aa r X i v : . [ m a t h . P R ] M a r Poisson cylinders in hyperbolic space
Erik I. Broman ∗ and Johan Tykesson † April 5, 2015
Abstract
We consider the Poisson cylinder model in d -dimensional hyperbolic space. Weshow that in contrast to the Euclidean case, there is a phase transition in theconnectivity of the collection of cylinders as the intensity parameter varies. We alsoshow that for any non-trivial intensity, the diameter of the collection of cylindersis infinite. In the recent paper [6], the authors considered the so-called Poisson cylinder model inEuclidean space. Informally, this model can be described as a Poisson process ω onthe space of bi-infinite lines in R d . The intensity of this Poisson process is u times anormalized Haar measure on this space of lines. One then places a cylinder c of radiusone around every line L ∈ ω, and with a slight abuse of notation, we say that c ∈ ω. Themain result of [6] was that for any 0 < u < ∞ and any two cylinders c , c ∈ ω, thereexists a sequence c , . . . c d − ∈ ω such that c ∩ c = ∅ , c ∩ c = ∅ , . . . c d − ∩ c = ∅ . Inwords, any two cylinders in the process is connected via a sequence of at most d − d − < u < ∞ , and thereforethere is no connectivity phase transition.This is in sharp contrast to what happens for other percolation models. For example,ordinary discrete percolation (see [8]), the Gilbert disc model (see [11]), and the Voronoipercolation model (see [5]) all have a connectivity phase transition. A common propertythat all the above listed models exhibit is something that we informally refer to as a ”lo-cality property” and can be described as follows. Having knowledge of the configurationin some region A, gives no, or almost no, information about the configuration in someother region B, as long as A and B are well separated. For instance, in ordinary discretepercolation the configurations are independent if the two regions A, B are disjoint, whilefor the Gilbert disc model with fixed disc radius r, the regions need to be at Euclideandistance at least 2 r in order to have independence. For Voronoi percolation, there is a ∗ Department of Mathematics, Uppsala University, Sweden. Research supported by the Swedish Re-search Council. E-mail: [email protected] † Department of Mathematics, Chalmers University of Technology and Gothenburg University, Swe-den. Research supported by the Knut and Alice Wallenberg foundation E-mail: [email protected] x and y decays exponentially in the distancebetween x and y. This is however not the case when dealing with the Poisson cylinder model in Eu-clidean space. Here, the dependency is polynomially decaying in that P E [ B ( x, ↔ B ( y, ∼ d E ( x, y ) − ( d − , (1.1)where the index E stresses that we are in the Euclidean case, and where ↔ denotes theexistence of a cylinder c ∈ ω connecting B ( x,
1) to B ( y, P H [ B ( x, ↔ B ( y, ∼ e − ( d − d H ( x,y ) , (1.2)(where P H stresses that we are in the hyperbolic case and d H denotes hyperbolic distance).Since the decay is now exponential, this is a form of locality property. Thus, by studyingthis model in hyperbolic space, we can study a model with unbounded percolation objects,but with a locality property. This is something that does not occur naturally in theEuclidean setting.Before we can present our main results, we will provide a short explanation of ourmodel, see Section 2 below for further details. Consider therefore the d -dimensionalhyperbolic space H d for any d ≥
2. We let A ( d,
1) be the set of geodesics in H d and let µ d, be the unique (up to scaling) measure on A ( d,
1) which is invariant under isometries.We will sometimes simply refer to the geodesics of A ( d,
1) as lines.Let ω be a Poisson point process on A ( d,
1) with intensity uµ d, , where u > L ∈ ω, we will let c ( L ) denote thecorresponding cylinder, and abuse notation somewhat in writing c ∈ ω. Let C := [ L ∈ ω c ( L ) , be the occupied set and let V := H d \ C be the vacant set. Furthermore, define u c = u c ( d ) := inf { u : C is a.s. a connected set } . We note that by Proposition 2.1 below, we have that P [ C is connected] ∈ { , } . Our main result is the following.
Theorem 1.1.
For any d ≥ we have u c ( d ) ∈ (0 , ∞ ) . Furthermore, for any u > u c , C is connected. Remarks:
Theorem 1.1 indicates that even though the cylinders are unbounded, theexponential decay of (1.2) seems to be the important feature in determining the existenceof a phase transition. The second part of the theorem is a monotonicity property, proving2hat when u is so large that C is a connected set, then we cannot have that C is againdisconnected for an even larger u. In [16] a result similar to Theorem 1.1 for the random interlacements model on certainnon-amenable graphs was proven. The random interlacements model (which was intro-duced in [15]) is a discrete percolation model exhibiting long-range dependence. However,the dependence structure for this model is very different from that of the Poisson cylindermodel. To see this, consider three points x, y, z ∈ H d (or R d in the Euclidean case). Ifwe know that there is a geodesic L ∈ ω such that x, y ∈ L, then this will determinewhether z ∈ L . For a random interlacement process, the objects studied are essentiallytrajectories of bi-infinite simple random walks, and so knowing that a trajectory containsthe points x, y ∈ Z d will give some information whether the trajectory contains z ∈ Z d , but not ”full” information. Thus, the dependence structure is in some sense more rigidfor the cylinder process.Knowing that C is connected, it is natural to consider the diameter of C defined asfollows. For any two cylinders c a , c b ∈ ω, let Cdist( c a , c b ) be the minimal number k ofcylinders c , . . . c k ∈ ω such that c a ∪ c b k [ i =1 c i is a connected set. If no such set exists, we say that Cdist( c a , c b ) = ∞ . We then definethe diameter of C as diam( C ) = sup { Cdist( c a , c b ) : c a , c b ∈ ω } . Our second main result is
Theorem 1.2.
For any u ∈ (0 , ∞ ) , we have that P [ diam ( C ) = ∞ ] = 1 . Remark:
Of course, the result is trivial for u < u c . When 0 < u < u c ( d ) , it is natural to ask about the number of unbounded components.Our next proposition addresses this. Proposition 1.3.
For any u ∈ (0 , u c ) the number of infinite connected components of C is a.s. infinite. One of the main tools will be the following discrete time particle process. Since webelieve that it may be of some independent interest, we present it here in the introduction,along with our main result concerning it. In essence, it behaves like a branching processwhere every particle gives rise to an infinite number of offspring whose types can takeany positive real value.Formally, let ξ , ( ξ k,n ) ∞ k,n =1 be an i.i.d. collection of Poisson processes on R withintensity measure ue min(0 ,x ) dx. Let ζ = { } , and we think of this as the single particlein generation 0. Then, let ζ = { x ≥ x ∈ ξ } be the particles of generation 1, and let Z , = min { x ∈ ζ } and inductively for any k ≥ , let Z k, = min { x ∈ ζ : x > Z k − , } .Thus Z , < Z , < · · · and { Z , , Z , , . . . } = ζ . We think of these as the offspring of3 , = { } . In general, if ζ n has been defined, and Z ,n < Z ,n < · · · are the points in ζ n , we let ζ k,n +1 = [ x ∈ ξ k,n : x + Z k,n ≥ { x + Z k,n } , (1.3)and ζ n +1 = S ∞ k =1 ζ k,n +1 . We think of ζ n +1 as the particles of generation n + 1 , and ζ k,n +1 as the offspring of Z k,n ∈ ζ n . From (1.3), we see that ζ k,n +1 ⊂ R + . Furthermore,conditioned on Z k,n = x, Z k,n gives rise to new particles in generation n + 1 accordingto a Poisson process with intensity measure dµ x = I ( y ≥ ue − ( x − y ) + dy (where I is anindicator function and ( x − y ) + = max(0 , x − y )). We let ζ = ( ζ n ) ∞ n =1 denote this particleprocess. We point out that in our definition, any enumeration of the particles of ζ n wouldbe as good as our ordering Z ,n < Z ,n < · · · , as long as the enumeration does not dependon ”the future”, i.e. ( ξ k,n +1 ) ∞ k =1 or such.Informally the above process can be described as follows. Thinking of a particle as apoint in R + corresponding to the type of that particle, it gives rise to new points with ahomogeneous rate forward of the position of the point, but at an exponentially decayingrate backward of the position of the point. Of course, since any individual gives rise toan infinite number of offspring, the process will never die out. However, it can still dieout weakly in the sense that for any R there will eventually be no new points of type R or smaller.For any n, let X n [ a,b ] = ∞ X k =1 I ( Z k,n ∈ [ a, b ]) . (1.4)Thus, X n [ a,b ] is the number of individuals in generation n of type between a and b. Wehave the following theorem
Theorem 1.4.
