+\eta}}$.

We can now work on the event that $\Xi_{a}^{z}\subset D_{z}$ for $z=x,y$. Let
$\mathcal{L}_{D,x,y}^{\theta}$ be the subset of $\mathcal{L}_{D}^{\theta}$
consisting of the loops that are not included in $D_{x}\cup D_{y}$. Let

$r_{z}=\inf\\{r>0:\exists\wp\in\mathcal{L}_{D,x,y}^{\theta}:\wp\cap
D(z,r)\neq\varnothing\\},\quad\quad z=x,y.$

By the restriction property of $\Xi_{a}^{x}$ (see Lemma 2.8),

$\displaystyle\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,\forall
z=x,y,\Xi_{a}^{z}\subset
D_{z}\right)\leq\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(\forall
z=x,y,z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z}}{\longleftrightarrow}\partial
D(z,r_{z}),\Xi_{a}^{z}\subset D_{z}\right)$
$\displaystyle=\frac{1}{Z_{\gamma}^{2}}\frac{\operatorname{CR}(x,D_{x})^{a}\operatorname{CR}(y,D_{y})^{a}}{\operatorname{CR}(x,D)^{a}\operatorname{CR}(y,D)^{a}}\mathbb{P}\left(\forall
z=x,y,z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\partial
D(z,r_{z})\right).$

In what follows, we will simply bound the ratio of conformal radii by 1 (which
is also a good approximation since $a\to 0$). By scale invariance of
$\mathcal{L}_{D_{z}}^{\theta}$ and $\Xi_{a}^{z,D_{z}}$ and by Lemma 2.11,

$\frac{1}{Z_{\gamma}}\mathbb{P}\left(z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\partial
D(z,r_{z})\Big{|}r_{z}\right)\leq
C\left(1+\log\frac{|x-y|}{2r_{z}}\right)^{1-\theta+\eta},\quad\quad z=x,y.$

Hence,

$\displaystyle\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,\forall
z=x,y,\Xi_{a}^{z}\subset D_{z}\right)\leq
C\mathbb{E}\left[\prod_{z=x,y}\left(\log\frac{|x-y|}{2r_{z}}\right)^{1-\theta+\eta}\right].$

In [JLQ23], we show that the above quantity is upper bounded by
$C|\log|x-y||^{2(1-\theta)+2\eta}$ (see (4.16) in Remark 4.5 therein)
concluding the proof of Lemma 4.2. ∎

We can now prove Proposition 4.3.

###### Proof of Proposition 4.3.

Let $x$ and $y$ be two distinct points in $D$. Compare to the upper bound, the
two points $x$ and $y$ are fixed and we wish to take the limit as
$\gamma_{1},\gamma_{2}\to 0$. It is therefore important for the proof to keep
in mind that $x$ and $y$ are at macroscopic distance to each other and that we
can assume $\gamma_{1}$ and $\gamma_{2}$ to be as small as desired. We start
by proving (4.3). We will explain at the end of the proof what needs to be
changed in order to get (4.4) and (4.6).

Let $\eta>0$ be small and let $D_{x}=D(x,\eta|x-y|/2)$ and
$D_{y}=D(y,\eta|x-y|/2)$. By taking $\eta>0$ small enough, we can ensure that
$D_{x}$ and $D_{y}$ are subsets of $D$. We will denote
$\mathcal{L}_{D,x,y}^{\theta}=\mathcal{L}_{D}^{\theta}\setminus(\mathcal{L}_{D_{x}}^{\theta}\cup\mathcal{L}_{D_{y}}^{\theta})$
the collection of loops which are not included in $D_{x}$ or $D_{y}$. We now
introduce the following events:

  * •

$E_{1}$: there is a loop in $\mathcal{L}_{D,x,y}^{\theta}$ that disconnects
$D_{x}$ from $D_{y}$;

  * •

$E_{2}$: for $z=x,y$, $\Xi_{a}^{z}\subset D_{z}$.

We first claim that we can work on $E_{1}\cap E_{2}$. Indeed, by Lemma 2.5
$\mathbb{P}\left(E_{1}^{c}\right)\to 0$ as $\eta\to 0$ and, by FKG inequality,

$\limsup_{\gamma\to
0}\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{1}^{c}\right)\to
0\quad\quad\text{as}\quad\eta\to 0.$

Moreover, in the proof of the upper bound, we showed that

$\lim_{\gamma\to
0}\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{2}^{c}\right)=0.$

See (4.10). We can therefore work on the event $E_{1}\cap E_{2}$. On the event
$E_{1}$, there is at most one cluster $\mathcal{C}_{x,y}$ of loops in
$\mathcal{L}_{D,x,y}^{\theta}$ that intersects both $D_{x}$ and $D_{y}$ (we
set $\mathcal{C}_{x,y}=\varnothing$ if there is no such cluster). By the
restriction property (2.16) of $\Xi_{a}^{z}$, we have

$\displaystyle\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{1}\cap
E_{2}\right)=\frac{\operatorname{CR}(x,D_{x})^{a}\operatorname{CR}(y,D_{y})^{a}}{\operatorname{CR}(x,D)^{a}\operatorname{CR}(y,D)^{a}}\mathbb{E}\left[\mathbf{1}_{E_{1}}\prod_{z=x,y}\mathbb{P}\left(z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\mathcal{C}_{x,y}\Big{|}\mathcal{C}_{x,y}\right)\right].$

The ratio of conformal radii converges to 1. Moreover, for $z=x$ or $y$, we
can rewrite

$\frac{1}{Z_{\gamma}}\mathbb{P}\Big{(}z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\mathcal{C}_{x,y}|\mathcal{C}_{x,y}\Big{)}=\mathbb{P}\Big{(}z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\mathcal{C}_{x,y}|\mathcal{C}_{x,y}\Big{)}\Big{/}\mathbb{P}\Big{(}z\overset{\mathcal{L}_{D_{z}}^{\theta}\cup\Xi_{a}^{z,D_{z}}}{\longleftrightarrow}\partial
D(z,e^{-1}\eta|x-y|/2)\Big{)}$

and we can replace $\mathcal{C}_{x,y}$ by the closure of
$(\mathcal{C}_{x,y}\cap D_{z})\cup\partial D_{z}$ which is a.s. a connected
compact subset of $\overline{D_{z}}\setminus\\{z\\}$. We are now in the
setting of Theorem 3.1 which shows that the above ratio converges as $a\to 0$,
a.s. with respect to $\mathcal{C}_{x,y}$. The upper bound (4.2) provides the
bound necessary to apply dominated convergence in order to exchange the
expectation w.r.t. $\mathcal{C}_{x,y}$ and the limit. We have proved that

$\lim_{\gamma\to
0}\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\left(x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y,E_{1}\cap
E_{2}\right)$ (4.12)

exists and that
$\frac{1}{Z_{\gamma}^{2}}\mathbb{P}\Big{(}x\overset{\mathcal{L}_{D}^{\theta}\cup\Xi_{a}^{x}\cup\Xi_{a}^{y}}{\longleftrightarrow}y\Big{)}$
converges as $\gamma\to 0$ towards the nondecreasing limit of (4.12) as
$\eta\to 0$. This shows (4.3).

The proof of (4.4) concerning the mixed case $(a_{1},a_{2})$ follows along the
same lines. The main difference comes from the fact that we only need a lower
bound, so we can add for free the extra events $E_{1}$ and $E_{2}$. The proof
of (4.6) is obtained similarly. Indeed, because the function $F$ does not
depend on loops that have diameter smaller than a given threshold
$\varepsilon$, $F$ does not depend on loops included in $D_{x}$ and $D_{y}$
provided $\eta$ is small enough. We then repeat the above procedure and use
Theorem 3.1.

Finally, the upper bound of (4.5) follows from Lemma 4.2. The lower bound is
much easier to prove and follows from Theorem 2.4 and FKG inequality. We omit
the details. ∎

##  5 $h_{\theta}$ as a sum of Minkowski measures: proof of Theorems 1.9 and
1.10

This section is devoted to the proofs of Theorems 1.9 and 1.10. We start by
proving Theorem 1.9. Recall the notations introduced above Theorem 1.9.

###### Proof of Theorem 1.9.

Let $k\geq 1$. Let $B\subset D$ be some Borel set. It is enough to show that
$(\mu_{k,r}(B))_{r>0}$ and $(\mu_{k,\gamma}(B))_{\gamma>0}$ are Cauchy in
$L^{2}$ and that $\mu_{k,r}(B)-\mu_{k,\gamma}(B)$ converges to 0 in $L^{2}$ as
$\gamma,r\to 0$. We already have all the ingredients to prove this. Indeed,

$\displaystyle\limsup_{r,r^{\prime}\to
0}\mathbb{E}\left[(\mu_{k,r}(B)-\mu_{k,r^{\prime}}(B))^{2}\right]=\limsup_{r,r^{\prime}\to
0}\frac{1}{Z_{r}Z_{r^{\prime}}}\int_{B\times
B}\Big{(}\mathbb{P}\left(\mathcal{C}_{k}\cap
D(x,r)\neq\varnothing,\mathcal{C}_{k}\cap D(y,r)\neq\varnothing\right)$
$\displaystyle~{}~{}~{}+\mathbb{P}\left(\mathcal{C}_{k}\cap
D(x,r^{\prime})\neq\varnothing,\mathcal{C}_{k}\cap
D(y,r^{\prime})\neq\varnothing\right)-2\mathbb{P}\left(\mathcal{C}_{k}\cap
D(x,r)\neq\varnothing,\mathcal{C}_{k}\cap
D(y,r^{\prime})\neq\varnothing\right)\Big{)}\mathrm{d}x\mathrm{d}y.$

The analogue of Lemma 4.2 with small discs instead of thick loops (and Fatou’s
lemma to deal with the mixed term $r$-$r^{\prime}$) provides the necessary
domination to exchange limit and integral. To show that the integrand
converges pointwise to zero, we then use the same approach as in the proof of
Proposition 4.3. Eventually, the proof of Theorem 1.9 boils down to Theorem
3.1. We omit the details.

The above line of argument also shows that the second moment of $\mu_{k}(D)$
is positive. In particular, the probability that $\mu_{k}(D)>0$ is positive.
We are going to show that this probability is actually equal to 1. Without
loss of generality assume that $D=[0,1]^{2}$. Let $\wp$ be some fixed
macroscopic loop of $\mathcal{L}_{D}^{\theta}$, e.g. the loop with the largest
diameter. Let $n\geq 1$ be a large integer and divide $D$ into disjoint
squares $Q_{i}$, $i=1,\dots,4^{n}$, of side length $2^{-n}$. For each
$i=1,\dots,4^{n}$, let $\mathcal{L}_{i}$ be the set of loops included in
$Q_{i}$. Conditionally on $\\{\text{diam}(\wp)>\sqrt{2}2^{-n}\\}$, the
collections $\mathcal{L}_{i}$, $i=1,\dots,4^{n}$, are independent from $\wp$
and independent loop soups in $D_{i}$. For $i=1,\dots,4^{n}$, let
$Q_{i}^{\prime}$ be the square with side length $2^{-n-1}$ with the same
centre as $Q_{i}$ and let $E_{i}$ be the event that there is a cluster
$\mathcal{C}_{i}$ of $\mathcal{L}_{i}$ with non-zero Minkowski content and
such that $\mathcal{C}_{i}$ disconnects $Q_{i}^{\prime}$ from $\partial
Q_{i}$. By scale invariance, $p=\mathbb{P}(E_{i})$ is positive and independent
of $n$ and $i$. Let $I=\\{i=1,\dots,4^{n}:\wp\cap
Q_{i}^{\prime}\neq\varnothing\\}$. On the event $E_{i}\cap\\{i\in I\\}$, the
cluster of $\wp$ contains $\mathcal{C}_{i}$. Hence, the probability that the
Minkowski content of the cluster of $\wp$ vanishes is at most

$\mathbb{P}\left(\text{diam}(\wp)\leq\sqrt{2}2^{-n}\right)+\mathbb{E}\left[(1-p)^{\\#I}\right]\xrightarrow[n\to\infty]{}0.$

This concludes the proof. ∎

The main item that remains to be checked is that, as stated in Theorem 1.10,
$\mu_{k}$ corresponds to the restriction of $|h_{\theta}|$ to
$\mathcal{C}_{k}$. The rest of this section is dedicated to showing this.

As usually, we will denote by $\overline{\mathcal{C}}_{k}$ the topological
closure of $\mathcal{C}_{k}$, seen as a compact subset of $D$. We will denote

$D_{k}=D\setminus\bigcup_{1\leq j\leq k}\overline{\mathcal{C}}_{j}.$

Then $D_{k}$ is an open subset of $D$ that has infinitely many connected
components, among which finitely many (at most $k$) are multiply connected and
the rest are simply connected, the total number of holes being $k$. We will
denote by

$\mathcal{L}^{\theta}_{D,k}=\\{\wp\in\mathcal{L}^{\theta}_{D}:\wp\subset
D_{k}\\}.$

We will denote by $\sigma_{k}$ the sign $\sigma_{\mathcal{C}_{k}}$ of the
cluster $\mathcal{C}_{k}$. We will denote by $\mathcal{F}_{k}$ the
$\sigma$-algebra generated by the loops contained in
$\mathcal{C}_{1},\dots,\mathcal{C}_{k}$ (including their time parametrization)
and the signs $\sigma_{1},\dots,\sigma_{k}$.

######  Lemma 5.1.

Fix $k\geq 1$. Let $O_{i}$, $i\geq 0$ be the connected components of $D_{k}$.
Conditionally on $\mathcal{F}_{k}$, we have:

  * •

The collections of loops $\\{\wp\in\mathcal{L}_{D}^{\theta}:\wp\subset
O_{i}\\},i\geq 0$, are independent;

  * •

For all $i\geq 0$ such that $O_{i}$ is simply connected,
$\\{\wp\in\mathcal{L}_{D}^{\theta}:\wp\subset O_{i}\\}$ has the law of a
Brownian loop soup in $O_{i}$ with intensity $\theta$;

  * •

For all $i\geq 0$ such that $O_{i}$ is not simply connected,
$\\{\wp\in\mathcal{L}_{D}^{\theta}:\wp\subset O_{i}\\}$ has the law of a
Brownian loop soup in $O_{i}$ with intensity $\theta$ conditioned on not
having clusters surrounding an inner hole of $O_{i}$.

###### Proof.

This result is folklore and has a very similar flavour as Lemma 8.5 and
follows from the same proof. We refer to Section 8 for more details. ∎

In particular, the collection of loops $\mathcal{L}^{\theta}_{D,k}$ is a
Brownian loop soup in $D_{k}$ conditioned on an event that has a positive
probability. So for $\gamma\in(0,2)$, one can define a multiplicative chaos
measure associated to $\mathcal{L}^{\theta}_{D,k}$. The fact that $D_{k}$ is
not simply connected is non essential: see Remark 2.2. The multiplicative
chaos measure associated to $\mathcal{L}^{\theta}_{D,k}$ is nothing else than

$\mathbf{1}_{\left\\{x\in
D_{k}\right\\}}e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k}}(x,x))}\mathcal{M}_{\gamma}(dx),$

where $G_{D}$ is the Dirichlet Green’s function on $D$ and $G_{D_{k}}$ is the
Dirichlet Green’s function on $D_{k}$. Note that
$G_{D}(x,x)-G_{D_{k}}(x,x)<+\infty$. We get a density and not just a
restriction to $D_{k}$ because our convention
$\mathbb{E}[\mathcal{M}_{\gamma}(dx)]=2dx$ makes the normalization for the
multiplicative chaos domain-dependent: how we normalize around a point $x$
depends also on the global shape of the domain and not just the Brownian loops
we see around $x$. With the normalisation used in [ABJL23], we would not get
this extra density: see (2.5) for the relation between these two
normalisations. One can further consider the measure
$\mathcal{M}_{\gamma}^{+}\mathbf{1}_{D_{k}}$ and the field

$h_{\theta,k}=\lim_{\gamma\to
0^{+}}\dfrac{1}{Z_{\gamma}}\mathbf{1}_{\left\\{x\in
D_{k}\right\\}}(e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k}}(x,x))}\mathcal{M}_{\gamma}^{+}(dx)-dx),$

which is a well defined random element of $H^{-\varepsilon}(\mathbb{C})$ for
$\varepsilon>0$ (simply by working in the domain $D_{k}$).

