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Poisson-Voronoi percolation in the hyperbolic plane
with small intensities

Benjamin T. Hansen Bernoulli Institute, Groningen University, The Netherlands. E-mail: [email protected]. Supported by the Netherlands Organisation for Scientific Research (NWO) under project no. 639.032.529.    Tobias Müller Bernoulli Institute, Groningen University, The Netherlands. E-mail: [email protected]. Supported in part by the Netherlands Organisation for Scientific Research (NWO) under project nos 612.001.409 and 639.032.529.
Abstract

We consider percolation on the Voronoi tessellation generated by a homogeneous Poisson point process on the hyperbolic plane. We show that the critical probability for the existence of an infinite cluster is asymptotically equal to πλ/3\pi\lambda/3 as λ0.\lambda\to 0. This answers a question of Benjamini and Schramm [9].

1 Introduction and statement of main result

We will study percolation on the Voronoi tessellation generated by a homogeneous Poisson point process on the hyperbolic plane 2\mathbb{H}^{2}. That is, with each point of a constant intensity Poisson process on 2\mathbb{H}^{2} we associate its Voronoi cell – which is the set of all points of the hyperbolic plane that are closer to it than to any other point of the Poisson process – and we colour each cell black with probability pp and white with probability 1p1-p, independently of the colours of all other cells. See Figure 1 for a depiction of computer simulations of this process.

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Figure 1: Simulations of hyperbolic Poisson-Voronoi percolation, depicted in the half-plane model. Top left: λ=1\lambda=1, top right: λ=110\lambda=\frac{1}{10}, bottom left: λ=125\lambda=\frac{1}{25}, bottom right: λ=150\lambda=\frac{1}{50}; and p=12p=\frac{1}{2} in all cases.

We say that percolation occurs if there is an infinite connected cluster of black cells. For each intensity λ>0\lambda>0 of the underlying Poisson process, the critical probability is defined as

pc(λ):=inf{p:λ,p(percolation)>0}.p_{c}(\lambda):=\operatorname*{\vphantom{p}inf}\{p:{\mathbb{P}}_{\lambda,p}(\text{percolation})>0\}.

To the best of our knowledge, hyperbolic Poisson-Voronoi percolation was first studied by Benjamini and Schramm in the influential paper [9]. Amongst other things, they showed that 0<pc(λ)<1/20<p_{c}(\lambda)<1/2 for all λ>0\lambda>0; that pc(λ)p_{c}(\lambda) is a continuous function of λ\lambda; and that pc(λ)0p_{c}(\lambda)\to 0 as λ0\lambda\searrow 0.

They also asked for the asymptotics of pc(λ)p_{c}(\lambda) as λ0\lambda\searrow 0, and specifically for “the derivative at 0”. Here we answer this question, by showing:

Theorem 1

pc(λ)=(π/3)λ+o(λ)p_{c}(\lambda)=(\pi/3)\cdot\lambda+o(\lambda) as λ0.\lambda\searrow 0.

For comparison, Benjamini and Schramm gave an upper bound pc(λ)1214πλ+2p_{c}(\lambda)\leq\frac{1}{2}-\frac{1}{4\pi\lambda+2}, whose asymptotics are πλ+o(λ)\pi\lambda+o(\lambda) as λ0\lambda\searrow 0.

The results of Benjamini and Schramm highlight striking differences between Poisson-Voronoi percolation in the hyperbolic plane and Poisson-Voronoi percolation in the ordinary, Euclidean plane. For starters, in the latter case it is known [50, 10] that the critical probability equals 1/21/2 for all values of the intensity parameter λ\lambda. More strikingly perhaps is the difference in the behaviour of the number of infinite, black clusters. In the Euclidean case there are no infinite black clusters when p1/2p\leq 1/2 and precisely one infinite black cluster otherwise (almost surely). For the hyperbolic case, Benjamini and Schramm showed that if ppc(λ)p\leq p_{c}(\lambda) all black clusters are finite; if p1pc(λ)p\geq 1-p_{c}(\lambda) then there is precisely one infinite black cluster; but if pc(λ)<p<1pc(λ)p_{c}(\lambda)<p<1-p_{c}(\lambda) then there are infinitely many, distinct, infinite, black clusters (almost surely).

Related work. Percolation theory is an active area of modern probability theory with a considerable history, going back to the work of Broadbent and Hammersley [12] in the late fifties. By now there is an immense amount of research articles, mostly centered on percolation on lattices. We direct the reader to the monographs [11, 22] for an overview.

Poisson-Voronoi tessellations (mostly in dd-dimensional, Euclidean space) are one of the central models studied in stochastic geometry. They are studied in connection with many different applications and have a long history going back at least to the work of Meijering [34] in the early fifties. For an overview, see the monographs [37, 39] and the references therein.

In the early nineties, Vahidi-Asl and Wierman [46] introduced first passage percolation (a notion related to, but distinct from percolation as we have defined it above) on planar, Euclidean Poisson-Voronoi tessellations. A few years later, Zvavitch [50] proved that in the Euclidean plane almost surely all black clusters are finite when p1/2p\leq 1/2 (and λ>0\lambda>0 arbitrary). About a decade after that Bollobás and Riordan were able to complement this result by showing that, almost surely, there exists an infinite black cluster when p>1/2p>1/2 (and λ>0\lambda>0 arbitrary) – thus establishing pc=1/2p_{c}=1/2 for planar, Euclidean Poisson-Voronoi percolation. Since then planar, Euclidean Poisson-Voronoi percolation, especially “at criticality”, has received a fair amount of additional attention. See e.g. [1, 2, 41, 49]. Poisson-Voronoi percolation on the projective plane was studied by Freedman [17], and Poisson-Voronoi percolation on more general two and three dimensional manifolds was studied by Benjamini and Schramm in [8]. Poisson-Voronoi percolation on higher dimensional Euclidean space has been studied as well, in [4, 3, 15].

Fuchsian groups can be seen as the analogue in the hyperbolic plane of lattices in the Euclidean plane. Lalley [29, 30] studied percolation on Fuchsian groups and amongst other things established that the critical probabilities for “existence” and “uniqueness” of infinite clusters are distinct. Works on continuum percolation models over Poisson processes in the hyperbolic plane include [5, 16, 42, 44, 45]. Aspects of hyperbolic Poisson-Voronoi tessellations besides percolation that have been studied include the (expected) combinatorial structure of their cells, random walks on them and anchored expansions – see for example [6, 7, 13, 19, 26, 27]. Percolation on hyperbolic Poisson-Voronoi tessellations was first studied specifically by Benjamini and Schramm in [9]. In the recent paper [24] the current authors showed that pc(λ)1/2p_{c}(\lambda)\to 1/2 as λ\lambda\to\infty for Poisson-Voronoi percolation on the hyperbolic plane, proving a conjecture from [9].

Comparing Theorem 1 with Isokawa’s formula (stated as Theorem 6 below), we see that our result can be rephrased as : as λ0\lambda\searrow 0, the critical probability is asymptotic to the reciprocal of the “typical degree” (defined precisely in Section 2.3 below). A similar phenomenon happens in several Euclidean percolation models when one sends the dimension to infinity. The most well-known result in this direction is probably that for both site and bond percolation on d\mathbb{Z}^{d}, we have that pc=(1+od(1))(2d)1p_{c}=(1+o_{d}(1))\cdot(2d)^{-1}, as was shown concurrently and independently by Gordon [21], Hara and Slade [25] and Kesten [28]. Prior to that there were non-rigorous derivations of the result in the physics literature and Cox and Durret [14] had proved the analogous result for oriented percolation (which is technically easier to analyze). Penrose [36] proved an analogous result for the Gilbert model on dd-dimensional Euclidean space as the dimension tends to infinity, and Meester, Penrose and Sarkar [33] extended this result to the random connection model. The analogous phenomenon was shown by Penrose [35] for spread out percolation in fixed dimension when the connections get more and more spread out.

Sketch of the main ideas used in the proof. The intuition guiding the proof is that when λ\lambda is small and pp is of the same order as λ\lambda then black clusters are “locally tree like”, in the sense that while there will be some short cycles in the black subgraph of the Delaunay graph, but these will be “rare”. (The Delaunay graph is the abstract combinatorial graph whose vertices are the Poisson points and whose edges are precisely those pairs of points whose Voronoi cells meet.) This is also the intuition behind the results on high-dimensional and spread-out percolation mentioned above. In fact, in several of the works cited it is in fact shown that if we scale pp as a constant μ\mu divided by the degree (so that the origin has μ\mu black neighbours in expectation) then the cluster of the origin behaves more and more like a Galton-Watson tree with a Poisson(μ)(\mu) offspring distribution as the dimension grows.

Before going further, it is instructive to informally discuss in a bit more detail the situation for the high-dimensional Gilbert model analyzed by Penrose in [36]. In that model, we build a random graph on a constant intensity Poisson process on d\mathbb{R}^{d} by joining any pair of Poisson points at distance <1<1 by an edge. We seek the critical intensity λc\lambda_{c} such that there is a.s. no percolation for λ<λc\lambda<\lambda_{c} and there is a positive probability of percolation when λ>λc\lambda>\lambda_{c}. We consider the scenario where dd is large and the intensity is λ=μπd1\lambda=\mu\cdot\pi_{d}^{-1} with πd\pi_{d} the volume of the dd-dimensional unit ball and μ>0\mu>0 a fixed constant. We add the origin oo to the Poisson process (note that this a.s. does not change whether or not there is an infinite component), and think of “exploring” the cluster of the origin. We do this in an iterative fashion : we first add the neighbours of the origin to the cluster, then we consider each of these neighbours in turn and add their neighbours to the cluster, and so on. Of course the neighbours of the origin are precisely those Poisson points that fall inside the dd-dimensional unit ball BB. In particular their number follows a Poisson distribution with mean μ\mu. Once we have already added some points to the cluster of the origin and we consider the neighbours of a previously added point uu, we add those Poisson points that fall inside the ball of radius one around uu from which we have removed the union of all radius one balls around points we have processed already. Because of the high-dimensional geometry, the volume of this set difference is typically very close to the volume of a unit radius ball with nothing removed – at least during the initial stages of the exploration. In other words, at least during the first few exploration steps, the number of new points that gets added in each exploration step is approximately distributed like a Poisson random variable with mean μ\mu. This naturally leads to the aforementioned connection with Galton-Watson trees with Poisson(μ)(\mu) offspring distribution. Essentially, the geometric reason why this works out is the “concentration of measure” for high dimensional balls (see for instance [32], Chapter 13) : in high-dimensional Euclidean space, the volume of the unit ball is concentrated near its boundary. From this one can derive that, in a sense that can be made precise, most pairs of points connected by an edge in the Gilbert graph will have distance close to one. Moreover, the mass of a dd-dimensional ball is also concentrated near its equator, from which one can derive that – in a sense that can be made precise – for most pairs of points with a common neighbour, their distance will be very close to 2\sqrt{2} and, more generally, most pairs with graph distance kk will have Euclidean distance close to k\sqrt{k} (for kk fixed).

A similar phenomenon, but maybe even more extreme in a sense, happens in the hyperbolic plane for disks of large radius rr : The area of a hyperbolic disk BB of large radius rr is concentrated near its boundary; and if we take two points at random from BB then typically their distance is close to 2r2r, the maximal possible distance between any two points in BB. In the current paper, we consider a Poisson point process 𝒵{\mathcal{Z}} on the hyperbolic plane with small intensity λ>0\lambda>0. We again include the origin and try to analyze the black cluster of the Voronoi cell of the origin in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\}, where the cell of the origin is coloured black and all other cells are each coloured black with probability pp and white with probability 1p1-p. By Isokawa’s formula (stated as Theorem 6 below) the expected number of cells that are adjacent to the cell of the origin is asymptotic to 3πλ\frac{3}{\pi\lambda} as λ\lambda tends to zero. Moreover, as can be seen either by looking more carefully at Isokawa’s computations or by reading some of our arguments below, it can be shown that most neighbours of the origin are at distance close to r:=2ln(1/λ)r:=2\ln(1/\lambda). Note that rr goes to infinity as λ0\lambda\searrow 0. Now suppose we take p=μ(π/3)λp=\mu\cdot(\pi/3)\cdot\lambda with μ\mu a constant and λ\lambda small, so that the expected number of black cells neighbouring the cell of the origin is close to μ\mu. If we follow an exploration process analogous to the one described above for the Euclidean Gilbert model, the black cluster will have the property that most pairs of points at graph distance kk have distance close to 2kr2kr in the hyperbolic metric.

For the upper bound, we will show that if p=(1+ε)(π/3)λp=(1+\varepsilon)\cdot(\pi/3)\cdot\lambda and λ\lambda is sufficiently small (and ε>0\varepsilon>0 is a fixed constant) then the size of the black cluster of the origin stochastically dominates the size of a supercritical Galton-Watson tree. We will use an exploration procedure similar to what we described above for the high-dimensional Gilbert model. During the exploration, when we consider some point zz that has already been added to the tree, we make sure to only add those black neighbours of zz whose distance to zz is within some large constant of rr. We also make sure to select a subset {z1,,zk}\{z_{1},\dots,z_{k}\} of these neighbours with the property that all angles zizzj\angle z_{i}zz_{j} are larger than some small constant, and moreover for each ziz_{i} there is a ball of radius some large (but fixed) constant that contains no other Poisson points. Essentially because of the geometric phenomena described earlier, it will turn out that in this version of the exploration procedure, for sufficiently small λ\lambda, the subgraph of the cluster of the origin we obtain follows the distribution of a certain supercritical Galton-Watson process exactly. In contrast, in the high-dimensional Gilbert model the correspondence between the exploration process and a supercritical Galton-Watson process will eventually break down, after a (large) number of exploration steps that depends on the dimension, so that additional techniques and arguments were needed by Penrose [36].

For hyperbolic Poisson-Voronoi percolation the lower bound is technically more involved than the upper bound. This is probably the most novel contribution in our paper, and we believe it might inspire similar arguments applicable to other problems in percolation, notably for high-dimensional models. For percolation on d\mathbb{Z}^{d} and the Gilbert model, a trivial (and “sharp” up to lower order corrections when dd\to\infty) lower bound is given by a comparison to branching processes : in the former model, the cluster of the origin is stochastically dominated by a Galton-Watson distribution with mean offspring p(2d1)p\cdot(2d-1) and in the latter with mean offspring μπd\mu\cdot\pi_{d}. So the critical probability satisfies pc1/(2d1)p_{c}\geq 1/(2d-1), respectively the critical intensity satisfies λc1/πd\lambda_{c}\geq 1/\pi_{d}. (For percolation on d\mathbb{Z}^{d} this was in fact already observed by Broadbent and Hammersley [12] and for the Gilbert model by Gilbert [18].) In our case a similar argument does not seem feasible. In the Gilbert model, imagine we have explored part of the cluster of the origin and in doing so have revealed the status of the Poisson process in some region. We now wish to add those neighbours that are not yet part of the cluster of a point that we have previously added to in the cluster. The number of new points added is stochastically dominated by the number of new points we add at the very start of the exploration, when add the neighbours of the origin. In the Poisson-Voronoi percolation model there is no obvious monotonicity of this kind. Once we have revealed the status of the Poisson point process in some region this can make new edges both more and less likely. As a side remark, let us mention that using the methods in our paper it ought to be possible to show that when p=(1ε)(π/3)λp=(1-\varepsilon)\cdot(\pi/3)\cdot\lambda then, as λ0\lambda\to 0, the size of black cluster of the origin will converge in distribution to the size of a Galton-Watson tree with Poisson(1ε)(1-\varepsilon) offspring distribution. We do not pursue this here however as it does not appear useful for bounding pc(λ)p_{c}(\lambda) : it will not exclude the possibility that – for any fixed, small λ>0\lambda>0 – there is an extremely small, but positive probability that the origin is in an infinite component.

A naive approach that one might try is to compute the expected number of black paths of length kk starting at the origin (for λ\lambda small and p=(1ε)(π/3)λp=(1-\varepsilon)\cdot(\pi/3)\cdot\lambda), and hope to show that this expectation tends to zero as kk\to\infty. Unfortunately this approach does not seem feasible either. Long paths might revisit the same area many times which introduces dependencies that are difficult to deal with.

In order to circumvent this issue, we introduce what we call linked sequences of chunks. As we wish to show percolation does not occur, it suffices to show no infinite black cluster exists with a more generous notion of adjacency, in the form of what we call pseudo-edges, that makes the analysis simpler. With each pseudo-edge we associate a “certificate”, being the region of the hyperbolic plane that needs to be examined to verify it is indeed a pseudo-edge. We’ll say a pseudo-edge on a path PP is good if it has length close to rr and does not make a small angle with the previous pseudo-edge. Otherwise it is bad. A linked sequence of chunks is a sequence of paths P1,,PnP_{1},\dots,P_{n} such that: 1) on each path, every pseudo-edge except the last is good and the last is bad, and 2) for each i2i\geq 2, either the first point of PiP_{i} is close to the certificate of some pseudo-edge of Pi1P_{i-1} or the certificate of the first pseudo-edge of PiP_{i} intersects one of the certificates of Pi1P_{i-1}, and for every other pseudo-edge of PiP_{i} its certificate is disjoint from all certificates of all pseudo-edges on P1,,Pi1P_{1},\dots,P_{i-1}. We’ll first show (Proposition 31 below) that if the cluster of the origin is infinite, then either there exists an infinite path all of whose edges are good, or there exists an infinite linked sequence of chunks starting from the origin. Unlike paths in general, it is technically feasible to give decent bounds on the expected number of good paths, respectively the number of linked sequences of chunks, of some given length nn. We are able to obtain bounds that tend to zero with nn, which implies no percolation occurs a.s.

Structure of the paper. In the next section, we collect some notations, definitions, facts and tools that we will need in our proofs. Section 3.1 contains some preparatory geometric observations needed in later sections. Section 3.2 contains the proof that pc(λ)(1+ε)(π/3)λp_{c}(\lambda)\leq(1+\varepsilon)\cdot(\pi/3)\cdot\lambda for small enough λ\lambda (and ε>0\varepsilon>0 fixed), and Section 3.3 contains the proof that pc(λ)(1ε)(π/3)λp_{c}(\lambda)\geq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda for small enough λ\lambda. We end the paper by suggesting some directions for further work in Section 4.

2 Notation and preliminaries

2.1 Ingredients from hyperbolic geometry

The hyperbolic plane 2\mathbb{H}^{2} is a two dimensional surface with constant Gaussian curvature 1-1. There are many models, i.e. coordinate charts, for 2\mathbb{H}^{2} including the Poincaré disk model, the Poincaré half-plane model, and the Klein disk model. A gentle introduction to Gaussian curvature, hyperbolic geometry and these representations of the hyperbolic plane can be found in [38].

Even though we used the Poincaré half-plane model for the visualizations in Figure 1, from now on we will exclusively use the Poincaré disk model. (A computer simulation of Poisson-Voronoi percolation depicted in the Poincaré disk model can be found on the second page of our earlier paper [24].) We briefly recollect its definition and some of the main facts we shall be using in our arguments below, and refer the reader to [38] for proofs and more information.

The Poincaré disk model is constructed by equipping the open unit disk 𝔻2\mathbb{D}\subseteq\mathbb{R}^{2} with an appropriate metric and area functional. For points u,v𝔻,u,v\in\mathbb{D}, the hyperbolic distance can be given explicitly by

dist2(u,v)=2arcsinh(uv(1u2)(1v2))\operatorname{dist}_{\mathbb{H}^{2}}(u,v)=2\operatorname{arcsinh}\left(\frac{\|u-v\|}{\sqrt{(1-\|u\|^{2})(1-\|v\|^{2})}}\right)

where \|\cdot\| denotes the Euclidean norm. Straightforward calculations show that in particular, for z𝔻z\in\mathbb{D}, the Euclidean and hyperbolic distance to the origin are related via:

z=tanh(dist2(o,z)/2).\|z\|=\tanh\left(\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2\right). (1)

We will use the notations

B2(p,r):={u𝔻:dist2(p,u)<r},B2(p,r):={u2:pu<r},B_{\mathbb{H}^{2}}(p,r):=\{u\in\mathbb{D}:\operatorname{dist}_{\mathbb{H}^{2}}(p,u)<r\},\quad B_{\mathbb{R}^{2}}(p,r):=\{u\in\mathbb{R}^{2}:\|p-u\|<r\},

for hyperbolic, respectively Euclidean, disks. A standard fact that we will rely on in the sequel is:

Fact 2

Every hyperbolic disk is also a Euclidean disk; and every Euclidean disk contained in the open unit disk 𝔻\mathbb{D} is also a hyperbolic disk.

(But, the centre and radius of a disk with respect to the hyperbolic metric do not coincide with the centre and radius with respect to the Euclidean metric.)

For any measurable subset A𝔻,A\subseteq\mathbb{D}, its hyperbolic area is given by

area2(A):=Af(z)dz,\text{area}_{\mathbb{H}^{2}}(A):=\int_{A}f(z)\operatorname{d}z,

where

f(u):=4(1u2)2.f(u):=\frac{4}{(1-\|u\|^{2})^{2}}. (2)

From the formulas for hyperbolic distance and hyperbolic area one can derive that:

area2(B2(p,r))=2π(coshr1).\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(p,r)\right)=2\pi\cdot(\cosh r-1). (3)

The hyperbolic polar coordinates (α,ρ)(\alpha,\rho) corresponding to a point z𝔻z\in\mathbb{D} are ρ:=dist2(o,z)\rho:=\operatorname{dist}_{\mathbb{H}^{2}}(o,z) and α[0,2π)\alpha\in[0,2\pi) is the (counterclockwise) angle between the positive xx-axis and the line segment ozoz. Put differently, z𝔻z\in\mathbb{D} and ρ[0,)\rho\in[0,\infty) and α[0,2π)\alpha\in[0,2\pi) are such that

z=(tanh(ρ/2)cosα,tanh(ρ/2)sinα).z=\left(\tanh(\rho/2)\cdot\cos\alpha,\tanh(\rho/2)\cdot\sin\alpha\right). (4)

In several computations in the paper we’ll use the change of variables to hyperbolic polar coordinates. By this we of course just mean applying the substitution (4). We will always apply it to integrals of the form 𝔻g(z)f(z)dz\int_{\mathbb{D}}g(z)f(z)\operatorname{d}z with ff as given by (2) above, to obtain:

𝔻g(z)f(z)dz=002πg(tanh(ρ/2)cosα,tanh(ρ/2)sinα)sinhρdαdρ.\int_{\mathbb{D}}g(z)f(z)\operatorname{d}z=\int_{0}^{\infty}\int_{0}^{2\pi}g\left(\tanh(\rho/2)\cdot\cos\alpha,\tanh(\rho/2)\cdot\sin\alpha\right)\sinh\rho\operatorname{d}\alpha\operatorname{d}\rho. (5)

A hyperbolic geodesic or hyperbolic line segment between two points a,b𝔻a,b\in\mathbb{D} is the shortest path between aa and bb with respect to the hyperbolic metric. If there is a (Euclidean) line passing through a,ba,b and the origin oo, then the hyperbolic geodesic between aa and bb is just the (ordinary) line segment between them. Otherwise, the hyperbolic geodesic between a,ba,b can be constructed geometrically as follows. We let C2C\subseteq\mathbb{R}^{2} be the unique (Euclidean) circle with a,bCa,b\in C and such that it hits the boundary 𝔻\partial\mathbb{D} of the unit disk at right angles. The points a,ba,b divide C{a,b}C\setminus\{a,b\} into two circle segments. The hyperbolic geodesic between aa and bb is the one contained in 𝔻\mathbb{D}.

A hyperbolic line 𝔻\ell\subseteq\mathbb{D} is either the intersection of an Euclidean line through the origin with the open unit disk 𝔻\mathbb{D}, or the intersection of 𝔻\mathbb{D} with a circle CC hitting 𝔻\partial\mathbb{D} at right angles.

If a,b,c𝔻a,b,c\in\mathbb{D} then we use abc\angle abc to denote the angle the hyperbolic line segment between aa and bb and the hyperbolic line segment between bb and cc make in the common point bb. (In general, when a,b,c,𝔻a,b,c,\in\mathbb{D} do not lie on a hyperbolic line, there will be two possible interpretations of this angle, one in (0,π)(0,\pi) and one in (π,2π)(\pi,2\pi). As is usual, we shall always take the smaller of the two.) A hyperbolic triangle is the set of the three hyperbolic line segments defined by three (non-collinear) points a,b,c𝔻a,b,c\in\mathbb{D}. A critical tool is hyperbolic law of cosines. A proof of this result can for instance be found in [43] (page 81).

Lemma 3

For a hyperbolic triangle, with sides of length a,b,ca,b,c and respective opposite angles α,β,γ\alpha,\beta,\gamma (see Figure 2):

cosh(c)=cosh(a)cosh(b)sinh(a)sinh(b)cos(γ).\cosh(c)=\cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma).
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Figure 2: A hyperbolic triangle with sides of length a,b,a,b, and cc and respective opposite angles α,β,\alpha,\beta, and γ.\gamma. The blue circle is the boundary of the Poincaré disk.

A hyperbolic ray is defined analogously to a ray in Euclidean geometry. That is, if \ell is a hyperbolic line then any pp\in\ell divides {p}\ell\setminus\{p\} into two connected components. Each of these is a ray emanating from pp. For distinct p,s𝔻p,s\in\mathbb{D} and ϑ(0,π)\vartheta\in(0,\pi) we let sect(p,s,ϑ)\operatorname{sect}\left(p,s,\vartheta\right) denote the set of all hyperbolic rays emanating from pp that make an angle of no more than ϑ\vartheta with the ray emanating from pp through ss. In particular, in the Poincaré disk model, if p=op=o is the origin then sect(p,s,ϑ)\operatorname{sect}\left(p,s,\vartheta\right) looks like a (Euclidean) disk sector of opening angle 2ϑ2\vartheta with bisector the ray emanating from pp through ss. See Figure 3. For any other pp, the set sect(p,s,ϑ)\operatorname{sect}\left(p,s,\vartheta\right) can be obtained by applying a suitable hyperbolic isometry to such a disk sector.

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Figure 3: The black segment is the ray emanating from pp through ss and the purple region is the set sect(p,s,ϑ).\operatorname{sect}\left(p,s,\vartheta\right). On the left, p=op=o. On the right, the sector after applying an isometry that maps pp away from the origin and ss to the origin. The blue circle is the boundary of the Poincaré disk.

A hyperbolic isometry is a bijection φ:𝔻𝔻\varphi:\mathbb{D}\to\mathbb{D} that preserves hyperbolic distance, i.e. dist2(u,v)=dist2(φ(u),φ(v))\operatorname{dist}_{\mathbb{H}^{2}}(u,v)=\operatorname{dist}_{\mathbb{H}^{2}}(\varphi(u),\varphi(v)) for all u,v𝔻u,v\in\mathbb{D}. For any two points in x,y𝔻x,y\in\mathbb{D}, there exists a unique hyperbolic line \ell through xx and yy, and there exists a hyperbolic isometry φ\varphi such that φ[]\varphi[\ell] is an open interval given by the line segment between (1,0)(-1,0) and (1,0)(1,0), φ(x)=o\varphi(x)=o is the origin and φ(y)\varphi(y) is on the positive x-axis. In addition to distance, hyperbolic isometries preserve angles, in the sense that if c1,c2𝔻c_{1},c_{2}\subseteq\mathbb{D} are curves (e.g. hyperbolic line segments, circles, …) that meet in the point pp at angle α\alpha, then the curves φ[c1],φ[c2]\varphi[c_{1}],\varphi[c_{2}] meet in the point φ(p)\varphi(p) at the same angle α\alpha. Hyperbolic isometries also preserve hyperbolic area. That is,

area2(φ[A])=area2(A),\operatorname{area}_{\mathbb{H}^{2}}(\varphi[A])=\operatorname{area}_{\mathbb{H}^{2}}(A), (6)

for all (measurable) A𝔻A\subseteq\mathbb{D} and every hyperbolic isometry φ\varphi. What is more, if φ:𝔻𝔻\varphi:\mathbb{D}\to\mathbb{D} is a hyperbolic isometry then, for any (integrable) g:𝔻g:\mathbb{D}\to\mathbb{R} we have

𝔻g(z)f(z)dz=𝔻g(φ(u))f(u)du,\int_{\mathbb{D}}g(z)f(z)\operatorname{d}z=\int_{\mathbb{D}}g(\varphi(u))f(u)\operatorname{d}u, (7)

with ff as defined by (2). (An easy way to see this is to first note it follows trivially from (6) when gg is the indicator function of some measurable A𝔻A\subseteq\mathbb{D}. From this it easily follows for step-functions g=i=1nai1Aig=\sum_{i=1}^{n}a_{i}1_{A_{i}}. Then it follows for an arbitrary measurable function, by approximating it arbitrarily closely by step functions.)

2.2 Hyperbolic Poisson point processes

In the rest of this paper 𝒵{\mathcal{Z}} will denote a homogeneous Poisson point process (PPP) on the hyperbolic plane. Analogously to homogeneous Poisson point processes on the ordinary, Euclidean plane, a homogeneous Poisson process 𝒵{\mathcal{Z}} of intensity λ\lambda on the hyperbolic plane is characterized completely by the properties that a) for each (measurable) set A𝔻A\subseteq\mathbb{D} the random variable |𝒵A||{\mathcal{Z}}\cap A| is Poisson distributed with mean λarea2(A)\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(A), and b) if A1,,Am𝔻A_{1},\dots,A_{m}\subseteq\mathbb{D} are (measurable and) disjoint then the random variables |𝒵A1|,,|𝒵Am||{\mathcal{Z}}\cap A_{1}|,\dots,|{\mathcal{Z}}\cap A_{m}| are independent. In the light of the formula for area2(.)\operatorname{area}_{\mathbb{H}^{2}}(.) above, we can alternatively view 𝒵{\mathcal{Z}} as an inhomogeneous Poisson point process on the ordinary, Euclidean plane 2\mathbb{R}^{2} with intensity function

uλ1𝔻(u)f(u),u\mapsto\lambda\cdot 1_{\mathbb{D}}(u)\cdot f(u),

with ff given by (2) above.

Throughout the remainder, we attach to each point of 𝒵{\mathcal{Z}} a randomly and independently chosen colour. (Black with probability pp and white with probability 1p1-p.) We let 𝒵b{\mathcal{Z}}_{\text{b}} denote the black points and 𝒵w{\mathcal{Z}}_{\text{w}} the white points of 𝒵{\mathcal{Z}}. In the language of for instance [31], we can view 𝒵{\mathcal{Z}} as a marked Poisson point process, the marks corresponding to the colours.

We will rely heavily on a specific case of the Slivniak-Mecke formula, which is our weapon of choice for counting tuples of points z1,,zk𝒵z_{1},\dots,z_{k}\in{\mathcal{Z}} satisfying a given property. Before stating it, we remind the reader that formally speaking a Poisson process on 2\mathbb{R}^{2} is a random variable that takes values in the space ΩPPP\Omega_{\text{PPP}} of locally finite subsets of 2\mathbb{R}^{2}, equipped with the sigma algebra generated by the family events of the form : a given Borel BB set contains precisely kk points.

Theorem 4 (Slivniak-Mecke formula)

Let 𝒵{\mathcal{Z}} be a homogeneous hyperbolic Poisson point process of intensity λ\lambda, and let Let g:𝔻k×ΩPPP[0,)g:\mathbb{D}^{k}\times\Omega_{\text{PPP}}\to[0,\infty) be a nonnegative, measurable function. Then

𝔼[z1,,zk𝒵distinctg(z1,,zk,𝒵)]=λk𝔻𝔻𝔼[g(x1,,xk,𝒵{x1,,xk})]f(x1)f(xk)dxkdx1,\begin{array}[]{c}\displaystyle{\mathbb{E}}\left[\sum_{\begin{subarray}{c}z_{1},...,z_{k}\in{\mathcal{Z}}\\ \text{distinct}\end{subarray}}g(z_{1},...,z_{k},{\mathcal{Z}})\right]\\ =\\ \displaystyle\lambda^{k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g(x_{1},...,x_{k},{\mathcal{Z}}\cup\{x_{1},...,x_{k}\})\right]f(x_{1})\dots f(x_{k})\operatorname{d}x_{k}\dots\operatorname{d}x_{1},\end{array}

with ff given by (2).

As mentioned above, the version we state here is a specific case of a more general result. The general version can for instance be found in [37], as Corollary 3.2.3. (The version we present here is the special case of Corollary 3.2.3 in [37] when the ambient space E=2E=\mathbb{R}^{2} and the intensity measure has density λ1𝔻f\lambda\cdot 1_{\mathbb{D}}\cdot f with ff as in (2).) The Slivniak-Mecke formula is sometimes also called Mecke formula or Campbell-Mecke formula in the literature.

We shall be applying the following consequence of the Slivniak-Mecke formula, that is tailored to our situation where 𝒵=𝒵b𝒵w{\mathcal{Z}}={\mathcal{Z}}_{\text{b}}\cup{\mathcal{Z}}_{\text{w}} and membership in 𝒵b{\mathcal{Z}}_{\text{b}} is determined via independent, pp-biased coin flips.

Corollary 5

For λ>0\lambda>0 and 0p10\leq p\leq 1, let 𝒵=𝒵b𝒵w{\mathcal{Z}}={\mathcal{Z}}_{\text{b}}\cup{\mathcal{Z}}_{\text{w}} be as above and let g:𝔻k×ΩPPP[0,)g:\mathbb{D}^{k}\times\Omega_{\text{PPP}}\to[0,\infty) be a nonnegative, measurable function. Then

𝔼[z1,,zk𝒵bdistinctg(z1,,zk,𝒵)]=(pλ)k𝔻𝔻𝔼[g(u1,,uk,𝒵{u1,,uk})]f(u1)f(uk)du1duk,\begin{array}[]{c}\displaystyle{\mathbb{E}}\left[\sum_{\begin{subarray}{c}z_{1},...,z_{k}\in{\mathcal{Z}}_{\text{b}}\\ \text{distinct}\end{subarray}}g(z_{1},...,z_{k},{\mathcal{Z}})\right]\\ =\\ \displaystyle\left(p\lambda\right)^{k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g(u_{1},...,u_{k},{\mathcal{Z}}\cup\{u_{1},...,u_{k}\})\right]f(u_{1})\dots f(u_{k})\operatorname{d}{u_{1}}\dots\operatorname{d}{u_{k}},\end{array}

with ff given by (2).

Proof. Let us write

S:=z1,,zk𝒵distinctg(z1,,zk,𝒵),Sb:=z1,,zk𝒵bdistinctg(z1,,zk,𝒵).S:=\sum_{\begin{subarray}{c}z_{1},...,z_{k}\in{\mathcal{Z}}\\ \text{distinct}\end{subarray}}g(z_{1},...,z_{k},{\mathcal{Z}}),\quad S_{\text{b}}:=\sum_{\begin{subarray}{c}z_{1},...,z_{k}\in{\mathcal{Z}}_{\text{b}}\\ \text{distinct}\end{subarray}}g(z_{1},...,z_{k},{\mathcal{Z}}).

We imagine first “revealing” the locations of the Poisson points 𝒵{\mathcal{Z}}, but not yet the colours/coin flips. For any fixed, locally finite 𝒰d{\mathcal{U}}\subseteq\mathbb{R}^{d} we have

𝔼(S|𝒵=𝒰)=u1,,uk𝒰,distinctf(u1,,uk,𝒰),{\mathbb{E}}\left(S|{\mathcal{Z}}={\mathcal{U}}\right)=\sum_{{u_{1},\dots,u_{k}\in{\mathcal{U}},}\atop\text{distinct}}f(u_{1},\dots,u_{k},{\mathcal{U}}),

while

𝔼(Sb|𝒵=𝒰)=u1,,uk𝒰,distinctf(u1,,uk,𝒰)(u1,,uk are coloured black)=pk(u1,,uk𝒰,distinctf(u1,,uk,𝒰))=pk𝔼(S|𝒵=𝒰).\begin{array}[]{rcl}{\mathbb{E}}\left(S_{\text{b}}|{\mathcal{Z}}={\mathcal{U}}\right)&=&\displaystyle\sum_{u_{1},\dots,u_{k}\in{\mathcal{U}},\atop\text{distinct}}f(u_{1},\dots,u_{k},{\mathcal{U}})\cdot{\mathbb{P}}(u_{1},\dots,u_{k}\text{ are coloured black})\\[17.22217pt] &=&\displaystyle p^{k}\cdot\left(\sum_{u_{1},\dots,u_{k}\in{\mathcal{U}},\atop\text{distinct}}f(u_{1},\dots,u_{k},{\mathcal{U}})\right)=p^{k}\cdot{\mathbb{E}}\left(S|{\mathcal{Z}}={\mathcal{U}}\right).\end{array}

This holds for every locally finite 𝒰{\mathcal{U}}, which implies

𝔼Sb=pk𝔼S.{\mathbb{E}}S_{\text{b}}=p^{k}\cdot{\mathbb{E}}S.

The result now follows by applying the Slivniak-Mecke formula to 𝔼S{\mathbb{E}}S. \blacksquare

2.3 Hyperbolic Poisson-Voronoi tessellations

For 𝒰𝔻{\mathcal{U}}\subseteq\mathbb{D} a countable point set and u𝒰u\in{\mathcal{U}} we will denote the corresponding hyperbolic Voronoi cell of uu by:

C(u;𝒰):={v𝔻:dist2(u,v)dist2(u,v) for all u𝒰}.\begin{array}[]{c}\displaystyle C(u;{\mathcal{U}}):=\{v\in\mathbb{D}:\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\leq\operatorname{dist}_{\mathbb{H}^{2}}(u^{\prime},v)\text{ for all $u^{\prime}\in{\mathcal{U}}$}\}.\end{array}

We will usually suppress the second argument, and just write C(u)C(u) for the Voronoi cell of uu; and 𝒰{\mathcal{U}} will usually be either a homogeneous hyperbolic Poisson process 𝒵{\mathcal{Z}} or 𝒵{o}{\mathcal{Z}}\cup\{o\}, such a Poisson process with the origin added in.

The hyperbolic Delaunay graph for 𝒰{\mathcal{U}} is the abstract combinatorial graph with vertex set 𝒰{\mathcal{U}} and an edge uuuu^{\prime} if the Voronoi cells C(u),C(u)C(u),C(u^{\prime}) meet. So we can alternatively view hyperbolic Poisson-Voronoi percolation as site-percolation on the hyperbolic Poisson-Delaunay graph. We say that u,u𝒰u,u^{\prime}\in{\mathcal{U}} are adjacent if they share an edge in the Poisson-Delaunay graph (in other words, the Voronoi cells C(u),C(u)C(u),C(u^{\prime}) have at least one point in common). In this case we will also say that u,uu,u^{\prime} are neighbours.

We point out that if vC(u)C(u)v\in C(u)\cap C(u^{\prime}) then it must hold that dist2(v,u)=dist2(v,u)\operatorname{dist}_{\mathbb{H}^{2}}(v,u)=\operatorname{dist}_{\mathbb{H}^{2}}(v,u^{\prime}) and B2(v,dist2(v,u))𝒰=B_{\mathbb{H}^{2}}(v,\operatorname{dist}_{\mathbb{H}^{2}}(v,u))\cap{\mathcal{U}}=\emptyset. In other words u,uu,u^{\prime} are adjacent if and only if there is hyperbolic disk BB such that u,uBu,u^{\prime}\in\partial B and B𝒰=B\cap{\mathcal{U}}=\emptyset. Although we shall not be using this in the present paper, it may be instructive to the reader to point out the following. By this last observation, Fact 2 and some relatively straightforward probabilistic considerations : if 𝒵{\mathcal{Z}} is a homogeneous, hyperbolic Poisson process then, almost surely, the Voronoi tessellations for 𝒵{\mathcal{Z}} with respect to the ordinary, Euclidean metric and the Voronoi tessellation with respect to the hyperbolic metric as defined above have the same combinatorial structure, in the sense that z,z𝒵z,z^{\prime}\in{\mathcal{Z}} are adjacent in the Euclidean tessellation if and only if they are adjacent in the hyperbolic tessellation. See Lemma 5 in our previous paper [24].