There exists a constant
C < ∞ such that for u < / , and any R < ∞ ∞ X n =1 E [ X n [0 ,R ] ] < C e uR − u < ∞ . That is, ζ dies out weakly. Furthermore, for any u > / , lim n →∞ E [ X n [0 ,R ] ] = ∞ . Theorem 1.4 will be used to prove that u c ( d ) > c passing through the origin o ∈ H d and a Poisson process of cylinders in H d as described above. Let c , , c , , . . . be the setof cylinders in this process that intersect c . These are the first generation of cylinders(and correspond to ζ ) . In the next step, we consider independent Poisson processes( ω k, ) ∞ k =1 and the collection of cylinders in ω k, that intersect c k, (these collections willcorrespond to ( ζ k, ) ∞ k =1 and the union of them corresponds to ζ ) . We then proceed forfuture generations in the obvious way. By a straightforward coupling of this ”independentcylinder process” and the original one described above (and since in every step we use anindependent process in the entire space H d ), we get that the set of cylinders connected to c through this procedure, will contain the set of cylinders in C ∪ c connected to c . With4ome work, the independent cylinder process can be compared to the particle process asindicated. By Theorem 1.4, for u < / , the latter dies out weakly. We will show thatthis implies that the number of cylinders (in the independent cylinder process) connectedto c and intersecting B ( o, R ) will be of order at most e ucR where c < ∞ . However, thenumber of cylinders in C intersecting B ( o, R ) must be of order e ( d − R , which of courseis strictly larger than e ucR for u > C is connected thenleads to a contradiction.We end the introduction with an outline of the rest of the paper. In Section 2 wegive some background on hyperbolic geometry and define the cylinder model. In Section3, we establish some preliminary results on connectivity probabilities that will be usefulin later sections. In Section 4, we prove that u c ( d ) < ∞ and the monotonicity part ofTheorem 1.1. In Section 5, we prove Theorem 1.4, which (as described) will be a keyingredient in proving u c ( d ) > , which is done in Section 6. In Sections 7 and 8 we proveTheorem 1.2 and Proposition 1.3 respectively. In this section we will start with some preliminaries of hyperbolic space which we willhave use for later, and proceed by defining the model. d -dimensional hyperbolic space There are many models for d -dimensional hyperbolic space (see for instance [2],[12] or[13]). In this paper, we prefer to consider the so-called Poincar´e ball model. Therefore,we consider the unit ball U d = { x ∈ R d : d E ( o, x ) < } (where d E denotes Euclideandistance) equipped with the hyperbolic metric d H ( x, y ) given by d H ( x, y ) = cosh − (cid:18) d E ( x, y ) (1 − d E ( o, x ) )(1 − d E ( o, y ) ) (cid:19) . (2.1)We refer to U d equipped with the metric d H as the Poincar´e ball model of d -dimensionalhyperbolic space, and denote it by H d .For future convenience, we now state two well known (see for instance Chapter 7.12 of[2]) rules from hyperbolic geometry. Here, we consider a triangle (consisting of segmentsof geodesics in H d ) with side lengths a, b, c and we let α, β, γ denote the angles oppositeof the segments corresponding to a, b and c respectively. Rule 1: cosh( c ) = cosh( a ) cosh( b ) − sinh( a ) sinh( b ) cos( γ ) (2.2) Rule 2: cosh( c ) = cos( α ) cos( β ) + cos( γ )sin( α ) sin( β ) (2.3)These rules are usually referred to as hyperbolic cosine rules.Let S d − denote the unit sphere in R d . We will identify ∂ H d with S d − . Any point x ∈ H d is then uniquely determined by the distance ρ = d H ( o, x ) of x from the origin o, and a point s ∈ S d − by going along the geodesic from o to s a distance ρ from o. If we let5 ν d − denote the solid angle element so that O d − = R S d − dν d − is the ( d − S d − , then the volume measure in H d can be expressed in hyperbolicspherical coordinates (see [13], Chapter 17) as dv d = sinh d − ( ρ ) dρdν d − . Thus, for any A ⊂ H d , the volume v d ( A ) can be written as v d ( A ) = Z A sinh d − ( ρ ) dρdν d − . (2.4) H d . Let A ( d,
1) be the set of all geodesics in H d . As mentioned in the introduction, a geodesic L ∈ A ( d,
1) will sometimes be referred to as a line. Although it will have no directrelevance to the paper, we note that it is well known (see [7], section 9) that in the Poincar´eball model, A ( d,
1) consists of diameters and boundary orthogonal circular segments ofthe unit ball U d . For any K ⊂ H d , we let L K := { L ∈ A ( d,
1) : L ∩ K = ∅} . If g is an isometry on H d (i.e. g is a M¨obius transform leaving U d invariant, see for instance [1] Chapters 2and 3), we define g L K := { gL : L ∈ A ( d, } (where of course gL = { gx : x ∈ L } ).There exists a unique measure µ d, on A ( d,
1) which is invariant under isometries (i.e. µ d, ( g L K ) = µ d, ( L K )), and normalized such that µ d, ( L B ( o, ) = O d − (see [13] Chapter17 or [3] Section 6).For any L ∈ A ( d,
1) we let a = a ( L ) be the point on L minimizing the distance to theorigin, and define ρ = ρ ( L ) = d H ( o, a ) . Note that ρ = d H ( o, L ) . Let L + K := { L ∈ L K : a ( L ) ∈ K } . According to (17.52) of [13], we have that µ d, ( L B ( o,r ) ) = µ d, ( L + B ( o,r ) ) (2.5)= ( d − O d − sinh d − (1) Z r cosh( ρ ) sinh d − ( ρ ) dρ = O d − sinh d − (1) sinh d − ( r ) . We consider the following space of point measures on A ( d, { ω = ∞ X i =0 δ L i where L i ∈ A ( d, ω ( L A ) < ∞ for all compact A ⊂ H d } . Here, δ L of course denotes Dirac’s point measure at L. We will often use the following standard abuse of notation: if ω is some point measure,then we will write ” L ∈ ω ” instead of ” L ∈ supp( ω )”. We will draw an element ω from Ωaccording to a Poisson point process with intensity measure uµ d, where u >
0. We call ω a (homogeneous) Poisson line process of intensity u in H d . If L ∈ A ( d, c ( L, s ) the cylinder of base radius s centered around L ,i.e. c ( L, s ) = { x ∈ H d : d H ( x, L ) ≤ s } . s = 1 we will simplify the notation and write c ( L,
1) = c ( L ). When convenient, we willwrite c ∈ ω instead of c ( L ) where L ∈ ω. Recall that the union of all cylinders is denotedby C , C = C ( ω ) = [ L ∈ ω c ( L ) , and that the vacant set V is the complement H d \ C . For an isometry g on H d and anevent B ⊂ Ω , we define gB := { ω ′ ∈ Ω : ω ′ = gω for some ω ∈ B } . We say that an event B ⊂ Ω, is invariant under isometries if gB = B for every isometry g. Furthermore, wehave the following 0 − Proposition 2.1.
Suppose that B is invariant under isometries. Then P [ B ] ∈ { , } . The proof of Proposition 2.1 is fairly standard, so we only give a sketch based on theproofs of Lemma 3 . ω B ( x,k ) denotes the restrictionof ω to L B ( x,k ) . Sketch of proof.
Let { z k } k ≥ ⊂ H d be such that for every k ≥ , d H ( o, z k ) = e k , andlet g k be an isometry mapping o to z k . Define I x,k = I ( ω ∈ { P [ B | ω B ( x,k ) ] > / } ), andnote that by L´evy’s 0-1 law, lim k →∞ I o,k = I B a.s.Using that B is invariant under isometries, it is straightforward to prove that the lawsof ( I B , I o,k ) and ( I B , I g k ( o ) ,k ) are the same, and so I g k ( o ) ,k converges in probability to I B .Thus, lim k →∞ P [ I o,k = I g k ( o ) ,k = 1 B ] = 1 . (2.6)The next step is to prove that I o,k and I g k ( o ) ,k are asymptotically independent, i.e.lim k →∞ | P [ I o,k = 1 , I g k ( o ) ,k = 1] − P [ I o,k = 1] P [ I g k ( o ) ,k = 1] | = 0 . (2.7)Essentially, (2.7) follows from the fact that when k is large, the probability that there isany cylinder in ω which intersects both B ( o, k ) and B ( g k ( o ) , k ) is very small (this is whywe choose d H ( o, z k ) to grow rapidly). For this, one uses the estimate of the measure oflines intersecting two distant balls, see Lemma 3.3 below.Since I o,k and I g k ( o ) ,k are asymptotically independent, we getlim k →∞ P [ I o,k = 1 , I g k ( o ) ,k = 0] = P [ B ](1 − P [ B ]) . (2.8)The only way both (2.6) and (2.8) can hold is if P [ B ] ∈ { , } .We note that the laws of the random objects ω , C and V are all invariant underisometries of H d . The purpose of this section is to establish some preliminary estimates on connectivityprobabilities, and in particular to establish (1.2). This result will then be used manytimes in the following sections. 7or any two sets
A, B ⊂ H d , we let A ↔ B denote the event that there exists acylinder c ∈ ω such that A ∩ c = ∅ and B ∩ c = ∅ . We have the following key estimate.
Lemma 3.1.
