Nearest Neighbor and Hard Sphere Models in Continuum Percolation Olle Haggstrom Department of Mathematics, Chalmers University of Technology, S-4 12 96 Goteborg, Sweden
Ronald Meester Department of Mathematics, University of Utrecht, P O . Box 80.010, 3508 TA Utrecht, The Netherlands
ABSTRACT Consider a Poisson process X in Rd with density 1. We connect each point of X to its k nearest neighbors by undirected edges. The number k is the parameter in this model. We show that, for k = 1, no percolation occurs in any dimension, while, for k = 2, percolation occurs when the dimension is sufficiently large. We also show that if percolation occurs, then there is exactly one infinite cluster. Another percolation model is obtained by putting balls of radius zero around each point of X and let the radii grow linearly in time until they hit another ball. We show that this model exists and that there is no percolation in the limiting configuration. Finally we discuss some general properties of percolation models where balls placed at Poisson points are not allowed to overlap (but are allowed to be tangent). 0 1996 John Wiley & Sons, Inc.
1. INTRODUCTION
In continuum percolation there are two models which have received a considerable amount of attention over the last 5 years, the Boolean model and the random connection model (RCM). For a description of these models, we start off with a Poisson point process X with density A > 0 in Rd,where d 2 1. This means that the number of points in a bounded, measurable set A is Poisson distributed with
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Random Structures and Algorithms, Vol. 9, No. 3 (1996) 01996 John Wiley & Sons, Inc. CCC 1042-98321961030295-21
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mean AV(A) [here V ( A ) is the d-dimensional volume of A], and the number of points in disjoint sets are independent. In a Boolean model, we consider i.i.d. nonnegative random variables p, p l , p2, . . . , which are also independent of X . We order the points of X according to some previously determined rule and put a closed Euclidean ball with radius pi around the ith point of X . In an RCM, we choose a so-called connection function g : Rd+ [0, 11, and we connect the points x and y of X with probability g( Ix - yl), independently of all other pairs of points, where 1.1 denotes Euclidean distance in Rd.For a detailed study of these models we refer to Meester and Roy [lo]. Here we shall confine ourselves to only a few basic properties of these models so as to provide the context for the related models that are the topic of the present paper. One of the most basic features of both the Boolean model and the random connection model is the existence, for d 2 2 and under otherwise very weak conditions, of a nontrivial phase transition. In a Boolean model, this means that there exists 0 < A,(p) < m (here we identify p with its distribution function) such that if A < h , ( p ) , then the part of space covered by balls consists of bounded connected components only a s . , while for A > A , ( p ) , the covered part contains at least one unbounded connected component a s . The number A,(p) is called the critical density of the Boolean model. In an RCM, the critical density can be defined similarly and is a function of the connection function g. It can be shown that all models under consideration are ergodic with respect to the group of all translations in space, and therefore the number of unbounded components is an a.s. constant. The a.s. existence of an unbounded component for A > A, is therefore equivalent to the existence with positive probability of such a component. It turns out [9] that, in each of the above models, the number of unbounded components is either 0 or 1. This is usually referred to as the uniqueness of the unbounded component. In general, if unbounded connected components arise, we say that percolation occurs. As far as the exact values of the critical densities are concerned, there seems to be no hope for exact calculations of A , ( p ) or A,(g) in any of these models, although some bounds are available. Certainly there are alternative, natural, ways to connect points of a Poisson point process X and the current paper introduces some of these. We shall now describe the two models that are the subject of this paper. One of the common features of these models is that the density of the underlying Poisson process is irrelevant (as we shall see), and therefore we can assume that X has density 1. Sometimes however, it will be convenient to assume that X has a very high or a very low density, and we will do so freely. First we define the model which we denote by NN(d, k ) , where NN stands for “nearest neighbor,” d for the dimension, and k E N is the parameter of the model. Similar models have been studied before (see, e.g., [ l l ] and [2]), although not from a percolation theoretic viewpoint. We shall abuse notation and write x E X if we want to say that there is a point of X at x E Rd. Given a realization of X , the connection rule is simple: We connect each x E X to its k nearest neighbors by an undirected edge. This is a.s. well defined by the fact that, in a Poisson process, no two interpoint distances are the same a.s. Note that the only randomness is in the position of the points in X . By changing the unit of length, we can vary the density of the underlying Poisson process without changing the connections, and therefore the density of X plays no role in this model. So with