Further, one can consider the conditional expectation
$\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]$. Since
$\mathbb{E}[\|h_{\theta}\|^{2}_{H^{-\varepsilon}(\mathbb{C})}]<+\infty$ for
$\varepsilon>0$, the conditional expectation
$\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]$ is a well defined random element of
$H^{-\varepsilon}(\mathbb{C})$, with

$\|\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]\|^{2}_{H^{-\varepsilon}(\mathbb{C})}\leq\mathbb{E}[\|h_{\theta}\|^{2}_{H^{-\varepsilon}(\mathbb{C})}|\mathcal{F}_{k}]<+\infty~{}~{}\text{a.s.}$

See [HvNVW16, Section 2.6] for a reference on conditional expectation of
random variables in a Banach space. We define
$\nu_{1}=\sigma_{1}\mathbb{E}[h_{\theta}|\mathcal{F}_{1}]$, and for $k\geq 2$,

$\nu_{k}=\sigma_{k}(\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]-\mathbb{E}[h_{\theta}|\mathcal{F}_{k-1}]),\quad\quad\text{so~{}that}\quad\quad\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]=\sum_{j=1}^{k}\sigma_{j}\nu_{j}.$

######  Lemma 5.2.

The following holds.

  1. 1.

For every $k\geq 1$, conditionally on $\mathcal{F}_{k}$, $h_{\theta,k}$ has
the same law as $-h_{\theta,k}$, and in particular
$\mathbb{E}[h_{\theta,k}|\mathcal{F}_{k}]=0$.

  2. 2.

A.s., for every $k\geq 1$ and every
$f\in\mathcal{C}^{\infty}_{c}(\mathbb{C})$, such that
$\operatorname{Supp}(f)\cap(D\setminus D_{k})=\varnothing$,
$(h_{\theta},f)=(h_{\theta,k},f)$.

  3. 3.

For every $k\geq 1$,
$h_{\theta}=\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]+h_{\theta,k}$ a.s.

  4. 4.

For every $\varepsilon>0$, as $k\to+\infty$,
$\|h_{\theta}-\mathbb{E}[h_{\theta}|\mathcal{F}_{k}]\|_{H^{-\varepsilon}(\mathbb{C})}\to
0$ a.s. and in $L^{2}$.

  5. 5.

For every $k\geq 2$, $h_{\theta,k-1}=\sigma_{k}\nu_{k}+h_{\theta,k}$ a.s. and

$\nu_{k}=\sigma_{k}\mathbb{E}[h_{\theta,k-1}|\mathcal{F}_{k}]~{}~{}\text{a.s.}$

  6. 6.

$\nu_{1}$ and $\sigma_{1}$ are independent, and for every $k\geq 2$, $\nu_{k}$
and $\sigma_{k}$ are independent conditionally on $\mathcal{F}_{k-1}$.

  7. 7.

A.s., for every $k\geq 1$ and every $f\in\mathcal{C}^{\infty}_{c}(\mathbb{C})$
such that $\operatorname{Supp}(f)\cap\overline{\mathcal{C}}_{k}=\varnothing$,
$(\nu_{k},f)=0$, that is to say $\nu_{k}$ is supported on
$\overline{\mathcal{C}}_{k}$.

###### Proof.

1\. The proof is the same as for $h_{\theta}$ (see Section 7.1).

2\. Note that we also deal with test functions that are not compactly
supported in $D$. Let $k\geq 1$ and $n\geq 1$. Let $\mathbb{C}_{k,n}$ be the
open subset of $\mathbb{C}$ formed by points $x\in\mathbb{C}$ such that
$d(x,D\setminus D_{k})>1/n$. By construction, for every
$f\in\mathcal{C}^{\infty}_{c}(\mathbb{C}_{k,n})$,

$(h_{\theta},f)=\lim_{\gamma\to
0^{+}}\dfrac{1}{Z_{\gamma}}\int_{D_{k}}f(x)(\mathcal{M}_{\gamma}^{+}(dx)-dx),$

where the convergence holds in $L^{2}$ for a given Sobolev norm
$H^{-\varepsilon}(\mathbb{C})$ (and thus for all test functions
simultaneously). Further, since $G_{D}(x,x)-G_{D_{k}}(x,x)$ is bounded on
$\mathbb{C}_{k,n}$, the convergence and the limit do no change if we replace
$\mathcal{M}_{\gamma}^{+}$ by

$e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k}}(x,x))}\mathcal{M}_{\gamma}^{+}(dx).$

3\. By 1., we only need to show that $h_{\theta}-h_{\theta,k}$ is measurable
w.r.t. $\mathcal{F}_{k}$. Actually will see that $h_{\theta}-h_{\theta,k}$ is
measurable w.r.t. the $\sigma$-algebra $\mathcal{F}_{k}$ augmented by the
negligible events (i.e. with probability $0$), which also implies the desired
result.

Consider the sets $\mathbb{C}_{k,n}$ defined above. Let $\mathcal{F}_{k,n}$ be
the $\sigma$-algebra generated by $\mathcal{F}_{k}$ as well as by the loops of
$\mathcal{L}^{\theta}_{D,k}$ that are in the clusters at distance at most
$1/n$ from $D\setminus D_{k}$, and the signs of these clusters. By definition,
$\mathcal{F}_{k}\subset\mathcal{F}_{k,n}$ and
$\mathcal{F}_{k,n+1}\subset\mathcal{F}_{k,n}$. Denote

$\mathcal{F}_{k,\infty}=\bigcap_{n\geq 1}\mathcal{F}_{k,n}.$

We have $\mathcal{F}_{k}\subset\mathcal{F}_{k,\infty}$.

For $k\geq 1$ and $n\geq 1$, let $\chi_{k,n}$ be a cutoff function such that
$\chi_{k,n}\in\mathcal{C}^{\infty}(\mathbb{C})$, $\chi_{k,n}$ takes values in
$[0,1]$, $\chi_{k,n}$ equals $0$ on $\mathbb{C}_{k,n}$ and $1$ on
$\mathbb{C}\setminus\mathbb{C}_{k,n+1}$. We also want $\chi_{k,n}$ to be, as a
random variable, measurable w.r.t. $D_{k}$. We omit the detailed construction
of such a cutoff function, which is standard. Fix $k\geq 1$ and $n\geq 1$. For
$m\geq 1$ we have $h_{\theta}=h_{\theta}\chi_{k,m}+h_{\theta}(1-\chi_{k,m})$,
and by 2., $h_{\theta}(1-\chi_{k,m})=h_{\theta,k}(1-\chi_{k,m})$. Moreover,
for $m\geq n$, $h_{\theta}\chi_{k,m}$ is measurable w.r.t.
$\mathcal{F}_{k,n}$. Further, for $\varepsilon>0$,

$\lim_{m\to+\infty}\mathbb{E}[\|h_{\theta,k}\chi_{k,m}\|_{H^{-\varepsilon}(\mathbb{C})}^{2}|\mathcal{F}_{k}]=0~{}~{}\text{a.s.}$

by a computation similar to (4.8). Therefore,

$\lim_{m\to+\infty}\mathbb{E}[\|(h_{\theta}-h_{\theta,k})-h_{\theta}\chi_{k,m}\|_{H^{-\varepsilon}(\mathbb{C})}^{2}|\mathcal{F}_{k}]=0~{}~{}\text{a.s.}$

So in particular, $h_{\theta}-h_{\theta,k}$ is the limit in probability of
$h_{\theta}\chi_{k,m}$ as $m\to+\infty$. Therefore, $h_{\theta}-h_{\theta,k}$
is measurable w.r.t. $\mathcal{F}_{k,n}$. By letting $n\to+\infty$, we get
that $h_{\theta}-h_{\theta,k}$ is measurable w.r.t. $\mathcal{F}_{k,\infty}$.

To conclude, it is enough to check that the $\sigma$-algebra
$\mathcal{F}_{k,\infty}$ is the same as $\mathcal{F}_{k}$ up to negligible
events. This is what is sketched below, by omitting some details.

  * •

Step 1: Let $\mathcal{F}^{\prime}_{k,n}$ be the $\sigma$-algebra generated by
$\mathcal{F}_{k}$ as well as by the loops of $\mathcal{L}^{\theta}_{D,k}$ that
are in the clusters at distance at most $1/n$ from $D\setminus D_{k}$.
Contrary to $\mathcal{F}_{k,n}$, $\mathcal{F}^{\prime}_{k,n}$ does not contain
any information on the sign of clusters of $\mathcal{L}^{\theta}_{D,k}$.
Denote

$\mathcal{F}^{\prime}_{k,\infty}=\bigcap_{n\geq 1}\mathcal{F}^{\prime}_{k,n}.$

By construction,
$\mathcal{F}_{k}\subset\mathcal{F}^{\prime}_{k,\infty}\subset\mathcal{F}_{k,\infty}$.
We further claim that $\mathcal{F}_{k,\infty}$ is the same as
$\mathcal{F}^{\prime}_{k,\infty}$ up to negligible events. Indeed, the signs
of clusters are sampled independently of the loops in
$\mathcal{L}^{\theta}_{D,k}$ and are i.i.d. Moreover, as $n$ increases,
$\mathcal{F}_{k,n}$ contains information about less clusters and their signs,
and in particular, $\mathcal{F}_{k,\infty}$ contains information only on the
tail of the sequence of signs. So essentially this follows from the $0-1$ law
of the tails of sequences of independent r.v.s.

  * •

Step 2: Let $\mathcal{F}^{\prime\prime}_{k,n}$ be the $\sigma$-algebra
generated by $\mathcal{F}_{k}$ as well as by the loops of
$\mathcal{L}^{\theta}_{D,k}$ that are at distance at most $1/n$ from
$D\setminus D_{k}$. So contrary to $\mathcal{F}^{\prime}_{k,n}$, we look only
at individual loops and do not consider how they are grouped into clusters. By
construction,
$\mathcal{F}^{\prime\prime}_{k,n}\subset\mathcal{F}^{\prime}_{k,n}$. Denote

$\mathcal{F}^{\prime\prime}_{k,\infty}=\bigcap_{n\geq
1}\mathcal{F}^{\prime\prime}_{k,n}.$

We claim that $\mathcal{F}^{\prime}_{k,\infty}$ is the same as
$\mathcal{F}^{\prime\prime}_{k,\infty}$ up to negligible events. For $m>n\geq
1$, let $A_{k,m,n}$ be the event that all the clusters of
$\mathcal{L}^{\theta}_{D,k}$ that are at distance at most $1/m$ from
$D\setminus D_{k}$, are actually made of loops that stay in the
$1/n$-neighborhood of $D\setminus D_{k}$. Then
$A_{k,m,n}\in\mathcal{F}^{\prime\prime}_{k,n}$ and for every
$A\in\mathcal{F}^{\prime}_{k,m}$, we have $A\cap
A_{k,m,n}\in\mathcal{F}^{\prime\prime}_{k,n}$. Moreover,

$\lim_{m\to+\infty}\mathbb{P}(A_{k,m,n}^{\rm c})=0.$

Thus, $\mathcal{F}^{\prime}_{k,\infty}$ is contained in
$\mathcal{F}^{\prime\prime}_{k,n}$ augmented by the negligible events,
whatever the value of $n$.

  * •

Step 3: We have that
$\mathcal{F}_{k}\subset\mathcal{F}^{\prime\prime}_{k,\infty}$. We claim that
$\mathcal{F}^{\prime\prime}_{k,\infty}$ is the same as $\mathcal{F}_{k}$ up to
negligible events. If the conditional distribution of
$\mathcal{L}^{\theta}_{D,k}$ given $\mathcal{F}_{k}$ were exactly Poisson,
this would have been immediate by $0-1$ law. However, by Lemma 5.1, this
conditional distribution is Poisson conditioned on an event with positive
probability, so the $0-1$ law still holds.

4\. This is simply due to the fact that the $\sigma$-algebra $\bigvee_{k\geq
1}\mathcal{F}_{k}$ generated by the union of the $\mathcal{F}_{k}$ is the
total $\sigma$-algebra generated by the loop soup $\mathcal{L}^{\theta}_{D}$
and the signs clusters.

5\. Since

$\mathbb{E}[h_{\theta}|\mathcal{F}_{k-1}]+h_{\theta,k-1}=\mathbb{E}[h_{\theta}|\mathcal{F}_{k-1}]+\sigma_{k}\nu_{k}+h_{\theta,k}=h_{\theta},$

we get that $\sigma_{k}\nu_{k}=h_{\theta,k-1}-h_{\theta,k}$. Since
$\sigma_{k}\nu_{k}$ is $\mathcal{F}_{k}$-measurable and
$\mathbb{E}[h_{\theta,k}|\mathcal{F}_{k}]=0$, we conclude by taking the
conditional expectation w.r.t. $\mathcal{F}_{k}$.

6\. Since $-h_{\theta}$ has the same law as $h_{\theta}$, we have that

$\nu_{1}\mathbf{1}_{\left\\{\sigma_{1}=-1\right\\}}=-\mathbb{E}[h_{\theta}|\mathcal{F}_{1}]\mathbf{1}_{\left\\{\sigma_{1}=-1\right\\}}=\mathbb{E}[-h_{\theta}|\mathcal{F}_{1}]\mathbf{1}_{\left\\{\sigma_{1}=-1\right\\}}\stackrel{{\scriptstyle\text{(d)}}}{{=}}\mathbb{E}[h_{\theta}|\mathcal{F}_{1}]\mathbf{1}_{\left\\{\sigma_{1}=1\right\\}}=\nu_{1}\mathbf{1}_{\left\\{\sigma_{1}=1\right\\}},$

which gives the independence between $\nu_{1}$ and $\sigma_{1}$. For $k\geq
2$, we use the same reasoning, by relying on points 5. and 1.

7\. From 3. and 2. follows that a.s. for every
$f\in\mathcal{C}^{\infty}_{c}(\mathbb{C}\setminus\overline{\mathcal{C}}_{1})$,
$(\mathbb{E}[h_{\theta}|\mathcal{F}_{1}],f)=0$. So this concludes in the case
$k=1$. Now take $k\geq 2$. By 5., for every
$f\in\mathcal{C}^{\infty}_{c}(\mathbb{C})$,

$\sigma_{k}(\nu_{k},f)=(h_{\theta,k-1}-h_{\theta,k},f).$

Then, similarly to 2., we get that $(h_{\theta,k-1},f)=(h_{\theta,k},f)$ for
every
$f\in\mathcal{C}^{\infty}_{c}(\mathbb{C}\setminus\overline{\mathcal{C}}_{k})$.
∎

By convention, we set $D_{0}=D$. Let $D_{k}^{\ast}$ be the open subset of
$D_{k}$ obtained as the union of the non-simply connected components of
$D_{k}$, which are finitely many.

######  Lemma 5.3.

The measure

$\dfrac{1}{Z_{\gamma}}(\mathbf{1}_{\left\\{x\in
D_{k}^{\ast}\right\\}}dx-\mathbb{E}[\mathbf{1}_{x\in
D_{k}^{\ast}}e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}_{\gamma}^{+}(dx))|\mathcal{F}_{k}])$
(5.1)

is a.s. a positive finite measure that converges a.s. to $0$ for the weak
topology of measures on $\overline{D_{k}^{\ast}}$ as $\gamma\to 0$.

###### Proof.

We start by proving that (5.1) is a positive measure. Let
$\mathcal{L}^{\theta}_{D_{k}^{\ast}}$ be the collection of Brownian loops,
distributed conditionally on $D_{k}$, as the Brownian loop soup in
$D_{k}^{\ast}$ with intensity $\theta$ (that is without any topological
conditioning). Let $\mathcal{M}_{D_{k}^{\ast},\gamma}$ denote the
multiplicative chaos of $\mathcal{L}^{\theta}_{D_{k}^{\ast}}$, normalized by
$\mathbb{E}[\mathcal{M}_{D_{k}^{\ast},\gamma}|D_{k}]=\mathbf{1}_{\left\\{x\in
D_{k}^{\ast}\right\\}}2dx$. Let $E_{k}$ be the event that no cluster of loops
in $\mathcal{L}^{\theta}_{D_{k}^{\ast}}$ surrounds an inner hole of
$D_{k}^{\ast}$, which is an event with positive probability. Then

$\mathbb{E}[\mathbf{1}_{x\in
D_{k}^{\ast}}e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}_{\gamma}^{+}(dx))|\mathcal{F}_{k}]=\dfrac{1}{2}\mathbb{E}\Big{[}e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}_{D_{k}^{\ast},\gamma}(dx)\Big{|}D_{k}^{\ast},E_{k}\Big{]}.$

Note that the Dirichlet Green’s function on $D_{k}^{\ast}$ is the restriction
of $G_{D_{k}}$ to $D_{k}^{\ast}\times D_{k}^{\ast}$. The event $E_{k}$ is a
decreasing event. Therefore, by FKG inequality for Poisson point processes,
a.s.