Let s ∈ (0 , ∞ ) . There exists two constants < c ( s ) < C ( s ) < ∞ such thatfor any x, y ∈ H d , and u ≤ /µ d, ( L B ( o,s +1) ) we have that c ( s ) u e − ( d − d H ( x,y ) ≤ P [ B ( x, s ) ↔ B ( y, s )] ≤ C ( s ) u e − ( d − d H ( x,y ) . Lemma 3.1 will follow easily from Lemmas 3.2 and 3.3 below, and we defer the proof ofLemma 3.1 till later.Recall that we identify S d − with ∂ H d in the Poincar´e ball model. Fix a half-line L / emanating from the origin. For 0 < θ < π, let L L / ,θ be the set of all half-lines L ′ / suchthat L ′ / emanates from the origin and such that the angle between L / and L ′ / is atmost θ . Let S θ ( L / ) be the set of all points s ∈ ∂ H d such that s is the limit point ofsome half-line in L L / ,θ . Then S θ ( L / ) is the intersection of ∂ H d with a hypersphericalcap of Euclidean height h = h ( θ ), where h ( θ ) = 1 − cos( θ ) . (3.1)The ( d − S θ is given by A ( θ ) = O d − I h − h (cid:18) d − , (cid:19) (3.2)where O d − (as above) is the ( d − S d − , and I h − h is a regularized incomplete beta function (this follows from [10], equation (1), by notingthat sin ( θ ) = 2 h − h ). Lemma 3.2.
There are constants < c < C < ∞ such that for any θ ≤ / , we have c θ d − ≤ A ( θ ) ≤ C θ d − . Proof.
First observe that if 0 ≤ θ ≤ /
10, then 1 − θ / ≤ cos( θ ) ≤ − θ / θ ≤ h ≤ θ ≤ θ ∈ [0 , / . (3.3)We have I h − h (cid:18) d − , (cid:19) = R h − h t ( d − / − (1 − t ) / − dt R t ( d − / − (1 − t ) / − dt . (3.4)The denominator in (3.4) is a dimension-dependent constant. Furthermore, if 0 ≤ h ≤ /
8, then 0 ≤ h − h ≤ /
4, and if 0 ≤ t ≤ /
4, then 1 ≤ / √ − t ≤
2. Hence, for h ≤ / C Z h − h t ( d − / − dt ≤ I h − h (cid:18) d − , (cid:19) ≤ C Z h − h t ( d − / − dt C (2 h − h ) d − ≤ I h − h (cid:18) d − , (cid:19) ≤ C (2 h − h ) d − . Hence, for h ≤ / C h ( d − / ≤ I h − h (cid:18) d − , (cid:19) ≤ C (2 h ) ( d − / . (3.5)The lemma now follows from (3.2), (3.3) and (3.5). Lemma 3.3.
Let s ∈ (0 , ∞ ) . There exists two constants < c ( s ) < C ( s ) < ∞ such thatfor any x, y ∈ H d , we have that c ( s ) e − ( d − d H ( x,y ) ≤ µ d, ( L B ( x,s ) ∩ L B ( y,s ) ) ≤ C ( s ) e − ( d − d H ( x,y ) . Proof.
For convenience, we perform the proof in the case s = 1. The general case isdealt with in the same way. The proof is somewhat similar to the proof of Lemma 3 . ∂ H d is identifiedwith S d − . Let R = d H ( x, y ) and without loss of generality assume that x = o and so y ∈ ∂B ( o, R ) . We can assume that
R > R ≤ c, C . For any R ∈ (0 , ∞ ] and A ⊂ ∂B ( o, R ), let τ R ( A ) := µ d, ( L B ( o, ∩ L A ) . The projection Π ∂ H d ( A ) of A onto ∂ H d is defined as the set of all points y in ∂ H d forwhich there is a half-line emanating from o , passing through A and with its end-point atinfinity at y .We now argue that µ d, ( L B ( o, ) σ R ( A ) ≤ τ R ( A ) ≤ µ d, ( L B ( o, ) σ R ( A ) , (3.6)where σ R is the unique rotationally invariant probability measure on ∂B ( o, R ). Here, σ ∞ is the rotationally invariant probability measure on ∂ H d , which is just a constant multipleof the Lebesgue measure on S d − . For A ⊂ ∂B ( o, R ), let N A ( ω ) denote the number ofpoints in A that are intersected by some line in L B ( o, ∩ ω . If L ∈ L B ( o, ∩ L A then L intersects A at one or two points. Hence N A ( ω ) / ≤ ω ( L B ( o, ∩ L A ) ≤ N A ( ω ) . (3.7)In addition, every line intersecting B ( o,
1) intersects ∂B ( o, R ) exactly twice. Hence, N ∂B ( o,R ) ( ω ) = 2 ω ( L B ( o, ) . (3.8)For A ⊂ ∂B ( o, R ) define ρ R ( A ) = E [ N A ( ω )]. Taking expectations in (3.7) we obtain ρ R ( A ) / ≤ u τ R ( A ) ≤ ρ R ( A ) . (3.9)9t is easily verified that ρ R ( A ) is invariant under rotations. Hence, ρ R is a constantmultiple of σ R . Taking expectations in (3.8), we obtain ρ R ( ∂B ( o, R )) = 2 uµ d, ( L B ( o, ) , from which it follows that ρ R ( · ) = 2 uµ d, ( L B ( o, ) σ R ( · ) . (3.10)Combining (3.9) and (3.10) we obtain (3.6). Since σ R ( A ) = σ ∞ (Π ∂ H d ( A )), this gives µ d, ( L B ( o, ) σ ∞ (Π ∂ H d ( A )) ≤ τ R ( A ) ≤ µ d, ( L B ( o, ) σ ∞ (Π ∂ H d ( A )) . (3.11)Having proved (3.6) and (3.11), we now proceed to prove the lower bound. We observethat L B ( o, ∩ L B ( y, ⊃ L B ( o, ∩ L B ( y, ∩ ∂B ( o,R ) . Hence, in view of (3.6), we need to estimate σ R ( E ) from below, where E = B ( y, ∩ ∂B ( o, R ). Let L be any line containing o and intersecting ∂B ( y, ∩ ∂B ( o, R ), and let L y be the line intersecting o and y. Denote the angle between L and L y by θ = θ ( R ).Observe that Π ∂ H d ( E ) is the intersection of ∂ H d and a hyperspherical cap of Euclideanheight 1 − cos( θ ), and so we need to find bounds on θ .Applying (2.2) to the triangle defined by L ∩ B ( o, R ), the line segment between o and y , and the line segment between L ∩ ∂B ( o, R ) and y , we havecosh(1) = cosh ( R ) − sinh ( R ) cos( θ ) . (3.12)Solving (3.12) for θ gives θ = arccos (cid:18) − (cid:18) cosh(1) − ( R ) (cid:19)(cid:19) . Observe that for any 0 ≤ x ≤ , arccos(1 − x ) = arcsin( √ x − x ) ≥ arcsin( √ x ) ≥ √ x. Hence for R ≥ θ ≥ C sinh( R ) ≥ Ce − R . (3.13)By Lemma 3.2 , we have σ ∞ (Π ∂ H d ( E )) ≥ c θ d − , (3.14)and so the lower bound follows by combining (3.11), (3.13) and (3.14).We turn to the upper bound. Let y ′ be the point on ∂B ( y,
1) closest to the origin,and let H be the ( d − L y and containing y ′ . Let Π ∂ H d ( H ) ⊂ ∂ H d be the projection of H onto ∂ H d . Since for any z ∈ H, d H ( y, z ) ≥ d H ( y, y ′ ) we get that L B ( o, ∩ L B ( y, ⊂ L B ( o, ∩ L Π ∂ H d ( H ) . Next we find an upper bound of σ ∞ (Π ∂ H d ( H )), which will imply the upper bound of µ d, ( L B ( o, ∩ L B ( y, ). Let L be any geodesic in H, and let s and s ′ be the two end-points10t infinity of L . Let L be the half-line between 0 and s , and let γ = γ ( R ) be the anglebetween L and L y . Applying (2.3) to the triangle defined by L , the half-line between s and y ′ and the line-segment between 0 and y ′ , we obtaincos(0) = − cos( π/
2) cos( γ ) + sin( π/
2) sin( γ ) cosh( R − , which gives 1 = sin( γ ) cosh( R − . Observe that we here applied (2.3) to an infinite triangle, which can be justified by alimit argument. Hence γ = arcsin (cid:18) R − (cid:19) . Observe that arcsin( x ) ≤ x for every 0 ≤ x ≤ , so that γ ≤ R − ≤ Ce − R . (3.15)We observe that Π ∂ H d ( H ) is the intersection between a hyperspherical cap of Euclideanheight 1 − cos( γ ) and ∂ H d . Hence, according to Lemma 3.2, σ ∞ (Π ∂ H d ( H )) ≤ cγ d − . (3.16)The upper bound follows by combining (3.11), (3.15) and (3.16), which concludes theproof. Lemma 3.4.
Suppose d H ( x, y ) = R and that r, s ∈ (0 , ∞ ) . There is a constant c ( d, s ) < ∞ such that if R > r + s , then µ d, ( L B ( x,s ) ∩ L B ( y,r ) ) ≤ c ( d, s ) exp( − ( d − R − r )) . Proof.
The proof is nearly identical to the proof of the upper bound in Lemma 3.3,and therefore we leave the details to the reader.We can now prove Lemma 3.1.
Proof of Lemma 3.1.
