$(\mathcal{A},d,\mu)$. Recall that $\mathrm{Mod}(\mathcal{A})$ is the modulus
of the planar annulus $A$.

######  Lemma 6.7.

The metric-measure space $(\mathcal{A},d,\mu)$ and quantum surface
$(A,\phi)/{\sim_{\gamma}}$ are measurable with respect to each other. In
particular, $\mathrm{Mod}(\mathcal{A})$ is measurable with respect to
$(\mathcal{A},d,\mu)$.

Lemma 6.7 follows from the measurability results in [MS21b, GMS20] for the
Brownian sphere and disk. See Appendix B for a proof. Our next lemma specifies
the law of $\mathcal{L}$ in Definition 1.1 of $\mathrm{BA}(a,b)^{\\#}$. It
follows from several known facts on the enumerations and scaling limits of
quadrangulations. We also give its proof in Appendix B. To draw the analog
with Section 4 we introduce the function

(6.2) $|\mathrm{Ball}_{1}(\ell)|=e^{-\frac{9\ell}{2}}\quad\textrm{for
}\ell>0.$

It plays the same role as $|\mathrm{LF}_{\mathbb{H}}^{\gamma,i}(\ell)|$ in the
description of $L^{\gamma}_{\phi}(\eta)$ in Theorem 4.2. In light of
Proposition B.4, $|\mathrm{Ball}_{1}(\ell)|$ can be thought of as the
partition function of the filled metric ball with radius $1$ and perimeter
$\ell$.

######  Lemma 6.8.

Fix $a>0$. Let $(\mathcal{D},p,d,\mu)$ be a Brownian disk sampled from
$\mathrm{BD}_{1,0}(a)^{\\#}$. On the event that $d(p,\partial\mathcal{D})>1$,
let $\mathcal{A}$ be the annulus in Definition 1.1 and let $\mathcal{L}$ be
the length of $\partial\mathcal{A}\setminus\partial\mathcal{D}$, which is also
the boundary length of the filled metric ball $\mathcal{B}^{\bullet}(p,1)$.
Then the law of $\mathcal{L}$ conditioned on $d(p,\partial\mathcal{D})>1$ is

(6.3)
$\frac{1}{Z(a)}\frac{1_{b>0}a^{3/2}b|\mathrm{Ball}_{1}(b)|\mathrm{d}b}{\sqrt{ab}(a+b)}\quad\textrm{where
}Z(a)=\int_{0}^{\infty}\frac{a^{3/2}b|\mathrm{Ball}_{1}(b)|\mathrm{d}b}{\sqrt{ab}(a+b)}.$

Moreover,
$\mathrm{BD}_{1,0}(a)^{\\#}[d(p,\partial\mathcal{D})>1]=Z(a)/Z(\infty)$ where
$Z(\infty)=\int_{0}^{\infty}b^{\frac{1}{2}}|\mathrm{Ball}_{1}(b)|\mathrm{d}b$.
Finally, conditioning on $\mathcal{B}^{\bullet}(p,1)$, the conditional law of
$\mathcal{A}$ only depends on $\mathcal{L}$.

We record the following Brownian sphere analog of Lemma 6.8, which we need in
Section 7. See Appendix B for the proof.

######  Lemma 6.9.

For $(\mathcal{S},x,y)$ sampled from $\mathrm{BS}_{2}$, let $\mathcal{L}_{x}$
and $\mathcal{L}_{y}$ be the boundary lengths of $\mathcal{B}^{\bullet}(x,1)$
and $\mathcal{B}^{\bullet}(y,1)$, respectively. Then under the restriction of
$\mathrm{BS}_{2}$ to the event $d(x,y)>2$, the joint law of
$(\mathcal{L}_{x},\mathcal{L}_{y})$ has density

$C\cdot
1_{a>0;b>0}\frac{\sqrt{ab}|\mathrm{Ball}_{1}(a)||\mathrm{Ball}_{1}(b)|}{a+b}\mathrm{d}a\mathrm{d}b\qquad\textrm{for
some constant }C\in(0,\infty).$

###  6.2. Liouville field description and the law of the modulus

Let

(6.4)
$\mathrm{BA}(a,b)=\frac{\mathrm{BA}(a,b)^{\\#}}{2\sqrt{ab}(a+b)}\quad\textrm{and}\quad\mathrm{BA}=\iint_{0}^{\infty}\mathrm{BA}(a,b)\,\mathrm{d}a\,\mathrm{d}b.$

In light of Lemma 6.7, we can view the measures $\mathrm{BA}$ and
$\mathrm{BA}(a,b)$ as laws of quantum surfaces. The proof of Theorem 6.12 is
parallel to that of Theorem 1.4. We first give the counterpart of Theorem 4.2.
This is essentially a reformation of Lemma 6.8 where we describe Brownian
disks in terms of Liouville fields.

######  Proposition 6.10.

Set $\gamma=\sqrt{8/3}$. Sample $\phi$ from
$\mathrm{LF}_{\mathbb{H}}^{\gamma,i}$. On the event
$E=\\{d_{\mathbb{H}}(i,\partial\mathbb{H})>1\\}$, let $A_{\eta}$ be the
annular connected component of $\mathbb{H}\setminus B_{d_{\mathbb{H}}}(i,1)$.
Let $\eta$ be the loop corresponding to $\partial
A_{\eta}\setminus\partial\mathbb{H}$ and $\mathcal{L}^{\gamma}_{\phi}(\eta)$
be its quantum length. Let $D_{\eta}$ be the Jordan domain bounded by $\eta$.
Let $f(b)=b|\mathrm{Ball}_{1}(b)|$. Then under the measure
$M=\frac{1_{E}}{f(\mathcal{L}^{\gamma}_{\phi}(\eta))}\cdot\mathrm{LF}^{\gamma,i}_{\mathbb{H}}$,
the law of $(A_{\eta},\phi)/{\sim_{\gamma}}$ is $C\cdot\mathrm{BA}$ for some
constant $C>0$. Moreover, conditioning on $(D_{\eta},\phi)/{\sim_{\gamma}}$,
the conditional law of $(A_{\eta},\phi)/{\sim_{\gamma}}$ only depends on
$\mathcal{L}^{\gamma}_{\phi}(\eta)$.