$\mathbb{E}[\mathbf{1}_{x\in
D_{k}^{\ast}}e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}_{\gamma}^{+}(dx))|\mathcal{F}_{k}]\\\
\leq\dfrac{1}{2}\mathbb{E}\Big{[}e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}_{D_{k}^{\ast},\gamma}(dx)\Big{|}D_{k}^{\ast}\Big{]}=\mathbf{1}_{\left\\{x\in
D_{k}^{\ast}\right\\}}e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}dx.$

Thus, for every $\gamma\in(0,2)$, (5.1) is a positive finite measure.

The rest of the proof is then dedicated to showing that (5.1) goes to zero as
$\gamma\to 0$. Define $f(x)=e^{2\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}$. We can
write the total mass of (5.1) as

$\dfrac{1}{Z_{\gamma}}\int_{D_{k}^{\ast}}(1-e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))})dx-\dfrac{1}{2Z_{\gamma}\mathbb{P}(E_{k})}\mathbb{E}\Big{[}\Big{(}(\mathcal{M}_{D_{k}^{\ast},\gamma},f^{\gamma^{2}/2})-2\int_{D_{k}^{\ast}}f^{\gamma^{2}/2}\Big{)}\mathbf{1}_{E_{k}}\Big{|}D_{k}^{\ast}\Big{]}.$
(5.2)

We are going to show separately that each of these two terms vanishes as
$\gamma\to 0$. By Cauchy–Schwarz and then by (2.13), the absolute value of the
second term is at most

$\displaystyle\dfrac{1}{2Z_{\gamma}\mathbb{P}(E_{k})^{1/2}}\mathbb{E}\Big{[}\Big{(}(\mathcal{M}_{D_{k}^{\ast},\gamma},f^{\gamma^{2}/2})-2\int_{D_{k}^{\ast}}f^{\gamma^{2}/2}\Big{)}^{2}\Big{|}D_{k}^{\ast}\Big{]}^{1/2}$
$\displaystyle=\frac{2}{Z_{\gamma}\mathbb{P}(E_{k})^{1/2}}\left(\int_{D_{k}^{\ast}\times
D_{k}^{\ast}}\big{(}(\gamma^{2}\pi
G_{D_{k}^{\ast}}(x,y))^{1-\theta}\Gamma(\theta)I_{\theta-1}(2\gamma^{2}\pi
G_{D_{k}^{\ast}}(x,y))-1\big{)}f(x)^{\gamma^{2}/2}f(y)^{\gamma^{2}/2}\,dx\,dy\right)^{1/2}$
$\displaystyle=O(\gamma^{2}/Z_{\gamma}).$

The last estimate can be obtained by expanding $I_{\theta-1}$ (see (2.11)).
Hence the second term of (5.2) vanishes as $\gamma\to 0$.

To conclude the proof, it remains to show that

$\dfrac{1}{Z_{\gamma}}\int_{D_{k}^{\ast}}(1-e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))})dx\to
0\quad\quad\text{as}\quad\gamma\to 0.$ (5.3)

For fixed $x\in D_{k}^{\ast}$, the density
$1-e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))})$ is of order
$\gamma^{2}$. Since $Z_{\gamma}\gg\gamma^{2}$, the density converges to $0$,
actually uniformly on compact subsets of $D_{k}^{\ast}$. However, this is not
enough for our purpose and we need to control what happens near the boundary
$\partial D_{k}^{\ast}$.

The boundary $\partial D_{k}^{\ast}$ has exactly $k+1$ connected components,
one of them being $\partial D$, and the $k$ others being the outer boundaries
of $\mathcal{C}_{1},\dots,\mathcal{C}_{k}$. $\partial D$ and
$\mathcal{C}_{1},\dots,\mathcal{C}_{k-1}$ lie outside both $D_{k-1}$ and
$D_{k}$. Thus $G_{D_{k-1}}(x,x)-G_{D_{k}}(x,x)$ stays bounded in a
neighborhood of $\partial D\cup\mathcal{C}_{1}\cup\dots\cup\mathcal{C}_{k-1}$.
Therefore, the uniform convergence of the density to $0$ is also true on
compact neighborhoods of these $k$ boundary components of $\partial
D_{k}^{\ast}$. On the other hand, $\mathcal{C}_{k}$ is contained in $D_{k-1}$
and lies outside of $D_{k}$. We will therefore have to treat this case
separately.

Let us denote by $\partial_{o}\mathcal{C}_{k}$ the outer boundary of
$\mathcal{C}_{k}$. Near $\partial_{o}\mathcal{C}_{k}$,
$G_{D_{k-1}}(x,x)-G_{D_{k}}(x,x)$ explodes to $+\infty$. But there is a
constant $c(D_{k-1})\in(0,1)$, depending on $D_{k-1}$, such that for every
$x\in D_{k}^{\ast}$,

$e^{2\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}\geq c(D_{k-1})\dfrac{d(x,\partial
D_{k}^{\ast})}{d(x,\partial D_{k-1})}.$

This is in a way a generalization of Koebe quarter theorem to domains with
holes; see [ALS20a, Lemma 5.13]. Fix
$\varepsilon_{0}\in(0,d(\partial_{o}\mathcal{C}_{k},\partial
D_{k}^{\ast}\setminus\partial_{o}\mathcal{C}_{k}))$. Then for every $x\in
D_{k}^{\ast}$ such that
$d(x,\partial_{o}\mathcal{C}_{k})<1\wedge(\varepsilon_{0}/2)$, we have

$\dfrac{1}{Z_{\gamma}}(1-e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))})\leq\dfrac{1}{Z_{\gamma}}\Big{(}1-c(D_{k-1})^{\gamma^{2}/2}\dfrac{d(x,\partial
D_{k}^{\ast})^{\gamma^{2}/2}}{d(x,\partial D_{k-1})^{\gamma^{2}/2}}\Big{)}\\\
\leq\dfrac{1}{Z_{\gamma}}\Big{(}1-c(D_{k-1})^{\gamma^{2}/2}\dfrac{d(x,\partial_{o}\mathcal{C}_{k})^{\gamma^{2}/2}}{\operatorname{diam}(D)^{\gamma^{2}/2}}\Big{)}=\dfrac{1}{Z_{\gamma}}(1-d(x,\partial_{o}\mathcal{C}_{k})^{\gamma^{2}/2})+\dfrac{1}{Z_{\gamma}}\Big{(}1-\dfrac{c(D_{k-1})^{\gamma^{2}/2}}{\operatorname{diam}(D)^{\gamma^{2}/2}}\Big{)}d(x,\partial_{o}\mathcal{C}_{k})^{\gamma^{2}/2}$

Since

$\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\Big{(}1-\dfrac{c(D_{k-1})^{\gamma^{2}/2}}{\operatorname{diam}(D)^{\gamma^{2}/2}}\Big{)}=0,$

we only need to consider the term
$(1-d(x,\partial_{o}\mathcal{C}_{k})^{\gamma^{2}/2})/Z_{\gamma}$. Further,

$\dfrac{1}{Z_{\gamma}}\int_{d(x,\partial_{o}\mathcal{C}_{k})<1}(1-d(x,\partial_{o}\mathcal{C}_{k})^{\gamma^{2}/2})\,dx=\dfrac{\gamma^{2}}{2Z_{\gamma}}\int_{0}^{1}dr\,r^{\gamma^{2}/2-1}\int_{d(x,\partial_{o}\mathcal{C}_{k})<r}dx$
(5.4)

Since $\partial_{o}\mathcal{C}_{k}$ is an SLEκ(θ) type curve, it is of
dimension $1+\kappa(\theta)/8\leq 3/2<2$ [Bef08]. In particular, for
$\beta\in(0,1/2)$, there is a.s. a (random) constant $c(\beta)>0$ such that
for every $r\in(0,1)$,

$\int_{d(x,\partial_{o}\mathcal{C}_{k})<r}dx\leq c(\beta)r^{\beta}.$

Thus, (5.4) is bounded by

$c(\beta)\dfrac{\gamma^{2}}{2Z_{\gamma}}\int_{0}^{1}r^{\gamma^{2}/2+\beta-1}\,dr,$

which tends to $0$ as $\gamma$ to $0$. This concludes. ∎

We now have all the ingredients to prove Theorem 1.10.

###### Proof of Theorem 1.10.

The only thing that remains to be proved is that the random generalised
function $\nu_{k}$ agrees a.s. with the measure $\mu_{k}$ from Theorem 1.9.
Let $f\in\mathcal{C}^{\infty}_{c}(\mathbb{C})$. It is enough to show that

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\int_{d(x,\mathcal{C}_{k})<1}f(x)(1-d(x,\mathcal{C}_{k})^{\gamma^{2}/2})\,dx,$
(5.5)

in probability. We have that

$(\nu_{k},f)=\sigma_{k}\mathbb{E}[(h_{\theta,k-1},f)|\mathcal{F}_{k}],$

where by convention, $h_{\theta,0}=h_{\theta}$. Since $\nu_{k}$ and
$\sigma_{k}$ are independent conditionally on $\mathcal{F}_{k}$, we get that

$(\nu_{k},f)=-\mathbb{E}[(h_{\theta,k-1},f)|\mathcal{F}_{k},\\{\sigma_{k}=-1\\}]$
(5.6)

Since

$(h_{\theta,k-1},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\int_{D_{k-1}}f(x)(e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}^{+}_{\gamma}(dx)-dx),$

where the convergence is a.s. in $L^{2}$ conditionally on $\mathcal{F}_{k-1}$
(that is to say $\lim_{\gamma\to
0}\mathbb{E}[(\cdot)^{2}|\mathcal{F}_{k-1}]=0$ a.s.). Therefore, in (5.6) one
can interchange the conditional expectation and the limit in $\gamma$ and get

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\Big{(}\int_{D_{k-1}}f(x)\,dx-\mathbb{E}\Big{[}\int_{D_{k-1}}f(x)e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}^{+}_{\gamma}(dx)\Big{|}\mathcal{F}_{k},\\{\sigma_{k}=-1\\}\Big{]}\Big{)},$

where the convergence is a.s. in $L^{2}$ conditionally on $\mathcal{F}_{k-1}$.
By Theorem 2.4, $\overline{\mathcal{C}}_{k}$ has a.s. zero Lebesgue measure
(the expectation of the Lebesgue measure of its $r$-neighbourhood vanishes as
$r\to 0$). So $\int_{D_{k-1}}f(x)\,dx=\int_{D_{k}}f(x)\,dx$ a.s. Moreover,
since on the event $\\{\sigma_{k}=-1\\}$,
$\mathbf{1}_{\left\\{D_{k-1}\right\\}}\mathcal{M}^{+}_{\gamma}=\mathbf{1}_{\left\\{D_{k}\right\\}}\mathcal{M}^{+}_{\gamma}$,
and since $\sigma_{k}$ is independent from
$\mathbf{1}_{\left\\{D_{k}\right\\}}\mathcal{M}^{+}_{\gamma}$ conditionally on
$\mathcal{F}_{k}$, we get that

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\Big{(}\int_{D_{k}}f(x)\,dx-\mathbb{E}\Big{[}\int_{D_{k}}f(x)e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}^{+}_{\gamma}(dx)\Big{|}\mathcal{F}_{k}\Big{]}\Big{)}.$

Now we partition $D_{k}=D_{k}^{0}\cup D_{k}^{\ast}$, where $D_{k}^{0}$ is the
union of the simply connected components of $D_{k}$ (there are infinitely many
of them) and $D_{k}^{\ast}$ is the union of the multiply connected components
of $D_{k}$ (there are finitely many of them). Then Lemma 5.3 ensures that

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\Big{(}\int_{D_{k}^{0}}f(x)\,dx-\mathbb{E}\Big{[}\int_{D_{k}^{0}}f(x)e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}^{+}_{\gamma}(dx)\Big{|}\mathcal{F}_{k}\Big{]}\Big{)}.$

Since conditionally on $\mathcal{F}_{k}$, the loops in $D_{k}^{0}$ form a
Poisson point process without conditioning, we get that

$\mathbb{E}\Big{[}\int_{D_{k}^{0}}f(x)e^{\gamma^{2}\pi(G_{D}(x,x)-G_{D_{k-1}}(x,x))}\mathcal{M}^{+}_{\gamma}(dx)\Big{|}\mathcal{F}_{k}\Big{]}=\int_{D_{k}^{0}}f(x)e^{\gamma^{2}\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}\,dx.$

The function $G_{D_{k-1}}(x,x)-G_{D_{k}}(x,x)$ is bounded for $x$ away from
$\mathcal{C}_{k}$ and explodes when $x$ approaches $\partial\mathcal{C}_{k}$.
Further, [ALS20a, Lemma 5.13] provides a comparison with the Euclidean
distance: there is a constant $c(D_{k-1})\in(0,1)$ depending on $D_{k-1}$ such
that for every $x\in D_{k}^{0}$,

$c(D_{k-1})\dfrac{d(x,\partial D_{k}^{0})}{d(x,\partial D_{k-1})}\leq
e^{2\pi(G_{D_{k}}(x,x)-G_{D_{k-1}}(x,x))}\leq 4\dfrac{d(x,\partial
D_{k}^{0})}{d(x,\partial D_{k-1})}.$

Then using these bounds, similarly to the proof of (5.3), we get that

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\int_{\begin{subarray}{c}x\in D_{k}^{0}\\\
d(x,\mathcal{C}_{k})<1\end{subarray}}f(x)(1-d(x,\mathcal{C}_{k})^{\gamma^{2}/2})\,dx.$

Since the dimension of the outer boundary of $\mathcal{C}_{k}$ is strictly
smaller than $2$, we get that (see the proof of (5.3))

$(\nu_{k},f)=\lim_{\gamma\to
0}\dfrac{1}{Z_{\gamma}}\int_{\begin{subarray}{c}d(x,\mathcal{C}_{k})<1\end{subarray}}f(x)(1-d(x,\mathcal{C}_{k})^{\gamma^{2}/2})\,dx,$

which concludes. ∎

##  6 Excursions from the boundary of a cluster

In this section, we will obtain estimates about the excursions induced by the
Brownian loops that touch the outer boundary of a loop soup cluster.

Let $\mathcal{L}$ be a loop soup with intensity $\theta\in(0,1/2]$ in the unit
disk $\mathbb{D}$, and let $\mathcal{C}_{0}$ be the outermost cluster in
$\mathcal{L}$ that encircles the origin. Let $O$ be the domain encircled by
the outermost boundary of $\mathcal{C}_{0}$. We will denote by
$\mathcal{L}_{O}$ the subset of $\mathcal{L}$ consisting in loops included in
$\overline{O}$. The set of loops in $\mathcal{L}_{O}$ that touch $\partial O$
induce a collection $\mathcal{E}(\mathcal{L}_{O})$ of excursions in $O$ with
endpoints in $\partial O$. If one maps $O$ onto $\mathbb{H}$ by some conformal
map $f$, then we say that $f(\mathcal{L}_{O})$ is a _loop soup in $\mathbb{H}$
with wired boundary conditions_. It was shown in [QW19] that the law of
$f(\mathcal{L}_{O})$ is conformally invariant, and does not depend on the
choice of $f$. Moreover, the following properties were proved in [QW19, QW18].

######  Theorem 6.1 ([QW19, QW18]).

The law of $f(\mathcal{L}_{O})$ is independent from $\partial O$ and the
distribution of $\mathcal{L}$ outside of $O$. Inside $f(\mathcal{L}_{O})$ (a
loop soup in $\mathbb{H}$ at intensity $\theta\in(0,1/2]$ with wired boundary
conditions), the set of loops that touch $\mathbb{R}$ are independent from the
set of loops that stay in $\mathbb{H}$. The latter set is distributed as a
loop soup in $\mathbb{H}$ (with free boundary conditions). Let $\mathcal{E}$
be the collection of excursions induced by the loops in $f(\mathcal{L}_{O})$
that touch $\mathbb{R}$. Then at $\theta=1/2$, $\mathcal{E}$ is distributed as
a Poisson point process of excursions of intensity $1/4$.