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the dimension d fixed, the only parameter is k . It is clear that if percolation occurs for some value of k , then percolation occurs for all larger values of k as well. It is therefore natural to define the critical point k , ( d ) as follows, writing U for the event that the resulting (random) graph contains an infinite component (which we shall call an infinite cluster), and P ( * )for the appropriate probability measure: k , ( d ) = min{k L 1 : P ( U ) > 0} . By ergodicity, we have k , ( d ) = min{k 2 1 : P ( U ) = l}. Now the situation is very different from the standard Boolean model in the sense that the parameter space is discrete; there is some hope that we can determine k , ( d ) exactly. It is easy to see that k , ( l ) = m, and we shall prove in the next section that 2 5 k , ( d ) < m, for all d L 2. In Section 3, we shall prove that k , ( d ) = 2 for all d sufficiently large. We do not really know how large “sufficiently large” is; the proof does not give us an explicit lower bound. In Section 4 we show that also in this model there can be at most one infinite cluster a.s. The second class of models discussed in this paper is the class of hard sphere Boolean models. A hare sphere Boolean model is a stationary (and ergodic) probability measure on the space of configurations of balls centered at Poisson points which assigns probability 1 to configurations of nonoverlapping (but possibly tangent) balls. Before we continue, we describe a particular model, the so-called dynamic lily-pond model. We start with a Poisson process in Rd with density 1. The configuration of balls in space is constructed dynamically as follows. At time 0, all points of X are the centre of a ball with radius 0. Then, as time t evolves, the radius of each ball grows linearly in t , and all radii grow with the same speed. As soon as a ball hits another ball, it stops growing forever. Thus at each time t the space is partitioned into a region that is covered by balls (the occupied region) and its complement, which we call the vacant region. Let C, be the occupied region at time 1. We are interested in the limiting configuration C defined as
c=uc, r20
Note that in this model, no isolated balls exist. The above construction seems clear enough, but in fact we have been quite careless in the description. In Section 5 we show that this model actually exists, and we also show that the set C does not contain an unbounded connected component a.s. in any dimension. In the last section of the paper we discuss some general properties of hard sphere Boolean models. We show for instance that any unbounded connected component in such a model necessarily has a tree structure.
2. PHASE TRANSITIONS IN THE MODEL NN(d,k)
In this section we show that for all d phase transition: Theorem 2.1. model.
2 2,
the NN(d, -) model exhibits a nontrivial
For any d 2 2, there is a s . no infinite cluster in the NN(d, 1)
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Theorem 2.2. For any d 2 2 , there exists some k < ~0 such that there a.s. is some infinite cluster in the N N ( d , k ) model. We can summarize these theorems as 2 Ik,(d) = 0) $) 1 - 7’($+)
2.
= l+PC
Let En,, be the event given by = E & f l Et,,, and for Z,n E Z, let El,“ be the obvious analogous event for the square [I, I + 11 X [n,n + 11. We have that the events { E l , n } l , n Eare Z independent with probabilities
W,,,) > 1 - (1 - P ( G , , ) ) - (1 - P ( G . 0 ) )= P c . For i = 0, . . . , 7 , let S, denote the square [q, +] X [$, f ] and note that S,, and P@/,,)
=
S, are centred at the same points as the squares [0, 11’ and [ l ,21 x [0, 11,
respectively. Suppose now that the events Eo,, and El,, occur. We then have that no point of X in f=, S, has more than m points within distance Two points x and y in S, and S,,, are at distance at most $ from each other. Since d < 3, this implies that, for all x E Si,y E S i + l , there is an edge between x and y in the “(2, m) model. This in turn implies that for all x E So, y E S,, there is a path from x to y . Similar statements hold whenever two events E,,n and E l + l , n(or El,n and E l , n + l )occur. A simple comparison with independent site percolation on the square lattice now shows that there is an infinite cluster a s . in the NN(2,m) model. 0
u
3.
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3. THE CRITICAL POINT IN HIGH DIMENSIONS
The results of the previous section do not tell us very much about the precise value of k,(d). Therefore, we have carried out Monte Carlo simulations which suggest that 3 k,(d)= (2
ford=2, fo r d L 3 .