###### Proof.

In this proof $C$ is a positive constant that could change from line to line.
By Theorem 6.5 and Remark 6.6, under $\mathrm{LF}_{\mathbb{H}}^{\gamma,i}$ the
law of $(\mathbb{H},\phi,i)/{\sim_{\gamma}}$ is $C\cdot\mathrm{BD}_{1,0}$.
Note that

$\frac{a^{3/2}b|\mathrm{Ball}_{1}(b)|}{\sqrt{ab}(a+b)}=C\frac{|\mathrm{BA}(a,b)|b|\mathrm{Ball}_{1}(b)|}{|\mathrm{BD}_{1,0}(a)|}.$

By Lemma 6.8, under $M$ the joint law of
$L_{\phi}^{\gamma}(\partial\mathbb{H})$ and $L_{\phi}^{\gamma}(\eta)$ is
$C1_{a>0,b>0}\cdot|\mathrm{BA}(a,b)|\,\mathrm{d}a\mathrm{d}b$. By the
definition of $\mathrm{BA}(a,b)^{\\#}$, under $M$ the law of
$(A_{\eta},\phi)/{\sim_{\gamma}}$ is
$C1_{a>0,b>0}\cdot|\mathrm{BA}(a,b)|\mathrm{BA}(a,b)^{\\#}=C\cdot\mathrm{BA}$
as desired. Moreover, the last assertion of Proposition 6.10 follows from the
last assertion of Lemma 6.8. ∎

######  Lemma 6.11.

Set $\gamma=\sqrt{8/3}$. There exists a measure $m(\mathrm{d}\tau)$ on
$(0,\infty)$ such that if we sample $(\tau,\phi)$ from
$\mathrm{LF}_{\tau}(d\phi)m(\mathrm{d}\tau)$, then the law of the quantum
surface $(\mathcal{C}_{\tau},\phi)/{\sim_{\gamma}}$ is $\mathrm{BA}$.

###### Proof.

The argument is almost identical to that of Proposition 4.4 so we only point
out the differences. Suppose we are in the setting of Proposition 6.10 where
$M$ is the reweighted measure. Let $\hat{\tau}$ be such that $A_{\eta}$ is
conformally equivalent to $\mathcal{C}_{\hat{\tau}}$, and let
$\psi:\mathcal{C}_{\hat{\tau}}\to A_{\eta}$ be the conformal map such that
$\psi(0)=0$. Let $\hat{\phi}(z)=\phi\circ\psi(z)+Q\log|\psi(z)^{\prime}|$ for
$z\in\mathcal{C}_{\hat{\tau}}$. Let $\hat{\phi}^{\mathrm{har}}$ be the
harmonic extension of
$\hat{\phi}|_{\mathcal{C}(\frac{2\hat{\tau}}{3},\hat{\tau})}$ onto
$\mathcal{C}(0,\frac{2\hat{\tau}}{3})$ with zero normal derivative on
$\\{0\\}\times[0,1]$. By Proposition 6.10, under $M$, the law of
$(A_{\eta},\phi)/{\sim_{\gamma}}$ is $C\cdot\mathrm{BA}$. Therefore, under $M$
the law of $(\hat{\phi},\hat{\tau})$ is a rotational invariant embedding of
$C\cdot\mathrm{BA}$. Now the statement of Lemma 4.6 still holds for
$(\phi,\eta)$ under $M$ with the same argument except the following
modification: when we apply the domain Markov property from Lemma 2.11, we are
in the second scenario of that lemma since the metric $d_{\mathbb{H}}$ is
determined by the field $\phi$ locally. This means that conditioning on
$\hat{\tau}$ and
$\hat{\phi}|_{\mathcal{C}(\frac{2\hat{\tau}}{3},\hat{\tau})}$, the conditional
law of $\hat{\phi}-\hat{\phi}^{\mathrm{har}}$ is a GFF on
$\mathcal{C}(0,\frac{2\hat{\tau}}{3})$ with the appropriate boundary
condition.

Another key ingredient for the proof of Proposition 4.4 is the symmetry of
$\mathrm{QA}$ under flipping the sides, namely Lemma 4.7. By Lemma 6.4, we
have $\mathrm{BA}(a,b)^{\\#}=\mathrm{BA}(b,a)^{\\#}$ which gives the desired
symmetry for $\mathrm{BA}$. The rest of the argument based on Proposition 2.13
is identical to that of Proposition 4.4. ∎

######  Theorem 6.12.

The measure $m(\mathrm{d}\tau)$ in Lemma 6.11 equals
$1_{\tau>0}2^{-1/2}\eta(2i\tau)d\tau$.

###### Proof.

The argument is essentially the same as that of Theorem 1.4. By (6.4)

$\displaystyle\mathrm{BA}[L_{1}e^{-L_{1}}L_{0}^{ix}]=\iint_{0}^{\infty}ae^{-a}b^{ix}|\mathrm{BA}(a,b)|dadb=\frac{1}{2}\iint_{0}^{\infty}\frac{ae^{-a}b^{ix}}{\sqrt{ab}(a+b)}dadb.$

By (4.9) we have
$\mathrm{BA}[L_{1}e^{-L_{1}}L_{0}^{ix}]=\frac{\pi\Gamma(1+ix)}{2\cosh(\pi
x)}$. Now the KPZ relation (Theorem 1.6) yields that

$\int_{0}^{\infty}e^{-\pi\gamma^{2}x^{2}\tau/4}m(\mathrm{d}\tau)=\frac{2\sinh(\frac{\gamma^{2}}{4}\pi
x)}{\pi\gamma
x\Gamma(1+ix)}\mathrm{BA}[L_{1}e^{-L_{1}}L_{0}^{ix}]=\frac{\sinh(\frac{\gamma^{2}}{4}\pi
x)}{\gamma x\cosh(\pi x)}\quad\text{for }x\in\mathbb{R}.$

By (A.4) in Appendix A, we have

$\int_{0}^{\infty}e^{-\frac{\pi\gamma^{2}x^{2}\tau}{4}}\eta(2i\tau)d\tau=\frac{\sqrt{3}}{2}\frac{\sinh(\frac{2\pi
x}{3})}{\cosh(\pi x)}\quad\textrm{for }x\in\mathbb{R}.$

Since $\gamma=\sqrt{8/3}$, comparing the last two equations we get
$m(\mathrm{d}\tau)=1_{\tau>0}2^{-1/2}\eta(2i\tau)\mathrm{d}\tau$. ∎

###### Proof of Theorems 1.2 and 1.3.

By Theorem 6.12 we get Theorem 1.3. For Theorem 1.2, note that $X_{\tau}$
agrees in law with
$\frac{\mathcal{L}_{h}^{\gamma}(\partial_{1}\mathcal{C}_{\tau})}{\mathcal{L}_{h}^{\gamma}(\partial_{0}\mathcal{C}_{\tau})}$
in Theorem 1.5 with $\gamma=\sqrt{\frac{8}{3}}$. By Lemma 2.12, when
$\gamma=\sqrt{\frac{8}{3}}$ we have

$\mathrm{LF}_{\tau}[f(\mathcal{L}^{\gamma}_{\phi}(\partial_{0}\mathcal{C}_{\tau}))g(\mathcal{L}^{\gamma}_{\phi}(\partial_{1}\mathcal{C}_{\tau}))]=\frac{2}{\gamma}\iint_{0}^{\infty}\ell^{-1}f(\ell)g(\ell
r)\rho_{\tau}(r)\mathrm{d}r\mathrm{d}\ell=\frac{2}{\gamma}\iint_{0}^{\infty}a^{-2}f(a)g(b)\rho_{\tau}(\frac{b}{a})\mathrm{d}b\mathrm{d}a.$

Therefore the joint density of the two boundary lengths $(L_{0},L_{1})$ under
$\mathrm{LF}_{\tau}$ is
$\frac{2}{\gamma}a^{-2}\rho_{\tau}(\frac{b}{a})\mathrm{d}a\mathrm{d}b$. Now by
Theorem 1.3, the law of $\mathrm{Mod}(\mathcal{A})$ under
$\mathrm{BA}(a,b)^{\\#}$ is proportional to
$1_{\tau>0}\rho_{\tau}(\frac{b}{a})\eta(2i\tau)\mathrm{d}\tau$. ∎

######  Remark 6.13 ($\mathrm{BA}$ as the limit of $\mathrm{QA}$).

Note that
$\lim_{\gamma\to\sqrt{8/3}}\frac{\pi}{2\cos(\pi(\frac{4}{\gamma^{2}}-1))}|\mathrm{QA}^{\gamma}(a,b)|=|\mathrm{BA}(a,b)|$
for $\mathrm{QA}^{\gamma}$ from Section 4.1. In fact Theorems 1.3 and 1.4 show
that $\frac{\pi}{2\cos(\pi(\frac{4}{\gamma^{2}}-1))}\mathrm{QA}^{\gamma}(a,b)$
converges to $\mathrm{BA}(a,b)$ at the quantum surface level, since both the
law of the modulus and the field under the $\mathcal{C}_{\tau}$ embedding
converge.

