Spherical functions andconformal densities on spherically symmetric CAT (�1){spacesby

Michel Coornaert and Athanase Papadopoulos *Institut de Recherche Math�ematique Avanc�eeUniversit�e Louis Pasteur et CNRS7, rue Ren�e Descartes, 67084 Strasbourg Cedex France

Abstract.| Let X be a CAT (�1)� space which is spherically symmetric around somepoint x0 2 X and whose boundary has �nite positive s�dimensional Hausdor� measure.Let � = (�x)x2X be a conformal density of dimension d > s=2 on @X. We prove that �x0 isa weak limit of measures supported on spheres centered at x0. These measures are expressedin terms of the total mass function of � and of the d�dimensional spherical function onX. In particular, this result proves that � is entirely determined by its dimension and itstotal mass function. The results of this paper apply in particular for symmetric spaces ofrank one and semi-homogeneous trees.

x0.|Introduction

Let X be a CAT (�1){space and let @X be its boundary. A conformal density ofdimension d on @X is a family � = (�x)x2X of �nite positive Borel measures on @X,such that for every x and y in X, �y is absolutely continuous with respect to �x, withRadon-Nikodym derivative given by d�yd�x (�) = jd(x; y; �) for all � 2 @X: Here, j(x; y; �)stands for the in�nitesimal distorsion at the point � of the visual metric on @X seen fromy with respect to the visual metrics seen from x (see Section 1 for the precise de�nitions).The total mass function of � is the function �� : X ! RI + de�ned by ��(x) = �x(@X).

For every x in X, we denote by Hsx the s�dimensional Hausdor� measure on @X withrespect to the visual metric seen from X.

* The second author is also supported by the Max-Plank-Institut f�ur Mathematik(Bonn).1

In this paper, we shall always assume that there exists a point x0 2 X and a realnumber s � 0 such that the following two properties are satis�ed:(0:1) X is spherically symmetric around x0 2 X. This means that the group Kx0 ofisometries of X which stabilizes x0 acts transitively on each sphere centered at this point.(0:2) For some (or, equivalently for every) x 2 X, we have

0 < Hsx(@X) <1:We note right away that condition (0:2) implies that s is the Hausdor� dimension of@X with respect to any of the visual metrics.If � is a discrete subgroup of X, conformal densities play a central role in the studyof the ergodic properties of the action of � on @X. The results of this paper concernconformal densities on the boundary of a space X satisfying conditions (0:1) and (0:2).For each d 2 RI , we de�ne the spherical function �d : X �X ! RI +, by the formula

�d(x; y) =Z

�2@Xjd(x; y; �) dHsx(�)

for every x and y in X.For every R � 0, let S(x0; R) denote the sphere in X of radius R centered at x0. Wede�ne !x0;R to be the natural measure on S(x0; R) which is induced by the Haar measureon Kx0 , the Haar measure being normalized to have total mass one.Let � be a conformal density of dimension d on @X. For every R � 0, we de�ne themeasure �x0;R on S(x0; R) by the formula

�x0;R = Hsx0(@X)cd;R ��!x0;R ;

where cd;R is the value taken by the function �d(x0; :) on S(x0; R). (The fact that X isspherically symmetric around x0 implies that the value of �d(x0; y), as a function of y,depends only on the distance from x to x0.)In this paper, we prove (Theorem 4.3) the following

Theorem A.| Let X be a CAT (�1)�space satisfying properties (0:1) and (0:2). Then,there exists a sequence (Rn) of real numbers, with Rn !1 as n!1, such that for everyconformal density � = (�x)x2X of dimension d on @X with d > s=2, the measure �x0 isthe limit, in the sense of weak convergence, of the sequence of measures �x0;Rn .Let us remark that the conformal measure � on @X is determined by its value at anarbitrary point x in X and the dimension d, so that the conformal density � is completelydetermined by �x0 and d.

2

We state now a few results which follow from Theorem A.>From previous work on the subject, it turns out that it is an interesting question toknow under which conditions a conformal density of a given dimension is determined byits total mass function, and there are several results in this direction. For example, bya result of D. Sullivan (see [Sul 3]), in the case where X = HI n (the hyperbolic n{space)a conformal density of dimension � (n � 1)=2 is uniquely determined by its total massfunction. The same result holds in the case where X is a homogeneous tree of degree k,provided that the dimension of the conformal density is � 1=2 log(k � 1) (see [CP 1]). Inthis paper, we obtain, as a consequence of Theorem A, the following

Corollary B.| Let X be a CAT (�1)�space satisfying (0:1) and (0:2) and let � =(�x)x2X and �0 = (�0x)x2X be two conformal densities on @X having the same dimen-sion d, with d > s=2. If the total mass functions of � and �0 are equal, then � = �0.Suppose now that X is a CAT (�1)�space which is spherically symmetric aroundeach of its points, that is, satisfying (0:1) for every point x0 in X. Suppose also that Xsatis�es condition (0:2). With these hypotheses, we have proved in [CP 2] the existenceof a transformation which for every conformal density � of dimension d < s=2 on @Xassociates a conformal density �+ of dimension s� d having the same total mass functionas �. As a consequence of Corollary B, we have the following

Corollary C.| The conformal density �+ is uniquely determined by the total mass func-tion of �.We shall prove also the following (Theorem 4.4)

Theorem D.| Suppose that X satis�es (0:1) and the following property:There exist four constants r0 > 0, C1 > 0, C2 > 0 and s � 0 such that for every� 2 @X and for every 0 � r � r0, we have

C1 � Hsx0�B(�; r)�rs � C2 ;where B(�; r) denotes the sphere in @X of center � and radius r with respect to the visualmetric seen from x0. (We note that this implies property (0:2), with the same s.) Then,for every conformal density � on @X of dimension d > s=2, we have

�x0 = limR!1�x0;Rin the sense of weak convergence of measures.