At present, we do not have sufficiently powerful mathematical techniques to even come close to proving this. The following result can be viewed as a first small step towards this goal.
Theorem 3.1.
There exists a do
)p ,
+E
for all (i, j ) E 9. Proof. Recall that 1 - 7.5 > p , . The probability that step (i, j ) fails due to (v) is independent of q , jand, by Lemma 3.5, less than E for all sufficiently large d. The same thing holds for failures of type (iv) by the definition of R , and for failures of type (iii) by Lemma 3.2 and the fact that the first Nu generations of the MBRW have boundedly many individuals. Next we consider the possible failure of step (i, j ) due to steps (i’, j ’ ) with M q ( i - i’)2 + ( j - j ’ ) * < R for some fixed R (to be determined later and independently of the dimension). By Lemma 3.3, the probability of this can be made less than E by taking d sufficiently large, because again only boundedly many points are involved. It remains to bound the probability of failure of type (i) or (ii) due to steps (i’, j ’ ) with
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Consider a failure of type (i) due to a previous step (i’, j ’ ) with
where 1x1 denotes the greatest integer smaller than o r equal to x . The total volume scanned in step (i’, j ’ ) is uniformly bounded by construction. Furthermore, it follows from Lemma 3.4 and the (faster than) exponential decay of the normal distribution that the fraction of this volume whose projection falls into S((Mi, M j ) , R o ) , is less than for all sufficiently large Q. [It is here where the point of (iv) comes in: The projected distance between a point in step ( i , j ) and a point in step ( i ’ , j ’ ) around which scanning has taken place is at least Q - 2R,.] Now the number of points (i’, j ’ ) with
3
LMq(i
-
ill2 + ( j -j’)21 = Q
is bounded by a constant times Q. Hence, for an S > 0, we can make the total such volume from all points (i’, j ’ ) with MY2 R smaller than 7rdS by choosing R sufficiently large. In particular, we can make this volume so small that each individual of the MBRW in step ( i , j ) has probability at most of landing in such previously scanned volume. Since there are at most 2N0t’ individuals in step ( i , j ) , the probability that this happens to at least one of them is at most E for large d . The probability of a failure of type (ii) due to steps ( i ’ ,j ’ ) with
+
Mv(i -
+( j
-
2
R
can be bounded by E similarly. Summing up the probabilities of all types of 0 failures we obtain the lemma.
4. UNIQUENESS OF THE INFINITE CLUSTER
In this section we prove uniqueness of the infinite cluster: Theorem 4.1. For any ( d , k ) , there can be at most one infinite cluster a s . in the model N N ( d , k ) .
We will prove uniqueness in a series of lemmas. In each lemma, it should be understood that we consider the N N ( d , k ) model for fixed d and k such that k 2 k , ( d ) . As usual in proving uniqueness of the infinite cluster for percolation processes, the first lemma is to show that there cannot exist an infinite cluster with three disjoint “branches” going off to infinity. This follows from by now fairly standard arguments (see [4]or [9, l o ] ) , so we omit the proof. Lemma 4.2. Pick r > 0 and x E Rd, and let E(x, r ) be the event that S(x, r ) , the ball with radius r centered at x , is intersected by an infinite cluster C in such a way that if all edges intersecting S(x, r) are removed, C falls apart into a number of components, of which exactly three are infinite. Then P(E(x, r ) ) = 0 .
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Remark. It follows that the event that C falls apart into four or more components under the same edge-removal also has probability 0, because then Lemma 4.2 would fail for some r' < r . Note that by ergodicity, the number of infinite clusters is (for fixed d and k ) an a.s. constant (possibly m). The next lemma rules out the possibility of this constant being greater than two.
Lemma 4.3.
The NN(d, k ) model has a s . at most two infinite clusters.
As preparation for the proof of this lemma, we now introduce a coupling, which will also be used in the subsequent proofs of Lemmas 4.4 and 4.5. Let X and X ' be two independent Poisson processes on Rd with density 1. Given r > 0 , construct a third point process X i , in the following way: On RdLS(O,3r), let X i , = X , while on S ( 0 , 3 r ) , include each point of X U X ' , each with probability +. It follows from the usual independence and thinning properties of Poisson processes that X l r is also a Poisson process on Rd with density 1. We will refer to the pair ( X , X i , ) as the special coupling.
