Abstract
Abstract. A class of negative definite kernels is defined in terms of measure spaces. Using this concept, property (T) for a countable group Γ is characterized in terms of measure preserving actions of Γ, as follows. If a set S is translated a finite amount by any fixed element of Γ, then there is a uniform bound on how far S is translated.
Introduction
A group G has Kazhdan's property (T) if the trivial representation is isolated in the unitary dual of G [K] . We refer to [HV] for an exposition of many of the remarkable properties of such groups.
Introduction
Groups with property (T) are characterized by the absence of unbounded negative definite functions [HV, Chapter 5, Theorem 20] . This characterization was obtained in [D] , [G] and, independently, in [AW] . We use this to prove the following dynamical characterization of property (T), which is the main result of this paper. All measure spaces are positive but not necessarily finite.
Introduction
A countable group Γ has property (T) if and only if it satisfies the following condition: for every measure-preserving action of Γ on a measure space (Ω, B, µ) and every set S ∈ B such that µ(S△gS) < ∞ for all g ∈ Γ, we have sup g∈Γ µ(S△gS) < ∞.
Introduction
Remark 0.1. To see how the condition above can fail to hold, suppose that there exists a homomorphism from Γ onto Z. (There is no such homomorphism, if Γ has property (T).) Take Ω = Z, µ = counting measure, S = {1, 2, 3, . . . } and let Γ act on Ω by translation. Then µ(S△gS) < ∞ for all g ∈ Γ but sup g∈Γ µ(S△gS) = ∞.
Introduction
Date: November 7, 1996. 1991 Mathematics Subject Classification. Primary 22D10. This research was supported by the Australian Research Council.
Introduction
Definition 0.1.1. Let Γ be a countable or finite set. Say that a function f : Γ × Γ −→ [0, ∞) is a measure definite kernel if there is a measure space (Ω, B, µ) which contains sets S x ∈ B for x ∈ Γ such that f (x, y) = µ(S x △S y ).
Introduction
Remark 0.2. If f is measure definite we can find a positive measure space (Ω, µ) which contains sets S x for x ∈ Γ such that f (x, y) = µ(S x \S y ) = µ(S y \S x ). For this, simply replace Ω by Ω ′ = Ω × {0, 1} with the product measure and replace each set
measure definite kernels
We refer to [HV, Chapter 5] for the definition and properties of negative definite kernels. A negative definite kernel f is assumed to be normalized, i.e. f (x, x) = 0. A universal construction of a negative definite kernel on a set Γ is to take a mapping x → v x of Γ into a real Hilbert space H and to let the kernel f (x, y) = v x − v y 2 . Moreover, for a given f , the embedding x → v x of Γ in H is unique up to rigid motions of the closed convex hull of {v x : x ∈ Γ} [HV, Chapter 5, Proposition 14].
Proposition 1.1. Every measure definite kernel is negative definite.
Proof: Let f be a measure definite kernel. Using the notation of Definition 0.1.1, fix x 0 ∈ Γ and define a function η :
Proposition 1.1. Every measure definite kernel is negative definite.
and so f is negative definite [HV, Chapter 5.13, Exemples] .
Proposition 1.1. Every measure definite kernel is negative definite.
Every measure definite kernel is a pseudometric. The negative definite kernel defined on R by f (x, y) = (x − y) 2 is clearly not a pseudometric. The converse of Proposition 1.1 is therefore false. However we shall show below that the square root of any negative definite kernel is measure definite.
Proposition 1.1. Every measure definite kernel is negative definite.
For the rest of this section let Γ be a fixed countable set. Let Ω 00 = {0, 1}
Proposition 1.1. Every measure definite kernel is negative definite.
Γ with the product topology. Then Ω 00 is either a Cantor set or finite. Let Ω 0 = Ω 00 \{(0, 0, 0, . . . ), (1, 1, 1, . . . )} and let B 0 be the σ-algebra of Borel sets in Ω 0 . Given x ∈ Γ, let S 0 x = {c ∈ Ω 0 : c(x) = 1}. Our next result shows that B 0 is a universal σ-algebra for measure definite kernels on Γ.
Proposition 1.1. Every measure definite kernel is negative definite.