##  7\. The annulus partition function of the SLE8/3 loop

In this section we prove Theorem 1.11. In Section 7.1 we recall the coupling
between the SLE8/3 loop and the conformally embedded Brownian sphere. In
Section 7.2 we use this coupling and the KPZ relation from Theorem 1.6 to
complete the proof.

###  7.1. The SLE8/3 loop zipper on the Brownian sphere

We first recall the $\sqrt{8/3}$-LQG construction of Miller and Sheffield
[MS20, MS21a, MS21b] for the conformally embedded Brownian sphere, and Zhan’s
construction of the $\mathrm{SLE}_{8/3}$ loop measure [Zha21]. These two
constructions (i.e. Theorem 7.1 and Definition 7.2) are included mainly for
concreteness. The only input we need is the conformal welding result from
[AHS22] for their coupling, which we recall as Theorem 7.4.

######  Theorem 7.1 ([MS20, MS21a, MS21b]).

Fix $\gamma=\sqrt{8/3}$ and $Q=\frac{2}{\gamma}+\frac{\gamma}{2}$. Let
$(B_{s})_{s\geq 0}$ be a standard Brownian motion conditioned on
$B_{s}-(Q-\gamma)s<0$ for all $s>0$, and let $(\widetilde{B}_{s})_{s\geq 0}$
be an independent copy of $(B_{s})_{s\geq 0}$. Let

$Y_{t}=\left\\{\begin{array}[]{ll}B_{t}-(Q-\gamma)t&\mbox{if }t\geq 0\\\
\widetilde{B}_{-t}+(Q-\gamma)t&\mbox{if }t<0\end{array}\right..$

Let $\mathcal{C}=\mathbb{R}\times[0,2\pi]/\sim$ be a horizontal cylinder. Let
$h^{1}(z)=Y_{\mathrm{Re}z}$ for each $z\in\mathcal{C}$. Let
$h^{2}_{\mathcal{C}}$ be independent of $h^{1}$ and have the law of the
lateral component of the Gaussian free field on $\mathcal{C}$. Let
$\hat{h}=h^{1}+h^{2}_{\mathcal{C}}$. Let $\mathbf{c}\in\mathbb{R}$ be sampled
from $\frac{\gamma}{2}e^{2(\gamma-Q)c}dc$ independently of $\hat{h}$ and set
$h=\hat{h}+\mathbf{c}$. Let $(\mathcal{S},x,y)$ be the (marked) metric-measure
space given by $(\mathcal{C},\infty,-\infty,c_{1}d_{h},c_{2}\mu_{h})$, with
the constants $c_{1}$ and $c_{2}$ from Theorem 6.5. Then there exists a
constant $C\in(0,\infty)$ such that the law of $(\mathcal{S},x,y)$ is
$C\cdot\mathrm{BS}_{2}$. Moreover, $(\mathcal{S},x,y)$ and the quantum surface
$(\mathcal{C},h,+\infty,-\infty)/{\sim_{\gamma}}$ are measurable with respect
to each other.

The quantum surface $(\mathcal{C},h,+\infty,-\infty)/{\sim_{\gamma}}$ is the
_two-pointed quantum sphere_ defined in [DMS21]. For more details of its
definition such as how to take the zero-probability conditioning when defining
$Y$, and the definition of the lateral component of the GFF on $\mathcal{C}$,
we refer to [DMS21, Section 4]. See also the preliminary sections of [AHS20b,
AHS21, AS21]. For Zhan’s construction of the $\mathrm{SLE}_{8/3}$ loop
measure, we first recall that for $\kappa\in(0,8)$ and two distinct points
$p,q\in\mathbb{C}$, the _two-sided whole plane SLE κ_ is the probability
measure on pairs of curves $(\eta_{1},\eta_{2})$ on $\mathbb{C}$ connecting
$p$ and $q$ where $\eta_{1}$ is a so-called whole-plane
$\mathrm{SLE}_{\kappa}(2)$ from $p$ to $q$, and conditioning on $\eta_{1}$,
the curve $\eta_{2}$ is a chordal $\mathrm{SLE}_{\kappa}$ on the complement of
$\eta_{1}$. We can view $\mathrm{SLE}^{p\rightleftharpoons q}_{\kappa}$ as a
measure on oriented loops by reversing the direction of $\eta_{2}$. Given a
loop $\eta$ sampled from $\mathrm{SLE}^{p\rightleftharpoons q}_{\kappa}$, with
probability 1 its $d_{\kappa}$-dimensional Minkowski content
$\mathrm{Cont}(\eta)$ exists [LR15], where $d_{\kappa}:=1+\frac{\kappa}{8}$ .

######  Definition 7.2.

Fix $\kappa=\frac{8}{3}$. Let $\mathrm{SLE}^{p\rightleftharpoons q}_{\kappa}$
be two-sided whole plane SLEκ between $p,q$. Zhan’s $\mathrm{SLE}_{8/3}$ loop
measure is the infinite measure on unmarked oriented loops defined by

(7.1)
$\mu_{\mathbb{C}}(\mathrm{d}\eta)=\mathrm{Cont}(\eta)^{-2}\iint_{\mathbb{C}\times\mathbb{C}}|p-q|^{-2(2-d_{\kappa})}\mathrm{SLE}^{p\rightleftharpoons
q}_{\kappa}(\mathrm{d}\eta)\,\mathrm{d}p\,\mathrm{d}q.$

For $\tau>0$, let $Z_{8/3}(\tau)=\mu_{\mathbb{C}}[E_{\tau}]$ where $E_{\tau}$
is the set of non-contractible loops in
$\mathbb{A}_{\tau}=\\{e^{-2\pi\tau}<|z|<1\\}$.

Zhan’s construction works for all $\kappa\in(0,8)$ and it was shown in [Zha21]
that modulo a multiplicative constant, the $\kappa=\frac{8}{3}$ case as in
Definition 7.2 gives the loop measure defined by Werner [Wer08]; see [Zha21,
Remark 5.3]. In particular, it has the following conformal restriction
property.