As one knows precisely the law of $\mathcal{E}$ for $\theta=1/2$, much less is
known for the subcritical intensities $\theta\in(0,1/2)$. However, [Qia19]
proved that the excursions attached to the boundary of a cluster satisfy a
conformal restriction property that we describe now (we give some background
on conformal restriction measures in Section 6.1). One can explore the outer
boundaries of the outermost clusters in $\mathcal{L}$ progressively, instead
of discovering an entire cluster at once, using the CLE exploration [SW12].
This exploration process allows one to define a _loop soup wired on a portion
of its boundary_ (see [Qia19] for a precise definition). Such a configuration
was shown to be conformally invariant (with two marked points on the
boundary). As a consequence, it is sufficient to consider a loop soup in the
upper half plane $\mathbb{H}$ which is wired on $\mathbb{R}^{-}$ and free on
$\mathbb{R}^{+}$.

######  Theorem 6.2 ([Qia19]).

Let $\mathcal{L}_{\mathbb{R}^{-}}$ be a loop soup in $\mathbb{H}$ with
intensity $\theta\in(0,1/2]$ which is wired on $\mathbb{R}^{-}$ and free on
$\mathbb{R}^{+}$. The set of loops in $\mathcal{L}_{\mathbb{R}^{-}}$ that
touch $\mathbb{R}^{-}$ are independent from the set of loops that stay in
$\mathbb{H}$. The latter set is distributed as a loop soup in $\mathbb{H}$
(with free boundary conditions). Let $\mathcal{E}_{\mathbb{R}^{-}}$ be the
collection of excursions induced by the loops in
$\mathcal{L}_{\mathbb{R}^{-}}$ that touch $\mathbb{R}$. Then the filling of
$\mathcal{E}_{\mathbb{R}^{-}}$ (namely the complement in $\mathbb{C}$ of the
unbounded connected component of $\mathbb{C}\setminus K$, where $K$ is the
closure of the union of all the excursions in $\mathcal{E}_{\mathbb{R}^{-}}$)
satisfies one-sided restriction in $\mathbb{H}$ with exponent
$\alpha(\theta):=(6-\kappa)/(2\kappa)$, where $\kappa\in(8/3,4]$ is related to
$\theta$ via (1.1).

Throughout this section, we use the following notations:

######  Notation 6.3.

For any interval $I\subset\mathbb{R}$, denote by $\mathcal{L}_{I}$ a loop soup
in $\mathbb{H}$ wired on $I$ and free elsewhere. We will use $\mathcal{E}_{I}$
to denote the boundary-touching excursions induced by $\mathcal{L}_{I}$.

For $z\in\mathbb{H}$, let $f_{z}(u)=(u-z)/(u-\bar{z})$ be the conformal map
from $\mathbb{H}$ onto $\mathbb{D}$ that sends $z$ to $0$. For all
$r\in(0,1)$, let $U(z,r):=f_{z}^{-1}(D(0,r))$.

Our main result on the boundary excursions in a cluster with any intensity
$\theta\in(0,1/2]$ is encapsulated in Proposition 6.4 below. This proposition
will be crucial to define and study the field $h_{\theta}$ with wired boundary
condition.

######  Proposition 6.4.

Let $\mathcal{L}_{\mathbb{R}}$ be a loop soup in $\mathbb{H}$ with intensity
$\theta\in(0,1/2]$ with wired boundary conditions. Let
$\mathcal{E}_{\mathbb{R}}$ be the Brownian excursions in $\mathbb{H}$ attached
on $\mathbb{R}$, induced by $\mathcal{L}_{\mathbb{R}}$. For all $\eta>0$,
there exists $C>0$, such that the following holds. For all $z\in\mathbb{H}$
and $r\in(0,1-\eta)$, we have

$\displaystyle\mathbb{P}[\mathcal{E}_{\mathbb{R}}\cap
U(z,r)\not=\emptyset]\leq C|\log r|^{-1}.$ (6.1)

For all $z_{1},z_{2}\in\mathbb{H}$ and $r_{1},r_{2}\in(0,1-\eta)$, we have

$\displaystyle\mathbb{P}[\mathcal{E}_{\mathbb{R}}\cap
U(z_{1},r_{1})\not=\emptyset,\mathcal{E}_{\mathbb{R}}\cap
U(z_{2},r_{2})\not=\emptyset]\leq C(1+G_{\mathbb{H}}(z_{1},z_{2}))|\log
r_{1}|^{-1}|\log r_{2}|^{-1}.$ (6.2)

In the special case of $\theta=1/2$, Proposition 6.4 can be obtained directly
using the Poisson point process description of $\mathcal{E}_{\mathbb{R}}$. We
would like to emphasise that, contrary to the restriction property stated in
Theorem 6.2, Proposition 6.4 is not concerned with the outer boundary of
$\mathcal{E}_{\mathbb{R}}$ and requires a deeper understanding of these
excursions. Our approach to prove Proposition 6.4 is to extract information
from the restriction property (Theorem 6.2), and transform this property about
the outer boundary of $\mathcal{E}_{\mathbb{R}}$ to a property about the
endpoints of the excursions in $\mathcal{E}_{\mathbb{R}}$. More concretely, we
deduce the following identities which are of independent interest.

######  Proposition 6.5.

For each excursion $e$, let $a(e)$ and $b(e)$ be the left and right endpoints
of $e$. For all $r>0$, we have

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e)-a(e))^{2}r^{2}}{(r-b(e))^{2}(r-a(e))^{2}}\right]=\alpha(\theta)$

where $\alpha(\theta)$ is the restriction exponent from Theorem 6.2.

A similar but more complicated “second-moment” result is also stated in
Proposition 6.16. Note that Propositions 6.5 and 6.16 are exact identities,
not only bounds. Nevertheless, a complete description of the laws of
$\mathcal{E}_{\mathbb{R}^{-}}$ and $\mathcal{E}_{\mathbb{R}}$ remains an open
question.

######  Remark 6.6.

In Section 6.3, we prove Propositions 6.5 and 6.16 using only two properties
of the set $\mathcal{E}_{\mathbb{R}^{-}}$: it satisfies conformal restriction
(Theroem 6.2) and is a locally finite point process of Brownian excursions
(Lemma 6.15). Therefore, Propositions 6.5 and 6.16 also hold for any locally
finite point process $\mathcal{E}$ of Brownian excursions in $\mathbb{H}$ with
endpoints on $\mathbb{R}^{-}$, such that the filling of $\mathcal{E}$
satisfies one-sided conformal restriction. Then in Sections 6.4 and 6.5, we
prove Proposition 6.4 using the additional property that we can obtain
$\mathcal{E}_{\mathbb{R}}$ by exploring the outer boundary of
$\mathcal{E}_{\mathbb{R}^{-}}$ in a Markovian way.

###  6.1 Properties and estimates about restriction measures

We will recall some properties of chordal restriction measures introduced in
[LSW03b], and deduce some estimates which will be useful later.

Fix a simply connected domain $D$ which is not the whole plane. Fix two points
$a,b\in\partial D$. Let $\Omega$ be the collection of all simply connected
compact sets $K\subset\overline{D}$, such that $K\cap\partial D=\\{a,b\\}$.
Let $\mathcal{Q}$ be the collection of all compact sets $A\subset\overline{D}$
such that $D\setminus A$ is simply connected and $a,b\not\in A$. For all
$A\in\mathcal{Q}$, let $f_{A}$ be a conformal map from $D\setminus A$ onto $D$
that leaves $a,b$ fixed. A probability law on $K\in\Omega$ is said to satisfy
chordal conformal restriction if

  1. 1.

The law of $K$ is invariant under conformal maps from $D$ to itself that fixes
$a,b$.

  2. 2.

For any $A\in\mathcal{Q}$, conditionally on $K\cap A=\emptyset$, $f_{A}(K)$ is
distributed as $K$.

It was shown in [LSW03b] that the family of chordal restriction measures is
characterized by one parameter $\alpha\geq 5/8$. A chordal restriction measure
in $D$ with parameter $\alpha$ satisfies

$\displaystyle\mathbb{P}(K\cap
A=\emptyset)=(f_{A}^{\prime}(a)f_{A}^{\prime}(b))^{\alpha}.$ (6.3)

The simplest example of chordal restriction measure is provided by:

######  Proposition 6.7 ([LSW03b], Proposition 4.1).

Let $D$ be a simply connected domain different from $\mathbb{C}$. Let
$a,b\in\partial D$ be two distinct boundary points. Then the law of the
filling of a Brownian excursion from $a$ to $b$ is the chordal restriction
measure in $D$ from $a$ to $b$ with exponent $\alpha=1$.

In the following, we obtain an estimate about a chordal restriction measure in
$\mathbb{H}$.

######  Lemma 6.8.

Let $a,b,r\in\mathbb{R}$ be three distinct boundary points. Let $K$ be a
chordal restriction measure in $\mathbb{H}$ between $a$ and $b$ with exponent
$\alpha$. We have

$\displaystyle\mathbb{P}[K\cap
D(r,\varepsilon)\not=\emptyset]=\alpha\frac{(a-b)^{2}}{(r-a)^{2}(r-b)^{2}}\varepsilon^{2}+O(1)\varepsilon^{3},$

where the $O(1)$ term is uniformly bounded for all $\varepsilon\leq 1$ by some
function of $a,b,r$. In particular, if $K$ is a Brownian excursion in
$\mathbb{H}$ between $a$ and $b$, then $\mathbb{P}[K\cap
D(r,\varepsilon)\not=\emptyset]$ is given by the above formula with
$\alpha=1$.

###### Proof.

Let $\varepsilon_{0}=\varepsilon_{0}(a,b,r)>0$ be small enough so that
$\overline{D(r,\varepsilon_{0})}$ does not contain $a$ or $b$. Since the
result is clear if $\varepsilon\in[\varepsilon_{0},1)$, we can assume that
$\varepsilon<\varepsilon_{0}$. We define the following conformal map
$f_{r,\varepsilon}$ from $\mathbb{H}\setminus D(r,\varepsilon)$ onto
$\mathbb{H}$, and compute its derivative

$\displaystyle f_{r,\varepsilon}(z)=z+\frac{\varepsilon^{2}}{z-r},\quad
f_{r,\varepsilon}^{\prime}(z)=1-\frac{\varepsilon^{2}}{(z-r)^{2}}.$ (6.4)

If we rescale $f_{r,\varepsilon}$ by
$(b-a)(f_{r,\varepsilon}(b)-f_{r,\varepsilon}(a))^{-1}$ and then do a proper
translation, then we obtain a conformal map from $\mathbb{H}\setminus
D(r,\varepsilon)$ onto $\mathbb{H}$ that fixes $a$ and $b$. By (6.3), we have

$\displaystyle\mathbb{P}[K\cap
D(r,\varepsilon)=\emptyset]=\left(f_{r,\varepsilon}^{\prime}(a)f_{r,\varepsilon}^{\prime}(b)\frac{(b-a)^{2}}{(f_{r,\varepsilon}(b)-f_{r,\varepsilon}(a))^{2}}\right)^{\alpha}=\left(1-\frac{(a-b)^{2}}{(r-a)^{2}(r-b)^{2}}\varepsilon^{2}+O(1)\varepsilon^{3}\right)^{\alpha},$

where the $O(1)$ term is bounded by some constant $C(a,b,r)$ for all
$\varepsilon\leq\varepsilon_{0}$. Since $\mathbb{P}[K\cap
D(r,\varepsilon)\not=\emptyset]=1-\mathbb{P}[B\cap
D(1,\varepsilon)=\emptyset]$, the lemma follows. The statement concerning
Brownian excursion follows from Proposition 6.7. ∎

Next, we need to obtain the following two-point estimate for chordal
restriction measures.

######  Lemma 6.9.

Let $a,b,r_{1},r_{2}\in\mathbb{R}$ be four distinct points. Let $K$ be a
chordal restriction measure in $\mathbb{H}$ between $a$ and $b$ with exponent
$\alpha$. Then

$\begin{split}\mathbb{P}[K\cap D(r_{1},\varepsilon)\not=\emptyset,K\cap
D(r_{2},\varepsilon)\not=\emptyset]=\frac{\alpha(\alpha-1)(a-b)^{4}\varepsilon^{4}}{(r_{1}-a)^{2}(r_{2}-a)^{2}(r_{1}-b)^{2}(r_{2}-b)^{2}}\\\
+\frac{\alpha(a-b)^{2}\varepsilon^{4}}{(r_{1}-b)^{2}(r_{2}-a)^{2}(r_{2}-r_{1})^{2}}+\frac{\alpha(a-b)^{2}\varepsilon^{4}}{(r_{1}-a)^{2}(r_{2}-b)^{2}(r_{2}-r_{1})^{2}}+O(1)\varepsilon^{5},\end{split}$
(6.5)

where the $O(1)$ term is uniformly bounded for all $\varepsilon\leq 1$ by some
function of $a,b,r_{1},r_{2}$.

There can be several ways to deduce Lemma 6.9. One can apply (6.3) to the
conformal maps that map out either one or both of the balls
$D(r_{1},\varepsilon)$ and $D(r_{2},\varepsilon)$. However, a direct
computation seems long and complicated. We opt for a proof of Lemma 6.9 that
relates to the trichordal restriction measures introduced in [Qia18], which
are measures on random sets in a simply connecte domain $D$ with three marked
points $a,b,c\in\partial D$. Let $\Omega_{3}$ be the collection of all simply
connected compact set $K\subset\overline{D}$, such that $K\cap\partial
D=\\{a,b,c\\}$. Let $\mathcal{Q}_{3}$ be the collection of all compact sets
$A\subset\overline{D}$ such that $D\setminus A$ is simply connected and
$a,b,c\not\in A$. For all $A\in\mathcal{Q}_{3}$, let $\widetilde{f}_{A}$ be
the unique conformal map from $D\setminus A$ onto $D$ that leaves $a,b,c$
fixed. A probability law on $K\in\Omega$ is said to satisfy trichordal
conformal restriction if for any $A\in\mathcal{Q}_{3}$, conditionally on
$K\cap A=\emptyset$, $\widetilde{f}_{A}(K)$ is distributed as $K$. It was
shown in [Qia18] that the family of trichordal restriction measures is
characterized by three parameter $\alpha,\beta,\gamma$. A trichordal
restriction measure in $D$ with parameters $\alpha,\beta,\gamma$ is
characterized by

$\displaystyle\mathbb{P}(K\cap
A=\emptyset)=\widetilde{f}_{A}^{\prime}(a)^{\alpha}\widetilde{f}_{A}^{\prime}(b)^{\beta}\widetilde{f}_{A}^{\prime}(c)^{\gamma}.$
(6.6)

The connection to trichordal restriction measures comes from the following
lemma.