We note that if a CAT (�1)�space satisfying (0:1) admits a discrete cocompact group� of isometries whose critical exponent �� is �nite, then the hypotheses of Theorem Dare satis�ed with s = �� (see [Coo] and [Bou]). In particular, this is the case if X is asymmetric space of rank one or a semi-homogeneous tree. Hence we obtain the following3

Corollary E.| Let X be a Riemannian symmetric space of rank one. Then, for everyconformal density � = (�x)x2X on @X of dimension d > s=2 , we have, for every point xin X,�x = limR!1�x;R

in the sense of weak convergence of measures.We note that if X is a Riemannian symmetric spaces of rank one, the visual metricson @X are Lipschitz-equivalent to the natural Carnot-Caratheodory metrics.It may be useful to recall the classi�cation of Riemannian symmetric spaces of rankone: These are the hyperbolic spaces HI nK where K denotes either the �eld of real numbers(with n being any integer � 1), or the �eld of complex numbers or of quaternions (andin these cases n � 2), or �nally the algebra of Cayley numbers (and here n = 2). Inall these cases, we have s = kn + k � 2, where k denotes the real dimension of K (theformula is in [Pan], and it is a consequence of a general formula in [Mit]). In a Riemanniansymmetric space of rank one, the total mass function of a conformal density of dimensiond is a positive eigenfunction of the Laplacian, with eigenvalue � = s(s� d).Let p and q be two integers which are � 2. Recall that the semi-homogeneous treeof type (p; q) is the simplicial tree T p;q where each vertex is of order p or q (that is,belongs respectively to p or q edges), and such that each vertex of order p (resp. q) isadjacent only to vertices of order q (resp. p). The tree T p;q satis�es (0:2) with s =12� log(p� 1)+ log(q� 1)� (see the example at the end of this paper). As a consequence ofTheorem D, we have therefore the following

Corollary F.| Let T p;q be a semi-homogeneous tree of type (p; q). Then, for everyconformal density � = (�x)x2X dimension d > 12� log(p�1)+ log(q�1)� on @X, we have,for every vertex x of X,�x = limR!1�x;R

in the sense of weak convergence.The outline of the paper is as follows.Section 1 contains background material on CAT (�1){spaces, with the visual metricon their boundary as de�ned in this setting by M. Bourdon.Section 2 introduces conformal densities on boundaries of CAT (�1){spaces.In Section 3, we introduce spherical functions on X and we study their basic prop-erties. In particular, we prove that they satisfy a multiplicative property (Corollary 3.6)which is analogous to a property satis�ed by the classical spherical functions associated tosymmetric spaces of rank one.In Section 4, we prove all the results stated above.In section 5, we study spherically symmetric trees, which give an interesting class ofexamples of spaces satisfying the hypotheses of Theorem D.

4

x1.|Preliminaries on CAT (�1){spaces

Let X be a metric space. The distance between two points x and y in X will bedenoted by jx�yj. A geodesic segment (resp. a geodesic ray) in X is a distance{preservingmap from [a; b] to X (resp. from [a;1[ to X), where [a; b] is a compact interval in RI .A triangle T in X consists in three points in X together with three geodesic segmentsjoining them two by two. The segments (or their images) are called the sides of the triangle.If T is a triangle in X, then a comparison triangle T 0 for T in HI 2 (the real hyperbolicplane) is a triangle in HI 2 together with a map fT : T ! T 0 which sends each side of Tisometrically onto a side of T 0.A triangle T is said to satisty the CAT (�1){inequality if for every x and y in T , wehave jx� yjX � jfT (x)� fT (y)jHI 2 ;

where fT is the map associated to a comparison triangle for T in HI 2. A metric space X isgeodesic if any two points in X can be joined by a geodesic segment. The space is said to beproper if all closed balls are compact. A CAT (�1){space is a proper geodesic metric spacein which every triangle satis�es the CAT (�1){inequality. The class of CAT (�1){spacesis a vast generalization of that of simply{connected Riemannian manifolds of sectionalcurvature � �1. It contains many singular spaces of combinatorial curvature � �1 (inparticular metric trees); see [Gro] and [Bou].CAT (�1){spaces are examples of Gromov{hyperbolic spaces, of which we recall nowthe de�nition. Let X be a metric space with basepoint x0. The Gromov product withrespect to x0 of two points x and y in X is de�ned as

(x:y)x0 = 12( jx0 � xj+ jx0 � yj � jx� yj ):

The space X is said to be Gromov{hyperbolic if there exists � � 0 such that thefollowing holds(1:1) (x:y)x0 � min �(x:z)x0 ; (y:z)x0 �� �for all x; y; z 2 X and for every choice of a basepoint x0.