In many of our arguments below we will be considering the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\}, a homogeneous, hyperbolic Poisson process with the origin added in. We refer to the origin as the typical point and to its Voronoi cell C(o)C(o) as the typical cell. See Figure 4 for computer simulations of the typical cell, shown in the Poincaré disk model, for various choices of λ\lambda.

Refer to caption
Refer to caption
Refer to caption
Figure 4: Computer simulations of the typical cell (highlighted), shown in the Poincaré disk model. Left: λ=1\lambda=1, middle : λ=110\lambda=\frac{1}{10}, right: λ=150\lambda=\frac{1}{50}.

We define the typical degree DD as

D:=|{z𝒵:o,z are neighbours in the Voronoi tessellation for 𝒵{o}}|.D:=\left|\left\{z\in{\mathcal{Z}}:\text{$o,z$ are neighbours in the Voronoi tessellation for ${\mathcal{Z}}\cup\{o\}$}\right\}\right|. (8)

The typical degree for homogeneous, hyperbolic Poisson processes has previously been studied by Isokawa [27]. She gave the following exact formula for its expectation, valid for all values of the intensity parameter λ>0\lambda>0.

Theorem 6 (Isokawa’s formula)

If DD is as given by (8) then

𝔼D=6+3πλ.{\mathbb{E}}D=6+\frac{3}{\pi\lambda}.

What is important for us is the asymptotics 𝔼D3πλ{\mathbb{E}}D\sim\frac{3}{\pi\lambda} as λ0\lambda\searrow 0. An independent, alternative derivation of these asymptotics, using conceptually simple arguments that are similar to the ones we will use in the present paper, is given in Chapter 3 of the doctoral thesis of the first author [23].

It may be instructive for the reader (but does not enter into the arguments in the present paper) to point out that as λ\lambda\to\infty the expected typical degree tends to 6, the value for the expected typical degree in planar, Euclidean Poisson-Voronoi tessellations [34].

It may also be instructive to have another look at the simulations in Figure 4 (and 1), which indeed suggest as λ\lambda decreases the typical cell tends to have more adjacencies.

3 Proofs

3.1 Drawing trees in the hyperbolic plane

We will find it useful to consider graphs embedded in the hyperbolic plane. From now on every graph TT in the rest of the paper, has vertices V(T)𝔻V(T)\subseteq\mathbb{D} that are points in the Poincaré disk, and we identify each edge uvE(T)uv\in E(T) with the (hyperbolic) geodesic line segment between uu and vv.

Definition 7

For ρ,w,ϑ>0\rho,w,\vartheta>0 we say that TT is a (ρ,w,ϑ)(\rho,w,\vartheta)-tree if it is connected and

  1. (i)

    For every edge uvE(T)uv\in E(T) we have ρw<dist2(u,v)<ρ+w\rho-w<\operatorname{dist}_{\mathbb{H}^{2}}(u,v)<\rho+w;

  2. (ii)

    If the edges uv,uwE(T)uv,uw\in E(T) have a common endpoint uu, then the angle vuw\angle vuw they make at uu is at least ϑ\vartheta.

We will speak of a (ρ,w,ϑ)(\rho,w,\vartheta)-path if the (ρ,w,ϑ)(\rho,w,\vartheta)-tree TT is a path.

The following proposition verifies that we are justified in using the word “tree” : a (ρ,w,ϑ)(\rho,w,\vartheta)-tree is also a tree in the graph theoretical sense, at least when the ρ\rho parameter is sufficiently large.

Proposition 8

For every w,ϑ>0w,\vartheta>0 there exists ρ0=ρ0(w,ϑ)\rho_{0}=\rho_{0}(w,\vartheta) such that, for all ρρ0\rho\geq\rho_{0}, every (ρ,w,ϑ)(\rho,w,\vartheta)-tree is acyclic.

We have to postpone the proof until some necessary preparations are out of the way. If TT is a (ρ,w,ϑ)(\rho,w,\vartheta)-tree then clearly the hyperbolic distance dist2(u,v)\operatorname{dist}_{\mathbb{H}^{2}}(u,v) between any two vertices u,vV(T)u,v\in V(T) is upper bounded by distT(u,v)(ρ+w)\operatorname{dist}_{T}(u,v)\cdot(\rho+w). Here and in the rest of the paper, distT\operatorname{dist}_{T} denotes the graph distance, i.e. the number of edges on the (unique) u,vu,v-path in TT.

The following proposition shows that for ρ\rho sufficiently large this trivial upper bound is close to being tight.

Proposition 9

For every w,ϑ>0w,\vartheta>0 there exists ρ0=ρ0(w,ϑ)\rho_{0}=\rho_{0}(w,\vartheta) and K=K(w,ϑ)K=K(w,\vartheta) such that, for all ρρ0\rho\geq\rho_{0}, every (ρ,w,ϑ)(\rho,w,\vartheta)-tree and for every two vertices u,vV(T)u,v\in V(T):

dist2(u,v)distT(u,v)(ρK).\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq\operatorname{dist}_{T}(u,v)\cdot(\rho-K).

Again we postpone the proof until we have made some additional observations.

If GG is a graph and uvE(G)uv\in E(G) then we denote by GuvG_{u\setminus v} the connected component of G{v}G\setminus\{v\} that contains uu. In other words, GuvG_{u\setminus v} is the subgraph induced on all nodes other than vv that can be reached from uu using a path that avoids vv.

The following observation is an important ingredient in the proof of our main result.

Proposition 10

For all w,ϑ1,ϑ2>0w,\vartheta_{1},\vartheta_{2}>0 there exist ρ0=ρ0(w,ϑ1,ϑ2)\rho_{0}=\rho_{0}(w,\vartheta_{1},\vartheta_{2}) and h=h(w,ϑ1,ϑ2)h=h(w,\vartheta_{1},\vartheta_{2}) such that, for all ρρ0,\rho\geq\rho_{0}, every (ρ,w,ϑ1)(\rho,w,\vartheta_{1})-tree TT and every edge uvE(T)uv\in E(T):

xV(Tuv)B2(x,ρ+w)B2(v,h)sect(v,u,ϑ2).\bigcup_{x\in V(T_{u\setminus v})}B_{\mathbb{H}^{2}}(x,\rho+w)\subseteq B_{\mathbb{H}^{2}}(v,h)\cup\operatorname{sect}\left(v,u,\vartheta_{2}\right).

Again we have to postpone the proof until we have made some more preparatory observations. The next few Lemma’s in this section are technical (hyperbolic) geometric considerations that are intermediate steps in the proof of the three propositions above. When reading the paper for the first time, the reader may wish to only read the lemma statements and skip over the proofs.

We start with a relatively straightforward, but for us useful, consequence of the hyperbolic cosine rule.

Lemma 11

For every γ0>0,\gamma_{0}>0, there exists K=K(γ0)K=K(\gamma_{0}) such that the following holds for every hyperbolic triangle. Let the lengths of the sides be a,b,a,b, and cc and the respective opposing angles α,β,γ\alpha,\beta,\gamma as in Figure 2. If γγ0\gamma\geq\gamma_{0} then

ca+bK.c\geq a+b-K.

Proof. We can assume without loss of generality that γ0π/2\gamma_{0}\leq\pi/2. For any hyperbolic triangle satisfying the hypothesis of the lemma, we have

ec\displaystyle e^{c} cosh(c)\displaystyle\geq\cosh(c)
=cosh(a)cosh(b)sinh(a)sinh(b)cos(γ)\displaystyle=\cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma)
cosh(a)cosh(b)sinh(a)sinh(b)cos(γ0)\displaystyle\geq\cosh(a)\cosh(b)-\sinh(a)\sinh(b)\cos(\gamma_{0})
ea+b4(1cos(γ0)).\displaystyle\geq\frac{e^{a+b}}{4}(1-\cos(\gamma_{0})).

The second line is the hyperbolic law of cosines (Lemma 3). The third line follows as 0<γ<π0<\gamma<\pi and cos(.)\cos(.) is decreasing on [0,π)[0,\pi). The fourth line uses that cosh(a)cosh(b)ea+b4\cosh(a)\cosh(b)\geq\frac{e^{a+b}}{4} and sinh(a)sinh(b)ea+b4\sinh(a)\sinh(b)\leq\frac{e^{a+b}}{4} and γ0π/2\gamma_{0}\leq\pi/2 (by assumption). Taking logs, we find

ca+b+ln(1cosγ04)=:a+bK.c\geq a+b+\ln\left(\frac{1-\cos\gamma_{0}}{4}\right)=:a+b-K.

\blacksquare

Next, we show that if u,vu,v are at least r+d0r+d_{0} apart with d0d_{0} a (large) constant and rr arbitrary then, from the point of view of vv (i.e. if we isometrically map vv to the origin of the Poincaré disk), a ball of large constant radius rr around uu will be contained in a sector of small opening angle.

Lemma 12

For every ϑ>0\vartheta>0 there exists a d0=d0(ϑ)>0d_{0}=d_{0}(\vartheta)>0 such that, for all u,v2u,v\in\mathbb{H}^{2} and r>0r>0, if dist2(u,v)>r+d0\operatorname{dist}_{\mathbb{H}^{2}}(u,v)>r+d_{0} then

B2(u,r)sect(v,u,ϑ).B_{\mathbb{H}^{2}}(u,r)\subseteq\operatorname{sect}\left(v,u,\vartheta\right).

(See Figure 5.)

Refer to caption
Figure 5: An example of B2(u,r)sect(v,u,ϑ)B_{\mathbb{H}^{2}}(u,r)\subseteq\operatorname{sect}\left(v,u,\vartheta\right).

Proof. We let d0d_{0} be a large constant, to be determined in the course of the proof. Applying a suitable isometry if needed, we can assume without loss of generality that v=ov=o is the origin and uu lies on the positive xx-axis. We recall that B:=B2(u,r)B:=B_{\mathbb{H}^{2}}(u,r) is also a Euclidean disk. Its Euclidean center must lie on the xx-axis. (This is for instance easily seen by noting that the reflection in the xx-axis is a hyperbolic isometry that fixes uu, and hence leaves BB invariant.)

By the triangle inequality and the assumption that dist2(u,v)>r+d0\operatorname{dist}_{\mathbb{H}^{2}}(u,v)>r+d_{0}, we have

dist2(v,z)>d0 for all zB\operatorname{dist}_{\mathbb{H}^{2}}(v,z)>d_{0}\text{ for all $z\in B$. }

By (1), for each zBz\in B we have

1>z>tanh(d0/2)1δ,1>\|z\|>\tanh(d_{0}/2)\geq 1-\delta,

where δ\delta is a small constant to be determined shortly, and the final inequality holds provided we have chosen d0d_{0} sufficiently large. Since BB is contained in the annulus 𝔻B2(o,1δ)\mathbb{D}\setminus B_{\mathbb{R}^{2}}(o,1-\delta), its Euclidean radius is no more than δ/2\delta/2.

As its Euclidean centre lies on the xx-axis, it is clear that δ=δ(ϑ)\delta=\delta(\vartheta) can be chosen so that Dsect(v,u,ϑ)D\subseteq\operatorname{sect}\left(v,u,\vartheta\right). \blacksquare

Our next observation is another intermediate step in the proofs of Propositions 89 and 10. It will be used only in the proof of Lemma 14 that immediately succeeds it in the text.

Lemma 13

Suppose the (Euclidean) circle CC intersects the unit circle 𝔻\partial\mathbb{D} at right angles, and intersects the xx-axis at an angle ϑ>0\vartheta>0. If rr denotes the (Euclidean) radius of CC and uu denotes the intersection point of CC and the xx-axis that falls inside 𝔻\mathbb{D}, then

r=1u22usinϑ.r=\frac{1-\|u\|^{2}}{2\|u\|\sin\vartheta}.

Proof. Let cc denote the centre of CC. Since CC hits the unit circle at a right angle, we have c=1+r2\|c\|=\sqrt{1+r^{2}}. In the Euclidean triangle with corners o,u,co,u,c, the angle at uu equals π/2+ϑ\pi/2+\vartheta. See Figure 6.

Refer to caption
𝔻\partial\mathbb{D}ϑ\vartheta

11

rr

rr

1+r2\sqrt{1+r^{2}}

oouuccxx-axisCC
Figure 6: Illustration of the proof of Lemma 13.

Applying the Euclidean cosine rule to the triangle with corners o,u,co,u,c we find

1+r2=u2+r22urcos(π/2+ϑ)=u2+r2+2ursin(ϑ).1+r^{2}=\|u\|^{2}+r^{2}-2\|u\|r\cos(\pi/2+\vartheta)=\|u^{2}\|+r^{2}+2\|u\|r\sin(\vartheta).

The claimed expression follows by reorganizing this last identity. \blacksquare

The next lemma makes the following perhaps rather counterintuitive observation. While 𝔻sect(u,v,ϑ)\mathbb{D}\setminus\operatorname{sect}\left(u,v,\vartheta\right) looks “large” from the vantage point of uu (i.e. if we isometrically map uu to the origin of the Poincaré disk, the sector looks only like a small “slice”), provided u,vu,v are sufficiently far apart, from the vantage point of vv the set 𝔻sect(u,v,ϑ)sect(v,u,ϑ)\mathbb{D}\setminus\operatorname{sect}\left(u,v,\vartheta\right)\subseteq\operatorname{sect}\left(v,u,\vartheta\right) is actually contained in a small “slice”.

Lemma 14

For all ϑ>0\vartheta>0 there exists d0=d0(ϑ)>0d_{0}=d_{0}(\vartheta)>0 such that, for all u,v𝔻u,v\in\mathbb{D}, if dist2(u,v)d0\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq d_{0}, then 𝔻\sect(u,v,ϑ)sect(v,u,ϑ)\mathbb{D}\backslash\operatorname{sect}\left(u,v,\vartheta\right)\subseteq\operatorname{sect}\left(v,u,\vartheta\right).

(See Figure 7.)

Refer to caption
Figure 7: The orange region is 𝔻\sect(u,v,ϑ)\mathbb{D}\backslash\operatorname{sect}\left(u,v,\vartheta\right). The union of the purple and the orange region is sect(v,u,ϑ)\operatorname{sect}\left(v,u,\vartheta\right).

Proof. Applying a suitable isometry if needed, we can assume without loss of generality that v=ov=o is the origin and uu lies on the positive xx-axis.

The boundary of sect(u,v,ϑ)\operatorname{sect}\left(u,v,\vartheta\right) consists of two rays emanating from uu that each make an angle ϑ\vartheta with the hyperbolic line through uu and vv at uu (and a circle segment that is part of the unit circle 𝔻\partial\mathbb{D}). In Euclidean terms, these two rays are circle segments of circles that each meet the xx-axis in uu at an angle of ϑ\vartheta, and intersect the boundary of the unit circle 𝔻\partial\mathbb{D} at right angles. (See Figure 7.) Let us call these circles C1,C2C_{1},C_{2}. Both circles meet the xx-axis in uu at angle ϑ\vartheta. Hence, by Lemma 13, both have Euclidean radius (1u2)/(2usinϑ)(1-\|u\|^{2})/(2\|u\|\sin\vartheta).

We have utanh(d0/2)\|u\|\geq\tanh(d_{0}/2) which approaches one as d0d_{0} approaches infinity. In particular, for any constant δ>0\delta>0, we can choose the constant d0d_{0} so that both C1,C2C_{1},C_{2} have Euclidean radius <δ<\delta, and both intersect the xx-axis at a point within distance δ\delta of (1,0)(1,0). In other words, we’ll have

𝔻sect(u,v,ϑ)B2((1,0),3δ)𝔻.\mathbb{D}\setminus\operatorname{sect}\left(u,v,\vartheta\right)\subseteq B_{\mathbb{R}^{2}}((1,0),3\delta)\cap\mathbb{D}.

Clearly, if we take δ=δ(ϑ)\delta=\delta(\vartheta) small enough, it now follows that 𝔻sect(u,v,ϑ)sect(u,v,ϑ)\mathbb{D}\setminus\operatorname{sect}\left(u,v,\vartheta\right)\subseteq\operatorname{sect}\left(u,v,\vartheta\right), as desired. \blacksquare

As our final preparation for the proofs of Propositions 8 and 9, we show that if TT is a (ρ,w,ϑ)(\rho,w,\vartheta)-tree and uvE(T)uv\in E(T) an edge then the vertices of TuvT_{u\setminus v} are all contained in a small sector from the vantage point of vv, provided ρ\rho is chosen sufficiently large.

Lemma 15

For every w,ϑ1,ϑ2>0w,\vartheta_{1},\vartheta_{2}>0 there exists a ρ0=ρ0(w,ϑ1,ϑ2)\rho_{0}=\rho_{0}(w,\vartheta_{1},\vartheta_{2}) such that, for every ρρ0\rho\geq\rho_{0}, every (ρ,w,ϑ1)(\rho,w,\vartheta_{1})-tree TT and every edge uvE(T)uv\in E(T):

V(Tuv)sect(v,u,ϑ2).V(T_{u\setminus v})\subseteq\operatorname{sect}\left(v,u,\vartheta_{2}\right).

Proof. We can and do assume, without loss of generality, that ϑ1<11000\vartheta_{1}<\frac{1}{1000} and ϑ2<ϑ1/2\vartheta_{2}<\vartheta_{1}/2; and we set ρ0:=w+d0\rho_{0}:=w+d_{0} with d0=d0(ϑ2)d_{0}=d_{0}(\vartheta_{2}) as provided by Lemma 14.

We will use induction on the number of edges m:=|E(T)|m:=|E(T)| of TT. The statement is clearly true when m1m\leq 1. Let us thus assume that, for some mm\in\mathbb{N}, the statement holds for all (ρ,w,ϑ1)(\rho,w,\vartheta_{1})-trees with <m<m edges, let TT be an arbitrary (ρ,w,ϑ1)(\rho,w,\vartheta_{1})-tree with mm edges, and pick an arbitrary edge uvE(T)uv\in E(T). If uu is a leaf then Tuv={u}T_{u\setminus v}=\{u\} and the statement is trivial. So we can and do assume uu has at least two neighbours. Let us denote the neighbours of uu as N(u)={v,x1,,xk}N(u)=\{v,x_{1},\dots,x_{k}\} and let TiT_{i} denote the connected component of TuxiT\setminus ux_{i} containing xix_{i}. By the induction hypothesis V(Ti)sect(u,xi,ϑ2)V(T_{i})\subseteq\operatorname{sect}\left(u,x_{i},\vartheta_{2}\right) for i=1,,ki=1,\dots,k. Since the line segments uvuv and uxiux_{i} make an angle ϑ1\geq\vartheta_{1}, it follows that sect(u,v,ϑ2)sect(u,xi,ϑ2)=\operatorname{sect}\left(u,v,\vartheta_{2}\right)\cap\operatorname{sect}\left(u,x_{i},\vartheta_{2}\right)=\emptyset. In other words

V(Tuv){u}i=1kV(Ti)𝔻sect(u,v,ϑ2)sect(v,u,ϑ2),V(T_{u\setminus v})\subseteq\{u\}\cup\bigcup_{i=1}^{k}V(T_{i})\subseteq\mathbb{D}\setminus\operatorname{sect}\left(u,v,\vartheta_{2}\right)\subseteq\operatorname{sect}\left(v,u,\vartheta_{2}\right),

where the last inclusion follows by Lemma 14 and our choice of ρ0\rho_{0}. (Using that dist2(v,u)ρ0wd0\operatorname{dist}_{\mathbb{H}^{2}}(v,u)\geq\rho_{0}-w\geq d_{0} by choice of ρ0\rho_{0}.) \blacksquare

Proof of Proposition 8. We let ρ0\rho_{0} be a sufficiently large constant, to be determined during the course of the proof. Let TT be an arbitrary (ρ,w,ϑ)(\rho,w,\vartheta)-tree for some ρρ0\rho\geq\rho_{0}, and suppose it contains a cycle CC. Observe that as CC is a connected subgraph of TT, it is also a (r,w,ϑ)(r,w,\vartheta)-tree. Let v1,v2,v3v_{1},v_{2},v_{3} be three vertices that are consecutive on CC. Clearly Cv1v2=Cv2C_{v_{1}\setminus v_{2}}=C\setminus v_{2} is just CC with v2v_{2} and both incident edges removed. We let ϑ>0\vartheta^{\prime}>0 be a sufficiently small constant, to be determined more precisely shortly. By the previous lemma,

V(C){v2}sect(v2,v1,ϑ),V(C)\setminus\{v_{2}\}\subseteq\operatorname{sect}\left(v_{2},v_{1},\vartheta^{\prime}\right), (9)

assuming we chose ρ0\rho_{0} sufficiently large. By symmetry (considering Cv3v2C_{v_{3}\setminus v_{2}}) we also have

V(C){v2}sect(v2,v3,ϑ).V(C)\setminus\{v_{2}\}\subseteq\operatorname{sect}\left(v_{2},v_{3},\vartheta^{\prime}\right). (10)

Since v1v2v3>ϑ\angle v_{1}v_{2}v_{3}>\vartheta, provided we chose ϑ\vartheta^{\prime} sufficiently small, we have that

sect(v2,v1,ϑ)sect(v2,v3,ϑ)=.\operatorname{sect}\left(v_{2},v_{1},\vartheta^{\prime}\right)\cap\operatorname{sect}\left(v_{2},v_{3},\vartheta^{\prime}\right)=\emptyset.

Combining this with (9) and (10), would imply that V(C)={v2}V(C)=\{v_{2}\}, contradicting our assumption that CC is a cycle. \blacksquare

Proof of Proposition 9. We let ρ0,K\rho_{0},K be sufficiently large constants, to be specified more precisely during the course of the proof. We will use induction on n:=distT(u,v)n:=\operatorname{dist}_{T}(u,v).

The base case, when n=1n=1 is trivial by definition of (ρ,w,ϑ)(\rho,w,\vartheta)-tree – provided we chose KwK\geq w.

Let us then assume the statement is true for n1n-1 and let u=v0,,vn=vu=v_{0},\dots,v_{n}=v be a (u,v)(u,v)-path in TT. By Proposition 8, having chosen ρ0\rho_{0} sufficiently large, we know there is precisely one path between any pair of vertices. By the induction hypothesis

dist2(v1,vn)(n1)(ρK),\operatorname{dist}_{\mathbb{H}^{2}}(v_{1},v_{n})\geq(n-1)\cdot(\rho-K),

and by Lemma 15, assuming we chose ρ0\rho_{0} sufficiently large, we have that

vV(Tv2v1)sect(v1,v2,ϑ/2).v\in V(T_{v_{2}\setminus v_{1}})\subseteq\operatorname{sect}\left(v_{1},v_{2},\vartheta/2\right).

By definition of (ρ,w,ϑ)(\rho,w,\vartheta)-tree we have uv1v2>ϑ\angle uv_{1}v_{2}>\vartheta. It follows that uv1v>ϑ/2\angle uv_{1}v>\vartheta/2. Applying Lemma 11, we find

dist2(u,v)dist2(u,v1)+dist2(v1,v)K,\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq\operatorname{dist}_{\mathbb{H}^{2}}(u,v_{1})+\operatorname{dist}_{\mathbb{H}^{2}}(v_{1},v)-K^{\prime},

for some constant K=K(ϑ/2)K^{\prime}=K^{\prime}(\vartheta/2) provided by Lemma 11. Hence,

dist2(u,v)(ρw)+(n1)(ρK)Kn(ρC),\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq(\rho-w)+(n-1)\cdot(\rho-K)-K^{\prime}\geq n\cdot(\rho-C),

assuming (without loss of generality) we have chosen K>w+KK>w+K^{\prime}. \blacksquare

The final ingredient we need for the proof of Proposition 10 is the following observation. It states that if the distance between uu and vv is between rwr-w and r+wr+w, where ww is constant and rr arbitrary, then a ball of radius r+wr+w around uu is contained in the union of a ball of constant radius hh around vv and a sector of constant opening angle ϑ\vartheta

Lemma 16

For every w,ϑ>0w,\vartheta>0 there exists h=h(w,ϑ)h=h(w,\vartheta) such that for all r>0r>0 and all u,v𝔻u,v\in\mathbb{D} with rw<dist2(u,v)<r+wr-w<\operatorname{dist}_{\mathbb{H}^{2}}(u,v)<r+w we have

B2(u,r+w)B2(v,h)sect(v,u,ϑ).B_{\mathbb{H}^{2}}(u,r+w)\subseteq B_{\mathbb{H}^{2}}(v,h)\cup\operatorname{sect}\left(v,u,\vartheta\right).

Proof. Applying a suitable isometry if needed, we can assume without loss of generality v=ov=o is the origin and that uu lies on the positive xx-axis.

Refer to caption
Figure 8: On the right, the disk with the black boundary is B2(u,r+w)B_{\mathbb{H}^{2}}(u,r+w). In the centre, the gold disk is DD. On the left, the purple sector is sect(v,u,ϑ)\operatorname{sect}\left(v,u,\vartheta\right) and the disk with the green boundary is B2(v,h)B_{\mathbb{H}^{2}}(v,h).

We recall that the hyperbolic disk B2(u,r+w)B_{\mathbb{H}^{2}}(u,r+w) is also a Euclidean disk B2(z,t)B_{\mathbb{R}^{2}}(z,t). The Euclidean center zz must lie on the xx-axis (an easy way to see this is that reflection in the xx-axis is a 2\mathbb{H}^{2}-isometry that leaves B2(u,r+w)B_{\mathbb{H}^{2}}(u,r+w) invariant). Let qmin=(xmin,0)q_{\min}=(x_{\min},0) and qmax=(xmax,0)q_{\max}=(x_{\max},0) be the points where the circle B2(u,r+w)\partial B_{\mathbb{H}^{2}}(u,r+w) intersects the xx-axis. See Figure 8, left. (So the Euclidean center of B2(u,r+w)B_{\mathbb{H}^{2}}(u,r+w) is the midpoint (qmin+qmax)/2(q_{\min}+q_{\max})/2 between these two points.) Since v=(0,0)B2(u,r+w)v=(0,0)\in B_{\mathbb{H}^{2}}(u,r+w) we have xmin0xmaxx_{\min}\leq 0\leq x_{\max}. As the points qmin,v=o,uq_{\min},v=o,u lie on the xx-axis, which is a hyperbolic line, we have

dist2(qmin,u)=dist2(qmin,v)+dist2(u,v),\operatorname{dist}_{\mathbb{H}^{2}}(q_{\min},u)=\operatorname{dist}_{\mathbb{H}^{2}}(q_{\min},v)+\operatorname{dist}_{\mathbb{H}^{2}}(u,v),

giving

dist2(qmin,o)=dist2(qmin,u)dist2(u,v)r+w(rw)=2w.\operatorname{dist}_{\mathbb{H}^{2}}(q_{\min},o)=\operatorname{dist}_{\mathbb{H}^{2}}(q_{\min},u)-\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\leq r+w-(r-w)=2w.

By (1) it follows that qmintanh(w)<1\|q_{\min}\|\leq\tanh(w)<1. In other words, tanh(w)xmin0-\tanh(w)\leq x_{\min}\leq 0.

We set

qmin:=(tanh(w),0),qmax:=(1,0),q_{\min}^{\prime}:=(-\tanh(w),0),\quad q_{\max}^{\prime}:=(1,0),

and let

D:=B2((1tanh(w)2,0),1+tanh(w)2),D:=B_{\mathbb{R}^{2}}\left(\left(\frac{1-\tanh(w)}{2},0\right),\frac{1+\tanh(w)}{2}\right),

be the Euclidean disk with center (qmin+qmax)/2=((1tanh(w))/2,0)(q_{\min}^{\prime}+q_{\max}^{\prime})/2=\left((1-\tanh(w))/2,0\right) and radius (1+tanh(w))/2(1+\tanh(w))/2. Put differently, DD is the disk with qmin,qmaxq_{\min}^{\prime},q_{\max}^{\prime} on its boundary, that meets the xx-axis at a right angle at both points. This is not a hyperbolic disk (it is what is called a horocycle), but we do have that B2(u,r+w)D𝔻B_{\mathbb{H}^{2}}(u,r+w)\subseteq D\subseteq\mathbb{D}. See Figure 8, middle.

Let R:=Dsect(v,u,ϑ)R:=D\setminus\operatorname{sect}\left(v,u,\vartheta\right) be the part of DD that is not contained in the sector sect(v,u,ϑ)\operatorname{sect}\left(v,u,\vartheta\right). Then

h:=sup{dist2(z,o):zR}<,h:=\sup\{\operatorname{dist}_{\mathbb{H}^{2}}(z,o):z\in R\}<\infty,

since all points of RR are at least some positive Euclidean distance away from the boundary of the unit disk. See Figure 8, right. Evidently we have

B2(u,r+w)DRsect(v,u,ϑ)B2(v,h)sect(v,u,ϑ),B_{\mathbb{R}^{2}}(u,r+w)\subseteq D\subseteq R\cup\operatorname{sect}\left(v,u,\vartheta\right)\subseteq B_{\mathbb{H}^{2}}(v,h)\cup\operatorname{sect}\left(v,u,\vartheta\right),

as desired. (Note that hh depends only on ww and ϑ\vartheta.) \blacksquare

Proof of Proposition 10. We let ρ0\rho_{0} and hh be large constants, to be determined in the course of the proof. Let ρρ0\rho\geq\rho_{0}, TT be an arbitrary (ρ,w,ϑ1)(\rho,w,\vartheta_{1})-tree, uvE(T)uv\in E(T) an arbitrary edge and xTuvx\in T_{u\setminus v} an arbitrary vertex.

We first assume that xux\neq u. Then distT(u,x)2\operatorname{dist}_{T}(u,x)\geq 2 by Proposition 8, assuming without loss of generality we have chosen rho0rho_{0} sufficiently large. And, assuming ρ0\rho_{0} was chosen appropriately, by Proposition 9 we have

dist2(v,x)2(ρK),\operatorname{dist}_{\mathbb{H}^{2}}(v,x)\geq 2(\rho-K),

where KK is as provided by the proposition.

By Lemma 15, provided we chose ρ0\rho_{0} appropriately, we have

xsect(v,u,ϑ2/2).x\in\operatorname{sect}\left(v,u,\vartheta_{2}/2\right).

Applying Lemma 12, provided we chose ρ0\rho_{0} sufficiently large, we have

B2(x,r+w)sect(v,x,ϑ2/2)sect(v,u,ϑ2).B_{\mathbb{H}^{2}}(x,r+w)\subseteq\operatorname{sect}\left(v,x,\vartheta_{2}/2\right)\subseteq\operatorname{sect}\left(v,u,\vartheta_{2}\right).

Let us thus consider the situation when x=ux=u. In this case we can apply Lemma 16 to show that, provided we chose ρ0\rho_{0} and hh appropriately large,

B2(u,r+w)B2(v,h)sect(v,u,ϑ2).B_{\mathbb{H}^{2}}(u,r+w)\subseteq B_{\mathbb{H}^{2}}(v,h)\cup\operatorname{sect}\left(v,u,\vartheta_{2}\right).

This concludes the proof. \blacksquare

3.2 The upper bound

Here we will show the following proposition, which constitutes half of our main result.

Proposition 17

For every ε>0\varepsilon>0, there exists a λ0=λ0(ε)>0\lambda_{0}=\lambda_{0}(\varepsilon)>0 such that for all 0<λ<λ00<\lambda<\lambda_{0}, we have pc(λ)(1+ε)(π/3)λp_{c}(\lambda)\leq(1+\varepsilon)\cdot(\pi/3)\cdot\lambda.

In order to prove this, we add the origin oo to the Poisson point process 𝒵{\mathcal{Z}} and colour it black, and consider the black component of oo in the Delaunay graph for 𝒵{o}{\mathcal{Z}}\cup\{o\}. It suffices to show that, when p(1+ε)(π/3)λp\geq(1+\varepsilon)\cdot(\pi/3)\cdot\lambda and λ\lambda is sufficiently small, with positive probability the origin will be in an infinite black component. This is because all edges not involving the origin are also in the Delaunay graph for 𝒵{\mathcal{Z}}, and the origin a.s. has only finitely many neighbours (by Isokawa’s formula), so if the origin is in an infinite component then removing the origin may split its component into several components but at least one of these will be infinite.

For u,v𝔻u,v\in\mathbb{D} and r,w,ϑ>0r,w,\vartheta>0 we define the region

C(u,v,r,w,ϑ):={x𝔻:rw<dist2(u,x)<r+w and vux>ϑ}.C(u,v,r,w,\vartheta):=\{x\in\mathbb{D}:r-w<\operatorname{dist}_{\mathbb{H}^{2}}(u,x)<r+w\text{ and }\angle vux>\vartheta\}.
Refer to captionoo
Figure 9: The C-shaped region C(o,(1/2,0),3,1/2,π/4)C(o,(1/2,0),3,1/2,\pi/4).

We choose to use the notation C(.,.,.,.)C(.,.,.,.) because the region is shaped like the letter C, at least for some choices of the parameters. See Figure 9.

Given uv𝔻u\neq v\in\mathbb{D} and r,w,ϑ,h>0r,w,\vartheta,h>0 we define 𝒳=𝒳(u,v,r,w,ϑ,h){\mathcal{X}}={\mathcal{X}}(u,v,r,w,\vartheta,h) and X=X(u,v,r,w,ϑ,h)X=X(u,v,r,w,\vartheta,h) by:

𝒳:={z𝒵b:zC(u,v,r,w,ϑ) and;zuz>ϑ for all zz𝒵bC(u,v,r,w,ϑ), and; 𝒵B2(z,h)={z}, and;  a disk B such that u,zB,B𝒵=,diam2(B)<r+w.},{\mathcal{X}}:=\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}z\in C(u,v,r,w,\vartheta)\text{ and;}\\ \text{$\angle z^{\prime}uz>\vartheta$ for all $z^{\prime}\neq z\in{\mathcal{Z}}_{b}\cap C(u,v,r,w,\vartheta)$, and; }\\ {\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(z,h)=\{z\},\text{ and; }\\ \text{$\exists$ a disk $B$ such that $u,z\in\partial B,B\cap{\mathcal{Z}}=\emptyset,\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w$.}\end{array}\right\},
X:=|𝒳|.X:=|{\mathcal{X}}|. (11)

We point out that the probability distribution of XX does not depend on the choice of uu and vv (as long as they are distinct – otherwise the definition does not make sense).

Proposition 18

For every w,ϑ>0w,\vartheta>0 there exist h=h(w,ϑ)h=h(w,\vartheta) and r0=r0(w,ϑ)r_{0}=r_{0}(w,\vartheta) such that, for all λ>0,0<p<1\lambda>0,0<p<1 and r>r0r>r_{0}, the size of the black cluster of oo stochastically dominates the size of a Galton-Watson branching process with offspring distribution XX.

Proof. We can assume without loss of generality ϑ<π\vartheta<\pi (otherwise X=0X=0 by definition, and the theorem holds trivially). We set ϑϑ\vartheta^{\prime}\ll\vartheta be a small constant to be determined during the course of the proof, let h=h(w,ϑ,ϑ)h=h(w,\vartheta,\vartheta^{\prime}) be as provided by Proposition 10, and we let r0r_{0} be a large constant, to be determined more precisely during the course of the proof.

We consider an “exploration process” that iteratively constructs a (r,w,ϑ)(r,w,\vartheta)-tree TT rooted at the origin. Recall that rr will be chosen larger than r0r_{0}. At every iteration there will be a (finite) number of nodes, some of which are explored and some unexplored. The node vv having been explored means that all children of vv have already been added to the tree.

For each iteration i1i\geq 1, let TiT_{i} denote the tree we have constructed at the end of iteration ii. We let i{\mathscr{E}}_{i} denote the set of nodes of TiT_{i} that have been explored at the end of iteration ii and 𝒰i{\mathscr{U}}_{i} be the set of nodes in TiT_{i} that have not yet been explored; and we set 0=,𝒰0={o}{\mathcal{E}}_{0}=\emptyset,{\mathcal{U}}_{0}=\{o\}.

If in any iteration there are no unexplored nodes, i.e. 𝒰i={\mathcal{U}}_{i}=\emptyset, then the construction of the tree stops and the final result is T=TiT=T_{i}. Otherwise, if in each iteration we always have at least one unexplored node, then we continue indefinitely in which case of course T:=iTiT:=\bigcup_{i}T_{i}.

In the first iteration we add the set 𝒳1:=𝒳(o,(tanh(r/2),0),r,w,ϑ,h){\mathcal{X}}_{1}:={\mathcal{X}}(o,(\tanh(r/2),0),r,w,\vartheta,h) defined above to the tree, as the children of the origin. To be more precise we define T1T_{1} by V(T1)={o}𝒳1,E(T1)={oz:z𝒳1}V(T_{1})=\{o\}\cup{\mathcal{X}}_{1},E(T_{1})=\{oz:z\in{\mathcal{X}}_{1}\}. We set 1={o},𝒰1=𝒳1{\mathcal{E}}_{1}=\{o\},{\mathscr{U}}_{1}={\mathcal{X}}_{1}. In particular, the number of children of the origin will be X1=dXX_{1}\hskip 0.86108pt\raisebox{-0.43057pt}{$=$}\raisebox{4.30554pt}{{$\scriptstyle d$}}\hskip 3.44444ptX.

In any subsequent iteration ii, assuming 𝒰i1{\mathscr{U}}_{i-1}\neq\emptyset (otherwise the construction process will have finished), we pick an arbitrary unexplored node ui𝒰i1u_{i}\in{\mathscr{U}}_{i-1}, and we denote by viv_{i} its parent in Ti1T_{i-1}. The children of uiu_{i} will be 𝒳i:=𝒳(ui,vi,r,w,ϑ,h){\mathcal{X}}_{i}:={\mathcal{X}}(u_{i},v_{i},r,w,\vartheta,h), and we let Xi:=|𝒳i|X_{i}:=|{\mathcal{X}}_{i}| denote the number of children of uiu_{i}. We update by defining TiT_{i} via V(Ti)=V(Ti1)𝒳i,E(Ti)=E(Ti1){uiz:z𝒳i}V(T_{i})=V(T_{i-1})\cup{\mathcal{X}}_{i},E(T_{i})=E(T_{i-1})\cup\{u_{i}z:z\in{\mathcal{X}}_{i}\}, and setting i=i1{ui},𝒰i=(𝒰i1{ui})𝒳i{\mathcal{E}}_{i}={\mathcal{E}}_{i-1}\cup\{u_{i}\},{\mathscr{U}}_{i}=({\mathscr{U}}_{i-1}\setminus\{u_{i}\})\cup{\mathcal{X}}_{i}.

We point out that, since TiT_{i} clearly is a (r,w,ϑ)(r,w,\vartheta)-tree, it follows from Proposition 8 that we do not include any point twice (put differently that 𝒳i𝒳j={\mathcal{X}}_{i}\cap{\mathcal{X}}_{j}=\emptyset for all iji\neq j), provided r0r_{0} was chosen sufficiently large.

It remains to see that the size of the tree TT constructed via this process dominates the size of a Galton-Watson tree with offspring distribution XX. In order to do this, we will first establish that X1,X2,X_{1},X_{2},\dots are independent, and then that X2,X3,X_{2},X_{3},\dots are i.i.d. and finally that X2X_{2} stochastically dominates X=X1X=X_{1}.

We will consider, for each iteration ii, a (random) region RiR_{i} that will be revealed by the exploration process during the ii-th iteration. Here we mean by “reveal” that the exploration process will use information about 𝒵Ri{\mathcal{Z}}\cap R_{i} in order to determine 𝒳i{\mathcal{X}}_{i}, but – crucially – after the ii-th iteration the exploration process will not have uncovered any information on the status of the Poisson point process 𝒵{\mathcal{Z}} outside of R1RiR_{1}\cup\dots\cup R_{i}.