Proof of Lemma 4.3. Suppose the NN(d, k ) model has at least three infinite clusters. We can then pick an r > 0 such that the event E*(O, r ) = {three infinite clusters intersect S(0, r ) } has positive probability. Let ( X , X i , ) be given by the special coupling, and suppose that E*(O, r ) occurs for the NN(d, k ) realisation obtained from X . What we will do now is to modify the configuration of points of X inside S ( 0 , 3 r ) by simply removing a number of points. (Note that this will affect the edge structure in the remaining points.) It is clear from the construction of ( X , X s , ) that this modified configuration has, conditional on X , positive probability of occurring for Xi,. So we are done if we can do the modification in such a way that E(O,3r) (defined in Lemma 4.2) will occur for some r', since this would contradict Lemma 4.2. Let C , , C,, and C, denote the three infinite clusters intersecting S(0, r ) . For i = 1,2,3, let x, be some point in Cj f l S(0,r ) . There is no edge between x i and x 2 , and lxl -x21 5 2 whence x1 has edges to at least k other points in C , f l S ( 0 , 3 r ) . We have k 2 2 since k 2 k,(d). Here is how we do the modification: Start by removing all points in S ( 0 , 3 r ) , which are not in C, U C , U C,. Note that when we remove a point, the effect is that this point together with all its edges are removed, but apart from this edges can only be added. Let x i and x i be two points in C, n S(0,3r), which have edges to xl. There must exist a path from x1 to infinity which either avoids x i or avoids x i (by a path t o infinity we mean a path which visits infinitely many vertices). So we may suppose there is a path from x , t o infinity avoiding x:. We can then remove x: without destroying this path to infinity. We can keep removing points from (Cl\{xl}) n S ( 0 , 3 r ) until an edge between x1 and some point in C, U C, is created. This will eventually happen, since when there is only one point left in ( C , \ { x , } )fl S ( 0 , 3 r ) , xI must have an edge to some other point in S ( 0 , 3 r ) (because Ix, -x21 s 2 r ) , and this point must be in either C, or C,. Now do the same procedure to C,as we did to C,, where C,
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is the one of C , and C , that is not yet connected to C , . This will also terminate, and we will end up with a configuration where (at least infinite parts of) C , , C , , and C , are connected to each other. Our point-removals will only add finitely many new edges, whence E(0, r ' ) occurs for some sufficiently large r', as 0 desired.
Now that Lemma 4.3 is established, we only have to rule out the possibility of exactly two infinite clusters in order to prove uniqueness. The next lemma does this partially, in that it rules out the coexistence of two infinite clusters, at least one of which has two disjoint branches to infinity. Lemma 4.4. Pick r > 0 and let E**(O,r) be the event that S(0,r ) is intersected by two infinite clusters C , and C,, where C, has the property that ifallpoints in S(0,r ) (and their edges) are removed, C , falls apart into a number of components of which exactly two are infinite. The P(E**(O,r)) = 0. Proof. We again use the coupling ( X , X i , ) . To get a contradiction, we assume that E**(O,r) occurs for the N N ( d , k ) process associated with X . Removing points, first from all finite clusters, then from C , , in the same manner as in the proof of Lemma 4.3 shows that the NN(d, k ) process associated with X l , has positive probability of having three (or more) disjoint branches to infinity, 0 contradicting Lemma 4.2 (or the remark following it). Now we are ready for the last lemma, which finally rules out the possibility of exactly two infinite clusters, thereby proving uniqueness.
Lemma 4.5.
The N N ( d , k ) model does a.s. not have exactly two infinite clusters.