Without going into the details, let us recall that a Gromov{hyperbolic space hasa canonical boundary, @X, called its Gromov boundary (see [Gro] Section 1.8 or [CDP]Chapter 2). In the case whereX is geodesic and proper, @X is realized as the set of geodesicrays up to the equivalence relation which identi�es two geodesic rays r1 : [0;1[! X andr2 : [0;1[! X if and only if supt�0 jr1(t)� r2(t)j <1 . We denote by r(1) the point in@X de�ned by the geodesic ray r, and we call it the endpoint of r.Let X be a CAT (�1){space. If x is in X and y in X (resp. in @X), then, up toreparametrization, there is a unique geodesic segment (resp. ray) joining x and y. This

5

segment (resp. ray) is denoted by [x; y] (resp. [x; y[). The Gromov product can be extendedto X [ @X �X [ @X by the following formula:(1:2) (�:�)x0 = lim(x;y)!(�;�)(x:y)x0 ;where � and � are points in X [ @X, with x converging to � and y to � respectively on[x0; �[ and [x0; �[. (To see that the limit exists, we note that if the point x0 (resp. y0) issituated beyond the point x (resp. y) on [x0; �[ (resp. [x0; �[ ) (as in Figure 1), then usingthe triangle inequality, we �nd that (x0:y0)x0 � (x:y)x0 .) The product (�:�)x0 is equal to+1 if and only if � = � 2 @X. The hyperbolicity relation (1.1) is valid for every x and yin X [ @X.

x0 xy

x0y0

�

�

Figure 1Recall that the Busemann function hr : X ! RI associated with a geodesic ray r :[0;1[! X is de�ned by hr(x) = limt!1(jx� r(t)j � t):Given x and y in X, � 2 @X, and a geodesic ray r : [0;1[! X satisfying r(1) = �,we de�ne the Busemann cocycle, as in [Bou] p. 17, by

(1:3) B(x; y; �) = hr(x)� hr(y):The value of B(x; y; �) does not depend on the choice of the geodesic ray r ending at�. (This is easily seen using the fact that if r and r0 are two geodesic rays ending at �, thedistance between the point r(t) and the image of the ray r0 tends to 0 as t tends to 1, asexplained in [Bou], p. 18.)There is a nice family of metrics (j jx)x2X on @X, which we shall call the visualmetrics, and which have been de�ned, in the setting of CAT (�1)�spaces, by Bourdon[Bou] using the formula

(1:4) j� � �jx = e�(�:�)xfor every x in X and for every � and � in @X.

The following formula is easy to prove (see [Bou] p. 26):(1:5) j� � �jy = j� � �jx e 12

�B(x;y;�)+B(x;y;�)�

6

for every x and y in X, and � and � in @X.We de�ne now, for every x and y in X and � in @X,

(1:6) j(x; y; �) = eB(x;y;�):As we shall see in Corollary 1.2, the function j(x; y; �) measures the in�nitesimal distorsionof the metric j jy with respect to the metric j jx. From the de�nitions and formula (1.5),we immediately deduce:Proposition 1.1.| For every x and y 2 X and for every � and � 2 @X, we have

(1:7) j� � �j2y = j(x; y; �)j(x; y; �)j� � �j2x:

Remark. For natural reasons, we call formula (1.7) the \ formula for the change of point ofview". It generalizes a formula in [Sul 1] for the case of hyperbolic space HI n and a formulain [Coo] which holds in the case of metric trees.Corollary 1.2.| For every x and y in X and for every � in @X, we have

j(x; y; �) = lim�!�� 6=�j� � �jyj� � �jx :

Proof.| The proof follows from Proposition 1.1, using the continuity of the functionB(x; y; �) with respect to �.The following formula is also useful:

Proposition 1.3.| For every x and y in X and for every � in @X, we havej(x; y; �) = e2(y:�)x�jx�yj:

Proof.| We use the de�nitionj(x; y; �) = ehr(x)�hr(y)

with r : [0;1[! X being the geodesic ray satisfying r(0) = x and r(1) = �. Therefore,hr(x) = 0. To compute hr(y), let us take a sequence of points xn on [x; �[ converging to�. We have: hr(y) = limn!1(jy � xnj � jx� xnj)= limn!1(�jx� xnj � jx� yj+ jy � xnj) + jx� yj= �2(y:�)x + jx� yj:

This proves the proposition.7

x2.|Conformal densities on @X

In the rest of this paper, X is a CAT (�1){space and @X is equipped with the classof visual metrics (j jx)x2X de�ned by formula (1.4).Let d be a real number. A conformal density of dimension d on @X is a family� = (�x)x2X of �nite positive Borel measures on @X such that for every x and y in X, themeasure �x is absolutely continuous with respect to �y, with Radon-Nikodym derivative

d�yd�x (�) = jd(x; y; �) 8� 2 @X:

We note that a conformal density is completely determined by its dimension and itsvalue at a given point of X, and that this value can be an arbitrary �nite positive Borelmeasure on @X.In the next proposition, we give an important example of a conformal density. Weshall make the hypothesis that the s�dimensional Hausdor� measure Hsx on @X (equippedwith the metric j jx) satis�es 0 < Hsx(@X) <1; and for this we note right away that sincefor all x and y in X, the metrics j jx and j jy on @X are Lipschitz{equivalent, then thiscondition does not depend on the choice of the basepoint x.