We define 𝒳i+𝒳i{\mathcal{X}}_{i}^{+}\supseteq{\mathcal{X}}_{i} by

𝒳i+:={z𝒵b:zC(ui,vi,r,w,ϑ) and;zuz>ϑ for all zz𝒵bC(ui,vi,r,w,ϑ).},{\mathcal{X}}_{i}^{+}:=\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}z\in C(u_{i},v_{i},r,w,\vartheta)\text{ and;}\\ \text{$\angle z^{\prime}uz>\vartheta$ for all $z^{\prime}\neq z\in{\mathcal{Z}}_{b}\cap C(u_{i},v_{i},r,w,\vartheta)$.}\end{array}\right\},

setting u1=o,v1:=(tanh(r/2),0)u_{1}=o,v_{1}:=(\tanh(r/2),0) so that the definition also applies for i=1i=1. We let Ti+T_{i}^{+} be the tree on vertex set {o}𝒳1+𝒳i+\{o\}\cup{\mathcal{X}}_{1}^{+}\cup\dots\cup{\mathcal{X}}_{i}^{+}, rooted at u1=ou_{1}=o and where the children of uju_{j} are 𝒳j+{\mathcal{X}}_{j}^{+} for each j=1,,ij=1,\dots,i. Clearly Ti+T_{i}^{+} is also a (r,w,ϑ)(r,w,\vartheta)-tree for each ii.

Given 𝒳1+,,𝒳i1+{\mathcal{X}}_{1}^{+},\dots,{\mathcal{X}}_{i-1}^{+} and uiu_{i}, we can determine 𝒳i+{\mathcal{X}}_{i}^{+} by revealing the status of the Poisson process inside C(ui,vi,r,w,ϑ)C(u_{i},v_{i},r,w,\vartheta). In order to now determine 𝒳i𝒳i+{\mathcal{X}}_{i}\subseteq{\mathcal{X}}_{i}^{+} we need to determine, for each z𝒳i+z\in{\mathcal{X}}_{i}^{+} whether B2(z,h)𝒵={z}B_{\mathbb{H}^{2}}(z,h)\cap{\mathcal{Z}}=\{z\} and there exists a disk BB such that ui,zB,𝒵B=u_{i},z\in\partial B,{\mathcal{Z}}\cap B=\emptyset and diam2(B)<r+w\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w. Any such BB is contained in B2(ui,r+w)B2(z,r+w)B_{\mathbb{H}^{2}}(u_{i},r+w)\cap B_{\mathbb{H}^{2}}(z,r+w). Hence, given 𝒳1+,,𝒳i1+{\mathcal{X}}_{1}^{+},\dots,{\mathcal{X}}_{i-1}^{+} and uiu_{i}, we can determine 𝒳i{\mathcal{X}}_{i} by revealing the status of the Poisson process inside the region

Ri:=zC(ui,vi,r,w,ϑ)(B2(ui,r+w)B2(z,r+w))z𝒳i+B2(z,h).R_{i}:=\bigcup_{z\in C(u_{i},v_{i},r,w,\vartheta)}\left(B_{\mathbb{H}^{2}}(u_{i},r+w)\cap B_{\mathbb{H}^{2}}(z,r+w)\right)\cup\bigcup_{z\in{\mathcal{X}}_{i}^{+}}B_{\mathbb{H}^{2}}(z,h).

By Proposition 10 (applied to the tree consisting of a single edge uizu_{i}z) and the choice of h,r0h,r_{0}, we have

B2(z,r+w)B2(ui,h)sect(ui,z,ϑ),B_{\mathbb{H}^{2}}(z,r+w)\subseteq B_{\mathbb{H}^{2}}(u_{i},h)\cup\operatorname{sect}\left(u_{i},z,\vartheta^{\prime}\right),

for all zC(ui,vi,r,w,ϑ)z\in C(u_{i},v_{i},r,w,\vartheta). It follows that

RiB2(ui,h)(zC(ui,vi,r,w,ϑ)sect(ui,z,ϑ))B2(ui,h)(𝔻sect(ui,vi,ϑϑ))=:Si,\begin{array}[]{rcl}R_{i}&\subseteq&\displaystyle B_{\mathbb{H}^{2}}(u_{i},h)\cup\left(\bigcup_{z\in C(u_{i},v_{i},r,w,\vartheta)}\operatorname{sect}\left(u_{i},z,\vartheta^{\prime}\right)\right)\\[17.22217pt] &\subseteq&\displaystyle B_{\mathbb{H}^{2}}(u_{i},h)\cup\left(\mathbb{D}\setminus\operatorname{sect}\left(u_{i},v_{i},\vartheta-\vartheta^{\prime}\right)\right)\\ &=:&\displaystyle S_{i},\end{array} (12)

(We point out that the arguments giving (12) also apply to the case when i=1i=1, showing that 𝒳1{\mathcal{X}}_{1} is determined completely by 𝒵S1{\mathcal{Z}}\cap S_{1}.)

On the other hand, we also have

RjB2(uj,r+w)z𝒳j+B2(z,h),R_{j}\subseteq B_{\mathbb{H}^{2}}(u_{j},r+w)\cup\bigcup_{z\in{\mathcal{X}}_{j}^{+}}B_{\mathbb{H}^{2}}(z,h),

for every jj. Hence

R1Ri1(j=1i1B2(ui,r+w))(v{o}𝒳1+𝒳i1+B2(v,h))B2(ui,h)(v{o}𝒳1+𝒳i1+,vuiB2(v,r+w))=B2(ui,h)(vV((Ti1+)viui)B2(v,r+w))B2(ui,h)sect(ui,vi,ϑ),\begin{array}[]{rcl}R_{1}\cup\dots\cup R_{i-1}&\subseteq&\displaystyle\left(\bigcup_{j=1}^{i-1}B_{\mathbb{H}^{2}}(u_{i},r+w)\right)\cup\left(\bigcup_{v\in\{o\}\cup{\mathcal{X}}_{1}^{+}\cup\dots\cup{\mathcal{X}}_{i-1}^{+}}B_{\mathbb{H}^{2}}(v,h)\right)\\[25.83325pt] &\subseteq&\displaystyle B_{\mathbb{H}^{2}}(u_{i},h)\cup\left(\bigcup_{v\in\{o\}\cup{\mathcal{X}}_{1}^{+}\cup\dots\cup{\mathcal{X}}_{i-1}^{+},\atop v\neq u_{i}}B_{\mathbb{H}^{2}}(v,r+w)\right)\\[25.83325pt] &=&\displaystyle B_{\mathbb{H}^{2}}(u_{i},h)\cup\left(\bigcup_{v\in V((T_{i-1}^{+})_{v_{i}\setminus u_{i}})}B_{\mathbb{H}^{2}}(v,r+w)\right)\\[25.83325pt] &\subseteq&\displaystyle B_{\mathbb{H}^{2}}(u_{i},h)\cup\operatorname{sect}\left(u_{i},v_{i},\vartheta^{\prime}\right),\end{array} (13)

where we apply Proposition 10 in the last line, and we assume we chose r0r_{0} sufficiently large.

Combining (12) and (13), having chosen ϑ\vartheta^{\prime} sufficiently small, we see that

(R1Ri1)SiB2(ui,h).(R_{1}\cup\dots\cup R_{i-1})\cap S_{i}\subseteq B_{\mathbb{H}^{2}}(u_{i},h).

Moreover, for i>1i>1, by construction of the exploration process we have

𝒵B2(ui,h)={ui}.{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(u_{i},h)=\{u_{i}\}.

Hence, for i>1i>1, given 𝒳1,𝒳1+,,𝒳i1,𝒳i1+,u1,,ui{\mathcal{X}}_{1},{\mathcal{X}}_{1}^{+},\dots,{\mathcal{X}}_{i-1},{\mathcal{X}}_{i-1}^{+},u_{1},\dots,u_{i}, the random set 𝒳i{\mathcal{X}}_{i} is completely determined by 𝒵Si{\mathcal{Z}}\cap S_{i}^{\prime} where

Si:=SiB2(ui,h)=𝔻(B2(ui,h)sect(ui,vi,ϑϑ)).S_{i}^{\prime}:=S_{i}\setminus B_{\mathbb{H}^{2}}(u_{i},h)=\mathbb{D}\setminus\left(B_{\mathbb{H}^{2}}(u_{i},h)\cup\operatorname{sect}\left(u_{i},v_{i},\vartheta-\vartheta^{\prime}\right)\right).

(For i=1i=1 it might be the case that B2(o,h)B_{\mathbb{H}^{2}}(o,h) contains point of 𝒵{\mathcal{Z}}. So we cannot say that 𝒳1{\mathcal{X}}_{1} is completely determined by 𝒵S1{\mathcal{Z}}\cap S_{1}^{\prime}. It is however completely determined by 𝒵S1{\mathcal{Z}}\cap S_{1}.)

We consider an isometry φ:𝔻𝔻\varphi:\mathbb{D}\to\mathbb{D} satisfying φ(ui)=o\varphi(u_{i})=o and that φ(vi)\varphi(v_{i}) lies on the positive xx-axis. Thus

φ[Si]=𝔻(B2(o,h)sect(o,v1,ϑϑ))=S1,\varphi\left[S_{i}^{\prime}\right]=\mathbb{D}\setminus\left(B_{\mathbb{H}^{2}}(o,h)\cup\operatorname{sect}\left(o,v_{1},\vartheta-\vartheta^{\prime}\right)\right)=S_{1}^{\prime},

and of course φ[B2(ui,h)]=B2(o,h)\varphi\left[B_{\mathbb{H}^{2}}(u_{i},h)\right]=B_{\mathbb{H}^{2}}(o,h).

Since SiS_{i}^{\prime} does not not intersect the areas R1,,Ri1R_{1},\dots,R_{i-1} revealed by previous iterations and it is isometric to S1S_{1}^{\prime} for each i>1i>1, we find that X1,X2,X_{1},X_{2},\dots are independent, and for i>1i>1 in fact

Xi=d(X1||𝒵B2(o,h)|=0) (for i>1.) X_{i}\hskip 0.86108pt\raisebox{-0.43057pt}{$=$}\raisebox{4.30554pt}{{$\scriptstyle d$}}\hskip 3.44444pt\left(X_{1}{\Big{|}}|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)|=0\right)\quad\text{ (for $i>1$.) }

(We write |B2(o,h)𝒵|=0|B_{\mathbb{H}^{2}}(o,h)\cap{\mathcal{Z}}|=0 and not B2(o,h)𝒵={o}B_{\mathbb{H}^{2}}(o,h)\cap{\mathcal{Z}}=\{o\} since o𝒵o\not\in{\mathcal{Z}}.)

If X1,X2,X_{1},X_{2},\dots were i.i.d. then TT would be a Galton-Watson tree with offspring distribution X1X_{1}. In our case, we can describe the situation by saying the tree TT consists of a root attached to X1X_{1} independent copies of a Galton-Watson tree with offspring distribution X1~=(X1|𝒵B2(o,h)=)\tilde{X_{1}}=\left(X_{1}|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)=\emptyset\right). We also point out that the sequence X1,X2,X_{1},X_{2},\dots completely determines the size of the tree TT, via

|V(T)|=inf{n:X1++Xnn1}.|V(T)|=\operatorname*{\vphantom{p}inf}\{n:X_{1}+\dots+X_{n}\leq n-1\}.

(See for instance Sections 1.5 and 1.6 of [20]. Note that while the discussion there focuses on X1,X2,X_{1},X_{2},\dots i.i.d., the above equation holds much more generally. See the remark following Definition 1.14 in [20].)

To conclude the proof, we note that, provided we chose r0r_{0} sufficiently large, B2(o,h)C(u1,v1,r,w,ϑ)=B_{\mathbb{H}^{2}}(o,h)\cap C(u_{1},v_{1},r,w,\vartheta)=\emptyset. Therefore, any point of 𝒵{\mathcal{Z}} in B2(o,h)B_{\mathbb{H}^{2}}(o,h) can only “prevent” the formation of edges between oo and some z𝒵bC(u1,v1,r,w,ϑ)z\in{\mathcal{Z}}_{b}\cap C(u_{1},v_{1},r,w,\vartheta). In particular, for any k,{0}k,\ell\in\mathbb{N}\cup\{0\} we have

(X1k||𝒵B2(o,h)|=)(X1k||𝒵B2(o,h)|=0)=(X1~k),{\mathbb{P}}\left(X_{1}\geq k{\Big{|}}|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)|=\ell\right)\leq{\mathbb{P}}\left(X_{1}\geq k{\Big{|}}|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)|=0\right)={\mathbb{P}}(\tilde{X_{1}}\geq k),

which gives

(X1k)==0(X1k||𝒵B2(o,h)|=)(|𝒵B2(o,h)|=)(X1~k).{\mathbb{P}}(X_{1}\geq k)=\sum_{\ell=0}^{\infty}{\mathbb{P}}\left(X_{1}\geq k{\Big{|}}|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)|=\ell\right){\mathbb{P}}\left(|{\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)|=\ell\right)\leq{\mathbb{P}}(\tilde{X_{1}}\geq k).

In other words, X1~\tilde{X_{1}} stochastically dominates X1X_{1}. By Strassen’s theorem ([40]; an elementary proof of the version we need can for instance be found in Section 2.3 of [47]) there is a coupling of the sequence X1,X2,X3,X_{1},X_{2},X_{3},\dots and an i.i.d. sequence Y1,Y2,Y_{1},Y_{2},\dots such that Yi=dX1Y_{i}\hskip 0.86108pt\raisebox{-0.43057pt}{$=$}\raisebox{4.30554pt}{{$\scriptstyle d$}}\hskip 3.44444ptX_{1} for all ii and XiYiX_{i}\geq Y_{i} almost surely. The sequence Y1,Y2,Y_{1},Y_{2},\dots can be used to generate a Galton-Watson tree TT^{\prime} with offspring distribution X1X_{1}. We have |V(T)|=inf{n1:Y1++Ynn1}|V(T^{\prime})|=\operatorname*{\vphantom{p}inf}\{n\geq 1:Y_{1}+\dots+Y_{n}\leq n-1\}. Hence, (under the coupling, almost surely) |V(T)||V(T)||V(T^{\prime})|\leq|V(T)|. In particular the size of the black cluster of the origin (which is at least |V(T)||V(T)|) stochastically dominates |V(T)||V(T^{\prime})|. This is what needed to be shown. \blacksquare

Having established Proposition 18, in order to prove Proposition 17 it suffices to show that, for λ\lambda sufficiently small and p=(1+ε)(π/3)λp=(1+\varepsilon)\cdot(\pi/3)\cdot\lambda, there is a choice of w,ϑ,h,rw,\vartheta,h,r such that (h=h(w,ϑ),rr0(w,ϑ)h=h(w,\vartheta),r\geq r_{0}(w,\vartheta) with h(.,.),r0(.,.)h(.,.),r_{0}(.,.) as specified in Proposition 18, and) 𝔼X>1{\mathbb{E}}X>1. More specifically, we’ll keep w,ϑw,\vartheta (and hh) constant, but we’ll let rr depend on λ\lambda.

We break the argument down in a series of relatively straightforward lemmas.

Lemma 19

For every w,λ>0w,\lambda>0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X𝐈:=|𝒵bB2(o,rw)|,X_{\bf{I}}:=\left|{\mathcal{Z}}_{b}\cap B_{\mathbb{H}^{2}}(o,r-w)\right|,

we have

𝔼X𝐈1000ew.{\mathbb{E}}X_{\bf{I}}\leq 1000e^{-w}.

Proof. If rw<0r-w<0 then clearly X𝐈=0X_{\bf{I}}=0 almost and we are done. So we can assume rw0r-w\geq 0. Clearly

𝔼X𝐈=pλarea2(B2(o,rw))=pλ2π(cosh(rw)1)1000λ2erw=1000ew,\begin{array}[]{rcl}{\mathbb{E}}X_{\bf{I}}&=&p\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r-w)\right)\\ &=&p\lambda\cdot 2\pi\left(\cosh(r-w)-1\right)\\ &\leq&1000\lambda^{2}e^{r-w}\\ &=&1000e^{-w},\end{array}

using that cosh(x)1ex\cosh(x)-1\leq e^{x} for x0x\geq 0, and the choice of rr. \blacksquare

Lemma 20

For every w,λ,ϑ>0w,\lambda,\vartheta>0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda) and letting v𝔻{o}v\in\mathbb{D}\setminus\{o\} be an arbitrary (fixed) point. Writing

X𝐈𝐈:=|𝒵bB2(o,r+w)sect(o,v,ϑ)|,X_{\bf{II}}:=\left|{\mathcal{Z}}_{b}\cap B_{\mathbb{H}^{2}}(o,r+w)\cap\operatorname{sect}\left(o,v,\vartheta\right)\right|,

we have

𝔼X𝐈𝐈1000ϑew.{\mathbb{E}}X_{\bf{II}}\leq 1000\vartheta e^{w}.

Proof. Clearly

𝔼X𝐈𝐈=pλarea2(B2(o,r+w)sect(o,v,ϑ))=pλ2ϑ2π2π(cosh(r+w)1)1000ϑλ2er+w=1000ϑew,\begin{array}[]{rcl}{\mathbb{E}}X_{\bf{II}}&=&p\lambda\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r+w)\cap\operatorname{sect}\left(o,v,\vartheta\right)\right)\\ &=&p\lambda\cdot\frac{2\vartheta}{2\pi}\cdot 2\pi\left(\cosh(r+w)-1\right)\\ &\leq&1000\vartheta\lambda^{2}e^{r+w}\\ &=&1000\vartheta e^{w},\end{array}

as before using that cosh(x)1ex\cosh(x)-1\leq e^{x} and the choice of rr. \blacksquare

Lemma 21

For every w,λ,ϑ>0w,\lambda,\vartheta>0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X𝐈𝐈𝐈:=|{(z1,z2)𝒵b×𝒵b:z1,z2B2(o,r+w), and;z1z2, and;,z1oz2<ϑ.}|,X_{\bf{III}}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}z_{1},z_{2}\in B_{\mathbb{H}^{2}}(o,r+w),\text{ and;}\\ z_{1}\neq z_{2},\text{ and;},\\ \angle z_{1}oz_{2}<\vartheta.\end{array}\right\}\right|,

we have

𝔼X𝐈𝐈𝐈1000ϑe2w.{\mathbb{E}}X_{\bf{III}}\leq 1000\vartheta e^{2w}.

Proof. By Corollary 5:

𝔼X𝐈𝐈𝐈=p2λ2𝔻𝔻𝔼[g(z1,z2,𝒵{z1,z2})]f(z1)f(z2)dz2dz2,{\mathbb{E}}X_{\bf{III}}=p^{2}\lambda^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{2},

where ff is as given by (2) and

g(u1,u2,𝒰)=1{u1,u2B2(o,r+w), and;u1u2, and;,u1ou2<ϑ.}.g(u_{1},u_{2},{\mathcal{U}})=1_{\left\{\begin{array}[]{l}u_{1},u_{2}\in B_{\mathbb{H}^{2}}(o,r+w),\text{ and;}\\ u_{1}\neq u_{2},\text{ and;},\\ \angle u_{1}ou_{2}<\vartheta.\end{array}\right\}}.

In other words

𝔼X𝐈𝐈𝐈=p2λ2B2(o,r+w)B2(o,r+w)sect(o,z1,ϑ)f(z1)f(z2)dz2dz1=p2λ2B2(o,r+w)area2(B2(o,r+w)sect(o,z1,ϑ))f(z1)dz1=p2λ2B2(o,r+w)(2ϑ2π)area2(B2(o,r+w))dz1=p2λ2(2ϑ2π)area2(B2(o,r+w))21000λ4ϑe2r+2w=1000ϑe2w,\begin{array}[]{rcl}{\mathbb{E}}X_{\bf{III}}&=&\displaystyle p^{2}\lambda^{2}\int_{B_{\mathbb{H}^{2}}(o,r+w)}\int_{B_{\mathbb{H}^{2}}(o,r+w)\cap\operatorname{sect}\left(o,z_{1},\vartheta\right)}f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{1}\\[8.61108pt] &=&\displaystyle p^{2}\lambda^{2}\int_{B_{\mathbb{H}^{2}}(o,r+w)}\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r+w)\cap\operatorname{sect}\left(o,z_{1},\vartheta\right)\right)f(z_{1})\operatorname{d}z_{1}\\[8.61108pt] &=&\displaystyle p^{2}\lambda^{2}\int_{B_{\mathbb{H}^{2}}(o,r+w)}\left(\frac{2\vartheta}{2\pi}\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r+w)\right)\operatorname{d}z_{1}\\[8.61108pt] &=&\displaystyle p^{2}\lambda^{2}\left(\frac{2\vartheta}{2\pi}\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r+w)\right)^{2}\\[8.61108pt] &\leq&\displaystyle 1000\lambda^{4}\vartheta e^{2r+2w}\\ &=&1000\vartheta e^{2w},\end{array}

using rotational symmetry of the hyperbolic area measure in the third line; that area2(B2(o,x))=2π(cosh(x)1)πex\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,x))=2\pi(\cosh(x)-1)\leq\pi e^{x} and the bound on pp in the fifth line; and the choice of rr in the last. \blacksquare

Lemma 22

For every w,λ,h>0w,\lambda,h>0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X𝐈𝐕:=|{(z1,z2)𝒵b×𝒵b:z1B2(o,r+w), and;z1z2, and;,dist2(z1,z2)<h.}|,X_{\bf{IV}}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}z_{1}\in B_{\mathbb{H}^{2}}(o,r+w),\text{ and;}\\ z_{1}\neq z_{2},\text{ and;},\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<h.\end{array}\right\}\right|,

we have

𝔼X𝐈𝐕1000λ2ew+h.{\mathbb{E}}X_{\bf{IV}}\leq 1000\lambda^{2}e^{w+h}.

Proof. Applying Corollary 5 again, we have

𝔼X𝐈𝐕=p2λ2𝔻𝔻𝔼[g(z1,z2,𝒵{z1,z2})]f(z1)f(z2)dz2dz2,{\mathbb{E}}X_{\bf{IV}}=p^{2}\lambda^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{2},

where ff is given by (2) and this time

g(u1,u2,𝒰)=1{u1B2(o,r+w), and;u1u2, and;,dist2(u1,u2)<h.}.g(u_{1},u_{2},{\mathcal{U}})=1_{\left\{\begin{array}[]{l}u_{1}\in B_{\mathbb{H}^{2}}(o,r+w),\text{ and;}\\ u_{1}\neq u_{2},\text{ and;},\\ \operatorname{dist}_{\mathbb{H}^{2}}(u_{1},u_{2})<h.\end{array}\right\}}.

In other words

𝔼X𝐈𝐕=p2λ2B2(o,r+w)B2(z1,h)f(z1)f(z2)dz2dz1=p2λ2area2(B2(o,r+w))area2(B2(o,h))1000λ2ew+h,\begin{array}[]{rcl}{\mathbb{E}}X_{\bf{IV}}&=&\displaystyle p^{2}\lambda^{2}\int_{B_{\mathbb{H}^{2}}(o,r+w)}\int_{B_{\mathbb{H}^{2}}(z_{1},h)}f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{1}\\ &=&\displaystyle p^{2}\lambda^{2}\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,r+w)\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,h)\right)\\ &\leq&\displaystyle 1000\lambda^{2}e^{w+h},\end{array}

again using that area2(B2(o,x))=2π(coshx1)πex\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,x))=2\pi\left(\cosh x-1\right)\leq\pi e^{x} and the choice of rr. \blacksquare

Lemma 23

There exists a λ0>0\lambda_{0}>0 such that, for all 0<λ<λ00<\lambda<\lambda_{0}, all w0w\geq 0 and all p10λp\leq 10\lambda, the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X𝐕:=|{z𝒵b:dist2(z,o)r+w, and; a disk B with o,zB and 𝒵B=.}|,X_{\bf{V}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z,o)\geq r+w,\text{ and;}\\ \text{$\exists$ a disk $B$ with $o,z\in\partial B$ and ${\mathcal{Z}}\cap B=\emptyset$.}\end{array}\right\}\right|,

we have

𝔼X𝐕1000eweew/2.{\mathbb{E}}X_{\bf{V}}\leq 1000e^{w}e^{-e^{w/2}}.

In the proof we’ll make use of the following definition and observations, that we’ll reuse later. For z1,z2𝔻z_{1},z_{2}\in\mathbb{D} the Gabriel disk is the disk BGab(z1,z2):=B2(c,dist2(z1,z2)/2)B_{\text{Gab}}(z_{1},z_{2}):=B_{\mathbb{H}^{2}}(c,\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})/2) whose center cc is the midpoint of the hyperbolic line segment between z1z_{1} and z2z_{2} and whose radius is half the hyperbolic distance between z1z_{1} and z2z_{2}. Put differently, among all disks that have both z1z_{1} and z2z_{2} on their boundary, BGab(z1,z2)B_{\text{Gab}}(z_{1},z_{2}) is the one of smallest hyperbolic radius.

The name “Gabriel disk” is in reference to the Gabriel graph, an object that has received some attention in the discrete and computational geometry literature. The Gabriel graph associated with a point set V2V\subseteq\mathbb{R}^{2} is the subgraph of the Delaunay graph whose edges are all pairs v1,v2Vv_{1},v_{2}\in V for which the Euclidean disk whose center is the midpoint between v1v_{1} and v2v_{2} and whose radius is half their distance contains no other points of VV.

The hyperbolic line segment between z1z_{1} and z2z_{2} splits BGab(z1,z2)B_{\text{Gab}}(z_{1},z_{2}) into two parts of equal hyperbolic area. We shall denote by BGab(z1,z2)B_{\text{Gab}}^{-}(z_{1},z_{2}) the part that is on the left as we travel from z1z_{1} to z2z_{2} along the hyperbolic line segment, and by BGab+(z1,z2)B_{\text{Gab}}^{+}(z_{1},z_{2}) the one that is on the right. See Figure 10.

Refer to caption
BGab(z1,z2)B_{\text{Gab}}^{-}(z_{1},z_{2})𝔻\partial\mathbb{D}BGab+(z1,z2)B_{\text{Gab}}^{+}(z_{1},z_{2})ooz1z_{1}z2z_{2}
Figure 10: The Gabriel disk BGab(z1,z2)B_{\text{Gab}}(z_{1},z_{2}), in the special case when z1,z2z_{1},z_{2} lie on the xx-axis and their midpoint is the origin.

We’ll repeatedly make use of the following straightforward observation.

  • (\spadesuit)

    For all z1,z2𝔻z_{1},z_{2}\in\mathbb{D} and every disk BB such that z1,z2Bz_{1},z_{2}\in\partial B, we have either BGab(z1,z2)BB_{\text{Gab}}^{-}(z_{1},z_{2})\subseteq B or BGab+(z1,z2)BB_{\text{Gab}}^{+}(z_{1},z_{2})\subseteq B (or both).

(This is easily seen by applying a suitable isometry that maps z1,z2z_{1},z_{2} to the xx-axis and their midpoint to the origin, so that BGab,BGab+B_{\text{Gab}}^{-},B_{\text{Gab}}^{+} are mapped to ordinary, Euclidean half-disks – as in Figure 10.)

In light of observation (\spadesuit), in order to prove Lemma 23 it suffices to prove the following statement, which we separate out as a lemma for convenient future reference.

Lemma 24

There exists a λ0>0\lambda_{0}>0 such that, for all 0<λ<λ00<\lambda<\lambda_{0}, all w0w\geq 0 and all p10λp\leq 10\lambda, the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X~𝐕:=|{z𝒵b:dist2(z,o)r+w, and;BGab(z,o)𝒵= or BGab+(z,o)𝒵=.}|,\tilde{X}_{\bf{V}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z,o)\geq r+w,\text{ and;}\\ \text{$B_{\text{Gab}}^{-}(z,o)\cap{\mathcal{Z}}=\emptyset$ or $B_{\text{Gab}}^{+}(z,o)\cap{\mathcal{Z}}=\emptyset$.}\end{array}\right\}\right|,

we have

𝔼X~𝐕1000eweew/2.{\mathbb{E}}\tilde{X}_{\bf{V}}\leq 1000e^{w}e^{-e^{w/2}}.

Proof of Lemma 24. We let λ0>0\lambda_{0}>0 be a small constant, to be determined in the course of the proof. By Corollary 5

𝔼X~𝐕=pλ𝔻𝔼[g(z,𝒵{z})]f(z)dz,{\mathbb{E}}\tilde{X}_{\bf{V}}=p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]f(z)\operatorname{d}z,

where ff is again given by (2) and

g(u,𝒰):=1{dist2(u,o)r+w, and;BGab(u,o)𝒰= or BGab+(u,o)𝒰=.}.g(u,{\mathcal{U}}):=1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(u,o)\geq r+w,\text{ and;}\\ \text{$B_{\text{Gab}}^{-}(u,o)\cap{\mathcal{U}}=\emptyset$ or $B_{\text{Gab}}^{+}(u,o)\cap{\mathcal{U}}=\emptyset$.}\end{array}\right\}}.

We have

𝔼X~𝐕=pλ𝔻B2(o,r+w)(BGab(o,z)𝒵= or BGab+(o,z)𝒵=)f(z)dz10λ2𝔻B2(o,r+w)2exp[12λarea2(BGab(o,z))]f(z)dz20λ2𝔻B2(o,r+w)exp[λedist2(o,z)/2]f(z)dz,\begin{array}[]{rcl}{\mathbb{E}}\tilde{X}_{\bf{V}}&=&\displaystyle p\lambda\int_{\mathbb{D}\setminus B_{\mathbb{H}^{2}}(o,r+w)}{\mathbb{P}}\left(B_{\text{Gab}}^{-}(o,z)\cap{\mathcal{Z}}=\emptyset\text{ or }B_{\text{Gab}}^{+}(o,z)\cap{\mathcal{Z}}=\emptyset\right)f(z)\operatorname{d}z\\[8.61108pt] &\leq&\displaystyle 10\lambda^{2}\int_{\mathbb{D}\setminus B_{\mathbb{H}^{2}}(o,r+w)}2\exp\left[-\frac{1}{2}\cdot\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\text{Gab}}(o,z)\right)\right]f(z)\operatorname{d}z\\[8.61108pt] &\leq&\displaystyle 20\lambda^{2}\int_{\mathbb{D}\setminus B_{\mathbb{H}^{2}}(o,r+w)}\exp\left[-\lambda e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2}\right]f(z)\operatorname{d}z,\end{array}

where in the last line we use that λ0\lambda_{0} was chosen sufficiently small; that 2π(coshx1)=(1+ox(1))πex2\pi(\cosh x-1)=(1+o_{x}(1))\pi e^{x} as xx\to\infty so that 2π(coshx1)2ex2\pi(\cosh x-1)\geq 2e^{x} for xx sufficiently large; that dist2(o,z)r+w\operatorname{dist}_{\mathbb{H}^{2}}(o,z)\geq r+w; and that r2ln(1/λ0)r\geq 2\ln(1/\lambda_{0}).

Switching to hyperbolic polar coordinates (i.e. z(α,ρ)=(cos(α)tanh(ρ/2),sin(α)tanh(ρ/2))z(\alpha,\rho)=(\cos(\alpha)\cdot\tanh(\rho/2),\sin(\alpha)\cdot\tanh(\rho/2))) we find

𝔼X~𝐕20λ2r+w02πeλeρ/2sinh(ρ)dαdρ=20λ2r+weλeρ/22πsinh(ρ)dρ20πλ2r+weλeρ/2eρdρ=40πλe(r+w)/2euudu=40π(λe(r+w)/2+1)exp[λe(r+w)/2]=40π(ew/2+1)eew/21000eweew/2,\begin{array}[]{rcl}{\mathbb{E}}\tilde{X}_{\bf{V}}&\leq&\displaystyle 20\lambda^{2}\int_{r+w}^{\infty}\int_{0}^{2\pi}e^{-\lambda e^{\rho/2}}\sinh(\rho)\operatorname{d}\alpha\operatorname{d}\rho\\[8.61108pt] &=&\displaystyle 20\lambda^{2}\int_{r+w}^{\infty}e^{-\lambda e^{\rho/2}}2\pi\sinh(\rho)\operatorname{d}\rho\\[8.61108pt] &\leq&\displaystyle 20\pi\lambda^{2}\int_{r+w}^{\infty}e^{-\lambda e^{\rho/2}}e^{\rho}\operatorname{d}\rho\\[8.61108pt] &=&\displaystyle 40\pi\int_{\lambda e^{(r+w)/2}}^{\infty}e^{-u}u\operatorname{d}u\\[8.61108pt] &=&\displaystyle 40\pi\left(\lambda e^{(r+w)/2}+1\right)\exp\left[-\lambda e^{(r+w)/2}\right]\\[8.61108pt] &=&\displaystyle 40\pi\left(e^{w/2}+1\right)e^{-e^{w/2}}\\[8.61108pt] &\leq&1000e^{w}e^{-e^{w/2}},\end{array}

using the substitution u=λeρ/2u=\lambda e^{\rho/2} (so that dρ=2duu\operatorname{d}\rho=\frac{2\operatorname{d}u}{u}) in the third line \blacksquare

Lemma 25

There exists a λ0>0\lambda_{0}>0 such that for all 0<λ<λ00<\lambda<\lambda_{0}, all w0w\geq 0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X𝐕𝐈:=|{z𝒵b:rw<dist2(z,o)<r+w, and; a disk B with o,zB,𝒵B= and diam2(B)r+w.}|,X_{\bf{VI}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}r-w<\operatorname{dist}_{\mathbb{H}^{2}}(z,o)<r+w,\text{ and;}\\ \text{$\exists$ a disk $B$ with $o,z\in\partial B,{\mathcal{Z}}\cap B=\emptyset$ and $\operatorname{diam}_{\mathbb{H}^{2}}(B)\geq r+w$.}\end{array}\right\}\right|,

we have

𝔼X𝐕𝐈1000eweew/2.{\mathbb{E}}X_{\bf{VI}}\leq 1000e^{w}e^{-e^{w/2}}.

In the proof we’ll make use of the following definition and observations, that we’ll reuse later. For z1,z2𝔻z_{1},z_{2}\in\mathbb{D} and ρ>0\rho>0 with dist2(z1,z2)<ρ\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<\rho, there exist precisely two disks BB such that z1,z2Bz_{1},z_{2}\in\partial B and diam2(B)=ρ\operatorname{diam}_{\mathbb{H}^{2}}(B)=\rho. We let DD(z1,z2,ρ)DD(z_{1},z_{2},\rho) denote the union of these two disks. (The notation DDDD stands for “double disk”.) The hyperbolic line segment between z1z_{1} and z2z_{2} splits DD(z1,z2,ρ)DD(z_{1},z_{2},\rho) into two parts of equal hyperbolic area. We denote by DD(z1,z2,ρ)DD^{-}(z_{1},z_{2},\rho) the part that is on the left as we travel from z1z_{1} to z2z_{2} along the hyperbolic line segment between them, and by DD+(z1,z2,ρ)DD^{+}(z_{1},z_{2},\rho) the part on the right. See Figure 11.

Refer to caption
z1z_{1}DD+(z1,z2,ρ)DD^{+}(z_{1},z_{2},\rho)ooDD(z1,z2,ρ)DD^{-}(z_{1},z_{2},\rho)𝔻\partial\mathbb{D}z2z_{2}
Figure 11: The set DD(z1,z2,ρ)DD(z_{1},z_{2},\rho), in the special case when z1,z2z_{1},z_{2} lie on the xx-axis with the origin oo as their midpoint.

Another observation that we’ll use repeatedly is:

  • (\clubsuit)

    For all z1z2𝔻z_{1}\neq z_{2}\in\mathbb{D} and ρ>dist2(z1,z2)\rho>\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2}) and every hyperbolic disk BB such that z1,z2Bz_{1},z_{2}\in\partial B and diam2(B)ρ\operatorname{diam}_{\mathbb{H}^{2}}(B)\geq\rho, we have either DD(z1,z2,ρ)BDD^{-}(z_{1},z_{2},\rho)\subseteq B or DD+(z1,z2,ρ)BDD^{+}(z_{1},z_{2},\rho)\subseteq B (or both).

(Again this is easily seen by applying a suitable isometry that maps z1,z2z_{1},z_{2} to the xx-axis and their midpoint to the origin, as in Figure 11. Fact 2 ensures that the image of the hyperbolic disk BB is a Euclidean disk with the images of z1,z2z_{1},z_{2} on its boundary.)

In light of observation (\clubsuit), in order to prove Lemma 25 it suffices to prove the following statement, which we separate out as a lemma for convenient future reference.

Lemma 26

There exists a λ0>0\lambda_{0}>0 such that for all 0<λ<λ00<\lambda<\lambda_{0}, all w0w\geq 0 and all p10λp\leq 10\lambda the following holds, setting r:=2log(1/λ)r:=2\log(1/\lambda). Writing

X~𝐕𝐈:=|{z𝒵b:rw<dist2(z,o)<r+w, and;DD(o,z,r+w)𝒵= or DD+(o,z,r+w)𝒵=.}|,\tilde{X}_{\bf{VI}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}r-w<\operatorname{dist}_{\mathbb{H}^{2}}(z,o)<r+w,\text{ and;}\\ \text{$DD^{-}(o,z,r+w)\cap{\mathcal{Z}}=\emptyset$ or $DD^{+}(o,z,r+w)\cap{\mathcal{Z}}=\emptyset$.}\end{array}\right\}\right|,

we have

𝔼X~𝐕𝐈1000eweew/2.{\mathbb{E}}\tilde{X}_{\bf{VI}}\leq 1000e^{w}e^{-e^{w/2}}.

Proof of Lemma 26. Applying Corollary 5, we have

𝔼X~𝐕𝐈=pλB2(o,r+w)(DD(o,z,r+w)𝒵= or DD(o,z,r+w)𝒵=)f(z)dz,{\mathbb{E}}\tilde{X}_{\bf{VI}}=p\lambda\int_{B_{\mathbb{H}^{2}}(o,r+w)}{\mathbb{P}}(DD^{-}(o,z,r+w)\cap{\mathcal{Z}}=\emptyset\text{ or }DD^{-}(o,z,r+w)\cap{\mathcal{Z}}=\emptyset)f(z)\operatorname{d}z,

where ff is as given by (2). Next we point out that, by symmetry

area2(DD(o,z,r+w))=area2(DD+(o,z,r+w))12area2(B2(o,(r+w)/2))e(r+w)/2,\operatorname{area}_{\mathbb{H}^{2}}(DD^{-}(o,z,r+w))=\operatorname{area}_{\mathbb{H}^{2}}(DD^{+}(o,z,r+w))\geq\frac{1}{2}\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,(r+w)/2))\geq e^{(r+w)/2},

where the last inequality holds provided we chose λ0\lambda_{0} sufficiently large, since r2ln(1/λ0)r\geq 2\ln(1/\lambda_{0}) and area2(B2(0,x))=2π(coshx1)=(1+ox(1))πex\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(0,x))=2\pi(\cosh x-1)=(1+o_{x}(1))\pi e^{x} as xx\to\infty.

Hence

𝔼X~𝐕𝐈pλarea2(B2(0,r+w))2eλe(r+w)/21000λ2er+weλe(r+w)/2=1000eweew/2,\begin{array}[]{rcl}{\mathbb{E}}\tilde{X}_{\bf{VI}}&\leq&p\lambda\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(0,r+w))\cdot 2e^{-\lambda e^{(r+w)/2}}\\ &\leq&1000\lambda^{2}e^{r+w}e^{-\lambda e^{(r+w)/2}}\\ &=&1000e^{w}e^{-e^{w/2}},\end{array}

using the choice of rr. \blacksquare

We are now ready to prove the following statement.