Proof. Suppose that the probability of exactly two infinite clusters is positive; the probability is then 1 by ergodicity. Pick r > O such that the event that both infinite clusters ( C , and C , , say) intersect S ( 0 , r) has positive probability. Suppose this happens for the NN(d, k ) process associated with X in the coupling ( X , Xi,). We then remove points from finite clusters and then from C , in the same way as above, until C , and C , connect. The resulting configuration then has positive conditional probability for X ; r . Now two things may have happened, either (a) we have only one single infinite cluster left, or (b) C , has fallen apart into (at least) two infinite clusters. Case (a) is impossible since the number of infinite clusters is an a.s. constant. Case (b), on the other hand, is also impossible since the new infinite cluster containing C, and parts of C , has two disjoint branches to infinity, which would 0 contradict Lemma 4.4. 5. THE DYNAMIC LILY-POND MODEL
The dynamic lily-pond model was introduced in the first section in an informal way. One can think of this model as a labeled point process, the label of a point
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being the radius of the ball centered at it. The first task in this section is to show that such a labeled point process with the required distribution actually exists. To this end, let us first see what happens if we run the process on a compact set rather than in the whole space. Let K be a compact subset of Rd, and let X , be a Poisson process restricted to K with density one. Denote the occurrences of X , by xl,. . . ,x,. Suppose n 2 2 (the case n = 0 is trivial while in the case n = 1 we have a single ball which keeps growing forever, so it gets radius 00). As long as no two growing balls hit each other, there is no problem, so we can run the process for at least t, := min{ [xi- xi(: 15 i # j 5 n } units of time (assuming that the speed of the radii is 1). At time t, the first two balls hit each other, and we can compute the appropriate distances between all balls at time t,. We then let the process evolve until the next collision takes place, and so on. Only finitely many balls are involved so after a finite amount of time, all balls have stopped growing, and the limiting configuration is well defined. Things are quite different if we run the process on the whole space. There is a s . no minimal interpoint distance, and we cannot describe the process as in the case of a compact set. To overcome this difficulty, we proceed as follows. Consider a Poisson process X on Rd with density 1. Let X , be the process X restricted to B, = [-n, nId (so that the occurrences of X , form a subset of the occurrences of X ) . Let C(n) be the (well defined) limiting configuration of the lily-pond model if we start off with X , rather than with X . Note that there is no monotonicity in the model: When n gets larger, the radius of a ball at any given point in the limiting configuration can both increase and decrease. However, we have the following result, where we assume that X has a point at the origin.
+
Lemma 5.1. There exists a (random) number N such that the radius of the ball at the origin in C(n) is constant for all n 2 N .
Proof. Let C,(n) be the configuration at time t if we start off with X,,, and let r,(n) be the radius of the ball at the origin in C,(n). We claim that there exists a sequence to(= 0), t , , t,, . . . with limn+=t, = ~0 such that for all i 2 1, there exists N, with the property that
for all n 2 N , . To see this, we choose the sequence { t , } by the requirements that for all i, the annulus A ( 2 t l + , ,2t,) := S(0,2t1+,)S(0,2t,) has volume *. (Note that with this choice it follows easily that t, = t..)Suppose the ball at the origin is still growing at time t,, and consider the set of balls which have direct interaction with this ball in the time interval [t,,t!+,];by this we mean the set of balls which still grow at time t, and which would overlap the ball at the origin by time f,+, if growth continued unrestricted during [t,,f , + l ] . These balls must be centred in A ( 2 t 1 + , , 2 t , ) ,and by the choice of the t , ’ ~ the , expected number of these “first generation interacting balls” is +. Next we examine the balls which have interaction in [t,,t r + , ]with this first generation of interacting balls, and we call this the second generation interacting balls. If we denote the centre of the first ball (if it exists) in the first generation by y , , the second generation balls interacting with this ball must be centred in ( y , + A ( 2 t , + I 2tl))\A(2t,+,, , 2t,). The
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expected number of such balls is less than i. Next , we examine the second ball (if it exists) in the first generation, centered at y 2 say. The second generation balls interacting with this ball (and which have not been previously seen) are centered , The expected number of in y 2 + A ( 2 t , + , ,2t,)\(A(2tI+,,24) U ( y , + A ( 2 t , + 12t,))). as before. We continue in the obvious way, each time such balls is at most scanning the Poisson process in a region that is disjoint from all regions inspected so far. This “branching process” can be dominated by a Galton-Watson process whose offspring distribution is Poisson with mean $, so it dies out a s . However, all the growing balls which can have interaction with the ball at the origin in the time interval [ t , ,t r + , ] must be centered at a point of the above-constructed branching process whence there are only finitely many such balls. This proves the claim. Our second claim is that there exists a time sn, say, such that for all n we have
+
rt(n) = r,,,(n)
7
for all t 2 sn. This is simpler than the first claim and follows from the fact that a ball at a point cannot grow beyond its nearest neighbor, so there exists a maximal radius, uniformly in n. Now we combine the two claims made above, thereby proving that, for all n large enough, lirn,+% r,(n) is independent of n which completes the proof of the 0 lemma. It follows immediately from Lemma 5.1 that given the realization of X , for each x E X , we can compute the radius of the ball at x in the limiting configuration C by looking at the radius of the ball at x in C(n) for large n. Thus the dynamic lily-pond model is well defined. Now that we have established the existence of the model, we can investigate its percolation properties. If the limiting configuration in the dynamic lily-pond model contains an unbounded component, we say that it percolates. The event that C percolates is invariant under translations and hence its probability is either 0 or 1. We show the following: Theorem 5.2. percolate.