######  Lemma 6.10.

Let $a,b,c\in\mathbb{R}$ be three distinct points. Suppose that $K$ is a
chordal restriction measure in $\mathbb{H}$ with marked points $a,b$ and
exponent $\alpha$, then the law of $K$ conditionally on $K\cap
D(c,\varepsilon)\not=\emptyset$ converges as $\varepsilon\to 0$ to that of a
trichordal restriction measure in $\mathbb{H}$ with marked points $a,b,c$ and
exponents $\alpha,\alpha,2$. Moreover, if $a<b<c<r$ and $A=D(r,\delta)$, then
there exist $\varepsilon_{0},\delta_{0}>0$ and a constant $C(a,b,c,r)$ such
that for all $\varepsilon\leq\varepsilon_{0},\delta\leq\delta_{0}$,

$\displaystyle\left|\mathbb{P}[K\cap A=\emptyset\mid K\cap
D(c,\varepsilon)\not=\emptyset]-\widetilde{f}_{A}^{\prime}(a)^{\alpha}\widetilde{f}_{A}^{\prime}(b)^{\alpha}\widetilde{f}_{A}^{\prime}(c)^{2}\right|\leq
C(a,b,c,r)\varepsilon\delta^{2}.$ (6.7)

###### Proof.

Here $D=\mathbb{H}$. For $A\in\mathcal{Q}_{3}$, we have

$\begin{split}&\mathbb{P}[K\cap A=\emptyset\mid K\cap
D(c,\varepsilon)\not=\emptyset]=\mathbb{P}[K\cap
D(c,\varepsilon)\not=\emptyset\mid K\cap A=\emptyset]\frac{\mathbb{P}[K\cap
A=\emptyset]}{\mathbb{P}[K\cap D(c,\varepsilon)\not=\emptyset]}\\\
&=\mathbb{P}[K\cap\widetilde{f}_{A}(D(c,\varepsilon))\not=\emptyset]\frac{\mathbb{P}[K\cap
A=\emptyset]}{\mathbb{P}[K\cap
D(c,\varepsilon)\not=\emptyset]}=\frac{\mathbb{P}[K\cap\widetilde{f}_{A}(D(c,\varepsilon))\not=\emptyset]}{\mathbb{P}[K\cap
D(c,\varepsilon)\not=\emptyset]}\widetilde{f}_{A}^{\prime}(a)^{\alpha}\widetilde{f}_{A}^{\prime}(b)^{\alpha}.\end{split}$
(6.8)

Since $\widetilde{f}_{A}$ is analytic in a neighbourhood of $c$, there exist
$\varepsilon_{0}>0$ and $C=\widetilde{f}_{A}^{\prime\prime}(c)$, such that for
all $\varepsilon\leq\varepsilon_{0}$,

$\displaystyle
B\big{(}c,\widetilde{f}_{A}^{\prime}(c)\varepsilon-C\varepsilon^{2}\big{)}\subset\widetilde{f}_{A}(D(c,\varepsilon))\subset
B\big{(}c,\widetilde{f}_{A}^{\prime}(c)\varepsilon+C\varepsilon^{2}\big{)}.$

By Lemma 6.8, we can deduce

$\displaystyle\frac{\mathbb{P}[K\cap\widetilde{f}_{A}(D(c,\varepsilon))\not=\emptyset]}{\mathbb{P}[K\cap
D(c,\varepsilon)\not=\emptyset]}=\widetilde{f}_{A}^{\prime}(c)^{2}+O(1)\varepsilon,$
(6.9)

where $O(1)$ is uniformly bounded for $\varepsilon\leq\varepsilon_{0}$ by a
function of $a,b,c,A$. A trichordal restriction measure $\widetilde{K}$ in
$\mathbb{H}$ with marked points $a,b,c$ and exponents $\alpha,\alpha,2$ is
characterized by the formula $\mathbb{P}[\widetilde{K}\cap
A=\emptyset]=f_{A}^{\prime}(a)^{\alpha}f_{A}^{\prime}(b)^{\alpha}f_{A}^{\prime}(c)^{2}$
for all $A\in\mathcal{Q}_{3}$. Plugging (6.9) back into (6.8) proves the weak
convergence of the law of $K$ conditionally on $K\cap
D(c,\varepsilon)\not=\emptyset$. Now, suppose $A=D(r,\delta)$, then
$C=C(a,b,c,r)\delta^{2}$ for all $\delta\leq(r-c)/2:=\delta_{0}$. One can
compute using Lemma 6.8 that the $O(1)$ term in (6.9) is bounded by
$2C(a,b,c,r)\delta^{2}$ for all $\varepsilon\leq\varepsilon_{0}$ and
$\delta\leq\delta_{0}$. ∎

######  Lemma 6.11.

Let $K$ be a trichordal restriction measure in $\mathbb{H}$ with marked points
$x_{1},x_{2},x_{3}\in\mathbb{R}$ and respective exponents
$\alpha_{1},\alpha_{2},\alpha_{3}$. Let $A\in\mathcal{Q}_{3}$ and
$f:\mathbb{H}\setminus A\to\mathbb{H}$ be any conformal map. Then

$\displaystyle\mathbb{P}[K\cap
A=\emptyset]=\prod_{i=1}^{3}\left[f^{\prime}(x_{i})f^{\prime}(x_{i+1})\frac{(x_{i}-x_{i+1})^{2}}{(f(x_{i})-f(x_{i+1}))^{2}}\right]^{\frac{\alpha_{i}+\alpha_{i+1}-\alpha_{i+2}}{2}},$
(6.10)

where we let $x_{4}=x_{1}$, $\alpha_{4}=\alpha_{1}$, $\alpha_{5}=\alpha_{2}$.

###### Proof.

Let $f_{A}$ be the unique conformal map from $\mathbb{H}\setminus A$ onto
$\mathbb{H}$ that fixes $x_{1},x_{2},x_{3}$. By (6.6), we have

$\displaystyle\mathbb{P}[K\cap
A=\emptyset]=f_{A}^{\prime}(x_{1})^{\alpha_{1}}f_{A}^{\prime}(x_{2})^{\alpha_{2}}f_{A}^{\prime}(x_{3})^{\alpha_{3}}=\prod_{i=1}^{3}\left(f_{A}^{\prime}(x_{i})f_{A}^{\prime}(x_{i+1})\right)^{\frac{\alpha_{i}+\alpha_{i+1}-\alpha_{i+2}}{2}}.$

Let $\varphi:\mathbb{H}\to\mathbb{H}$ be the unique conformal map that maps
$f(x_{i})$ to $x_{i}$, $i=1,2,3$, so that $f_{A}=\varphi\circ f$. With
explicit computations, we obtain that for $i=1,2,3$,

$\varphi^{\prime}(f(x_{i}))=\frac{(x_{i}-x_{i+1})(x_{i}-x_{i+2})(f(x_{i+1})-f(x_{i+2}))}{(x_{i+1}-x_{i+2})(f(x_{i})-f(x_{i+1})(f(x_{i})-f(x_{i+2}))}.$

We thus get that, for $i=1,2,3$,

$\displaystyle
f_{A}^{\prime}(x_{i})f_{A}^{\prime}(x_{i+1})=f^{\prime}(x_{i})f^{\prime}(x_{i+1})\varphi^{\prime}(f(x_{i}))\varphi^{\prime}(f(x_{i+1}))=f^{\prime}(x_{i})f^{\prime}(x_{i+1})\frac{(x_{i}-x_{i+1})^{2}}{(f(x_{i})-f(x_{i+1}))^{2}}.$

This completes the proof. ∎

We can now return to the proof of Lemma 6.9.

###### Proof of Lemma 6.9.

As in the proof of Lemma 6.8, we can assume that $\varepsilon_{1}$ and
$\varepsilon_{2}$ are small enough so that
$\overline{D(r_{1},\varepsilon_{1})}\cup\overline{D(r_{2},\varepsilon_{2})}$
does not contain $a$ or $b$. Note that $\mathbb{P}[K\cap
D(r_{1},\varepsilon_{1})\not=\emptyset,K\cap
D(r_{2},\varepsilon_{2})\not=\emptyset]$ is equal to

$\displaystyle\mathbb{P}[K\cap
D(r_{1},\varepsilon_{1})\not=\emptyset]\mathbb{P}[K\cap
D(r_{2},\varepsilon_{2})\not=\emptyset\mid K\cap
D(r_{1},\varepsilon_{1})\not=\emptyset].$ (6.11)

By Lemma 6.8, we have

$\displaystyle\mathbb{P}[K\cap
D(r_{1},\varepsilon_{1})\not=\emptyset]=\alpha\frac{(a-b)^{2}}{(r_{1}-a)^{2}(r_{1}-b)^{2}}\varepsilon_{1}^{2}+O(1)\varepsilon_{1}^{3}.$
(6.12)

By Lemma 6.10, we know that the law of $K$ conditioned on $K\cap
D(r_{1},\varepsilon_{1})\not=\emptyset$ converges as $\varepsilon_{1}\to 0$ to
the law of a trichordal restriction measure $\widetilde{K}$ in $\mathbb{H}$
with marked points $a,b,r_{1}$ and respective exponents $\alpha,\alpha,2$.
More quantitatively, there exist $\varepsilon_{0},\delta_{0}\geq 0$, such that

$\displaystyle\mathbb{P}[K\cap D(r_{2},\varepsilon_{2})\not=\emptyset\mid
K\cap D(r_{1},\varepsilon_{1})\not=\emptyset]=\mathbb{P}[\widetilde{K}\cap
D(r_{2},\varepsilon_{2})\not=\emptyset]+O(1)\varepsilon_{1}\varepsilon_{2}^{2},$
(6.13)

where $O(1)$ is bounded by $C(a,b,r_{1},r_{2})$ for all for all
$\varepsilon_{1}\leq\varepsilon_{0}$ and $\varepsilon_{2}\leq\delta_{0}$. Let
$f_{r_{2},\varepsilon}:\mathbb{H}\setminus D(r_{2},\varepsilon)\to\mathbb{H}$
be the map defined in (6.4). By (6.4), we have

$\displaystyle
f^{\prime}_{r_{2},\varepsilon}(a)f^{\prime}_{r_{2},\varepsilon}(b)\frac{(a-b)^{2}}{(f_{r_{2},\varepsilon}(a)-f_{r_{2},\varepsilon}(b))^{2}}=1-\frac{(a-b)^{2}}{(r_{2}-a)^{2}(r_{2}-b)^{2}}\varepsilon^{2}+O(1)\varepsilon^{4},$
$\displaystyle
f^{\prime}_{r_{2},\varepsilon}(a)f^{\prime}_{r_{2},\varepsilon}(r_{1})\frac{(a-r_{1})^{2}}{(f_{r_{2},\varepsilon}(a)-f_{r_{2},\varepsilon}(r_{1}))^{2}}=1-\frac{(a-r_{1})^{2}}{(r_{2}-a)^{2}(r_{2}-r_{1})^{2}}\varepsilon^{2}+O(1)\varepsilon^{4},$
$\displaystyle
f^{\prime}_{r_{2},\varepsilon}(r_{1})f^{\prime}_{r_{2},\varepsilon}(b)\frac{(r_{1}-b)^{2}}{(f_{r_{2},\varepsilon}(r_{1})-f_{r_{2},\varepsilon}(b))^{2}}=1-\frac{(r_{1}-b)^{2}}{(r_{2}-r_{1})^{2}(r_{2}-b)^{2}}\varepsilon^{2}+O(1)\varepsilon^{4}.$

Plugging them into Lemma 6.11, we can compute $\mathbb{P}[\widetilde{K}\cap
D(r_{2},\varepsilon_{2})=\emptyset]$, and then deduce that

$\displaystyle\mathbb{P}[\widetilde{K}\cap
D(r_{2},\varepsilon_{2})\not=\emptyset]=\frac{(\alpha-1)(a-b)^{2}}{(r_{2}-a)^{2}(r_{2}-b)^{2}}\varepsilon_{2}^{2}+\frac{(a-r_{1})^{2}}{(r_{2}-a)^{2}(r_{2}-r_{1})^{2}}\varepsilon_{2}^{2}+\frac{(r_{1}-b)^{2}}{(r_{2}-r_{1})^{2}(r_{2}-b)^{2}}\varepsilon_{2}^{2}+O(1)\varepsilon_{2}^{4}.$
(6.14)

Combining (6.11), (6.12), (6.13), (6.14) and letting
$\varepsilon_{1}=\varepsilon_{2}$ proves (6.5) where the $O(1)$ term therein
is uniformly bounded for $\varepsilon\leq\min(\varepsilon_{0},\delta_{0})$. It
is clear that the $O(1)$ term is also bounded for
$\varepsilon\in[\min(\varepsilon_{0},\delta_{0}),1]$. This completes the proof
of the lemma. ∎

We immediately get the following estimate on Brownian excursions.

######  Corollary 6.12.

Let $a,b,r_{1},r_{2}\in\mathbb{R}$ be four distinct points. Let $K$ be a
Brownian excursion in $\mathbb{H}$ between $a$ and $b$. Then

$\displaystyle\mathbb{P}[K\cap D(r_{1},\varepsilon)\not=\emptyset,K\cap
D(r_{2},\varepsilon)\not=\emptyset]$ $\displaystyle=$
$\displaystyle\frac{(a-b)^{2}}{(r_{1}-a)^{2}(r_{2}-b)^{2}(r_{1}-r_{2})^{2}}\varepsilon^{4}+\frac{(a-b)^{2}}{(r_{2}-a)^{2}(r_{1}-b)^{2}(r_{1}-r_{2})^{2}}\varepsilon^{4}+O(1)\varepsilon^{5},$

where the $O(1)$ term is uniformly bounded for all $\varepsilon\leq 1$ by some
function of $a,b,r_{1},r_{2}$.

###### Proof.

The formula follows from (6.5) applied to $\alpha=1$. In the case of a
Brownian motion, one can alternatively compute $\mathbb{P}[K\cap
D(r_{1},\varepsilon)\not=\emptyset,K\cap D(r_{2},\varepsilon)\not=\emptyset]$
using the heat kernel. The latter computation will show that $\mathbb{P}[K\cap
D(r_{1},\varepsilon)\not=\emptyset,K\cap D(r_{2},\varepsilon)\not=\emptyset]$
is analytic in $\varepsilon$ near $0$. Therefore the $o(\varepsilon^{4})$ term
in (6.5) is in fact $O(\varepsilon^{5})$. ∎

###  6.2 Estimates about Brownian excursions

In this subsection, we aim to obtain estimates about the probability that a
Brownian excursion gets close to one interior point (Lemma 6.13) and two
interior points (Lemma 6.14). We derive these standard estimates for ease of
reference. First recall some definitions from [Law05, Section 5].

##### Green’s function and Poisson kernel

For any domain $D$, let $G_{D}$ be the Green’s function in $D$ as defined in
Section 2.1. In our normalisation,

$\forall
x,y\in\mathbb{D},G_{\mathbb{D}}(x,y)=\frac{1}{2\pi}\log\frac{|1-x\bar{y}|}{|x-y|}\quad\text{and}\quad\forall
x,y\in\mathbb{H},G_{\mathbb{H}}(x,y)=\frac{1}{2\pi}\log\frac{|x-\bar{y}|}{|x-y|}.$
(6.15)

Suppose that $\partial D$ is smooth. Then for $x\in D$ and $z\in\partial D$,
the Poisson kernel is given by $H_{D}(x,z)=\lim_{\varepsilon\to
0}\varepsilon^{-1}G(x,z+\varepsilon\mathbf{n}_{z}),$ where $\mathbf{n}_{z}$ is
the inward unit normal vector of $\partial D$ at $z$. We have

$\forall x\in\mathbb{D},\forall
z\in\partial\mathbb{D},H_{\mathbb{D}}(x,z)=\frac{1}{2\pi}\frac{1-|x|^{2}}{|x-z|^{2}}\quad\text{and}\quad\forall
x\in\mathbb{H},\forall
z\in\partial\mathbb{H},H_{\mathbb{H}}(x,z)=\frac{\mathrm{Im}(x)}{\pi|x-z|^{2}}.$
(6.16)

For $z,w\in\partial D$, let the boundary Poisson kernel be
$H_{D}(z,w)=\lim_{\varepsilon\to
0}\varepsilon^{-1}H_{D}(z+\varepsilon\mathbf{n}_{z},w).$ Under this
normalisation, we have

$\forall
z,w\in\partial\mathbb{D},H_{\mathbb{D}}(z,w)=\frac{1}{\pi|z-w|^{2}}\quad\text{and}\quad\forall
z,w\in\partial\mathbb{H},H_{\mathbb{H}}(z,w)=\frac{1}{\pi(z-w)^{2}}.$ (6.17)

Finally recall the definitions introduced in Notation 6.3 of the conformal map
$f_{z}:\mathbb{H}\to\mathbb{D}$ mapping $z$ to $0$ and
$U(z,r)=f_{z}^{-1}(D(0,r))$.