Proposition 27

For every 0<ε<110000<\varepsilon<\frac{1}{1000} there exist w,ϑ>0w,\vartheta>0 such that the following holds. For every h>0h>0 there exists a λ0=λ0(ε,w,ϑ,h)\lambda_{0}=\lambda_{0}(\varepsilon,w,\vartheta,h) such that, for all 0<λ<λ00<\lambda<\lambda_{0}, setting p:=(1+ε)(π/3)λp:=(1+\varepsilon)(\pi/3)\lambda and r:=2ln(1/λ)r:=2\ln(1/\lambda), we have that

𝔼X>1.{\mathbb{E}}X>1.

(With XX as defined in (11).)

Proof. As pointed out right after the definition of XX, its probability distribution is the same for any choice of uv𝔻u\neq v\in\mathbb{D}. For definiteness we take u=ou=o and v=(1/2,0)v=(1/2,0). We let h>0h>0 be an arbitrary constant. We let w,λ0>0w,\lambda_{0}>0 be large constants, and ϑ>0\vartheta>0 a small constant, to be determined in the course of the proof.

Let us denote by DbD_{b} the number of black cells adjacent to the typical cell. (Recall that “typical cell” refers to the cell of the origin in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\}.) By Isokawa’s formula

𝔼Db=p𝔼D=p(6+3πλ)>1+ε,{\mathbb{E}}D_{b}=p{\mathbb{E}}D=p\cdot\left(6+\frac{3}{\pi\lambda}\right)>1+\varepsilon,

using the choice of pp in the inequality. Next, we point out that

XDb(X𝐈+X𝐈𝐈+X𝐈𝐈𝐈+X𝐈𝐕+X𝐕+X𝐕𝐈).X\geq D_{b}-\left(X_{\bf{I}}+X_{\bf{II}}+X_{\bf{III}}+X_{\bf{IV}}+X_{\bf{V}}+X_{\bf{VI}}\right).

with X𝐈X_{\bf{I}}X𝐕𝐈X_{\bf{VI}} as defined in Lemmas 1925. Applying these lemmas we see that, provided we chose λ0\lambda_{0} sufficiently small,

𝔼X>1+ε1000(ew+ϑew+ϑe2w+λ02ew+h+2eweew/2)>1+ε/2,{\mathbb{E}}X>1+\varepsilon-1000\cdot\left(e^{-w}+\vartheta e^{w}+\vartheta e^{2w}+\lambda_{0}^{2}e^{w+h}+2e^{w}e^{-e^{w/2}}\right)>1+\varepsilon/2,

where the second inequality holds provided we chose the constant ww sufficiently large and the constants ϑ,λ0\vartheta,\lambda_{0} sufficiently small. (To be more explicit : we can for instance first choose ww such that ew,eweew/2<ε/105e^{-w},e^{w}e^{-e^{w/2}}<\varepsilon/10^{5} and then we choose ϑ\vartheta such that ϑe2w<ε/105\vartheta e^{2w}<\varepsilon/10^{5} and finally choose λ0\lambda_{0} such that λ02ew+h<ε/105\lambda_{0}^{2}e^{w+h}<\varepsilon/10^{5} – and at the same time λ0\lambda_{0} is small enough so that Lemma 23 and 25 apply.) \blacksquare

As pointed out earlier, Proposition 17, the upper bound in our main theorem, follows directly from Propositions 18 and 27.

3.3 The lower bound

Here we will show the following proposition, which constitutes the remaining half of our main result.

Proposition 28

For every ε>0\varepsilon>0, there exists a λ0=λ0(ε)>0\lambda_{0}=\lambda_{0}(\varepsilon)>0 such that for all 0<λ<λ00<\lambda<\lambda_{0}, we have pc(λ)(1ε)(π/3)λp_{c}(\lambda)\geq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda.

In the proof of Proposition 17 we could in a sense afford to “ignore” some adjacencies in the Voronoi tessellation. The supercritical Galton-Watson tree we constructed in the proof of Proposition 18 uses only adjacencies of convenient lengths and is such that if two edges share an endpoint, the angle they make is not too small.

Now we need to show that no infinite black component exists almost surely, for λ\lambda sufficiently small and p=(1ε)(π/3)λp=(1-\varepsilon)\cdot(\pi/3)\cdot\lambda. We cannot a priori exclude the possibility that an infinite component exists but every infinite connected subgraph contains (has to contain) “unusual” edges. In particular, a hypothetical infinite component might exist that does not contain an infinite (r,w,ϑ)(r,w,\vartheta)-tree.

To make our life easier, we adopt a more generous notion of adjacency, in the form of pseudopaths.

Definition 29

Given parameters r,w1,w2,ϑ>0r,w_{1},w_{2},\vartheta>0, we say a (finite or infinite) sequence u0,u1,𝔻u_{0},u_{1},\dots\in\mathbb{D} of distinct points is a pseudopath (wrt. r,w1,w2,ϑr,w_{1},w_{2},\vartheta and 𝒵{\mathcal{Z}}) if one of the following holds for each i1i\geq 1:

  1. I

    rw1<dist2(ui1,ui)<r+w2,ui2ui1uiϑr-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})<r+w_{2},\angle u_{i-2}u_{i-1}u_{i}\geq\vartheta and one of the following holds. There exists a disk BB such that ui1,uiB,B𝒵=u_{i-1},u_{i}\in\partial B,B\cap{\mathcal{Z}}=\emptyset and diam2(B)<r+w2\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w_{2}, or; DD(ui1,ui,r+w2)𝒵=DD^{-}(u_{i-1},u_{i},r+w_{2})\cap{\mathcal{Z}}=\emptyset, or; DD+(ui1,ui,r+w2)𝒵=DD^{+}(u_{i-1},u_{i},r+w_{2})\cap{\mathcal{Z}}=\emptyset.

  2. II

    dist2(ui1,ui)rw1\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})\leq r-w_{1};

  3. III

    rw1<dist2(ui1,ui)<r+w2r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})<r+w_{2} and ui2ui1ui<ϑ\angle u_{i-2}u_{i-1}u_{i}<\vartheta;

  4. IV

    dist2(ui1,ui)r+w2\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})\geq r+w_{2} and either BGab(ui1,ui)𝒵=B_{\text{Gab}}^{-}(u_{i-1},u_{i})\cap{\mathcal{Z}}=\emptyset, or BGab+(ui1,ui)𝒵=B_{\text{Gab}}^{+}(u_{i-1},u_{i})\cap{\mathcal{Z}}=\emptyset, (or both).

(Of course the demand that ui2ui1uiϑ\angle u_{i-2}u_{i-1}u_{i}\geq\vartheta in I only applies when i2i\geq 2, and similarly the case III can only occur when i2i\geq 2.) The length of a finite pseudopath P=u0,u1,,unP=u_{0},u_{1},\dots,u_{n} is nn, the number of points minus one, and we use the term pseudo-edge for a consecutive pair of points ui1,uiu_{i-1},u_{i} on a pseudopath. In particular pseudopaths of length one are pseudo-edges, but not all pseudo-edges are pseudopaths of length one (because of III which can only apply when the pseudopath has length 2\geq 2). As in the case of ordinary paths, we denote by V(P)V(P) the set of vertices of the pseudopath PP.

With each pair of points u,v𝔻u,v\in\mathbb{D} of a pseudopath we associate a “certificate” cert(u,v)𝔻\operatorname{cert}(u,v)\subseteq\mathbb{D}. This will be a region such that in order to verify that uvuv is a pseudo-edge, in addition to the location of the previous point of the pseudopath if uvuv is part of a pseudopath of length 2\geq 2, only 𝒵cert(u,v){\mathcal{Z}}\cap\operatorname{cert}(u,v) is relevant. Specifically, we set

cert(u,v):={DD(u,v,r+w2) if rw1<dist2(u,v)<r+w2BGab(u,v) if dist2(u,v)r+w2 otherwise. \operatorname{cert}(u,v):=\begin{cases}DD(u,v,r+w_{2})&\text{ if $r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u,v)<r+w_{2}$; }\\ B_{\text{Gab}}(u,v)&\text{ if $\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq r+w_{2}$; }\\ \emptyset&\text{ otherwise. }\end{cases}

(Note that if dist2(u,v)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(u,v)<r+w_{2} and BB is a disk with u,vBu,v\in\partial B and diam2(B)<r+w2\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w_{2} then BDD(u,v,r+w2)B\subseteq DD(u,v,r+w_{2}). Also note that if dist2(u,v)rw1\operatorname{dist}_{\mathbb{H}^{2}}(u,v)\geq r-w_{1} then cert(u,v)BGab(u,v)\operatorname{cert}(u,v)\supseteq B_{\text{Gab}}(u,v).) For notational convenience we also set

cert(u0,,uk):=cert(u0,u1)cert(uk1,uk),\operatorname{cert}(u_{0},\dots,u_{k}):=\operatorname{cert}(u_{0},u_{1})\cup\dots\cup\operatorname{cert}(u_{k-1},u_{k}),

for every sequence u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D}.

Pseudo-edges of type I will be called good, and all other types of pseudo-edges are bad. A pseudopath is good if all its pseudo-edges are good.

A finite pseudopath P=u0,,ukP=u_{0},\dots,u_{k} is called a chunk if the final pseudo-edge uk1uku_{k-1}u_{k} is bad and all other pseudo-edges are good. So a chunk can for instance consist of a single bad pseudo-edge.

Definition 30

A linked sequence of chunks is a (finite or infinite) sequence P1,P2,P_{1},P_{2},\dots of chunks such that

  1. (i)

    V(Pi)V(Pj)=cert(Pi)cert(Pj)=V(P_{i})\cap V(P_{j})=\operatorname{cert}(P_{i})\cap\operatorname{cert}(P_{j})=\emptyset if |ij|>1|i-j|>1, and;

  2. (ii)

    For each i2i\geq 2, writing Pi1=u0i1,,uki1i1P_{i-1}=u_{0}^{i-1},\dots,u_{k_{i-1}}^{i-1} and Pi=u0i,,ukiiP_{i}=u_{0}^{i},\dots,u_{k_{i}}^{i}, we have

    {u1i,,ukii}V(Pi1)=,cert(u1i,,ukii)cert(Pi1)=,dist2(uji,cert(Pi1))r1000 for all 1jki,\begin{array}[]{l}\{u_{1}^{i},\dots,u_{k_{i}}^{i}\}\cap V(P_{i-1})=\emptyset,\\ \operatorname{cert}(u_{1}^{i},\dots,u_{k_{i}}^{i})\cap\operatorname{cert}(P_{i-1})=\emptyset,\\ \operatorname{dist}_{\mathbb{H}^{2}}(u_{j}^{i},\operatorname{cert}(P_{i-1}))\geq\frac{r}{1000}\text{ for all $1\leq j\leq k_{i}$},\end{array}

    and in addition one of the following holds:

    • a)

      u0i=uki1i1u_{0}^{i}=u_{k_{i-1}}^{i-1}, or

    • b)

      u0iV(Pi1)u_{0}^{i}\not\in V(P_{i-1}) and cert(u0i,u1i)cert(Pi1)\operatorname{cert}(u_{0}^{i},u_{1}^{i})\cap\operatorname{cert}(P_{i-1})\neq\emptyset, or

    • c)

      u0iV(Pi1),cert(u0i,u1i)cert(Pi1)=u_{0}^{i}\not\in V(P_{i-1}),\operatorname{cert}(u_{0}^{i},u_{1}^{i})\cap\operatorname{cert}(P_{i-1})=\emptyset and dist2(u0i,cert(Pi1))<r/1000\operatorname{dist}_{\mathbb{H}^{2}}(u_{0}^{i},\operatorname{cert}(P_{i-1}))<r/1000.

We’ll say a linked sequence of chunks is infinite if the sum of the lengths of the chunks is infinite. In other words, an infinite linked sequence of chunks either consists of infinitely many finite chunks, or consists of finitely many chunks of which the last chunk has infinite length.

Proposition 31

For every λ>0,0p1\lambda>0,0\leq p\leq 1 and r,w1,w2,ϑ>0r,w_{1},w_{2},\vartheta>0, almost surely, one of the following holds.

  1. (i)

    The black component of oo in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\} is finite, or;

  2. (ii)

    There is a black, infinite, good pseudopath, or;

  3. (iii)

    There is a black, infinite linked sequence of chunks starting from oo.

For clarity, we emphasize that the infinite pseudopath mentioned in item (ii) does not need to contain the origin oo.

Our strategy for the proof of Proposition 28 is of course to show, after having established Proposition 31, that there is a choice of r,w1,w2,ϑr,w_{1},w_{2},\vartheta such that, for all small enough λ\lambda and all p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, almost surely options (ii) and (iii) of Proposition 31 do not occur. This will imply that the black component of the origin in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\} is a.s. finite. A short argument will then show this also implies that, almost surely, all black components in the Voronoi tessellation for 𝒵{\mathcal{Z}} are finite.

For the proof of Proposition 31 we will use the following observation that is a straightforward consequence of the Slivniak-Mecke formula and Isokawa’s formula. We provide a proof for completeness.

Lemma 32

For every λ>0\lambda>0 and 0p10\leq p\leq 1, almost surely, every Voronoi cell is adjacent to a finite number of other Voronoi cells.

Proof. Let us write

I:=|{z𝒵:deg(z)=}|.I:=\left|\left\{z\in{\mathcal{Z}}:\text{deg}(z)=\infty\right\}\right|.

By the Slivniak-Mecke formula

𝔼I=λ𝔻𝔼[g(z,𝒵{z})]f(z)du,{\mathbb{E}}I=\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]f(z)\operatorname{d}u,

where ff is as given by (2) and g(u,𝒰):=1{deg(u;𝒰)=}g(u,{\mathcal{U}}):=1_{\left\{\text{deg}(u;{\mathcal{U}})=\infty\right\}} with deg(u;𝒰)\text{deg}(u;{\mathcal{U}}) denoting the degree of uu in the Delaunay graph of 𝒰{\mathcal{U}}. By symmetry considerations, for every (fixed) u𝔻u\in\mathbb{D} we have

𝔼[deg(u;𝒵{u})]=𝔼[deg(o,𝒵{o})]=𝔼D=6+3πλ<,{\mathbb{E}}\left[\text{deg}(u;{\mathcal{Z}}\cup\{u\})\right]={\mathbb{E}}\left[\text{deg}(o,{\mathcal{Z}}\cup\{o\})\right]={\mathbb{E}}D=6+\frac{3}{\pi\lambda}<\infty,

where DD denotes the typical degree and we apply Isokawa’s formula. It follows that, for every u𝔻u\in\mathbb{D}:

𝔼[g(u,𝒵{u})]=(deg(u;𝒵{u})=)=0.{\mathbb{E}}\left[g(u,{\mathcal{Z}}\cup\{u\})\right]={\mathbb{P}}\left(\text{deg}(u;{\mathcal{Z}}\cup\{u\})=\infty\right)=0.

Hence also 𝔼I=0{\mathbb{E}}I=0, which implies that I=0I=0 almost surely. \blacksquare

We’ll also need the following observation in the proof of Proposition 31.

Lemma 33

For every λ>0\lambda>0, almost surely, for every x>0x>0 the number of pseudo-edges z1z2z_{1}z_{2} with z1,z2𝒵z_{1},z_{2}\in{\mathcal{Z}} for which cert(z1,z2)B2(o,x)\operatorname{cert}(z_{1},z_{2})\cap B_{\mathbb{H}^{2}}(o,x)\neq\emptyset is finite.

Proof. Fix an arbitrary x>0x>0. Clearly the number of pseudo-edges z1z2z_{1}z_{2} with z1,z2𝒵z_{1},z_{2}\in{\mathcal{Z}} for which dist2(z1,z2)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2} and cert(z1,z2)B2(o,x)\operatorname{cert}(z_{1},z_{2})\cap B_{\mathbb{H}^{2}}(o,x)\neq\emptyset is bounded by X2X^{2} where X:=𝒵B2(o,x+2r+2w2)X:={\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,x+2r+2w_{2}). Since 𝔼X=λarea2(B2(o,x+2r+2w2))<{\mathbb{E}}X=\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,x+2r+2w_{2}))<\infty, the random variable XX is finite almost surely.

Pseudo-edges z1z2z_{1}z_{2} not counted by X2X^{2} must satisfy dist2(z1,z2)r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})\geq r+w_{2} and hence cert(z1,z2)=BGab(z1,z2)\operatorname{cert}(z_{1},z_{2})=B_{\text{Gab}}(z_{1},z_{2}) and either BGab(z1,z1)𝒵=B_{\text{Gab}}^{-}(z_{1},z_{1})\cap{\mathcal{Z}}=\emptyset or BGab+(z1,z2)𝒵=B_{\text{Gab}}^{+}(z_{1},z_{2})\cap{\mathcal{Z}}=\emptyset (or both). By symmetry considerations, it suffices to count

Y:=|{(z1,z2)𝒵2:dist2(z1,o)>dist2(z2,o), and; BGab(z1,z2)B2(o,x), and; BGab(z1,z1)𝒵= or BGab+(z1,z2)𝒵=.}|.Y:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}^{2}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},o)>\operatorname{dist}_{\mathbb{H}^{2}}(z_{2},o),\text{ and; }\\ B_{\text{Gab}}(z_{1},z_{2})\cap B_{\mathbb{H}^{2}}(o,x)\neq\emptyset,\text{ and; }\\ \text{$B_{\text{Gab}}^{-}(z_{1},z_{1})\cap{\mathcal{Z}}=\emptyset$ or $B_{\text{Gab}}^{+}(z_{1},z_{2})\cap{\mathcal{Z}}=\emptyset$.}\end{array}\right\}\right|.

For each pair (z1,z2)(z_{1},z_{2}) counted by YY there is an nn\in\mathbb{N} such that n1dist2(o,z1)nn-1\leq\operatorname{dist}_{\mathbb{H}^{2}}(o,z_{1})\leq n. We must also have that dist2(z1,z2)>dist2(o,z1)x\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})>\operatorname{dist}_{\mathbb{H}^{2}}(o,z_{1})-x, since diam(BGab(z1,z2))=dist2(z1,z2)\operatorname{diam}(B_{\text{Gab}}(z_{1},z_{2}))=\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2}) and BGab(z1,z2)B2(o,x)B_{\text{Gab}}(z_{1},z_{2})\cap B_{\mathbb{H}^{2}}(o,x)\neq\emptyset.

Writing

Yn:=|{(z1,z2)𝒵2:dist2(z1,o),dist2(z2,o)n and; dist2(z1,z2)>nx1, and; BGab(z1,z1)𝒵= or BGab+(z1,z2)𝒵=.}|,Y_{n}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}^{2}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},o),\operatorname{dist}_{\mathbb{H}^{2}}(z_{2},o)\leq n\text{ and; }\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})>n-x-1,\text{ and; }\\ \text{$B_{\text{Gab}}^{-}(z_{1},z_{1})\cap{\mathcal{Z}}=\emptyset$ or $B_{\text{Gab}}^{+}(z_{1},z_{2})\cap{\mathcal{Z}}=\emptyset$.}\end{array}\right\}\right|,

we thus have

𝔼Yn𝔼Yn.{\mathbb{E}}Y\leq\sum_{n}{\mathbb{E}}Y_{n}.

Applying the Slivniak-Mecke formula, we find that

𝔼Yn=λ2𝔻𝔻𝔼[g(z1,z2,𝒵{z1,z2})]f(z1)f(z2)dz1dz2,{\mathbb{E}}Y_{n}=\lambda^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]f(z_{1})f(z_{2})\operatorname{d}z_{1}\operatorname{d}z_{2},

where ff is as given by (2) and

g(u1,u2,𝒰):=1{dist2(u1,o),dist2(u2,o)n,dist2(u1,u2)>nx1,BGab(u1,u1)𝒰= or BGab+(u1,u2)𝒰=.}.g(u_{1},u_{2},{\mathcal{U}}):=1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(u_{1},o),\operatorname{dist}_{\mathbb{H}^{2}}(u_{2},o)\leq n,\\ \operatorname{dist}_{\mathbb{H}^{2}}(u_{1},u_{2})>n-x-1,\\ \text{$B_{\text{Gab}}^{-}(u_{1},u_{1})\cap{\mathcal{U}}=\emptyset$ or $B_{\text{Gab}}^{+}(u_{1},u_{2})\cap{\mathcal{U}}=\emptyset$.}\end{array}\right\}}.

For every z1,z2𝔻z_{1},z_{2}\in\mathbb{D} have

𝔼[g(z1,z2,𝒵{z1,z2})]1{dist2(z1,o),dist2(z2,o)n,dist2(z1,z2)>nx1}2eλarea2(BGab(z1,z2))/21{dist2(z1,o),dist2(z2,o)n}2ecen/2,\begin{array}[]{rcl}{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]&\leq&\displaystyle 1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},o),\operatorname{dist}_{\mathbb{H}^{2}}(z_{2},o)\leq n,\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})>n-x-1\end{array}\right\}}\cdot 2e^{-\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(z_{1},z_{2}))/2}\\ &\leq&\displaystyle 1_{\left\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},o),\operatorname{dist}_{\mathbb{H}^{2}}(z_{2},o)\leq n\right\}}\cdot 2e^{-ce^{n/2}},\end{array}

for a suitably chosen small constant c=c(x,λ)c=c(x,\lambda). (The last step in some more detail : for all z1,z2z_{1},z_{2} for which dist2(z1,z2)>nx1\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})>n-x-1, we have area2(BGab(z1,z2))2π(cosh((nx1)/2)1)\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(z_{1},z_{2}))\geq 2\pi(\cosh((n-x-1)/2)-1). Since cosh(t)=(1/2+ot(1))et\cosh(t)=(1/2+o_{t}(1))e^{t} as tt\to\infty, there is an n0n_{0} such that 2π(cosh((nx1)/2)1)/2e(nx1)/22\pi(\cosh((n-x-1)/2)-1)/2\geq e^{(n-x-1)/2} for all nn0n\geq n_{0}. We can now choose a small 0<c<(λ/2)e(x+1)/20<c<(\lambda/2)\cdot e^{-(x+1)/2} such that 2ecen/212e^{-ce^{n/2}}\geq 1 for all nn0n\leq n_{0}.) It follows that

𝔼Yn(λarea2(B2(o,n)))22ecen/2=λ2(2π)2(coshn1)22ecen/22π2λ2e2necen/2,\begin{array}[]{rcl}{\mathbb{E}}Y_{n}&\leq&\left(\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,n))\right)^{2}\cdot 2e^{-ce^{n/2}}\\ &=&\lambda^{2}\cdot(2\pi)^{2}\cdot\left(\cosh n-1\right)^{2}\cdot 2e^{-ce^{n/2}}\\ &\leq&2\pi^{2}\lambda^{2}e^{2n}e^{-ce^{n/2}},\end{array}

and hence

𝔼Y2π2λ2ne2necen/2<.{\mathbb{E}}Y\leq 2\pi^{2}\lambda^{2}\sum_{n}e^{2n}e^{-ce^{n/2}}<\infty.

So YY is finite almost surely.

We have now shown that for every fixed x>0x>0 we have (Nx<)=1{\mathbb{P}}(N_{x}<\infty)=1 where

Nx:=|{pseudo-edges z1,z2𝒵 for which cert(z1,z2)B2(o,x)}|.N_{x}:=\left|\left\{\text{pseudo-edges $z_{1},z_{2}\in{\mathcal{Z}}$ for which $\operatorname{cert}(z_{1},z_{2})\cap B_{\mathbb{H}^{2}}(o,x)\neq\emptyset$}\right\}\right|.

Since NxNyN_{x}\leq N_{y} whenever x<yx<y we also have

(x>0{Nx<})=limx(Nx<)=1,{\mathbb{P}}\left(\bigcap_{x>0}\{N_{x}<\infty\}\right)=\lim_{x\to\infty}{\mathbb{P}}(N_{x}<\infty)=1,

which is what needed to be shown. \blacksquare

Proof of Proposition 31. We fix arbitrary λ>0,0p1\lambda>0,0\leq p\leq 1 and r,w1,w2,ϑ>0r,w_{1},w_{2},\vartheta>0. Since the typical degree has finite expectation by Isokawa’s formula, almost surely, the cell of the origin in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\} is adjacent to at most finitely many other cells. For every point of 𝒵{\mathcal{Z}}, the number of cells it is adjacent to in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\} is at most one more than the number of cells it is adjacent to in the tessellation for 𝒵{\mathcal{Z}}. In particular, by Lemma 32, also in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\}, almost surely, each cell is adjacent to at most finitely many other cells.

We consider an arbitrary realization of the Poisson process 𝒵{\mathcal{Z}} (in other words, a locally finite point set 𝒵𝔻{\mathcal{Z}}\subseteq\mathbb{D}, partitioned into two parts 𝒵=𝒵b𝒵w{\mathcal{Z}}={\mathcal{Z}}_{\text{b}}\uplus{\mathcal{Z}}_{\text{w}}) for which each Voronoi cell of 𝒵{o}{\mathcal{Z}}\cup\{o\} is adjacent to finitely many others, and such that, for every x>0x>0, the ball B2(o,x)B_{\mathbb{H}^{2}}(o,x) intersects at most finitely many certificates of pseudo-edges. We will show that in such a realization either the black cluster of the origin is finite, or there exists an infinite, good, black pseudopath, or there exists an infinite linked sequence of chunks all of whose points are black (or more than one of the three options occurs).

Suppose thus that the origin oo lies in an infinite black component (otherwise we are done). Since each cell is adjacent to finitely many others, there must exist an infinite (ordinary) path P=z0,z1,z2,P=z_{0},z_{1},z_{2},\dots with z0=oz_{0}=o and z1,z2,𝒵bz_{1},z_{2},\dots\in{\mathcal{Z}}_{b} (distinct). This is of course also a pseudo-path. If only finitely many pseudo-edges of PP are bad then we are done, as PP will contain a black, infinite, good pseudopath. Hence from now on we will assume that PP has infinitely many bad edges. In particular, we can find a t11t_{1}\geq 1 such that the pseudo-edges z0z1,z1z2,,zt12zt11z_{0}z_{1},z_{1}z_{2},\dots,z_{t_{1}-2}z_{t_{1}-1} are good and zt11zt1z_{t_{1}-1}z_{t_{1}} is bad. We set P1:=z0,z1,,zt1P_{1}:=z_{0},z_{1},\dots,z_{t_{1}}.

Let us say that an index ii interacts with a subpath Q=zs,zs+1,,ztQ=z_{s},z_{s+1},\dots,z_{t} of PP if one of the following holds: sits\leq i\leq t or dist2(zi,cert(Q))<r/1000\operatorname{dist}_{\mathbb{H}^{2}}(z_{i},\operatorname{cert}(Q))<r/1000 or cert(zi,zi+1)cert(Q)\operatorname{cert}(z_{i},z_{i+1})\cap\operatorname{cert}(Q)\neq\emptyset.

We now let s2t1s_{2}\geq t_{1} be the largest index that interacts with P1P_{1}. That s2s_{2} exists (is finite) follows from the fact that

ucert(P1)B2(u,r1000)B2(o,x),\bigcup_{u\in\operatorname{cert}(P_{1})}B_{\mathbb{H}^{2}}\left(u,\frac{r}{1000}\right)\subseteq B_{\mathbb{H}^{2}}(o,x),

for some x>0x>0. Since PP contains infinitely many bad pseudo-edges, there exists a t2>s2t_{2}>s_{2} such that the pseudo-edges zs2zs2+1,,zt22zt21z_{s_{2}}z_{s_{2}+1},\dots,z_{t_{2}-2}z_{t_{2}-1} are good and zt21zt2z_{t_{2}-1}z_{t_{2}} is bad. We set P2:=zs2,zs2+1,,zt2P_{2}:=z_{s_{2}},z_{s_{2}+1},\dots,z_{t_{2}}. (A minor subtlety needs to be added here. When we say zt21zt2z_{t_{2}-1}z_{t_{2}} is bad, we mean that it is bad wrt. the path zs2,zs2+1,,zt2z_{s_{2}},z_{s_{2}+1},\dots,z_{t_{2}}. We do this in order to handle the case where zs2zs2+1z_{s_{2}}z_{s_{2}+1} is a type III pseudo-edge of PP. If we took this single pseudo-edge as the path P2P_{2}, then P2P_{2} would not be a chunk but a good pseudopath.)

We now let s3t2s_{3}\geq t_{2} be the largest index that interacts with P2P_{2}. Again s3s_{3} exists as ucert(P2)B2(u,r/1000)\bigcup_{u\in\operatorname{cert}(P_{2})}B_{\mathbb{H}^{2}}(u,r/1000) is contained in B2(o,x)B_{\mathbb{H}^{2}}(o,x) for some x>0x>0. Since PP contains infinitely many bad edges, there exists t3>s3t_{3}>s_{3} such that the pseudo-edges zs3zs3+1,,zt32zt31z_{s_{3}}z_{s_{3}+1},\dots,z_{t_{3}-2}z_{t_{3}-1} are good and zt31zt3z_{t_{3}-1}z_{t_{3}} is bad (wrt. the path zs3,zs3+1,,zt3z_{s_{3}},z_{s_{3}+1},\dots,z_{t_{3}}). We set P3:=zs3,zs3+1,,zt3P_{3}:=z_{s_{3}},z_{s_{3}+1},\dots,z_{t_{3}}.

We continue defining s4,s5,s_{4},s_{5},\dots and t4,t5,t_{4},t_{5},\dots and P4,P5,P_{4},P_{5},\dots analogously, producing an infinite linked sequence of chunks P1,P2,P_{1},P_{2},\dots.

(We point out that the choice of sis_{i} guarantees that no index j>sij>s_{i} interacts with any of P1,,Pi1P_{1},\dots,P_{i-1} while sis_{i} interacts with Pi1P_{i-1} but not P1,,Pi2P_{1},\dots,P_{i-2}. From this it easily follows that the demands (i), (ii) of Definition 30 are met.)

Since we considered an arbitrary realization of 𝒵{\mathcal{Z}} that satisfies two conditions that both hold almost surely, it follows that almost surely either the black cluster of the origin is finite or there exists an infinite linked sequence of chunks starting from the origin all of whose points are black. \blacksquare

The main thing that now remains, in order to prove Proposition 28, is to show that for a suitable choice of r,w1,w2,ϑr,w_{1},w_{2},\vartheta and for λ\lambda sufficiently small and p=(1ε)(π/3)λp=(1-\varepsilon)(\pi/3)\lambda with 0<ε<10<\varepsilon<1 a small constant, almost surely there is no black, infinite, good pseudopath and no black, infinite linked sequence of chunks starting from oo.

As in the proof of the upper bound, from now on we will always take r:=2ln(1/λ)r:=2\ln(1/\lambda). The parameters w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 will be constants that depend on ε\varepsilon and that we will choose appropriately in the end. We will use an inductive approach bounding the expected number of good pseudopaths, respectively linked sequences of chunks, of length nn starting from the origin. With this in mind, we will prove a number of lemmas designed to deal with the different ways in which an additional point can be added to an existing good pseudopath, respectively linked sequence of chunks, of length n1n-1. But, first we need to derive some additional facts about hyperbolic geometry. Again, when reading the paper for the first time, the reader may wish to only read the definitions and lemma statements and skip over the proofs.

3.3.1 More geometric ingredients

Let us say that a sequence of points u0,u1,,uk𝔻u_{0},u_{1},\dots,u_{k}\in\mathbb{D} is a pre-chunk (wrt. r,w1,w2,ϑr,w_{1},w_{2},\vartheta) if they are placed in such a way that they will form a chunk whenever the random point set 𝒵{\mathcal{Z}} turns out favourably. That is, u0,u1,,uk𝔻u_{0},u_{1},\dots,u_{k}\in\mathbb{D} is a pre-chunk if rw1<dist2(ui1,ui)<r+w2r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})<r+w_{2} for i=1,,k1i=1,\dots,k-1 and ui2ui1ui>ϑ\angle u_{i-2}u_{i-1}u_{i}>\vartheta for all 2ik12\leq i\leq k-1 and in addition dist2(uk1,uk)rw1\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})\leq r-w_{1} or dist2(uk1,uk)r+w2\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})\geq r+w_{2} or rw1<dist2(uk1,uk)<r+w2r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})<r+w_{2} and uk2uk1ukϑ\angle u_{k-2}u_{k-1}u_{k}\leq\vartheta. In particular, a pre-chunk is such that u0,,uk1u_{0},\dots,u_{k-1} forms a (r,w,ϑ)(r,w,\vartheta)-path with w:=max(w1,w2)w:=\max(w_{1},w_{2}).

If rw1<dist2(ui1,ui)<r+w2r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{i-1},u_{i})<r+w_{2} and ui2ui1ui>ϑ\angle u_{i-2}u_{i-1}u_{i}>\vartheta (in case i2i\geq 2) for i=1,,ki=1,\dots,k then we speak of a good pre-pseudopath.

The first geometric observation in this section will be helpful for the situation when we want to add a new pseudo-edge to an existing pseudopath, and will also be used by later proofs in this section. It tells us that if we want to add a new pseudo-edge to a good pre-pseudopath, then either the intersection of the certificate of the new pseudo-edge and the certificates of the existing path is contained in a ball of constant radius (and hence has rather small area), or the new edge makes a small angle with the last edge of the existing pseudo-path. This will translate into useful bounds once we start estimating the expected number of linked sequences of chunks later on.

Lemma 34

For every w1,w2,ϑ1,ϑ2>0w_{1},w_{2},\vartheta_{1},\vartheta_{2}>0 there are r0=r0(w1,w2,ϑ1,ϑ2)r_{0}=r_{0}(w_{1},w_{2},\vartheta_{1},\vartheta_{2}) and h=h(w1,w2,ϑ1,ϑ2)h=h(w_{1},w_{2},\vartheta_{1},\vartheta_{2}) such that the following holds for all rr0r\geq r_{0}. If u0,,uku_{0},\dots,u_{k} is such that u0,,uk1u_{0},\dots,u_{k-1} is a good pre-pseudopath wrt. r,w1,w2,ϑ1r,w_{1},w_{2},\vartheta_{1} then either

cert(u0,,uk1)cert(uk1,uk)B2(uk1,h),\operatorname{cert}(u_{0},\dots,u_{k-1})\cap\operatorname{cert}(u_{k-1},u_{k})\subseteq B_{\mathbb{H}^{2}}(u_{k-1},h),

or

uk2uk1uk<ϑ2,\angle u_{k-2}u_{k-1}u_{k}<\vartheta_{2},

(or both).

Before giving the proof, we point out that by setting ϑ1=ϑ2\vartheta_{1}=\vartheta_{2} we obtain:

Corollary 35

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there are r0=r0(w1,w2,ϑ)r_{0}=r_{0}(w_{1},w_{2},\vartheta) and h=h(w1,w2,ϑ)h=h(w_{1},w_{2},\vartheta) such that the following holds for all rr0r\geq r_{0}. If u0,,uku_{0},\dots,u_{k} is a good pre-pseudopath (wrt. r,w1,w2,ϑr,w_{1},w_{2},\vartheta) then

cert(u0,,uk1)cert(uk1,uk)B2(uk1,h).\operatorname{cert}(u_{0},\dots,u_{k-1})\cap\operatorname{cert}(u_{k-1},u_{k})\subseteq B_{\mathbb{H}^{2}}(u_{k-1},h).

Proof of Lemma 34. We fix 0<ϑ<min(ϑ2/2,1/1000)0<\vartheta^{\prime}<\min(\vartheta_{2}/2,1/1000). For any r>0r>0, if P=u0,,ukP=u_{0},\dots,u_{k} is a pre-chunk wrt. r,w1,w2,ϑ1r,w_{1},w_{2},\vartheta_{1} then P=u0,,uk1P^{\prime}=u_{0},\dots,u_{k-1} is a (r,w,ϑ1)(r,w,\vartheta_{1})-path, setting w:=max(w1,w2)w:=\max(w_{1},w_{2}). We also point out that, since PP^{\prime} is a good pre-pseudopath, cert(ui1,ui)=DD(ui1,ui,r+w2)B2(ui1,r+w)B2(ui,r+w)\operatorname{cert}(u_{i-1},u_{i})=DD(u_{i-1},u_{i},r+w_{2})\subseteq B_{\mathbb{H}^{2}}(u_{i-1},r+w)\cap B_{\mathbb{H}^{2}}(u_{i},r+w) for i=1,,k1i=1,\dots,k-1. By Proposition 10 there exist constants r0=r0(w,ϑ1,ϑ)r_{0}=r_{0}(w,\vartheta_{1},\vartheta^{\prime}) and h=h(w,ϑ1,ϑ)h=h(w,\vartheta_{1},\vartheta^{\prime}) such that, whenever rr0r\geq r_{0}, we have

cert(u0,,uk1)i=0k2B2(ui,r+w)=vPuk2uk1B2(v,r+w)B2(uk1,h)sect(uk1,uk2,ϑ).\begin{array}[]{rcl}\operatorname{cert}(u_{0},\dots,u_{k-1})&\subseteq&\bigcup_{i=0}^{k-2}B_{\mathbb{H}^{2}}(u_{i},r+w)=\bigcup_{v\in P_{u_{k-2}\setminus u_{k-1}}}B_{\mathbb{H}^{2}}(v,r+w)\\ &\subseteq&B_{\mathbb{H}^{2}}(u_{k-1},h)\cup\operatorname{sect}\left(u_{k-1},u_{k-2},\vartheta^{\prime}\right).\end{array}

Let us write s:=max(dist2(uk1,uk),r+w)s:=\max\left(\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k}),r+w\right). By Lemma 16 (applied with r=sr=s), we can assume without loss of generality that the constant hh is such that:

cert(uk1,uk)B2(uk,s)B2(uk1,h)sect(uk1,uk,ϑ).\begin{array}[]{rcl}\operatorname{cert}(u_{k-1},u_{k})\subseteq B_{\mathbb{H}^{2}}(u_{k},s)\subseteq B_{\mathbb{H}^{2}}(u_{k-1},h)\cup\operatorname{sect}\left(u_{k-1},u_{k},\vartheta^{\prime}\right).\end{array}

Since ϑ2>2ϑ\vartheta_{2}>2\vartheta^{\prime}, we have that if uk2uk1ukϑ2\angle u_{k-2}u_{k-1}u_{k}\geq\vartheta_{2} then sect(uk1,uk2,ϑ)sect(uk1,uk,ϑ)=\operatorname{sect}\left(u_{k-1},u_{k-2},\vartheta^{\prime}\right)\cap\operatorname{sect}\left(u_{k-1},u_{k},\vartheta^{\prime}\right)=\emptyset. It follows that either uk2uk1uk<ϑ2\angle u_{k-2}u_{k-1}u_{k}<\vartheta_{2} or cert(u0,,uk1)cert(uk1,uk)B2(uk1,h)\operatorname{cert}(u_{0},\dots,u_{k-1})\cap\operatorname{cert}(u_{k-1},u_{k})\subseteq B_{\mathbb{H}^{2}}(u_{k-1},h), as claimed. \blacksquare

Our next observation provides an upper bound on the number of edges in a pseudopath whose certificates intersect a given ball. This will be of use to us later on several times, for instance when we want to estimate or bound the expected number of pseudo-edges that can act as the start of a new chunk extending a given sequence of chunks.

Lemma 36

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exists r0=r0(w1,w2,ϑ)>0r_{0}=r_{0}(w_{1},w_{2},\vartheta)>0 such that, for every rr0r\geq r_{0}, every pre-chunk u0,u1,,uku_{0},u_{1},\dots,u_{k} and every ball B=B2(c,s)B=B_{\mathbb{H}^{2}}(c,s), it holds that

|{1ik:cert(ui1,ui)B}|4sr+10.|\{1\leq i\leq k:\operatorname{cert}(u_{i-1},u_{i})\cap B\neq\emptyset\}|\leq\frac{4s}{r}+10.