Proposition 2.1.| Suppose that 0 < Hsx(@X) <1: Then Hs = (Hsx)x2X is a conformaldensity of dimension s on @X.Proof.| The proof follows easily from the de�nition of the Hausdor� measure, using Propo-sition 1.2 and the continuity of the in�nitesimal distorsion j(x; y; �) as a function of �.

Given a conformal density � = (�x)x2X of dimension d, we recall that its total massfunction �� : X ! RI + is de�ned, for every x 2 X, by the formula��(x) = �x(@X):

>From the de�nition of a conformal density, we may also write, for every y 2 X,

(2:1) ��(y) =Z

�2@Xjd(x; y; �) d�x(�):

x3.|Spherical functions

Let X be a CAT (�1)�space satisfying 0 < Hsx(@X) <1; for some s � 0.8

We de�ne, for each d 2 RI , the spherical function �d : X �X ! RI + by the formula

�d(x; y) =Z

�2@Xjd(x; y; �) dHsx(�):

In other words, for a �xed x, �d(x; :) is the total mass function of the unique conformaldensity of dimension d whose value at x is the Hausdor� measure Hsx.Let x0 be a �xed point of X and Kx0 the group of isometries of X that �x x0. Thisgroup is equipped with the topology of uniform convergence on compact subsets. It is easyto see that Kx0 is compact. Note also that Kx0 acts on @X as a group of isometries withrespect to the metric j jx0 . We shall say that the space X is spherically symmetric aroundx0 if the group Kx0 acts transitively on every sphere centered at x0.We begin with the following:

Proposition 3.1.| Suppose that X is spherically symmetric around x0. Then, the groupKx0 acts transitively on @X. Converesely, suppose that Kx0 acts transitively on @X andthat every point in X belongs to some geodesic ray originating at x0. Then X is sphericallysymmetric around x0.Proof.| Suppose that X is spherically symmetric around x0 and let � and � be twoarbitrary points in @X. Consider the two geodesic rays [x0; �[ and [x0; �[, and for everyn = 0; 1; 2; :::, let xn (resp. yn) be the point of intersection of [x0; �[ (resp. [x0; �[) withthe sphere S(x0; n). For each n, there exists an isometry gn 2 Kx0 which sends xn to yn.Since Kx0 is compact, there is a subsequence of gn converging to an isometry g in Kx0 ,and it is clear that g sends � to �.

Conversely, suppose that Kx0 acts transitively on @X. Given R > 0, let x and y betwo points on the sphere S(x0; R). Consider two geodesic rays [x0; �[ and [x0; �[, containingrespectively x and y, and let g be an isometry in Kx0 sending � to �. Then g(x) = y. Thiscompletes the proof of the proposition.

We study now a few elementary properties of �d(x; :) that will be useful for us in thesequel.Proposition 3.2.| For every x and y in X and for every isometry g of X, we have

�d(x; y) = �d(gx; gy):

Proof.| This is clear from the de�nitions.Corollary 3.3.| If X is spherically symmetric around x0, then for every y and z in Xsuch that jx0 � yj = jx0 � zj, we have �d(x0; y) = �d(x0; z):

9

For every R � 0, let y be an arbitrary point on the sphere S(x0; r) and consider theorbit map � : Kx0 ! S(x0; R) by �(g) = g(y) and we let !x0;R be the image by � of theHaar measure mx0 on Kx0 (normalized by mx0(Kx0) = 1). It is clear that the de�nitionof !x0;R does not depend on the choice of the point y. We have the following:Proposition 3.4.| Assume that X is spherically symmetric with respect to x0 and let y bean arbitrary point in X, with r = jy�x0j. Then, for every conformal density � = (�x)x2Xof dimension d on @X, we have

��(x0) = Hsx0(@X)�d(x0; y)

Zz2S(x0;R)

��(z)d!x0;R(z):

Proof.|We haveZ

z2S(x0;R)��(z)d!x0;R(z) =

Z 2Kx0

��( y)dmx0( )

=Z

2Kx0

Z�2@X

jd(x0; y; �)d�x0(�)dmx0( )

=Z

2Kx0

Z�2@X

jd(x0; y; �1�)d�x0(�)dmx0( ):

We use now the following lemma:Lemma 3.5.| The measure Hsx0 is, up to a constant factor, the unique �nite positiveBorel measure on @X which is invariant by the group Kx0 .Proof.|The proof follows from general facts in measure theory. Suppose that �1 and �2are two Kx0�invariant �nite positive Borel measures on @X, and consider the measure� = �1 + �2. For every Borel subset A � @X, we can write �(A) � �1(A). Therefore,�1 is absolutely continuous with respect to � and the Radon-Nikodym derivative d�1d� is afunction on @X which is invariant under the action of Kx0 . Since Kx0 acts transitively on@X, d�1d� is constant. Therefore, the measure �1 is itself proportional to �, which impliesthat �2 is proportional to �1. This proves Lemma 3.5.

Now we can �nish the proof of Proposition 3.4. The measure on @X de�ned byf 7!