######  Theorem 7.3 ([Wer08, Zha21]).

For each domain $\Omega\subset\mathbb{C}$, let $\mu_{\Omega}$ be the
restriction of $\mu_{\mathbb{C}}$ to loops in $\Omega$. Then for any conformal
map $f:\Omega\to\Omega^{\prime}$ between two domains, we have
$f\circ\mu_{\Omega}=\mu_{\Omega^{\prime}}$. Namely, for $\eta$ sampled from
$\mu_{\Omega}$, the law of $f\circ\eta$ is $\mu_{\Omega^{\prime}}$.

We now recall a conformal welding result for $\mu_{\mathbb{C}}$ proved in
[AHS22]. This particular variant was proved as [AS21, Proposition 2.20] based
on [AHS22].

######  Theorem 7.4.

Fix $\gamma=\sqrt{8/3}$. Let $\mathbb{F}_{\mathbb{C}}$ be the law of the field
$\phi=h\circ\log-Q\log|\cdot|$ on $\mathbb{C}$ where $h$ as in Theorem 7.1 and
$\log$ is the conformal map from $\mathbb{C}$ to $\mathcal{C}$. Let
$\hat{\mu}_{\mathbb{C}}$ be the restriction of $\mu_{\mathbb{C}}$ to the set
of loops separating $0$ and $\infty$. Now sample $(\phi,\eta)$ from
$\mathbb{F}_{\mathbb{C}}\times\hat{\mu}_{\mathbb{C}}$. Let $D_{0}$ and
$D_{\infty}$ be the two connected components of $\mathbb{C}\setminus\eta$ such
that $0\in D_{0}$. Then the joint law of $(D_{0},\phi,0)/{\sim_{\gamma}}$ and
$(D_{\infty},\phi,\infty)/{\sim_{\gamma}}$ is

$C\int_{0}^{\infty}\ell\mathrm{BD}_{1,0}(\ell)\times\mathrm{BD}_{1,0}(\ell)\,d\ell\quad\text{
for some constant }C\in(0,\infty).$

Here $\mathrm{BD}_{1,0}(\ell)$ is identified as a law of quantum surface in
$\sqrt{8/3}$-LQG using Theorem 6.5.

###  7.2. The annulus partition function from the KPZ relation

The idea behind our proof of Theorem 1.11 is that if we glue together two
samples of $\mathrm{BA}$, this gives a sample of $\mathrm{BA}$ decorated by a
non-contractible $\mathrm{SLE}_{8/3}$ loop, where the law of the modulus of
the annulus is reweighted by the partition function $Z_{8/3}$. To make it
rigorous, we realize the Brownian annulus as a subset of the Brownian sphere
and apply Theorem 7.4.

######  Lemma 7.5.

Let $\phi$ be a sample from $\mathbb{F}_{\mathbb{C}}$ in Theorem 7.4. Let $E$
be the event that $d(0,\infty)>2$, where $d=c_{1}d_{\phi}$ is the Brownian
metric. On the event $E$, let $B^{\bullet}(0,1)$ and $B^{\bullet}(\infty,1)$
be the filled metric ball of radius 1 around $0$ and $\infty$, respectively.
Let $\mathcal{L}_{1}$ and $\mathcal{L}_{\infty}$ be the boundary lengths of
$B^{\bullet}(0,1)$ and $B^{\bullet}(\infty,1)$, respectively. Let
$A=\mathbb{C}\setminus(B^{\bullet}(0,1))\cup B^{\bullet}(\infty,1))$. Let
$f(b)=b|\mathrm{Ball}_{1}(b)|$ as in Proposition 6.10. Let $M$ be
$\frac{1}{f(\mathcal{L}_{0})f(\mathcal{L}_{\infty})}\mathbb{F}$ restricted to
$E$. Then under $M$ the law of $(A,\phi)/{\sim_{\gamma}}$ is
$C\cdot\mathrm{BA}$ for some constant $C>0$.

###### Proof.

The result follows from Lemmas 6.3 and 6.9 via the same argument as in
Proposition 6.10. ∎

Recall from Lemma 6.11 and Theorem 6.12 that if we sample $(\phi,\tau)$ from
$\mathrm{LF}_{\tau}(d\phi)m(\mathrm{d}\tau)$ with

(7.2) $m(\mathrm{d}\tau)=1_{\tau>0}2^{-1/2}\eta(2i\tau)d\tau,$

then the law of the quantum surface
$(\mathcal{C}_{\tau},\phi)/{\sim_{\gamma}}$ is $\mathrm{BA}$.

######  Lemma 7.6.

Let $(\phi,\eta)$ be a sample from
$\mathbb{F}_{\mathbb{C}}\times\hat{\mu}_{\mathbb{C}}$ in Theorem 7.4. Let
$E=\\{d(0,\infty)>2\\}$, $f(\ell)=\ell|\mathrm{Ball}_{1}(\ell)|$, and
$A=\mathbb{C}\setminus(B^{\bullet}(0,1))\cup B^{\bullet}(\infty,1))$ as in
Lemma 7.5. Let $F$ be the event that $\eta\subset A$ and $\eta$ is non-
contractible within $A$. Let $\hat{M}$ be the measure
$\frac{1}{f(\mathcal{L}_{0})f(\mathcal{L}_{\infty})}\mathbb{F}\times\hat{\mu}_{\mathbb{C}}$
restricted to $E\cap F$. Then under $\hat{M}$ the law of
$(A,\phi)/{\sim_{\gamma}}$ is that of
$(\mathcal{C}_{\tau},\phi)/{\sim_{\gamma}}$ with $(\phi,\tau)$ sampled from
$C\mathrm{LF}_{\tau}(\mathrm{d}\phi)Z_{8/3}(\tau)m(\mathrm{d}\tau)$ for some
conctant $C>0$.

###### Proof.

By the conformal restriction property Theorem 7.3, given $A$, the
$\hat{\mu}_{\mathbb{C}}$-mass of the event $F$ is $Z_{8/3}(\mathrm{Mod}(A))$,
where $\mathrm{Mod}(A)$ is the modulus of $A$. By Lemma 7.5, we obtain the
desired marginal law of $(A,\phi)/{\sim_{\gamma}}$ under $\hat{M}$ after
integrating over $\eta$. ∎

Similar to $|\mathrm{QA}^{\gamma}_{2}(a,b)|$ in (5.6), for $a>0$ and $b>0$ we
let

(7.3)
$|\mathrm{BA}_{2}(a,b)|=\int_{0}^{\infty}|\mathrm{BA}(a,\ell)|\ell|\mathrm{BA}(\ell,b)|\mathrm{d}\ell.$

Now the following lemma together with Lemma 7.6 will allow us to apply the KPZ
relation in Theorem 1.6.