Proof. By Proposition 9 and the remark following the definition of pre-chunk, there exists an r0,Kr_{0},K such that if rr0r\geq r_{0} and 0i,jk10\leq i,j\leq k-1 then

dist2(ui,uj)|ij|(rK)|ij|(r/2),\operatorname{dist}_{\mathbb{H}^{2}}(u_{i},u_{j})\geq|i-j|\cdot(r-K)\geq|i-j|\cdot(r/2),

for all sufficiently large rr and all 1i,jk1\leq i,j\leq k.

For 1k11\leq\ell\leq k-1, let us denote by c+,cc_{\ell}^{+},c_{\ell}^{-} the centers of the two disks whose union is cert(u1,u)=DD(u1,u,r+w2)\operatorname{cert}(u_{\ell-1},u_{\ell})=DD(u_{\ell-1},u_{\ell},r+w_{2}). If 1i<jk11\leq i<j\leq k-1 then

dist2(ci+,cj+),dist2(ci+,cj),dist2(ci,cj+),dist2(ci,cj)dist2(ui,uj+1)2(r+w22)(j+1i)(r/2)2r=|ij|(r/2)(3/2)r,\begin{array}[]{rcl}\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{+},c_{j}^{+}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{+},c_{j}^{-}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{-},c_{j}^{+}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{-},c_{j}^{-})&\geq&\operatorname{dist}_{\mathbb{H}^{2}}(u_{i},u_{j+1})-2\left(\frac{r+w_{2}}{2}\right)\\ &\geq&(j+1-i)\cdot(r/2)-2r\\ &=&|i-j|\cdot(r/2)-(3/2)\cdot r,\end{array}

(the second inequality holding for rr sufficiently large).

Now observe Bcert(u1,u)B\cap\operatorname{cert}(u_{\ell-1},u_{\ell})\neq\emptyset if and only if min(dist2(c,c+),dist2(c,c))<(r+w2)/2+s\min\left(\operatorname{dist}_{\mathbb{H}^{2}}(c,c_{\ell}^{+}),\operatorname{dist}_{\mathbb{H}^{2}}(c,c_{\ell}^{-})\right)<(r+w_{2})/2+s. So if Bcert(ui1,ui)B\cap\operatorname{cert}(u_{i-1},u_{i})\neq\emptyset and Bcert(uj1,uj)B\cap\operatorname{cert}(u_{j-1},u_{j})\neq\emptyset then

2r+2s2(r+w22+s)min(dist2(ci+,cj+),dist2(ci+,cj),dist2(ci,cj+),dist2(ci,cj))|ij|(r/2)(3/2)r,\begin{array}[]{rcl}2r+2s&\geq&2\cdot\left(\frac{r+w_{2}}{2}+s\right)\\ &\geq&\min\left(\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{+},c_{j}^{+}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{+},c_{j}^{-}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{-},c_{j}^{+}),\operatorname{dist}_{\mathbb{H}^{2}}(c_{i}^{-},c_{j}^{-})\right)\\ &\geq&|i-j|\cdot(r/2)-(3/2)\cdot r,\end{array}

the first inequality holding provided r0r_{0} was chosen sufficiently large. Rearranging, we find

|ij|4sr+7.|i-j|\leq\frac{4s}{r}+7.

This shows that the number of 1ik11\leq i\leq k-1 such that cert(ui1,ui)B\operatorname{cert}(u_{i-1},u_{i})\cap B\neq\emptyset is at most 4sr+8\frac{4s}{r}+8. Including also cert(uk1,uk)\operatorname{cert}(u_{k-1},u_{k}), we obtain 4sr+9\frac{4s}{r}+9 which is smaller than the bound in the statement of the lemma. \blacksquare

Using the previous lemma, we can now show that, for any ball whose radius is only a constant larger than (r+w2)/2(r+w_{2})/2 (the radius of the two balls whose union is the certificate of a good pseudoedge), most of the area of the ball lies outside the certificate of any given pseudopath.

Lemma 37

For every ε,w1,w2,ϑ>0\varepsilon,w_{1},w_{2},\vartheta>0 there exists r0=r0(ε,w1,w2,ϑ)>0r_{0}=r_{0}(\varepsilon,w_{1},w_{2},\vartheta)>0 such that, for every rr0r\geq r_{0}, every good pre-pseudopath u0,u1,,uku_{0},u_{1},\dots,u_{k} and every ball B=B2(c,s)B=B_{\mathbb{H}^{2}}(c,s) with radius s(r+w2)/2+10ln10lnεs\geq(r+w_{2})/2+10\ln 10-\ln\varepsilon, it holds that

area2(Bcert(u0,,uk))(1ε)area2(B).\operatorname{area}_{\mathbb{H}^{2}}(B\setminus\operatorname{cert}(u_{0},\dots,u_{k}))\geq(1-\varepsilon)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B).

Proof. We let r0r_{0} be a large constant, to be chosen more precisely during the proof.

By Lemma 36, we have

area2(Bcert(u0,,uk))(8sr+20)area2(B2(o,(r+w2)/2))(8sr+20)πe(r+w2)/2.\begin{array}[]{rcl}\operatorname{area}_{\mathbb{H}^{2}}(B\cap\operatorname{cert}(u_{0},\dots,u_{k}))&\leq&\left(\frac{8s}{r}+20\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,(r+w_{2})/2))\\ &\leq&\left(\frac{8s}{r}+20\right)\cdot\pi e^{(r+w_{2})/2}.\end{array}

As area2(B)=2π(coshs1)=(1+os(1))πes\operatorname{area}_{\mathbb{H}^{2}}(B)=2\pi(\cosh s-1)=(1+o_{s}(1))\cdot\pi e^{s}, we have, provided we chose r0r_{0} sufficiently large, that area2(B)12πes\operatorname{area}_{\mathbb{H}^{2}}(B)\geq\frac{1}{2}\pi e^{s}. Hence

area2(Bcert(u0,,uk))area2(B)((16/r)ses+40es)e(r+w2)/2.\frac{\operatorname{area}_{\mathbb{H}^{2}}(B\cap\operatorname{cert}(u_{0},\dots,u_{k}))}{\operatorname{area}_{\mathbb{H}^{2}}(B)}\leq\left((16/r)se^{-s}+40e^{-s}\right)\cdot e^{(r+w_{2})/2}.

Since the RHS is decreasing in ss for s1s\geq 1, we have

area2(Bcert(u0,,uk))area2(B)(16(r+w2)/2+10ln10lnεr+40)e10ln10+lnε(56+16w2/2+10ln10lnεr0)1010ε<ε,\begin{array}[]{rcl}\frac{\operatorname{area}_{\mathbb{H}^{2}}(B\cap\operatorname{cert}(u_{0},\dots,u_{k}))}{\operatorname{area}_{\mathbb{H}^{2}}(B)}&\leq&\left(16\frac{(r+w_{2})/2+10\ln 10-\ln\varepsilon}{r}+40\right)\cdot e^{-10\ln 10+\ln\varepsilon}\\ &\leq&\left(56+16\frac{w_{2}/2+10\ln 10-\ln\varepsilon}{r_{0}}\right)\cdot 10^{-10}\cdot\varepsilon\\ &<&\varepsilon,\end{array}

provided we chose r0r_{0} sufficiently large. \blacksquare

Next we present another observation that will be helpful for the situation when we want to add a new pseudo-edge to an existing pseudopath. It tells us that one of three things must happen, each of which will translate into useful bounds once we start estimating the number of linked sequences of chunks later on.

Corollary 38

For every ε,w1,w2,ϑ1,ϑ2>0\varepsilon,w_{1},w_{2},\vartheta_{1},\vartheta_{2}>0 there exists r0=r0(ε,w1,w2,ϑ1,ϑ2)>0r_{0}=r_{0}(\varepsilon,w_{1},w_{2},\vartheta_{1},\vartheta_{2})>0 such that, for every rr0r\geq r_{0} and every pre-chunk u0,u1,,uku_{0},u_{1},\dots,u_{k} wrt. r,w1,w2,ϑ1r,w_{1},w_{2},\vartheta_{1} we have that either

area2(BGab(uk1,uk)cert(u0,,uk1))(1ε)area2(BGab(uk1,uk)),\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k-1},u_{k})\setminus\operatorname{cert}(u_{0},\dots,u_{k-1}))\geq(1-\varepsilon)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k-1},u_{k})),

or

dist2(uk1,uk)rw1,\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})\leq r-w_{1},

or

rw1<dist2(uk1,uk)<r+w2+20ln102lnε and uk2uk1uk<ϑ2.r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})<r+w_{2}+20\ln 10-2\ln\varepsilon\text{ and }\angle u_{k-2}u_{k-1}u_{k}<\vartheta_{2}.

(or more than one of the above hold).

Proof. We let r0r_{0} be a large constant, to be determined in the course of the proof and we let rr0r\geq r_{0} be arbitrary and we let u0,,uku_{0},\dots,u_{k} be an arbitrary pre-chunk wrt. r,w1,w2,ϑ1r,w_{1},w_{2},\vartheta_{1}.

Of course there is nothing to prove if dist2(uk1,uk)rw1\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})\leq r-w_{1}. If dist2(uk1,uk)r+w2+20ln102lnε\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})\geq r+w_{2}+20\ln 10-2\ln\varepsilon then BGab(uk1,uk)B_{\text{Gab}}(u_{k-1},u_{k}) has radius (r+w2)/2+10ln10lnε\geq(r+w_{2})/2+10\ln 10-\ln\varepsilon and we are done by Lemma 37 – assuming without loss of generality we chose r0r_{0} larger that the bound provided in that lemma.

Let us thus assume rw1<dist2(uk1,uk)<r+w2+20ln10lnε=:r+w2r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})<r+w_{2}+20\ln 10-\ln\varepsilon=:r+w_{2}^{\prime}. By Lemma 34, there are constants r0=r0(w1,w2,ϑ1,ϑ2),h=h(w1,w2,ϑ1,ϑ2)r_{0}^{\prime}=r_{0}^{\prime}(w_{1},w_{2}^{\prime},\vartheta_{1},\vartheta_{2}),h=h(w_{1},w_{2}^{\prime},\vartheta_{1},\vartheta_{2}) such that if rr0r\geq r_{0}^{\prime} then either

BGab(uk1,uk)cert(u0,,uk1)cert(uk1,uk)cert(u0,,uk1)B2(uk1,h),B_{\text{Gab}}(u_{k-1},u_{k})\cap\operatorname{cert}(u_{0},\dots,u_{k-1})\subseteq\operatorname{cert}(u_{k-1},u_{k})\cap\operatorname{cert}(u_{0},\dots,u_{k-1})\subseteq B_{\mathbb{H}^{2}}(u_{k-1},h),

or

uk2uk1uk<ϑ2.\angle u_{k-2}u_{k-1}u_{k}<\vartheta_{2}.

In the latter case we are clearly done. In the former case we note that

area2(BGab(uk1,uk)cert(u0,,uk1))area2(BGab(uk1,uk))coshh1cosh(r0w12)1<ε,\frac{\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k-1},u_{k})\cap\operatorname{cert}(u_{0},\dots,u_{k-1}))}{\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k-1},u_{k}))}\leq\frac{\cosh h-1}{\cosh\left(\frac{r_{0}-w_{1}}{2}\right)-1}<\varepsilon,

provided r0r0r_{0}\geq r_{0}^{\prime} was chosen sufficiently large. \blacksquare

For some of the remaining proofs needed for the lower bound, we’ll use the following notion which may be of independent interest. For sets A,B𝔻A,B\subseteq\mathbb{D} we let

ahd(A,B):=supxBdist2(x,A)=supxBinfyAdist2(x,y).\operatorname{ahd}(A,B):=\sup_{x\in B}\operatorname{dist}_{\mathbb{H}^{2}}(x,A)=\sup_{x\in B}\operatorname*{\vphantom{p}inf}_{y\in A}\operatorname{dist}_{\mathbb{H}^{2}}(x,y).

The abbreviation ahd\operatorname{ahd} stands for “asymmetric Hausdorff distance”. As the reader may already have recognized, the Hausdorff distance between A,BA,B equals max(ahd(A,B),ahd(B,A))\max(\operatorname{ahd}(A,B),\operatorname{ahd}(B,A)).

The next two lemmas are a preparation for the third one, Lemma 41. That lemma allows us to bound the area of intersection of a given ball with the certificates of a given (pre-)pseudopath. This will be of use to us when we estimate the expected number of linked sequences of chunks later on.

Lemma 39

For every ε>0\varepsilon>0 there exists an a=a(ε)a=a(\varepsilon) such that the following holds. Let B=B2(c1,r1)B=B_{\mathbb{H}^{2}}(c_{1},r_{1}) be a disk and C=B2(c2,r2)C=\partial B_{\mathbb{H}^{2}}(c_{2},r_{2}) a hyperbolic circle such that ahd(B,C)a\operatorname{ahd}(B,C)\geq a. Then we have BCsect(c2,c1,ε)B\cap C\subseteq\operatorname{sect}\left(c_{2},c_{1},\varepsilon\right).

Proof. The statement is obviously true if BC=B\cap C=\emptyset. Similarly it is also true if BB and CC intersect in a single point (the common point would line on the line joining c1c_{1} and c2c_{2}). So from now on we can assume this is not the case.

Let x1,x2x_{1},x_{2} denote the two intersection points of the circles B\partial B and CC. It suffices to show that, provided ahd(B,C)a\operatorname{ahd}(B,C)\geq a and aa is sufficiently large, α:=c1c2x1(=c1c2x2)ε\alpha:=\angle c_{1}c_{2}x_{1}(=\angle c_{1}c_{2}x_{2})\leq\varepsilon. For convenience we write d:=dist2(c1,c2)d:=\operatorname{dist}_{\mathbb{H}^{2}}(c_{1},c_{2}). We next point out that

ahd(B,C)=r2r1+d.\operatorname{ahd}(B,C)=r_{2}-r_{1}+d.

(To see this we first note that by the triangle inequality dist2(z,c1)dist2(z,c2)+dist2(c1,c2)=r2+d\operatorname{dist}_{\mathbb{H}^{2}}(z,c_{1})\leq\operatorname{dist}_{\mathbb{H}^{2}}(z,c_{2})+\operatorname{dist}_{\mathbb{H}^{2}}(c_{1},c_{2})=r_{2}+d for all zCz\in C. In other words, dist2(z,B)=max(0,dist2(z,c1)r1)max(0,r2r1+d)\operatorname{dist}_{\mathbb{H}^{2}}(z,B)=\max(0,\operatorname{dist}_{\mathbb{H}^{2}}(z,c_{1})-r_{1})\leq\max(0,r_{2}-r_{1}+d) for all zCz\in C. Applying a suitable isometry if needed, we can assume without loss of generality that c2=oc_{2}=o is the origin and c1c_{1} lies on the negative xx-axis. The point p=(tanh(r2/2),0)Cp=(\tanh(r_{2}/2),0)\in C satisfies dist2(z,B)=dist2(z,c1)r1=r2r1+d\operatorname{dist}_{\mathbb{H}^{2}}(z,B)=\operatorname{dist}_{\mathbb{H}^{2}}(z,c_{1})-r_{1}=r_{2}-r_{1}+d, as c1,c2,pc_{1},c_{2},p lies on the xx-axis which is a hyperbolic line.)

By the hyperbolic law of cosines,

er1\displaystyle e^{r_{1}} cosh(r1)\displaystyle\geq\cosh(r_{1})
=cosh(r2)cosh(d)sinh(r2)sinh(d)cos(α)\displaystyle=\cosh(r_{2})\cosh(d)-\sinh(r_{2})\sinh(d)\cos(\alpha)
er2+d4(1cos(α)).\displaystyle\geq\frac{e^{r_{2}+d}}{4}\cdot\left(1-\cos(\alpha)\right).

Hence

cos(α)14er1r2d=14eahd(B,C).\cos(\alpha)\geq 1-4e^{r_{1}-r_{2}-d}=1-4e^{-\operatorname{ahd}(B,C)}.

The statement follows by taking a>ln(1cosε4)a>-\ln\left(\frac{1-\cos\varepsilon}{4}\right). \blacksquare

Lemma 40

We have

limainf{area2(B2\B1)area2(B2):B1,B2𝔻 (hyperbolic) disks, ahd(B1,B2)a}=1.\lim_{a\to\infty}\operatorname*{\vphantom{p}inf}\left\{\frac{\operatorname{area}_{\mathbb{H}^{2}}(B_{2}\backslash B_{1})}{\operatorname{area}_{\mathbb{H}^{2}}(B_{2})}:\text{$B_{1},B_{2}\subseteq\mathbb{D}$ (hyperbolic) disks, }\operatorname{ahd}(B_{1},B_{2})\geq a\right\}=1.

Proof. It suffices to show that for every ε>0\varepsilon>0 there exists a0a_{0} such that ahd(B1,B2)a0\operatorname{ahd}(B_{1},B_{2})\geq a_{0} implies that area2(B2B1)(1ε)area2(B2)\operatorname{area}_{\mathbb{H}^{2}}(B_{2}\setminus B_{1})\geq(1-\varepsilon)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{2}). We thus let a0a_{0} be a large constant, to be determined in the course of the proof, and we let B1=B2(c1,r1),B2=B2(c2,r2)B_{1}=B_{\mathbb{H}^{2}}(c_{1},r_{1}),B_{2}=B_{\mathbb{H}^{2}}(c_{2},r_{2}) be arbitrary disks with a:=ahd(B1,B2)a0a:=\operatorname{ahd}(B_{1},B_{2})\geq a_{0}.

If r2a/2r_{2}\leq a/2 then B1B_{1} and B2B_{2} are disjoint and we are done. For the remainder of the proof, we therefore assume r2>a/2r_{2}>a/2.

Each of the circles Cs:=B2(c2,s)C_{s}:=\partial B_{\mathbb{H}^{2}}(c_{2},s) with r2a1000sr2r_{2}-\frac{a}{1000}\leq s\leq r_{2} satisfies ahd(B1,Cs)(9991000)a\operatorname{ahd}(B_{1},C_{s})\geq\left(\frac{999}{1000}\right)a. Applying Lemma 39, having chosen a0a_{0} sufficiently large, we can assume that CsB1sect(c2,c1,ε1000)C_{s}\cap B_{1}\subseteq\operatorname{sect}\left(c_{2},c_{1},\frac{\varepsilon}{1000}\right) for all r2a1000sr2r_{2}-\frac{a}{1000}\leq s\leq r_{2}. It follows that

B1B2B2(c2,r2a1000)sect(c2,c1,ε1000).B_{1}\cap B_{2}\subseteq B_{\mathbb{H}^{2}}\left(c_{2},r_{2}-\frac{a}{1000}\right)\cup\operatorname{sect}\left(c_{2},c_{1},\frac{\varepsilon}{1000}\right).

Next we note that, by rotational symmetry of the hyperbolic area measure

area2(sect(c2,c1,ε1000)B2)area2(B2)=ε1000π.\frac{\operatorname{area}_{\mathbb{H}^{2}}\left(\operatorname{sect}\left(c_{2},c_{1},\frac{\varepsilon}{1000}\right)\cap B_{2}\right)}{\operatorname{area}_{\mathbb{H}^{2}}(B_{2})}=\frac{\varepsilon}{1000\cdot\pi}. (14)

We also have

area2(B2(c2,r2a1000)area2(c2,r2)=cosh(r2a1000)1cosh(r2)1cosh(r2a1000)cosh(r2)cosh(4991000a)cosh(a/2)ε1000,\begin{array}[]{rcl}\displaystyle\frac{\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(c_{2},r_{2}-\frac{a}{1000})}{\operatorname{area}_{\mathbb{H}^{2}}(c_{2},r_{2})}&=&\displaystyle\frac{\cosh(r_{2}-\frac{a}{1000})-1}{\cosh(r_{2})-1}\\[8.61108pt] &\leq&\displaystyle\frac{\cosh(r_{2}-\frac{a}{1000})}{\cosh(r_{2})}\\[8.61108pt] &\leq&\displaystyle\frac{\cosh(\frac{499}{1000}a)}{\cosh(a/2)}\\[8.61108pt] &\leq&\displaystyle\frac{\varepsilon}{1000},\end{array} (15)

where the second inequality follows from the fact that xcosh(xy)/cosh(x)x\mapsto\cosh(x-y)/\cosh(x) is decreasing in xx for xyx\geq y and the last inequality holds provided we chose a0a_{0} sufficiently large, and uses that cosh(4991000a)/cosh(a/2)0\cosh(\frac{499}{1000}a)/\cosh(a/2)\to 0 as aa\to\infty.

Combining (14) and (15), we see that

area2(B1B2)<εarea2(B2),\operatorname{area}_{\mathbb{H}^{2}}(B_{1}\cap B_{2})<\varepsilon\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{2}),

provided we chose a0a_{0} sufficiently large. This clearly implies the statement of the lemma. \blacksquare

Lemma 41

For every ε,w1,w2,ϑ>0\varepsilon,w_{1},w_{2},\vartheta>0 there exist r0=r0(ε,w1,w2,ϑ),a=a(ε,w,ϑ)>0r_{0}=r_{0}(\varepsilon,w_{1},w_{2},\vartheta),a=a(\varepsilon,w,\vartheta)>0 such that, for every pre-chunk u0,,uku_{0},\dots,u_{k} and every disk BB with ahd(cert(u0,,uk),B)a\operatorname{ahd}(\operatorname{cert}(u_{0},\dots,u_{k}),B)\geq a we have

area2(Bcert(u0,,uk))(1ε)area2(B).\operatorname{area}_{\mathbb{H}^{2}}\left(B\setminus\operatorname{cert}(u_{0},\dots,u_{k})\right)\geq(1-\varepsilon)\operatorname{area}_{\mathbb{H}^{2}}(B).

Proof. We fix a large constant K>0K>0, to be determined more precisely during the course of the proof. For convenience we’ll write C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and B=B2(c,s)B=B_{\mathbb{H}^{2}}(c,s). We let B1=B2(c1,r1),,BN=B2(cN,rN)B_{1}=B_{\mathbb{H}^{2}}(c_{1},r_{1}),\dots,B_{N}=B_{\mathbb{H}^{2}}(c_{N},r_{N}) be the balls that feature in the definition of cert(u0,u1),,cert(uk1,uk)\operatorname{cert}(u_{0},u_{1}),\dots,\operatorname{cert}(u_{k-1},u_{k}). That is each BiB_{i} is either BGab(uk1,uk)B_{\text{Gab}}(u_{k-1},u_{k}) or a disk of diameter r+w2r+w_{2} with uj1,ujBiu_{j-1},u_{j}\in\partial B_{i} for some 1jk1\leq j\leq k. In particular C=B1BNC=B_{1}\cup\dots\cup B_{N} and N2kN\leq 2k.

By Lemma 40 we can take the constant aa such that the assumption that ahd(C,B)a\operatorname{ahd}(C,B)\geq a implies that area2(BBi)(1ε/K)area2(B)\operatorname{area}_{\mathbb{H}^{2}}(B\setminus B_{i})\geq(1-\varepsilon/K)\operatorname{area}_{\mathbb{H}^{2}}(B) for each i=1,,Ni=1,\dots,N.

We set I:={i:BiB}I:=\{i:B_{i}\cap B\neq\emptyset\}. If |I|K|I|\leq K then the statement clearly holds. For the remainder of the proof we thus assume |I|>K|I|>K.

By Lemma 36, assuming r0r_{0} was chosen sufficiently large, we have |I|8(s/r)+20|I|\leq 8(s/r)+20 (each certificate is either a single ball or the union of two balls). Hence, provided we chose r0r_{0} and KK sufficiently large, |I|>K|I|>K implies s(K/10)rs\geq(K/10)\cdot r. We have

area2(BC)area2(B)iIarea2(BBi)area2(B)(εK)area2(B)iI:BiBGab(uk1,uk)area2(Bi)(1ε/K)area2(B)(8(s/r)+20)area2(B2(o,(r+w2)/2))(1ε/K)area2(B)sarea2(B2(o,(r+w2)/2)),\begin{array}[]{rcl}\operatorname{area}_{\mathbb{H}^{2}}\left(B\setminus C\right)&\geq&\operatorname{area}_{\mathbb{H}^{2}}(B)-\sum_{i\in I}\operatorname{area}_{\mathbb{H}^{2}}(B\cap B_{i})\\ &\geq&\operatorname{area}_{\mathbb{H}^{2}}(B)-\left(\frac{\varepsilon}{K}\right)\operatorname{area}_{\mathbb{H}^{2}}(B)-\sum_{i\in I:B_{i}\neq B_{\text{Gab}}(u_{k-1},u_{k})}\operatorname{area}_{\mathbb{H}^{2}}(B_{i})\\ &\geq&(1-\varepsilon/K)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B)-\left(8(s/r)+20\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,(r+w_{2})/2)\right)\\ &\geq&(1-\varepsilon/K)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B)-s\cdot\operatorname{area}_{\mathbb{H}^{2}}\left(B_{\mathbb{H}^{2}}(o,(r+w_{2})/2)\right),\end{array}

the last inequality holding assuming we have chosen r0r_{0} and KK sufficiently large. Next we point out that

sarea(B2(o,(r+w2)/2))area2(B)=s(cosh((r+w2)/2)1)coshs11000e(r+w2)/2ses1000e(r+w2)/2(K/10)re(K/10)r=1000ew2/2(K/10)re(K510)r1000ew2/2(K/10)r0e(K510)r0ε/2,\begin{array}[]{rcl}\frac{s\cdot\operatorname{area}\left(B_{\mathbb{H}^{2}}(o,(r+w_{2})/2)\right)}{\operatorname{area}_{\mathbb{H}^{2}}(B)}&=&\frac{s\cdot(\cosh((r+w_{2})/2)-1)}{\cosh s-1}\\ &\leq&1000\cdot e^{(r+w_{2})/2}\cdot s\cdot e^{-s}\\ &\leq&1000\cdot e^{(r+w_{2})/2}\cdot(K/10)\cdot r\cdot e^{-(K/10)r}\\ &=&1000\cdot e^{w_{2}/2}\cdot(K/10)\cdot r\cdot e^{-\left(\frac{K-5}{10}\right)r}\\ &\leq&1000\cdot e^{w_{2}/2}\cdot(K/10)\cdot r_{0}\cdot e^{-\left(\frac{K-5}{10}\right)r_{0}}\\ &\leq&\varepsilon/2,\end{array}

provided r0r_{0} and KK were chosen sufficiently large, using in the second line that coshx1=(1+ox(1))12ex\cosh x-1=(1+o_{x}(1))\cdot\frac{1}{2}\cdot e^{x} as xx\to\infty; in the third line that ssess\mapsto se^{-s} is decreasing for s>1s>1; and in the last line that recr0r\cdot e^{-cr}\to 0 as rr\to\infty (for every c>0c>0). It follows that if |I|>K|I|>K then also

area2(BC)>(1ε)area2(B),\operatorname{area}_{\mathbb{H}^{2}}\left(B\setminus C\right)>(1-\varepsilon)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B),

as claimed. \blacksquare

We conclude this subsection, with two lemmas giving conditions under which hyperbolic angles are small. Again, we will use these when estimating the expected number of linked sequences of chunks later on.

Lemma 42

There is a universal constant K>0K>0 such that the following holds. If B1=B2(c1,r1),B2=B2(c2,r2)B_{1}=B_{\mathbb{H}^{2}}(c_{1},r_{1}),B_{2}=B_{\mathbb{H}^{2}}(c_{2},r_{2}) and pB2p\in\partial B_{2} are such that B1B2B_{1}\cap B_{2}\neq\emptyset then either dist2(p,c1)<r1+K\operatorname{dist}_{\mathbb{H}^{2}}(p,c_{1})<r_{1}+K or c1pc2<10exp[(r1dist2(p,c1))/2]\angle c_{1}pc_{2}<10\exp\left[(r_{1}-\operatorname{dist}_{\mathbb{H}^{2}}(p,c_{1}))/2\right].

Proof. Let K=K(π/2)K=K(\pi/2) be as provided by Lemma 11. For notational convenience we write α:=c1pc2,ρ:=dist2(c1,p)\alpha:=\angle c_{1}pc_{2},\rho:=\operatorname{dist}_{\mathbb{H}^{2}}(c_{1},p) and d:=dist2(c1,c2)d:=\operatorname{dist}_{\mathbb{H}^{2}}(c_{1},c_{2}).

The disks B1B_{1} and B2B_{2} intersect if and only if d<r1+r2d<r_{1}+r_{2}. If απ/2\alpha\geq\pi/2 then Lemma 11 tells us that dρ+r2Kd\geq\rho+r_{2}-K. It follows that if απ/2\alpha\geq\pi/2 and B1B2B_{1}\cap B_{2}\neq\emptyset then ρ<r1+K\rho<r_{1}+K.

Let us then assume α<π/2\alpha<\pi/2. If B1B2B_{1}\cap B_{2}\neq\emptyset then the hyperbolic cosine rule gives

er1+r2>coshd=cosh(ρ)cosh(r2)cos(α)sinh(ρ)sinh(r2)14eρ+r2(1cosα)14πeρ+r2α2,\begin{array}[]{rcl}e^{r_{1}+r_{2}}&>&\cosh d\\ &=&\cosh(\rho)\cosh(r_{2})-\cos(\alpha)\sinh(\rho)\sinh(r_{2})\\ &\geq&\frac{1}{4}e^{\rho+r_{2}}(1-\cos\alpha)\\ &\geq&\frac{1}{4\pi}e^{\rho+r_{2}}\alpha^{2},\end{array}

where we have used that 1cos(α)α2π1-\cos(\alpha)\geq\frac{\alpha^{2}}{\pi} for all 0<α<π/20<\alpha<\pi/2. Rearranging and taking square roots gives α<4πe(r1ρ)/2<10e(r1ρ)/2\alpha<\sqrt{4\pi}\cdot e^{(r_{1}-\rho)/2}<10e^{(r_{1}-\rho)/2}. \blacksquare

We’ll use another rather straightforward consequence of the hyperbolic cosine rule:

Lemma 43

For all x1,x2𝔻x_{1},x_{2}\in\mathbb{D} we have

2πexp[12(dist2(x1,x2)dist2(o,x1)dist2(o,x2))]>x1ox2.2\pi\cdot\exp\left[\frac{1}{2}\cdot\left(\operatorname{dist}_{\mathbb{H}^{2}}(x_{1},x_{2})-\operatorname{dist}_{\mathbb{H}^{2}}(o,x_{1})-\operatorname{dist}_{\mathbb{H}^{2}}(o,x_{2})\right)\right]>\angle x_{1}ox_{2}.

Proof. For notational convenience, write ρ1=dist2(o,x1),ρ2=dist2(o,x2)\rho_{1}=\operatorname{dist}_{\mathbb{H}^{2}}(o,x_{1}),\rho_{2}=\operatorname{dist}_{\mathbb{H}^{2}}(o,x_{2}) and γ=x1ox2\gamma=\angle x_{1}ox_{2}. It suffices to show that edist2(x1,x2)>14π2eρ1+ρ2γ2e^{\operatorname{dist}_{\mathbb{H}^{2}}(x_{1},x_{2})}>\frac{1}{4\pi^{2}}\cdot e^{\rho_{1}+\rho_{2}}\cdot\gamma^{2}.

If γ<π/2\gamma<\pi/2 then the hyperbolic cosine rule gives that

edist2(x1,x2)>cosh(dist2(x1,x2))=cosh(ρ1)cosh(ρ2)cos(γ)sinh(ρ1)sinh(ρ2)14eρ1+ρ2(1cosγ)14πeρ1+ρ2γ2.\begin{array}[]{rcl}e^{\operatorname{dist}_{\mathbb{H}^{2}}(x_{1},x_{2})}&>&\cosh(\operatorname{dist}_{\mathbb{H}^{2}}(x_{1},x_{2}))\\ &=&\cosh(\rho_{1})\cosh(\rho_{2})-\cos(\gamma)\sinh(\rho_{1})\sinh(\rho_{2})\\ &\geq&\frac{1}{4}\cdot e^{\rho_{1}+\rho_{2}}\cdot(1-\cos\gamma)\\ &\geq&\frac{1}{4\pi}\cdot e^{\rho_{1}+\rho_{2}}\cdot\gamma^{2}.\end{array}

(Using that 1cos(γ)α2π1-\cos(\gamma)\geq\frac{\alpha^{2}}{\pi} for 0<γ<π/20<\gamma<\pi/2 in the third line.)

If on the other hand γ[π/2,π]\gamma\in[\pi/2,\pi] then similarly

edist2(x1,x2)>cosh(ρ1)cosh(ρ2)cos(γ)sinh(ρ1)sinh(ρ2)cosh(ρ1)cosh(ρ2)14eρ1+ρ214π2eρ1+ρ2γ2.\begin{array}[]{rcl}e^{\operatorname{dist}_{\mathbb{H}^{2}}(x_{1},x_{2})}&>&\cosh(\rho_{1})\cosh(\rho_{2})-\cos(\gamma)\sinh(\rho_{1})\sinh(\rho_{2})\\ &\geq&\cosh(\rho_{1})\cosh(\rho_{2})\\ &\geq&\frac{1}{4}\cdot e^{\rho_{1}+\rho_{2}}\\ &\geq&\frac{1}{4\pi^{2}}\cdot e^{\rho_{1}+\rho_{2}}\cdot\gamma^{2}.\end{array}

\blacksquare

3.3.2 Counting good pseudopaths

The next point of order is to bound the expected number of black, good pseudopaths starting from the origin. We plan to use an inductive approach and begin with a couple of lemmas designed for the situation where we add a good pseudo-edge to an existing, good pseudo-path.

Lemma 44

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 and 0<ε<10<\varepsilon<1 there exist λ0=λ0(ε,w1,w2,ϑ)\lambda_{0}=\lambda_{0}(\varepsilon,w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and all p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uku_{0},\dots,u_{k} be an arbitrary good pre-pseudopath and, writing C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}), define

YI-i:=|{z𝒵b:rw1dist2(uk,z)r+w2, and;uk1ukz>ϑ, and; a disk B with 𝒵BC=,uk,zB and diam2(B)<r+w2.}|.Y_{\text{\bf{I-i}}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}r-w_{1}\leq\operatorname{dist}_{\mathbb{H}^{2}}(u_{k},z)\leq r+w_{2},\text{ and;}\\ \angle u_{k-1}u_{k}z>\vartheta,\text{ and;}\\ \text{$\exists$ a disk $B$ with }{\mathcal{Z}}\cap B\setminus C=\emptyset,\\ u_{k},z\in\partial B\text{ and }\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w_{2}.\end{array}\right\}\right|.

We have

𝔼YI-i(1ε/2).{\mathbb{E}}Y_{\text{\bf{I-i}}}\leq(1-\varepsilon/2).

Before proceeding with the proof, let us clarify that in the above lemma we also allow k=0k=0. In that case the demand uk1ukz>ϑ\angle u_{k-1}u_{k}z>\vartheta of course does not apply.

Proof. We let λ0\lambda_{0} be a small constant to be determined in the course of the proof, and take an arbitrary λ<λ0\lambda<\lambda_{0} and p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda. Once again we apply Corollary 5 to find that

𝔼YI-i=pλ𝔻𝔼[g(z,𝒵{z})]f(z)dz,{\mathbb{E}}Y_{\text{\bf{I-i}}}=p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]f(z)\operatorname{d}z,

where ff is again as given by (2) and this time

g(u,𝒰):=1{rw1dist2(uk,u)r+w2, and;uk1uku>ϑ, and; a disk B with 𝒰BC=,uk,uB and diam2(B)<r+w2.}.g(u,{\mathcal{U}}):=1_{\left\{\begin{array}[]{l}r-w_{1}\leq\operatorname{dist}_{\mathbb{H}^{2}}(u_{k},u)\leq r+w_{2},\text{ and;}\\ \angle u_{k-1}u_{k}u>\vartheta,\text{ and;}\\ \text{$\exists$ a disk $B$ with }{\mathcal{U}}\cap B\setminus C=\emptyset,\\ u_{k},u\in\partial B\text{ and }\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w_{2}.\end{array}\right\}}.

By Corollary 35, assuming λ0\lambda_{0} was chosen sufficiently small, if z𝔻z\in\mathbb{D} is such that 𝔼[g(z,𝒵{z})]0{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]\neq 0 then Ccert(uk,z)B2(uk,h)C\cap\operatorname{cert}(u_{k},z)\subseteq B_{\mathbb{H}^{2}}(u_{k},h).

Applying a suitable isometry if needed (which leaves the distribution of 𝒵{\mathcal{Z}} invariant), we can assume without loss of generality that uk=ou_{k}=o is the origin. Recall that any disk BB satisfying o,zBo,z\in\partial B and diam2(B)<r+w\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w satisfies BDD(o,z,r+w)B\subseteq DD(o,z,r+w). So if z𝔻z\in\mathbb{D} is such that rw1dist2(0,z)r+w2r-w_{1}\leq\operatorname{dist}_{\mathbb{H}^{2}}(0,z)\leq r+w_{2} and uk1oz>ϑ\angle u_{k-1}oz>\vartheta then

𝔼[g(z,𝒵{z})]=( a disk B with 𝒵B=,o,zB and diam2(B)r+w2|𝒵C=)=( a disk B with 𝒵B=,o,zB and diam2(B)r+w2|𝒵Ccert(o,z)=)( a disk B with 𝒵B= and o,zB)(𝒵Ccert(o,z)=)( a disk B with 𝒵B= and o,zB)(𝒵B2(o,h)=)=𝔼[g^(z,𝒵{z})](𝒵B2(o,h)=),\begin{array}[]{rcl}{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]&=&{\mathbb{P}}\left(\begin{array}[]{l}\text{$\exists$ a disk $B$ with }{\mathcal{Z}}\cap B=\emptyset,\\ o,z\in\partial B\text{ and }\operatorname{diam}_{\mathbb{H}^{2}}(B)\leq r+w_{2}\end{array}{\Big{|}}{\mathcal{Z}}\cap C=\emptyset\right)\\[8.61108pt] &=&{\mathbb{P}}\left(\begin{array}[]{l}\text{$\exists$ a disk $B$ with }{\mathcal{Z}}\cap B=\emptyset,\\ o,z\in\partial B\text{ and }\operatorname{diam}_{\mathbb{H}^{2}}(B)\leq r+w_{2}\end{array}{\Big{|}}{\mathcal{Z}}\cap C\cap\operatorname{cert}(o,z)=\emptyset\right)\\[8.61108pt] &\leq&\displaystyle\frac{{\mathbb{P}}\left(\text{$\exists$ a disk $B$ with }{\mathcal{Z}}\cap B=\emptyset\text{ and }o,z\in\partial B\right)}{{\mathbb{P}}\left({\mathcal{Z}}\cap C\cap\operatorname{cert}(o,z)=\emptyset\right)}\\[8.61108pt] &\leq&\displaystyle\frac{{\mathbb{P}}\left(\text{$\exists$ a disk $B$ with }{\mathcal{Z}}\cap B=\emptyset\text{ and }o,z\in\partial B\right)}{{\mathbb{P}}\left({\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)=\emptyset\right)}\\[8.61108pt] &=&\displaystyle\frac{{\mathbb{E}}\left[\hat{g}(z,{\mathcal{Z}}\cup\{z\})\right]}{{\mathbb{P}}({\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)=\emptyset)},\end{array}

where g^(u,𝒰)=1\hat{g}(u,{\mathcal{U}})=1 if ouou is an edge of the Delaunay graph for 𝒰{o}{\mathcal{U}}\cup\{o\} and g^(u,𝒰)=0\hat{g}(u,{\mathcal{U}})=0 otherwise. It follows that

𝔼YI-ipλ(𝒵B2(o,h)=)𝔻𝔼[g^(z,𝒵{z})]f(z)dz=p(𝒵B2(o,h)=)(6+3πλ)(1ε)(1+2πλ)1λarea2(B2(o,h)),\begin{array}[]{rcl}{\mathbb{E}}Y_{\text{\bf{I-i}}}&\leq&\displaystyle\frac{p\lambda}{{\mathbb{P}}({\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)=\emptyset)}\cdot\int_{\mathbb{D}}{\mathbb{E}}\left[\hat{g}(z,{\mathcal{Z}}\cup\{z\})\right]f(z)\operatorname{d}z\\[8.61108pt] &=&\displaystyle\frac{p}{{\mathbb{P}}({\mathcal{Z}}\cap B_{\mathbb{H}^{2}}(o,h)=\emptyset)}\cdot\left(6+\frac{3}{\pi\lambda}\right)\\[8.61108pt] &\leq&\displaystyle\frac{(1-\varepsilon)\cdot(1+2\pi\lambda)}{1-\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,h))},\end{array}

where the second line follows by Isokawa’s formula (and the Slivniak-Mecke formula to phrase the typical degree as an integral).

