# Percolation probability and critical exponents
for site percolation on the UIPT

Laurent Ménard111New York University Shanghai, 200122 Shanghai, China, and
Modal’X, UMR CNRS 9023, UPL, Univ. Paris Nanterre, F92000 Nanterre, France.
<EMAIL_ADDRESS>
(January 27, 2022)

###### Abstract

We derive three critical exponents for Bernoulli site percolation on the on
the Uniform Infinite Planar Triangulation (UIPT). First we compute explicitly
the probability that the root cluster is infinite. As a consequence, we show
that the off-critical exponent for site percolation on the UIPT is
$\beta=1/2$. Then we establish an integral formula for the generating function
of the number of vertices in the root cluster. We use this formula to prove
that, at criticality, the probability that the root cluster has at least $n$
vertices decays like $n^{-1/7}$. Finally, we also derive an expression for the
law of the perimeter of the root cluster and use it to establish that, at
criticality, the probability that the perimeter of the root cluster is equal
to $n$ decays like $n^{-4/3}$. Among these three exponents, only the last one
was previously known. Our main tools are the so-called gasket decomposition of
percolation clusters, generic properties of random Boltzmann maps, as well as
analytic combinatorics.

MSC 2010 Classification: 05A15, 05A16, 05C12, 05C30, 60C05, 60D05, 60K35.

##  1 Introduction

Percolation on random planar maps has been studied intensively since the
pioneering work of Angel [3]. The main feature of random planar maps making
this study so fruitful is the spatial Markov property. It can be used with two
different approaches. The first approach is to perform an exploration process
of percolation interfaces with the so-called peeling process. This is the
approach developed by Angel [3] to prove that the threshold for Bernoulli site
percolation on the Uniform Infinite Planar Triangulation (UIPT, the limit in
law of large uniform random triangulations for the local topology [5]) is
$1/2$. This approach has been later used by several authors to study other
models of percolation on maps, see for example [4, 17, 23, 26, 28] and the
references therein. The second approach is more global and consists on
decomposing the map into the cluster of the root vertex and pieces filling the
faces of this cluster. Such a decomposition is often called the Gasket
decomposition, see for instance the works of Borot, Bouttier, Duplantier and
Guitter [10, 11, 12]. This second approach has been used very recently to
study percolation on random finite triangulations by Bernardi, Curien and
Miermont [7], following the previous work by Curien and Kortchemski [18]; and
on other natural models of random finite planar maps by Curien and
Richier[19].

This work builds on the article by Bernardi-Curien-Miermont [7] to study site
percolation on the UIPT. Our first main result is an explicit formula for the
probability that percolation from the root occurs.

######  Theorem 1.

Let $\mathbb{P}_{\infty}^{p}$ denote the law of the UIPT, with vertices
colored black with probability $p$ and white with probability $1-p$, and
conditioned on the event where the root edge has both end vertices colored
black. Let $\mathfrak{C}$ denote the site percolation cluster of the root
vertex under $\mathbb{P}_{\infty}^{p}$. Then, for every $p\in[0,1]$, we have

$\displaystyle\mathbb{P}_{\infty}^{p}(|\mathfrak{C}|=\infty)$
$\displaystyle=2\,\frac{\sqrt{2p-1}\,\left(\sqrt{3}-\cos^{3}\left(\frac{2}{3}\mathrm{arccos}(\sqrt{p})\right)\right)\,\left(\cos\left(\frac{2}{3}\mathrm{arccos}(\sqrt{p})\right)\right)^{3/2}+p(2p-1)}{p\,(2\sqrt{3}-3(1-p))}\,\mathbf{1}_{p\geqslant
1/2}.$

In particular, the critical exponent is $\beta=1/2$: as $p\to 1/2^{+}$ one has

$\displaystyle\mathbb{P}_{\infty}^{p}(|\mathfrak{C}|=\infty)$
$\displaystyle=3^{1/4}\,\frac{15}{26}\,\left(1+\frac{4\sqrt{3}}{3}\right)\sqrt{p-\frac{1}{2}}+\mathcal{O}\left(p-\frac{1}{2}\right).$

Figure 1: Plot of the probability
$\mathbb{P}_{\infty}^{p}(|\mathfrak{C}|=\infty)$ for $p\in[0,1]$.

To the best of our knowledge, this is the first formula of this type and the
first calculation of the critical exponent $\beta$ for percolation on the
UIPT. A similar formula was obtained for percolation on the Uniform Infinite
Half-Plane Triangulation (UIHPT) by Angel and Curien [4] using the peeling
process. They also calculate explicitly the probability that percolation from
the root occurs and then obtain $\beta=1$ in the UIHPT setting. The relation
between these two exponents remains quite mysterious, and we do not know of
any strategy to obtain the exponent $\beta$ without first computing explicitly
the probability of percolation.

Our second main result gathers estimates for the volume and perimeter of
critical percolation clusters.

######  Theorem 2.

With the notation of Theorem 1 we have for an explicit constant $\kappa>0$:

$\displaystyle\mathbb{P}_{\infty}^{1/2}\left(|V(\mathfrak{C})|\geqslant
n\right)\underset{n\to\infty}{\sim}\kappa\,n^{-1/7}.$

Furthermore, if $\partial\mathfrak{C}$ denotes the root face of
$\mathfrak{C}$, then there is an explicit constant $\kappa^{\prime}>0$ such
that

$\displaystyle\mathbb{P}_{\infty}^{1/2}\left(|V(\partial\mathfrak{C})|=n\right)\underset{n\to\infty}{\sim}\kappa^{\prime}\,n^{-4/3}.$

The perimeter exponent $4/3$ was established by Curien and Kortchemski [18]
using the gasket decomposition but with a different approach than the present
work. The exponent $1/7$ was conjectured in [23] using heuristics for the
peeling process, and the present work is the first time it is established
rigorously. Previous works that computed volume exponents for critical
percolation models on infinite random planar maps did so for cluster hulls
(part of the maps separated from infinity by a percolation interface). The
reason being that all these works use variations around the peeling process,
which is particularly well suited to study percolation interfaces – and
therefore hulls – but not so useful to study the geometry of the clusters
themselves. For instance, let $\mathfrak{H}$ denote the hull of the root
cluster $\mathfrak{C}$ in the UIPT; i.e. $\mathfrak{H}$ is the complement of
the only infinite connected component of
$\mathbf{T}_{\infty}\setminus\mathfrak{C}$, where $\mathbf{T}_{\infty}$ is the
UIPT with a critical site percolation configuration. Gorny, Maurel-Segala and
Singh [23] proved that

$\mathbb{P}\left(|V(\mathfrak{H})|\geqslant n\right)\asymp
n^{-1/8}\quad\text{and}\quad\mathbb{P}\left(|V(\partial\mathfrak{H})|\geqslant
n\right)\asymp n^{-1/6},$

where $u_{n}\asymp v_{n}$ means that $u_{n}/v_{n}$ is bounded.

#### Main ingredients and organization of the paper.

For every site percolated finite rooted planar triangulation $\mathfrak{t}$,
we denote by $\mathrm{v}_{\circ}(\mathfrak{t})$ and
$\mathrm{v}_{\bullet}(\mathfrak{t})$ its number of white (resp. black)
vertices, and by $\mathrm{e}(\mathfrak{t})$ its number of edges. For
$p\in(0,1)$ and for $t>0$ small enough, we can consider a random finite
percolated triangulation $\mathbf{t}$ whose law gives a probability
proportional to
$t^{\mathrm{e}(\mathfrak{t})}\,p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}$
to every finite triangulation $\mathfrak{t}$. Let us denote by $\mathcal{Z}$
the partition function of this model and by $\mathbb{P}^{p}$ the corresponding
probability:

$\mathcal{Z}(p,t):=\sum_{\mathfrak{t}}t^{\mathrm{e}(\mathfrak{t})}\,p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}<+\infty,\quad\text{and}\quad\mathbb{P}^{p}\left(\mathbf{t}=\mathfrak{t}\right)=\frac{t^{\mathrm{e}(\mathfrak{t})}\,p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}}{\mathcal{Z}(p,t)}.$

The partition function $\mathcal{Z}$ and its generalizations to triangulations
with a boundary (with additional parameters counting respectively the number
of boundary vertices and boundary edges) are studied in Section 2. In
particular, we establish a rational parametrization for the generalized
partition function. This parametrization significantly simplifies the study of
its analytic properties started in [7].

The gasket decomposition resides in the following statement. There exists an
explicit sequence of positive numbers $(q_{k}(p,t))_{k\geqslant 1}$ such that,
for every finite non atomic map $\mathfrak{m}$, the probability that the root
cluster $\mathfrak{C}(\mathbf{t})$ of the random triangulation $\mathbf{t}$ is
equal to $\mathfrak{m}$ is given by

$\mathbb{P}^{p}\left(\mathfrak{C}(\mathbf{t})=\mathfrak{m}\right)=\frac{\prod_{f\in\mathrm{Faces}(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t)}{\mathcal{Z}(p,t)}.$

In this sense, the root cluster is a Boltzmann random planar map associated to
the weight sequence $(q_{k}(p,t))_{k\geqslant 1}$. The properties of such
random maps depend on the asymptotic behavior in $k$ of the weight sequence,
which can be inferred from the generating function of the weights. In our
case, the weight generating function is closely related to the generalized
partition function of percolated triangulations counted by edges, boundary
edges, and boundary vertices. A crucial consequence of this relation is that
it allows to calculate explicitly the so-called pointed disk generating
function of the root cluster in terms of the singularities of the generalized
partition function of percolated triangulations in Equations (12) and (13).
See Section 3 for details.

To study the origin cluster of the UIPT, we can condition the random
triangulation $\mathbf{t}$ to have $3n$ edges. By continuity for the local
topology, this gives

$\mathbb{P}^{p}\left(\mathfrak{C}(\mathbf{t})=\mathfrak{m}\middle|\mathrm{e}(\mathbf{t})=3n\right)=\frac{[t^{3n}]\prod_{f\in\mathrm{Faces}(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t)}{[t^{3n}]\mathcal{Z}(p,t)}\underset{n\to\infty}{\rightarrow}\mathbb{P}_{\infty}^{p}(\mathfrak{C}=\mathfrak{m}).$
(1)

With a careful study of the dependency in $t$ of the weight sequence
$(q_{k}(p,t))_{k\geqslant 1}$ performed in Section 6.1, we are able to compute
the above limit, giving the law of the root cluster in the UIPT on the event
where it is finite.

We are then able to establish an integral formula for the sum of the limiting
probability over every finite map, which is the probability that the root
cluster of the UIPT is finite. See Proposition 7. In particular, this
calculation uses the explicit universal form of the pointed disk generating
functions and cylinder generating functions of Boltzmann maps. With the help
of our rational parametrizations, we can then calculate explicitely our
integral formula for the probability that the root cluster is finite and
establish Theorem 1. This is performed in Section 4.

The formula we obtain for the limit (1) is quite easy to sum over maps with
the same perimeter. As a consequence we are also able to calculate explicitely
the law of the perimeter of the root cluster and obtain the second statement
of Theorem 2 on the perimeter of the root cluster. This is done at the
beginning of Section 5.2.

Finally, to compute the tail probability of the number of vertices in the root
cluster at criticality ($p=1/2$), we establish an integral formula for the
generating function
$\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]$ in Section 5.2.
See in particular identity (25), which also originates from our explicit
formula for the limit (1). The expression we get involves two quantities. The
first is the generating series derived from the asymptotics of the
coefficients in $t$ of the weights $(q_{k}(p,t))_{k\geqslant 1}$ studied in
Section 6.1 for which we have an explicit parametric expression. The second
quantity involves the pointed disk generating functions of Boltzmann maps with
modified weight sequence $(g^{(k-2)/2}q_{k}(p,t))_{k\geqslant 1}$. We can
analyze the behavior as $g\to 1^{-}$ of these modified pointed disk generating
functions using the bivariate generating functions associated to the Bouttier-
Di Francesco-Guitter bijection [15] presented in Section 5.1 and their
singularities obtained in Section 6.2. We put all that together in Section 5.2
to study the singular behavior as $g\to 1^{-}$ of
$\mathbb{E}_{\infty}^{1/2}\left[g^{|V(\mathfrak{C})|}\right]$ and prove
Theorem 2.

#### On the robustness of our approach.

We believe that it should be possible to adapt the strategy of this paper to
other models of random planar maps and other statistical models. Indeed, what
is needed first is the gasket decomposition; which exists for example for
percolation on other models of maps [19], or for the $\mathcal{O}(n)$ model on
maps [10, 11]. We then need information on the generalized generating series
of maps with a boundary and their singularities. In the present work we have
explicit parametric expressions to simplify calculations, but we really only
need to identify the nature of the singularities for the critical exponents.
Any model for which such information is available should fall into the scope
of our method.

In another article [2], we derive several critical exponents for the sign
clusters in finite and infinite planar triangulations coupled with an Ising
model. In particular, we establish in [2] counterparts for the main results of
the present work. Theorem 1 and its counterpart for Ising model are unrelated,
but Theorem 2 is a particular case of its Ising version at infinite
temperature.

#### Links with other critical exponents for percolation on the UIPT.

The volume exponent $1/7$ of Theorem 2 suggests that the scaling limit of a
large critical percolation cluster in the UIPT should be of quantum dimension
$7/8$. There is also very strong evidence (see for instance the works of
Bernardi-Holden-Sun [9] and Holden-Sun [9]) that in some sense, this scaling
limit is a CLE6 on an independent pure Liouville Quantum Gravity surface. The
quantum dimension $7/8$ of the scaling limit of the root cluster and the
dimension $91/48$ of the gasket of a CLE6 agree with the KPZ [25] relation:

$1-\frac{1}{2}\frac{91}{48}=\frac{2}{3}\left(1-\frac{7}{8}\right)^{2}+\frac{1}{3}\left(1-\frac{7}{8}\right).$

In a similar fashion, the perimeter exponent $4/3$ of Theorem 2 agrees with
the KPZ relation and the dimension $\frac{7}{4}$ of a SLE6 curve [6]

The value of the critical exponent $\beta$ agrees heuristically with known
quantities of the UIPT and Kesten’s scaling relations. Indeed, Bernardi-
Holden-Sun [9] and Holden-Sun [24] established that the the number of
percolation pivotal points in a random triangulation of size $n$ is of order
$n^{1/4}$. In this sense, the quantum dimension of the set of pivotal points
of critical percolation on the UIPT should be $1$ (the map itself having
dimension $4$). This dimension can also be predicted with the KPZ relation and
the dimension $3/4$ of critical percolation pivotal points in Euclidean
geometry [22]. On the other hand, the one arm exponent $\alpha_{1}$ of
critical percolation on the UIPT should be $1/2$ from the quantum dimension
$7/8$ of the the clusters. Kesten’s scaling relation then states that the
dimension of pivotal points should be $\alpha_{1}/\beta$, giving $1$ with the
exponent of Theorem 1.