Z 2Kx0

Z�2@X

f( �1�)d�x0(�)dmx0( );

and which appears in the last integral, is Kx0�invariant, and is therefore a constantmultiple of Hsx0 . Its total mass is equal to �x0(@X), since the total mass of the measuremx0 on Kx0 is equal to one. Therefore, we obtain:10

Zz2S(x0;R)

��(z)d!x0;R(z) = �x0(@X)Hsx0(@X)

Z�2@X

jd(x0; y; �)dHsx0(�)

= ��(x0)Hx0(@X)�d(x0; y):

This proves Proposition 3.4.Corollary 3.6 (the multiplicative property of spherical functions).| We have,for every x in X and for every y 2 S(x0; R),

Zz2S(x0;R)

�d(x; z)d!x0;R(z) = 1Hsx0(@X)�d(x; x0)�d(x0; y)

Proof.|Let � be the conformal density of dimension d on @X which at the point x isequal to the measure Hsx. Then ��(w) is equal to �d(x;w) for every w 2 X, and the prooffollows from Proposition 3.4.Remark.| Let G denote the isometry group of X. We suppose that we have a G�invariantsubset V of X such that whenever (x; y) and (x0; y0) are two elements in V � V withjx� yj = jx0 � y0j, then there exists an element g in G such that g(x) = x0 and g(y) = y0.This implies in particular that G acts transitively on V and therefore the value Hsx0(@X)does not depend on the choice of x0 2 V . Thus, we can normalize this value to be equalto one. Let us note also that by Proposition 3.2, the value of �d(x; y), for x and y in V ,depends only on the distance jx� yj. Assume also that X is spherically symmetric aroundsome (and hence any) point in V . Homogeneous trees and symmetric spaces of rank onesatisfy all the required properties, with V being the set of vertices in the �rst case andwith V being the whole space in the second case. Let x0 be a basepoint in V . Considerthe function f : G! RI + de�ned, for all g 2 G, by

f(g) = �d(x0; gx0):Let us note that f(g) = f(g�1) for every g in G, since �d(:; :) on V � V dependsonly on the distance between the two variables. Note also that f is Kx0�bi-invariant. ByCorollary 3.6, we have (taking x = g�1x0 and y = hx0)

Z 2Kx0

�d(g�1x0; hx0)dmx0( ) = f(g)f(h)

which implies Z 2Kx0

�d(x0; g hx0)dmx0( ) = f(g)f(h)

11

and �nally Z 2Kx0

f(g h) = f(g)f(h);

for every g and h in G.In the case where X is a symmetric space of rank one, taking V = X, we obtaintherefore a formula in [Hel], Chapter IV, Proposition 2.2, which characterizes sphericalfunctions.The following properties of our symmetric functions will also be useful:

Lemma 3.7.| For every x, y and z in X, we have�d(x; y) � e�djy�zj�d(x; z):

Proof.| Let � be a point in @X and h the Busemann function associated with somegeodesic ray ending at �. Using the triangle inequality, we see thath(y) � h(z) + jy � zj:

Therefore,j(x; y; �) = eh(x)�h(y)

� e�jy�zjj(x; z; �):It follows that

�d(x; y) =Z

�2@Xjd(x; y; �) dHsx(�)

� e�djy�zjZ

�2@Xjd(x; z; �) dHsx(�)

= e�djy�zj�d(x; z):This proves the lemma.Proposition 3.8.| Asume that X is spherically symmetric around x0. Then, for everyy in X and for every � in @X, we have

�d(x0; y) = Hsx0(@X)Z

2Kx0jd(x0; y; �) dmx0( ):

Proof.| By Proposition 3.2, we can write, for every in Kx0 ,12

�d(x0; y) =Z

�2@Xjd(x0; y; �)dHsx0(�):

Integrating with respect to the measure mx0 on Kx0 , we obtain:�d(x0; y) =

Z 2Kx0

Z�2@X

jd(x0; y; �) dHsx0(�)dmx0( )

=Z

�2@XZ

2Kx0jd(x0; y; �) dmx0( )dHsx0(�):

Using the transitivity of the action of Kx0 on @X and the left-invariance of the Haar-measure, we see that the value of the integralZ

2Kx0jd(x0; y; �) dmx0( )

does not depend on the choice of the point �. Therefore, we can separate the variables,and we obtain�d(x0; y) =

Z 2Kx0

jd(x0; y; �) dmx0( )Z

�2@XdHsx0(�);

which proves the lemma.

x4.|Proof of the results.

In this section, we prove the results stated in the introduction. In all the section, wesuppose that X is a CAT (�1)�space which is spherically symmetric around some pointx0, and that the s-dimensional Hausdor� measure Hsx0 of @X satis�es0 < Hsx0(@X) <1;

for some s � 0.Proposition 4.1.| There exists a sequence (yn) of points in X with jx0 � ynj ! 1 asn ! 1, such that for every d 2 RI , there exists a constant C = C(d) such that for everyn = 0; 1; 2; :::, we have �d(x0; yn) � C e�(s�d)jx0�ynj:

Proof.| By Proposition 5.1 of [Fal], there exists a constant C 0 > 0 such that forHsx0�almost every �0 in @X, we have13

(4:1) lim supr!0Hsx0�B(�0; r)�rs � C 0;

where B(�0; r) is as usual the closed ball centered at �0 with radius r, with respect to themetric j jx0 .Note that by the symmetry hypothesis, (4.1) is valid for every point �0 in @X. let�0 be now a �xed point in @X. Then there exists a sequence of real numbers rn > 0(n = 0; 1; 2; :::), with rn ! 0, such that for every n, we have:

Hsx0�B(�0; rn)�rsn � C 00;where C 00 is a positive constant (slightly smaller than C 0).