In the dynamic lily-pond model, the limiting set C does a.s. not
In the proof of this theorem, it will be convenient to define a graph T as follows. The vertices of T are the occurrences of X and two points x and y are neighbors (to be denoted by x y ) if the balls centered at x and y are tangent (or, equivalently, have nonempty intersection). To say that C percolates is the same as to say that T contains an infinite component (in the usual graph-theoretical sense). We say that the point x is smaller than y (notation: x < y ) if the ball centered at x has smaller radius than the ball centered at y . We define the relation “ 5 ” between points of X in the obvious way.
-
Lemma 5.3. With probability 1 , each point x of X has at most one neighbor y for which y I x .
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Proof. The only way for a ball to get a neighbor with the same radius is to hit each other while both are still growing. The fact that a s . no two interpoint distances are the same implies that a ball can have at most one such neighbor and in such case has no smaller neighbors. The only way for a ball to get a smaller neighbor is to hit a ball which already stopped growing before. Hence it suffices to show that it is a.s. impossible that a growing ball hits two or more balls simultaneously. This is quite obvious, and one way of seeing this is the following. Select two points x and y of X and wait until they both stop growing. Suppose this happens at time to. Consider the union of the components in C,o containing x and y and denote this union by W. Given W , the point process X outside the region W' := {x E R d : Ix - WI 5 t o } is still unconditioned. But all potential balls that might hit W at two different balls at the same time are centered outside W'. Note that W consists of the union of finitely many balls, and the set of points outside W' that have the same distance to two or more balls of W has Lebesgue measure zero. Hence the probability that the balls 0 associated with x and y are hit simultaneously by a larger ball is zero. It follows from Lemma 5.3 that T is almost surely a forest, i.e., its components are a.s. trees. To see this, note that if T contains a circuit, then this circuit has to contain a largest point (i.e., a point with largest associated radius) which leads to a contradiction if we consider the two neighbors of this point in the circuit. Furthermore, two tangent balls in C have the same radius if and only if they stop growing at the same time, i.e., when they hit each other. We call two such balls a root. It follows from Lemma 5.3 that any component in C can obtain at most one root a.s. To see this, suppose there is a component with two roots. It is obvious that the two roots have different associated radii a s . The balls in the larger root can, according to Lemma 5.3, only have larger neighbors and such a neighbor can again only have larger further neighbors and so on. Hence a path t o the smaller root cannot exist a s . Proof of Theorem 5.2. We distinguish between two cases. First, suppose that with positive probability (and hence with probability one by ergodicity), C contains an unbounded component W with a root. As remarked above, W contains exactly one root a x in this case. By ergodicity there must be a positive density of roots that belong to an unbounded component, and we call such a root a special root. We construct a subset L of X by choosing one center from each special root with equal probability and independent of all other special roots. For each x in L and for each i E N , we choose a subset K J x ) of X by choosing the i closest points to x which belong to the component of x. Note that by construction, K,(x)n K i ( y ) = 0 for x # y and it then follows from Lemma 2.3 that L must have density zero and hence special roots cannot exist. Secondly, we rule out the possibility of unbounded components without a root. First note that an unrooted component cannot contain a smallest point as this point would be one of a root. Also, a point cannot have only neighbors that are all strictly larger than the point itself. Hence every point has at least one neighbor that is strictly smaller than the point itself, and we conclude that any unbounded unrooted component contains an infinite sequence of tangent balls with strictly
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decreasing radii. The radii in such a sequence approach a limit a , say (which is random). We now show that this is impossible. We can assume that there is a Poisson point at the origin. Consider the numbers ti introduced in the proof of Lemma 5.1, and suppose that the radius of the ball at the origin is contained in [t,,ti+,]and that this ball is one of an infinite chain of tangent balls with decreasing radii converging to a limit a E [ti,ti+,].The branching process argument in the proof of Lemma 5.1 shows that the nth ball in this sequence has to be a member of the nth generation of a nonsurviving branching process. Hence such an infinite chain does not exist, and the proof is 0 complete.