#### Acknowledgments.

The author thanks Pierre Nolin for stimulating discussions on critical
exponents. This work is partially supported by the ANR grant ProGraM (Projet-
ANR-19-CE40-0025) and the Labex MME-DII (ANR11-LBX-0023-01).

###### Contents

  1. 1 Introduction
  2. 2 Generating series
  3. 3 The root cluster as a Boltzmann map
    1. 3.1 The weight sequence from [7] with the edge parameter
    2. 3.2 Generating series of the weights and of the associated Boltzmann maps
  4. 4 Percolation probability
  5. 5 Cluster volume and perimeter
    1. 5.1 Admissibility equations and volume-modified weight sequence
    2. 5.2 Proof of Theorem 2
  6. 6 Technical Lemmas
    1. 6.1 Dependency in $t$ of the weights
    2. 6.2 BDFG functions
    3. 6.3 Perimeter asymptotics at criticality

##  2 Generating series

Let $T$ be the generating series of rooted triangulations with a (not
necessarily simple) boundary counted by edges (variable $t$), boundary length
(parameter $y$), and boundary vertices (parameter $p$). That is, we define:

$T(p,t,y)=\sum_{k\geqslant
1}\sum_{\mathfrak{t}\in\mathcal{T}_{k}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\mathrm{out}}(\mathfrak{t})}y^{k}=\sum_{k\geqslant
1}T_{k}(p,t)\,y^{k},$

where $\mathcal{T}_{k}$ is the set of all rooted triangulations with a
boundary face of degree $k$, and where ${e(\mathfrak{t})}$ and
$\mathrm{v}_{\mathrm{out}}(\mathfrak{t})$ denote respectively the total number
of edges and number of boundary vertices of the triangulation with a boundary
$\mathfrak{t}$.

From [7, Lemma 3.1], we have the following equation fo $T(p,t,y)$:

$T(p,t,y)=p+y^{2}tT^{2}(p,t,y)+(p-1)t\frac{(T(p,t,y)-p)^{2}}{pyT(p,t,y)}+\frac{t}{py}\left(T(p,t,y)-p-y\,T_{1}(p,t)\right).$
(2)

Using the quadratic method, the authors of [7] establish the following
algebraic equation for $T_{1}\equiv T_{1}(p,t)$ that will be our starting
point:

$64\,T_{1}^{3}\,t^{5}-27\,p^{3}\,t^{5}-96\,T_{1}^{2}\,p\,t^{4}+30\,T_{1}\,p^{2}\,t^{3}+p^{3}\,t^{2}+T_{1}^{2}\,p\,t-T_{1}\,p^{2}=0.$

Up to a multiplicative constant $p$, the series $T_{1}$ is simply the
generating series of triangulations of the $1$-gon counted by edges and admits
a proper rational parametrization:

######  Lemma 3.

Let $U$ be the unique power series in $t^{3}$ having constant term $0$ and
satisfying

$t^{3}=\hat{w}(U):=\frac{1}{2}U(1-U)(1-2U).$

The series $t\,T_{1}(p,t)$ seen as a series in $t^{3}$ admits the following
proper rational parametrization in $U$:

$t\,T_{1}(p,t)=\hat{T}_{1}(p,U):=p\,U\,\frac{1-3U}{1-2U}.$

Furthermore, the series $U(t^{3})$ has a unique dominant singularity at
$t^{3}=(t_{c})^{3}:=\frac{\sqrt{3}}{36}$ with the following singular behavior:

$\begin{split}U(t^{3})=\frac{3-\sqrt{3}}{6}-\frac{\sqrt{2}}{6}\left(1-\left(\frac{t}{t_{c}}\right)^{3}\right)^{1/2}&+\frac{\sqrt{3}}{54}\left(1-\left(\frac{t}{t_{c}}\right)^{3}\right)\\\
&-\frac{5\sqrt{2}}{648}\left(1-\left(\frac{t}{t_{c}}\right)^{3}\right)^{3/2}+\mathcal{O}\left(1-\left(\frac{t}{t_{c}}\right)^{3}\right)^{2}.\end{split}$
(3)

###### Proof.

This result could be qualified as folklore since the series $T_{1}$ is just
the generating series of triangulations of the $1$-gon multiplied by $p$. The
rational parametrization given is also the classical one. The fact that
$U(t^{3})$ is unique comes from the Lagrangean form of the equation that
defines it. We can also see that the algebraic equation satisfied by
$t\,T_{1}(t)$ has a unique solution that is power series with constant term
$0$. By composition and unicity we can see that indeed
$tT_{1}(t)=\hat{T}_{1}(p,U(t^{3}))$ as power series in $t$ since both verify
the algebraic equation and have constant term $0$. The singular behavior of
$U$ is also very classical and without difficulties. See the Maple companion
file [1] for details. ∎

Injecting the parametrization of $t^{3}$ and $t\,T_{1}$ by $U$ in the equation
for $T$ then allows to establish a proper rational parametrization for $T$:

######  Lemma 4.

Recall the definition of the power series $U\equiv U(t^{3})$ from Lemma 3. Let
$V\equiv V(p,U,y)$ be the unique power series in $\mathbb{Q}[p,U]\llbracket
y\rrbracket\subset\mathbb{Q}[p]\llbracket t^{3},y\rrbracket$ with constant
term in $y$ equal to $0$ satisfying:

$y=\hat{y}(p,U,V):=\frac{2\,V\,(2-4\,U-V)}{4\,p\,U\,(1-U)\,(1-2U)+2\,U\,(1-3U)\,V+2\,(1-3U)V^{2}-V^{3}}.$

The series $T(p,t,ty)$ seen as a series in $t^{3}$ and $y$ is algebraic and
admits the following proper rational parametrization in $U$ and $V$

$T(p,t,ty)=\hat{T}(p,U,V):=\frac{4\,p\,U\,(1-U)\,(1-2U)+2\,U\,(1-3U)\,V+2\,(1-3U)V^{2}-V^{3}}{4\,U\,(1-U)\,(1-2U)}.$

In addition, for any $p\in(0,1]$ and any fixed $t\in(0,t_{c}]$, the series
$T(p,t,ty)$ seen as a series in $y$ has radius of convergence $y_{+}(p,t)>1$
where it is singular. Furthermore, the series $T(p,t,ty)$ can be analytically
continued in the domain
$\mathbb{C}\setminus\left((-\infty,y_{-}(p,t)]\cup[y_{+}(p,t),+\infty)\right)$,
where $y_{-}(p,t)$ is on the negative real line ($-\infty$ included) and is
such that $y_{-}(p,t)<-y_{+}(p,t)$.

###### Proof.

All computations are available in the Maple companion file [1].

The fact that $V$ is uniquely defined as a power series comes from the
Lagrangean form of the equation $V=y\times R(p,U,V)$ with $R$ a rational
fraction such that $R(p,U,0)\neq 0$. This form also implies by inductive
calculation of the coefficients in $y$ of $V$ that they are all rational in
$p$ and $U$. Similarly, the equation (2) verified by $T(p,t,ty)$ takes the
form

$\begin{split}p(T(p,t,ty)-p)&(T(p,t,ty)-(1-p))\\\
&=y\,T(p,t,ty)\,\left(tT_{1}(p,t)-p(T(p,t,ty)-p)-y^{2}\,p\,t^{3}\,T(p,t,ty)^{2}\right).\end{split}$

Here again, by inductive calculation of the coefficients, we can see that this
last equation has a unique solution that is a power series in $t^{3}$ and $y$
with constant term in $y$ equal to $p$. Note that since the series
$tT_{1}(p,t)$ is algebraic, this equation also ensures that $T(p,t,ty)$ is
algebraic. By composition, $\hat{T}\left(p,U(t^{3}),V(p,U(t^{3}),y)\right)$ is
a power series in $t^{3}$ and $y$ with constant term in $y$ equal to $p$. We
can verify that it satisfies the same algebraic equation as $T(p,t,ty)$, and
therefore the two power series are identical.

Now, fix $t^{3}\in(0,t_{c}]$. The function $V\mapsto\hat{y}(p,U(t^{3}),V)$ has
poles and stationary points that we can locate. Let us start with the poles.
The denominator of $\hat{y}(p,U(t^{3}),V)$ is a polynomial of degree three in
$V$. It is positive with positive derivative at $V=0$, and changes signs
between $1-2U$ and $2(1-2U)$. Since the coefficient of $V^{3}$ is $-1$, this
leaves two possibilites for the poles of $\hat{y}(p,U(t^{3}),V)$: there is
always a pole between $1-2U$ and $2(1-2U)$, and either there is no additional
real pole or there are two negative poles.

The stationary points of $\hat{y}(p,U(t^{3}),V)$ are the roots of the
polynomial
$-2V^{4}+(-16U+8)V^{3}-4(3U-1)(3U-2)V^{2}-16Up(2U-1)(U-1)V-16Up(U-1)(2U-1)^{2}$,
where $U$ stands for $U(t^{3})$. By computing the values of the polyomial at
$0$, $1-2U$ and $2(1-2U)$, we can see that it has $4$ real roots
$V_{-}(p,U)<0<V_{+}(p,U)\leqslant 1-2U\leqslant V_{l}(p,U)<2(1-2U)<V_{r}(p,U)$
(the case of a double root at $1-2U$ only happens when $U=U_{c}$ and is
treated separately in the Maple file). It is also easy to see that
$\partial_{V}\hat{y}(p,U(t^{3}),0)>0$, and therefore
$y_{+}(p,t):=\hat{y}(p,U(t^{3}),V_{+}(p,U(t^{3})))>0$. We can define
$y_{-}(p,t):=\hat{y}(p,U(t^{3}),V_{-}(p,U(t^{3})))<0$ when
$\hat{y}(p,U(t^{3}),V)$ has no pole between $V_{-}(p,U(t^{3}))$ and $0$, and
$y_{-}(p,t):=-\infty$ when it has such a pole. By singular inversion, the
inverse function $V(p,U(t_{3}),y)$ of $\hat{y}(p,U(t^{3}),V)$ is analytic in
the domain
$\mathbb{C}\setminus\left((-\infty,y_{-}(p,t)]\cup[y_{+}(p,t),+\infty)\right)$
and singular at both points $y_{\pm}(p,t)$ when they are finite.

By composition $y\mapsto T(p,t,ty)$ is analytic in the same domain. Since it
has nonnegative coefficients, we know that it is singular at its radius of
convergence. Therefore $y_{+}(p,t)$ is its radius of convergence and
$y_{-}(p,t)\leqslant-y_{+}(p,t)$. Checking that $y_{-}(p,t)<-y_{+}(p,t)$ just
with our current material is cumbersome but could be done. However, we do not
need to go through this since this inequality is a direct consequence of
equations (13) established with no computations at the end of Section 3.
Finally, to see that $y_{+}(p,t)>1$, we first note that for $p$ fixed,
$y_{+}(p,t)$ is non increasing with respect to $|t|$ and then check
$y_{+}(p,t_{c})>1$ by direct computation. ∎

##  3 The root cluster as a Boltzmann map

In this whole section, $p\in(0,1)$ and $t\in(0,t_{c}]$ are fixed.

###  3.1 The weight sequence from [7] with the edge parameter

In [7, Section 2.2] it was established that for a random site-percolated
triangulation (conditioned on the event where both ends of the root edge are
colored in black), the cluster of the root is a random Boltzmann map with
weight sequence $\mathbf{q}(p,t)=\left(q_{k}(p,t)\right)_{k\geqslant 1}$ given
by

$q_{k}(p,t)=\frac{1}{p}\left((pt)^{3/2}\delta_{\\{k=3\\}}+(pt^{3})^{k/2}\sum_{l\geqslant
0}\binom{k+l-1}{k-1}[y^{l}]T(1-p,t,ty)\right),$ (4)

for $k\geqslant 1$ (see [7, Equation (9)]). We briefly recall here what this
statment means and how to obtain it.

Figure 2: Gasket decomposition: A face of the root cluster (black vertices in
the figure) is filled with a triangulation with arbitrary white boundary
(upper right figure, grey vertices can be either black or white) and a
necklace (lower right figure) triangulating the region between the two.

For every percolated rooted planar triangulation $\mathfrak{t}$, we denote
respectively by $\mathrm{v}_{\circ}(\mathfrak{t})$ and
$\mathrm{v}_{\bullet}(\mathfrak{t})$ the number of white (resp. black)
vertices. Let $\mathcal{T}^{\mathrm{perc}}$ be the set of all percolated
rooted triangulations with both end vertices of the root edge black. For every
$\mathfrak{t}\in\mathcal{T}^{\mathrm{perc}}$, we denote by
$\mathfrak{C}(\mathfrak{t})$ the cluster of its root edge viewed as a planar
rooted map (whose root is the root edge of the triangulation). The fact that
the root cluster is a random Boltzmann map comes from the following identity
(see Equation (7) of [7]): for every non atomic map $\mathfrak{m}$ one has

$\sum_{\mathfrak{t}\in\mathcal{T}^{\mathrm{perc}}\,:\,\mathfrak{C}(\mathfrak{t})=\mathfrak{m}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}=p^{2}\,\cdot\prod_{f\in
F(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t).$ (5)

This identity stems from the classical gasket decomposition [11] sometimes
called island decomposition [7] and we briefly explain how it is obtained in
the following lines (see Figure 2 for an illustration of these arguments). Fix
$\mathfrak{m}$ a non atomic map and $\mathfrak{t}$ a site-percolated
triangulation with root cluster $\mathfrak{m}$ colored black. For such a
triangulation $\mathfrak{t}$, each face of its root cluster is filled with a
triangulation with an arbitrary boundary of white vertices, and a _necklace_
of triangles with no additional vertices between this triangulation with a
boundary and the cluster. For each cluster face of degree $k\geqslant 1$
filled with a triangulation with perimeter $l\geqslant 1$, there are
$\binom{k+l-1}{k-1}$ different possible necklaces, and each of these necklaces
require an additional $k+l$ edges. Taking into account the case where the
cluster face has degree $3$ and can be part of the triangulation, this writes:

$\displaystyle\sum_{\mathfrak{t}\in\mathcal{T}^{\mathrm{perc}}\,:\,\mathfrak{C}(\mathfrak{t})=\mathfrak{m}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}$
$\displaystyle\quad=p^{|V(\mathfrak{m})|}\,t^{|e(\mathfrak{m})|}\prod_{f\in
F(\mathfrak{m})}\left(\mathbf{1}_{\mathrm{deg}(f)=3}+\sum_{l\geqslant
0}\binom{\mathrm{deg}(f)+l-1}{\mathrm{deg}(f)-1}t^{\mathrm{deg}(f)+l}[y^{l}]T(1-p,t,y)\right),$
$\displaystyle\quad=p^{2}\cdot\prod_{f\in
F(\mathfrak{m})}\frac{1}{p}(pt)^{\mathrm{deg}(f)/2}\left(\mathbf{1}_{\mathrm{deg}(f)=3}+\sum_{l\geqslant
0}\binom{\mathrm{deg}(f)+l-1}{\mathrm{deg}(f)-1}t^{\mathrm{deg}(f)+l}[y^{l}]T(1-p,t,y)\right),$

where we used Euler’s relation $|V(\mathfrak{m})|-2=\sum_{f\in
F(\mathfrak{m})}\left(\mathrm{deg}(f)/2-1\right)$ in the last display. The
expression (5) follows easily and we refer the reader to [7, Section 2.2] for
additional details. Note for future reference that for any $k\geqslant 1$, the
weight $q_{k}(p,t)$ is $1/p$ times the generating series of all triangulations
of the $k$-gon with a weight $t$ per inner edge, a weight $\sqrt{pt}$ per
boundary edge, and a weight $1-p$ per inner vertex adjacent to the boundary.