We suppose now, without loss of generality, that rn < 1 for every n, and we considerthe sequence of points yn on the geodesic ray [x0; �0[ satisfying, for every n = 0; 1; 2; :::,jx0 � ynj = � log(rn):

By Proposition 1.3, we can write, for every � 2 B(�0; rn),(4:2) j(x0; yn; �) = e2(yn:�)�jx0�ynj:

Let us �nd an upper bound for (yn:�). As the space X is Gromov-hyperbolic, thereexists � � 0 such that(4:3) (yn:�) � min �(yn:�0); (�:�0)�� �:

>From the de�nition of the metric j jx0 , we have(�:�0) � � log rn = jx0 � ynj:

which implies, using the de�nition of the point yn,(yn:�0) = jx0 � ynj � (�:�0):

Therefore, the min in formula (4:3) is attained by (yn:�0) and we can write(yn:�) � (yn:�0)� � = jx0 � ynj � �:

Thus, formula (4:2) impliesj(x0; yn; �) � ejx0�ynj�2�:

14

Summing up, we have, for all n � 0,�d(x0; yn) �

Z�2B(�0;rn)

jd(x0; yn; �)dHsx0(�)

� edjx0�ynj�2�dZ

�2B(�0;rn)dHsx0(�)

� edjx0�ynj�2�dC 00rsn= edjx0�ynj�2�dC 00e�sjx0�ynj:

Therefore, we have(4:4) �d(x0; yn) � Ce�(s�d)jx0�ynj;

where C is the constant C 00e�2�d. This proves Proposition 4.1.Proposition 4.2.| Let d be a real number satisfying d > s=2 and f : X [ @X ! RI acontinuous function. For every n, de�ne the function gn : @X ! RI by

gn(�) = Hsx0(@X)�d(x0; yn)

Z 2Kx0

jd(x0; yn; �)f( yn)dmx0( ) :

where (yn) is the sequence of points de�ned by Proposition 4.1. Then the sequence (gn)converges uniformly to f on @X, as n!1.Proof.| By Proposition 3.8, we have

Hsx0(@X)�d(x0; yn)

Z 2Kx0

jd(x0; yn; �)dmx0( ) = 1:

Therefore, for every � in @X, the value of gn(�)� f(�) is equal to

Hsx0(@X)�d(x0; yn)

� Z 2Kx0

f( yn)jd(x0; yn; �)dmx0( )�Z

2Kx0f(�)jd(x0; yn; �)dmx0( ) �

= Hsx0(@X)�d(x0; yn)

Z 2Kx0

�f( yn)� f(�)�jd(x0; yn; �)dmx0( ):

The function f being continuous on X [ @X, we can �nd K � 0 such that(4:5) jf( yn)�f(�)j � �

2 whenever ( yn:�)x0 � K:15

De�ne now the subset W � Kx0 asW = f 2 Kx0 such that ( yn:�)x0 � Kg:

Then we havejgn(�)� f(�)j � Hsx0(@X)

�d(x0; yn)Z

2Kx0jf( yn)� f(�)j jd(x0; yn; �) dmx0( )

= Hsx0(@X)�d(x0; yn)

Z 2W

jf( yn)� f(�)j jd(x0; yn; �) dmx0( )

+ Hsx0(@X)�d(x0; yn)

Z 2Kx0nW

jf( yn)� f(�)j jd(x0; yn; �) dmx0( ):

Let us estimate separately each of the two terms in the last expression.For each 2W , we have

jf( yn)� f(�)j � �2 :

Therefore, we can writeHsx0(@X)�d(x0; yn)

Z 2W

jf( yn)� f(�)j jd(x0; yn; �)dmx0( )

� �2Hsx0(@X)�d(x0; yn)

Z 2W

jd(x0; yn; �)dmx0( )

� �2Hsx0(@X)�d(x0; yn)

Z 2Kx0

jd(x0; yn; �)dmx0( ):

Using Proposition 3.8, we obtainHsx0(@X)�d(x0; yn)

Z 2W

jf( yn)� f(�)j jd(x0; yn; �)dmx0( ) � �2 :

To estimate the other term, letjjf jj1 = supX[@X jf j:

Then, jf(y)� f(�)j � 2jjf jj116

for all y in X [ @X and � 2 @X.For every 2 Kx0 nW , we have, using Proposition 1.3,j(x0; yn; �) � e2K�jx0� ynj = e2K�jx0�ynj:

Therefore, we obtainjgn(�)� f(�)j � �

2 + 2Hsx0(@X)�d(x0; yn) jjf jj1ed(2K�jx0�ynj):

Using Proposition 4.1, we obtain nowjgn(�)� f(�)j � �

2 + 2CHsx0(@X)jjf jj1ed(2K�jx0�ynj)+(s�d)jx0�ynj

� �2 + 2

CHsx0(@X)jjf jj1e2Kde(s�2d)jx0�ynj:As s� 2d < 0, we have, since jx0� ynj ! 1 as n!1, e(s�2d)jx0�ynj ! 0 as n!1.Therefore there exists an integer N0 such that for all n � N0, we have

jgn(�)� f(�)j � �2 + �

2 = �:This completes the proof of Proposition 4.2.Let � be a conformal density on @X, whose dimension d satis�es d > s=2. For everyR � 0, de�ne the measure �x0;R on the sphere S(x0; R) by the formula