######  Lemma 6.13 (Getting close to one interior point).

For all $\eta\in(0,1)$, there exists $C=C(\eta)>0$ such that the following
holds. Let $a,b\in\mathbb{R}$ be two distinct boundary points and let $B$ be a
Brownian excursion in $\mathbb{H}$ between $a$ and $b$. For all
$r\in(0,1-\eta)$ and $z=x+yi\in\mathbb{H}$,

$\displaystyle\mathbb{P}[B\cap U(z,r)\not=\emptyset]\leq
C\frac{(a-b)^{2}y^{2}}{|a-z|^{2}|b-z|^{2}}\left|\log r\right|^{-1}.$ (6.18)

###### Proof.

Consider the conformal map $f_{z}:\mathbb{H}\to\mathbb{D}$ that sends $z$ to
0. Let $\beta_{0}\in[0,2\pi)$ be the angle between $f_{z}(a)$ and $f_{z}(b)$.
By conformal invariance, the left hand side of (6.18) is equal to the
probability $p(\beta_{0},r)$ that a Brownian excursion in $\mathbb{D}$ between
$1$ and $e^{i\beta_{0}}$ visits $D(0,r)$. Let $\mu$ be the infinite measure on
excursions $B$ in $\mathbb{D}$ started from $1$, defined by
$\mu=\int_{0}^{2\pi}H_{\mathbb{D}}(1,e^{i\beta})\mu_{\mathbb{D}}^{\\#}(1,e^{i\beta})d\beta$.
Let $T$ be the total time length of $B$. We have

$\displaystyle p(\beta,r)H_{\mathbb{D}}(1,e^{i\beta})d\beta=\mu(B\cap
D(0,r)\not=\emptyset,\arg B_{T}\in(\beta,\beta+d\beta))$
$\displaystyle=\mathbb{P}\left(\arg B_{T}\in(\beta,\beta+d\beta)\mid B\cap
D(0,r)\not=\emptyset\right)\mu(B\cap D(0,r)\not=\emptyset).$

Moreover,

$\displaystyle\mu(B\cap D(0,r)\not=\emptyset)=\lim_{\varepsilon\to
0}\frac{1}{\varepsilon}\mathbb{P}_{1-\varepsilon}\left(\tau_{D(0,r)}<\tau_{\partial\mathbb{D}}\right)=|\log
r|^{-1},$

where the probability above stands for the probability that a Brownian motion
starting at $1-\varepsilon$ hits $D(0,r)$ before $\partial\mathbb{D}$. For
fixed $\eta\in(0,1)$ and $r\in(0,\eta)$, for an excursion $B$ started from $1$
and conditioned to hit $D(0,r)$, there exists a constant $C(\eta)>0$, such
that

$\mathbb{P}\left(\arg B_{T}\in(\beta,\beta+d\beta)\mid B\cap
D(0,r)\not=\emptyset\right)\leq C(\eta)d\beta.$

Therefore

$\displaystyle p(\beta,r)\leq C(\eta)H_{\mathbb{D}}(1,e^{i\beta})^{-1}|\log
r|^{-1}\quad\text{ for all }r\leq 1-\eta.$

Finally, by (6.17),

$H_{\mathbb{D}}(1,e^{i\beta_{0}})=\frac{1}{\pi|f_{z}(a)-f_{z}(b)|^{2}}=\frac{|a-z|^{2}|b-z|^{2}}{4\pi
y^{2}(a-b)^{2}}.$

This concludes the proof. ∎

######  Lemma 6.14 (Getting close to two interior points).

For all $\eta\in(0,1)$, there exist $\nu=\nu(\eta)>0$ and $C=C(\eta)>0$ such
that the following holds. Let $a,b\in\mathbb{R}$ be two distinct boundary
points and let $B$ be a Brownian excursion in $\mathbb{H}$ between $a$ and
$b$. For all $z_{j}=x_{j}+y_{j}i\in\mathbb{H}$, $j=1,2$, and
$r_{1},r_{2}\in(0,1-\eta)$ such that
$r_{j}\mathrm{Im}(z_{j})\leq\nu|z_{1}-z_{2}|$, $j=1,2$, we have

$\displaystyle\mathbb{P}[B\cap U(z_{1},r_{1})\not=\emptyset,B\cap
U(z_{2},r_{2})\not=\emptyset]$ $\displaystyle\leq
C\left(\frac{(a-b)^{2}y_{1}y_{2}}{|a-z_{1}|^{2}|b-z_{2}|^{2}}+\frac{(a-b)^{2}y_{1}y_{2}}{|a-z_{2}|^{2}|b-z_{1}|^{2}}\right)(1+G_{\mathbb{H}}(z_{1},z_{2}))\left|\log
r_{1}\right|^{-1}\left|\log r_{2}\right|^{-1}.$

The condition that $r_{j}\mathrm{Im}(z_{j})\leq\nu|z_{1}-z_{2}|$, $j=1,2$
ensures that $U(z_{1},r_{1})$ and $U(z_{2},r_{2})$ do not intersect and are at
distance of order $|z_{1}-z_{2}|$ to each other.

###### Proof.

We start by recording two facts for ease of future reference:

1\. For all $w\in\partial\mathbb{D}$ and $r\in(0,1)$,

$\int_{\partial D(0,r)}H_{\mathbb{D}\setminus D(0,r)}(w,z)\mathrm{d}z=|\log
r|^{-1}.$ (6.19)

Indeed, the left hand side is equal to

$\displaystyle\lim_{\varepsilon\to 0}\frac{1}{\varepsilon}\int_{\partial
D(0,r)}H_{\mathbb{D}\setminus
D(0,r)}(w+\varepsilon\mathbf{n}_{w},z)\mathrm{d}z=\lim_{\varepsilon\to
0}\frac{1}{\varepsilon}\mathbb{P}_{w+\varepsilon\mathbf{n}_{w}}\left(\tau_{D(0,r)}<\tau_{\partial\mathbb{D}}\right)=\lim_{\varepsilon\to
0}\frac{\log(1-\varepsilon)}{\varepsilon\log r}=\frac{1}{|\log r|},$

where the probability appearing in the above display stands for the
probability that a Brownian motion starting from $w+\varepsilon\mathbf{n}_{w}$
hits $D(0,r)$ before hitting $\partial\mathbb{D}$.

2\. There exists $C_{0}>0$ such that for all $z_{3}\in\partial U(z_{1},r_{1})$
and $z_{4}\in\partial U(z_{2},r_{2})$,

$G_{\mathbb{H}}(z_{3},z_{4})\leq C_{0}(1+G_{\mathbb{H}}(z_{1},z_{2})).$ (6.20)

This follows from the fact that there exists $\nu=\nu(\eta)>0$ such that for
all $z_{2j}\in U(z_{j},r_{j})$,
$|z_{2j}-z_{j}|\leq(10\nu)^{-1}r_{j}\mathrm{Im}(z_{j})$, $j=1,2$. The
assumption that $r_{j}\mathrm{Im}(z_{j})\leq\nu|z_{1}-z_{2}|$, $j=1,2$, then
ensures that the points of $U(z_{1},r_{1})$ stay much closer to $z_{1}$ than
to any other point of $U(z_{2},r_{2})$. A direct computation using (6.15) then
concludes the proof of (6.20).

Having collected these two initial facts, we can now proceed with the proof of
Lemma 6.14. The event $B\cap U(z_{1},r_{1})\not=\emptyset,B\cap
U(z_{2},r_{2})\not=\emptyset$ is the disjoint union of the two events $E_{1}$
and $E_{2}$:

  * •

Let $E_{1}$ be the event that $B$ first visits $U(z_{1},r_{1})$ before
$U(z_{2},r_{2})$.

  * •

Let $E_{2}$ be the event that $B$ first visits $U(z_{2},r_{2})$ before
$U(z_{1},r_{1})$.

On $E_{1}$, we can decompose $B$ into an excursion from $a$ to
$z_{3}\in\partial U(z_{1},r_{1})$ in $\mathbb{H}\setminus(U(z_{1},r_{1})\cup
U(z_{2},r_{2}))$, an excursion from $z_{3}$ to $z_{4}\in\partial
U(z_{2},r_{2})$ in $\mathbb{H}$ and an excursion from $z_{4}$ to $b$ in
$\mathbb{H}\setminus U(z_{2},r_{2})$. This results in the following identity

$\displaystyle H_{\mathbb{H}}(a,b)\mathbb{P}[E_{1}]=\int_{\partial
U(z_{1},r_{1})}\mathrm{d}z_{3}\int_{\partial
U(z_{2},r_{2})}\mathrm{d}z_{4}H_{\mathbb{H}\setminus(U(z_{1},r_{1})\cup
U(z_{2},r_{2}))}(a,z_{3})G_{\mathbb{H}}(z_{3},z_{4})H_{\mathbb{H}\setminus
U(z_{2},r_{2})}(b,z_{4}).$ (6.21)

Increasing the domain can only increase the pointwise values of the Poisson
kernel. We can thus bound $H_{\mathbb{H}\setminus(U(z_{1},r_{1})\cup
U(z_{2},r_{2}))}(a,z_{3})\leq H_{\mathbb{H}\setminus U(z_{1},r_{1})}(a,z_{3})$
for all $z_{3}\in\partial U(z_{1},r_{1})$. Using further (6.20), we deduce
that

$\displaystyle H_{\mathbb{H}}(a,b)\mathbb{P}[E_{1}]\leq
C_{0}(1+G_{\mathbb{H}}(z_{1},z_{2}))\int_{\partial
U(z_{1},r_{1})}\mathrm{d}z_{3}H_{\mathbb{H}\setminus
U(z_{1},r_{1})}(a,z_{3})\times\int_{\partial
U(z_{2},r_{2})}\mathrm{d}z_{4}H_{\mathbb{H}\setminus
U(z_{2},r_{2})}(b,z_{4}).$

With a change of variable and conformal covariance of the boundary Poisson
kernel, this is further equal to

$\displaystyle
C_{0}(1+G_{\mathbb{H}}(z_{1},z_{2}))f_{z_{1}}^{\prime}(a)f_{z_{2}}^{\prime}(b)\int_{\partial
D(0,r_{1})}H_{\mathbb{D}\setminus
D(0,r_{1})}(f_{z_{1}}(a),w_{3})\mathrm{d}w_{3}\int_{\partial
D(0,r_{2})}H_{\mathbb{D}\setminus
D(0,r_{2})}(f_{z_{2}}(b),w_{4})\mathrm{d}w_{4}.$

By (6.19), the product of the last two integrals equals $|\log
r_{1}|^{-1}|\log r_{2}|^{-1}.$ Finally, using that (see (6.17))

$\displaystyle H_{\mathbb{H}}(a,b)=\frac{1}{\pi(a-b)^{2}},\quad
f_{z_{1}}^{\prime}(a)=\frac{2y_{1}}{|a-z_{1}|^{2}},\quad
f_{z_{2}}^{\prime}(b)=\frac{2y_{2}}{|b-z_{2}|^{2}},$

we obtain that

$\displaystyle\mathbb{P}[E_{1}]\leq\pi^{-1}C_{1}(1+G_{\mathbb{H}}(z_{1},z_{2}))\frac{(a-b)^{2}y_{1}y_{2}}{|a-z_{1}|^{2}|b-z_{2}|^{2}}\left|\log
r_{1}\right|^{-1}\left|\log r_{2}\right|^{-1}.$

To bound $\mathbb{P}[E_{2}]$ simply exchange the roles of $z_{1}$ and $z_{2}$.
This concludes the proof. ∎

###  6.3 Boundary excursions in a loop soup with mixed boundary conditions

In this subsection, we aim to prove Propositions 6.5 and 6.16 which provide
quantitative controls on the number of excursions in
$\mathcal{E}_{\mathbb{R}^{-}}$. Proposition 6.5 can be understood as a first
moment estimate, and Propositions 6.16 can be understood as a second moment
estimate.

The proofs of Propositions 6.5 and 6.16 rely essentially on the restriction
property of $\mathcal{E}$ (Theorem 6.2) and the independence of the excursions
conditionally on their endpoints (Lemma 6.15). Lemma 6.15 is a variation of
[QW18, Lemma 9] which states that $\mathcal{E}_{\mathbb{R}}$ is a locally
finite point process of Brownian excursions. Changing
$\mathcal{E}_{\mathbb{R}}$ to $\mathcal{E}_{\mathbb{R}^{-}}$ only modifies the
setup of the proof, not the essential arguments, so we omit the proof of Lemma
6.15.

######  Lemma 6.15.

The collection $\mathcal{E}_{\mathbb{R}^{-}}$ is a locally finite point
process of Brownian excursions in $\mathbb{H}$.

The “point process of Brownian excursions” property means that, conditionally
on the endpoints of the excursions, $\mathcal{E}_{\mathbb{R}^{-}}$ is
distributed as the collection of independent Brownian excursions between the
given endpoints. Throughout, for each excursion $e$, we assign an i.i.d.
orientation with probability $1/2$. Let $a(e)$ and $b(e)$ be respectively the
starting and ending endpoints of an excursion $e$. The “locally finite”
property means that for each $\delta>0$, the collection of excursions
$e\in\mathcal{E}_{\mathbb{R}^{-}}$ with $|a(e)-b(e)|\geq\delta$ is a.s.
finite.

Let us first prove Proposition 6.5.

###### Proof of Proposition 6.5.

We write $\mathcal{E}$ for $\mathcal{E}_{\mathbb{R}^{-}}$ for simplicity. We
define for each $e\in\mathcal{E}$ and $r>0$,

$\displaystyle p_{r}(e):=\frac{(b(e)-a(e))^{2}}{(r-b(e))^{2}(r-a(e))^{2}}.$

By Lemma 6.15 and Lemma 6.8, we know that

$\displaystyle\mathbb{P}\left[e\cap D(r,\varepsilon)\not=\emptyset\mid
a(e),b(e)\right]=p_{r}(e)\varepsilon^{2}+O(1)\varepsilon^{4},$ (6.22)

where the $O(1)$ term depends on $a(e),b(e)$ and $r$, but is uniformly bounded
for $\varepsilon\leq 1$. Note that for all $a(e)<b(e)<0$ and $r>0$, we have
$p_{r}(e)\leq r^{-2}$. For each $M>0$, let $\mathcal{E}^{M}$ denote the
collection of the $M$ excursions in $\mathcal{E}$ with the largest $p_{r}(e)$.
Then $\mathcal{E}=\cup_{M>0}\mathcal{E}^{M}$. Let us first show that

$\displaystyle\mathbb{P}\left[\mathcal{E}^{M}\cap
D(r,\varepsilon)\not=\emptyset\mid\\{(a(e),b(e))\\}_{e\in\mathcal{E}^{M}}\right]=\sum_{e\in\mathcal{E}^{M}}p_{r}(e)\varepsilon^{2}+O(1)\varepsilon^{4},$
(6.23)

where the $O(1)$ term depends on $r,M$ and the random set
$\\{(a(e),b(e))\\}_{e\in\mathcal{E}^{M}}$, but is uniformly bounded for
$\varepsilon\leq 1$. The $\leq$ direction in (6.23) directly follows from
Lemma 6.15 and (6.22). To prove the $\geq$ direction, note that by the
inclusion-exclusion principle, the left-hand side of (6.23) is at least

$\displaystyle\sum_{e\in\mathcal{E}^{M}}\mathbb{P}[e\cap
D(r,\varepsilon)\not=\emptyset]-\sum_{e_{1}\not=e_{2}\in\mathcal{E}^{M}}\mathbb{P}[e_{1}\cap
D(r,\varepsilon)\not=\emptyset,e_{2}\cap D(r,\varepsilon)\not=\emptyset]$
$\displaystyle=$
$\displaystyle\sum_{e\in\mathcal{E}^{M}}p_{r}(e)\varepsilon^{2}-\sum_{e_{1}\not=e_{2}\in\mathcal{E}^{M}}p_{r}(e_{1})p_{r}(e_{2})\varepsilon^{4}+O(1)\varepsilon^{4}=\sum_{e\in\mathcal{E}^{M}}p_{r}(e)\varepsilon^{2}+O(1)\varepsilon^{4}.$

In the last equality, we have used the bound
$\sum_{e_{1}\not=e_{2}\in\mathcal{E}^{M}}p_{r}(e_{1})p_{r}(e_{2})\leq
M^{2}r^{-4}$. This completes the proof of (6.23). Note that the $O(1)$ term in
(6.23) has finite expectation, since $\mathbb{P}\left[\mathcal{E}^{M}\cap
D(r,\varepsilon)\not=\emptyset\right]\in[0,1]$. Taking expectations on both
sides of (6.23), we get

$\displaystyle\mathbb{P}\left[\mathcal{E}^{M}\cap
D(r,\varepsilon)\not=\emptyset\right]=\mathbb{E}\bigg{[}\sum_{e\in\mathcal{E}^{M}}p_{r}(e)\bigg{]}\varepsilon^{2}+O(1)\varepsilon^{4},$
(6.24)

where the $O(1)$ term depends on $M$ and $r$, but is uniformly bounded for
$\varepsilon\leq 1$.