We can define the partition function of our percolated triangulations by:

$\mathcal{Z}(p,t)=\frac{1}{p^{2}}\cdot\sum_{\mathfrak{t}\in\mathcal{T}^{\mathrm{perc}}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}.$
(6)

From identity (5), denoting by $\mathcal{M}$ the set of all non atomic rooted
planar maps, we see that

$\displaystyle\mathcal{Z}(p,t)=\sum_{\mathfrak{m}\in\mathcal{M}}\prod_{f\in
F(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t).$ (7)

Notice that this sum is finite when $p\in(0,1)$ and $t\in(0,t_{c}]$, meaning
that the weight sequence $\mathbf{q}(p,t)$ is admissible in the sense of of
[27]. We will need the asymptotic behavior of the coefficients in $t^{3}$ of
the series $\mathcal{Z}(p,t)$:

######  Proposition 5.

Fix $p\in(0,1)$, we have

$[t^{3n}]\mathcal{Z}(p,t)\underset{n\to\infty}{\sim}\frac{\sqrt{2}\,\left(3p-3+2\sqrt{3}\right)}{2p\sqrt{\pi}}\,t_{c}^{-3n}\,n^{-5/2}.$

###### Proof.

By opening the root edge of the triangulations appearing in the sum (6), we
can see that, for $p\in(0,1)$ and $|t|\leqslant t_{c}$, the partition function
$\mathcal{Z}(p,t)$ is given by

$\mathcal{Z}(p,t)=\frac{1}{p^{2}\,t}T_{2}(p,t)=\frac{1}{p^{2}\,t^{3}}\,t^{2}T_{2}(p,t)=\frac{1}{t^{3}}tT_{1}(p,t)\left(1+\frac{1-p}{p^{3}}tT_{1}(p,t)\right).$
(8)

From Lemma 3, The function $tT_{1}(t)$ seen as a series in $t^{3}$ has a
unique dominant singularity at $t^{3}_{c}$ and we can obtain its asymptotic
expansion at $t_{c}^{3}$ from the expansion of $U$. As a consequence, the
function $\mathcal{Z}(p,t)$ also has a unique dominant singularity at
$t_{c}^{3}$ ans we can easily obtain the following asymptotic expansion:

$\displaystyle\mathcal{Z}(p,t)$
$\displaystyle=\frac{3\sqrt{3}\,\left(7-4\sqrt{3}\right)+p}{4p}-\frac{\sqrt{3}\,\left(3p-27+16\sqrt{3}\right)}{4p}\,\left(1-t^{3}/t_{c}^{3}\right)$
$\displaystyle\qquad+\frac{2\sqrt{2}\,\left(3p-3+2\sqrt{3}\right)}{3p}\,\left(1-(t/t_{c})^{3})^{3/2}\right)+\mathcal{O}\left(\left(1-t^{3}/t_{c}^{3})^{2}\right)\right).$

The asymptotic behavior of $[t^{3n}]\mathcal{Z}(p,t)$ then follows from the
classical transfer Theorem [21, Theorem VI.4]. ∎

###  3.2 Generating series of the weights and of the associated Boltzmann maps

The generating series of the weight sequence $\mathbf{q}(p,t)$ is
straightforward to compute. As shown by the following lines, the expression we
obtain is valid for every $p\in(0,1)$, $t$ in the greater domain of analycity
of $U(t^{3})$, and $z$ close enough to $0$. More precisely we need
$\sqrt{pt^{3}}z\in\mathbb{C}\setminus[1,+\infty)$ and
$\left(1-\sqrt{pt^{3}}z\right)^{-1}$ must belong to the domain of analycity in
$y$ of $T(t,1-p,ty)$ which was studied in the previous Section. Under these
conditions for $p$, $t$ and $z$, the computation of the weight sequence
generating function is as follows:

$\displaystyle F_{\mathbf{q}(p,t)}(z):$ $\displaystyle=\sum_{k\geqslant
1}q_{k}(p,t)z^{k},$
$\displaystyle=\frac{1}{p}(pt)^{3/2}z^{3}+\frac{1}{p}\sum_{l\geqslant
0}[y^{l}]T(1-p,t,ty)\sum_{k\geqslant 1}\binom{k+l-1}{k-1}(pt^{3})^{k/2}z^{k},$
$\displaystyle=\frac{1}{p}(pt)^{3/2}z^{3}+\frac{1}{p}\sqrt{pt^{3}}z\sum_{l\geqslant
0}[y^{l}]T(1-p,t,ty)\,\left(1-\sqrt{pt^{3}}z\right)^{-l-1},$
$\displaystyle=\frac{1}{p}(pt)^{3/2}z^{3}+\frac{1}{p}\frac{\sqrt{pt^{3}}z}{1-\sqrt{pt^{3}}z}T\left(1-p,t,\frac{t}{1-\sqrt{pt^{3}}z}\right).$
(9)

We will need expressions for the pointed and unpointed disk generating
functions associated to the weight sequence $\mathbf{q}(p,t)$. For every
$l\geqslant 1$, let $\mathcal{M}^{l}$ denote the set of all rooted planar maps
with root face of degree $l$ (for $l=1$, this set contains only the atomic
map). The unpointed disk generating function is defined as follows for $|z|$
large enough:

$W_{\mathbf{q}(p,t)}(z):=\sum_{l\geqslant
1}\,\left(\sum_{\mathfrak{m}\in\mathcal{M}^{l}}\prod_{f\in
F(\mathfrak{m})\setminus\\{f_{r}\\}}q_{\mathrm{deg}(f)}(p,t)\right)\,z^{-l-1},$

where we denote the root face of a planar map by $f_{r}$. From our discussion
establishing (5), we can compute the coefficients of these series. Indeed,
from Equation (5) and the fact that $q_{l}$ is $1/p$ times the generating
series of triangulations with a boundary of perimeter $l$ counted with a
weight $t$ per inner edge, a weight $\sqrt{pt}$ per boundary edge, and a
weight $1-p$ per inner vertex adjacent to the boundary, we have:

$\displaystyle\sum_{\mathfrak{m}\in\mathcal{M}^{l}}\,\sum_{\mathfrak{t}\in\mathcal{T}^{\mathrm{perc}}\,:\,\mathfrak{C}(\mathfrak{t})=\mathfrak{m}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\bullet}(\mathfrak{t})}(1-p)^{\mathrm{v}_{\circ}(\mathfrak{t})}$
$\displaystyle=p\sqrt{p\,t}^{-l}\,q_{l}(p,t)\,\sum_{\mathfrak{t}\in\mathcal{T}_{l}}t^{e(\mathfrak{t})}p^{\mathrm{v}_{\mathrm{out}}(\mathfrak{t})}.$

Comparing with the right hand side of (5), this gives for every $l\geqslant
1$:

$\sum_{\mathfrak{m}\in\mathcal{M}^{l}}\prod_{f\in
F(\mathfrak{m})\setminus\\{f_{r}\\}}q_{\mathrm{deg}(f)}(p,t)=\frac{1}{p\,\sqrt{pt}^{l}}\,[y^{l}]T(p,t,y),$
(10)

and thus

$W_{\mathbf{q}(p,t)}(z)=\frac{1}{p\,z}\,T\left(p,t,\frac{t}{\sqrt{pt^{3}}z}\right).$
(11)

The pointed disk generating function is defined similarly:

$W_{\mathbf{q}(p,t)}^{\bullet}(z)=\sum_{l\geqslant
1}\,\left(\sum_{\mathfrak{m}\in\mathcal{M}^{l}}|V(\mathfrak{m})|\,\prod_{f\in
F(\mathfrak{m})\setminus\\{f_{r}\\}}q_{\mathrm{deg}(f)}(p,t)\right)\,z^{-l-1}.$

It has the following universal form: For $p\in(0,1)$ and and $t\in(0,t_{c}]$
fixed, there exists real numbers $c_{+}(p,t)>2$ and
$c_{-}(p,t)\in(-c_{+}(p,t),c_{+}(p,t))$ (the lower bound is excluded since our
maps are not bipartite) such that for
$z\in\mathbb{C}\setminus[c_{-}(p,t),c_{+}(p,t)]$ one has

$W_{\mathbf{q}(p,t)}^{\bullet}(z)=\frac{1}{\sqrt{(z-c_{+}(p,t))(z-c_{-}(p,t))}}.$
(12)

This expression, sometimes called the one-cut Lemma, appears in several
references. See for example [16, Proposition 12] or [13, Section 6.1]. It is
also established in these articles that the two disk generating functions $W$
and $W^{\bullet}$ have the same domain of analycity. Comparing our two
expressions (11) and (12), we see that as a consequence

$\displaystyle c_{\pm}(p,t)=\frac{1}{\sqrt{p\,t^{3}}\,y_{\pm}(p,t)},$ (13)

where $y_{\pm}(p,t)$ are the respective positive and negative singularities in
$y$ of the series $T(p,t,ty)$ defined in Lemma 4. Note that, as mentioned in
the proof of Lemma 4, this directly implies that $y_{-}(p,t)<-y_{+}(p,t)$.

##  4 Percolation probability

Recall that for any triangulation with a site percolation configuration,
$\mathfrak{C}$ denotes the percolation cluster of its root vertex. Also recall
that $\mathbb{P}^{p}_{\infty}$ denotes the law of the UIPT with vertices
colored independently black with probability $p$ and white with probability
$1-p$, conditioned on the event where the root edge has both end vertices
colored black. We start by identifying the law of $\mathfrak{C}$ under
$\mathbb{P}^{p}_{\infty}$ on the event where it is finite:

######  Proposition 6.

For every $p\in(0,1)$ and every non atomic rooted finite map $\mathfrak{m}$ we
have

$\mathbb{P}^{p}_{\infty}\left(\mathfrak{C}=\mathfrak{m}\right)=\left(\prod_{f\in
F(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t_{c})\right)\cdot\sum_{f\in
F(\mathfrak{m})}\frac{(pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)}{q_{\mathrm{deg}(f)}(p,t_{c})},$

where the numbers $\left(\delta_{k}(p)\right)_{k\geqslant 1}$ are defined in
Lemma 8.

###### Proof.

For $p\in(0,1)$ and $n\geqslant 1$, we denote by $\mathbb{P}_{n}^{p}$ the law
of a uniform triangulation of the sphere with $3n$ edges and vertices colored
independently black with probability $p$ and white with probability $1-p$,
conditioned on the event where both end vertices of the root edge are colored
black. From equation (7), we can write that for every finite non atomic rooted
map $\mathfrak{m}$ we have

$\mathbb{P}_{n}^{p}\left(\mathfrak{C}=\mathfrak{m}\right)=\frac{[t^{3n}]\prod_{f\in
F(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t)}{[t^{3n}]\mathcal{Z}(p,t)}.$ (14)

Since the event $\\{\mathfrak{C}=\mathfrak{m}\\}$ is continuous for the local
topology, the limit as $n\to\infty$ of the previous display gives access to
the annealed law of the percolation cluster of the root in the UIPT.

As can be seen from Proposition 5 and Lemma 8, the asymptotic behaviors of
$[t^{3n}]\mathcal{Z}(p,t)$ and of $[t^{3n}](pt^{3})^{-k/2}q_{k}(p,t)$ for
every fixed $k\geqslant 1$ are all of the form $\mathrm{Cst}\cdot
t_{c}^{-3n}n^{-5/2}$. It is then very classical to establish the limit

$\displaystyle\lim_{n\to\infty}\frac{[t^{3n}]\prod_{f\in
F(\mathfrak{m})}q_{\mathrm{deg}(f)}(p,t)}{[t^{3n}]\mathcal{Z}(p,t)}$
$\displaystyle=\lim_{n\to\infty}\frac{[t^{3n}](pt^{3})^{\mathrm{e}(\mathfrak{m})}\prod_{f\in
F(\mathfrak{m})}(pt^{3})^{-\mathrm{deg}(f)/2}q_{\mathrm{deg}(f)}(p,t)}{[t^{3n}]\mathcal{Z}(p,t)},$
$\displaystyle=\sum_{f\in
F(\mathfrak{m})}(pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m}),f^{\prime}\neq f}q_{\mathrm{deg}(f)}(p,t_{c}),$

proving the proposition. ∎

Our next result is an integral formula for the probability that the root
cluster $\mathfrak{C}$ of the UIPT is infinite:

######  Proposition 7.

Recall the definition of the quantities $c_{+}(p,t_{c})$ and $c_{+}(p,t_{c})$
given in equations (12) and (13). For every $p\in(0,1)$ one has

$\begin{split}&\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)\\\
&=\frac{1}{2\pi}\int_{c_{-}(p,t_{c})}^{c_{+}(p,t_{c})}\frac{dz}{z}\Delta\left(p,\sqrt{pt^{3}_{c}}\,z\right)\left(z+\frac{c_{+}(p,t_{c})+c_{-}(p,t_{c})}{2}\right)\sqrt{(c_{+}(p,t_{c})-z)(z-c_{-}(p,t_{c}))},\end{split}$
(15)

where the function $\Delta(p,z)$ is defined in Lemma 8.