We perform the proof in the case s = 1 as the general casefollows similarly. First observe that { B ( x, ↔ B ( y, } = { ω ( L B ( x, ∩ L B ( y, ) ≥ } . Using that 1 − e − x ≤ x for x ≥ , we have that P [ B ( x, ↔ B ( y, − P [ B ( x, B ( y, − e − uµ d, ( L B ( x, ∩L B ( y, ) ≤ uµ d, ( L B ( x, ∩ L B ( y, ) ≤ Cue − ( d − d H ( x,y ) by Lemma 3.3 with C as in the same lemma.Using that 1 − e − x ≥ x/ x ≤ , and that uµ d, ( L B ( x, ∩ L B ( y, ) ≤ uµ d, ( L B ( x, ) = uµ d, ( L B ( o, ) ≤ P [ B ( x, ↔ B ( y, ≥ uµ d, ( L B ( x, ∩ L B ( y, )2 ≥ cue − ( d − d H ( x,y ) by again using Lemma 3.3 and letting c be half of that of Lemma 3.3.11 Proof of u c < ∞ and monotonicity of uniqueness We start by proving the monotonicity of uniqueness. For convenience, in this section wedenote by ω u a Poisson line process with intensity u . In addition, we will let E and P denote expectation and probability measure for several Poisson processes simultaneously.Recall also that A ( d,
1) is the set of all geodesics in H d . Lemma 4.1.
If for u > P [ C ( ω u ) is connected ] = 1 , then P [ C ( ω u ) is connected ] = 1 for every u > u . Proof.
It is straightforward to show that for any L ∈ A ( d, µ d, ( L c ( L ) ) = ∞ . Hence,for any L ∈ A ( d, , P [ c ( L ) ∩ C ( ω u ) = ∅ ] = 1 . (4.1)Let u ′ = u − u and let ω u ′ be a Poisson line process of intensity u ′ , independent of ω u . Bythe Poissonian nature of the process, C ( ω u ) has the same law as C ( ω u ) ∪ C ( ω u ′ ). Henceit suffices to show that the a.s. connectedness of C ( ω u ) implies the a.s. connectednessof C ( ω u ) ∪ C ( ω u ′ ). To show this, it suffices to show that a.s., every line in ω u ′ intersects C ( ω u ). To this end, for L ∈ A ( d, S ( L ) = { c ( L ) ∩ C ( ω u ) = ∅} . Thenlet D := ∩ L ∈ ω u ′ S ( L ) . We will show that P [ D c ] = 0 and we start by observing that P [ D c ] = P (cid:2) ∪ L ∈ ω u ′ S ( L ) c (cid:3) ≤ E X L ∈ ω u ′ I ( S ( L ) c ) . For clarity, we let E ω u ′ and E ω u denote expectation with respect to the processes ω u ′ and ω u respectively, and we will let E denote expectation with respect to ω u ′ ∪ ω u . We usesimilar notation for probability. We then have that, E X L ∈ ω u ′ I ( S ( L ) c ) = E ω u ′ E ω u X L ∈ ω u ′ I ( S ( L ) c ) (cid:12)(cid:12)(cid:12)(cid:12) ω u ′ = E ω u ′ X L ∈ ω u ′ E ω u [ I ( S ( L ) c ) | ω u ′ ] = E ω u ′ X L ∈ ω u ′ P ω u [ S ( L ) c | ω u ′ ] = E ω u ′ X L ∈ ω u ′ P ω u [ S ( L ) c ] = 0 , where we use the independence between ω u ′ and ω u in the penultimate equality and that P [ S ( L ) c ] = 0 which follows from (4.1). This finishes the proof of the proposition.The aim of the rest of this section is to prove the following proposition, which is apart of Theorem 1.1 Proposition 4.2.
For any d ≥ , u c ( d ) < ∞ .
12n order to prove Proposition 4.2, we will need some preliminary results and termi-nology. Recall the definition of L + A for A ⊂ H d and the definitions of a ( L ) and ρ ( L ) , allfrom Section 2.2. Using the line process ω , we define a point process τ in H d as follows: τ = τ ( ω ) := X L ∈ ω δ a ( L ) . In other words, τ is the point process induced by the points that minimize the distancebetween the origin and the lines of ω . We observe that since ω is a Poisson process,it follows that τ is also a Poisson process (albeit inhomogeneous). We will considera percolation model with balls in place of cylinders, using τ as the underlying pointprocess. Our aim is to prove that V does not percolate for u < ∞ large enough byanalyzing this latter model. For this, we will need Lemma 4.4, which provides a uniformbound (in z ∈ H d ) of the probability that a point of τ falls in the ball of radius 1 / z ∈ H d . Before that, we present the following lemma, which will be useful onseveral occasions.
Lemma 4.3.
There exists a set D of points in H d with the following properties:1. d H ( z, D ) ≤ / for all z ∈ H d .2. If x, y ∈ D and x = y , then d H ( x, y ) ≥ / .Furthermore, for any such set, there exist constants < c ( d ) < c ( d ) < ∞ so that forany x ∈ H d , and r ≥ ,c ( d ) v d ( B ( o, r )) ≤ | D ∩ B ( x, r ) | ≤ c ( d ) v d ( B ( o, r + 1)) . (4.2) Proof.
We give an explicit construction of the set D . First let D = { o } and E = { x ∈ H d : d H ( o, x ) = 1 / } , and define D = D ∪ { x } where x is any point in E . Inductively, having defined D n , we let E n = { x ∈ H d : d H ( D n , x ) = 1 / } and define D n +1 = D n ∪ { x n } where x n is any point in E n such that d H ( o, x n ) = d H ( o, E n ) whichexists by compactness of the set E n . Finally we let D = ∪ ∞ n =1 D n . By construction, anytwo points in D will then satisfy condition 2. Assume now that there exists a point z ∈ H d such that d H ( z, D ) > / , and let m be any integer such that d H ( o, x m ) ≥ d H ( o, z ) . Since d H ( z, D m ) ≥ d H ( z, D ) > / d H z, [ x ∈ D m B ( x, / ! > . (4.3)Let S z be the line segment from o to z, and observe that since o ∈ S x ∈ D m B ( x, / , theremust be some point s = s ( E m , z ) belonging to S z ∩ E m . Because of (4.3), we see that forsome ǫ > , we have that d H ( o, z ) = d H ( o, s ) + ǫ and so we get that d H ( o, z ) = d H ( o, s ) + ǫ ≥ d H ( o, E m ) + ǫ = d H ( o, x m ) + ǫ > d H ( o, x m ) , leading to a contradiction.We now turn to (4.2) and start with the upper bound. Let y , . . . , y N be an enu-meration of D ∩ B ( x, r ). By construction, the balls B ( y k , /
5) are all disjoint, and13o
N v d ( B ( o, / ≤ v d ( B ( o, r + 1)) from which the upper bound follows with c =1 /v d ( B ( o, / . For the lower bound, it suffices to observe that from the construction we have that B ( o, r ) ⊂ N [ k =1 B ( y k , , so that N ≥ v d ( B ( o, r )) /v d ( B ( o, c = 1 /v d ( B ( o, . Lemma 4.4.
There is a constant c ( d ) > such that for any z ∈ H d , µ d, ( L + B ( z, / ) ≥ c. Proof.
We first claim that there is a constant c = c ( d ) ∈ (0 , ∞ ) such that for any r ≥ B ( o, r + 1 / \ B ( o, ( r − / + ) can be covered by at most N r = ⌈ c e ( d − r ⌉ ballsof radius 1 / ∂B ( o, r ). For this, we observe that by modifying the proof ofLemma 4.3, we can obtain a set of points E ⊂ H d with the properties that d ( x, E ) ≤ / x ∈ H d and | E ∩ B ( o, r + 1 / | ≤ cν d ( B ( o, r + 3 / c < ∞ andall r ≥
1. Let E r = E ∩ ( B ( o, r + 1 / \ B ( o, ( r − / + ). Since d ( x, E ) ≤ / x ∈ H d we have B ( o, r + 1 / \ B ( o, ( r − / + ) ⊂ [ x ∈ E r B ( x, / . For x ∈ E r let x ′ be the point on ∂B ( o, r ) minimizing the distance between x and ∂B ( o, r ) , and let E ′ r ⊂ ∂B ( o, r ) denote the collection of all such x ′ . Since d ( x, x ′ ) ≤ / B ( x, / ⊂ B ( x ′ , / B ( o, r + 1 / \ B ( o, ( r − / + ) ⊂ [ x ′ ∈ E ′ r B ( x ′ , / . The claim follows, since | E ′ r | ≤ | E ∩ B ( o, r + 1 / | ≤ cν d ( B ( o, r + 3 / ≤ c ′ e ( d − r .Now fix z ∈ H d and let r := d H ( o, z ). The µ d, -measure of lines that have their closestpoint to the origin inside the shell B ( o, r + 1 / \ B ( o, ( r − / + ) is given by µ d, ( L B ( o,r +1 / \ L B ( o, ( r − / + ) ) (4.4)= µ d, ( L B ( o,r +1 / ) − µ d, ( L B ( o, ( r − / + ) )= C ( d )(sinh d − ( r + 1 / − sinh d − (( r − / + ) ≥ C ′ ( d ) e ( d − r , where the second equality uses (2.5) with C ( d ) = ( d − / sinh d − (1) . Let ( x i ) N r i =1 be acollection of points in ∂B ( o, r ) such that B ( o, r + 1 / \ B ( o, ( r − / + ) ⊂ ∪ N r i =1 B ( x i , / . (4.5)From (4.4) and (4.5) we obtain C ′ ( d ) e ( d − r ≤ N r X i =1 µ d, ( L + B ( x i , / ) = N r µ d, ( L + B ( z, / ) , (4.6)14here we used that µ d, is invariant under rotations in the last equality. From (4.6) weconclude that µ d, ( L + B ( z, / ) ≥ C ′ ( d ) e ( d − r /N r ≥ c ( d ) > , finishing the proof of the lemma. Proposition 4.5.