�x0;R = Hsx0(@X)cd;R ��!x0;R ;

where cd;R is the common value taken by the function �d(x0; :) on S(x0; R). (Note that�� is a continuous function on X and that the formula above means that �x0;R is ab-solutely continuous with respect to the measure !x0;R, with Radon-Nikodym derivativeHsx0 (@X)cd;R ��:)Theorem 4.3.| There exists a sequence (Rn)n�0 with Rn > 0 and Rn !1 as n!1such that for every conformal density � of dimension d > s=2 on @X, we have

�x0 = limn!1�x0;Rnin the sense of weak convergence of measures.Proof.| Consider the sequence of points (yn) given in Proposition 4.1, and let Rn =jx0�ynj. For any continuous function f : X [@X ! RI , let gn : @X ! RI be the associatedsequence of functions de�ned by Proposition 4.2. We have, for every n 2 NI ,

17

�x0;Rn(f) = Hsx0(@X)�d(x0; yn)

Z 2Kx0

��( y) f( y) dmx0( )

= Hsx0(@X)�d(x0; yn)

Z 2Kx0

Z�2@X

jd(x0; yn; �) d�x0(�) f( yn) dmx0( )

= Hsx0(@X)�d(x0; yn)

Z�2@X

Z 2Kx0

jd(x0; yn; �) f( yn) dmx0( )d�x0(�)

=Z

�2@Xgn(�) d�x0(�)

= �x0(gn):Now, by Proposition 4.2, the sequence (gn) converges uniformly to f on @X, as n!1.Therefore, �x0;Rn(f) converges to �x0(f). Thus, the sequence �x0;Rn converges weakly to�x0 . This proves Theorem 4.3.

Remark.|Let A be an ��dense subset of RI +. Using Lemma 3.7, we can �nd a sequence ofpoints (yn) satisfying the properties of Proposition 4.1 with jx0� ynj 2 A for all n. Hence,in Theorem 4.3, we can take Rn in A for all n.Theorem 4.4.| Suppose that X is a CAT (�1)�space which is spherically symmetricaround x0 and that the following condition holds: For some s � 0, there exist three con-stants r0 > 0, C1 > 0, and C2 > 0 such that for every � 2 @X and for every 0 < r � r0,we have(4:6) C1 � Hsx0�B(�; r)�rs � C2:

Then, for every conformal density � = (�x)x2X of dimension d > s=2, we have�x0 = limR!1�x0;R

in the sense of weak convergence of measures.Remark.| Note that condition (4.6) clearly implies that 0 < Hsx0(@X) < 1. Conversely,the condition 0 < Hsx0(@X) <1 implies that there exists r0 > 0 and C2 > 0 such that forevery 0 < r � r0, we have

Hsx0�B(�; r)�rs � C2 :for Hsx0�almost every � 2 @X (see [Fal], Proposition 5.1).

18

Proof of Theorem 4.4.|The same proof as for Theorem 4.3 goes through with the followingmodi�cations: Instead of Proposition 4.1, we use Proposition 4.5 below, and the statementof Proposition 4.2 is replaced by the corresponding statement valid for every sequence ofpoints yn tending to in�nity.Proposition 4.5.| Suppose that X satis�es the hypotheses of Theorem 4.4. Then, forevery d 2 RI , there exists a constant C = C(d) such that for every y 2 X, we have

�d(x0; y) � C e�(s�d)jx0�yj:

Proof of Proposition 4.5.| The proof follows that of Proposition 4.1, with the di�erncethat here, inequality (4:1) is, by hypothesis, valid for all r in the interval [0; r0].We deduce from Theorem 4.4 the following corollary:

Corollary 4.6| Let X be CAT (�1)�space which is spherically symmetric around somepoint x0, and admitting a cocompact isometric action of a group � such that the criticalexponent �� is �nite. Then, for every conformal density � = (�x)x2X of dimension d >��=2 on @X, we have, for every x in X,�x = limR!1�x;R

in the sense of weak convergence of measures.Proof.| By a result in [Coo] and [Bou], the space X satis�es condition (4:6) with s = ��.

As was said in x0, the hypotheses of Corollary 4.6 are satis�ed by Riemannian sym-metric spaces of rank one and semi-homogeneous trees.

x5.|Spherically symmetric trees

We conclude the paper with a description of a class of examples satisfying the hy-potheses of Theorem 4.4:Consider a simplicial locally �nite tree X which is spherically symmetric around somevertex x0. The degree of a vertex v of X therefore depends only on the distance jx0 � vj.Let p0 denote the degree of x0 and let 1 + pn be the degree of a vertex at distance n � 1from x0. It is known (see [Lyo], p. 935) that the Hausdor� dimension of @X is equal tolimn!1 1n+1Pnk=0 log pi:

Proposition 5.1.| Condition (4:6) of Theorem 4.4 is satis�ed if and only if the followingholds:(5:1) There exists C1 > 0 and C2 > 0 such that C1 � ens

p0p1:::pn � C2;19

for every n = 0; 1; :::.Proof.| Suppose �rst that (4.6) is satis�ed. We know that the Hausdor� measure Hsx0on @X, which by hypothesis is �nite and positive, is Kx0�invariant. For every vertex vin X, let B(v) denote the set of � in @X such that v 2 [x0; �[. Note that B(v) is a closedball of radius e�(n+1), where n+ 1 = jx0 � vj, with respect to the visual metric j jx0 . Bysymmetry, the Hsx0�mass of B(v) depends only on the distance jx0 � vj. Therefore, thevalue of Hsx0�B(v)� is equal to Hsx0(@X)=kn, where kn = p0:::pn is the number of pointsat distance n+ 1 from x0. Condition (4.6) now clearly implies (5.1).