On the other hand, by Theorem 6.2, we know that $\mathcal{E}$ is distributed
as a restriction measure of exponent $\alpha(\theta)$. Applying Lemma 6.8 to
$a=\infty$ and $b=0$, we get

$\displaystyle\mathbb{P}\left[\mathcal{E}\cap
D(r,\varepsilon)\not=\emptyset\right]=\alpha(\theta)\varepsilon^{2}/r^{2}+O(1)\varepsilon^{4}.$
(6.25)

Since $\mathbb{P}\left[\mathcal{E}^{M}\cap
D(r,\varepsilon)\not=\emptyset\right]\leq\mathbb{P}\left[\mathcal{E}\cap
D(r,\varepsilon)\not=\emptyset\right]$, together with (6.24), we deduce that

$\displaystyle\mathbb{E}\bigg{[}\sum_{e\in\mathcal{E}^{M}}p_{r}(e)\bigg{]}\leq\alpha(\theta)/r^{2}.$

Letting $M\to\infty$, by monotone convergence, we have

$\displaystyle\mathbb{E}\bigg{[}\sum_{e\in\mathcal{E}}p_{r}(e)\bigg{]}\leq\alpha(\theta)/r^{2}.$
(6.26)

Let $M\to\infty$ in both sides of (6.24). The left-hand side of (6.24) tends
to $\mathbb{P}\left[\mathcal{E}\cap D(r,\varepsilon)\not=\emptyset\right]$
which is less than $1$. In the right-hand side of (6.24), the coefficient in
front of $\varepsilon^{2}$ is increasing as $M\to\infty$, and tends to the
same expected sum over $\mathcal{E}$ instead of $\mathcal{E}^{M}$, which is
less than $\alpha(\theta)/r^{2}$ by (6.26). Therefore, the $O(1)$ term in
(6.24) should also be bounded as $M\to\infty$. This implies

$\displaystyle\mathbb{P}\left[\mathcal{E}\cap
D(r,\varepsilon)\not=\emptyset\right]=\mathbb{E}\bigg{[}\sum_{e\in\mathcal{E}}p_{r}(e)\bigg{]}\varepsilon^{2}+O(1)\varepsilon^{4},$

where the $O(1)$ term depends only on $r$. Comparing with (6.25) yields the
lemma. ∎

Let us now deduce a “second moment” estimate.

######  Proposition 6.16.

For all $0<r_{1}<r_{2}$, we have

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e)-a(e))^{2}}{(r_{1}-b(e))^{2}(r_{2}-a(e))^{2}(r_{1}-r_{2})^{2}}\right]+\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e)-a(e))^{2}}{(r_{1}-a(e))^{2}(r_{2}-b(e))^{2}(r_{1}-r_{2})^{2}}\right]$
$\displaystyle+\mathbb{E}\left[\sum_{e_{1}\not=e_{2}\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e_{1})-a(e_{1}))^{2}}{(r_{1}-b(e_{1}))^{2}(r_{1}-a(e_{1}))^{2}}\frac{(b(e_{2})-a(e_{2}))^{2}}{(r_{2}-b(e_{2}))^{2}(r_{2}-a(e_{2}))^{2}}\right]$
(6.27)
$\displaystyle=\frac{\alpha(\theta)}{r_{1}^{2}(r_{2}-r_{1})^{2}}+\frac{\alpha(\theta)}{r_{2}^{2}(r_{2}-r_{1})^{2}}+\frac{\alpha(\theta)(\alpha(\theta)-1)}{r_{1}^{2}r_{2}^{2}}.$

######  Remark 6.17.

When $\theta=1/2$, the endpoints of $\mathcal{E}_{\mathbb{R}^{-}}$ are
distributed according to a Poisson point process with intensity $1/4$ (see
Theorem 6.1 for $\mathcal{E}_{\mathbb{R}}$ instead of
$\mathcal{E}_{\mathbb{R}^{-}}$). At this value of $\theta$, the restriction
exponent $\alpha(\theta)$ equals $1/4$ and thus also corresponds to the
intensity of the point process. With a simple Poisson point process
computation, one can deduce from Proposition 6.5 that the third expectation on
the left hand side of (6.27) is equal to the third term on the right hand
side. There is a priori no reason to believe that this still holds for other
values of $\theta$. However, we will only need to upper bound the terms on the
left hand side of (6.27).

###### Proof.

We write $\mathcal{E}$ for $\mathcal{E}_{\mathbb{R}^{-}}$ for simplicity. By
Theorem 6.2 and Lemma 6.9 applied to $a=\infty$ and $b=0$, we have

$\displaystyle\mathbb{P}\left[\mathcal{E}\cap
D(r_{1},\varepsilon)\not=\emptyset,\mathcal{E}\cap
D(r_{2},\varepsilon)\not=\emptyset\right]=\frac{\alpha(\theta)(\alpha(\theta)-1)\varepsilon^{4}}{r_{1}^{2}r_{2}^{2}}+\frac{\alpha(\theta)\varepsilon^{4}}{r_{1}^{2}(r_{2}-r_{1})^{2}}+\frac{\alpha(\theta)\varepsilon^{4}}{r_{2}^{2}(r_{2}-r_{1})^{2}}+O(1)\varepsilon^{5}.$
(6.28)

The event $\\{\mathcal{E}\cap
D(r_{1},\varepsilon)\not=\emptyset,\mathcal{E}\cap
D(r_{2},\varepsilon)\not=\emptyset\\}$ is equal to the union of the two
following events:

  * •

Let $E_{1}$ be the event that $D(r_{1},\varepsilon)$ and
$D(r_{2},\varepsilon)$ are visited by a same excursion in $\mathcal{E}$.

  * •

Let $E_{2}$ be the event that $D(r_{1},\varepsilon)$ and
$D(r_{2},\varepsilon)$ are visited by two different excursions in
$\mathcal{E}$.

We first evaluate $\mathbb{P}[E_{1}]$. Let

$\displaystyle
s_{r_{1},r_{2}}(e):=\frac{(b(e)-a(e))^{2}}{(r_{1}-b(e))^{2}(r_{2}-a(e))^{2}(r_{1}-r_{2})^{2}}+\frac{(b(e)-a(e))^{2}}{(r_{1}-a(e))^{2}(r_{2}-b(e))^{2}(r_{1}-r_{2})^{2}}.$

By Lemma 6.15 and Corollary 6.12, we have

$\displaystyle\mathbb{P}[e\cap D(r_{1},\varepsilon)\not=\emptyset,e\cap
D(r_{2},\varepsilon)\not=\emptyset]=s_{r_{1},r_{2}}(e)\varepsilon^{4}+O(1)\varepsilon^{5},$

where the $O(1)$ term depends on $a(e),b(e)$ and $r_{1},r_{2}$, but is
uniformly bounded for $\varepsilon\leq 1$. Note that for all $a(e)<b(e)<0$ and
$r_{1},r_{2}>0$, we have
$p_{r}(e)\leq(r_{1}-r_{2})^{-2}(r_{1}^{-2}+r_{2}^{-2})$.

For each $M>0$, let $\mathcal{E}^{M}$ be the set of the $M$ excursions in
$\mathcal{E}$ with the largest $s_{r_{1},r_{2}}(e)$. Then
$\mathcal{E}=\cup_{M>0}\mathcal{E}^{M}$. Let $E_{1}^{M}$ and $E_{2}^{M}$ be
defined respectively as $E_{1}$ and $E_{2}$ where we replace $\mathcal{E}$ by
$\mathcal{E}^{M}$ in their definitions. By Lemma 6.15, Corollary 6.12, and a
use of the inclusion-exclusion principle as in the proof of 6.5, we can deduce

$\displaystyle\mathbb{P}\left[E_{1}^{M}\mid\\{(a(e),b(e))\right]=\sum_{e\in\mathcal{E}^{M}}s_{r_{1},r_{2}}(e)\varepsilon^{4}+O(1)\varepsilon^{5},$

where the $O(1)$ term depends on $r_{1},r_{2}$ and the random set
$\\{(a(e),b(e))\\}_{e\in\mathcal{E}^{M}}$, but is uniformly bounded for
$\varepsilon\leq 1$. Moreover, this $O(1)$ term has finite expectation, since
$\mathbb{P}[E_{1}^{M}]\in[0,1]$. A similar reasoning as in the proof of
Proposition 6.5 will allow us to take the expectations on both sides of the
previous equality and then let $M\to\infty$. We will get that

$\displaystyle\mathbb{P}[E_{1}]=\mathbb{E}\left[\sum_{e\in\mathcal{E}}s_{r_{1},r_{2}}(e)\right]\varepsilon^{4}+O(1)\varepsilon^{5}.$

Let us now evaluate $\mathbb{P}[E_{2}]$. By Lemmas 6.15 and 6.8, we have

$\displaystyle\mathbb{P}\left[E_{2}^{M}\mid\\{(a(e),b(e))\right]=\sum_{e_{1}\not=e_{2}\in\mathcal{E}}\frac{(b(e_{1})-a(e_{1}))^{2}\varepsilon^{2}}{(r_{1}-b(e_{1}))^{2}(r_{1}-a(e_{1}))^{2}}\frac{(b(e_{2})-a(e_{2}))^{2}\varepsilon^{2}}{(r_{2}-b(e_{2}))^{2}(r_{2}-a(e_{2}))^{2}}+O(\varepsilon^{5}).$

The same reasoning will again allow us to conclude that $\mathbb{P}[E_{2}]$ is
equal to the line (6.27) times $\varepsilon^{4}$ plus $O(1)\varepsilon^{5}$.

Finally, one can see that $\mathbb{P}[E_{1}\cap E_{2}]$ is of smaller order
than $\varepsilon^{4}$, because on $E_{1}\cap E_{2}$, there needs to be one
excursion $e_{1}\in\mathcal{E}$ that visits both $D(r_{1},\varepsilon)$ and
$D(r_{2},\varepsilon)$ and a different excursion $e_{2}\in\mathcal{E}$ that
visits $D(r_{1},\varepsilon)$ or $D(r_{2},\varepsilon)$. This implies that

$\displaystyle\mathbb{P}\left[\mathcal{E}\cap
D(r_{1},\varepsilon)\not=\emptyset,\mathcal{E}\cap
D(r_{2},\varepsilon)\not=\emptyset\right]=\mathbb{P}[E_{1}]+\mathbb{P}[E_{2}]+o(\varepsilon^{4}).$

Combining with (6.28) and the previous evaluations of $\mathbb{P}[E_{1}]$ and
$\mathbb{P}[E_{2}]$, we obtain the lemma. ∎

###  6.4 Boundary excursions in a loop soup with wired boundary conditions

In this section, we will use the results from Section 6.3 to obtain
quantitative controls on excursions in $\mathcal{E}_{\mathbb{R}}$, i.e., from
a loop soup in $\mathbb{H}$ which is wired on the entire boundary. The main
step is to prove Lemma 6.18 below. After that, we will deduce Lemmas 6.19 and
6.20 as consequences.

######  Lemma 6.18.

We have

$\displaystyle
S_{1}:=\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}\right]<\infty.$
(6.29) $\displaystyle
S_{2}:=\mathbb{E}\left[\sum_{e_{1},e_{2}\in\mathcal{E}_{\mathbb{R}};\,a(e_{1}),b(e_{1}),a(e_{2}),b(e_{2})\in[0,1]}(b(e_{1})-a(e_{1}))^{2}(b(e_{2})-a(e_{2}))^{2}\right]<\infty.$
(6.30)

###### Proof.

We use $\mathcal{E}$ to denote $\mathcal{E}_{\mathbb{R}}$ for simplicity. We
apply the conformal map $z\in\mathbb{H}\mapsto 1/(1-z)\in\mathbb{H}$ that maps
$\mathbb{R}^{-}$ to $[0,1]$ to Proposition 6.5 (with $r=1$), and get

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{[0,1]}}(a(e)-b(e))^{2}\right]=\alpha(\theta).$

By scaling invariance and translation invariance, for all $R>0$, we have

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{[-R,R]}}(a(e)-b(e))^{2}\right]=4R^{2}\alpha(\theta).$

Therefore

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}\right]\leq
4R^{2}\alpha(\theta).$

In a loop soup $\mathcal{L}_{[-R,R]}$ in $\mathbb{H}$ wired on $[-R,R]$ and
free elsewhere, let $\eta$ be the curve from $-R$ to $R$ which is the outer
boundary of the cluster glued at $[-R,R]$. Then $\eta$ is distributed as a
SLEκ from $-R$ to $R$. Let $g_{t}$ be the conformal map from
$\mathbb{H}\setminus\eta([0,t])$ onto $\mathbb{H}$ that fixes $0,1,\infty$.
Let $E$ be the event that $\eta\cap D(0,10)=\emptyset$. We can take $R$ big
enough so that $\mathbb{P}(E)\geq 1/2$. Let us first prove that on the event
$E$,

$\displaystyle\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1]}(g_{t}(a(e))-g_{t}(b(e)))^{2}\leq
C\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}.$ (6.31)

There is an infinite measure $\mu$ on Brownian excursions in $\mathbb{H}$
between a point $x\in\mathbb{R}$ and $\infty$, so that the endpoint $x$ is
distributed according to the Lebesgue measure [LSW03b]. For a given $g_{t}$,
there is a constant $C$ (which depends only on $g_{t}$) such that for all
$-R<a<b<R$, $g_{t}(b)-g_{t}(a)$ is equal to $C$ times the $\mu$-measure of
excursions in $\mathbb{H}$ between $\infty$ and an endpoint in $[a,b]$ which
do not intersect $\eta([0,t])$. On the event $E$, an excursion in
$\mathbb{H}\setminus\eta([0,t])$ from $\infty$ to $[a,b]$ can further be
decomposed as the concatenation of an excursion $\xi_{1}$ between
$z_{0}\in\partial D(0,10)$ in $\mathbb{H}\setminus\eta([0,t])$ and an
excursion $\xi_{2}$ between $z_{0}$ and a point in $[a,b]$ in $D(0,10)$. The
distribution of $z_{0}$ depends on $\eta([0,t])$, and we denote it by
$\mathbb{P}_{0}$. However, given $z_{0}$, the distribution of $\xi_{2}$ is
independent from $\eta([0,t])$. For all $0\leq a<b\leq 1$, the ratio

$\displaystyle\frac{g_{t}(b)-g_{t}(a)}{g_{t}(1)-g_{t}(0)}=\mathbb{E}_{0}\left[\frac{\int_{a}^{b}H_{D(0,10)}(z_{0},x)dx}{\int_{0}^{1}H_{D(0,10)}(z_{0},x)dx}\right].$
(6.32)

Since $\partial D(0,10)$ has positive distance from $[0,1]$, there exist
$C_{1}>c_{1}>0$ such that for all $z_{0}\in\partial D(0,10)$ and $x\in[0,1]$,
we have $c_{1}\leq H_{D(0,10)}(z_{0},x)\leq C_{1}$. Therefore (6.32) is at
most $(C_{1}/c_{1})(b-a)$. This implies that $g_{t}(b)-g_{t}(a)\leq C(b-a)$
for all $0\leq a<b\leq 1$ where $C$ is some universal constant, hence (6.31)
follows.

As $t\to\infty$, $g_{t}$ converges to the conformal map $g_{\infty}$ from the
bounded connected component of $\mathbb{H}\setminus\eta([0,\infty])$ onto
$\mathbb{H}$ that sends $0,1,R$ to $0,1,\infty$. The left-hand side of (6.31)
is at least

$\displaystyle\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1],e\cap\eta=\emptyset}(g_{t}(a(e))-g_{t}(b(e)))^{2},$

which converges a.s. to

$\displaystyle\sum_{e\in\mathcal{E},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2},$ (6.33)

where $\mathcal{E}$ is the set of excursions induced from
$g_{\infty}(\mathcal{L}_{[-R,R]})$, a loop soup in $\mathbb{H}$ with wired
boundary conditions. By Theorem 6.1, the law of (6.33) is independent from
$\eta$, hence remains unchanged if we condition on $E$. Taking the conditional
expectation on $E$ of both sides of (6.31), letting $t\to\infty$, and using
the dominated convergence where the dominator is given by the right-hand side
of (6.31), we get

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}\right]\leq
C\mathbb{E}\left[\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}\mid
E\right]$ (6.34) $\displaystyle\leq
C\mathbb{E}\left[\sum_{e\in\mathcal{E}_{[-R,R]},a(e),b(e)\in[0,1]}(a(e)-b(e))^{2}\right]\mathbb{P}(E)^{-1}\leq
8CR^{2}\alpha(\theta).$ (6.35)

This completes the proof of (6.29).