For any d ≥ , the set V does not percolate if u is large enough. Proof.
The proof follows the proof of Lemma 6 . W := [ x ∈ τ B ( x, ! c . Then it is clear that
W ⊃ V so it suffices to show that W does not percolate when u islarge.For z ∈ H d let Q ( z ) be the event that z is within distance 1 / W . Then Q ( z )is determined by τ ∩ B ( z, /
2) so that Q ( z ) and Q ( z ′ ) are independent if d H ( z, z ′ ) ≥ A be the event that o belongs to an infinite component of W . If A occurs, then thereexists an infinite continuous curve γ : [0 , ∞ ) → W with the properties that γ (0) = o and d H ( o, γ ( t )) → ∞ as t → ∞ . Let t = 0 and y = o , and for k ≥ t k = sup { t : d H ( γ ( t ) , y k − ) = 6 } and y k = γ ( t k ). For each k , let y ′ k be a point in D which minimizes the distance to y k . By definition d H ( y j , y k ) ≥ j = k , andsince d H ( y j , y ′ j ) ≤ / d H ( y k , y ′ k ) ≤ /
2, we get d H ( y ′ j , y ′ k ) ≥ j = k . Since d H ( y k , y k +1 ) = 6 we also have d H ( y ′ k , y ′ k +1 ) ≤
7. Observe that since y k ∈ W , the event Q ( y ′ k ) occurs.Let D be as in Lemma 4.3, and let X n be the set of sequences x , ..., x n of points in D such that d H ( o, x ) ≤ / d H ( x n , x n +1 ) ≤ d H ( x j , x k ) ≥ j = k . Furthermore,let N n denote the number of such sequences. We have that P [ A ] ≤ X ( x ,...,x n ) ∈X n P [ Q ( x ) ∩ ... ∩ Q ( x n )] , (4.7)and that N n ≤ sup {| D ∩ B ( z, | n +1 : z ∈ H d } (4.2) ≤ c ( d ) n +1 (4.8)for some constant c ( d ) < ∞ . By independence, P [ Q ( x ) ∩ ... ∩ Q ( x n )] = Π ni =0 P [ Q ( x i )] . (4.9)Observe that if τ ( B ( z, / ≥
1, then B ( z, / ⊂ W c . Hence we have P [ Q ( z )] = P [ B ( z, / ∩ W 6 = ∅ ] (4.10) ≤ P [ τ ( B ( z, / e − uµ d, ( L + B ( z, / ) ≤ e − uc ( d ) , where the last inequality follows from Lemma 4.4. From (4.7), (4.8), (4.9), and (4.10)it follows that P [ A ] ≤ ( c ( d ) e − uc ( d ) ) n +1 → . (4.11)15s n → ∞ if u < ∞ is large enough. We conclude that P [ A ] = 0 for u large enough butfinite.We can now prove Proposition 4.2. Proof of Proposition 4.2: If C is disconnected, then it consists of more than oneinfinite connected component. Since any two disjoint infinite components of C must beseparated by some infinite component of V , we get that the disconnectedness of C impliesthat V percolates. According to Proposition (4.5), there is no percolation in V when u islarge enough. Hence C is connected when u is large enough. Before we can prove Theorem 1.4, we will need to do some preliminary work. To thatend, let { c k,n } n ≥ , − ≤ k ≤ n be defined by letting c , = c , = c , = 1 and c − ,n = 0 forevery n and then inductively for every 0 ≤ k ≤ n letting c k,n := n − X l = k − c l,n − , (5.1)where we define c n +1 ,n = 0 . Note that by this definition, c k,n = c k − ,n − + c k +1 ,n . Thesenumbers constitute (a version) of the Catalan triangle, and it is easy to verify that c k,n = (2 n − k )!( k + 1)( n − k )!( n + 1)! = k + 1 n + 1 (cid:18) n − kn (cid:19) (5.2)for every n and 0 ≤ k ≤ n. This follows by using that if (5.2) holds for c k − ,n − and c k +1 ,n , we get that c k,n = c k − ,n − + c k +1 ,n = (2 n − k − k ( n − k )! n ! + (2 n − k − k + 2)( n − k − n + 1)!= (2 n − k − k ( n + 1) + (2 n − k − k + 2)( n − k )( n − k )!( n + 1)!= (2 n − k − kn − k + 2 n − k )( n − k )!( n + 1)! = (2 n − k )!( k + 1)( n − k )!( n + 1)! . By an induction argument, we see that (5.2) holds for every 0 ≤ k ≤ n. Consider the following sequence { g n ( x ) } n ≥ of functions such that g n : R + → R + forevery n. Let g ( x ) ≡ , and define g ( x ) , g ( x ) , . . . inductively by letting g n +1 ( x ) = Z x g n ( y ) dy + Z ∞ x e x − y g n ( y ) dy, (5.3)for every n ≥ . roposition 5.1. With definitions as above, we have that g n ( x ) = n X k =0 c k,n x k k ! . Proof.
We start by noting that g ( x ) = Z x dy + e x Z ∞ x e − y dy = x + 1 , and since c , = c , = 1 the statement holds for n = 1 . Assume therefore that it holdsfor n − c − ,n = 0, g n ( x ) = Z x g n − ( y ) dy + e x Z ∞ x g n − ( y ) e − y dy = n − X k =0 c k,n − (cid:18)Z x y k k ! dy + e x Z ∞ x y k k ! e − y dy (cid:19) = n − X k =0 c k,n − (cid:18) x k +1 ( k + 1)! + x k k ! + · · · + 1 (cid:19) = n − X k =0 c k,n − k +1 X l =0 x l l ! = n X k =0 x k k ! n − X l = k − c l,n − . By using (5.1), we conclude the proof.Our next result provides a link between the particle process ζ defined in the intro-duction, and the functions g n ( x ) . Recall the interpretation that a particle at position Z k,n = x, independently gives rise to new particles according to a Poisson process withintensity measure dµ x = I ( y ≥ ue − ( y − x ) + dy , so that in particular the entire process isrestricted to R + . Recall also the definition of X n [ a,b ] in (1.4). Proposition 5.2.
Let F n ( R ) = E [ X n [0 ,R ] ] . For any u < ∞ , F n ( R ) is differentiable with respect to R , and we have that with f n ( R ) := F ′ n ( R ) , f n ( R ) = u n n − X k =0 c k,n − R k k ! = u n g n − ( R ) , for every n ≥ . Proof of Proposition 5.2.
We will prove the statement by induction, and so we startby noting that F ( R ) = E [ X ,R ] ] = uR, which follows since Z , is of type 0. Therefore, the statement holds for n = 1.Assume now that the statement holds for some fixed n ≥ . Let R, ∆ R > F n +1 ( R + ∆ R ) − F n +1 ( R ) = E [ X n +1[ R,R +∆ R ] ] . n of type smaller than R gives rise to individuals in [ R, R +∆ R ](in generation n + 1) at rate u. Furthermore, any individual of type x ∈ [ R, R + ∆ R ] givesrise to individuals in [ R, R + ∆ R ] at most at rate u while individuals of type x > R + ∆ R produce individuals in [ R, R + ∆ R ] at rate at most ue R +∆ R − x . We therefore get thefollowing upper bound E [ X n +1[ R,R +∆ R ] ] ≤ u ∆ R E [ X n [0 ,R +∆ R ] ] + ∞ X k =0 E [ X n [ R +∆ R + k/N,R +∆ R +( k +1) /N ] ] e − k/N ! , (5.4)where N is an arbitrary number. By assumption, F n ( R ) is differentiable, and by themean value theorem, E [ X n [ R +∆ R + k/N,R +∆ R +( k +1) /N ] ] ≤ f n ( R + ∆ R + ( k + 1) /N ) N , since f n ( x ) is increasing. Thus, we conclude from (5.4) that E [ X n +1[ R,R +∆ R ] ] ≤ lim sup N →∞ u ∆ R F n ( R + ∆ R ) + ∞ X k =0 f n ( R + ∆ R + ( k + 1) /N ) N e − k/N ! ≤ lim sup N →∞ u ∆ R (cid:18) F n ( R + ∆ R ) + Z ∞ f n ( R + ∆ R + ( y + 1) /N ) N e − ( y − /N dy (cid:19) = lim sup N →∞ u ∆ R (cid:18) F n ( R + ∆ R ) + e /N Z ∞ f n ( R + ∆ R + z + 1 /N ) e − z dz (cid:19) = u ∆ R (cid:18) F n ( R + ∆ R ) + Z ∞ f n ( R + ∆ R + z ) e − z dz (cid:19) , by the dominated convergence theorem. Hence, we conclude thatlim sup ∆ R → F n +1 ( R + ∆ R ) − F n +1 ( R )∆ R (5.5) ≤ u (cid:18) F n ( R ) + Z ∞ f n ( R + z ) e − z dz (cid:19) = u (cid:18)Z R f n ( z ) dz + Z ∞ R f n ( z ) e R − z dz (cid:19) , again by the dominated convergence theorem.Similarly, we get the following lower bound E [ X n +1[ R,R +∆ R ] ] (5.6) ≥ u ∆ R lim inf N →∞ E [ X n [0 ,R ] ] + ∞ X k =0 E [ X n [ R + k/N,R +( k +1) /N ] ] e − ( k +1) /N ! ≥ u ∆ R lim inf N →∞ F n ( R ) + ∞ X k =0 f n ( R + k/N ) N e − ( k +1) /N ! ≥ u ∆ R lim inf N →∞ (cid:18) F n ( R ) + Z ∞ f n ( R + ( y − /N ) N e − ( y +1) /N dy (cid:19) = u ∆ R (cid:18) F n ( R ) + Z ∞ f n ( R + z ) e − z dz (cid:19) , ∆ R → F n +1 ( R + ∆ R ) − F n +1 ( R )∆ R = u (cid:18)Z R f n ( z ) dz + Z ∞ R f n ( z ) e R − z dz (cid:19) . Thus, we conclude that F n +1 ( R ) is differentiable and that f n +1 ( R ) = u n +1 (cid:18)Z R g n − ( z ) dz + Z ∞ R g n − ( z ) e R − z dz (cid:19) = u n +1 g n ( R ) , where the last equality follows from (5.3). Remarks:
The proof shows that for u = 1, the functions f n ( x ) = F ′ n ( x ) satisfies (5.3),which is of course why (5.3) is introduced in the first place.For future reference, we observe that F n ( R ) in fact depends on u, and we sometimesstress this by writing F n ( R, u ). Furthermore, it is easy to see that for any 0 < u < ∞ ,we have that F n ( R, u ) = u n F n ( R,
1) for every n ≥ . We have the following result
Proposition 5.3.