6. GENERAL HARD SPHERE BOOLEAN MODELS
So far we have only considered one example of a hard sphere Boolean model, the dynamic lily-pond model. It is not hard to come up with some variations on the same theme: One extension is to assign i.i.d. random speeds to the radii. Another possible model is the following: Each point of X has a nonnegative random variable associated with it, its waiting time. Waiting times are i.i.d., and when the waiting time for a particular point is over, it “explodes” and creates a ball as big as possible without overlapping existing balls and points. It is possible to show, along the lines of the proofs in Section 5 , that both these models exist and that there is no percolation in either of them. Our experience with hard sphere Boolean models so far raises the following natural question to which we do not know the answer:
Question 6.1. Is it the case that no hard sphere Boolean model percolates? For d = 1 this is certainly the case, but the answer may of course be dimension dependent. We end the section with some general properties of hard sphere Boolean models. Some of the statements below might turn out to be empty (if the answer to Question 6.1 is yes for all d ) . We say that (xl, x 2 , . . . , x , ) forms a circuit if the balls S ( x j ) around xi have the property that S(x,) and S ( x , + , ) are tangent for i = 1,. . . , k - 1 and S, and S, are also tangent. When (x,, . . . ,x,) forms a circuit, we call k the length of the circuit. Proposition 6.2.
(i) There are a.s. no circuits of even length in any hard sphere Boolean model. (ii) Each component in a hard sphere Boolean model contains at most one circuit a .s. (iii) A n unbounded component in a hard sphere Boolean model contains no circuit a .s. Proof. To prove (i), we consider circuits of length 4, as the argument can be easily generalized to other even lengths. Consider an ordered triple { x l , x , , x3} of points of X and suppose we want to create a circuit (xl, x 2 , x 3 , y ) for some y E X . Denote the ball around x i by Si. Once the radius of S, is given, the radii of S, and
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S, are completely determined. (Note that the position of the points could be such that no circuit is possible to begin with.) Increasing the radius of S , by r means that the radius of S, decreases by r, and the radius of S, increases by r. This means that independently of the radius of S , , a circuit (x, ,x2, x 3 , y ) is only possible (if at all) if y lies on a particular (d - 1)-dimensional curve. Hence, for each triple of Poisson points, the conditional probability that these three points are part of a circuit of length 4 is zero whence no circuits of length 4 exist. To prove (ii), note that a similar argument shows that for a given ordered set of points { x l , . . . ,x ~ ~ +either ~ } , no circuit consisting of these points is possible or a circuit is possible for exactly one choice of the radii. It now follows essentially as in the first part of the proof of Theorem 5.2 that no two circuits can belong to the same connected component. For (iii), suppose that there is a positive probability that a circuit is part of an unbounded component. Then there is an integer k such that circuits of length k appear in unbounded components with positive probability, and hence there must be a positive density of such circuits in space. For each such circuit, we pick one point at random (with equal probability) thereby obtaining a stationary subset L of X . For each x E L , K i ( x ) is defined to be the set of the i nearest points to x which are in the same (unbounded!) component. An application of Lemma 2.3 0 now yields the desired contradiction.
ACKNOWLEDGMENT
We would like to thank Mathew Penrose for a number of valuable comments and suggestions, in particular the suggestion to use the comparison with branching random walk and oriented percolation to prove Theorem 3.1.
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[lo] R. W. J . Meester and R. Roy, Continuum Percolation, Cambridge University Press, Cambridge, 1996. [ll] C. M. Newman, Y. Rinott, and A. Tversky, Nearest neighbors and Voronoi regions in certain point processes, A d v . Appl. Probab., 15, 726-751 (1983). [12] M. D. Penrose, Continuum percolation and Euclidean minimal spanning trees in high dimensions, Appl. Probab. (1996), to appear. Received March 8, 1995 Accepted April 3, 1996