######  Lemma 7.7.

In the setting of Lemma 7.6, the joint law of
$(\mathcal{L}_{0},\mathcal{L}_{\infty})$ under $\hat{M}$ has density

$C\cdot
1_{a>0;b>0}|\mathrm{BA}_{2}(a,b)|\mathrm{d}a\mathrm{d}b\qquad\textrm{for some
constant }C\in(0,\infty).$

###### Proof.

By Theorem 7.4 and Lemma 6.8, under $\mathbb{F}\times\hat{\mu}_{\mathbb{C}}$
restricted to $E\cap F$ the joint law of
$(\mathcal{L}_{0},\mathcal{L}_{\infty})$ is

$C\cdot
1_{a>0;b>0}|\mathrm{Ball}_{1}(a)|a|\mathrm{BA}_{2}(a,b)|b|\mathrm{Ball}_{1}(b)|\mathrm{d}a\mathrm{d}b\qquad\textrm{for
some constant }C\in(0,\infty).$

Since $f(\ell)=\ell|\mathrm{Ball}_{1}(\ell)|$, we get the desired law of
$(\mathcal{L}_{0},\mathcal{L}_{\infty})$ under $\hat{M}$. ∎

###### Proof of Theorem 1.11..

By Theorem 1.6 and Lemmas 7.6– 7.7, we have

(7.4)
$\displaystyle\int_{0}^{\infty}e^{-\frac{\pi\gamma^{2}x^{2}\tau}{4}}Z_{8/3}(\tau)m(\mathrm{d}\tau)=\frac{2\sinh(\frac{\gamma^{2}}{4}\pi
x)}{\pi\gamma
x\Gamma(1+ix)}\iint_{0}^{\infty}ae^{-a}b^{ix}|\mathrm{BA}_{2}(a,b)|\mathrm{d}a\mathrm{d}b\quad\textrm{for
}x\in\mathbb{R},$

where $\gamma=\sqrt{8/3}$ and
$m(\mathrm{d}\tau)=1_{\tau>0}2^{-1/2}\eta(2i\tau)d\tau$ is as in (7.2). By the
same calculation in the proof of Lemma 5.5 we get

$\iint_{0}^{\infty}ae^{-a}b^{ix}|\mathrm{BA}_{2}(a,b)|\mathrm{d}a\mathrm{d}b=\frac{C_{1}}{\cosh(\pi
x)}\mathrm{BA}[L_{1}e^{-L_{1}}L_{0}^{ix}]\quad\textrm{for some constant
}C_{1}>0.$

Recall from the proof of Theorem 6.12 that
$\mathrm{BA}[L_{1}e^{-L_{1}}L_{0}^{ix}]=\frac{\pi\Gamma(1+ix)}{2\cosh(\pi
x)}$. Therefore

(7.5) $\int_{0}^{\infty}e^{-\frac{2\pi
x^{2}}{3}\tau}\eta(2i\tau)Z_{8/3}(\tau)\,\mathrm{d}\tau=C_{2}\frac{\sinh(\frac{2\pi
x}{3})}{x\cosh(\pi x)^{2}}\qquad\text{ for }x\in\mathbb{R}\textrm{ and
}C_{2}=\gamma^{-1}C_{1}.$

Now we take the inverse Laplace transform to compute $Z_{\mathrm{8/3}}(\tau)$.
Let $\mathcal{L}$ and $\mathcal{L}^{-1}$ denote the Laplace transform and its
inverse. Let $F(s):=\frac{\sinh(\alpha\sqrt{s})}{\sqrt{s}\cosh(\sqrt{s})^{2}}$
with $\alpha=\frac{2}{3}$. Since
$\frac{1}{\cosh(u)^{2}}=\sum_{n=1}^{\infty}(-1)^{n-1}4ne^{-2nu}$, we have

$F(s)=\frac{e^{\alpha\sqrt{s}}-e^{-\alpha\sqrt{s}}}{2\sqrt{s}\cosh(\sqrt{s})^{2}}=2\sum_{n=1}^{\infty}(-1)^{n-1}n\frac{e^{-(2n-\alpha)\sqrt{s}}-e^{-(2n+\alpha)\sqrt{s}}}{\sqrt{s}}.$

Since
$\mathcal{L}^{-1}[\frac{e^{-a\sqrt{s}}}{\sqrt{s}}](t)=\frac{e^{-\frac{a^{2}}{4t}}}{\sqrt{\pi
t}}$ for $a\geq 0$, we have

$\mathcal{L}^{-1}[F(s)](t)=2\sum_{n=1}^{\infty}(-1)^{n-1}n\frac{e^{-\frac{(2n-\alpha)^{2}}{4t}}-e^{-\frac{(2n+\alpha)^{2}}{4t}}}{\sqrt{\pi
t}}=\frac{2}{\sqrt{\pi
t}}\sum_{k\in\mathbb{Z}}^{\infty}k(-1)^{k-1}{e^{-\frac{1}{t}(k-\frac{1}{3})^{2}}}.$

By (7.5) we have
$\mathcal{L}[\eta(2i\tau)Z_{8/3}(\tau)](\frac{2s}{3\pi})=C_{2}F(s)$. Therefore

(7.6)
$\eta(2i\tau)Z_{8/3}(\tau)=\frac{2C_{2}}{3\pi}\mathcal{L}^{-1}[F(s)](\frac{2\tau}{3\pi})=\frac{4C_{2}}{\pi\sqrt{3}}\cdot(2\tau)^{-1/2}\sum_{k\in\mathbb{Z}}k(-1)^{k-1}e^{-\frac{3\pi}{2\tau}(k-\frac{1}{3})^{2}}.$

On the other hand, recall from (5.20) that
$\eta(2i\tau)=(2\tau)^{-1/2}{q}^{\frac{1}{24}}\prod_{r=1}^{\infty}(1-q^{r})$
with $q=e^{-\pi/\tau}$. Therefore

(7.7)
$Z_{\mathrm{Cardy}}(q)\eta(2i\tau)=(2\tau)^{-1/2}{q}^{\frac{1}{24}}\sum_{k\in\mathbb{Z}}k(-1)^{k-1}q^{\frac{3k^{2}}{2}-k+\frac{1}{8}}=(2\tau)^{-1/2}\sum_{k\in\mathbb{Z}}k(-1)^{k-1}q^{\frac{3}{2}(k-\frac{1}{3})^{2}}.$

Comparing (7.6) and (7.7), we see that $Z_{8/3}=CZ_{\mathrm{Cardy}}$ where
$C=\frac{4C_{2}}{\pi\sqrt{3}}$. ∎

##  Appendix A Background on special functions

The Dedekind eta function with imaginary argument is given by

(A.1) $\eta(i\tau)=e^{-\frac{\pi\tau}{12}}\prod_{k=1}^{\infty}(1-e^{-2\pi
k\tau})\quad\textrm{with }\tau>0.$

We use the following convention for the Jacobi theta function with imaginary
argument:

(A.2)
$\theta_{1}\\!\left(x,i\tau\right):=-i{e}^{-\pi\tau/4}\sum_{n=-\infty}^{\infty}{\left(-1\right)}^{n}{e}^{-n\left(n+1\right)\pi\tau}{e}^{(2n+1)\pi
ix}\quad\textrm{for }x\in\mathbb{R}\textrm{ and }\tau\in(0,\infty).$

By Euler’s pentagonal identity
$\prod_{k=1}^{\infty}(1-x^{k})=\sum_{n\in\mathbb{Z}}(-1)^{n}x^{n(3n-1)/2}$, we
have $\theta_{1}(\frac{1}{3},i\tau)={\sqrt{3}\eta(3i\tau)}$.