Let u < / , then for every x ≥ , ∞ X n =1 f n ( x ) ≤ u e ux − u . Proof.
By Propositions 5.1 and 5.2, ∞ X n =1 f n ( x ) = ∞ X n =1 u n g n − ( x ) = ∞ X n =0 u n +1 g n ( x ) (5.7)= ∞ X n =0 u n +1 n X k =0 c k,n x k k ! = ∞ X k =0 x k k ! ∞ X n = k u n +1 c k,n . Furthermore, by using that (cid:0) mn (cid:1) is increasing in m ≥ n, we see that c k,n = k + 1 n + 1 (cid:18) n − kn (cid:19) ≤ (cid:18) nn (cid:19) ≤ n X l =0 (cid:18) nl (cid:19) = 4 n . (5.8)Combining (5.7) and (5.8), we see that for u < / , ∞ X n =1 f n ( x ) ≤ ∞ X k =0 x k k ! u ∞ X n = k (4 u ) n = u − u ∞ X k =0 (4 ux ) k k ! = u e ux − u , (5.9)finishing the proof. Remark:
As pointed out to us by an anonymous referee, a variant of Proposition 5.3can be proved along the following lines. Let T be the integral operator defined by T ( g ) = Z x g ( y ) dy + Z ∞ x e x − y g ( y ) dy.
19t is easy to check that g ( x ) = ( x + 2) e x/ is an eigenfunction of T satisfying T ( g ) = 4 g. Thus, since g ( x ) ≡ ≤ g ( x ) we get that g = T ( g ) ≤ T ( g ) = 4 g, and iterating wesee that g n +1 = T ( g n ) ≤ T (4 n g ) = 4 n +1 g. This can then be used in conjunction withProposition 5.2 to prove the desired result.The justification for obtaining and using the explicit forms of g n , f n and F n is twofold.Firstly, these forms will be convenient when proving the second part of Theorem 1.4 andalso when proving Lemma 7.1 below. Secondly, we believe that the infinite type branchingprocess ζ is of independent interest, and therefore a detailed analysis is intrinsically ofvalue.We can now prove Theorem 1.4. Proof of Theorem 1.4.
We have that F n ( R ) = Z R f n ( x ) dx, so that ∞ X n =1 F n ( R ) = ∞ X n =1 Z R f n ( x ) dx. Furthermore, for u < /
4, we can use Proposition 5.3 and the dominated convergencetheorem to conclude that ∞ X n =1 F n ( R ) ≤ Z R u e ux − u dx ≤ e uR − u ) . We can now use Propositions 5.1 and 5.2 to get that F n +1 ( R ) = u n +1 Z R g n ( x ) dx = u n +1 Z R n X k =0 c k,n x k k ! dx = u n +1 n X k =0 c k,n R k +1 ( k + 1)! ≥ u n +1 c ,n R = u n +1 n + 1 (cid:18) nn (cid:19) R ≥ u n +1 n n + 1) R, by using that 2( n + 1) (cid:0) nn (cid:1) ≥ P nl =0 (cid:0) nl (cid:1) = 4 n which follows since l = n maximizes (cid:0) nl (cid:1) . We see that if u > /
4, the right hand side diverges, and so the statement follows. u c ( d ) > . The aim of this section is to prove the lower bound of Theorem 1.1. We will do this byestablishing a link between the cylinder process ω and the particle process of Section 5.As an intermediate step, we will in Section 6.1 consider particle processes with offspringdistributions that can be weakly bounded above by ζ . In Section 6.2, these new particleprocesses and the cylinder process in H d will be compared. Thereafter, this link is usedin Section 6.3 to obtain the required lower bound.20 .1 Particle processes weakly dominated by ζ Recall that dµ x ( y ) = 1 ( y ≥ ue − ( x − y ) + dy and suppose that ( ν x ) x ∈ R + is a family of measureswith the following property: there is a constant c ∈ (0 , ∞ ) such that for all integers k, l ≥ x ∈ ( l,l +1] ν x (( k, k + 1]) ≤ c inf x ∈ ( l,l +1] µ x (( k, k + 1]) , (6.1)and moreover, ν x ( { } ) = 0 for all x ≥
0. This last assumption is made only for con-venience; if one allows the measures to have an atom at 0 what follows below can bemodified fairly easy to get similar conclusions. The particle processes that we considerhere are defined as the one in Theorem 1.4, but using ν x as the offspring distributionin place of µ x for a particle of type x. Recall that we think of the position of a particlein R + as being the type of that particle. Of course, we still assume that every particleproduces offspring independently. For this process, let ˜ X nD be the number of individualsin generation n of type in D ⊂ R + . Furthermore let˜ F n ( R ) = E [ ˜ X n [0 ,R ] ] . Lemma 6.1.
With c ∈ (0 , ∞ ) as in (6.1) , we have that for every R ∈ N + ˜ F n ( R ) ≤ c n F n ( R ) . Proof.