Conversely, suppose that condition (5.1) is satis�ed. There is a canonical measure �on @X which is characterized by the fact that for a vertex v at distance n + 1 from x0,we have ��B(v)� = 1p0p1:::pn . Condition (5.1) implies now that there exists two constantsC 01 > 0 and C 02 > 0 such that for every closed ball B(�; r) in (X; j jx0) centered at � and ofradius r, we have, for every r � 1,

C 01 � ��B(�; r)�rs � C 02:

Let us �x a real number � 2]0; 1], and consider a covering of (@X; j jx0) by a set ofclosed balls (Bi) of radius ri � �. By the ultrametric property of the space (@X; j jx0), wecan suppose (up to taking a subcover) that all the Bi's are disjoint. We can write now:1C 02�(@X) = 1

C 02Xi �(Bi) �Xi rsi � 1

C 01Xi �(Bi) = 1

C 01�(@X):

Therefore, we have 0 < 1C02�(@X) � Hsx0 � 1C01�(@X) < 1 which implies, usingLemma 3.5, that Hsx0 is, up to a constant factor, equal to the measure �. Therefore,condition (4.6) is satis�ed.

An interesting special case of simplicial trees satisfying condition (5.1) is the case wherethe sequence p0; :::; pn is preperiodic. By this we mean that there exist two integers r � 1and l � 1 such that for every k � r, we have pk+l = pk. Then condition (5.1) is satis�edwith s = 1l Pli=1 log(pr+i). In particular, a semi-homogeneous tree of degree (p; q) (usingthe notations of x0), satis�es (5.1) with s = 12� log(p� 1)+ log(q� 1)�. Semi-homogeneoustrees admit many discrete cocompact group actions, but the construction above providesus also with examples of simplicial trees which satisfy condition (5.1) and which do notadmit any cocompact group action. For example, consider spherically symmetric treessuch that the sequence p0; p1; p2::: is preperiodic, with x0 being the unique vertex of Xhaving order p0. In this case, the isometry group of X is reduced to Kx0 . In fact, itis possible to classify spherically symmetric trees which admit discrete cocompact groupactions. Indeed, one can easily see that for such a tree, the sequence p1; p2; ::: is periodicof period N with pN = p0 � 1 and pN�k = pk for all k = 1; :::; N � 1.

20

References[Bou] M. Bourdon, Actions quasi-convexes d'un groupe hyperbolique, ot g�eod�esique,Th�ese, Universit�e de Paris-Sud, centre d'Orsay, (1993).[Coo] M. Coornaert, Mesures de Patterson{Sullivan sur le bord d'un espace hyperboliqueau sens de Gromov, Paci�c Journal of Mathematics, 159, No. 2, 241-270 (1993).[CDP] M. Coornaert, T. Delzant, A. Papadopoulos, \G�eom�etrie et th�eorie des groupes",Lecture Notes in Mathematics,vol. 1441, Springer-Verlag (1990).[CP 1] M. Coornaert and A. Papadopoulos, Positive ��harmonic functions and conformaldensities on homogeneous trees, preprint no. 94-68 , Max-Plank-Institut f�ur Math., Bonn(1994).[CP 2] M. Coornaert and A. Papadopoulos, Sur une formule de transformation pour lesdensit�es conformes au bord des CAT (�1){espaces, Comptes Rendus Acad. Sci. Paris,(1995).[Fal] K. Falconer, Fractal geometry, Mathematical Foundations and Applications, J. Wiley& Sons Ltd. (1990).[Gr] M. Gromov, \Hyperbolic groups", MSRI Publications 8, p. 75-263, Springer Verlag(1987).[Hel] S. Helgason, \Groups and geometric analysis ", Academic press, (1984).[Lyo] R. Lyons, Random walks and percolation on trees, Annals of probability, vol. 18No. 3, p. 931-958 (1990).[Mit] J. Mitchell, Carnot-Carath�eodory metrics, J. Di�erential Geometry, 21, p. 35-45(1985).[Pan] P. Pansu, Dimension conforme et sph�ere �a l'in�ni des vari�et�es �a courbure n�egative,Annales Acad. Scient. Fennicae, Series A. I. Mathematica, vol. 14, p. 177-212 (1989).[Sul 1] D. Sullivan, The density at in�nity of a discrete group of hyperbolic motions, Publ.Math. IHES 50, 171-202 (1979).[Sul 2] D. Sullivan, Discrete conformal groups, Bulletin Am. Math. Soc. (N.S.) 6, 57-73(1982).[Sul 3] D. Sullivan, Related aspects of positivity in Riemannian manifolds, J. Di�. Geom.25, 327-351 (1987) .

21