###### Proof.

We start with the formula established in Proposition 6. Summing over every
finite non atomic map gives for every $p\in(0,1)$:

$\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)=\sum_{\mathfrak{m}\in\mathcal{M}}\,\sum_{f\in
F(\mathfrak{m})}(pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}).$

By opening the root edges of the maps $\mathfrak{m}$ involved in the previous
display, the sum transforms into a sum over maps in $\mathcal{M}^{2}$, the set
of all rooted planar maps with root face of degree $2$:

$\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)=\sum_{\mathfrak{m}\in\mathcal{M}^{2}}\,\sum_{f\in
F(\mathfrak{m})\setminus\\{f_{r}\\}}(pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f,f_{r}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}),$

where $f_{r}$ denotes the root face of a map. By rooting the non root face $f$
involved in the last sum (so that $f$ lies on the right hand side of this
additional root) we can transform the sum into a sum over maps of the
cylinder. For $l_{1}$, $l_{2}\geqslant 1$, let $\mathcal{M}^{(l_{1},l_{2})}$
denote the set of all planar maps with two distinct marked rooted faces
$f_{1},\,f_{2}$ of respective degree $l_{1},\,l_{2}$ (rooted so that the
corresponding marked faces lie on the right hand side of their respective
root). Taking into account the $\mathrm{deg}(f)$ possible roots for the face
$f$ in the sum we have:

$\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)=\sum_{k\geqslant
1}\,\frac{1}{k}\,\sum_{\mathfrak{m}\in\mathcal{M}^{(2,k)}}\,(pt^{3}_{c})^{k/2}\delta_{k}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f_{1},f_{2}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}).$
(16)

Setting for every $k\geqslant 0$

$\varphi_{k}(p)=\frac{1}{k}\,\sum_{\mathfrak{m}\in\mathcal{M}^{(2,k)}}\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f_{1},f_{2}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}),$

and denote the generating series of these numbers by

$\Phi(p,z)=\sum_{k\geqslant 1}\varphi_{k}(p)z^{k},$

the sum in (16) takes the form of the Hadamard product of
$\Delta(p,\sqrt{pt^{3}_{c}}z)$ and $\Phi(p,z)$ evaluated at $z=1$:

$\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)=\sum_{k\geqslant
1}(pt^{3}_{c})^{k/2}\delta_{k}(p)\cdot\varphi_{k}(p)=\Delta\left(p,\sqrt{pt^{3}_{c}}\,z\right)\odot\Phi(p,z)|_{z=1}.$

We will compute this Hadamard product with the help of its contour integral
representation:

$\mathbb{P}_{\infty}^{p}\left(|\mathfrak{C}|<\infty\right)=\frac{1}{2i\pi}\oint_{\gamma}\frac{dz}{z}\Delta\left(p,\sqrt{pt^{3}_{c}}\,z\right)\,\Phi(p,1/z),$

where the contour $\gamma$ must lie in a region enclosing $0$ where both
functions $z\mapsto\Delta\left(p,\sqrt{pt^{3}_{c}}\,z\right)$ and
$z\mapsto\Phi(p,1/z)$ are analytic (see for example [21, Section VI.10.2]).

To compute this integral, we first have to compute $\Phi$. To this end, we
introduce the cylinder generating functions of Boltzmann planar maps. It is
defined for $|z_{1}|$ and $|z_{2}|$ large enough by

$W^{\mathrm{cyl}}_{\mathbf{q}(p,t)}(z_{1},z_{2})=\sum_{l_{1},l_{2}\geqslant
0}\frac{\sum_{\mathfrak{m}\in\mathcal{M}^{(l_{1},l_{2})}}\,\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f_{1},f_{2}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t)}{z_{1}^{l_{1}+1}\,z_{2}^{l_{2}+1}}.$

In a similar fashion than the pointed disk generating function $W^{\bullet}$
of Section 2, this series has a universal form involving the two constants
$c_{\pm}(p,t)$:

$\begin{split}&W^{\mathrm{cyl}}_{\mathbf{q}(p,t)}(z_{1},z_{2})\\\
&=\frac{1}{2(z_{1}-z_{2})^{2}}\left(W^{\bullet}_{\mathbf{q}(p,t)}(z_{1})W^{\bullet}_{\mathbf{q}(p,t)}(z_{2})\left(z_{1}z_{2}-\frac{c_{+}(p,t)+c_{-}(p,t)}{2}(z_{1}+z_{2})+c_{+}(p,t)c_{-}(p,t)\right)-1\right).\end{split}$

This formula appears in [14], [20] and a multitude of other references. We
refer to [2] for a proof and a review of the literature on this matter.

Of particular interest to us will be the generating function where the first
root face has degree 2:

$\displaystyle[z_{1}^{-3}]$ $\displaystyle
W^{\mathrm{cyl}}_{\mathbf{q}(p,t)}(z_{1},z)$ $\displaystyle=\sum_{k\geqslant
0}\,z^{-k-1}\,\sum_{\mathfrak{m}\in\mathcal{M}^{(2,k)}}\,\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f_{1},f_{2}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t),$
$\displaystyle=-z+W^{\bullet}_{\mathbf{q}(p,t)}(z)\left(z^{2}-\frac{c_{+}(p,t)+c_{-}(p,t)}{2}z+\frac{1}{4}c_{+}(p,t)c_{-}(p,t)-\frac{1}{8}\left(c_{+}^{2}(p,t)+c^{2}_{-}(p,t)\right)\right).$

The antiderivative of this function has a simple expression giving the
identity:

$\displaystyle\Phi(p,1/z)$ $\displaystyle=\sum_{k\geqslant
0}\,\frac{1}{k}\,z^{-k}\,\sum_{\mathfrak{m}\in\mathcal{M}^{(2,k)}}\,\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f_{1},f_{2}\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}),$
$\displaystyle=\frac{z^{2}}{2}-\frac{z+\frac{c_{+}(p,t_{c})+c_{-}(p,t_{c})}{2}}{2}\sqrt{(z-c_{+}(p,t_{c}))(z-c_{-}(p,t_{c}))}.$

From this expression, we see that the function $z\mapsto\Phi(p,z)$ is analytic
on $\mathbb{C}\setminus[c_{-}(p,t_{c}),c_{+}(p,t_{c})]$. From Lemma 8, we know
that $z\mapsto\Delta(p,\sqrt{pt^{3}_{c}}z)$ is analytic on
$\mathbb{C}\setminus\left[c_{+}(p,t_{c}),+\infty\right)$. We cannot directly
pick an appropriate contour $\gamma$ to compute our Hadamard product of series
evaluated at $1$, however, if we take $w\in(0,1)$ we have

$\displaystyle\Delta\left(p,\sqrt{pt^{3}_{c}}\,z\right)\odot\Phi(p,z)|_{z=w}=\frac{1}{2i\pi}\oint_{\gamma(w)}\frac{dz}{z}\Delta\left(p,w\,\sqrt{pt^{3}_{c}}\,z\right)\,\Phi(p,1/z),$
(17)

where the contour $\gamma(w)$ encloses the interval
$[c_{-}(p,t_{c}),c_{+}(p,t_{c})]$ and crosses the positive real line at some
point inside the interval $(c_{+}(p,t_{c}),c_{+}(p,t_{c})+\varepsilon)$ for
some $\varepsilon>0$. Note that the in the last display, the function
$z\mapsto\Delta(p,w\sqrt{pt_{c}^{3}}z)/z$ is well defined and continuous at
$z=0$. Using the fact that $\Delta(p,w\,\sqrt{pt^{3}_{c}}\,z)$ is analytic
inside $\gamma(w)$ and deforming the contour gives, for every $w\in(0,1)$,

$\displaystyle\Delta$
$\displaystyle\left(p,\sqrt{pt^{3}_{c}}\,z\right)\odot\Phi(p,z)|_{z=w}$
$\displaystyle=\frac{1}{2i\pi}\oint_{\gamma(w)}\frac{dz}{z}\Delta\left(p,w\,\sqrt{pt^{3}_{c}}\,z\right)\,\left(\frac{z^{2}}{2}-\frac{z+\frac{c_{+}(p,t_{c})+c_{-}(p,t_{c})}{2}}{2}\sqrt{(z-c_{+}(p,t_{c}))(z-c_{-}(p,t_{c}))}\right),$
$\displaystyle=-\frac{1}{4i\pi}\oint_{\gamma(w)}\frac{dz}{z}\Delta\left(p,w\,\sqrt{pt^{3}_{c}}\,z\right))\left(z+\frac{c_{+}(p,t_{c})+c_{-}(p,t_{c})}{2}\right)\sqrt{(z-c_{+}(p,t_{c}))(z-c_{-}(p,t_{c}))},$
$\displaystyle=\frac{1}{2\pi}\int_{c_{-}(p,t_{c})}^{c_{+}(p,t_{c})}\frac{dz}{z}\Delta\left(p,w\,\sqrt{pt^{3}_{c}}\,z\right)\left(z+\frac{c_{+}(p,t_{c})+c_{-}(p,t_{c})}{2}\right)\sqrt{(c_{+}(p,t_{c})-z)(z-c_{-}(p,t_{c}))}.$

Taking the limit as $w\to 1$ finaly gives the proposition ∎

We are now ready to prove our first main result:

###### Proof of Theorem 1.

The proof basically consists on computing the integral (15), which still
requires some work. The change of variables $z=(1-1/y)(pt_{c}^{3})^{-1/2}$
gives:

$\displaystyle\mathbb{P}_{\infty}^{\nu}(|\mathfrak{C}|<\infty)$
$\displaystyle=\frac{1}{2\pi\,p\,t_{c}^{3}}\int_{\frac{y_{-}(p,t_{c})}{y_{-}(p,t_{c})-1}}^{\frac{y_{+}(p,t_{c})}{y_{+}(p,t_{c})-1}}\,\frac{dy}{y(y-1)}\,\hat{\Delta}\left(p,V(1-p,U(t_{c}^{3}),y)\right)\,\left(\frac{y-1}{y}+\frac{1}{2y_{-}(p,t_{c})}+\frac{1}{2y_{+}(p,t_{c})}\right)$
$\displaystyle\qquad\qquad\qquad\times\sqrt{\left(\frac{1}{y_{+}(p,t_{c})}-\frac{y-1}{y}\right)\left(\frac{y-1}{y}-\frac{1}{y_{-}(p,t_{c})}\right)},$

where $\hat{\Delta}$ is defined in Lemma 8 and $V(1-p,U,y)$ is defined in
Lemma 4. To calculate this new integral, we want to do the change of variables
$y=\hat{y}(1-p,U(t_{c}^{3}),2\sqrt{3}/3-V)$, where $\hat{y}$ is defined in
Lemma 4 (as we will see shortly, this change of variables instead of simply
$y=\hat{y}(1-p,U(t_{c}^{3}),V)$ simplifies calculations a bit more).

Recall that, from Lemma 4 and its proof, the function
$V\mapsto\hat{y}(1-p,U(t_{c}^{3}),V)$ is an increasing bijection from
$[V_{-}(1-p,U(t_{c}^{3})),V_{+}(1-p,U(t_{c}^{3}))]$ onto
$[y_{-}(1-p,t_{c}),y_{+}(1-p,t_{c})]$, and is analytic on
$(V_{-}(1-p,U(t_{c}^{3})),V_{+}(1-p,U(t_{c}^{3})))$. In view of this, for our
change of variables, we want $2\sqrt{3}/3-V$ to be in a sub-interval of
$[V_{-}(1-p,U(t_{c}^{3})),V_{+}(1-p,U(t_{c}^{3}))]$. This will also enable us
to use the rational parametrization of the series $\Delta$ of Lemma 8. The
details of the calculations that follow are available in the Maple file [1].

We have to solve for $V$ the two equations

$\hat{y}\left(1-p,U(t_{c}^{3}),\frac{2\sqrt{3}}{3}-V\right)=\frac{y_{\pm}(p,t_{c})}{y_{\pm}(p,t_{c})-1}.$

There is an interesting symmetry to exploit: for every $p\in(0,1)$ and every
$V\in\mathbb{C}$ one has

$\hat{y}\left(p,U(t_{c}^{3}),V\right)=\frac{\hat{y}(1-p,U(t_{c}^{3}),\frac{2\sqrt{3}}{3}-V)}{\hat{y}(1-p,U(t_{c}^{3}),\frac{2\sqrt{3}}{3}-V)-1},$

therefore, we want to solve for $V$ the two equations

$\hat{y}\left(p,U(t_{c}^{3}),V\right)=y_{\pm}(p,t_{c}).$ (18)

Since $\hat{y}\left(p,U(t_{c}^{3}),V\right)$ is a rational fraction in $V$ of
degree $2$ for its numerator and $3$ for its denominator, the solutions of
equation (18) are the roots of a polynomial of degree $3$. By definition, one
of the root of this polynomial is $V_{\pm}(p,U(t_{c}^{3}))$, and it will even
be a double root except for $V_{-}(p,U(t_{c}^{3}))$ when it is a negative pole
of $\hat{y}$. Fortunately, we can compute explicitly these values. Indeed, the
stationary points of $V\mapsto\hat{y}(p,U(t_{c}^{3}),V)$ are the roots of the
polynomial

$\left(9V^{3}-9V^{2}\sqrt{3}+4p\sqrt{3}\right)\,\left(V-\sqrt{3}/3\right).$

The roots of the polynomial of degree $3$ are given by

$\displaystyle V_{m}(p)$
$\displaystyle=-\frac{\sqrt{3}}{3}\,\frac{\sqrt{p}}{\cos\left(\frac{\mathrm{arccos}(\sqrt{p})}{3}\right)}<0,$
$\displaystyle V_{l}(p)$
$\displaystyle=\frac{\sqrt{3}}{3}\,\frac{\sqrt{p}}{\cos\left(\frac{\mathrm{arccos}(\sqrt{p})}{3}-\frac{\pi}{3}\right)}\in[0,2\sqrt{3}/3],$
$\displaystyle V_{r}(p)$ $\displaystyle=\frac{2\sqrt{3}}{3}-V_{m}(p)\geqslant
2\sqrt{3}/3.$

From there, we see that $V_{-}(p,U(t_{c}^{3}))=V_{m}(p)$ when $\hat{y}$ has no
negative pole, which is the case when $p>\frac{1}{2}-\frac{5\sqrt{3}}{18}\sim
0.018$. Since we are interested in $p\geqslant 1/2$, the case when $\hat{y}$
has negative poles will not bother us. The positive singularity is given by

$V_{+}(p,U(t_{c}^{3}))=\begin{cases}\frac{\sqrt{3}}{3}&\text{when $p\geqslant
1/2$},\\\ V_{l}(p)&\text{when $p\leqslant 1/2$}.\end{cases}$

We can compute the third root of equation (18) from the constant term of the
polynomial, which is always $-\frac{2\sqrt{3}p}{9}$. When
$p>\frac{1}{2}-\frac{5\sqrt{3}}{18}$, this root is given by

$V^{i}_{\pm}(p)=\frac{2\sqrt{3}p}{9V_{\pm}(p,U(t_{c}^{3}))^{2}}.$

Since $V_{\pm}(p,U(t_{c}^{3}))\leqslant\sqrt{3}/3$, we can see that
$2\sqrt{3}/3-V_{\pm}(p,U(t_{c}^{3}))$ is always larger than $\sqrt{3}/3$, and
therefore $V_{+}(1-p,U(t_{c}^{3}))$. From the explicit expression of
$V_{\pm}(p,U(t_{c}^{3}))$, we can also check that $2\sqrt{3}/3-V_{\pm}^{i}(p)$
is always between $0$ and $V_{+}(1-p,U(t_{c}^{3}))$. The correct bounds for
our change of variables are then $V^{i}_{\pm}(p)$ since they are the only
solutions inside the interval
$[V_{-}(1-p,U(t_{c}^{3})),V_{+}(1-p,U(t_{c}^{3}))]$. Note that we have
$0<V_{+}^{i}(p)<V_{-}^{i}(p)$. This discussion also provides the following
nice factorizations:

$\displaystyle\frac{\hat{y}(1-p,U(t_{c}^{3}),\frac{2\sqrt{3}}{2}-V))-1}{\hat{y}(1-p,U(t_{c}^{3}),\frac{2\sqrt{3}}{2}-V))}-\frac{1}{y_{\pm}(p,t_{c})}$
$\displaystyle=\frac{1}{\hat{y}(p,U(t_{c}^{3}),V))}-\frac{1}{y_{\pm}(p,t_{c})},$
$\displaystyle=\frac{\left(V-V_{\pm}^{i}(p)\right)\left(V-V_{\pm}(p,U(t_{c}^{3}))\right)^{2}}{2\,V\left(V-\frac{2\sqrt{3}}{3}\right)}.$

With the explicit expression for $\hat{\Delta}$ in terms of $V$ given in Lemma
8, it is easy to check:

$\hat{\Delta}\left(p,V\right)\,\frac{\partial_{V}\hat{y}(1-p,U(t_{c}^{3}),V)}{\hat{y}(1-p,U(t_{c}^{3}),V)\,(\hat{y}(1-p,U(t_{c}^{3}),V)-1)}=\frac{1}{3\,(2\sqrt{3}-3(1-p))\left(V-\frac{\sqrt{3}}{3}\right)^{2}}.$

Using the change of variables $y=\hat{y}(1-p,U(t_{c}^{3}),2\sqrt{3}/3-V)$ we
finally get:

$\displaystyle\mathbb{P}_{\infty}^{\nu}(|\mathfrak{C}|<\infty)$
$\displaystyle=\frac{1}{12\,p\,t_{c}^{3}\,(2\sqrt{3}-3(1-p))\,\pi}\int_{V_{+}^{i}(p)}^{V_{-}^{i}(p)}dV\,\sqrt{(V_{-}^{i}(p)-V)(V-V_{+}^{i}(p))}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\frac{(V-V_{+}(p,U(t_{c}^{3})))(V-V_{-}(p,U(t_{c}^{3})))}{V\,\left(\frac{2\sqrt{3}}{3}-V\right)\,\left(V-\frac{\sqrt{3}}{3}\right)^{2}}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\left(\frac{1}{\hat{y}(p,U(t_{c}^{3}),V)}+\frac{1}{2}\left(\frac{1}{y_{+}(p,t_{c})}+\frac{1}{y_{-}(p,t_{c})}\right)\right).$

Embarrassingly, we were not able to simplify directly this integral (it is
equal to $1$ when $p<1/2$ since we know that the critical percolation
threshold is $1/2$ !). Fortunately, when $p>1/2$, Maple is able to give an
expression in terms of $V_{\pm}(p,U(t_{c}^{3}))$. Using basic trigonometric
identities, we can further simplify this expression into the one given in the
theorem after injecting the expressions for $V_{\pm}(p,U(t_{c}^{3}))$. ∎

##  5 Cluster volume and perimeter

###  5.1 Admissibility equations and volume-modified weight sequence

We review in this Section additional background on Boltzmann planar maps that
we will use in the proof of Theorem 2. We refer the reader to the references
[7, 16, 27] for details.

For $p\in(0,1)$ and $t\in(0,t_{c}]$, consider the two following bivariate
power series in $(z_{1},z_{2})$:

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\sum_{k,k^{\prime}\geqslant
0}z_{1}^{k}z_{2}^{k^{\prime}}\binom{2k+k^{\prime}+1}{k+1,k,k^{\prime}}q_{2+2k+k^{\prime}}(p,t),$
$\displaystyle f^{\diamond}(p,t;z_{1},z_{2})$
$\displaystyle=\sum_{k,k^{\prime}\geqslant
0}z_{1}^{k}z_{2}^{k^{\prime}}\binom{2k+k^{\prime}}{k,k,k^{\prime}}q_{1+2k+k^{\prime}}(p,t).$

These two functions are linked with Boltzmann maps with weight sequence
$(q_{k}(p,t_{c}))_{k\geqslant 1}$ by the Bouttier-Di Francesco-Guitter
bijection [15] and will be instrumental in the remaining of this work. In our
case, we can compute alternative expressions for these functions that will be
more amenable to analysis, see Lemma 10.

Recall the parameters $c_{+}(p,t)$ and $c_{-}(p,t)$ defined in Equation (13).
Since the weight sequence $\mathbf{q}(p,t)$ is admissible, the two functions
defined above are well defined at least in the domain $|z_{1}|\leqslant
z^{+}(p,t)$ and $|z_{2}|\leqslant z^{\diamond}(p,t)$ where $z^{+}(p,t)$ and
$z^{\diamond}(p,t)$ are positive real numbers defined by

$c_{\pm}(p,t)=z^{\diamond}(p,t)\pm 2\sqrt{z^{+}(p,t)}.$ (19)

Furthermore, from Proposition 4.2 and Lemma 4.4 of [7],
$(z^{+}(p,t),z^{\diamond}(p,t))$ is the minimal solution of the system of
equations

$\begin{cases}f^{\bullet}(p,t;z^{+}(p,t),z^{\diamond}(p,t))&=1-\frac{1}{z^{+}(p,t))},\\\
f^{\diamond}(p,t;z^{+}(p,t),z^{\diamond}(p,t))&=z^{\diamond}(p,t).\end{cases}$
(20)

In addition, when $p=1/2$ and $t=t_{c}$, the weight sequence is critical (see
Theorem 1.1 of [7]), and $(z^{+}(1/2,t_{c}),z^{\diamond}(1/2,t_{c}))$ is the
unique solution of the the system of equations (20) such that

$\left(\partial_{z_{2}}+\sqrt{z_{1}}\partial_{z_{1}}\right)f^{\diamond}(p,t;z^{+}(p,t),z^{\diamond}(p,t))=1.$
(21)

For $g\in(0,1]$, let $(z^{+}(p,t;g),z^{\diamond}(p,t;g))$ be the unique
solution in $(0,z^{+}(p,t)]\times(0,z^{\diamond}(p,t)]$ of the system of
equations

$\begin{cases}f^{\bullet}(p,t;z^{+}(p,t;g),z^{\diamond}(p,t;g))&=1-\frac{g}{z^{+}(p,t;g))},\\\
f^{\diamond}(p,t;z^{+}(p,t;g),z^{\diamond}(p,t;g))&=z^{\diamond}(p,t;g).\end{cases}$
(22)

Define

$c_{\pm}(p,t;g)=\frac{1}{g}\left(z^{\diamond}(p,t;g)\pm
2\sqrt{z^{+}(p,t;g)}\right).$

From e.g. [7, Equations (4.3) and (4.4)], the pointed disk generating function
of Boltzmann maps with modified weight sequence
$\mathbf{q}(p,t;g):=\left(g^{(k-2)/2}\,q_{k}(p,t)\right)_{k\geqslant 1}$
defined as in (10) is given by

$W_{\mathbf{q}(p,t;g)}^{\bullet}(z)=\frac{1}{\sqrt{(z-c_{+}(p,t;g))(z-c_{-}(p,t;g)-z)}}.$
(23)

###  5.2 Proof of Theorem 2

We start with the perimeter exponent as it is the easiest of the two. From
Equations (5) and (10), we can write

$\mathbb{P}_{n}^{p}\left(|V(\partial\mathfrak{C})|=k\right)=\frac{[t^{3n}]\left(q_{k}(p,t)\cdot[z^{-(k+1)}]W_{\mathbf{q}(p,t)}(z)\right)}{[t^{3n}]\mathcal{Z}(p,t)}=\frac{1}{p}\frac{[t^{3n}]\left(\sqrt{pt^{3}}^{-k}q_{k}(p,t)\cdot[y^{k}]T(p,t,ty)\right)}{[t^{3n}]\mathcal{Z}(p,t)}.$

Applying Lemmas 8 and 9 directly give the limit, which is the probability that
$|V(\partial\mathfrak{C})|=k$ in the UIPT by continuity for the local
topology.

$\mathbb{P}_{\infty}^{p}\left(|V(\partial\mathfrak{C})|=k\right)=\frac{1}{p}\left(\delta_{k}(p)\,t_{c}^{k}T_{k}(p,t_{c})+\sqrt{pt_{c}^{3}}^{-k}q_{k}(p,t_{c})\,\theta_{k}(p)\right).$

Using the asymptotics (35), (38), (40) and (42) derived in Section 6.3 gives
that for $p=1/2$,

$\mathbb{P}_{\infty}^{1/2}\left(|V(\partial\mathfrak{C})|=k\right)\underset{k\to\infty}{\sim}\kappa^{\prime}\,k^{-4/3}$

with

$\kappa^{\prime}=-8\,\frac{1}{\Gamma(4/3)}\left(\frac{8\,3^{\frac{5}{6}}}{351}+\frac{2\,3^{\frac{1}{3}}}{117}\right)\,\frac{3^{5/6}}{2\Gamma(-2/3)}\simeq
0.454,$ (24)

proving the second statement of Theorem 2.

We now turn on the first statement of the Theorem on the volume of the root
cluster. Although we are only interested in the case $p=1/2$, we start the
proof with a generic $p\in(0,1)$ as it will be easier to follow. From
proposition 6 we have for every $g\leqslant 1$

$\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]=\sum_{\mathfrak{m}\in\mathcal{M}}\,g^{|V(\mathfrak{m})|}\,\sum_{f\in
F(\mathfrak{m})}(pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f\\}}q_{\mathrm{deg}(f^{\prime})}(p,t_{c}).$

Applying Euler’s formula then gives

$\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]=g\,\sum_{\mathfrak{m}\in\mathcal{M}}\,\sum_{f\in
F(\mathfrak{m})}(g\,pt^{3}_{c})^{\mathrm{deg}(f)/2}\delta_{\mathrm{deg}(f)}(p)\,\prod_{f^{\prime}\in
F(\mathfrak{m})\setminus\\{f\\}}g^{(\mathrm{deg}(f^{\prime})-2)/2}\,q_{\mathrm{deg}(f^{\prime})}(p,t_{c}).$

Using the exact same line of reasoning as in the proof of Theorem 1, we can
compute this sum as the Hadamard product evaluated at $z=1$ of the functions
$z\mapsto\Delta(p,\sqrt{gpt_{c}^{3}})$ and the function $z\mapsto\Phi(p,z)$
where $c_{+}(p,t_{c})$ and $c_{-}(p,t_{c})$ are replaced by the two constants
$c_{+}(p,t_{c};g)$ and $c_{-}(p,t_{c};g)$ associated to the pointed disk
generating function
$\left((z-c_{+}(p,t_{c};g))(z-c_{-}(p,t_{c};g)-z)\right)^{-1/2}$ of Boltzmann
maps with weight sequence
$\mathbf{q}(p,t_{c};g)=\left(g^{(k-2)/2}\,q_{k}(p,t_{c})\right)_{k\geqslant
1}$ introduced in Section 5.1. Indeed, the first function should be obvious,
and the second comes from the antiderivative of cylinder generating functions
associated to the weight sequence $\mathbf{q}(p,t_{c};g)$ instead of the
weight sequence $\mathbf{q}(p,t_{c};1)$, which has the same universal form. Of
course, it remains to calculate these two constants, which are less explicit
than their counterparts $c_{+}(p,t_{c})$ and $c_{-}(p,t_{c})$. Nevertheless,
after mimicking the part of the proof of Theorem 1 leading to (15), we arrive
at the formula

$\displaystyle\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]=\frac{g}{2\pi}\int_{c_{-}(p,t_{c};g)}^{c_{+}(p,t_{c};g)}\frac{dz}{z}$
$\displaystyle\Delta\left(p,\sqrt{g\,pt^{3}_{c}}\,z\right)\left(z+\frac{c_{+}(p,t_{c};g)+c_{-}(p,t_{c};g)}{2}\right)$
$\displaystyle\times\sqrt{(c_{+}(p,t_{c};g)-z)(z-c_{-}(p,t_{c};g))}.$ (25)

We want to study the asymptotic behavior of the integral (25) when $g\to
1^{-}$. To do so, we must first study the dependency in $g$ of the two
constants $c_{\pm}(p,t_{c};g)$. We do not have a simple formula for the
unpointed or pointed disk generating function of the Boltzmann maps as was the
previously the case. However, the two constants can be studied via the
solution the system of equations (22). Let us denote
$z^{+}(g)=z^{+}(1/2,t_{c};g)$ and $z^{\diamond}(g)=z^{\diamond}(1/2,t_{c};g)$
the solution of this system. We can calculate an expansion as $g\to 1^{-}$ of
these two quantities with Lemma 11. Indeed, the development of $f^{\diamond}$
gives

$\displaystyle z^{+}-z^{+}(g)$
$\displaystyle=\sqrt{z^{+}}\,\left(z^{\diamond}-z^{\diamond}(g)\right)+\frac{\kappa^{\diamond}}{\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})}\,\left((z^{+}-z^{+}(g))+\sqrt{z^{+}}(z^{\diamond}-z^{\diamond}(g))\right)^{7/6}$
$\displaystyle\qquad+o\left(\left((z^{+}-z^{+}(g))+\sqrt{z^{+}}(z^{\diamond}-z^{\diamond}(g))\right)^{7/6}\right).$

Therefore we have

$\displaystyle z^{\diamond}-z^{\diamond}(g)$
$\displaystyle=\frac{1}{\sqrt{z^{+}}}\,\left(z^{+}-z^{+}(g)\right)$
$\displaystyle\qquad-\frac{\kappa^{\diamond}}{\sqrt{z^{+}}\,\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})}\,\left(2(z^{+}-z^{+}(g))\right)^{7/6}+o\left(\left(z^{+}-z^{+}(g)\right)^{7/6}\right).$
(26)

Now the development of $f^{\bullet}$ gives

$\displaystyle\frac{1}{z^{+}}-\frac{g}{z^{+}(g)}$
$\displaystyle=-\frac{1}{(z^{+})^{2}}\left(z^{+}-z^{+}(g)\right)$
$\displaystyle\qquad+\left(\frac{\kappa^{\diamond}}{\sqrt{z^{+}}}+\kappa^{\bullet}\right)\left(2(z^{+}-z^{+}(g))\right)^{7/6}+o\left(\left(2(z^{+}-z^{+}(g))\right)^{7/6}\right).$

Expanding the left hand side of this identity then gives

$\displaystyle\frac{1-g}{z^{+}}+\frac{1-g}{(z^{+})^{2}}\left(z^{+}-z^{+}(g)\right)=\left(\frac{\kappa^{\diamond}}{\sqrt{z^{+}}}+\kappa^{\bullet}\right)\left(2(z^{+}-z^{+}(g))\right)^{7/6}+o\left(\left(2(z^{+}-z^{+}(g))\right)^{7/6}\right),$

which in turns gives

$\displaystyle
z^{+}(g)=z^{+}-\frac{1}{2}\,\left(\kappa^{\diamond}\sqrt{z^{+}}+\kappa^{\bullet}z^{+}\right)^{-6/7}\left(1-g\right)^{6/7}+o\left(\left(1-g\right)^{6/7}\right).$

Plugging this expression in equation (5.2) gives

$\displaystyle
z^{\diamond}(g)=z^{\diamond}-\frac{1}{2\sqrt{z^{+}}}\,\left(\kappa^{\diamond}\sqrt{z^{+}}+\kappa^{\bullet}z^{+}\right)^{-6/7}\left(1-g\right)^{6/7}+o\left(\left(1-g\right)^{6/7}\right).$