Let R ∈ N + . It suffices to show that with c as in (6.1), and any integers n ≥ k ≥
0, we have E [ ˜ X n ( k,k +1] ] ≤ c n E [ X n ( k,k +1] ] . (6.2)Since ˜ F n (0) = F n (0) = 0, the claim of the lemma will then follow by summing the twosides of (6.2) from k = 0 to k = R −
1. We proceed by induction in n . For any k ∈ N wehave that E [ ˜ X k,k +1] ] = ν (( k, k + 1]) (6.1) ≤ c µ (( k, k + 1]) = c E [ X k,k +1] ] , so that (6.2) holds for n = 1 . Assume therefore that (6.2) holds for some n ≥ k ≥ . Let ˜ Y nk,l denote the number of individuals in generation n of type in ( k, k + 1] withparents of type in ( l, l + 1]. We have E [ ˜ X n +1( k,k +1] ] (6.3)= ∞ X l =0 E [ ˜ Y n +1 k,l ] ≤ ∞ X l =0 sup x ∈ ( l,l +1] ν x (( k, k + 1]) E [ ˜ X n ( l,l +1] ] (6.2) ≤ ∞ X l =0 c inf x ∈ ( l,l +1] µ x (( k, k + 1]) c n E [ X n ( l,l +1] ] ≤ c n +1 E [ X n +1( k,k +1] ] . This finishes the proof of the lemma. 21 .2 The independent cylinder process
We now turn to the independent cylinder process discussed in the introduction. Westart by defining the process itself, and the coupling with the ordinary line process ω. Thereafter we establish a link between our independent cylinder process and the particleprocess studied in Section 6.1.Formally, we define the independent cylinder process as follows. Let ω, ( ω k,n ) ∞ k,n =1 bean i.i.d. collection of Poisson line processes with intensity uµ d, . We use ω to define C ( ω ).Fix any (deterministic) line L , such that o ∈ L , . This is the single line of generation0 . Let η := { L ∈ ω : c ( L ) ∩ c ( L , ) = ∅} . Thus η is simply the collection of lines in ω such that the corresponding cylinders inter-sect c ( L , ) . Recall the definition of ρ ( L ) for L ∈ A ( d,
1) from Section 2.2. Let L , , L , , . . . be the enumeration of the lines in η satisfying ρ ( L k, ) < ρ ( L k +1 , ) for every k ≥ . Aswhen we defined the particle process ζ in the introduction, the particular choice of enu-meration is somewhat arbitrary. Define η , := { L ∈ ω \ η : c ( L ) ∩ c ( L , ) = ∅} ∪ { L ∈ ω , : c ( L ) ∩ c ( L , ) = ∅ , c ( L ) ∩ c ( L , ) = ∅} . The first set of lines corresponds to the cylinders in ω that intersect c ( L , ) but arenot included in the definition of η (i.e. intersect c ( L , )) . The second set of lines is anindependent copy of the set of lines in ω that intersects both c ( L , ) and c ( L , ) . Thus, wesee that η , and η are created in the same way, i.e. by considering the set of cylindersof a Poisson cylinder process intersecting c ( L , ) and c ( L , ) respectively. For any k ≥ , let η k, := { L ∈ ω \ (cid:0) η ∪ k − l =1 η l, (cid:1) : c ( L ) ∩ c ( L k, ) = ∅}∪{ L ∈ ω k, : c ( L ) ∩ c ( L k, ) = ∅ , c ( L ) ∩ (cid:0) c ( L , ) ∪ k − l =1 c ( L l, ) (cid:1) = ∅} . We think of η k, as being created from ω where ω has not already been used, and from ω k, where ω has been used. From this construction it is obvious that given η , the sequence η , , η , , . . . is independent. We let η = ∪ ∞ k =1 η k, , and let L , , L , , . . . be the enumeration of the lines in η satisfying ρ ( L , ) < ρ ( L , ) < · · · . These are the lines belonging to generation 2.We proceed in the obvious way, defining η k,n and η n for k, n ≥ . Finally, let η := ∪ ∞ n =1 η n , and C ( η ) = c ( L , ) [ L ∈ η c ( L ) . (6.4)Consider the set c ( L , ) ∪ C ( ω ) , and define C ( ω ) ⊂ c ( L , ) ∪ C ( ω ) to be the maximallyconnected component of the origin. By our construction, C ( ω ) ⊂ C ( η ) . (6.5)22ndeed, any cylinder in C ( ω ) intersecting L , is by definition in η , and in general, anycylinder in C ( ω ) separated from L , by k other cylinders will belong to η k +1 . From η , a particle process ¯ ζ is induced in the following way. With L ,n , L ,n , . . . beingthe enumeration of η n satisfying ρ ( L ,n ) < ρ ( L ,n ) < · · · , we let ¯ Z k,n = ¯ Z k,n ( L k,n ) := ρ ( L k,n ) for every k = 1 , , . . . . Furthermore, ¯ ζ k,n = { ¯ Z l,n ( L l,n ) : L l,n ∈ η k,n } , and of course¯ ζ n = S ∞ k =1 ¯ ζ k,n . Since given η n , the sequence η ,n +1 , η ,n +1 , . . . is independent, it followsthat ¯ ζ ,n +1 , ¯ ζ ,n +1 , . . . are independent given ¯ ζ n . Therefore, ¯ ζ = ( ¯ ζ n ) ∞ n =1 has the desiredindependence properties. We note the similarities between ζ and ¯ ζ , and that the onlyessential difference lies in the offspring distributions, which we address next.Fix some L k,n ∈ η n and consider an offspring L ∈ η k,n +1 . If L is at distance between l and l + 1 from the origin, then this corresponds to an offspring ¯ Z ∈ ¯ ζ k,n +1 of ¯ Z k,n suchthat ¯ Z ∈ ( l, l + 1] . Furthermore, the expected number of offspring (of L k,n ) belongingto L B ( o,l +1) \ L B ( o,l ) equals uµ d, (cid:0) L c ( L k,n , ∩ ( L B ( o,l +1) \ L B ( o,l ) ) (cid:1) , and so we see that theparticle process ¯ ζ can be described using the intensity measures { τ x } x ≥ where τ x (( l, l + 1]) = uµ d, (cid:0) L c ( L x , ∩ ( L B ( o,l +1) \ L B ( o,l ) ) (cid:1) , (6.6)with L x satisfying x = d H ( o, L x ) . Our next result will be used to prove that { τ x } x ≥ satisfies (6.1) for some c < ∞ . Lemma 6.2.
Let x ∈ R + and L ∈ A ( d, be such that d H ( o, L ) = x. There exists aconstant C ( d ) ∈ (0 , ∞ ) such that for any k ≥ ,µ d, ( L c ( L, ∩ (cid:0) L B ( o,k +1) \ L B ( o,k ) ) (cid:1) ≤ C exp( − ( d − x − k ) + ) . Proof.
Fix k ∈ N . Suppose that L = { γ ( t ) : −∞ < t < ∞} where the parametriza-tion of γ is chosen to be unit speed and so that d H ( o, L ) = d H ( o, γ (0)). For i ∈ Z let y i = γ ( i ) and B i = B ( y i , c ( L, ⊂ ∪ B i since any point in c ( L,
2) is at distanceat most 2 from L, and any point in L is at distance at most 1 / y i . We nowclaim that d H ( o, y i +1 ) − d H ( o, y i ) ≥ c (6.7)for every i ≥ , and some constant c >
0. To see this, assume that i ≥ , and observethat by (2.2), d H (0 , y i ) = cosh − (cosh( x ) cosh( i )) , since the angle between L and the geodesic from o to L is π/ . Equivalently, we get that d H (0 , y i ) = log (cid:18) cosh( x ) cosh( i ) + q cosh ( x ) cosh ( i ) − (cid:19) . d H (0 , y i +1 ) − d H (0 , y i )= log cosh( x ) cosh( i + 1) + q cosh ( x ) cosh ( i + 1) − x ) cosh( i ) + q cosh ( x ) cosh ( i ) − ≥ log cosh( x ) cosh( i ) cosh(1) + q cosh ( x ) cosh ( i ) cosh (1) − x ) cosh( i ) + q cosh ( x ) cosh ( i ) − ≥ log cosh( x ) cosh( i ) cosh(1) + q cosh ( x ) cosh ( i ) cosh (1) − cosh (1)cosh( x ) cosh( i ) + q cosh ( x ) cosh ( i ) − = log(cosh(1)) , where we use that cosh( i + 1) ≥ cosh( i ) cosh(1) which holds since i ≥ . Hence, (6.7)follows with c = log(cosh(1)).Assume first that k < x . From (6.7), we get that d ( y i , ≥ x + c | i | for every i usingsymmetry. We get that µ d, ( L c ( L, ∩ (cid:0) L B ( o,k +1) \ L B ( o,k ) ) (cid:1) ≤ µ d, ( L c ( L, ∩ (cid:0) L B ( o,k +1) ) (cid:1) ≤ X i ∈ Z µ d, ( L B i ∩ L B ( o,k +1) ) ≤ X i ∈ Z C exp( − ( d − x + c | i | − k ) ≤ C ′ exp( − ( d − x − k )) , where the penultimate inequality follows from Lemma 3.4. Now assume instead that x ≤ k . Let p = inf {| i | : d H ( o, y i ) ≥ k − } . Using the union bound and that µ d, ( L B i ∩ (cid:0) L B ( o,k +1) \ L B ( o,k ) ) (cid:1) = 0 when | i | ≤ p , we get µ d, ( L c ( L, ∩ (cid:0) L B ( o,k +1) \ L B ( o,k ) ) (cid:1) ≤ X i ∈ Z µ d, ( L B i ∩ (cid:0) L B ( o,k +1) \ L B ( o,k ) ) (cid:1) ≤ ∞ X i = p µ d, ( L B i ∩ L B ( o,k +1) ) ≤ ∞ X i = p c exp( − ( d − c ( i − p ))) ≤ C, where again we use Lemma 3.4.Let ¯ Z n [0 ,R ] = ∞ X k =1 I ( ¯ Z k,n ≤ R ) , and let H n ( R, u ) = E [ ¯ Z n [0 ,R ] ] . It is not hard to show that, similarly to the observation afterthe proof of Proposition 5.2, H n ( R, u ) = u n H n ( R, . This follows from the Poissoniannature of the process.We can now use Lemma 6.2 to show the following result (recall the definition of F n ( R ) = F n ( R, u ) from Section 5). 24 emma 6.3.
There is a constant c ( d ) ∈ (0 , ∞ ) such that for every R ∈ N + , and
Since H n ( R, u ) = E [ ¯ Z n [0 ,R ] ], and the particle process ¯ ζ uses { τ x } x ≥ as intensitymeasures, it suffices in view of Lemma 6.1, to show that there is a constant c < ∞ suchthat for every integer k, l ≥ x ∈ ( k,k +1] τ x (( l, l + 1]) ≤ c inf x ∈ ( k,k +1] µ x (( l, l + 1]) . (6.8)From Lemma 6.2, we havesup x ∈ ( k,k +1] τ x (( l, l + 1]) (6.9) ≤ cu sup x ∈ ( k,k +1] exp( − ( d − x − l ) + ) ≤ c ′ u exp( − ( d − k − l ) + ) . On the other handinf x ∈ ( k,k +1] µ x (( l, l + 1]) = inf x ∈ ( k,k +1] u Z l +1 l exp( − ( x − y ) + ) dy ≥ cu exp( − ( k − l ) + ) . (6.10)Equations (6.9) and (6.10) establishes (6.8), and the lemma follows.In what follows, we drop the explicit dependence on u from the notation and simply write H n ( R ) and F n ( R ) . We now have all the ingredients to prove our main result.
Proof of Theorem 1.1.