By e.g. [BS02, Section 11, Appendix 2], we have the following Laplace
transforms:

(A.3)
$\displaystyle\int_{0}^{\infty}{e}^{-a\tau}\theta_{1}\\!\left(x,ib\tau\right)\,d\tau$
$\displaystyle=\sqrt{\frac{\pi}{ab}}\frac{\sinh\\!\left(2x\sqrt{\frac{\pi
a}{b}}\right)}{\cosh\\!\left(\sqrt{\frac{\pi a}{b}}\right)}\quad$
$\displaystyle\textrm{for
}a>0,b>0\;\mathbin{\operatorname{and}}\;x\in\left[-1/2,1/2\right].$ (A.4)
$\displaystyle\int_{0}^{\infty}{e}^{-a\tau}\eta(i\tau)d\tau$
$\displaystyle=\sqrt{\frac{\pi}{a}}\frac{\sinh\\!\left(2\sqrt{\frac{\pi
a}{3}}\right)}{\cosh\\!\left(\sqrt{3\pi a}\right)}$ $\displaystyle\textrm{for
}a>0.$

##  Appendix B Background on Brownian surfaces

In this appendix we provide more background on Brownian surfaces and give
proofs for Lemmas 6.7– 6.9. Some of these lemmas may already exist in the
literature, but we cannot locate them hence include a proof for completeness.
We first recall a scaling limit result for the Brownian sphere due to [Le 13,
Mie13].

######  Theorem B.1 ([Le 13, Mie13]).

Let $\mathcal{S}_{n}$ be uniformly sampled from the set of quadrangulations of
the sphere with $n$ faces and two distinguished directed edges. View
$\mathcal{S}_{n}$ as a two-pointed metric-measure space as follows. The metric
is the graph distance on the vertex set rescaled by
$(\frac{9}{8})^{1/4}n^{-1}$. The measure is the probability measure on its
vertex set where the weight of a vertex is proportional to its degree. Then
$\mathcal{S}_{n}$ converges in law to the _unit area Brownian sphere_ , with
respect to the Gromov-Hausdorff-Prokhorov topology as compact metric-measure
spaces with two marked points.

We now recall a scaling limit result for $\mathrm{BD}_{0,1}(L)^{\\#}$ proved
by Gwynne and Miller [GM19]; see [AHS20a] for the triangulation case. The
topology of convergence is the natural one for compact metric-measure spaces
decorated with a continuous curve, which is called the Gromov-Hausdorff-
Prohkorov-uniform (GHPU) topology; see [GM19, Section 1.2.3] for its
definition. To state the result, for a positive integer $a$ we let
$\mathcal{Q}^{\mathrm{disk}}_{0,1}(2a)$ be the set of quadrangulations with
simple boundary, having boundary length $2a$, and having a marked directed
boundary edge. For later use we also let
$\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a)$ be the set of such quadrangulations
where we further mark a directed interior edge (i.e. an edge whose endpoints
are both interior vertices).

######  Theorem B.2 ([GM19]).

Let $L>0$ and let $(a_{n})$ be a sequence of integers satisfying
$\lim_{n\to\infty}\frac{2a_{n}}{3n^{2}}=L$. Let
$\mathrm{BD}^{n}_{0,1}(2a_{n})^{\\#}$ be the probability measure on
$\mathcal{Q}^{\mathrm{disk}}_{0,1}(2a_{n})$ where each map with $v$ vertices
is assigned mass proportional to $\rho^{v}$ with $\rho=\frac{1}{12}$. Let
$(\mathcal{D}_{n},e_{n})$ be a sample from
$\mathrm{BD}^{n}_{0,1}(2a_{n})^{\\#}$ where $e_{n}$ is the boundary marked
edge of $\mathcal{D}_{n}$. View $\mathcal{D}_{n}$ as a metric measure space
where the metric is graph distance rescaled by $n^{-1}$, and the mass at each
vertex $v$ with degree $\mathrm{deg}(v)$ is
$\frac{2}{9}n^{-4}\mathrm{deg}(v)$. We also view the boundary of
$\mathcal{D}_{n}$ as a continuous curve starting and ending at $e_{n}$. Then
with respect to the GHPU topology, as $n\to\infty$, the law of
$(\mathcal{D}_{n},e_{n})$ converges weakly to $\mathrm{BD}_{0,1}(L)^{\\#}$
defined in (6.1).

Let $\mathcal{Q}_{\mathrm{ann}}(2a,2b)$ be the set of annular quadrangulations
with labelled boundaries, where the number of edges on the two boundaries are
$2a$ and $2b$, respectively. The weak limit of the critical Boltzman measure
on $\mathcal{Q}_{\mathrm{ann}}(a,b)$ should be $\mathrm{BA}(a,b)$. Although
this scaling limit result is not needed for our paper, we will use the
following enumeration asymptotic, which explains why we set
$|\mathrm{BA}(a,b)|$ to be proportional to $\frac{1}{\sqrt{ab}(a+b)}$.

######  Lemma B.3.

Let $\rho=\frac{1}{12}$, $\theta=\frac{1}{54}$. Let
$\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a;v)\subset\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a)$
be the subset of maps with $v$ interior vertices. Then

$\sum_{v\geq
2}\\#\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a;v)\rho^{v}\theta^{a}\sim\frac{1}{\sqrt{27\pi
a}},\quad\textrm{where $\sim$ indicates the ratio tends to 1 as }a\to\infty.$

Likewise, let $\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b)$ be the set of such
quadrangulations in $\mathcal{Q}_{\mathrm{ann}}(2a,2b)$ where each boundary
has a marked edge. Let
$\mathcal{Q}_{\mathrm{ann}}(2a,2b;v)\subset\mathcal{Q}_{\mathrm{ann}}(2a,2b)$
and
$\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b;v)\subset\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b)$
be the subset of maps with $v$ interior vertices, respectively. Then with
$\sim$ indicating the ratio tends to 1 as $a,b\to\infty$, we have

$\sum_{v\geq
0}\\#\mathcal{Q}_{\mathrm{ann}}(2a,2b;v)\rho^{v}\theta^{a+b}\sim\frac{1}{\pi\sqrt{ab}(a+b)},\quad\textrm{and}\quad\sum_{v\geq
0}\\#\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b;v)\rho^{v}\theta^{a+b}\sim\frac{\sqrt{ab}}{\pi(a+b)}.$

###### Proof.

The key is the following enumeration from [BF18, Theorem 1.2]:

(B.1)
$\\#\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b;v)=\frac{3^{v}(3(a+b)+2v-1)!}{v!(3(a+b)+v)!}\cdot
4ab\binom{3a}{a}\binom{3b}{b}.$

We claim that

$\sum_{v\geq
0}\\#\mathcal{Q}^{2}_{\mathrm{ann}}(2a,2b;v)\rho^{v}=\frac{4ab}{3(a+b)}8^{a+b}\binom{3a}{a}\binom{3b}{b},\quad\textrm{and}\quad\sum_{v\geq
2}\\#\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a;v)\rho^{v}=\frac{2a}{9(a+1)}8^{a}\binom{3a}{a}.$

The first identity follows from (B.1) and [GKP94, (5.68), (5.70)]. The second
is obtained by identifying the interior marked edge as a boundary of length 2,
then using the first identity with $b=1$. Stirling’s approximation then gives
both asymptotics. Finally the asymptotic for $\\#\mathcal{Q}_{\mathrm{ann}}$
follow from that of $\\#\mathcal{Q}^{2}_{\mathrm{ann}}$ by dropping the
boundary marked edges. ∎

We now turn to the proof of Lemmas 6.8 and 6.9. We first state the asymptotic
of quadrangulation metric balls in Proposition B.4 below. For positive
integers $r,a$ we let $\mathcal{Q}_{\mathrm{ball}}(r;2a)$ be the set of
quadrangulations with the disk topology with boundary length $2a$, with a
distinguished oriented edge called the _root edge_ , such that the four
vertices of every quadrangle sharing an edge with the boundary are at
distances $r-1,r,r+1,r$ from the root vertex, respectively.