We finally get an expansion for $c_{\pm}(g)$:

$\displaystyle c_{+}(g)$
$\displaystyle=z^{\diamond}(g)+2\sqrt{z^{+}(g)}=c_{+}(1)-\frac{1}{\sqrt{z^{+}}}\,\left(\kappa^{\diamond}\sqrt{z^{+}}+\kappa^{\bullet}z^{+}\right)^{-6/7}\left(1-g\right)^{6/7}+o\left(\left(1-g\right)^{6/7}\right),$
$\displaystyle c_{-}(g)$
$\displaystyle=z^{\diamond}(g)-2\sqrt{z^{+}(g)}=c_{-}(1)+o\left(\left(1-g\right)^{6/7}\right).$

The change of variable
$z=\phi(g,\xi):=c_{-}(g)+\left(c_{+}(g)-c_{-}(g)\right)\xi$ in Equation (25)
gives

$\displaystyle\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]$
$\displaystyle=\frac{g(c_{+}(g)-c_{-}(g))^{2}}{2\pi}\int_{0}^{1}d\xi\sqrt{\xi(1-\xi)}\frac{\Delta\left(1/2,\sqrt{g\,t^{3}_{c}/2}\,\phi(g,\xi)\right)}{\phi(g,\xi)}\left(\phi(g,\xi)+\frac{c_{+}(g)+c_{-}(g)}{2}\right),$
$\displaystyle=\frac{(c_{+}(1)-c_{-}(1))^{2}}{2\pi}\int_{0}^{1}d\xi\sqrt{\xi(1-\xi)}\frac{\Delta\left(1/2,\sqrt{g\,t^{3}_{c}/2}\,\phi(g,\xi)\right)}{\sqrt{g}\,\phi(g,\xi)}\left(\phi(1,\xi)+\frac{c_{+}(1)+c_{-}(1)}{2}\right)$
$\displaystyle\qquad\qquad+\mathcal{O}\left((1-g)^{6/7}\right)$

where we used the developments of $c_{\pm}(g)$ to obtain the second equality.
In a similar fashion than in the proof of Lemma 11, the dominant singular term
of the integral comes for the singularity at $g=1$ and $\xi=1$ of the term
$\frac{\Delta\left(1/2,\sqrt{g\,t^{3}_{c}/2}\,\phi(g,\xi)\right)}{\sqrt{g}\,\phi(g,\xi)}$.
The asymptotic expansion (39) established in Section 6.3 gives

$\frac{\Delta\left(1/2,z\right)}{z}\underset{z\to
1/2}{\sim}\left(\frac{16\,3^{\frac{5}{6}}}{351}+\frac{4\,3^{\frac{1}{3}}}{117}\right)\,\left(1-2z\right)^{-4/3}.$

Applying the same techniques as in the proof of Lemma 11, we see that the main
singular term of the expansion of
$\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]$ comes from the
integral

$\displaystyle\frac{\tilde{\kappa}}{\pi}\int_{0}^{1}d\xi\sqrt{\xi(1-\xi)}$
$\displaystyle\left(1-2\sqrt{g\,t^{3}_{c}/2}\,\phi(g,\xi)\right)^{-4/3}$
$\displaystyle=\frac{\tilde{\kappa}}{8\left(1-2\sqrt{g\,t^{3}_{c}/2}c_{-}(g)\right)^{4/3}}\,_{2}F_{1}\left(\frac{4}{3},\frac{3}{2};3;\frac{2\sqrt{g\,t^{3}_{c}/2}\left(c_{+}(g)-c_{-}(g)\right)}{1-2\sqrt{g\,t^{3}_{c}/2}c_{-}(g)}\right),$

with

$\tilde{\kappa}=\frac{(c_{+}(1)-c_{-}(1))^{2}\left(\phi(1,1)+\frac{c_{+}(1)+c_{-}(1)}{2}\right)\sqrt{t_{c}^{3}/2}}{2}\left(\frac{4\,3^{\frac{5}{6}}}{351}+\frac{3^{\frac{1}{3}}}{117}\right).$

Using the singular expansion of the hypergeometric function at $1$ and the
developments of $c_{\pm}(g)$, we finally get

$\displaystyle\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]=1-$
$\displaystyle\frac{\tilde{\kappa}}{8\left(1-2\sqrt{\,t^{3}_{c}/2}c_{-}(1)\right)^{4/3}}\,\frac{36\sqrt{3}\,\Gamma\\!\left(\frac{5}{6}\right)\,\Gamma\\!\left(\frac{2}{3}\right)}{\pi^{\frac{3}{2}}}$
$\displaystyle\qquad\left(\frac{2\sqrt{t_{c}^{3}/2}}{1-2\sqrt{t_{c}^{3}/2}c_{-}(1)}\frac{1}{\sqrt{z^{+}}}\,\left(\kappa^{\diamond}\sqrt{z^{+}}+\kappa^{\bullet}z^{+}\right)^{-6/7}\left(1-g\right)^{6/7}\right)^{1/6}+o\left(1-g\right)^{1/7}.$

This expansion is unfortunately not enough to extract the asymptotic behavior
of the probabilities
$\mathbb{P}_{\infty}^{1/2}\left(|V(\mathfrak{C})|=n\right)$ as $n\to\infty$.
Indeed, the generating series of these probabilities is
$\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]$, but this function
could have singularities of modulus $1$ other than $1$ contributing to the
asymptotic. However, we do not have this problem for tail probabilities. For
every $n$, denote
$p_{n}=\mathbb{P}_{\infty}^{1/2}\left(|V(\mathfrak{C})|\geqslant n\right)$. A
simple computation gives

$\sum_{n\geqslant
0}p_{n}g^{n}=\frac{1-g\,\mathbb{E}_{\infty}^{p}\left[g^{|V(\mathfrak{C})|}\right]}{1-g}\underset{g\to
1^{-}}{\sim}\kappa_{1}\,\left(1-g\right)^{-6/7},$

with

$\kappa_{1}=\frac{\tilde{\kappa}}{8\left(1-2\sqrt{\,t^{3}_{c}/2}c_{-}(1)\right)^{4/3}}\,\frac{36\sqrt{3}\,\Gamma\\!\left(\frac{5}{6}\right)\,\Gamma\\!\left(\frac{2}{3}\right)}{\pi^{\frac{3}{2}}}\left(\frac{2\sqrt{t_{c}^{3}/2}}{1-2\sqrt{t_{c}^{3}/2}c_{-}(1)}\frac{1}{\sqrt{z^{+}}}\right)^{1/6}\,\left(\kappa^{\diamond}\sqrt{z^{+}}+\kappa^{\bullet}z^{+}\right)^{-8/7}.$

From there, a classical Tauberian Theorem (see e.g. Theorem VI.13 of [21] and
the following discussion) establishes the asymptotic behavior of $p_{n}$ given
in the Theorem with

$\kappa=\frac{\kappa_{1}}{\Gamma(8/7)}=\frac{63\left(3^{\frac{1}{3}}+\frac{4\,3^{\frac{5}{6}}}{3}\right)3^{\frac{17}{21}}7^{\frac{1}{7}}137^{\frac{6}{7}}\Gamma\\!\left(\frac{2}{3}\right)^{\frac{18}{7}}2^{\frac{3}{14}}5^{\frac{13}{14}}}{56992\pi^{\frac{12}{7}}\Gamma\\!\left(\frac{1}{7}\right)}\simeq
0.278.$ (27)

##  6 Technical Lemmas

###  6.1 Dependency in $t$ of the weights

######  Lemma 8.

Fix $p\in(0,1)$ and let $V_{c}^{z}$ be the power series in $z$ defined be
$V_{c}^{z}=V(1-p,U(t_{c}^{3}),1/(1-z))$, where $V$ is defined in Lemma 4. Let
$\Delta(p,z)$ be the power series in $z$ defined by:

$\displaystyle\Delta(p,z)$
$\displaystyle=\hat{\Delta}(p,V_{c}^{z}):=3\,\frac{V_{c}^{z}\,(2\sqrt{3}-3V_{c}^{z})\,\left(9(V_{c}^{z})^{3}-9(\sqrt{3}+1)(V_{c}^{z})^{2}+3(3+2\sqrt{3})V_{c}^{z}-2(1-p)\sqrt{3}\right)}{(3(p-1)+2\sqrt{3})\,\left(\sqrt{3}-3V_{c}^{z}\right)^{3}\,\left(9(V_{c}^{z})^{3}-9(V_{c}^{z})^{2}\sqrt{3}+4(1-p)\sqrt{3}\right)}.$
(28)

For every $k\geqslant 1$ one has

$\frac{[t^{3n}]q_{k}(p,t)\sqrt{pt^{3}}^{-k}}{[t^{3n}]\mathcal{Z}(p,t)}\underset{n\to\infty}{\rightarrow}\delta_{k}(p)$
(29)

where the generating series of the numbers $\delta_{k}(p)$ is given by
$\Delta(p,z)$, which is analytic in the domain
$\mathbb{C}\setminus\left[1-1/y_{+}(p,t_{c}),+\infty\right)$.

###### Proof.

In the whole proof, $p\in(0,1)$ is fixed. All calculations are available in
the Maple companion file [1]. We start by proving that, for every $k\geqslant
1$, the series

$\tilde{q}_{k}(p,t)=\sqrt{pt^{3}}^{-k}\cdot\left(q_{k}(p,t)-(pt)^{3/2}\mathbf{1}_{k=3}\right)$

seen as a series in $t^{3}$ is algebraic and has a unique dominant singularity
at $t_{c}^{3}$. In view of (9), the generating series of these modified
weights is given by

$\tilde{F}(p,t,z)=\sum_{k\geqslant
1}\tilde{q}_{k}(p,t)\,z^{k-1}=\frac{1}{p}\,\frac{1}{1-z}T\left(1-p,t,\frac{t}{1-z}\right).$
(30)

Injecting this into the algebraic equation verified by $T$, we get an
algebraic equation verified by $\tilde{F}$:

$\left(p\,\tilde{F}(p,t,z)-T(1-p,t,t)\right)\,\mathrm{Pol}_{1}(p\,\tilde{F}(p,t,z),p,t^{3},T(1-p,t,t),tT_{1}(p,t))=z\cdot\mathrm{Pol}_{2}(p\,\tilde{F}(p,t,z),p,z),$

where $\mathrm{Pol}_{1}$ and $\mathrm{Pol}_{2}$ are explicit polynomials. The
form of this equation has the following consequences. First, using Lemma 4,
the series $\tilde{F}(p,t,z)$ is algebraic. Second, it is the unique solution
of this equation with constant term in $z$ equal to $T(1-p,t,t)/p$, and we can
compute its coefficients in $z$ inductively. These coefficients are the
modified weights $\tilde{q}_{k}(p,t)$ and their expressions are then rational
fractions in $p$, $t^{3}$, $tT_{1}(p,t)$ and $T(1-p,t,t)$, whose denominator
(up to a factor $p$) are the $k$-th power of

$\displaystyle\mathrm{Pol}_{1}(\tilde{F}(p,t,0),p,t^{3},T(1-p,t,t),tT_{1}(p,t))$
$\displaystyle=\mathrm{Pol}_{1}(T(1-p,t,t),p,t^{3},T(1-p,t,t),tT_{1}(p,t)),$
$\displaystyle=3t^{3}(p-1)\,T^{2}(1-p,t,t)+p(p-1)+tT_{1}(t).$

A quick glance at equation (2) with $y=1$ shows that the last display is the
derivative of the algebraic equation verified by $T(1-p,t,t)$ with respect to
$T$. Therefore this quantity can only be $0$ at singularities of $T(1-p,t,t)$,
which leaves only $t^{3}_{c}$ according to Lemma 4. As a consequence, we just
proved that the series $\tilde{q}(p,t)$ are all algebraic series in $t^{3}$
and all have a unique dominant singularity at $t_{c}^{3}$.

Now that we know that $\tilde{q}_{k}(p,t)$ has a unique dominant singularity
at $t_{c}^{3}$, it will follow from the general form of Puiseux expansions of
algebraic series near their singularities (see [21, Theorem VII.7, p. 49])
that $[t^{3n}]\tilde{q}_{k}(p,t)$ has the same asymptotic behavior as
$[t^{3n}]\mathcal{Z}(p,t)$ if we can prove that there exists two positive
constants $c,c^{\prime}$, depending on $k$ and $p$ such that for $n$ large
enough:

$c\cdot[t^{3n}]\mathcal{Z}(p,t)\leqslant[t^{3n}]\tilde{q}_{k}(p,t)\leqslant
c^{\prime}\cdot[t^{3n}]\mathcal{Z}(p,t).$

The upper bound is easily obtained by putting a triangulation of the $1$-gon
with a white boundary vertex inside a cycle of $k$ black vertices and summing
over every possible necklace between the two. The lower bound is obtained
similarly starting from cycle of $k$ black vertices by putting a single white
vertex inside the cycle and an edge joining this additional vertex to every
boundary vertex, and putting an arbitrary triangulation with white boundary
together with a matching necklace on the outside of the cycle. To sum up, we
have proved that for every $k$, $\tilde{q}_{k}(p,t)$ seen as a series in
$t^{3}$ has a unique dominant singularity at $t_{c}^{3}$ and that its
asymptotic expansion at $t_{c}^{3}$ is of the form

$\displaystyle\tilde{q}_{k}(p,t)=\tilde{q}_{k}(p,t_{c})-\tilde{a}_{k}(p)\left(1-\frac{t^{3}}{t^{3}_{c}}\right)+\tilde{b}_{k}(p)\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2}+o\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2}.$
(31)

This finishes the proof proof of the first statement (29) of the Lemma and we
now have to identify the generating series of the numbers
$\delta_{k}(p)=\tilde{b}_{k}(p)/\kappa(p)$, where $\kappa(p)$ is the
coefficient of the term $\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2}$ in the
asymptotic expansion of $\mathcal{Z}(p,t)$ calculated in Proposition 5.

Using the rational parametrization of Lemma 4, we can find a rational
parametrization for $\tilde{F}$. Indeed, if $V=V(1-p,U,y)$ is the power series
in $\mathbb{Q}[p,U]\llbracket y\rrbracket\subset\mathbb{Q}[p]\llbracket
t^{3},y\rrbracket$ defined in Lemma 4, we have

$\begin{split}\tilde{F}(p,t,z)&=\frac{1}{p}\hat{y}\left(1-p,U(t^{3}),V\Big{(}1-p,U(t^{3}),1/(1-z)\Big{)}\right)\\\
&\qquad\times\hat{T}\left(1-p,U(t^{3}),V\Big{(}1-p,U(t^{3}),1/(1-z)\Big{)}\right),\end{split}$

where $\hat{y}$ and $\hat{T}$ are rational fractions defined in Lemma 4. For
$z$ fixed such that $|1/(1-z)|<y_{+}(1-p,t_{c})$ (which includes a
neighborhood of $0$ for $z$ since $y_{+}(1-p,t_{c})>1$, the series
$\tilde{F}(p,t,z)$ seen as a series in $t^{3}$ has non negative coefficients
and has radius of convergence $t^{3}_{c}$. In addition, this implies that
$(t^{3},z)\mapsto\tilde{F}(p,t,z)$ is analytic in the larger domain
$\mathcal{D}(0,t^{3}_{c})\times\mathcal{D}(0,y_{+}(1-p,t_{c})/(y_{+}(1-p,t_{c})-1))$.
We will produce an asymptotic expansion of $\tilde{F}(p,t,z)$ near $t^{3}_{c}$
using our rational parametrization. To do so, we start by computing the
asymptotic expansion of $V\Big{(}p,U(t^{3}),1/(1-z)\Big{)}$ near $t^{3}_{c}$,
with $z$ fixed.