Using Lemma 4.1 and Proposition 4.2, we only need to prove that u c ( d ) >
0. To thatend, let V ( R ) = E [ |{ L ∈ C ( ω ) : L ∩ B ( o, R ) = ∅}| ] , that is, V ( R ) is the expected number of cylinders in C ( ω ) which intersect B ( o, R ).Recall that C ( ω ) is the maximally connected component of c ( L , ) ∪ C ( ω ) , and recallalso the definition of C ( η ) from (6.4). By (6.5) we can couple C ( ω ) and C ( η ) so that C ( ω ) ⊂ C ( η ) , and so we have as in the proof of Theorem 1.4, that for u < / (4 c ) with c = c ( d ) as in Lemma 6.3, V ( R ) ≤ ∞ X n =0 H n ( R ) ≤ ∞ X n =0 c n F n ( R ) ≤ Ce ucR − cu )by using Lemma 6.3 in the second inequality.Hence, we see that when 0 < u < / (4 c ), V ( R ) grows at most exponentially in R ata rate which is strictly smaller than 1. On the other hand, we have that E [ | L ∈ ω : L ∩ B ( o, R ) | ] = uµ d, ( L B ( o,R ) ) = Cu sinh d − ( R ) , by (2.5). This grows exponentially at rate ( d − R and so we see that with probabilityone, C ( ω ) is a strict subset of c ( L , ) ∪ C ( ω ) . We conclude that C ( ω ) is a.s. not connectedfor this choice of u. Proof of Theorem 1.2.
Similar to the notation of Section 3 we let A m ↔ B denote the event that there exists1 ≤ l ≤ m and a sequence of cylinders c , · · · , c l ∈ ω such that A ∩ c = ∅ , c ∩ c = ∅ , . . . , c l ∩ B = ∅ . That is, the sequence c , · · · , c l connects A to B in l steps. We observethat { A ↔ B } = { A ↔ B } .We start with the following lemma. Lemma 7.1.
There exists a constant D ( m ) < ∞ (depending only on u and m ) such thatfor any x, y ∈ H d we have that P [ B ( x, m ↔ B ( y, ≤ D ( m ) d H ( x, y ) m e − ( d − d H ( x,y ) . (7.1) Proof.
Assume without loss of generality that x = o and let d H ( o, y ) = R. As inSection 6.2, the expected number of cylinders in η up to generation m that intersect theball B ( o, R ) is bounded by m X n =1 H n ( R ) ≤ m X n =1 c ( d ) n F n ( R )= m X n =1 c ( d ) n Z R f n ( x ) dx = m X n =1 c ( d ) n u n Z R g n − ( x ) dx = m X n =1 c ( d ) n u n Z R n − X k =0 c k,n − x k k ! dx ≤ m X n =1 c ( d ) n u n C n Z R n − X k =0 x k k ! dx = m X n =1 c ( d ) n u n C n n X k =1 R k k ! = m X k =1 R k k ! C m X n = k (4 cu ) n ≤ D ( m ) R m , for some constant D ( m ) < ∞ .There exists a collection B R of balls of radius 1 / ∂B ( o, R ) such that |B R | ≥ ce ( d − R for some c > B ( o, R ) intersectsat most c ( d ) < ∞ balls in B R . To construct such a collection B R , we consider first D asin Lemma 4.3. Let G R = D ∩ ( B ( o, R + 3 / \ B ( o, ( R − / + )). By a slight modificationof the lower bound in (4.2), we get that | G R | ≥ ce ( d − R /ν d ( B ( o, / c ′ e ( d − R . For x ∈ G R , let x ′ be defined as the point on ∂B ( o, R ) such that x ′ minimizes the distanceto ∂B ( o, R ), and let G ′ R be the collection of all such x ′ . Obviously, the collection ofballs B R := { B ( x, / } x ∈ G ′ R satisfies |B R | ≥ c ′ e ( d − R . Now let L be a line intersecting B ( o, R + 5 /
4) (only cylinders centered around such lines might intersect some ball in B R ).Using (6.7), there is a universal constant c < ∞ and two points x , x ∈ ∂B ( o, R ) (thesepoints depend on L ) such that c ( L ) ∩ ( B ( o, R + 1 / \ B ( o, ( R − / + )) ⊂ B ( x , c ) ∪ B ( x , c ). Hence the number of balls from B R intersecting c ( L ) is bounded by the numberof points in D ∩ ( B ( x , c + 2) ∪ B ( x , c + 2)) . This in turn is bounded by some constant c ( d ) < ∞ , by the upper bound of (4.2). Hence, the existence of the B R is verified.Using B R , we see that the probability that a fixed ball at distance R from o will beintersected by any cylinder in η of generation less than or equal to m is bounded by c |B R | m X n =1 H n ( R ) ≤ D ( m ) R m e ( d − R ,
26y possibly increasing the value of D ( m ) . The statement follows by using that C ( ω ) ⊂ C ( η ) and noting that any cylinder thatintersects B ( o,
1) must also intersect the cylinder c ( L , ) . Proof of Theorem 1.2.
Let m ∈ N + and fix ǫ ∈ (0 , r = r ( m, ǫ ) < ∞ so large that the probability that any two fixed cylindersseparated by distance r will be connected in at most m steps is less than ǫ. Indeed, take r so large that the probability that B ( o,
1) and B ( y,
1) (where y ∈ ∂B ( o, r )) are connectedin at most m + 2 steps is less than ǫ . Consider then two cylinders c , c separated bydistance r, and assume without loss of generality that c ∩ B ( o, = ∅ and c ∩ B ( y, = ∅ . Then, if the probability that c , c are connected in at most m steps is larger than ǫ, thiswould lead to a contradiction.For lines L , L ∈ A ( d, E m ( L , L ) be the event that c ( L ) and c ( L ) areconnected in at most m steps. Define the event H = X ( L ,L ) ∈ ω = I ( E m ( L , L ) c ) ≥ , where the union is over all 2-tuples of distinct lines in ω . In words, H is the event thatthere is at least one pair of lines in ω whose corresponding cylinders are not connected inat most m steps. We now let E L ,L denote the expectation with respect to ω + δ L + δ L .Using the Slivnyak-Mecke formula ([14] Corollary 3 . .
3) we get that E X ( L ,L ) ∈ ω = I ( E m ( L , L ) c ) = u Z A ( d, Z A ( d, E L ,L [ I ( E m ( L , L ) c )] µ d, ( dL ) µ d, ( dL )= u Z A ( d, Z A ( d, P L ,L [ E m ( L , L ) c ] µ d, ( dL ) µ d, ( dL )= u Z A ( d, Z A ( d, P [ E m ( L , L ) c ] µ d, ( dL ) µ d, ( dL ) ≥ u Z A ( d, Z A ( d, I ( d H ( L , L ) ≥ r ) P [ E m ( L , L ) c ] µ d, ( dL ) µ d, ( dL ) ≥ u Z A ( d, Z A ( d, I ( d H ( L , L ) ≥ r )(1 − ǫ ) µ d, ( dL ) µ d, ( dL ) . Obviously, the expression on the right hand side diverges for any 0 < u < ∞ , so that E X ( L ,L ) ∈ ω = I ( E m ( L , L ) c ) = ∞ , and from this, it follows that P [ H ] >
0. Since the event H is invariant under isometriesof H d , it follows from Proposition 2.1 that P [ H ] = 1. Hence for any m ∈ N + we have P [diam( C ) ≥ m −
2] = 1, from which we conclude that P [diam( C ) = ∞ ] = 1.27 Proof of Proposition 1.3
In this section, we prove that when u < u c ( d ), there are a.s. infinitely many connectedcomponents in C . Let N ( ω ) =the number of connected components in C . Proof of Proposition 1.3
Obviously, the event N ( ω ) = k is invariant under isome-tries of H d and so using Proposition 2.1, we have that for any u there is k = k ( u ) ∈ N ∪{∞} such that P [ N ( ω ) = k ] = 1. Suppose u < u c ( d ) and suppose that 1 < k ( u ) < ∞ . It isnot hard to show that there exist points y , ..., y k ∈ H d such that the event A := {∪ ki =1 y i intersects all components of C ( ω ) } ∩ { ω ( L B ( o, ) = 0 } has positive probability.Now let ω be the restriction of ω to L B ( o, , and let ω be the restriction of ω to( L B ( o, ) c . Since P [ A ] >
0, it follows that B := {∪ ki =1 y i intersects all components of C ( ω ) } has positive probability. Define the event C := {∪ ki =1 y i ⊂ C ( ω ) } ∩ {C ( ω ) is connected } . It is easy to see that P [ C ] > ω might consist of k lines L ,..., L k such that c ( L i ) contains o and y i ). Since ω and ω are independent, it follows that B and C areindependent and hence P [ B ∩ C ] >
0. The event B ∩ C implies that N ( ω ) = 1, whence P [ N ( ω ) = 1] > P [ N ( ω ) = k ] = 1. We conclude that N ( ω ) ∈ { , ∞} , and since u < u c by assumption, it follows that a.s. N ( ω ) = ∞ . Acknowledgements:
We would like to thank I. Benjamini for suggesting some ofthe problems dealt with in the paper. We would also like to thank the anonymous refereesfor many useful suggestions and comments.
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