Since hh is a constant we have λarea2(B2(o,h))0\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,h))\to 0 as λ0\lambda\searrow 0. So, provided we chose λ0\lambda_{0} sufficiently small, we have

𝔼YI-i<1ε/2,{\mathbb{E}}Y_{\text{\bf{I-i}}}<1-\varepsilon/2,

as claimed. \blacksquare

Lemma 45

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exists a λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all λ<λ0\lambda<\lambda_{0} and all p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uku_{0},\dots,u_{k} be an arbitrary good pre-pseudopath and, writing C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}), define

YI-ii:=|{z𝒵b:rw1<dist2(uk,z)<r+w2, and;uk1ukz>ϑ, and;either 𝒵DD(uk,z,r+w)C=,or 𝒵DD+(uk,z,r+w)C=,(or both).}|.Y_{\text{\bf{I-ii}}}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(u_{k},z)<r+w_{2},\text{ and;}\\ \angle u_{k-1}u_{k}z>\vartheta,\text{ and;}\\ \text{either }{\mathcal{Z}}\cap DD^{-}(u_{k},z,r+w)\setminus C=\emptyset,\\ \text{or }{\mathcal{Z}}\cap DD^{+}(u_{k},z,r+w)\setminus C=\emptyset,\\ \text{(or both).}\end{array}\right\}\right|.

We have

𝔼YI-ii104ew2eew2/2.{\mathbb{E}}Y_{\text{\bf{I-ii}}}\leq 10^{4}\cdot e^{w_{2}}\cdot e^{-e^{w_{2}/2}}.

Proof. The proof is nearly identical to the proof of the previous lemma. Reasoning as the proof of Lemma 44, there is a constant h=h(w1,w2,ϑ)h=h(w_{1},w_{2},\vartheta) such that

𝔼YI-ii𝔼X~𝐕𝐈1λarea2(B2(o,h)),{\mathbb{E}}Y_{\text{\bf{I-ii}}}\leq\frac{{\mathbb{E}}\tilde{X}_{\bf{VI}}}{1-\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,h))},

for all sufficiently small λ\lambda, where X~𝐕𝐈\tilde{X}_{\bf{VI}} is as in Lemma 26.

Since 1λarea2(B2(o,h))>1/101-\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,h))>1/10 provided we chose λ0\lambda_{0} sufficiently small, the result follows from Lemma 26 \blacksquare

These last two lemmas allow us to prove the following statement using an inductive approach. Here and in the rest of the paper we write

Gk:=|{black, good pseudopaths of length k starting from o}|.G_{k}:=\left|\left\{\text{black, good pseudopaths of length $k$ starting from $o$}\right\}\right|.
Lemma 46

For every 0<ε<10<\varepsilon<1 and w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exists a λ0=λ0(ε,w1,w2,ϑ)\lambda_{0}=\lambda_{0}(\varepsilon,w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and all p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, setting r:=2ln(1/λ)r:=2\ln(1/\lambda).

𝔼Gk(1ε/2+104ew2eew2/2)k.{\mathbb{E}}G_{k}\leq\left(1-\varepsilon/2+10^{4}e^{w_{2}}e^{-e^{w_{2}/2}}\right)^{k}.

Proof. By Corollary 5 we have

𝔼Gk=(pλ)n𝔻𝔻𝔼[gk(z1,,zk;𝒵{z1,,zk})]f(z1)f(zk)dzkdz1,{\mathbb{E}}G_{k}=(p\lambda)^{n}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k}(z_{1},\dots,z_{k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})\right]f(z_{1})\dots f(z_{k})\operatorname{d}z_{k}\dots\operatorname{d}z_{1}, (16)

where ff is as given by (2) and gkg_{k} is the indicator function of the event that o,z1,,zko,z_{1},\dots,z_{k} form a good pseudopath (wrt. the parameters r,w1,w2,ϑr,w_{1},w_{2},\vartheta and the point set 𝒵{o,z1,,zk}{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{k}\}). We note that

gk(z1,,zk;𝒵{z1,,zk})gk1(z1,,zk1;𝒵{z1,,zk1})gYI(z1,,zk,𝒵),\begin{array}[]{rcl}g_{k}(z_{1},\dots,z_{k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})&\leq&g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\cdot\\ &&g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}}),\end{array} (17)

where

gYI(z1,,zk,𝒵)=1{rw1<dist2(zk1,zk)<r+w2, and;zn2zk1zk>ϑ, and; either  a disk B with zk1,zkB,diam2(B)<r+w2, 𝒵Bcert(o,z1,,zk1)= or 𝒵DD(zk1,zk,r+w2)cert(o,z1,,zk1)= or 𝒵DD+(zk1,zk,r+w2)cert(o,z1,,zk1)=(or several of the above three hold.)}.g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})=1_{\left\{\begin{array}[]{l}r-w_{1}<\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})<r+w_{2},\text{ and;}\\ \angle z_{n-2}z_{k-1}z_{k}>\vartheta,\text{ and; }\\ \text{either $\exists$ a disk $B$ with $z_{k-1},z_{k}\in\partial B$,$\operatorname{diam}_{\mathbb{H}^{2}}(B)<r+w_{2},$}\\ \text{ \hskip 15.06943pt ${\mathcal{Z}}\cap B\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1})=\emptyset$, }\\ \text{ \hskip 6.02777pt or ${\mathcal{Z}}\cap DD^{-}(z_{k-1},z_{k},r+w_{2})\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1})=\emptyset$, }\\ \text{ \hskip 6.02777pt or ${\mathcal{Z}}\cap DD^{+}(z_{k-1},z_{k},r+w_{2})\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1})=\emptyset$, }\\ \text{(or several of the above three hold.)}\end{array}\right\}}.

Next, we note that if o,z1,,zko,z_{1},\dots,z_{k} is not a good pre-pseudopath then gk(z1,,zk,𝒵{z1,,zk})=0g_{k}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})=0 and whenever o,z1,,zko,z_{1},\dots,z_{k} is a good pre-pseudopath then

𝔼[gk(z1,,zk1;𝒵{z1,,zk1})gYI(z1,,zk,𝒵)]=𝔼[gk1(z1,,zk1;𝒵{z1,,zk1})]𝔼[gYI(z1,,zk,𝒵)],\begin{array}[]{c}{\mathbb{E}}\left[g_{k}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\cdot g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]\\ =\\ {\mathbb{E}}\left[g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot{\mathbb{E}}\left[g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right],\end{array} (18)

as 𝒵cert(o,z1,,zk1){\mathcal{Z}}\cap\operatorname{cert}(o,z_{1},\dots,z_{k-1}) and 𝒵cert(o,z1,,zk1){\mathcal{Z}}\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1}) are independent; and the former determines the value of gk(z1,,zk1;𝒵{z1,,zk1})g_{k}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\}), while the latter determines gYI(z1,,zk,𝒵)g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}}).

Next, notice that by Corollary 5 and Lemmas 44 and 45, we have

pλ𝔻𝔼[gYI(z1,,zk,𝒵)]f(zk)dzk1ε/2+104ew2eew2/2,p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]f(z_{k})\operatorname{d}z_{k}\leq 1-\varepsilon/2+10^{4}e^{w_{2}}e^{-e^{w_{2}/2}}, (19)

for every z1,,zk1z_{1},\dots,z_{k-1} such that o,z1,,zk1o,z_{1},\dots,z_{k-1} is a good pre-pseudopath. If o,z1,,zk1o,z_{1},\dots,z_{k-1} is not a good pre-pseudopath then gk1(z1,,zk1;𝒵{z1,,zk1})g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\}) ==
gk(z1,,zn;𝒵{z1,,zn})g_{k}(z_{1},\dots,z_{n};{\mathcal{Z}}\cup\{z_{1},\dots,z_{n}\}) =0=0. Combining (16)–(19), we see that

𝔼Gk(pλ)n𝔻𝔻𝔼[gk1(z1,,zk1;𝒵{z1,,zk1})]𝔼[gYI(z1,,zk,𝒵)]f(z1)f(zk)dzkdz1=(pλ)n1𝔻𝔻𝔼[gk1(z1,,zk1;𝒵{z1,,zk1})](pλ𝔻𝔼[gYI(z1,,zk,𝒵)]f(zk)dzk)f(z1)f(zk1)dzk1dz1(1ε/2+104ew2eew2/2)(pλ)n1𝔻𝔻𝔼[gk1(z1,,zk1;𝒵{z1,,zk1})]f(z1)f(zk1)dzk1dz1=(1ε/2+104ew2eew2/2)𝔼Gk1.\begin{array}[]{rcl}{\mathbb{E}}G_{k}&\leq&\displaystyle(p\lambda)^{n}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot\\ &&\hskip 64.58313pt{\mathbb{E}}\left[g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]f(z_{1})\dots f(z_{k})\operatorname{d}z_{k}\dots\operatorname{d}z_{1}\\ &=&\displaystyle(p\lambda)^{n-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot\\ &&\displaystyle\hskip 64.58313pt\left(p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g_{Y_{\text{\bf{I}}}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]f(z_{k})\operatorname{d}z_{k}\right)f(z_{1})\dots f(z_{k-1})\operatorname{d}z_{k-1}\dots\operatorname{d}z_{1}\\ &\leq&\displaystyle\left(1-\varepsilon/2+10^{4}e^{w_{2}}e^{-e^{w_{2}/2}}\right)\cdot\\ &&\displaystyle(p\lambda)^{n-1}\cdot\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k-1}(z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot\\ &&\displaystyle\hskip 73.19421ptf(z_{1})\dots f(z_{k-1})\operatorname{d}z_{k-1}\dots\operatorname{d}z_{1}\\[8.61108pt] &=&\displaystyle\left(1-\varepsilon/2+10^{4}e^{w_{2}}e^{-e^{w_{2}/2}}\right)\cdot{\mathbb{E}}G_{k-1}.\end{array}

The lemma follows by iterating the recursive inequality. \blacksquare

The previous lemma immediately gives that, provided we chose w2w_{2} a sufficiently large constant (how large we need to take it depends on ε\varepsilon), for all sufficiently small intensities λ\lambda, almost surely, there are no infinite, black, good pseudopaths starting from the origin. In fact something slightly stronger holds:

Corollary 47

For every 0<ε<10<\varepsilon<1 there exists a c=c(ε)c=c(\varepsilon) such that for all w1,ϑ>0w_{1},\vartheta>0 and w2>cw_{2}>c the following holds.

There exists a λ0=λ0(ε,w1,w2,ϑ)\lambda_{0}=\lambda_{0}(\varepsilon,w_{1},w_{2},\vartheta) such that, when 0<λ<λ00<\lambda<\lambda_{0} and p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, setting r:=2ln(1/λ)r:=2\ln(1/\lambda), almost surely there are no infinite, black, good pseudopaths.

Proof. As pointed out earlier, we can choose c=c(ε)c=c(\varepsilon) such that for all w2>cw_{2}>c and w1,ϑ>0w_{1},\vartheta>0 there is a λ0\lambda_{0} such that, whenever when 0<λ<λ00<\lambda<\lambda_{0} and p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, almost surely there is no infinite, black, good pseudopath starting from the origin.

Let x>0x>0 be arbitrary and write

Nx:=|{z𝒵bB2(o,x):there is an infinite, black, goodpseudopath starting from z }|.N_{x}:=\left|\left\{z\in{\mathcal{Z}}_{b}\cap B_{\mathbb{H}^{2}}(o,x):\begin{array}[]{l}\text{there is an infinite, black, good}\\ \text{pseudopath starting from $z$ }\end{array}\right\}\right|.

By Corollary 5

𝔼Nx=pλ𝔻𝔼[g(z,𝒵{z})]f(z)dz,{\mathbb{E}}N_{x}=p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]f(z)\operatorname{d}z,

where ff is given by (2) as usual and g(u,𝒰)=1g(u,{\mathcal{U}})=1 if uB2(o,x)u\in B_{\mathbb{H}^{2}}(o,x) and u𝒰u\in{\mathcal{U}} and there is an infinite, black, good, pseudopath (wrt. r,w1,w2,ϑr,w_{1},w_{2},\vartheta and the point set 𝒰{\mathcal{U}}) starting from uu; and g(u,𝒰)=0g(u,{\mathcal{U}})=0 otherwise. If zB2(o,x)z\not\in B_{\mathbb{H}^{2}}(o,x) then g(z,𝒵{z})=0g(z,{\mathcal{Z}}\cup\{z\})=0 by definition of gg, and if zB2(o,x)z\in B_{\mathbb{H}^{2}}(o,x) then

𝔼[g(z,𝒵{z})]=𝔼[g(o,𝒵{o})]=0,{\mathbb{E}}\left[g(z,{\mathcal{Z}}\cup\{z\})\right]={\mathbb{E}}\left[g(o,{\mathcal{Z}}\cup\{o\})\right]=0,

since the distribution of 𝒵{\mathcal{Z}} is invariant under the action of isometries of the Poincaré disk (𝒵=dφ[𝒵]{\mathcal{Z}}\hskip 0.86108pt\raisebox{-0.43057pt}{$=$}\raisebox{4.30554pt}{{$\scriptstyle d$}}\hskip 3.44444pt\varphi[{\mathcal{Z}}] if φ\varphi is an isometry), and in particular under an isometry that maps zz to oo; and we know that the origin a.s. is not on any infinite, back, good, pseudopath as remarked just before the statement of the lemma. It follows that 𝔼Nx=0{\mathbb{E}}N_{x}=0 and hence also Nx=0N_{x}=0 a.s., for each x>0x>0 separately. Thus

( infinite, black, good pseudopath)=(x>0{Nx0})=limx(Nx0)=0,{\mathbb{P}}\left(\text{$\exists$ infinite, black, good pseudopath}\right)={\mathbb{P}}\left(\bigcup_{x>0}\{N_{x}\neq 0\}\right)=\lim_{x\to\infty}{\mathbb{P}}(N_{x}\neq 0)=0,

as {Nx0}{Ny0}\{N_{x}\neq 0\}\subseteq\{N_{y}\neq 0\} whenever x<yx<y. \blacksquare

Recall that our strategy for the proof that pc(1ε)(π/3)λp_{c}\geq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda for small λ\lambda is to prove that almost surely no black, infinite, good pseudopath and no black, infinite, linked sequence of chunks starting from the origin will exist (for an appropriate choice of w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 and r=2ln(1/λ)r=2\ln(1/\lambda)). In the light of the last corollary, we now turn our attention to linked sequences of chunks.

3.3.3 Counting chunks starting from the origin

Next, we bound the number of black chunks starting from the origin. We will exploit our results on good pseudopaths, using that a chunk is just a good pseudopath with a single bad pseudo-edge stuck to the end. We begin with a few lemmas designed for adding a last, bad edge to an existing good pseudopath. The names of the random variables described by the lemmas are LL with some subscript, where LL stands for “last” and the subscript corresponds to the type of edge under consideration.

Lemma 48

For all w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)>0\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta)>0 such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and all p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be arbitrary, and set C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and

LIV-i,d:=|{z𝒵b:dist2(uk,z)d, and;area2(BGab(uk,z)C)110area2(BGab(uk,z)), and;𝒵BGab(uk,z)C= or 𝒵BGab+(uk,z)C=.}|.L_{\text{\bf{IV-i}},\geq d}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(u_{k},z)\geq d,\text{ and;}\\ \operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k},z)\cap C)\leq\frac{1}{10}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k},z)),\text{ and;}\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(u_{k},z)\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(u_{k},z)\setminus C=\emptyset.\end{array}\right\}\right|.

We have

𝔼LIV-i,r+v1000ev/2eev/2,{\mathbb{E}}L_{\text{\bf{IV-i}},\geq r+v}\leq 1000\cdot e^{v/2}\cdot e^{-e^{v/2}},

for all v0v\geq 0.

Proof. We let λ0>0\lambda_{0}>0 be a small constant to be determined in the course of the proof, and take an arbitrary λ<λ0\lambda<\lambda_{0} and p10λp\leq 10\lambda. Without loss of generality (applying a suitable isometry if needed) we can assume uk=ou_{k}=o is the origin.

If zz satisfies dist2(o,z)r\operatorname{dist}_{\mathbb{H}^{2}}(o,z)\geq r and area2(BGab(o,z)C)110area2(BGab(o,z))\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z)\cap C)\leq\frac{1}{10}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z)) then of course also

area2(BGab(o,z)C)(12110)area2(BGab(o,z))edist2(o,z)/2,\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}^{-}(o,z)\setminus C)\geq\left(\frac{1}{2}-\frac{1}{10}\right)\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z))\geq e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2},

using that

area2(BGab(o,z))=2π(cosh(dist2(o,z)/2)1)=(1+oλ0(1))πedist2(o,z)/2,\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z))=2\pi\left(\cosh\left(\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2\right)-1\right)=(1+o_{\lambda_{0}}(1))\pi e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2},

(the oλ0(1)o_{\lambda_{0}}(1) term referring to the situation where λ0\lambda_{0} tends to zero) and assuming that we have chosen λ0\lambda_{0} sufficiently small (so that dist2(o,z)r2ln(1/λ0)\operatorname{dist}_{\mathbb{H}^{2}}(o,z)\geq r\geq 2\ln(1/\lambda_{0}) is large). Analogously we also have

area2(BGab+(o,z)C)edist2(o,z)/2,\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}^{+}(o,z)\setminus C)\geq e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2},

for all such zz. By Corollary 5 and this last observation:

𝔼LIV-i,r+v=pλ𝔻B2(o,r+v)1{area2(BGab(o,z)C)110area2(BGab(o,z))}(𝒵BGab(o,z)C= or 𝒵BGab+(o,z)C=)f(z)dzpλ𝔻B2(o,r+v)2eλedist2(o,z)/2f(z)dz,\begin{array}[]{rcl}{\mathbb{E}}L_{\text{\bf{IV-i}},\geq r+v}&=&\displaystyle p\lambda\int_{\mathbb{D}\setminus B_{\mathbb{H}^{2}}(o,r+v)}1_{\{\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z)\cap C)\leq\frac{1}{10}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z))\}}\cdot\\[8.61108pt] &&\displaystyle\hskip 43.05542pt{\mathbb{P}}\left({\mathcal{Z}}\cap B_{\text{Gab}}^{-}(o,z)\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(o,z)\setminus C=\emptyset\right)f(z)\operatorname{d}z\\[8.61108pt] &\leq&\displaystyle p\lambda\int_{\mathbb{D}\setminus B_{\mathbb{H}^{2}}(o,r+v)}2e^{-\lambda e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2}}f(z)\operatorname{d}z,\end{array}

with ff given by (2).

Applying a change to hyperbolic polar coordinates to zz we find

𝔼LIV-i,r+v2pλr+v02πeλeρ/2sinh(ρ)dαdρ200λ2r+veλeρ/2eρdρ.\begin{array}[]{rcl}{\mathbb{E}}L_{\text{\bf{IV-i}},\geq r+v}&\leq&\displaystyle 2p\lambda\int_{r+v}^{\infty}\int_{0}^{2\pi}e^{-\lambda e^{\rho/2}}\sinh(\rho)\operatorname{d}\alpha\operatorname{d}\rho\\ &\leq&\displaystyle 200\lambda^{2}\int_{r+v}^{\infty}e^{-\lambda e^{\rho/2}}e^{\rho}\operatorname{d}\rho.\end{array}

Applying the substitution t=λeρ/2t=\lambda e^{\rho/2} (so that dρ=2dtt\operatorname{d}\rho=\frac{2\operatorname{d}t}{t}), we find

𝔼LIV-i,r+v200λ2λe(r+v)/2et(tλ)22dtt=400ev/2tetdt=400(ev/2+1)eev/21000ev/2eev/2.\begin{array}[]{rcl}{\mathbb{E}}L_{\text{\bf{IV-i}},\geq r+v}&\leq&\displaystyle 200\lambda^{2}\int_{\lambda e^{(r+v)/2}}^{\infty}e^{-t}\cdot\left(\frac{t}{\lambda}\right)^{2}\frac{2\operatorname{d}t}{t}\\[8.61108pt] &=&\displaystyle 400\int_{e^{v/2}}^{\infty}te^{-t}\operatorname{d}t\\[8.61108pt] &=&\displaystyle 400(e^{v/2}+1)e^{-e^{v/2}}\\[8.61108pt] &\leq&\displaystyle 1000\cdot e^{v/2}\cdot e^{-e^{v/2}}.\end{array}

\blacksquare

Corollary 49

For all w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)>0\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta)>0 such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and all p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary good pre-pseudopath, and set C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and

LIV,d:=|{z𝒵b:dist2(uk,z)d, and; 𝒵BGab(uk,z)C= or 𝒵BGab+(uk,z)C=.}|.L_{\text{\bf{IV}},\geq d}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(u_{k},z)\geq d,\text{ and; }\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(u_{k},z)\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(u_{k},z)\setminus C=\emptyset.\end{array}\right\}\right|.

We have

𝔼LIV,r+v104ev/2eev/2,{\mathbb{E}}L_{\text{\bf{IV}},\geq r+v}\leq 10^{4}\cdot e^{v/2}\cdot e^{-e^{v/2}},

for all v0v\geq 0.

Proof. We let λ0,ϑ>0\lambda_{0},\vartheta^{\prime}>0 be small constants to be determined in the course of the proof, and take an arbitrary 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda. Without loss of generality (applying a suitable isometry if needed) we can assume uk=ou_{k}=o is the origin.

Since v0v\geq 0, by Corollary 38 with ε:=110\varepsilon:=\frac{1}{10} and ϑ1=ϑ,ϑ2=ϑ\vartheta_{1}=\vartheta,\vartheta_{2}=\vartheta^{\prime}, assuming we have chosen λ0\lambda_{0} sufficiently small, we have

LIV,r+vLIV-i,r+v+LIV-ii,L_{\text{\bf{IV}},\geq r+v}\leq L_{\text{\bf{IV-i}},\geq r+v}+L_{\text{\bf{IV-ii}}},

where LIV-i,r+vL_{\text{\bf{IV-i}},\geq r+v} is as defined in Lemma 48 above and

LIV-ii:=|𝒵bB2(o,r+w)sect(o,uk1,ϑ)|.L_{\text{\bf{IV-ii}}}:=\left|{\mathcal{Z}}_{b}\cap B_{\mathbb{H}^{2}}(o,r+w)\cap\operatorname{sect}\left(o,u_{k-1},\vartheta^{\prime}\right)\right|.

setting w:=w2+22ln10w:=w_{2}+22\ln 10.

In particular LIV,r+v=LIV-i,r+vL_{\text{\bf{IV}},\geq r+v}=L_{\text{\bf{IV-i}},\geq r+v} if vwv\geq w. In this case we are clearly done by Lemma 48.

To prove it also for 0v<w0\leq v<w, we also need to bound 𝔼LIV-ii{\mathbb{E}}L_{\text{\bf{IV-ii}}}. Lemma 20 shows that

𝔼LIV-ii1000ϑew1000ev/2eev/2,{\mathbb{E}}L_{\text{\bf{IV-ii}}}\leq 1000\vartheta^{\prime}e^{w}\leq 1000\cdot e^{v/2}\cdot e^{-e^{v/2}},

having chosen ϑ\vartheta^{\prime} sufficiently small for the second inequality. (To be precise, having chosen ϑ2<ewmin0xwex/2eex/2\displaystyle\vartheta_{2}<e^{-w}\cdot\min_{0\leq x\leq w}e^{x/2}\cdot e^{-e^{x/2}}.)

Adding the bounds on 𝔼LIV-i,r+v{\mathbb{E}}L_{\text{\bf{IV-i}},\geq r+v} and 𝔼LIV-ii{\mathbb{E}}L_{\text{\bf{IV-ii}}} proves the result for 0vw0\leq v\leq w. \blacksquare

We will denote

CkII:=|{black chunks of length k, starting from o, with final pseudo-edge of type II}|,C_{k}^{\text{\bf{II}}}:=\left|\left\{\text{black chunks of length $k$, starting from $o$, with final pseudo-edge of type {\bf{II}}}\right\}\right|,
CkIII:=|{black chunks of length k, starting from o, with final pseudo-edge of type III}|,C_{k}^{\text{\bf{III}}}:=\left|\left\{\text{black chunks of length $k$, starting from $o$, with final pseudo-edge of type {\bf{III}}}\right\}\right|,

and, for all vw2v\geq w_{2}:

CkIV,v:=|{black chunks of length k, starting from o, with final pseudo-edge of type IV, and the final pseudo-edge having length [r+v,r+v+1)}|.C_{k}^{\text{\bf{IV}},v}:=\left|\left\{\begin{array}[]{l}\text{black chunks of length $k$, starting from $o$, with final pseudo-edge of type {\bf{IV}}},\\ \text{ and the final pseudo-edge having length $\in[r+v,r+v+1)$}\end{array}\right\}\right|.

Recall that GkG_{k} denotes the number of black, good, pseudopaths starting from the origin of length kk.

Corollary 50

For every w1,w2,ϑw_{1},w_{2},\vartheta there exists a λ0=λ0(w1,w2,ϑ)>0\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta)>0 such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2ln(1/λ)r:=2\ln(1/\lambda).

We have

𝔼CkII103ew1𝔼Gk1,{\mathbb{E}}C_{k}^{\text{\bf{II}}}\leq 10^{3}\cdot e^{-w_{1}}\cdot{\mathbb{E}}G_{k-1}, (20)
𝔼CkIII103ϑew2𝔼Gk1,{\mathbb{E}}C_{k}^{\text{\bf{III}}}\leq 10^{3}\cdot\vartheta\cdot e^{w_{2}}\cdot{\mathbb{E}}G_{k-1}, (21)

and

𝔼CkIV,v104ev/2eev/2𝔼Gk1,{\mathbb{E}}C_{k}^{\text{\bf{IV}},v}\leq 10^{4}\cdot e^{v/2}\cdot e^{-e^{v/2}}\cdot{\mathbb{E}}G_{k-1}, (22)

for all vw2v\geq w_{2}.

Proof. We will argue analogously to Lemma 46. We start by considering CkIIC_{k}^{\text{\bf{II}}}. By Corollary 5

𝔼CkII=(pλ)k𝔻𝔻𝔼[gkII(z1,,zk,𝒵{z1,,zk})]f(z1)f(zk)dzkdz1,{\mathbb{E}}C_{k}^{\text{\bf{II}}}=\left(p\lambda\right)^{k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k}^{\text{\bf{II}}}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})\right]\cdot f(z_{1})\cdot\dots\cdot f(z_{k})\operatorname{d}z_{k}\dots\operatorname{d}z_{1},

where ff is given by (2) and gkIIg_{k}^{\text{\bf{II}}} is the indicator function that o,z1,,zko,z_{1},\dots,z_{k} is a chunk (wrt. the point set 𝒵{o,z1,,zk}{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{k}\}), whose last pseudo-edge zk1zkz_{k-1}z_{k} is of type II.

Let hk1h_{k-1} denote the indicator function that o,z1,,zk1o,z_{1},\dots,z_{k-1} is a good pseudo-path wrt. the point set 𝒵{o,z1,,zk1}{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{k-1}\}. We have

gkII(z1,,zk,𝒵{z1,,zk})hk1(z1,,zk1,𝒵{z1,,zk1})1{dist2(zk1,zk)rw1}.\begin{array}[]{rcl}g_{k}^{\text{\bf{II}}}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})&\leq&\displaystyle h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\cdot\\ &&\displaystyle 1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})\leq r-w_{1}\}}.\end{array}

It follows that

𝔼CkII(pλ)k1𝔻𝔻𝔼[hk1(z1,,zk1,𝒵{z1,,zk1})]f(z1)f(zk1)(pλ𝔻1{dist2(zk1,zk)rw1}f(zk)dzk)dzk1dz1=pλarea2(B2(o,rw1))(pλ)k1𝔻𝔻𝔼[hk1(z1,,zk1,𝒵{z1,,zk1})]f(z1)f(zk1)dzk1dz1=pλarea2(B2(o,rw1))𝔼Gk1,\begin{array}[]{rcl}{\mathbb{E}}C_{k}^{\text{\bf{II}}}&\leq&\displaystyle\left(p\lambda\right)^{k-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot f(z_{1})\cdot\dots\cdot f(z_{k-1})\cdot\\ &&\displaystyle\hskip 64.58313pt\left(p\lambda\int_{\mathbb{D}}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})\leq r-w_{1}\}}f(z_{k})\operatorname{d}z_{k}\right)\operatorname{d}z_{k-1}\dots\operatorname{d}z_{1}\\ &=&\displaystyle p\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,r-w_{1}))\cdot\\ &&\displaystyle(p\lambda)^{k-1}\cdot\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot\\ &&\displaystyle\hskip 73.19421ptf(z_{1})\cdot\dots\cdot f(z_{k-1})\operatorname{d}z_{k-1}\dots\operatorname{d}z_{1}\\ &=&\displaystyle p\lambda\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,r-w_{1}))\cdot{\mathbb{E}}G_{k-1},\end{array}

Applying Lemma 19 gives (20).

The proof of (21) is analogous, using the indicator function 1{dist2(zk1,zk)<r+w2,zk2zk1zkϑ}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})<r+w_{2},\angle z_{k-2}z_{k-1}z_{k}\leq\vartheta\}} in place of 1{dist2(zk1,zk)rw1}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})\leq r-w_{1}\}}, replacing area2(B2(o,rw1))\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,r-w_{1})) with area2(B2(o,r+w2)sect(o,v,ϑ))\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,r+w_{2})\cap\operatorname{sect}\left(o,v,\vartheta\right)) (for vov\neq o arbitrary) and using Lemma 20 in place of Lemma 19.

For the proof of (22), we use that if gkIV,vg_{k}^{\text{\bf{IV}},v} denotes the indicator function that o,z1,,zko,z_{1},\dots,z_{k} form a chunk (wrt. 𝒵{o,z1,,zk}{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{k}\}) whose last pseudo-edge has length [r+v,r+v+1)\in[r+v,r+v+1) then

gkIV,v(z1,,zk,𝒵{z1,,zk})hk1(z1,,zk1,𝒵{z1,,zk1})gLIV,r+v(z1,,zk,𝒵),\begin{array}[]{rcl}g_{k}^{\text{\bf{IV}},v}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})&\leq&h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\cdot\\ &&g_{L_{\text{\bf{IV}},\geq r+v}}(z_{1},\dots,z_{k},{\mathcal{Z}}),\end{array}

where

gLIV,r+v(z1,,zk,𝒵):=1{dist2(zk1,zk)r+v, and; either 𝒵BGab(zk1,zk)cert(o,z1,,zk1)=, or 𝒵BGab+(zk1,zk)cert(o,z1,,zk1)=,(or both).}.g_{L_{\text{\bf{IV}},\geq r+v}}(z_{1},\dots,z_{k},{\mathcal{Z}}):=1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{k-1},z_{k})\geq r+v,\text{ and;}\\ \text{ either }{\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z_{k-1},z_{k})\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1})=\emptyset,\\ \text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z_{k-1},z_{k})\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1})=\emptyset,\\ \text{(or both).}\end{array}\right\}}.

Since 𝒵cert(o,z1,,zk1){\mathcal{Z}}\cap\operatorname{cert}(o,z_{1},\dots,z_{k-1}) and 𝒵cert(o,z1,,zk1){\mathcal{Z}}\setminus\operatorname{cert}(o,z_{1},\dots,z_{k-1}) are independent for any choice of z1,,zk1z_{1},\dots,z_{k-1}, we have

𝔼[gkIV,v(z1,,zk,𝒵{z1,,zk})]𝔼[hk1(z1,,zk1,𝒵{z1,,zk1})]𝔼[gLIV,r+v(z1,,zk,𝒵)].\begin{array}[]{rcl}{\mathbb{E}}\left[g_{k}^{\text{\bf{IV}},v}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})\right]&\leq&{\mathbb{E}}\left[h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot\\ &&{\mathbb{E}}\left[g_{L_{\text{\bf{IV}},\geq r+v}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right].\end{array}

If z1,,zk1z_{1},\dots,z_{k-1} do not form a good pre-pseudopath then hk1(z1,,zk1,𝒵{z1,,zk1})=0h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})=0. Otherwise we can apply Corollary 49 (and Corollary 5) to show that

pλ𝔻𝔼[gLIV,r+v(z1,,zk,𝒵)]f(zk)dzk104ev/2eev/2.p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g_{L_{\text{\bf{IV}},\geq r+v}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]f(z_{k})\operatorname{d}z_{k}\leq 10^{4}e^{v/2}e^{-e^{v/2}}.

In other words,

𝔼CkIV,v=(pλ)k𝔻𝔻𝔼[gkIV,r+v(z1,,zk,𝒵{z1,,zk})]f(z1)f(zk)dzkdz1(pλ)k1𝔻𝔻𝔼[hk1(z1,,zk1,𝒵{z1,,zk1})]f(z1)f(zk1)(pλ𝔻𝔼[gLIV,r+v(z1,,zk,𝒵)]f(zk)dzk)dzk1dz1𝔼Gk1104ev/2eev/2,\begin{array}[]{rcl}{\mathbb{E}}C_{k}^{\text{\bf{IV}},v}&=&\displaystyle\left(p\lambda\right)^{k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{k}^{\text{\bf{IV}},\geq r+v}(z_{1},\dots,z_{k},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k}\})\right]\cdot f(z_{1})\cdot\dots\cdot f(z_{k})\operatorname{d}z_{k}\dots\operatorname{d}z_{1}\\ &\leq&\displaystyle\left(p\lambda\right)^{k-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[h_{k-1}(z_{1},\dots,z_{k-1},{\mathcal{Z}}\cup\{z_{1},\dots,z_{k-1}\})\right]\cdot f(z_{1})\cdot\dots\cdot f(z_{k-1})\cdot\\ &&\displaystyle\hskip 64.58313pt\left(p\lambda\int_{\mathbb{D}}{\mathbb{E}}\left[g_{L_{\text{\bf{IV}},\geq r+v}}(z_{1},\dots,z_{k},{\mathcal{Z}})\right]f(z_{k})\operatorname{d}z_{k}\right)\operatorname{d}z_{k-1}\dots\operatorname{d}z_{1}\\ &\leq&\displaystyle{\mathbb{E}}G_{k-1}\cdot 10^{4}e^{v/2}e^{-e^{v/2}},\end{array}

as desired. \blacksquare

3.3.4 Counting linked sequences of chunks

Next, we turn attention to counting the number of linked sequences of chunks. The first few lemmas are designed for dealing with the first pseudo-edge of a new chunk that will be linked to an existing chunk. The names of the random variables described in these lemmas are FF with some subscript, where FF stands for “first” and the subscript corresponds to the type of the new pseudo-edge to be added and the type of link under consideration. (The link types being a, b, c corresponding to the cases listed in part (ii) of Definition 30.)

Lemma 51

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary pre-chunk and define

FI-b:=|{(z1,z2)𝒵b×𝒵b:0<dist2(z1,z2)<r+w2, and;cert(z1,z2)cert(u0,,uk).}|.F_{\text{\bf{I-b}}}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}0<\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2},\text{ and;}\\ \operatorname{cert}(z_{1},z_{2})\cap\operatorname{cert}(u_{0},\dots,u_{k})\neq\emptyset.\end{array}\right\}\right|.

Then we have

𝔼FI-b104kexp[32w2+max(w2,dist2(uk1,uk)r)/2].{\mathbb{E}}F_{\text{\bf{I-b}}}\leq 10^{4}\cdot k\cdot\exp\left[\frac{3}{2}w_{2}+\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2\right].

Proof. We can write cert(u0,,uk)=B1Bm\operatorname{cert}(u_{0},\dots,u_{k})=B_{1}\cup\dots\cup B_{m} with m2km\leq 2k and B1=B2(c1,s1),,Bm=B2(cm,sm)B_{1}=B_{\mathbb{H}^{2}}(c_{1},s_{1}),\dots,B_{m}=B_{\mathbb{H}^{2}}(c_{m},s_{m}) balls whose radii are either (r+w2)/2(r+w_{2})/2 or dist2(uk1,uk)/2\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})/2.

For i=1,,mi=1,\dots,m we define

Fi:=|{(z1,z2)𝒵b×𝒵b:0<dist2(z1,z2)<r+w2, and;B2(z1,r+w)Bi, and; B2(z2,r+w)Bi.}|.F_{i}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}0<\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2},\text{ and;}\\ B_{\mathbb{H}^{2}}(z_{1},r+w)\cap B_{i}\neq\emptyset,\text{ and; }\\ B_{\mathbb{H}^{2}}(z_{2},r+w)\cap B_{i}\neq\emptyset.\end{array}\right\}\right|.

Observe that if dist2(z1,z2)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2} then cert(z1,z2)\operatorname{cert}(z_{1},z_{2}) either equals DD(z1,z2,r+w2)DD(z_{1},z_{2},r+w_{2}) or \emptyset. In particular we have cert(z1,z2)B2(z1,r+w)B2(z2,r+w)\operatorname{cert}(z_{1},z_{2})\subseteq B_{\mathbb{H}^{2}}(z_{1},r+w)\cap B_{\mathbb{H}^{2}}(z_{2},r+w).

We see that

FI-bF1++Fm.F_{\text{\bf{I-b}}}\leq F_{1}+\dots+F_{m}.

We can thus focus on bounding the individual expectations 𝔼Fi{\mathbb{E}}F_{i}. For the moment, we pick some 1im1\leq i\leq m. Without loss of generality (applying a suitable isometry if needed) the center cic_{i} of BiB_{i} is the origin oo. Applying Corollary 5 and a (double) switch to hyperbolic polar coordinates:

𝔼Fi=p2λ20002π02πg(z1(α1,ρ1),z2(α2,ρ2))sinh(ρ1)sinh(ρ2)dα2dα1dρ2dρ1100λ40002π02πg(z1(α1,ρ1),z2(α2,ρ2))eρ1+ρ2dα2dα1dρ2dρ1,\begin{array}[]{rcl}{\mathbb{E}}F_{i}&=&\displaystyle p^{2}\lambda^{2}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{2\pi}g(z_{1}(\alpha_{1},\rho_{1}),z_{2}(\alpha_{2},\rho_{2}))\sinh(\rho_{1})\sinh(\rho_{2})\operatorname{d}\alpha_{2}\operatorname{d}\alpha_{1}\operatorname{d}\rho_{2}\operatorname{d}\rho_{1}\\ &\leq&\displaystyle 100\lambda^{4}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{2\pi}g(z_{1}(\alpha_{1},\rho_{1}),z_{2}(\alpha_{2},\rho_{2}))e^{\rho_{1}+\rho_{2}}\operatorname{d}\alpha_{2}\operatorname{d}\alpha_{1}\operatorname{d}\rho_{2}\operatorname{d}\rho_{1},\end{array}

where

g(u1,u2):=1{0<dist2(u1,u2)<r+w2, and;B2(u1,r+w2)Bi, and; B2(u2,r+w2)Bi.}.g(u_{1},u_{2}):=1_{\left\{\begin{array}[]{l}0<\operatorname{dist}_{\mathbb{H}^{2}}(u_{1},u_{2})<r+w_{2},\text{ and;}\\ B_{\mathbb{H}^{2}}(u_{1},r+w_{2})\cap B_{i}\neq\emptyset,\text{ and; }\\ B_{\mathbb{H}^{2}}(u_{2},r+w_{2})\cap B_{i}\neq\emptyset.\end{array}\right\}}.