######  Proposition B.4 ([CLG17]).

Let $\rho=\frac{1}{12}$ and $\theta=\frac{1}{54}$ as in Lemma B.3. Let
$\mathbb{P}^{\mathrm{ball}}(n)$ be the probability measure on
$\bigcup_{a=1}^{\infty}\mathcal{Q}_{\mathrm{ball}}(n;2a)$ where each element
is assigned weight proportional to $\rho^{V}(\rho/\theta)^{L}$, where $V$ is
the number of internal vertices and $L$ is the number of boundary vertices.
Let $\mathcal{L}_{n}$ be the number of boundary vertices of a sample from
$\mathbb{P}^{\mathrm{ball}}(n)$. Then $(3n^{2})^{-1}\mathcal{L}_{n}$ converges
in law as $n\to\infty$ to the random variable with density proportional to
$1_{\ell>0}|\mathrm{Ball}_{1}(\ell)|d\ell$, where
$|\mathrm{Ball}_{1}(\ell)|=e^{-\frac{9\ell}{2}}$ as defined in (6.2).

###### Proof.

Recall that the Uniform Infinite Planar Quadrangulation (UIPQ) is the local
limit of uniform quadrangulations; see e.g. [CLG17]. Let
$\mathbb{P}^{\mathrm{ball}}_{\mathrm{UIPQ}}(n)$ be the law of the filled
metric ball around the root at radius $n$ on a UIPQ. Here a filled metric ball
is the union of the ball itself and all its bounded complementary components.
Let $\mathcal{Q}^{\mathrm{disk}}_{0,1}(2a;v)$ be the set of quadrangulations
with simple boundary, $2a$ boundary vertices and $n$ interior vertices, and a
marked boundary edge. As explained in [CLG17, Section 6.2],

$\\#\mathcal{Q}^{\mathrm{disk}}_{0,1}(2a;v)=\frac{3^{v-1}(3a)!(3a-3+2v)!}{v!a!(2a-1)!(v+3a-1)!}\qquad\textrm{and}\qquad\\#\mathcal{Q}^{\mathrm{disk}}_{0,1}(2a;v)\sim
C^{\square}(a)\rho^{-v}v^{-5/2}\textrm{ as }v\to\infty$

for $C^{\square}(a)=\frac{8^{a-1}(3a)!}{3\sqrt{\pi}a!(2a-1)!}$. Therefore, by
the definition of UIPQ, the probability measure
$\mathbb{P}^{\mathrm{ball}}_{\mathrm{UIPQ}}(n)$ is proportional to
$\theta^{\mathcal{L}_{n}/2}C^{\square}(\mathcal{L}_{n}/2)\mathbb{P}^{\mathrm{ball}}(n)$.
By the main result of [CLG17], the law of $(3n^{2})^{-1}\mathcal{L}_{n}$ under
$\mathbb{P}^{\mathrm{ball}}_{\mathrm{UIPQ}}(n)$ weakly converges to the
probability measure with density proportional to
$1_{\ell>0}\ell^{1/2}e^{-9\ell/2}d\ell$, which is the law of the boundary
length of the filled metric ball of radius 1 around the root on the Brownian
plane. This statement for triangulations (namely UIPT) was proved as Theorem 2
in [CLG17] and then extended to the UIPQ case in Section 6.2 there. Since
$C^{\square}(a)\sim\frac{1}{8\sqrt{3}\pi}\theta^{-a}\sqrt{a}$, the law of
$(3n^{2})^{-1}\mathcal{L}_{n}$ under $\mathbb{P}^{\mathrm{ball}}(n)$ weakly
converges to the probability measure with density proportional to
$1_{\ell>0}\ell^{-1/2}\cdot\ell^{1/2}e^{-9\ell/2}d\ell=1_{\ell>0}|\mathrm{Ball}_{1}(\ell)|d\ell$.
∎

###### Proof of Lemma 6.8.

Let $\rho=\frac{1}{12}$ and $\theta=\frac{1}{54}$ as in Lemma B.3 and
Proposition B.4. Let $a_{n}=\lfloor\frac{3}{2}n^{2}a\rfloor$ and
$\mathrm{BD}^{n}_{1,1}(2a_{n})^{\\#}$ be the probability measure on
$\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a_{n})$ where each element is assigned
weight proportional to $\rho^{V}$, with $V$ being the number of internal
vertices. Let $(\mathcal{D}_{n},p_{n},e_{n})$ be a sample from
$\mathrm{BD}^{n}_{1,1}(2a_{n})^{\\#}$, where $p_{n}$ is the interior marked
edge and $e_{n}$ is the boundary one. Let $E^{n}$ be the event that the graph
distance between $p_{n}$ and $\partial\mathcal{D}_{n}$ is at least $n$. On the
event $E^{n}$, let $\mathcal{B}^{\bullet}_{1}(p_{n})$ be the filled metric
ball of $\mathcal{D}_{n}$ around $p_{n}$ of radius $n$. Here a filled metric
ball is the union of the ball itself and all its complementary components
except the one touching the boundary. Let $\mathcal{L}_{n}$ be the number of
edges on the boundary of $\mathcal{B}^{\bullet}_{1}(p_{n})$. Let
$M^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})$ be the (non-probability) measure
describing the law of $\mathcal{B}^{\bullet}_{1}(p_{n})$ under the restriction
of $\mathrm{BD}^{n}_{1,1}(2a_{n})^{\\#}$ to $E^{n}$. Let
$Z_{n}=\sum\rho^{V}(\rho/\theta)^{L}$ where the summation is over
$\bigcup_{x=1}^{\infty}\mathcal{Q}_{\mathrm{ball}}(n;2x)$ as in the definition
of $\mathbb{P}^{\mathrm{ball}}(n)$ in Proposition B.4. Then with
$\mathcal{Q}_{\mathrm{ann}}^{2}(2a_{n},\mathcal{L}_{n};v)$ and
$\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a_{n};v)$ defined in Lemma B.3, we have

(B.2)
$\displaystyle\mathbb{M}^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})=\frac{\sum_{v}\\#\mathcal{Q}_{\mathrm{ann}}^{2}(2a_{n},\mathcal{L}_{n};v)\rho^{v}\theta^{a_{n}+b}}{\sum_{v}\\#\mathcal{Q}^{\mathrm{disk}}_{1,1}(2a_{n};v)\rho^{v}\theta^{a_{n}}}\cdot
Z_{n}\mathbb{P}^{\mathrm{ball}}(n).$