First, writing

$\begin{split}\frac{1}{1-z}&=\hat{y}\left(1-p,U(t^{3}),V\Big{(}1-p,U(t^{3}),1/(1-z)\Big{)}\right)\\\
&=\hat{y}\left(1-p,U(t^{3}_{c}),V\Big{(}1-p,U(t^{3}_{c}),1/(1-z)\Big{)}\right)\end{split}$

we get an algebraic equation between
$V_{c}^{z}=V\Big{(}1-p,U(t^{3}_{c}),1/(1-z)\Big{)}$,
$V\Big{(}1-p,U(t^{3}),1/(1-z)\Big{)}$ and $U(t^{3})$. Plugging the asymptotic
expansion (3) of $U(t^{3})$ in this equation, we obtain an asymptotic
expansion for $V\Big{(}p,U(t^{3}),1/(1-z)\Big{)}$ of the form:

$\displaystyle V\Big{(}1-p,U(t^{3}),1/(1-z)\Big{)}$
$\displaystyle=V_{c}^{z}-a_{1}(V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{1/2}+a_{2}(V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)$
$\displaystyle\qquad\qquad\qquad+a_{3}(V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2}+o\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2},$

where the $a_{i}$’s are explicit rational fractions whose expressions are
given in the Maple companion file [1]. Injecting in turn the asymptotic
expansions in $t^{3}$ of $U$ and $V$ in $\hat{T}$ and $\hat{y}$ we find an
asymptotic expansion for $\tilde{F}$ of the form:

$\tilde{F}(p,t,z)=\tilde{F}(p,t_{c},z)+\tilde{A}(p,V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)+\tilde{B}(p,V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)+o\left(1-\frac{t^{3}}{t^{3}_{c}}\right)^{3/2},$

where $\tilde{A}$ and $\tilde{B}$ are explicit rational functions, and are
analytic on the disk $\mathcal{D}(0,y_{+}(1-p,t_{c})/(y_{+}(1-p,t_{c})-1)$
(this is obvious from their expressions: $z\mapsto V_{c}^{z}$ is analytic in
this region, and the poles of $\tilde{A}$ and $\tilde{B}$ fall outside it, see
the Maple companion file [1]). Note that the error term in the previous
expansion is _a priori_ not uniform in $z$. To ensure that $\tilde{A}$ and
$\tilde{B}$ are the respective generating series of the numbers
$\tilde{a}_{k}$ and $\tilde{b}_{k}$, we see that, as power series in
$(z,t^{3})$ we have

$\tilde{A}(p,V_{c}^{z})=\lim_{t\to
t_{c}}\left(\tilde{F}(p,t,z)-\tilde{F}(p,t_{c},z)\right)\cdot(1-t^{3}/t_{c}^{3})^{-1},$

and

$\tilde{B}(p,V_{c}^{z})=\lim_{t\to
t_{c}}\left(\tilde{F}(p,t,z)-\tilde{F}(p,t_{c},z)-\tilde{A}(p,V_{c}^{z})\left(1-\frac{t^{3}}{t^{3}_{c}}\right)\right)\cdot(1-t^{3}/t_{c}^{3})^{-3/2}.$

Combined with the analycity properties of these series, this ensures that
$\tilde{A}(p,V_{c}^{z})$ and $\tilde{B}(p,V_{c}^{z})$ are indeed the
generating series of the numbers $\tilde{a}_{k}$ and $\tilde{b}_{k}$.

Finally, the generating series of the numbers $\delta_{k}$ is then given by

$\Delta(p,z)=\frac{z}{\kappa(p)\,(1-z)}\,\tilde{B}(p,V_{c}^{z})=\frac{\hat{y}\left(1-p,U(t^{3}_{c}),V_{c}^{z}\right)-1}{\kappa(p)}\,\tilde{B}(p,V_{c}^{z}),$

giving the expression of the Lemma. See the Maple file [1] for detailed
computations. ∎

Applying the same proof to the function $T(p,t,ty)$ instead of
$\tilde{F}(p,t,z)$ defined in Equation (30) allows to establish the asymptotic
behavior in $n$ of $[t^{3n}]t^{k}T_{k}(p,t)$ for every $k\geqslant 0$. We do
not reproduce the proof as it will be almost exactly the same as the proof of
Lemma 8, with no additional difficulties but with the function

$\frac{1-p}{y}\,\tilde{F}\left(1-p,t,1-\frac{1}{y}\right)$

instead of $\tilde{F}(p,t,z)$. The statement is as follows.

######  Lemma 9.

Fix $p\in(0,1)$ and $k\geqslant 1$. One has

$\frac{[t^{3n}]t^{k}T_{k}(p,t)}{[t^{3n}]\mathcal{Z}(p,t)}\underset{n\to\infty}{\rightarrow}\theta_{k}(p)$
(32)

where the generating series of the numbers $\theta_{k}(p)$ is given by

$\Theta(p,y):=\sum_{k\geqslant
0}\theta_{k}(p)\,y^{k}=\frac{1-p}{y}\,\Delta\left(1-p,1-\frac{1}{y}\right),$
(33)

which is analytic on $\mathbb{C}\setminus\left[y_{+}(p,t_{c}),+\infty\right)$.

###  6.2 BDFG functions

We start with an integral formula for the functions $f^{\bullet}$ and
$f^{\diamond}$.

######  Lemma 10.

Fix $p\in(0,1)$ and $t\in(0,t_{c}]$. Then for $0<z_{1}\leqslant z^{+}(p,t)$
and $0<z_{2}\leqslant z^{\diamond}(p,t)$ we have

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,2z_{2}+\frac{t^{3}}{\pi}\int_{\left(1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)^{-1}}^{\left(1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)^{-1}}{dz}\,\frac{1-\sqrt{pt^{3}}z_{2}-\frac{1}{z}}{2pt^{3}\,z_{1}}$
$\displaystyle\qquad\qquad\cdot\frac{T(1-p,t,tz)}{\sqrt{\left(1-z\left(1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)\right)\left(z\left(1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)-1\right)}},$
$\displaystyle f^{\diamond}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,(2z_{2}+z_{1}^{2})+\frac{\sqrt{t^{3}/p}}{\pi}\,\int_{\left(1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)^{-1}}^{\left(1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)^{-1}}dz$
$\displaystyle\qquad\qquad\cdot\frac{T(1-p,t,tz)}{\sqrt{\left(1-z\left(1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)\right)\left(z\left(1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)-1\right)}}.$

###### Proof.

Fix $p\in(0,1)$ and $t\in(0,t_{c}]$. Replacing the weights by their expression
(4) in the definitions of the two functions gives

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\frac{1}{p}2(pt)^{3/2}z_{2}+\frac{pt^{3}}{p}\sum_{k,k^{\prime},l\geqslant
0}\binom{2k+k^{\prime}+l+1}{k+1,k,k^{\prime},l}(pt^{3}\,z_{1})^{k}\,(\sqrt{pt^{3}}\,z_{2})^{k^{\prime}}\,[y^{l}]T(1-p,t,ty),$
$\displaystyle f^{\diamond}(p,t;z_{1},z_{2})$
$\displaystyle=\frac{1}{p}(pt)^{3/2}(2z_{2}+z_{1}^{2})+\frac{\sqrt{pt^{3}}}{p}\sum_{k,k^{\prime},l\geqslant
0}\binom{2k+k^{\prime}+l}{k,k,k^{\prime},l}(pt^{3}\,z_{1})^{k}\,(\sqrt{pt^{3}}\,z_{2})^{k^{\prime}}\,[y^{l}]T(1-p,t,ty).$

We can express these two functions as Hadamard products. Indeed, define the
trivariate power series in $(z_{1},z_{2};z)$:

$\displaystyle h^{\bullet}(z_{1},z_{2};z)$
$\displaystyle:=\sum_{k,k^{\prime},l\geqslant
0}\binom{2k+k^{\prime}+l+1}{k+1,k,k^{\prime},l}(pt^{3}\,z_{1})^{k}\,(\sqrt{pt^{3}}\,z_{2})^{k^{\prime}}\,z^{l},$
$\displaystyle=\frac{1}{2pt^{3}\,z_{1}}\left(\frac{1-z-\sqrt{pt^{3}}z_{2}}{\sqrt{\left(1-z-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)\left(1-z-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)}}-1\right);$

and

$\displaystyle h^{\diamond}(z_{1},z_{2};z)$
$\displaystyle:=\sum_{k,k^{\prime},l\geqslant
0}\binom{2k+k^{\prime}+l}{k,k,k^{\prime},l}(pt^{3}\,z_{1})^{k}\,(\sqrt{pt^{3}}\,z_{2})^{k^{\prime}}\,z^{l},$
$\displaystyle=\frac{1}{\sqrt{\left(1-z-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)\left(1-z-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)}}.$

Then we have

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,2z_{2}+t^{3}\,T(1-p,t,tz)\odot
h^{\bullet}(z_{1},z_{2};z)|_{z=1},$ $\displaystyle
f^{\diamond}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,(2z_{2}+z_{1}^{2})+\frac{\sqrt{pt^{3}}}{p}\,T(1-p,t,tz)\odot
h^{\diamond}(z_{1},z_{2};z)|_{z=1}.$

We can calculate these two Hadamard products as contour integrals in a similar
fashion than in the proof of Theorem 1 where we established (17). For
$(z_{1},z_{2})\in(0,z^{+}(p,t)]\times(0,z^{\diamond}(p,t)]$ the functions
$h^{\bullet}$ and $h^{\diamond}$ are analytic in the domain
$|z|<1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})$ that contains the domain
$|z|<1-\sqrt{pt^{3}}c_{+}(p,t)=1-\frac{1}{y_{+}(p,t)}$. This last domain is
not empty since $y_{+}(p,t)>1$ from Lemma 4. Therefore, we can represent the
Hadamard product as a contour integral similar to (17) if
$\left(1-\frac{1}{y_{+}(p,t)}\right)^{-1}\leqslant y_{+}(1-p,t)$. We can check
that this is the case since we have explicit formulas for $y_{+}(p,t_{c})$ and
$y_{+}(p,t)$ is increasing in $t$ (see the Maple file [1] for details).
Therefore, for a contour $\gamma$ enclosing $0$ and a point in the interval
$\left[\left(1-\frac{1}{y_{+}(p,t)}\right)^{-1},y_{+}(1-p,t)\right]$, we have

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,2z_{2}+\frac{t^{3}}{2i\pi}\oint_{\gamma}\frac{dz}{z}T(1-p,t,tz)\,h^{\bullet}(z_{1},z_{2},1/z),$
$\displaystyle f^{\diamond}(p;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,(2z_{2}+z_{1}^{2})+\frac{\sqrt{t_{c}^{3}/p}}{2i\pi}\oint_{\gamma}\frac{dz}{z}T(1-p,t,tz)\,h^{\diamond}(z_{1},z_{2},1/z).$

We obtain the expressions of the Lemma after simplifications and taking
contours converging to the cut
$\left[{\left(1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})\right)^{-1}},{\left(1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})\right)^{-1}}\right]$.
∎

When $p=1/2$ and $t=t_{c}$, we can compute explicitly an asymptotic expansion
of $f^{\bullet}$ and $f^{\diamond}$ at the point
$(z^{+}(p,t),z^{\diamond}(p,t))$ that will be used in the proof of Theorem 2.

######  Lemma 11.

Write $f^{\bullet}(z_{1},z_{2})=f^{\bullet}(1/2,t_{c};z_{1},z_{2})$,
$f^{\diamond}(z_{1},z_{2})=f^{\diamond}(1/2,t_{c};z_{1},z_{2})$,
$z^{+}=z^{+}(1/2,t_{c})$ and $z^{\diamond}=z^{\diamond}(1/2,t_{c})$. Then
$z_{+}=\frac{27\sqrt{3}}{32}$ and $z^{\diamond}=\frac{3^{1/4}\sqrt{2}}{4}$ and
we have the following asymptotic expansions at $(z^{+},z^{\diamond})^{-}$:

$\displaystyle f^{\bullet}(z_{1},z_{2})$
$\displaystyle=1-\frac{1}{z^{+}}+\frac{1}{z^{+}}\left(\frac{1}{z^{+}}-\sqrt{z_{+}}\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})\right)\,\left(z_{1}-z^{+}\right)+\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})\,\left(z_{2}-z^{\diamond}\right)$
$\displaystyle\qquad+\kappa^{\bullet}\,\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}+o\left(\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}\right),$
$\displaystyle f^{\diamond}(z_{1},z_{2})$
$\displaystyle=z^{\diamond}+\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})\,\left(z_{1}-z^{+}\right)+\left(1-\sqrt{z^{+}}\partial_{z_{1}}f^{\diamond}(z^{+},z^{\diamond})\right)\,\left(z_{2}-z^{\diamond}\right)$
$\displaystyle\qquad+\kappa^{\diamond}\,\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}+o\left(\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}\right),$

where

$\displaystyle\kappa^{\diamond}=\frac{4\,2^{\frac{1}{6}}\Gamma\\!\left(\frac{2}{3}\right)^{3}3^{\frac{2}{3}}\sqrt{5}}{63\pi^{2}}\quad\text{and}\quad\kappa^{\bullet}=\frac{512\,3^{\frac{1}{4}}\,\sqrt{2}}{81}\kappa^{\diamond}.$

###### Proof.

The respective values of $z^{+}$ and $z^{\diamond}$ are computed from (13) and
the explicit values of the singularities $y_{+}(1/2,t_{c})=2$ and
$y_{-}(1/2,t_{c})=-4$ of the function $y\mapsto T(1/2,t_{c},t_{c}y)$ (see the
Maple file [1] for details):

$\displaystyle z^{+}$
$\displaystyle=\frac{c_{+}(1/2,t_{c})+c_{-}(1/2,t_{c})}{2}=\frac{27\sqrt{3}}{32},$
$\displaystyle z^{\diamond}$
$\displaystyle=\left(\frac{c_{+}(1/2,t_{c})-c_{-}(1/2,t_{c})}{4}\right)^{2}=\frac{3^{3/4}\sqrt{2}}{4}.$

Now, the change of variable

$z=\phi(z_{1},z_{2};\xi):=\frac{1}{1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})}+\left(\frac{1}{1-\sqrt{pt^{3}}(z_{2}+2\sqrt{z_{1}})}-\frac{1}{1-\sqrt{pt^{3}}(z_{2}-2\sqrt{z_{1}})}\right)\xi$

in the expressions of Lemma 10 gives

$\displaystyle f^{\bullet}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,2z_{2}+\frac{1}{2pz_{1}}\frac{1}{\pi}\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\,\frac{\phi(z_{1},z_{2};\xi)(1-\sqrt{pt^{3}}z_{2})-1}{\phi(z_{1},z_{2};\xi)}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot
T\left(1-p,t,t\,\phi(z_{1},z_{2};\xi)\right),$ $\displaystyle
f^{\diamond}(p,t;z_{1},z_{2})$
$\displaystyle=\sqrt{pt^{3}}\,(2z_{2}+z_{1}^{2})+\sqrt{\frac{t^{3}}{p}}\frac{1}{\pi}\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\,T\left(1-p,t,t\,\phi(z_{1},z_{2};\xi)\right).$

We will see that, when $p=1/2$ and $t=t_{c}$, the main term in the expansion
of these functions stems from the singular behavior of $T$ at $y_{+}=2$.