Note that B2(zi,r+w2)BiB_{\mathbb{H}^{2}}(z_{i},r+w_{2})\cap B_{i}\neq\emptyset implies that ρi=dist2(o,zi)<si+r+w2\rho_{i}=\operatorname{dist}_{\mathbb{H}^{2}}(o,z_{i})<s_{i}+r+w_{2}. Applying Lemma 43, we see that dist2(z1,z1)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{1})<r+w_{2} implies that z1oz2=|α1α2|2π<2πe(r+w2(ρ1+ρ2))/2\angle z_{1}oz_{2}=|\alpha_{1}-\alpha_{2}|_{2\pi}<2\pi e^{(r+w_{2}-(\rho_{1}+\rho_{2}))/2}. (Here and elsewhere |x|2π:=min(|x|,2π|x|)|x|_{2\pi}:=\min(|x|,2\pi-|x|) for 2πx2π-2\pi\leq x\leq 2\pi.) Therefore

g(z1,z2)1{ρ1,ρ2<si+r+w2,|α1α2|2π<2πe(r+w2(ρ1+ρ2))/2}.g(z_{1},z_{2})\leq 1_{\left\{\begin{array}[]{l}\rho_{1},\rho_{2}<s_{i}+r+w_{2},\\ |\alpha_{1}-\alpha_{2}|_{2\pi}<2\pi e^{(r+w_{2}-(\rho_{1}+\rho_{2}))/2}\end{array}\right\}}.

It follows that

𝔼Fi100λ40si+r+w20si+r+w202π02π1{|α1α2|2π<2πe(r+w2ρ1ρ2)/2}eρ1+ρ2dα2dα1dρ2dρ1.\begin{array}[]{rcl}{\mathbb{E}}F_{i}&\leq&\displaystyle 100\lambda^{4}\int_{0}^{s_{i}+r+w_{2}}\int_{0}^{s_{i}+r+w_{2}}\int_{0}^{2\pi}\int_{0}^{2\pi}1_{\left\{|\alpha_{1}-\alpha_{2}|_{2\pi}<2\pi e^{(r+w_{2}-\rho_{1}-\rho_{2})/2}\right\}}\cdot\\ &&\displaystyle\hskip 150.69397pte^{\rho_{1}+\rho_{2}}\operatorname{d}\alpha_{2}\operatorname{d}\alpha_{1}\operatorname{d}\rho_{2}\operatorname{d}\rho_{1}.\end{array} (23)

By symmetry considerations

02π02π1{|α1α2|2π<2πe(r+w2(ρ1+ρ2))/2}dα2dα18π2e(r+w2(ρ1+ρ2))/2.\int_{0}^{2\pi}\int_{0}^{2\pi}1_{\left\{|\alpha_{1}-\alpha_{2}|_{2\pi}<2\pi e^{(r+w_{2}-(\rho_{1}+\rho_{2}))/2}\right\}}\operatorname{d}\alpha_{2}\operatorname{d}\alpha_{1}\leq 8\pi^{2}\cdot e^{(r+w_{2}-(\rho_{1}+\rho_{2}))/2}.

(We have an inequality and not an equality to account for the possibility that 2πer+w2(ρ1+ρ2)π2\pi e^{r+w_{2}-(\rho_{1}+\rho_{2})}\geq\pi.) Filling this back into (23), we obtain

𝔼Fi104λ4e(r+w2)/20si+r+w20si+r+w2e(ρ1+ρ2)/2dρ2ρ1=14104λ4e(r+w2)/2esi+r+w2=14104exp[32w2+sir/2]14104exp[32w2+max(w2,dist2(uk1,uk)r)/2].\begin{array}[]{rcl}{\mathbb{E}}F_{i}&\leq&\displaystyle 10^{4}\lambda^{4}e^{(r+w_{2})/2}\int_{0}^{s_{i}+r+w_{2}}\int_{0}^{s_{i}+r+w_{2}}e^{(\rho_{1}+\rho_{2})/2}\operatorname{d}\rho_{2}\rho_{1}\\ &=&\frac{1}{4}\cdot 10^{4}\lambda^{4}e^{(r+w_{2})/2}\cdot e^{s_{i}+r+w_{2}}\\ &=&\frac{1}{4}\cdot 10^{4}\cdot\exp\left[\frac{3}{2}w_{2}+s_{i}-r/2\right]\\ &\leq&\frac{1}{4}\cdot 10^{4}\cdot\exp\left[\frac{3}{2}w_{2}+\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2\right].\end{array}

using in the last line that 2si2s_{i}, the diameter of BiB_{i}, is either r+w2r+w_{2} or dist2(uk1,uk)\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k}). Adding the bounds on 𝔼Yi{\mathbb{E}}Y_{i}^{\prime} and using the inequality m2km\leq 2k gives

𝔼FI-b𝔼F1++𝔼Fm2k14104exp[32w2+max(w2,dist2(uk1,uk)r)/2]104kexp[32w2+max(w2,dist2(uk1,uk)r)/2],\begin{array}[]{rcl}{\mathbb{E}}F_{\text{\bf{I-b}}}&\leq&{\mathbb{E}}F_{1}+\dots+{\mathbb{E}}F_{m}\\ &\leq&2k\cdot\frac{1}{4}\cdot 10^{4}\cdot\exp\left[\frac{3}{2}w_{2}+\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2\right]\\ &\leq&10^{4}\cdot k\cdot\exp\left[\frac{3}{2}w_{2}+\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2\right],\end{array}

as claimed in the lemma statement. \blacksquare

Lemma 52

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary pre-chunk and define

FI-c:=|{(z1,z2)𝒵b×𝒵b:0<dist2(z1,z2)<r+w2, and;z1z2 is a pseudoedge, and; cert(z1,z2)cert(u0,,uk)=, and;dist2(z1,cert(u0,,uk))<r/1000.}|.F_{\text{\bf{I-c}}}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}0<\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2},\text{ and;}\\ z_{1}z_{2}\text{ is a pseudoedge, and; }\\ \operatorname{cert}(z_{1},z_{2})\cap\operatorname{cert}(u_{0},\dots,u_{k})=\emptyset,\text{ and;}\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},\operatorname{cert}(u_{0},\dots,u_{k}))<r/1000.\end{array}\right\}\right|.

Then we have

𝔼FI-ckexp[max(w2,dist2(uk1,uk)r)/2].{\mathbb{E}}F_{\text{\bf{I-c}}}\leq k\cdot\exp\left[\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2\right].

Proof. As usual, we let λ0>0\lambda_{0}>0 be a small constant, to be determined in the course of the proof. As in the previous proof, we write C:=cert(u0,,uk)=B1BmC:=\operatorname{cert}(u_{0},\dots,u_{k})=B_{1}\cup\dots\cup B_{m} with m2km\leq 2k and B1=B2(c1,s1),,Bm=B2(cm,sm)B_{1}=B_{\mathbb{H}^{2}}(c_{1},s_{1}),\dots,B_{m}=B_{\mathbb{H}^{2}}(c_{m},s_{m}) balls whose radii are either (r+w2)/2(r+w_{2})/2 or dist2(uk1,uk)/2\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})/2. Let us write C:={u𝔻:dist2(u,C)<r/1000}C^{\prime}:=\{u\in\mathbb{D}:\operatorname{dist}_{\mathbb{H}^{2}}(u,C)<r/1000\}. By Corollary 5

𝔼FI-c(pλ)2𝔻𝔻1{z1C1˙{z1z2 pseudoedge,dist2(z1,z2)<r+w2}f(z1)f(z2)dz1dz2=(pλ)2area2(C)B2(o,r+w2)1{0z pseudoedge}f(z)dzpλarea2(C)(p𝔼D+𝔼XI+𝔼X~VI)10λ22kπemax(r+w2,dist2(uk1,uk))/2+r/1000(10λ(6+3πλ)+1000ew2+1000ew2eew2/2)=kemax(w2,dist2(uk1,uk)r)/2e(4991000)r(120πλ+60+2000πew2+2000πew2eew2/2)kemax(w2,dist2(uk1,uk)r)/2,\begin{array}[]{rcl}{\mathbb{E}}F_{\text{\bf{I-c}}}&\leq&\displaystyle(p\lambda)^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}1_{\{z_{1}\in C^{\prime}}\dot{1}_{\left\{\begin{array}[]{l}z_{1}z_{2}\text{ pseudoedge,}\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})<r+w_{2}\end{array}\right\}}f(z_{1})f(z_{2})\operatorname{d}z_{1}\operatorname{d}z_{2}\\ &=&\displaystyle(p\lambda)^{2}\operatorname{area}_{\mathbb{H}^{2}}(C^{\prime})\cdot\int_{B_{\mathbb{H}^{2}}(o,r+w_{2})}1_{\{0z\text{ pseudoedge}\}}f(z)\operatorname{d}z\\ &\leq&p\lambda\operatorname{area}_{\mathbb{H}^{2}}(C^{\prime})\cdot\left(p{\mathbb{E}}D+{\mathbb{E}}X_{\text{\bf{I}}}+{\mathbb{E}}\tilde{X}_{\text{\bf{VI}}}\right)\\ &\leq&10\cdot\lambda^{2}\cdot 2k\cdot\pi e^{\max(r+w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k}))/2+r/1000}\cdot\\ &&\displaystyle\quad\quad\left(10\lambda\cdot\left(6+\frac{3}{\pi\lambda}\right)+1000e^{-w_{2}}+1000e^{w_{2}}e^{-e^{w_{2}/2}}\right)\\ &=&\displaystyle k\cdot e^{\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2}\cdot e^{-\left(\frac{499}{1000}\right)r}\cdot\\ &&\displaystyle\quad\quad\cdot\left(120\pi\lambda+60+2000\pi e^{-w_{2}}+2000\pi e^{w_{2}}e^{-e^{w_{2}/2}}\right)\\ &\leq&k\cdot e^{\max\left(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r\right)/2},\end{array}

where DD is the “typical degree” as given by (8), XIX_{\text{\bf{I}}} is as defined in Lemma 19 and X~VI\tilde{X}_{\text{\bf{VI}}} as defined in Lemma 26, and; we apply Isokawa’s formula (Theorem 6) and Lemmas 19 and 26 and we use that p10λp\leq 10\lambda and that area2(B2(u,s))πes\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(u,s))\leq\pi e^{s} for all u𝔻,s>0u\in\mathbb{D},s>0 to obtain the fourth line. In the fifth line we use that r=2ln(1/λ)r=2\ln(1/\lambda) and in the last line that we chose λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) sufficiently small (so that rr is large). \blacksquare

Lemma 53

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary pre-chunk and write C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and

FIV-a,d:=|{z𝒵b:dist2(z,uk)d, and; dist2(z,C)r/1000, and; 𝒵BGab(z,uk)C= or 𝒵BGab+(z,uk)C=.}|.F_{\text{\bf{IV-a}},\geq d}:=\left|\left\{z\in{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z,u_{k})\geq d,\text{ and; }\\ \operatorname{dist}_{\mathbb{H}^{2}}(z,C)\geq r/1000,\text{ and; }\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z,u_{k})\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z,u_{k})\setminus C=\emptyset.\end{array}\right\}\right|.

Then we have

𝔼FIV-a,r+v103ev/2eev/2,{\mathbb{E}}F_{\text{\bf{IV-a}},\geq r+v}\leq 10^{3}\cdot e^{v/2}\cdot e^{-e^{v/2}},

for all vw2v\geq w_{2}.

Proof. The demand that dist2(z,C)>r/1000\operatorname{dist}_{\mathbb{H}^{2}}(z,C)>r/1000 implies that

ahd(C,BGab(uk,z))>r/1000,\operatorname{ahd}(C,B_{\text{Gab}}(u_{k},z))>r/1000,

as well. Provided λ0\lambda_{0} is sufficiently small (so that rr is sufficiently large), Lemma 41 (with ε=1/10\varepsilon=1/10) implies that

area2(BGab(uk,z)C)110area2(BGab(uk,z)).\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k},z)\cap C)\leq\frac{1}{10}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(u_{k},z)).

The result now follows from immediately Lemma 48. \blacksquare

Lemma 54

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary pre-chunk and write C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and

FIV-b,d:=|{(z1,z2)𝒵b×𝒵b:dist2(z1,z2)d, and; dist2(z1,C)r/1000, and;BGab(z1,z2)C, and; 𝒵BGab(z1,z2)C= or 𝒵BGab+(z1,z2)C=.}|.F_{\text{\bf{IV-b}},\geq d}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})\geq d,\text{ and; }\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},C)\geq r/1000,\text{ and;}\\ B_{\text{Gab}}(z_{1},z_{2})\cap C\neq\emptyset,\text{ and; }\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z_{1},z_{2})\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z_{1},z_{2})\setminus C=\emptyset.\end{array}\right\}\right|.

Then we have

𝔼FIV-b,r+v106keveev/2emax(w2,dist2(uk1,uk)r)/2,{\mathbb{E}}F_{\text{\bf{IV-b}},\geq r+v}\leq 10^{6}\cdot k\cdot e^{v}\cdot e^{-e^{v/2}}\cdot e^{\max(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r)/2},

for all vw2v\geq w_{2}.

Proof. As usual, we let λ0>0\lambda_{0}>0 be a small positive constant, to be determined during the course of the proof, and we fix an arbitrary vw2v\geq w_{2}, 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda.

We can write cert(u0,,uk)=B1Bm\operatorname{cert}(u_{0},\dots,u_{k})=B_{1}\cup\dots\cup B_{m} with m2km\leq 2k and B1=B2(c1,s1),,Bm=B2(cm,sm)B_{1}=B_{\mathbb{H}^{2}}(c_{1},s_{1}),\dots,B_{m}=B_{\mathbb{H}^{2}}(c_{m},s_{m}) balls whose radii are either (r+w2)/2(r+w_{2})/2 or dist2(uk1,uk)/2\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})/2.

For i=1,,mi=1,\dots,m we define

Fi:=|{(z1,z2)𝒵b×𝒵b:dist2(z1,z2)r+v, and; dist2(z1,C)>r/1000, and;BGab(z1,z2)Bi, and; 𝒵BGab(z1,z2)C= or 𝒵BGab+(z1,z2)C.}|,F_{i}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})\geq r+v,\text{ and; }\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},C)>r/1000,\text{ and;}\\ B_{\text{Gab}}(z_{1},z_{2})\cap B_{i}\neq\emptyset,\text{ and; }\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z_{1},z_{2})\setminus C=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z_{1},z_{2})\setminus C\neq\emptyset.\end{array}\right\}\right|,

and we point out that

FIV-b,r+vF1++Fm.F_{\text{\bf{IV-b}},\geq r+v}\leq F_{1}+\dots+F_{m}.

In particular it suffices to bound each expectation 𝔼Fi{\mathbb{E}}F_{i} separately. For the moment we fix 1im1\leq i\leq m. Applying a suitable isometry if needed, we assume without loss of generality that the center of BiB_{i} is ci=oc_{i}=o the origin.

By Corollary 5:

𝔼Fi=p2λ2𝔻𝔻𝔼[g(z1,z2,𝒵{z1,z2})]f(z1)f(z2)dz2dz1,{\mathbb{E}}F_{i}=p^{2}\lambda^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{1},

with ff given by (2) and

g(u1,u1,𝒰):=1{dist2(u1,C)>r/1000, and;dist2(u1,u2)r+v, and;BGab(u1,u2)Bi, and; 𝒰BGab(u1,u2)C= or 𝒰BGab+(z1,z2)C.}.g(u_{1},u_{1},{\mathcal{U}}):=1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(u_{1},C)>r/1000,\text{ and;}\\ \operatorname{dist}_{\mathbb{H}^{2}}(u_{1},u_{2})\geq r+v,\text{ and;}\\ B_{\text{Gab}}(u_{1},u_{2})\cap B_{i}\neq\emptyset,\text{ and; }\\ {\mathcal{U}}\cap B_{\text{Gab}}^{-}(u_{1},u_{2})\setminus C=\emptyset\text{ or }{\mathcal{U}}\cap B_{\text{Gab}}^{+}(z_{1},z_{2})\setminus C\neq\emptyset.\end{array}\right\}}.

Applying a change to hyperbolic polar coordinates to z1z_{1} we find

𝔼Fi=p2λ2s+r/100002π𝔻𝔼[g(z1(α1,ρ1),z2,𝒵{z1(α1,ρ1),z2})]f(z2)dz2dα1sinh(ρ1)dρ1.\begin{array}[]{rcl}{\mathbb{E}}F_{i}&=&\displaystyle p^{2}\lambda^{2}\int_{s+r/1000}^{\infty}\int_{0}^{2\pi}\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1}(\alpha_{1},\rho_{1}),z_{2},{\mathcal{Z}}\cup\{z_{1}(\alpha_{1},\rho_{1}),z_{2}\})\right]\cdot\\ &&\displaystyle\hskip 107.63855ptf(z_{2})\operatorname{d}z_{2}\operatorname{d}\alpha_{1}\sinh(\rho_{1})\operatorname{d}\rho_{1}.\end{array} (24)

Next, we consider the inner integral for some (fixed for the moment) z1=(cos(α)tanh(ρ1/2),sin(α)tanh(ρ1/2))z_{1}=(\cos(\alpha)\cdot\tanh(\rho_{1}/2),\sin(\alpha)\cdot\tanh(\rho_{1}/2)) satisfying dist2(z1,C)>r/1000\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},C)>r/1000. Let φ:𝔻𝔻\varphi:\mathbb{D}\to\mathbb{D} be a hyperbolic isometry that maps z1z_{1} to the origin and ci=oc_{i}=o to the negative xx-axis. (So that φ(ci)=(tanh(r1/2),0)\varphi(c_{i})=(-\tanh(r_{1}/2),0).) Note that

𝔼[g(z1,z2,𝒵{z1,z2})]=h(φ(z2)),{\mathbb{E}}\left[g(z_{1},z_{2},{\mathcal{Z}}\cup\{z_{1},z_{2}\})\right]=h(\varphi(z_{2})),

where

h(z):=1{B2(φ(ci),si)BGab(o,z),dist2(o,z)>r+v,dist2(o,C)r/1000}( either 𝒵BGab(o,z)φ[C]=, or 𝒵BGab+(o,z)φ[C]=, (or both). ).h(z):=1_{\left\{\begin{array}[]{l}B_{\mathbb{H}^{2}}(\varphi(c_{i}),s_{i})\cap B_{\text{Gab}}(o,z)\neq\emptyset,\\ \operatorname{dist}_{\mathbb{H}^{2}}(o,z)>r+v,\\ \operatorname{dist}_{\mathbb{H}^{2}}(o,C)\geq r/1000\end{array}\right\}}\cdot{\mathbb{P}}\left(\begin{array}[]{l}\text{ either }{\mathcal{Z}}\cap B_{\text{Gab}}^{-}(o,z)\setminus\varphi[C]=\emptyset,\\ \text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(o,z)\setminus\varphi[C]=\emptyset,\\ \text{ (or both). }\end{array}\right).

We point out that if B2(φ(c),si)BGab(o,z)B_{\mathbb{H}^{2}}(\varphi(c),s_{i})\cap B_{\text{Gab}}(o,z)\neq\emptyset then it must certainly hold that dist2(o,z)>dist2(o,φ(c))si=ρ1s\operatorname{dist}_{\mathbb{H}^{2}}(o,z)>\operatorname{dist}_{\mathbb{H}^{2}}(o,\varphi(c))-s_{i}=\rho_{1}-s. By Lemma 42, using that dist2(o,φ(c))=dist2(z1,c)=ρ1>si+r/1000\operatorname{dist}_{\mathbb{H}^{2}}(o,\varphi(c))=\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},c)=\rho_{1}>s_{i}+r/1000 by assumption, we must also have that

φ(c)oz<10e(siρ1)/2.\angle\varphi(c)oz<10e^{(s_{i}-\rho_{1})/2}.

We see that

1{B2(φ(c),si)BGab(o,z),dist2(o,z)>r+v.}1{dist2(o,z)>max(ρ1si,r+v),φ(c)oz<10e(siρ1)/2}.1_{\left\{\begin{array}[]{l}B_{\mathbb{H}^{2}}(\varphi(c),s_{i})\cap B_{\text{Gab}}(o,z)\neq\emptyset,\\ \operatorname{dist}_{\mathbb{H}^{2}}(o,z)>r+v.\end{array}\right\}}\leq 1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(o,z)>\max(\rho_{1}-s_{i},r+v),\\ \angle\varphi(c)oz<10e^{(s_{i}-\rho_{1})/2}\end{array}\right\}}.

Since dist2(o,φ[C])=dist2(z1,C)>r/1000\operatorname{dist}_{\mathbb{H}^{2}}(o,\varphi[C])=\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},C)>r/1000, we certainly have

ahd(φ[C],BGab(0,z))>r/1000.\operatorname{ahd}(\varphi[C],B_{\text{Gab}}(0,z))>r/1000.

Assuming λ0\lambda_{0} was chosen sufficiently small, we can apply Lemma 41 to see that

area2(BGab(o,z)φ[C])11000area2(BGab(o,z)),\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z)\cap\varphi[C])\leq\frac{1}{1000}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(o,z)),

and hence

area2(BGab(o,z)φ[C])4991000area2(BGab(0,z))edist2(o,z)/2,\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}^{-}(o,z)\setminus\varphi[C])\geq\frac{499}{1000}\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}(0,z))\geq e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2},

where the last inequality holds assuming we chose λ0\lambda_{0} sufficiently small, and assuming that dist2(o,z)>r+v\operatorname{dist}_{\mathbb{H}^{2}}(o,z)>r+v (and using r=2ln(1/λ)r=2\ln(1/\lambda)). Completely analogously, area2(BGab+(o,z)φ[C])edist2(o,z)/2\operatorname{area}_{\mathbb{H}^{2}}(B_{\text{Gab}}^{+}(o,z)\setminus\varphi[C])\geq e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2} as well.

We see that

h(z)1{dist2(o,z)>max(ρ1si,r+v),φ(c)oz<10e(siρ1)/2}2eλedist2(o,z)/2=:ψ(z).h(z)\leq 1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(o,z)>\max(\rho_{1}-s_{i},r+v),\\ \angle\varphi(c)oz<10e^{(s_{i}-\rho_{1})/2}\end{array}\right\}}\cdot 2e^{-\lambda e^{\operatorname{dist}_{\mathbb{H}^{2}}(o,z)/2}}=:\psi(z).

Applying (7), we find

𝔻𝔼[g(z1(α,r1),z2,𝒵)]f(z2)dz2𝔻ψ(φ(z2))f(z2)dz2=𝔻ψ(u)f(u)du.\int_{\mathbb{D}}{\mathbb{E}}\left[g(z_{1}(\alpha,r_{1}),z_{2},{\mathcal{Z}})\right]f(z_{2})\operatorname{d}z_{2}\leq\int_{\mathbb{D}}\psi(\varphi(z_{2}))f(z_{2})\operatorname{d}z_{2}=\int_{\mathbb{D}}\psi(u)f(u)\operatorname{d}u.

Changing to hyperbolic coordinates, i.e. u(α2,ρ2)=(cos(α2)tanh(ρ2/2),sin(α2)tanh(ρ2/2))u(\alpha_{2},\rho_{2})=(\cos(\alpha_{2})\cdot\tanh(\rho_{2}/2),\sin(\alpha_{2})\cdot\tanh(\rho_{2}/2)) gives

𝔻ψ(u)f(u)du=002πψ(u(α2,ρ2))dα2sinh(ρ2)dρ2max(ρ1si,r+v)02π1{|α2π|<10e(siρ1)/2}2eλeρ2/2dα2sinh(ρ2)dρ2=max(ρ1si,r+v)20e(siρ1)/22eλeρ2/2sinh(ρ2)dρ2=40e(sρ1)/2max(ρ1si,r+v)eλeρ2/2sinh(ρ2)dρ220e(siρ1)/2max(ρ1si,r+v)eλeρ2/2eρ2dρ2\begin{array}[]{rcl}\displaystyle\int_{\mathbb{D}}\psi(u)f(u)\operatorname{d}u&=&\displaystyle\int_{0}^{\infty}\int_{0}^{2\pi}\psi(u(\alpha_{2},\rho_{2}))\operatorname{d}\alpha_{2}\sinh(\rho_{2})\operatorname{d}\rho_{2}\\[8.61108pt] &\leq&\displaystyle\int_{\max(\rho_{1}-s_{i},r+v)}^{\infty}\int_{0}^{2\pi}1_{\{|\alpha_{2}-\pi|<10e^{(s_{i}-\rho_{1})/2}\}}2e^{-\lambda e^{\rho_{2}/2}}\operatorname{d}\alpha_{2}\sinh(\rho_{2})\operatorname{d}\rho_{2}\\[8.61108pt] &=&\displaystyle\int_{\max(\rho_{1}-s_{i},r+v)}^{\infty}20e^{(s_{i}-\rho_{1})/2}2e^{-\lambda e^{\rho_{2}/2}}\sinh(\rho_{2})\operatorname{d}\rho_{2}\\[8.61108pt] &=&\displaystyle 40e^{(s-\rho_{1})/2}\int_{\max(\rho_{1}-s_{i},r+v)}^{\infty}e^{-\lambda e^{\rho_{2}/2}}\sinh(\rho_{2})\operatorname{d}\rho_{2}\\[8.61108pt] &\leq&\displaystyle 20e^{(s_{i}-\rho_{1})/2}\int_{\max(\rho_{1}-s_{i},r+v)}^{\infty}e^{-\lambda e^{\rho_{2}/2}}e^{\rho_{2}}\operatorname{d}\rho_{2}\end{array}

We next apply the substitution t:=λeρ2/2t:=\lambda e^{\rho_{2}/2} (so that dρ2=2dtt\operatorname{d}\rho_{2}=\frac{2\operatorname{d}t}{t}) to obtain

𝔻ψ(u)f(u)du20e(siρ1)/2λ2λemax(ρ1si,r+v)/2tetdt=20e(siρ1)/2λ2(λemax(ρ1si,r+v)/2+1)eλemax(ρ1si,r+v)/240e(siρ1)/2λ2λemax(ρ1si,r+v)/2eλemax(ρ1si,r+v)/2,\begin{array}[]{rcl}\displaystyle\int_{\mathbb{D}}\psi(u)f(u)\operatorname{d}u&\leq&\displaystyle\frac{20e^{(s_{i}-\rho_{1})/2}}{\lambda^{2}}\cdot\int_{\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}}^{\infty}te^{-t}\operatorname{d}t\\[8.61108pt] &=&\displaystyle\frac{20e^{(s_{i}-\rho_{1})/2}}{\lambda^{2}}\cdot\left(\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}+1\right)\cdot e^{-\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}}\\[8.61108pt] &\leq&\displaystyle\frac{40e^{(s_{i}-\rho_{1})/2}}{\lambda^{2}}\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}e^{-\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}},\end{array}

using in the last line that λer/2=1\lambda e^{r/2}=1 and vw>0v\geq w>0.

Filling this back into (24) we find

𝔼Fip2λ2si+r/100002π40πe(siρ1)/2λ2λemax(ρ1si,r+v)/2eλemax(ρ1si,r+v)/2dα1sinh(ρ1)dρ1=80πp2λesi/2s+r/1000eρ1/2emax(ρ1si,r+v)/2eλemax(ρ1si,r+v)/2sinh(ρ1)dρ1104λ3esi/2si+r/1000eρ1/2emax(ρ1si,r+v)/2eλemax(ρ1si,r+v)/2dρ1=104λ3esi/2(s+r/1000si+r+veρ1/2e(r+v)/2eλe(r+v)/2dρ1+si+r+veρ1/2e(ρ1si)/2eλe(ρ1si)/2dρ1)=:104λ3esi/2(I1+I2),\begin{array}[]{rcl}{\mathbb{E}}F_{i}&\leq&\displaystyle p^{2}\lambda^{2}\int_{s_{i}+r/1000}^{\infty}\int_{0}^{2\pi}\frac{40\pi e^{(s_{i}-\rho_{1})/2}}{\lambda^{2}}\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}e^{-\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}}\operatorname{d}\alpha_{1}\sinh(\rho_{1})\operatorname{d}\rho_{1}\\[8.61108pt] &=&\displaystyle 80\pi p^{2}\lambda e^{s_{i}/2}\int_{s+r/1000}^{\infty}e^{-\rho_{1}/2}\cdot e^{\max(\rho_{1}-s_{i},r+v)/2}e^{-\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}}\sinh(\rho_{1})\operatorname{d}\rho_{1}\\[8.61108pt] &\leq&\displaystyle 10^{4}\lambda^{3}e^{s_{i}/2}\int_{s_{i}+r/1000}^{\infty}e^{\rho_{1}/2}\cdot e^{\max(\rho_{1}-s_{i},r+v)/2}e^{-\lambda e^{\max(\rho_{1}-s_{i},r+v)/2}}\operatorname{d}\rho_{1}\\ &=&\displaystyle 10^{4}\lambda^{3}e^{s_{i}/2}\cdot\left(\int_{s+r/1000}^{s_{i}+r+v}e^{\rho_{1}/2}\cdot e^{(r+v)/2}e^{-\lambda e^{(r+v)/2}}\operatorname{d}\rho_{1}\right.\\[8.61108pt] &&\displaystyle\left.\hskip 43.05542pt+\int_{s_{i}+r+v}^{\infty}e^{\rho_{1}/2}\cdot e^{(\rho_{1}-s_{i})/2}e^{-\lambda e^{(\rho_{1}-s_{i})/2}}\operatorname{d}\rho_{1}\right)\\[8.61108pt] &=:&\displaystyle 10^{4}\lambda^{3}e^{s_{i}/2}\cdot(I_{1}+I_{2}),\end{array}

using that sinh(ρ1)12eρ1\sinh(\rho_{1})\leq\frac{1}{2}e^{\rho_{1}} and p10λp\leq 10\lambda in the third line.

Now

104λ3esi/2I12104λ3esi/2e(si+r+v)/2e(r+v)/2eλe(r+v)/2=2104esir/2eveev/2,\begin{array}[]{rcl}10^{4}\lambda^{3}e^{s_{i}/2}I_{1}&\leq&2\cdot 10^{4}\lambda^{3}e^{s_{i}/2}\cdot e^{(s_{i}+r+v)/2}\cdot e^{(r+v)/2}e^{-\lambda e^{(r+v)/2}}\\ &=&2\cdot 10^{4}\cdot e^{s_{i}-r/2}\cdot e^{v}\cdot e^{-e^{v/2}},\end{array} (25)

using that r=2ln(1/λ)r=2\ln(1/\lambda).

We also have

104λ3esi/2I2=104λ3esisi+r+veρ1sieλe(ρ1si)/2dρ1=104λ3esir+veueλeu/2du=2104λesiev/2tetdt4104λesiev/2eev/24104esir/2eveev/2,\begin{array}[]{rcl}10^{4}\lambda^{3}e^{s_{i}/2}I_{2}&=&\displaystyle 10^{4}\lambda^{3}e^{s_{i}}\int_{s_{i}+r+v}^{\infty}e^{\rho_{1}-s_{i}}e^{-\lambda e^{(\rho_{1}-s_{i})/2}}\operatorname{d}\rho_{1}\\[8.61108pt] &=&\displaystyle 10^{4}\lambda^{3}e^{s_{i}}\int_{r+v}^{\infty}e^{u}e^{-\lambda e^{u/2}}\operatorname{d}u\\[8.61108pt] &=&\displaystyle 2\cdot 10^{4}\lambda e^{s_{i}}\int_{e^{v/2}}^{\infty}te^{-t}\operatorname{d}t\\[8.61108pt] &\leq&\displaystyle 4\cdot 10^{4}\lambda e^{s_{i}}e^{v/2}e^{-e^{v/2}}\\[8.61108pt] &\leq&\displaystyle 4\cdot 10^{4}\cdot e^{s_{i}-r/2}\cdot e^{v}\cdot e^{-e^{v/2}},\end{array} (26)

using the substitution u=ρ1siu=\rho_{1}-s_{i} in the second line, the substitution t:=λeu/2t:=\lambda e^{u/2} in the third line and that r=2ln(1/λ)r=2\ln(1/\lambda). Adding (25) and (26) and multiplying by 2k2k gives the result. \blacksquare

Lemma 55

For every w1,w2,ϑ>0w_{1},w_{2},\vartheta>0 there exist λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that the following holds for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2log(1/λ)r:=2\log(1/\lambda).

Let u0,,uk𝔻u_{0},\dots,u_{k}\in\mathbb{D} be an arbitrary pre-chunk and write C:=cert(u0,,uk)C:=\operatorname{cert}(u_{0},\dots,u_{k}) and

FIV-c,d:=|{(z1,z2)𝒵b×𝒵b:dist2(z1,z2)d, and; dist2(z1,C)r/1000, and;BGab(z1,z2)C=, and; 𝒵BGab(z1,z2)= or 𝒵BGab+(z1,z2)=.}|.F_{\text{\bf{IV-c}},\geq d}:=\left|\left\{(z_{1},z_{2})\in{\mathcal{Z}}_{b}\times{\mathcal{Z}}_{b}:\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})\geq d,\text{ and; }\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{1},C)\leq r/1000,\text{ and;}\\ B_{\text{Gab}}(z_{1},z_{2})\cap C=\emptyset,\text{ and; }\\ {\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z_{1},z_{2})=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z_{1},z_{2})=\emptyset.\end{array}\right\}\right|.

Then we have

𝔼FIV-c,r+v103keveev/2emax(w2,dist2(uk1,uk)r)/2,{\mathbb{E}}F_{\text{\bf{IV-c}},\geq r+v}\leq 10^{3}\cdot k\cdot e^{v}\cdot e^{-e^{v/2}}\cdot e^{\max(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r)/2},

for all vw2v\geq w_{2}.

Proof. As usual, we let λ0>0\lambda_{0}>0 be a small constant, to be chosen appropriately during the course of the proof. Writing C:={u𝔻:dist2(u,C)<r/1000}C^{\prime}:=\{u\in\mathbb{D}:\operatorname{dist}_{\mathbb{H}^{2}}(u,C)<r/1000\} and applying Corollary 5, we have

𝔼FIV-b,r+v(pλ)2𝔻𝔻1{z1C,dist2(z1,z2)r+v}(𝒵BGab(z1,z2)= or 𝒵BGab+(z1,z2)=)f(z1)f(z2)dz2dz1pλarea2(C)103eveev/2,\begin{array}[]{c}{\mathbb{E}}F_{\text{\bf{IV-b}},\geq r+v}\\ \leq\\ \displaystyle(p\lambda)^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}1_{\left\{z_{1}\in C^{\prime},\operatorname{dist}_{\mathbb{H}^{2}}(z_{1},z_{2})\geq r+v\right\}}\cdot{\mathbb{P}}({\mathcal{Z}}\cap B_{\text{Gab}}^{-}(z_{1},z_{2})=\emptyset\text{ or }{\mathcal{Z}}\cap B_{\text{Gab}}^{+}(z_{1},z_{2})=\emptyset)\\ \displaystyle\quad\quad f(z_{1})f(z_{2})\operatorname{d}z_{2}\operatorname{d}z_{1}\\[8.61108pt] \leq\\ p\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(C^{\prime})\cdot 10^{3}\cdot e^{v}\cdot e^{-e^{v/2}},\end{array}

applying Lemma 24. Next we remark

pλarea2(C)10λ22kπemax(r+w2,dist2(uk1,uk))/2+r/100020πkemax(w2,dist2(uk1,uk)r)/2(4991000)rkemax(w2,dist2(uk1,uk)r)/2,\begin{array}[]{rcl}p\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(C^{\prime})&\leq&10\lambda^{2}\cdot 2k\cdot\pi e^{\max(r+w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k}))/2+r/1000}\\ &\leq&20\pi\cdot k\cdot e^{\max(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r)/2-\left(\frac{499}{1000}\right)r}\\ &\leq&k\cdot e^{\max(w_{2},\operatorname{dist}_{\mathbb{H}^{2}}(u_{k-1},u_{k})-r)/2},\end{array}

the last inequality holding provided we chose λ0\lambda_{0} sufficiently small. \blacksquare

Recall that CkII,CkIII,CkIV,vC_{k}^{\text{\bf{II}}},C_{k}^{\text{\bf{III}}},C_{k}^{\text{\bf{IV}},v} denote the number of black chunks starting from the origin, with length kk and final pseudo-edge of type II, respectively type III, respectively type IV and final pseudo-edge of length [r+v,r+v+1)\in[r+v,r+v+1).

We now need to introduce analogous notations to deal with linked sequences of chunks. We set

Sn,kII:=|{black linked sequences of chunks starting from o, consisting of precisely n chunks,having a last chunk of length k and final pseudo-edge of type II}|.S_{n,k}^{\text{\bf{II}}}:=\left|\left\{\begin{array}[]{l}\text{black linked sequences of chunks starting from $o$, consisting of precisely $n$ chunks,}\\ \text{having a last chunk of length $k$ and final pseudo-edge of type $\text{\bf{II}}$}\end{array}\right\}\right|.

We let Sn,kIIIS_{n,k}^{\text{\bf{III}}} and Sn,kIV,vS_{n,k}^{\text{\bf{IV}},v} be defined analogously. Of course

S1,kII=CkII,S1,kIII=CkIII,S1,kIV,v=CkIV,v.S_{1,k}^{\text{\bf{II}}}=C_{k}^{\text{\bf{II}}},\quad S_{1,k}^{\text{\bf{III}}}=C_{k}^{\text{\bf{III}}},\quad S_{1,k}^{\text{\bf{IV}},v}=C_{k}^{\text{\bf{IV}},v}. (27)

In particular, Corollary 50 provides bounds on the expectations 𝔼Sn,kII,𝔼Sn,kIII{\mathbb{E}}S_{n,k}^{\text{\bf{II}}},{\mathbb{E}}S_{n,k}^{\text{\bf{III}}} and Sn,kIV,vS_{n,k}^{\text{\bf{IV}},v} when n=1n=1. The next lemma provides a system of recursive inequalities that will allow us to also bound these expectations for n2n\geq 2. From now on, it will be convenient to assume the parameter w2w_{2} is an integer. Note that so far the only result that puts any restrictions on the value of w2w_{2} is Corollary 47 which just states it has to be taken sufficiently large.

Lemma 56

For every w1,ϑ>0w_{1},\vartheta>0 and w2w_{2}\in\mathbb{N} there exists a λ0=λ0(w1,w2,ϑ)\lambda_{0}=\lambda_{0}(w_{1},w_{2},\vartheta) such that for all 0<λ<λ00<\lambda<\lambda_{0} and p10λp\leq 10\lambda, setting r:=2ln(1/λ)r:=2\ln(1/\lambda), the following holds.