The proof of (6.30) follows a similar line. From Proposition 6.16, we can
deduce that for $0<r_{1}<r_{2}<1$,

$\displaystyle\mathbb{E}\left[\sum_{e_{1}\not=e_{2}\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e_{1})-a(e_{1}))^{2}}{(1-b(e_{1}))^{2}(1-a(e_{1}))^{2}}\frac{(b(e_{2})-a(e_{2}))^{2}}{(1-b(e_{2}))^{2}(1-a(e_{2}))^{2}}\right]$
(6.36)
$\displaystyle\leq\mathbb{E}\left[\sum_{e_{1}\not=e_{2}\in\mathcal{E}_{\mathbb{R}^{-}}}\frac{(b(e_{1})-a(e_{1}))^{2}}{(r_{1}-b(e_{1}))^{2}(r_{1}-a(e_{1}))^{2}}\frac{(b(e_{2})-a(e_{2}))^{2}}{(r_{2}-b(e_{2}))^{2}(r_{2}-a(e_{2}))^{2}}\right]<\infty.$

Under the conformal map $f(z)=1/(1-z)$, $f(\mathcal{E}_{\mathbb{R}^{-}})$ is
distributed as $\mathcal{E}_{[0,1]}$, and (6.36) becomes

$\displaystyle\mathbb{E}\left[\sum_{e_{1}\not=e_{2}\in\mathcal{E}_{[0,1]}}(b(e_{1})-a(e_{1}))^{2}(b(e_{2})-a(e_{2}))^{2}\right]<\infty.$

Following the same strategy as before, we deduce that

$\displaystyle\mathbb{E}\left[\sum_{e_{1}\not=e_{2}\in\mathcal{E}_{\infty};\,a(e_{1}),b(e_{1}),a(e_{2}),b(e_{2})\in[0,1]}(b(e_{1})-a(e_{1}))^{2}(b(e_{2})-a(e_{2}))^{2}\right]<\infty.$

To deal with the case $e_{1}=e_{2}$, we simply bound
$(b(e)-a(e))^{4}\leq(b(e)-a(e))^{2}$ and rely on (6.29). This completes the
proof of (6.30). ∎

######  Lemma 6.19.

For all $\varepsilon\in[0,1]$, let

$\displaystyle
X(\varepsilon):=\sum_{e\in\mathcal{E}_{\mathbb{R}};\,a(e)\in[0,\varepsilon);b(e)\in[0,1]}(b(e)-a(e))^{2}.$

Then there exists a universal constant $C>0$ such that
$\mathbb{E}\left[X(\varepsilon)\right]\leq C\varepsilon$ and
$\mathbb{E}\left[X(\varepsilon)^{2}\right]\leq C\varepsilon$.

###### Proof.

Without loss of generality, assume that $\varepsilon=1/N$ for some integer
$N\geq 1$. Note that $[0,1)$ is the disjoint union of the sets $I_{n}$ for
$0\leq n\leq N-1$, where $I_{n}:=[n/N,(n+1)/N)$. Let

$\displaystyle
X_{n}(\varepsilon):=\sum_{e\in\mathcal{E}_{\mathbb{R}};\,a(e)\in
I_{n};\,b(e)\in[n/N,n/N+1]}(b(e)-a(e))^{2}.$

By translation invariance, $X_{n}(\varepsilon)$ has the same distribution as
$X(\varepsilon)$. Hence

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}};\,a(e),b(e)\in[0,2]}(b(e)-a(e))^{2}\right]\geq\sum_{n=0}^{N-1}\mathbb{E}\left[X_{n}(\varepsilon)\right]\geq
N\mathbb{E}[X(\varepsilon)].$

By (6.29), this implies that $\mathbb{E}[X(\varepsilon)]\leq
CN^{-1}=C\varepsilon$. The upper bound on $\mathbb{E}[X(\varepsilon)^{2}]$ is
obtained similarly using (6.30) and

$\displaystyle\mathbb{E}\left[\left(\sum_{e\in\mathcal{E}_{\mathbb{R}};\,a(e),b(e)\in[0,2]}(b(e)-a(e))^{2}\right)^{2}\right]\geq\mathbb{E}\left[\left(\sum_{n=0}^{N-1}X_{n}(\varepsilon)\right)^{2}\right]\geq\sum_{n=0}^{N-1}\mathbb{E}[X_{n}(\varepsilon)^{2}]\geq
N\mathbb{E}[X(\varepsilon)^{2}].$

∎

######  Lemma 6.20.

For all intervals $I_{1},I_{2}\subset\mathbb{R}$ with respective lengths
$r_{1}$ and $r_{2}$, there is a universal constant $C>0$, such that

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}_{\mathbb{R}},\,a(e)\in
I_{1},b(e)\in I_{2}}(b(e)-a(e))^{2}\right]\leq Cr_{1}r_{2}.$

###### Proof.

By decomposing $I_{1}=(I_{1}\setminus I_{2})\cup(I_{1}\cap I_{2})$, it is
enough to treat two cases: $I_{1}\subset I_{2}$ and $I_{1}\cap
I_{2}=\varnothing$.

Case 1. $I_{1}\subset I_{2}$. We can decompose $I_{2}$ into the union of two
subintervals sharing an endpoint with $I_{1}$. We can therefore assume that
$I_{2}$ shares an endpoint with $I_{1}$. By scaling and translation, this case
reduces to Lemma 6.19.

Case 2. $I_{1}\cap I_{2}=\varnothing$. By scaling and translation, it is
enough to consider the case where $I_{1}=[0,x]$ and $I_{2}=[y,1]$, where
$0<x\leq y<1$. If $I_{1}$ or $I_{2}$ has length at least $1/4$, then the
result is a direct consequence of Lemma 6.19. We can thus suppose that both
$I_{1}$ and $I_{2}$ have length less than $1/4$. We apply the conformal map
$f:z\in\mathbb{H}\mapsto\frac{(1-x)z}{(1-2x)z+x}\in\mathbb{H}$ that fixes $0$
and $1$ and sends $x$ to $1/2$. Then

$\displaystyle\mathbb{E}\left[\sum_{a(e)\in[0,x],b(e)\in[y,1]}(b(e)-a(e))^{2}\right]=\mathbb{E}\left[\sum_{a(e)\in[0,f(x)],b(e)\in[f(y),1]}(f^{-1}(b(e))-f^{-1}(a(e)))^{2}\right].$
(6.37)

Note that

$\displaystyle y-x+2x(1-y)\geq y-x\geq 1/2\quad\text{and}\quad
1-f(y)=\frac{x(1-y)}{y-x+2x(1-y)}\leq 2x(1-y)\leq 1/8.$

This implies that for all $a\in[0,x]$ and $b\in[y,1]$, we have $f(b)-f(a)\geq
3/8\geq 3/8(b-a)$. Conversely, for all $a\in[0,f(x)]$ and $b\in[f(y),1]$, we
have $f^{-1}(b)-f^{-1}(a)\leq 8/3(b-a)$. Therefore (6.37) is at most

$\displaystyle\frac{64}{9}\mathbb{E}\left[\sum_{a(e)\in[0,f(x)],b(e)\in[f(y),1]}(b(e)-a(e))^{2}\right]\leq\frac{64}{9}\mathbb{E}\left[\sum_{a(e)\in[0,1],b(e)\in[f(y),1]}(b(e)-a(e))^{2}\right]$

which is bounded by $C(1-f(y))\leq 2Cx(1-y)$ by Lemma 6.19. This concludes
Case 2 and the proof of Lemma 6.20. ∎

###  6.5 Proof of Proposition 6.4

We have now gathered all the ingredients to prove Proposition 6.4. In this
section, we will only deal with $\mathcal{E}_{\mathbb{R}}$, so will write
$\mathcal{E}$ for $\mathcal{E}_{\mathbb{R}}$. Let us first prove the one-point
estimate (6.1).

###### Proof of (6.1).

By conformal invariance, it is enough to prove (6.1) for $z=i$. Fix $\eta>0$.
By Lemma 6.13, there exists $C>0$, such that for all $r\in(0,1-\eta)$,

$\displaystyle\mathbb{P}[\mathcal{E}\cap U(i,r)\not=\emptyset]\leq
C\mathbb{E}\left[\sum_{e\in\mathcal{E}}\frac{(a(e)-b(e))^{2}}{|a(e)-i|^{2}|b(e)-i|^{2}}\right]|\log
r|^{-1}.$ (6.38)

Let $I_{-1}=[-1,1]$ and for each $n\geq 0$, let
$I_{n}:=[-2^{n+1},-2^{n})\cup(2^{n},2^{n+1}]$ so that
$\mathbb{R}=\bigcup_{n\geq-1}I_{n}$. Note that for all $n,m\geq-1$, if
$a(e)\in I_{n}$ and $b(e)\in I_{m}$, then $|a(e)-i|\geq 2^{n}$ and
$|b(e)-i|\geq 2^{m}$. We can thus bound the expectation in (6.38) by

$\displaystyle\sum_{n,m\geq-1}2^{-2n-2m}\mathbb{E}\left[\sum_{e\in\mathcal{E},a(e)\in
I_{n},b(e)\in I_{m}}(a(e)-b(e))^{2}\right].$

By Lemma 6.20, the expectation above is bounded by $C2^{n+m}$. This shows that
the expectation in (6.38) is bounded by some constant concluding the proof. ∎

Let us now prove the two-point estimate (6.2).

###### Proof of (6.2).

Suppose $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$. Let $\nu=\nu(\eta)$ be
as in Lemma 6.14. Let us first assume that the condition
$r_{i}y_{i}\leq\nu|z_{1}-z_{2}|$, $i=1,2$, from Lemma 6.14 is violated. For
instance, assume that $r_{1}y_{1}>\nu|z_{1}-z_{2}|$. In that case, we simply
bound

$\displaystyle\mathbb{P}[\mathcal{E}\cap
U(z_{1},r_{1})\not=\emptyset,\mathcal{E}_{\mathbb{R}}\cap
U(z_{2},r_{2})\not=\emptyset]\leq\mathbb{P}[\mathcal{E}_{\mathbb{R}}\cap
U(z_{2},r_{2})\not=\emptyset]\leq C|\log r_{2}|^{-1}$

by (6.1). Since $r_{1}y_{1}>\nu|z_{1}-z_{2}|$, we further have

$|\log r_{2}|^{-1}\leq C\log\frac{y_{1}}{|z_{1}-z_{2}|}|\log r_{1}|^{-1}|\log
r_{2}|^{-1}\leq CG_{\mathbb{H}}(z_{1},z_{2})|\log r_{1}|^{-1}|\log
r_{2}|^{-1}.$

This concludes the proof of (6.2) in this case.

Let us now assume that $z_{1}$ and $z_{2}$ are not too close to each other so
that $r_{i}y_{i}\leq\nu|z_{1}-z_{2}|$, $i=1,2$. This will allow us to apply
Lemma 6.14. The event $\mathcal{E}\cap
U(z_{1},r_{1})\not=\emptyset,\mathcal{E}\cap U(z_{2},r_{2})\not=\emptyset$ is
the union of the following two events:

  * •

Let $E_{1}$ be the event that $U(z_{1},r_{1})$ and $U(z_{2},r_{2})$ are both
visited by a same excursion in $\mathcal{E}$.

  * •

Let $E_{2}$ be the event that $U(z_{1},r_{1})$ and $U(z_{2},r_{2})$ are
visited by two different excursions in $\mathcal{E}$.

We are going to show that

$\mathbb{P}[E_{1}]\leq C(1+G_{\mathbb{H}}(z_{1},z_{2}))|\log r_{1}|^{-1}|\log
r_{2}|^{-1}\quad\text{and}\quad\mathbb{P}[E_{2}]\leq C|\log r_{1}|^{-1}|\log
r_{2}|^{-1}$

which will conclude the proof of (6.2).

Contribution of $E_{1}$. By Lemma 6.14, $\mathbb{P}[E_{1}]$ is at most

$\displaystyle
C\mathbb{E}\left[\sum_{e\in\mathcal{E}}\left(\frac{(a(e)-b(e))^{2}y_{1}y_{2}}{|a(e)-z_{1}|^{2}|b(e)-z_{2}|^{2}}+\frac{(a(e)-b(e))^{2}y_{1}y_{2}}{|a(e)-z_{2}|^{2}|b(e)-z_{1}|^{2}}\right)\right](1+G_{\mathbb{H}}(z_{1},z_{2}))|\log
r_{1}|^{-1}|\log r_{2}|^{-1}.$

By symmetry, it is enough to prove that there exists a universal constant
$C>0$ such that

$\displaystyle\mathbb{E}\left[\sum_{e\in\mathcal{E}}\frac{(a(e)-b(e))^{2}y_{1}y_{2}}{|a(e)-z_{1}|^{2}|b(e)-z_{2}|^{2}}\right]\leq
C.$ (6.39)

For $j=1,2$ and $n\geq 0$, let

$I_{j,-1}:=[x_{j}-y_{j},x_{j}+y_{j}]\quad\text{and}\quad
I_{j,n}:=[x_{j}-y_{j}2^{n+1},x_{j}-y_{j}2^{n})\cup(x_{j}+y_{j}2^{n},x_{j}+y_{j}2^{n+1}].$
(6.40)

As in the proof of (6.1), we can bound the left hand side of (6.39) by

$\displaystyle
y_{1}y_{2}\sum_{n,m\geq-1}(y_{1}2^{n})^{-2}(y_{2}2^{n})^{-2}\mathbb{E}\left[\sum_{e\in\mathcal{E};a(e)\in
I_{1,n},b(e)\in I_{2,m}}(a(e)-b(e))^{2}\right]\leq C\sum_{n,m\geq-1}2^{-n-m}$

using Lemma 6.20 in the last inequality.

Contribution of $E_{2}$. By Lemma 6.13, $\mathbb{P}[E_{2}]$ is upper bounded
by a constant times

$\displaystyle\mathbb{E}\left[\sum_{e_{1}\in\mathcal{E}}\sum_{e_{2}\in\mathcal{E}}\frac{(b(e_{1})-a(e_{1}))^{2}y_{1}^{2}}{|z_{1}-b(e_{1})|^{2}|z_{1}-a(e_{1})|^{2}}\frac{(b(e_{2})-a(e_{2}))^{2}y_{2}^{2}}{|z_{2}-b(e_{2})|^{2}|z_{2}-a(e_{2})|^{2}}\right]|\log
r_{1}|^{-1}|\log r_{2}|^{-1}.$ (6.41)

Let $I_{j,n}$, $j=1,2$, $n\geq-1$, be as in (6.40). We can bound the above
expectation by

$\displaystyle
y_{1}^{-2}y_{2}^{-2}\sum_{\begin{subarray}{c}n_{1},m_{1}\geq-1\\\
n_{2},m_{2}\geq-1\end{subarray}}2^{-2(n_{1}-m_{1}-n_{2}-m_{2})}\mathbb{E}\left[\sum_{\begin{subarray}{c}e_{1}\in\mathcal{E};a(e_{1})\in
I_{1,n_{1}},b(e_{1})\in I_{1,m_{1}}\\\ e_{2}\in\mathcal{E}_{2};a(e_{2})\in
I_{2,n_{2}},b(e_{2})\in
I_{2,m_{2}}\end{subarray}}(b(e_{1})-a(e_{1}))^{2}(b(e_{2})-a(e_{2}))^{2}\right].$

By Cauchy-Schwarz, the expectation inside the sum is at most
$\sqrt{X_{1}X_{2}}$ where

$X_{j}:=\mathbb{E}\left[\left(\sum_{e\in\mathcal{E};a(e)\in
I_{j,n_{j}},b(e)\in I_{j,m_{j}}}(b(e)-a(e))^{2}\right)^{2}\right],\quad
j=1,2.$

It is enough to show that $X_{j}\leq Cy_{j}^{4}2^{3n_{j}+3m_{j}}$ to conclude
the proof. If $n=m$, then by scaling, $X_{j}$ is at most
$Cy_{j}^{4}2^{4n_{j}}S_{2}$, where $S_{2}$ is defined in (6.30). By symmetry,
we can now suppose that $m_{j}>n_{j}$. By scaling, we have

$\displaystyle X_{j}\leq
y_{j}^{4}2^{4m_{j}}\mathbb{E}\left[\left(\sum_{e\in\mathcal{E};a(e)\in[-2^{n_{j}-m_{j}},2^{n_{j}-m_{j}}],b(e)\in[-2,2]}(b(e)-a(e))^{2}\right)^{2}\right]\leq
Cy_{j}^{4}2^{4m_{j}}2^{n_{j}-m_{j}}$

by Lemma 6.19. This shows that $X_{j}\leq Cy_{j}^{4}2^{3m_{j}+n_{j}}\leq
Cy_{j}^{4}2^{3m_{j}+3n_{j}}$ as desired. ∎

The previous proofs already indicate that the probability of
$\mathcal{E}_{\mathbb{R}}$ intersecting a neighborhood of $z\in\mathbb{H}$ is
mainly contributed by the excursions in $\mathcal{E}_{\mathbb{R}}$ which are