From Lemma 12 and the discussion that follows, we know that the function
$y\mapsto T(1/2,t_{c},t_{c}\,y)$ is analytic for $y\in[0,2)$ and has the
following expansion as $y\to 2^{-}$:

$T(1/2,t_{c},t_{c}\,y)=\frac{\sqrt{3}}{2}-\frac{3^{5/6}}{2}\left(1-\frac{y}{2}\right)^{2/3}+\frac{\sqrt{3}}{2}\left(1-\frac{y}{2}\right)-3^{1/6}\left(1-\frac{y}{2}\right)^{4/3}+\mathcal{O}\left(\left(1-\frac{y}{2}\right)^{5/3}\right).$

The function

$\varphi(y):=T(1/2,t_{c},t_{c}\,y)-\left(\frac{\sqrt{3}}{2}-\frac{3^{5/6}}{2}\left(1-\frac{y}{2}\right)^{2/3}+\frac{\sqrt{3}}{2}\left(1-\frac{y}{2}\right)+3^{1/6}\left(1-\frac{y}{2}\right)^{4/3}\right)$

is twice differentiable on $[0,2)$. In addition, for $\xi\in(0,1)$ and if
$(z_{1},z_{2})\in[0,z^{+}]\times[0,z^{\diamond}]$, the quantity
$\phi(z_{1},z_{2};\xi)$ varies in a subset of
$(0,(1-1/y_{+}(1/2,t_{c}))^{-1})=(0,2)$. We then have

$\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\left|\varphi^{\prime}\left(\phi(z_{1},z_{2};\xi)\right)\right|<+\infty$

and

$\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\left|\varphi^{\prime\prime}\left(\phi(z_{1},z_{2};\xi)\right)\right|<+\infty.$

As a consequence, the function

$\Phi(z_{1},z_{2})=\frac{1}{\pi}\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\varphi\left(\phi(z_{1},z_{2};\xi)\right)$

is twice differentiable on $[0,z^{+}]\times[0,z^{\diamond}]$ and has the
following asymptotic expansion when $(z_{1},z_{2})\to(z^{+},z^{\diamond})$:

$\displaystyle\Phi(z_{1},z_{2})$
$\displaystyle=\Phi(z^{+},z^{\diamond})+\nabla\Phi(z^{+},z^{\diamond})\cdot\left(z_{1}-z^{+},z_{2}-z^{\diamond}\right)+\mathcal{O}\left((z_{1}-z^{+})^{2}+(z_{2}-z^{\diamond})^{2}\right).$

The singular parts of the expansions of $f^{\bullet}$ and $f^{\diamond}$ come
from the singularities of the form $(1-y/2)^{\alpha}$ in the development of
$T(1/2,t_{c},t_{c}\,y)$ at $y=2$ for $\alpha\in\\{2/3,4/3\\}$ (it is
straightforward to check that the linear term $(1-y/2)$ contributes only to
non singular parts in the expansion). Indeed, for $\alpha\in\\{2/3,4/3\\}$,
set

$\displaystyle I_{\alpha}(z_{1},z_{2})$
$\displaystyle=\frac{1}{\pi}\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}\left(1-\frac{\phi(z_{1},z_{2};\xi)}{2}\right)^{\alpha},$
$\displaystyle=\left(1-\frac{1}{2(1-\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}}))}\right)^{\alpha}\,\frac{1}{\pi}\int_{0}^{1}d\xi\,\xi^{-1/2}(1-\xi)^{-1/2}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\left(1-\xi\frac{\frac{1-\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}{1-\sqrt{t_{c}^{3}/2}(z_{2}+2\sqrt{z_{1}})}-1}{1-2\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}\right)^{\alpha},$
$\displaystyle=\left(1-\frac{1}{2(1-\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}}))}\right)^{\alpha}\,_{2}F_{1}\left(-\alpha,\frac{1}{2};1;\frac{\frac{1-\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}{1-\sqrt{t_{c}^{3}/2}(z_{2}+2\sqrt{z_{1}})}-1}{1-2\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}\right),$

where we used Euler’s integral representation of the hypergeometric function
${}_{2}F_{1}$ in the last line. This last equality is valid when
$0<z_{1}\leqslant z^{+}$ and $0<z_{2}\leqslant z^{\diamond}$ since in this
case $\sqrt{t_{c}^{3}/2}(z_{2}+2\sqrt{z_{1}})\leqslant 1/2$ and the variable
in the hypergeometric function is in $(0,1]$. Furthermore, using the values of
$z^{+}$ and $z^{\diamond}$, a simple computation done in the Maple file [1]
gives:

$\frac{\frac{1-\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}{1-\sqrt{t_{c}^{3}/2}(z_{2}+2\sqrt{z_{1}})}-1}{1-2\sqrt{t_{c}^{3}/2}(z_{2}-2\sqrt{z_{1}})}=1-\frac{20\sqrt{3}}{81}\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)+\mathcal{O}\left((z^{+}-z_{1})^{2}+(z^{\diamond}-z_{2})^{2}\right),$

Using the standard asymptotic development of hypergeometric functions at $1$
we see that the first singular term in the development of $I_{2/3}$ is

$-\kappa\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}\,\text{with}\,\kappa=\frac{18\,2^{\frac{1}{3}}\,\Gamma\\!\left(\frac{2}{3}\right)^{3}}{7\pi^{2}}\,\left(\frac{3}{5}\right)^{2/3}\,\left(\frac{20\sqrt{3}}{81}\right)^{7/6}.$

The first singular term of $I_{4/3}(z_{1},z_{2})$ is similar, but with
exponent $11/6$ instead of $7/6$. This means that the first singular term in
the development of $f^{\diamond}(1/2,z_{1},z_{2})$ is from
$-\frac{3^{5/6}}{2}\,\sqrt{t_{c}^{3}/p}\,I_{2/3}(z_{1},z_{2})$ and we have

$\displaystyle f^{\diamond}(z_{1},z_{2})$
$\displaystyle=f^{\diamond}(z^{+},z^{\diamond})+\nabla
f^{\diamond}(z^{+},z^{\diamond})\cdot\left(z_{1}-z^{+},z_{2}-z^{\diamond}\right)$
$\displaystyle+\frac{3^{5/6}}{2}\,\sqrt{2t_{c}^{3}}\,\kappa\,\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}+o\left(\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}\right).$

The statement for $f^{\diamond}$ follows using the fact that
$f^{\diamond}(z^{+},z^{\diamond})=z^{\diamond}$ and the criticality equation
(21).

The expansion for $f^{\bullet}(1/2,z_{1},z_{2})$ is obtained similarly by
replacing $\varphi$ by

$\displaystyle\frac{1-\sqrt{pt_{c}^{3}}z_{2}-\frac{1}{y}}{2pt_{c}^{3}\,z_{1}}T(1/2,t_{c},t_{c}\,y)$
$\displaystyle-\frac{1-\sqrt{t_{c}^{3}/2}z_{2}-\frac{1}{2}}{t_{c}^{3}\,z_{1}}\left(\frac{\sqrt{3}}{2}-\frac{3^{5/6}}{2}\left(1-\frac{y}{2}\right)^{2/3}+\frac{\sqrt{3}}{2}\left(1-\frac{1}{2t_{c}^{3}z_{1}}\right)\left(1-\frac{y}{2}\right)-3^{1/6}\left(1-\frac{y}{2}\right)^{4/3}\right).$

The first singular term in the development of $f^{\bullet}$ is then the one
from
$-\frac{3^{5/6}}{2}\,\frac{1-\sqrt{t_{c}^{3}/2}z^{\diamond}-\frac{1}{2}}{t_{c}^{3}\,z^{+}}\frac{1}{z^{+}}I_{2/3}$
and we have

$\displaystyle f^{\bullet}(z_{1},z_{2})$
$\displaystyle=f^{\bullet}(z^{+},z^{\diamond})+\nabla
f^{\bullet}(z^{+},z^{\diamond})\cdot\left(z_{1}-z^{+},z_{2}-z^{\diamond}\right)$
$\displaystyle\qquad+\frac{3^{5/6}}{2}\,\frac{1-2\sqrt{t_{c}^{3}/2}z^{\diamond}}{2t_{c}^{3}\,(z^{+})^{2}}\,\kappa\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}$
$\displaystyle\qquad+o\left(\left((z^{+}-z_{1})+\sqrt{z^{+}}(z^{\diamond}-z_{2})\right)^{7/6}\right).$

The statement for $f^{\bullet}$ the follows from
$f^{\bullet}(z^{+},z^{\diamond})=1-\frac{1}{z^{+}}$ and from the generic
properties $\partial_{z_{2}}f^{\bullet}=\partial_{z_{1}}f^{\diamond}$ and
$z_{1}\partial_{z_{1}}f^{\bullet}+f^{\bullet}=\partial_{z_{2}}f^{\diamond}$. ∎

###  6.3 Perimeter asymptotics at criticality

We gather in this Section asymptotics of several quantities appearing in this
paper when $p=1/2$ and $t=t_{c}$. They are all consequences of the following
Lemma characterizing the singularities of the function $y\mapsto
V(1/2,U(t_{c}^{3}),y)$ defined in Lemma 4.

######  Lemma 12.

The function $y\mapsto V(1/2,U(t_{c}^{3}),y)$ is analytic on
$\mathbb{C}\setminus\left((-\infty,-4]\cup[2,+\infty)\right)$. in addition it
has the following asymptotic expansion in a slit neighborhood of $2$:

$V(1/2,U(t_{c}^{3}),y)=\frac{\sqrt{3}}{3}-\frac{1}{3^{\frac{1}{3}}}\left(1-\frac{y}{2}\right)^{\frac{1}{3}}+\frac{1}{3}\left(1-\frac{y}{2}\right)-\frac{1}{3^{\frac{4}{3}}}\left(1-\frac{y}{2}\right)^{\frac{4}{3}}+\mathcal{O}\left(\left(1-\frac{y}{2}\right)^{\frac{5}{3}}\right).$
(34)

###### Proof.

From Lemma 4, we already know that $y\mapsto V(1/2,U(t_{c}^{3}),y)$ is
analytic on the domain
$\mathbb{C}\setminus\left((-\infty,y_{-}(1/2,t_{c})]\cup[y_{+}(1/2,t_{c}),+\infty)\right)$.
We also know that $y_{\pm}(1/2,t_{c})$ are the values of
$\hat{y}(1/2,U(t_{c}^{3}),V)$ at $V=V_{\pm}(1/2,t_{c})$ the two stationary
points of $\hat{y}$ enclosing $0$. We can easily compute the corresponding
values in the Maple file [1]. The expansion is then easily obtained by
singular inversion. ∎

This Lemma combined with the rational expressions in terms of $V$ that we have
allow to immediately compute asymptotic expansions. Indeed, the expressions
$\hat{T}(1/2,U(t_{c}^{3}),V)$ and $\hat{\Delta}(1/2,V)$ defined respectively
in Lemma 4 and Lemma 8 are singular only when $V$ is singular, and we can get
an asymptotic expansion at their unique dominant singularity by plugging the
expansion of $V$ in their expression. As usual, calculations are available in
the Maple companion file [1].

We get the following expansion for $T$:

$T(1/2,t_{c},t_{c}\,y)=\frac{\sqrt{3}}{2}-\frac{3^{5/6}}{2}\left(1-\frac{y}{2}\right)^{2/3}+\frac{\sqrt{3}}{2}\left(1-\frac{y}{2}\right)-3^{1/6}\left(1-\frac{y}{2}\right)^{4/3}+\mathcal{O}\left(\left(1-\frac{y}{2}\right)^{5/3}\right),$
(35)

and as a consequence

$t_{c}^{k}T_{k}(1/2,t_{c})\underset{k\to\infty}{\sim}\frac{-3^{5/6}}{2\Gamma(-2/3)}\,2^{k}\,k^{-5/3}.$
(36)

Similarly, using the expression (30) gives the asymptotic expansion of the
weights $\sqrt{t_{c}^{3}/2}^{-k}q_{k}(1/2,t_{c})$:

$\tilde{F}(1/2,t_{c},z)=2\,\sqrt{3}-2\,3^{5/6}\left(1-2z\right)^{2/3}++\mathcal{O}\left(\left(1-2z\right)^{5/3}\right),$
(37)

and as a consequence

$\sqrt{t_{c}^{3}/2}^{-k}q_{k}(1/2,t_{c})\underset{k\to\infty}{\sim}\frac{-2\,3^{5/6}}{\Gamma(-2/3)}\,2^{-k}\,k^{-5/3}.$
(38)

Note that these two singular expansions and the corresponding asymptotics were
established in [8] using different, and more involved techniques. The rational
parametrization that we have simplifies this analysis a lot.

Using $\hat{\Delta}$, we get the following expansion:

$\Delta\left(1/2,z\right)\underset{z\to
1/2}{\sim}\left(\frac{32\,3^{\frac{5}{6}}}{351}+\frac{8\,3^{\frac{1}{3}}}{117}\right)\,\left(1-2z\right)^{-4/3},$
(39)

giving

$\delta_{k}(1/2)\underset{k\to\infty}{\sim}\frac{1}{\Gamma(4/3)}\left(\frac{32\,3^{\frac{5}{6}}}{351}+\frac{8\,3^{\frac{1}{3}}}{117}\right)2^{-k}\,k^{1/3}.$
(40)

Finally, using the expression (33), we get

$\Theta\left(1/2,y\right)\underset{y\to
2}{\sim}\left(\frac{8\,3^{\frac{5}{6}}}{351}+\frac{2\,3^{\frac{1}{3}}}{117}\right)\,\left(1-y/2\right)^{-4/3},$
(41)

giving

$\theta_{k}(1/2)\underset{k\to\infty}{\sim}\frac{1}{\Gamma(4/3)}\left(\frac{8\,3^{\frac{5}{6}}}{351}+\frac{2\,3^{\frac{1}{3}}}{117}\right)2^{k}\,k^{1/3}.$
(42)

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