For all n1n\geq 1, writing

Σn:=107=1(e2w2𝔼Sn,II+e2w2𝔼Sn,III+v=w2e2v+2𝔼Sn,IV,v),\Sigma_{n}:=10^{7}\cdot\sum_{\ell=1}^{\infty}\ell\cdot\left(e^{2w_{2}}\cdot{\mathbb{E}}S_{n,\ell}^{\text{\bf{II}}}+e^{2w_{2}}\cdot{\mathbb{E}}S_{n,\ell}^{\text{\bf{III}}}+\sum_{v=w_{2}}^{\infty}e^{2v+2}\cdot{\mathbb{E}}S_{n,\ell}^{\text{\bf{IV}},v}\right),

we have

𝔼Sn+1,1IIΣn103ew1𝔼Sn+1,2IIIΣn103ϑe2w2𝔼Sn+1,1IV,vΣn106eveev/2\begin{array}[]{rcl}{\mathbb{E}}S_{n+1,1}^{\text{\bf{II}}}&\leq&\Sigma_{n}\cdot 10^{3}\cdot e^{-w_{1}}\\[8.61108pt] {\mathbb{E}}S_{n+1,2}^{\text{\bf{III}}}&\leq&\Sigma_{n}\cdot 10^{3}\cdot\vartheta\cdot e^{2w_{2}}\\[8.61108pt] {\mathbb{E}}S_{n+1,1}^{\text{\bf{IV}},v}&\leq&\Sigma_{n}\cdot 10^{6}\cdot e^{v}\cdot e^{-e^{v/2}}\end{array}

and

𝔼Sn+1,kτΣn𝔼Ck1τ,{\mathbb{E}}S_{n+1,k}^{\tau}\leq\Sigma_{n}\cdot{\mathbb{E}}C_{k-1}^{\tau},

for τ\tau one of II,III\text{\bf{II}},\text{\bf{III}} or (IV,v)(\text{\bf{IV}},v) with vw2v\geq w_{2}, and for all k3k\geq 3 when τ=III\tau=\text{\bf{III}} and all k2k\geq 2 otherwise.

Proof. We’ll need to introduce even more notation. We let

𝒯:={II,III,(IV,v):v=w2,w2+1,},:={a,b,c},{\mathscr{T}}:=\{\text{\bf{II}},\text{\bf{III}},(\text{\bf{IV}},v):v=w_{2},w_{2}+1,\dots\},\quad{\mathscr{L}}:=\{a,b,c\},

denote the possible types of the final pseudo-edges of chunks, respectively types of links between consecutive chunks – corresponding to the cases described in part (ii) of Definition 30.

For nn\in\mathbb{N} and k¯n,τ¯𝒯n,t¯n1\underline{k}\in\mathbb{N}^{n},\underline{\tau}\in{\mathscr{T}}^{n},\underline{t}\in{\mathscr{L}}^{n-1} we write Sk¯,τ¯,t¯S_{\underline{k},\underline{\tau},\underline{t}} for the number of black, linked sequences of chunks P1,,PnP_{1},\dots,P_{n} starting from oo, such that PiP_{i} has length kik_{i} and final pseudo-edge of type τi\tau_{i} (for all i=1,,ni=1,\dots,n), and such that the link between PiP_{i} and Pi+1P_{i+1} corresponds to case tit_{i} of part (ii) of Definition 30 (for i=1,,n=1i=1,\dots,n=1). Clearly we can write for each n,k1n,k\geq 1 and τ𝒯\tau\in{\mathscr{T}}:

Sn,kτ=k¯n,τ¯𝒯n,t¯n1,τn=τ,kn=kSk¯,τ¯,t¯.S_{n,k}^{\tau}=\sum_{{\underline{k}\in\mathbb{N}^{n},\underline{\tau}\in{\mathscr{T}}^{n},\underline{t}\in{\mathscr{L}}^{n-1},}\atop\tau_{n}=\tau,k_{n}=k}S_{\underline{k},\underline{\tau},\underline{t}}. (28)

We will first derive the last inequality claimed in the lemma statement. Let us fix n,k,n,k,\ell\in\mathbb{N} and σ,τ𝒯\sigma,\tau\in{\mathscr{T}} such that either k2k\geq 2 in case τIII\tau\neq\text{\bf{III}} or k3k\geq 3 in case τ=III\tau=\text{\bf{III}}. We pick vectors k¯n,τ¯𝒯n,t¯n1\underline{k}\in\mathbb{N}^{n},\underline{\tau}\in{\mathscr{T}}^{n},\underline{t}\in{\mathscr{L}}^{n-1} satisfying kn1=,kn=kk_{n-1}=\ell,k_{n}=k and τn1=σ,τn=τ\tau_{n-1}=\sigma,\tau_{n}=\tau.

The parameters k¯,τ¯,t¯\underline{k},\underline{\tau},\underline{t} determine the number of vertices in any of the linked sequences of pseudopaths counted by Sk¯,τ¯,t¯S_{\underline{k},\underline{\tau},\underline{t}}. For any such sequence of pseudopaths we have V(P1)V(Pn)={o,z1,,zm}V(P_{1})\cup\dots\cup V(P_{n})=\{o,z_{1},\dots,z_{m}\} with z1,,zm𝒵bz_{1},\dots,z_{m}\in{\mathcal{Z}}_{\text{b}} and k1++knmk1++kn+n1k_{1}+\dots+k_{n}\leq m\leq k_{1}+\dots+k_{n}+n-1. (To be precise m=k1++kn+|{1in1:tia}|m=k_{1}+\dots+k_{n}+|\{1\leq i\leq n-1:t_{i}\neq a\}| as V(Pi)=ki+1V(P_{i})=k_{i}+1 for all ii but every time ti=at_{i}=a the last vertex of PiP_{i} coincides with the first vertex of Pi+1P_{i+1}.) By Corollary 5 we can write, for m=m(k¯,τ¯,t¯)m=m(\underline{k},\underline{\tau},\underline{t}) as determined by the parameters

𝔼Sk¯,τ¯,t¯=(pλ)m𝔻𝔻𝔼[gk¯,τ¯,t¯(z1,,zm;𝒵{z1,,zm})]f(z1)f(zm)dz1dzm,{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}=(p\lambda)^{m}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{\underline{k},\underline{\tau},\underline{t}}(z_{1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m}\})\right]f(z_{1})\dots f(z_{m})\operatorname{d}z_{1}\dots\operatorname{d}z_{m},

where ff is given by (2) and gk¯,τ¯,t¯g_{\underline{k},\underline{\tau},\underline{t}} is the indicator function that o,z1,,zmo,z_{1},\dots,z_{m} form a linked sequence of pseudopaths of the required kind wrt. the point set 𝒵{o,z1,,zm}{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{m}\}.

The vertices of PnP_{n} are zmk,,zmz_{m-k},\dots,z_{m}. If tn1=at_{n-1}=a then zmz_{m} is both the first vertex of PnP_{n} and also the last vertex of Pn1P_{n-1}. Recall that cert(zmk+1,,zm)\operatorname{cert}(z_{m-k+1},\dots,z_{m}) needs to be disjoint from cert(P1)cert(Pn1)\operatorname{cert}(P_{1})\cup\dots\cup\operatorname{cert}(P_{n-1}) (and this set is completely determined by z1,,zmkz_{1},\dots,z_{m-k}), otherwise gk¯,τ¯,t¯g_{\underline{k},\underline{\tau},\underline{t}} will equal zero. So, provided tn1=at_{n-1}=a we can write, for every z1,,zm𝔻z_{1},\dots,z_{m}\in\mathbb{D}:

𝔼[gk¯,τ¯,t¯(z1,,zm;𝒵{z1,,zm})]𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk})]𝔼[gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})]1{dist2(zmk,zmk+1)<r+w2},\begin{array}[]{c}{\mathbb{E}}\left[g_{\underline{k},\underline{\tau},\underline{t}}(z_{1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m}\})\right]\\ \leq\\ {\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})\right]\cdot\\ {\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})\right]\cdot\\ 1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2}\}},\end{array} (29)

where gCk1τg_{C^{\tau}_{k-1}} is the indicator function that zmk+1,,zmz_{m-k+1},\dots,z_{m} forms a chunk of length k1k-1 and with final edge of type τ\tau (with respect to the point set 𝒵{zmk+1,,zm}{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\}). Here we use that gk¯,τ¯,t¯=0g_{\underline{k},\underline{\tau},\underline{t}}=0 unless the position of z1,,zmz_{1},\dots,z_{m} is such that cert(zmk+1,,zm)cert(Pi)=\operatorname{cert}(z_{m-k+1},\dots,z_{m})\cap\operatorname{cert}(P_{i})=\emptyset for i=1,,n1i=1,\dots,n-1; and if z1,,zmz_{1},\dots,z_{m} are such that all these intersections are empty then the event that g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk})=1g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})=1 and the event that gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})=1g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})=1 are independent.

We can write

𝔼Sk¯,τ¯,t¯(pλ)mk𝔻𝔻𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk})](pλ𝔻1{dist2(zmk,zmk+1)<r+w2}((pλ)k1𝔻𝔻𝔼[gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})]f(zmk+1)f(zm)dzmdzmk+1)f(zmk+1)dzmk+1)f(z1)f(zmk)dzmkdz1\begin{array}[]{rcl}{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}&\leq&\displaystyle(p\lambda)^{m-k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}\\[8.61108pt] &&\displaystyle{\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})\right]\cdot\\[8.61108pt] &&\displaystyle{\bigg{(}}p\lambda\int_{\mathbb{D}}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2}\}}\cdot{\bigg{(}}(p\lambda)^{k-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}\\[8.61108pt] &&\displaystyle\hskip 21.52771pt{\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})\right]\cdot\\ &&\displaystyle\hskip 21.52771ptf(z_{m-k+1})\dots f(z_{m})\operatorname{d}z_{m}\dots\operatorname{d}z_{m-k+1}{\bigg{)}}f(z_{m-k+1})\operatorname{d}z_{m-k+1}{\bigg{)}}\cdot\\[8.61108pt] &&\displaystyle\hskip 21.52771ptf(z_{1})\dots f(z_{m-k})\operatorname{d}z_{m-k}\dots\operatorname{d}z_{1}\end{array} (30)

For every fixed z1,,zmk+1z_{1},\dots,z_{m-k+1}, applying an isometry that maps zmk+1z_{m-k+1} to oo and Corollary 5, we have

(pλ)k1𝔻𝔻𝔼[gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})]f(zmk+2)f(zm)dzmdzmk+2=(pλ)k1𝔻𝔻𝔼[gCk1τ(o,z1,,zk1;𝒵{o,z1,,zk1})]f(z1)f(zk1)dz1dzk1=𝔼Ck1τ.\begin{array}[]{c}\displaystyle(p\lambda)^{k-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})\right]\cdot\\ \displaystyle\hskip 21.52771ptf(z_{m-k+2})\dots f(z_{m})\operatorname{d}z_{m}\dots\operatorname{d}z_{m-k+2}\\ =\\ \displaystyle(p\lambda)^{k-1}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}{\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(o,z_{1},\dots,z_{k-1};{\mathcal{Z}}\cup\{o,z_{1},\dots,z_{k-1}\})\right]\cdot\\ \displaystyle\hskip 21.52771ptf(z_{1})\dots f(z_{k-1})\operatorname{d}z_{1}\dots\operatorname{d}z_{k-1}\\ =\\ {\mathbb{E}}C_{k-1}^{\tau}.\end{array} (31)

Of course, for every fixed zmk𝔻z_{m-k}\in\mathbb{D}, we have

pλ𝔻1{dist2(zmk,zmk+1)<r+w2}f(zmk+1)dzmk+1=pλarea2(B2(o,r+w2))103ew2,\begin{array}[]{rcl}\displaystyle p\lambda\int_{\mathbb{D}}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2}\}}f(z_{m-k+1})\operatorname{d}z_{m-k+1}&=&\displaystyle p\lambda\cdot\operatorname{area}_{\mathbb{H}^{2}}(B_{\mathbb{H}^{2}}(o,r+w_{2}))\\ &\leq&10^{3}\cdot e^{w_{2}},\end{array}

by Lemma 20. It follows that:

𝔼Sk¯,τ¯,t¯103ew2𝔼Ck1τ(pλ)mk𝔻𝔻𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk})]f(z1)f(zmk)dzmkdz1=103ew2𝔼Ck1τ𝔼S(k1,,kn),(τ1,,τn1),(t1,,tn2).\begin{array}[]{rcl}{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}&\leq&\displaystyle 10^{3}\cdot e^{w_{2}}\cdot{\mathbb{E}}C_{k-1}^{\tau}\cdot(p\lambda)^{m-k}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}\\[8.61108pt] &&\displaystyle{\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})\right]\cdot\\[8.61108pt] &&\displaystyle f(z_{1})\dots f(z_{m-k})\operatorname{d}z_{m-k}\dots\operatorname{d}z_{1}\\[8.61108pt] &=&\displaystyle 10^{3}\cdot e^{w_{2}}\cdot{\mathbb{E}}C_{k-1}^{\tau}\cdot{\mathbb{E}}S_{(k_{1},\dots,k_{n}),(\tau_{1},\dots,\tau_{n-1}),(t_{1},\dots,t_{n-2})}.\end{array} (32)

(Provided tn1=at_{n-1}=a.)

If tn1=bt_{n-1}=b then zmkz_{m-k}, the first vertex of PnP_{n}, does not lie on Pn1P_{n-1} and we need that dist2(zmk,zmk+1)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2} and cert(zmk,zmk+1)\operatorname{cert}(z_{m-k},z_{m-k+1}) intersects cert(Pn1)=cert(zmkn1k2,,zmk1)\operatorname{cert}(P_{n-1})=\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1}). So instead of (29) we can now write

𝔼[gk¯,τ¯,t¯(z1,,zm;𝒵{z1,,zm})]𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn1)(z1,,zmk;𝒵{z1,,zmk})]𝔼[gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})]1{cert(zmk,zmk+1)cert(zmkn1k2,,zmk1),dist2(zmk,zmk+1)<r+w2}\begin{array}[]{c}{\mathbb{E}}\left[g_{\underline{k},\underline{\tau},\underline{t}}(z_{1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m}\})\right]\\ \leq\\ {\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-1})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})\right]\cdot\\ {\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})\right]\cdot\\ 1_{\left\{\begin{array}[]{l}\operatorname{cert}(z_{m-k},z_{m-k+1})\cap\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1})\neq\emptyset,\atop\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2}\end{array}\right\}}\end{array} (33)

Arguing as in (30) and reusing (31) we find that

𝔼Sk¯,τ¯,t¯𝔼Ck1τ(pλ)m(k+1)𝔻𝔻𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk1})]((pλ)2𝔻𝔻1{cert(zmk,zmk+1)cert(zmkn1k2,,zmk1),dist2(zmk,zmk+1)<r+w2}f(zmk)f(zmk+1)dzmkdzmk+1)f(z1)f(zmk)dzmkdz1\begin{array}[]{rcl}{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}&\leq&\displaystyle{\mathbb{E}}C_{k-1}^{\tau}\cdot(p\lambda)^{m-(k+1)}\int_{\mathbb{D}}\dots\int_{\mathbb{D}}\\[8.61108pt] &&\displaystyle{\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k-1}\})\right]\cdot\\[8.61108pt] &&\displaystyle{\bigg{(}}(p\lambda)^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}1_{\left\{\begin{array}[]{l}\operatorname{cert}(z_{m_{k}},z_{m-k+1})\cap\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1})\neq\emptyset,\atop\operatorname{dist}_{\mathbb{H}^{2}}(z_{m_{k}},z_{m-k+1})<r+w_{2}\end{array}\right\}}\cdot\\[8.61108pt] &&\displaystyle f(z_{m-k})f(z_{m-k+1})\operatorname{d}z_{m-k}\operatorname{d}z_{m-k+1}{\bigg{)}}\cdot\\ &&\displaystyle f(z_{1})\dots f(z_{m-k})\operatorname{d}z_{m-k}\dots\operatorname{d}z_{1}\end{array}

For any fixed z1,,zmk1𝔻z_{1},\dots,z_{m-k-1}\in\mathbb{D} we have that if zmkn1k2,,zmk1z_{m-k_{n-1}-k-2},\dots,z_{m-k-1} is not a pre-chunk then g(k1,,kn1);(τ1,,τn1);(t1,,tn2)(z1,,zmk;𝒵{z1,,zmk1})=0g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k-1}\})=0 and otherwise we can apply Lemma 51 to get

(pλ)2𝔻𝔻1{cert(zmk,zmk+1)cert(zmkn1k2,,zmk1),dist2(zmk,zmk+1)<r+w2}f(zmk)f(zmk+1)dzmkdzmk+1105kn1e2v(τn1)+2106kn1e2v(τn1),\begin{array}[]{c}\displaystyle(p\lambda)^{2}\int_{\mathbb{D}}\int_{\mathbb{D}}1_{\left\{\begin{array}[]{c}\operatorname{cert}(z_{m_{k}},z_{m-k+1})\cap\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1})\neq\emptyset,\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{m_{k}},z_{m-k+1})<r+w_{2}\end{array}\right\}}\cdot\\[12.91663pt] \displaystyle f(z_{m-k})f(z_{m-k+1})\operatorname{d}z_{m-k}\operatorname{d}z_{m-k+1}\\ \leq\\ 10^{5}\cdot k_{n-1}\cdot e^{2v(\tau_{n-1})+2}\\ \leq\\ 10^{6}\cdot k_{n-1}\cdot e^{2v(\tau_{n-1})},\end{array}

where v(σ)=w2v(\sigma)=w_{2} if σ{II,III}\sigma\in\{\text{\bf{II}},\text{\bf{III}}\} and otherwise v(σ)v(\sigma) is determined via σ=:(IV,v(σ)1)\sigma=:(\text{\bf{IV}},v(\sigma)-1). (That is, if σ=(IV,x)\sigma=(\text{\bf{IV}},x) for some xw2x\geq w_{2} then v(σ)=x+1v(\sigma)=x+1.) We conclude that if tn1=bt_{n-1}=b then

𝔼Sk¯,τ¯,t¯106kn1e2v(τn1)𝔼Ck1τ𝔼S(k1,,kn1);(τ1,,τn1);(t1,,tn2).{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}\leq 10^{6}\cdot k_{n-1}\cdot e^{2v(\tau_{n-1})}\cdot{\mathbb{E}}C_{k-1}^{\tau}\cdot{\mathbb{E}}S_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}. (34)

If tn1=ct_{n-1}=c then we replace (33) with

𝔼[gk¯,τ¯,t¯(z1,,zm;𝒵{z1,,zm})]𝔼[g(k1,,kn1);(τ1,,τn1);(t1,,tn1)(z1,,zmk;𝒵{z1,,zmk})]𝔼[gCk1τ(zmk+1,,zm;𝒵{zmk+1,,zm})]1{dist2(zmk,cert(zmkn1k2,,zmk1))<r/1000,dist2(zmk,zmk+1)<r+w2,cert(zmk,zmk+1)cert(zmkn1k2,,zmk1)=,zmkzmk+1 is a pseudoedge. }.\begin{array}[]{c}{\mathbb{E}}\left[g_{\underline{k},\underline{\tau},\underline{t}}(z_{1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m}\})\right]\\ \leq\\ {\mathbb{E}}\left[g_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-1})}(z_{1},\dots,z_{m-k};{\mathcal{Z}}\cup\{z_{1},\dots,z_{m-k}\})\right]\cdot\\ {\mathbb{E}}\left[g_{C^{\tau}_{k-1}}(z_{m-k+1},\dots,z_{m};{\mathcal{Z}}\cup\{z_{m-k+1},\dots,z_{m}\})\right]\cdot\\ 1_{\left\{\begin{array}[]{l}\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1}))<r/1000,\\ \operatorname{dist}_{\mathbb{H}^{2}}(z_{m-k},z_{m-k+1})<r+w_{2},\\ \operatorname{cert}(z_{m-k},z_{m-k+1})\cap\operatorname{cert}(z_{m-k_{n-1}-k-2},\dots,z_{m-k-1})=\emptyset,\\ z_{m-k}z_{m-k+1}\text{ is a pseudoedge. }\end{array}\right\}}.\end{array} (35)

Arguing as before, but using Lemma 52 in place of Lemma 51 gives:

𝔼Sk¯,τ¯,t¯kn1ev(τn1)/2𝔼Ck1τ𝔼S(k1,,kn1);(τ1,,τn1);(t1,,tn2).{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}\leq k_{n-1}\cdot e^{v(\tau_{n-1})/2}\cdot{\mathbb{E}}C_{k-1}^{\tau}\cdot{\mathbb{E}}S_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}. (36)

(Provided tn1=ct_{n-1}=c.)

Combining (28) and (32), (34), (36) gives

𝔼Sn,kτ=k¯n,τ¯𝒯n,t¯n1,τn=τ,kn=k𝔼Sk¯,τ¯,t¯k1,,kn1τ1,,τn1𝒯t1,,tn2𝔼Ck1τ(103ew2+106kn1e2v(τn1)+kn1ev(τn1)/2)𝔼S(k1,,kn1);(τ1,,τn1);(t1,,tn2)k1,,kn1τ1,,τn1𝒯t1,,tn2𝔼Ck1τ107kn1e2v(τn1)𝔼S(k1,,kn1);(τ1,,τn1);(t1,,tn2)=107𝔼Ck1τ=1σ𝒯e2v(σ)k1,,kn1τ1,,τn1𝒯t1,,tn2kn1=,τn1=σ𝔼S(k1,,kn1);(τ1,,τn1);(t1,,tn2)=𝔼Ck1τ107=1σ𝒯e2v(σ)𝔼Sn1,σ=𝔼Ck1τ107=1(e2w2𝔼Sn1,II+e2w2𝔼Sn1,III+vw2e2v+2𝔼Sn1,IV,v)=𝔼Ck1τΣn,\begin{array}[]{rcl}{\mathbb{E}}S^{\tau}_{n,k}&=&\displaystyle\sum_{{\underline{k}\in\mathbb{N}^{n},\underline{\tau}\in{\mathscr{T}}^{n},\underline{t}\in{\mathscr{L}}^{n-1},}\atop\tau_{n}=\tau,k_{n}=k}{\mathbb{E}}S_{\underline{k},\underline{\tau},\underline{t}}\\[21.52771pt] &\leq&\displaystyle\sum_{{k_{1},\dots,k_{n-1}\in\mathbb{N}}\atop{{\tau_{1},\dots,\tau_{n-1}\in{\mathscr{T}}}\atop t_{1},\dots,t_{n-2}\in{\mathscr{L}}}}{\mathbb{E}}C_{k-1}^{\tau}\cdot\left(10^{3}e^{w_{2}}+10^{6}\cdot k_{n-1}\cdot e^{2v(\tau_{n-1})}+k_{n-1}\cdot e^{v(\tau_{n-1})/2}\right)\cdot\\ &&\displaystyle\hskip 86.11084pt{\mathbb{E}}S_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}\\[21.52771pt] &\leq&\displaystyle\sum_{{k_{1},\dots,k_{n-1}\in\mathbb{N}}\atop{{\tau_{1},\dots,\tau_{n-1}\in{\mathscr{T}}}\atop t_{1},\dots,t_{n-2}\in{\mathscr{L}}}}{\mathbb{E}}C_{k-1}^{\tau}\cdot 10^{7}\cdot k_{n-1}\cdot e^{2v(\tau_{n-1})}\cdot{\mathbb{E}}S_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}\\[21.52771pt] &=&\displaystyle 10^{7}\cdot{\mathbb{E}}C_{k-1}^{\tau}\cdot\sum_{\ell=1}^{\infty}\ell\cdot\sum_{\sigma\in{\mathscr{T}}}e^{2v(\sigma)}\cdot\sum_{{k_{1},\dots,k_{n-1}\in\mathbb{N}}\atop{{{\tau_{1},\dots,\tau_{n-1}\in{\mathscr{T}}}\atop t_{1},\dots,t_{n-2}\in{\mathscr{L}}}\atop k_{n-1}=\ell,\tau_{n-1}=\sigma}}{\mathbb{E}}S_{(k_{1},\dots,k_{n-1});(\tau_{1},\dots,\tau_{n-1});(t_{1},\dots,t_{n-2})}\\[21.52771pt] &=&\displaystyle{\mathbb{E}}C_{k-1}^{\tau}\cdot 10^{7}\cdot\sum_{\ell=1}^{\infty}\ell\cdot\sum_{\sigma\in{\mathscr{T}}}e^{2v(\sigma)}\cdot{\mathbb{E}}S_{n-1,\ell}^{\sigma}\\[21.52771pt] &=&\displaystyle{\mathbb{E}}C_{k-1}^{\tau}\cdot 10^{7}\cdot\sum_{\ell=1}^{\infty}\ell\cdot\left(e^{2w_{2}}\cdot{\mathbb{E}}S_{n-1,\ell}^{\text{\bf{II}}}+e^{2w_{2}}\cdot{\mathbb{E}}S_{n-1,\ell}^{\text{\bf{III}}}+\sum_{v\geq w_{2}}e^{2v+2}\cdot{\mathbb{E}}S_{n-1,\ell}^{\text{\bf{IV}},v}\right)\\[21.52771pt] &=&\displaystyle{\mathbb{E}}C_{k-1}^{\tau}\cdot\Sigma_{n},\end{array}

establishing the last inequality in the lemma statement.

The first inequality (the case when k=1k=1 and τ=II\tau=\text{\bf{II}}) follows analogously, replacing gCk1τg_{C_{k-1}^{\tau}} in (29), (33), (35) by 1{dist2(zm1,zm)<rw1}1_{\{\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-1},z_{m})<r-w_{1}\}} and applying Lemma 20 to the innermost integral in the analogues of (30).

The second inequality (the case when k=2k=2 and τ=III\tau=\text{\bf{III}}) follows in the same way, now using the indicator function that dist2(zm2,zm1),dist2(zm1,zm)<r+w2\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-2},z_{m-1}),\operatorname{dist}_{\mathbb{H}^{2}}(z_{m-1},z_{m})<r+w_{2} and zm2zm1zm<ϑ\angle z_{m-2}z_{m-1}z_{m}<\vartheta and applying Lemma 21.

For the case when k=1k=1 and τ=(IV,v)\tau=(\text{\bf{IV}},v) we replace gCk1τg_{C_{k-1}^{\tau}} by the indicator function corresponding to FIV-a,vF_{\text{IV-a},\geq v} from Lemma 53 in case tn1=at_{n-1}=a; by the indicator function corresponding to FIV-b,vF_{\text{IV-b},\geq v} from Lemma 54 in case tn1=bt_{n-1}=b; and the indicator function corresponding to FIV-c,vF_{\text{IV-c},\geq v} form Lemma 55 in case tn1=ct_{n-1}=c; and then apply Lemma 53, respectively Lemma 54, respectively Lemma 55 to the innermost integral in the analogues of (30). \blacksquare

Lemma 57

For every 0<ε<10<\varepsilon<1 and c>0c>0 there exist λ0,w1,ϑ>0\lambda_{0},w_{1},\vartheta>0 and w2>cw_{2}>c such that w2w_{2}\in\mathbb{N} and for all 0<λ<λ00<\lambda<\lambda_{0} and p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, setting r:=2ln(1/λ)r:=2\ln(1/\lambda), almost surely there are no infinite, black, linked sequences of chunks starting from the origin oo.

Proof. We will make use of the bounds provided by Lemma 46, Corollary 50 and Lemma 56. We will choose the integer w2>cw_{2}>c so large that 104ew2ew2/2<ε/610^{4}e^{w_{2}}e^{-w_{2}/2}<\varepsilon/6. So in particular, for λ\lambda sufficiently small, the expected number of good, black pseudopaths starting from the origin satisfies

𝔼Gk(1ε/3)k.{\mathbb{E}}G_{k}\leq(1-\varepsilon/3)^{k}.

Let Σn\Sigma_{n} be as in the statement of Lemma 56. By the observation (27) and Corollary 50, we have

Σ1=107=1(e2w2𝔼CII+e2w2𝔼CIII+vw2e2v𝔼CIV,v)107(103e2w2(ew1+ϑew2)+104v=w2e(5/2)veev/2)(=1(1ε/3)1)<.\begin{array}[]{rcl}\Sigma_{1}&=&\displaystyle 10^{7}\cdot\sum_{\ell=1}^{\infty}\ell\cdot\left(e^{2w_{2}}\cdot{\mathbb{E}}C_{\ell}^{\text{\bf{II}}}+e^{2w_{2}}\cdot{\mathbb{E}}C_{\ell}^{\text{\bf{III}}}+\sum_{v\geq w_{2}}e^{2v}\cdot{\mathbb{E}}C_{\ell}^{\text{\bf{IV}},v}\right)\\ &\leq&\displaystyle 10^{7}\cdot\left(10^{3}\cdot e^{2w_{2}}\cdot\left(e^{-w_{1}}+\vartheta e^{w_{2}}\right)+10^{4}\cdot\sum_{v=w_{2}}^{\infty}e^{(5/2)v}e^{-e^{v/2}}\right)\cdot\left(\sum_{\ell=1}^{\infty}\ell\cdot(1-\varepsilon/3)^{\ell-1}\right)\\ &<&\infty.\end{array}

The recursive relations given by Lemma 56 show that for all n1n\geq 1:

Σn+1Σn107(e2w2103ew1+2e2w2103ϑew2+106v=w2e3veev/2+=2e2w2𝔼C1II+=3e2w2𝔼C1III+=2e2v𝔼C1IV,v)Σn107(103e2w2w1+2103ϑe3w2+106v=w2e3veev/2+(103e2w2w1+103ϑe3w2+104v=w2e(5/2)veev/2)(=2(1ε/3)2))Σn1013(e2w2w1+ϑe3w2+v=w2e3veev/2)(1+=2(1ε/3)2)=Σn1013(e2w2w1+ϑe3w2+v=w2e3veev/2)(9+3ε+ε2ε2)Σn1015ε2(e2w2w1+ϑe3w2+v=w2e3veev/2)\begin{array}[]{rcl}\Sigma_{n+1}&\leq&\displaystyle\Sigma_{n}\cdot 10^{7}\cdot\left(e^{2w_{2}}\cdot 10^{3}\cdot e^{-w_{1}}+2e^{2w_{2}}\cdot 10^{3}\cdot\vartheta\cdot e^{w_{2}}+10^{6}\cdot\sum_{v=w_{2}}e^{3v}e^{-e^{v/2}}+\right.\\ &&\displaystyle\hskip 43.05542pt\left.\sum_{\ell=2}^{\infty}e^{2w_{2}}\cdot\ell\cdot{\mathbb{E}}C_{\ell-1}^{\text{\bf{II}}}+\sum_{\ell=3}^{\infty}e^{2w_{2}}\cdot\ell\cdot{\mathbb{E}}C_{\ell-1}^{\text{\bf{III}}}+\sum_{\ell=2}^{\infty}e^{2v}\cdot\ell\cdot{\mathbb{E}}C_{\ell-1}^{\text{\bf{IV}},v}\right)\\[8.61108pt] &\leq&\displaystyle\Sigma_{n}\cdot 10^{7}\cdot\left(10^{3}\cdot e^{2w_{2}-w_{1}}+2\cdot 10^{3}\cdot\vartheta\cdot e^{3w_{2}}+10^{6}\cdot\sum_{v=w_{2}}^{\infty}e^{3v}e^{-e^{v/2}}+\right.\\ &&\displaystyle\left.\left(10^{3}\cdot e^{2w_{2}-w_{1}}+10^{3}\cdot\vartheta\cdot e^{3w_{2}}+10^{4}\cdot\sum_{v=w_{2}}^{\infty}e^{(5/2)v}\cdot e^{-e^{v/2}}\right)\cdot\left(\sum_{\ell=2}^{\infty}\ell\cdot(1-\varepsilon/3)^{\ell-2}\right)\right)\\[8.61108pt] &\leq&\displaystyle\Sigma_{n}\cdot 10^{13}\cdot\left(e^{2w_{2}-w_{1}}+\vartheta\cdot e^{3w_{2}}+\sum_{v=w_{2}}^{\infty}e^{3v}e^{-e^{v/2}}\right)\cdot\left(1+\sum_{\ell=2}^{\infty}\ell\cdot(1-\varepsilon/3)^{\ell-2}\right)\\[8.61108pt] &=&\displaystyle\Sigma_{n}\cdot 10^{13}\cdot\left(e^{2w_{2}-w_{1}}+\vartheta\cdot e^{3w_{2}}+\sum_{v=w_{2}}^{\infty}e^{3v}e^{-e^{v/2}}\right)\cdot\left(\frac{9+3\varepsilon+\varepsilon^{2}}{\varepsilon^{2}}\right)\\[8.61108pt] &\leq&\displaystyle\Sigma_{n}\cdot 10^{15}\cdot\varepsilon^{-2}\cdot\left(e^{2w_{2}-w_{1}}+\vartheta\cdot e^{3w_{2}}+\sum_{v=w_{2}}^{\infty}e^{3v}e^{-e^{v/2}}\right)\end{array}

For the sake of the presentation, we introduce an additional small constant δ>0\delta>0. Without loss of generality we can assume chose w2>cw_{2}>c large enough so that

v=w2e3veev/2<δ.\sum_{v=w_{2}}^{\infty}e^{3v}\cdot e^{-e^{v/2}}<\delta.

We can also assume w1,ϑw_{1},\vartheta are such that e2w2w1,ϑe3w2<δe^{2w_{2}-w_{1}},\vartheta e^{3w_{2}}<\delta. Filling these bounds into the bound on Σn+1\Sigma_{n+1} gives

Σn+1Σn1015(3δε2)Σn(12),\Sigma_{n+1}\leq\Sigma_{n}\cdot 10^{15}\cdot\left(\frac{3\delta}{\varepsilon^{2}}\right)\leq\Sigma_{n}\cdot\left(\frac{1}{2}\right),

the last line holding because we have chosen δ\delta appropriately.

To recap, we can choose w1,w2,ϑw_{1},w_{2},\vartheta in such a way that Σn(12)n1Σ1\Sigma_{n}\leq\left(\frac{1}{2}\right)^{n-1}\cdot\Sigma_{1} for all nn; and Σ1<\Sigma_{1}<\infty. In particular Σn0\Sigma_{n}\to 0 as nn\to\infty.

Let us denote by SnS_{n} the total number of black, linked sequences of chunks starting from the origin and consisting of precisely nn chunks. Obviously

𝔼Sn=k=1𝔼Sn,kII+k=1𝔼Sn,kIII+k=1v=w2𝔼Sn,kIV,vΣn.{\mathbb{E}}S_{n}=\sum_{k=1}^{\infty}{\mathbb{E}}S_{n,k}^{\text{\bf{II}}}+\sum_{k=1}^{\infty}{\mathbb{E}}S_{n,k}^{\text{\bf{III}}}+\sum_{k=1}^{\infty}\sum_{v=w_{2}}^{\infty}{\mathbb{E}}S_{n,k}^{\text{\bf{IV}},v}\leq\Sigma_{n}.

But then we also have that

𝔼Snn0.{\mathbb{E}}S_{n}\xrightarrow[n\to\infty]{}0.

It follows that, almost surely, there are is no infinite, black, linked sequence of chunks starting from oo. \blacksquare

Combining Proposition 31, Corollary 47 and Lemma 57 immediately gives:

Corollary 58

For every ε>0\varepsilon>0 there is a λ0=λ0(ε)\lambda_{0}=\lambda_{0}(\varepsilon) such that for all 0<λ<λ00<\lambda<\lambda_{0} and all p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, almost surely, the black cluster of oo in the Voronoi tessellation for 𝒵{o}{\mathcal{Z}}\cup\{o\} is finite.

For completeness we point out how Proposition 28 follows from Corollary 58.

Proof of Proposition 28. This follows from Corollary 58 by a near verbatim repeat of the proof of Corollary 47. We only mention the changes that need to be made. Now, we let NxN_{x} be the number of z𝒵bB2(o,x)z\in{\mathcal{Z}}_{b}\cap B_{\mathbb{H}^{2}}(o,x) that are part of an infinite black component of the Voronoi tessellation, and we let gg be the indicator function that zB2(o,x)z\in B_{\mathbb{H}^{2}}(o,x) and that zz lies in an infinite, black cluster in the Voronoi tessellation for 𝒵{z}{\mathcal{Z}}\cup\{z\}. \blacksquare

4 Suggestions for further work

Our Theorem 1 states that pc(λ)=(π/3)λ+o(λ)p_{c}(\lambda)=(\pi/3)\lambda+o(\lambda) as λ0\lambda\searrow 0, answering a question of Benjamini and Schramm [9]. A natural direction for research is to try and find more terms in the expansion.

Problem 59

Determine a constant cc such that pc(λ)=(π/3)λ+cλ2+o(λ2)p_{c}(\lambda)=(\pi/3)\lambda+c\lambda^{2}+o(\lambda^{2}) as λ0\lambda\searrow 0, or show no such constant exists.

In our proofs we have either taken p(1ε)(π/3)λp\geq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, for the upper bound on pc(λ)p_{c}(\lambda), or p(1ε)(π/3)λp\leq(1-\varepsilon)\cdot(\pi/3)\cdot\lambda, for the lower bound. We have always taken ε\varepsilon constant in our paper and have not made any effort to see how fast we can send ε\varepsilon to zero as a function of λ\lambda before our proof technique breaks down. Of course many of our bounds are rather crude and it seems likely that more fine-grained proof techniques will need to be developed.

As mentioned in the introduction, the critical value for percolation in d\mathbb{Z}^{d} for high dimension dd is related, at least on an intuitive level, to our result. And, as also mentioned, a trivial comparison to branching processes shows the critical value for bond percolation on d\mathbb{Z}^{d} satisfies pc(d)12d1>12dp_{c}(\mathbb{Z}^{d})\geq\frac{1}{2d-1}>\frac{1}{2d}, the reciprocal of the degree. (It was in fact shown by Van der Hofstad and Slade [48] that pc(d)=12d+14d2+716d3+O(d4)p_{c}(\mathbb{Z}^{d})=\frac{1}{2d}+\frac{1}{4}d^{-2}+\frac{7}{16}d^{-3}+O(d^{-4}). So we even have pc(d)>12d1p_{c}(\mathbb{Z}^{d})>\frac{1}{2d-1} for large enough dd.) Inspired by this, and the fact that the typical degree is actually strictly larger that 3πλ\frac{3}{\pi\lambda}, we offer the following conjecture:

Conjecture 60

There exists a λ0>0\lambda_{0}>0 such that pc(λ)>(π/3)λp_{c}(\lambda)>(\pi/3)\cdot\lambda for all 0<λ<λ00<\lambda<\lambda_{0}.

Of course the answer to Problem 59 is likely to also tell us whether this conjecture holds or not. But, perhaps Conjecture 60 can be settled via a different route.

We reiterate a natural conjecture of Benjamini and Schramm.

Conjecture 61 ([9])

pc(λ)p_{c}(\lambda) is strictly increasing.

Another natural conjecture in the same vein is:

Conjecture 62

pc(λ)p_{c}(\lambda) is differentiable.

We were tempted to write “smooth” in place of “differentiable”, but opted against it to give whoever attempts to prove the conjecture the best chances.

Of course Poisson-Voronoi percolation can also be defined on dd-dimensional hyperbolic space d\mathbb{H}^{d}, and we expect that the main result of the current paper will generalize. In two dimensions, the typical degree is the same as the number of 11-faces of the typical cell. In dd-dimensions the relevant corresponding quantity is the number of (d1)(d-1)-faces of the typical cell.

Conjecture 63

For every dd, as the intensity λ0\lambda\searrow 0, the critical value for Poisson-Voronoi percolation on d\mathbb{H}^{d} is asymptotically equal to the reciprocal of the expected number of (d1)(d-1)-faces of the typical cell.

It might be possible to leverage some of the existing work on the expected ff-vectors of the typical cell in [13, 19, 26]. But, of course it might also be possible to prove or disprove the conjecture without knowing the expected number of (d1)(d-1)-faces precisely.

In a recent separate paper, we proved the conjecture of Benjamini and Schramm that pc(λ)1/2p_{c}(\lambda)\to 1/2 as λ\lambda\to\infty for planar, hyperbolic Poisson-Voronoi percolation. It seems natural to expect that this result generalizes to arbitrary dimensions.

Conjecture 64

For every dd, as the intensity λ\lambda\to\infty, the critical value for Poisson-Voronoi percolation on d\mathbb{H}^{d} tends to the critical value for Poisson-Voronoi percolation on d\mathbb{R}^{d}.

A complicating issue here is that for Poisson-Voronoi percolation on d\mathbb{R}^{d} there is a large gap between the best known lower and upper bounds [3, 4]. But, again, it may be possible to prove the conjecture without first determining the precise critical value in the Euclidean case.

Acknowledgements

We thank the anonymous referee for many helpful comments that have greatly improved the paper.

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