By definition the probability $\mathrm{BD}^{n}_{1,1}(2a_{n})^{\\#}(E^{n})$ is
the total mass $|\mathbb{M}^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})|$ of
$\mathbb{M}^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})$. By Theorem B.2,
$\mathrm{BD}^{n}_{1,1}(2a_{n})^{\\#}(E^{n})$ converges to
$\mathrm{BD}_{1,1}(a)^{\\#}[d(p,\partial\mathcal{D})>1]$. Recall that
$|\mathrm{BA}(a,b)|=\frac{1}{2\sqrt{ab}(a+b)}$ and
$|\mathrm{BD}_{1,1}(a)|=a^{-1/2}$. By the asymptotics in Lemma B.3 and
Proposition B.4, we have

$|\mathbb{M}^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})|=C_{1}(1+o_{n}(1))Z_{n}\int_{0}^{\infty}\frac{ab|\mathrm{BA}(a,b)||\mathrm{Ball}_{1}(b)|}{|\mathrm{BD}_{1,1}(a)|}\mathrm{d}b$

for some constant $C_{1}>0$. Therefore $Z_{n}$ converges to a constant
$C_{2}>0$ such that

$\mathrm{BD}_{1,1}(a)^{\\#}[d(p,\partial\mathcal{D})>1]=C_{1}C_{2}\int\frac{ab|\mathrm{BA}(a,b)||\mathrm{Ball}_{1}(b)|}{|\mathrm{BD}_{1,1}(a)|}\mathrm{d}b=CZ(a)$

for some constant $C>0$ not depending on $a$ and $Z(a)$ as defined in Lemma
6.8. Therefore

$\mathrm{BD}_{1,0}(a)^{\\#}[d(p,\partial\mathcal{D})>1]=BD_{1,0}(a)^{\\#}[d(p,\partial\mathcal{D})>1]=CZ(a).$

Sending $a\to\infty$ we get $C=Z(\infty)^{-1}$ as desired. Now combining
(B.2), the asymptotics in Lemma B.3 and Proposition B.4, and that $Z_{n}\to
C_{2}$, we see that the law of $(3n^{2})^{-1}\mathcal{L}_{n}$ under the
probability measure proportional to
$M^{\mathrm{ball}}_{\mathrm{disk}}(n;2a_{n})$ weakly converges to the law of
$\mathcal{L}$ described in Lemma 6.8. Finally, the conditional independence of
the annulus $\mathcal{A}$ and the filled metric ball
$\mathcal{B}^{\bullet}(p,1)$ is an instance of the domain Markov property of
the peeling process; see e.g. [MS20, MS21a] for this property from the QLE
perspective. ∎

###### Proof of Lemma 6.9.

We claim that under $\mathrm{BS}_{2}$ restricted to $d(x,y)>1$, the law of
$\mathcal{L}_{x}$ has density

$1_{\ell>0}|\mathrm{BD}_{1,0}(\ell)|\ell|\mathrm{Ball}_{1}(\ell)\mathrm{d}\ell|.$

This follows from the same argument as in the proof of Lemma 6.8 with $a_{n}$
being $1$ for all $n$ instead. We omit the detailed proof. Now Lemma 6.9
follows from Proposition 6.2 and Lemma 6.8. ∎

We finally explain how the measurability result Lemma 6.7 follows from known
results.

###### Proof of Lemma 6.7.

Suppose $(\mathcal{S},x,y)$ is a Brownian sphere with two marked points
conditioned on the event that $d(x,y)>2$. Sample a point $q$ on
$\partial\mathcal{B}^{\bullet}(x,1)$ according to the boundary length measure.
We claim that the filled metric ball $(\mathcal{B}^{\bullet}(x,1),q)$ and the
Brownian disk $(\mathcal{S}\setminus\mathcal{B}^{\bullet}(x,1),y,q)$ as marked
metric-measure spaces together determine $(\mathcal{S},x,y,q)$. To see this,
from the Brownian snake construction of the Brownian map, $(\mathcal{S},x)$ is
determined by the geodesic tree $\mathcal{T}_{x}$ rooted at $x$ and its dual
tree $\tilde{\mathcal{T}}_{x}$ mated together as a pair of rooted planar
trees. Here $\tilde{\mathcal{T}}_{x}$ is a Brownian continuum random tree
(CRT) and the contour function of $\mathcal{T}_{x}$ is a Brownian snake
indexed by the CRT; see [LP08] for more on this mating-of-trees perspective
for the Brownian map. For each point $z\in\mathcal{S}$, if $z\in
B^{\bullet}(x,1)$, then the geodesic from $z$ to $x$ must be inside
$\mathcal{B}^{\bullet}(x,1)$. Otherwise, the geodesic from $z$ to $x$ is a
concatenation of two internal geodesics, one in
$\mathcal{S}\setminus\mathcal{B}^{\bullet}(x,1)$ and the other in
$\mathcal{B}^{\bullet}(x,1)$. Therefore, knowing
$(\mathcal{B}^{\bullet}(x,1),q)$ and
$(\mathcal{S}\setminus\mathcal{B}^{\bullet}(x,1),q)$, we can determine how
$\mathcal{T}_{x}$ and $\tilde{\mathcal{T}}_{x}$ are mated together to form
$(\mathcal{S},x)$. The point $q$ is added is to fix how boundary points on
$\mathcal{B}^{\bullet}(x,1)$ and
$\mathcal{S}\setminus\mathcal{B}^{\bullet}(x,1)$ are identified.

The Brownian disk $\mathcal{D}$ with an interior point $p$ can also be viewed
as the mating of the geodesic tree rooted at $p$ and its dual, which is a
planar forest rooted at $\partial\mathcal{D}$; see [LG19]. On the event
$d(p,\partial\mathcal{D})>1$, we let $\mathcal{A}$ be the annulus bounded by
$\partial\mathcal{B}^{\bullet}(p,1)$ and $\partial\mathcal{D}$. Sample a point
$q$ on $\partial\mathcal{B}^{\bullet}(p,1)$ from its length measure. Then the
same argument as above shows that $(\mathcal{A},q)$ and
$(\partial\mathcal{B}^{\bullet}(p,1),q)$ together determine
$(\mathcal{D},p,q)$. This further determines the conformal structure of
$(\mathcal{D},p,q)$ and hence that of $(\mathcal{A},q)$. On the other hand,
let $\mathcal{L}$ be the boundary length of $(\mathcal{B}^{\bullet}(p,1),q)$.
By the last assertion of Lemma 6.8, once we rescale the metric and measure of
$(\mathcal{B}^{\bullet}(p,1),q)$ by $\mathcal{L}^{-1/2}$ and
$\mathcal{L}^{-2}$ respectively, then the rescaled metric measure space is
independent of the conformal structure of $(\mathcal{A},q)$. Therefore the
conformal structure of $(\mathcal{A},q)$ is determined by the metric-measure
structure of $(\mathcal{A},q)$ alone. Forgetting the extra marked point $q$ we
are done. ∎

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