Proof: Suppose that a measure space (Ω, B, µ) is given, which contains sets S x , such that µ(S x △S y ) is finite for x, y ∈ Γ. We can and will delete from Ω all points belonging to all (respectively none) of the sets S x without changing µ(S x △S y ). Define a map φ :
Proposition 1.1. Every measure definite kernel is negative definite.
Since φ is measurable, we may define a corresponding measure
Proposition 1.1. Every measure definite kernel is negative definite.
. In this way any measure definite kernel may be expressed as µ 0 (S [R, Theorem 2.18 ], these two assertions imply that the measure µ 0 is regular.
Proposition 1.1. Every measure definite kernel is negative definite.
Let C denote the family of all cylinder sets of the form
Proposition 1.1. Every measure definite kernel is negative definite.
where {x 1 , . . . , x m , y 1 , . . . , y n } ⊂ Γ and m, n ≥ 1. Each set in C has finite measure since µ 0 (S
Proposition 1.1. Every measure definite kernel is negative definite.
follows from the fact that the family C is a countable base for the topology of Ω 0 consisting of compact open sets. Assertion (b) follows because every compact subset of Ω 0 is contained in a finite union of sets in C, and hence is contained in a finite union of sets of finite measure. on Ω 0 give rise to measure definite kernels f j which converge pointwise to f . We must show that f is measure definite. For fixed x, y ∈ Γ, f j (x, y) → f (x, y), so the restricted measures µ j | S 0
Proposition 1.1. Every measure definite kernel is negative definite.
x △S 0 y are positive and of uniformly bounded finite mass. Since the set of positive Borel measures of total mass bounded by a given fixed constant form a compact metric space, we can, by passing to a subsequence, assume that µ j | S 0
Proposition 1.1. Every measure definite kernel is negative definite.
x △S 0 y approaches some positive, finite mass Borel measure on S 0 x △S 0 y . Since Γ is countable, we may pass to a subsequence where this happens for all x, y ∈ Γ.
Proposition 1.1. Every measure definite kernel is negative definite.
Write Ω 0 = ∞ k=1 T k , where µ j | T k has a limit for each T k and where each S 0 x △S 0 y is contained in a finite union of T k 's. To see that this is possible, suppose that the elements of Γ are labelled, x 1 , x 2 , etc. Let all sets of the two forms {c ∈ Ω 0 : (c(x 1 ), . . . , c(x n−1 ), c(x n )) = (1, . . . , 1, 0)} and {c ∈ Ω 0 : (c(x 1 ), . . . , c(x n−1 ), c(x n )) = (0, . . . , 0, 1)} make up the collection (T k ) k . This collection will be countable or finite as Γ is countable or finite. The disjoint union of the T k gives all of Ω 0 . Put the limit measures on the T k together to form a measure µ on Ω 0 . For any x, y ∈ Γ, we have
Proposition 1.1. Every measure definite kernel is negative definite.
This shows that f is measure definite and so proves the result.
Proposition 1.1. Every measure definite kernel is negative definite.
′ be two tuples of vectors in two real Hilbert spaces. We say that the two tuples are in the same configuration
Proposition 1.1. Every measure definite kernel is negative definite.
′ have the same dimension, then the two tuples are in the same configuration if and only if there is an affine isometry from H to H ′ sending u j to u Proof: Suppose the result has been proved for finite Γ. In the infinite case, think of Γ as the increasing union of a sequence of finite sets, and for each of these finite sets construct a measure µ 0 satisfying the conditions of the statement insofar as they apply to elements of that set. Take limits as in the proof of Proposition 1.3 to obtain a measure µ 0 satisfying the conditions for all of Γ.
Proposition 1.1. Every measure definite kernel is negative definite.
Suppose, therefore, that Γ is finite. Replace H by the linear subspace which the the v x generate. This guarantees that H is finite dimensional. Let Ω be the homogeneous space of all half-spaces of H. Then Ω may be realized as E n /E n−1 , where E n is the (unimodular) group of rigid motions of the n-dimensional real Hilbert space H and the stabilizer of a given half space is isomorphic to E n−1 . Endow Ω with its usual measure µ, which is invariant under rigid motions of H. We can normalize µ so that µ{M ∈ Ω : v ∈ M, w / ∈ M} = v − w .
Proposition 1.1. Every measure definite kernel is negative definite.
(This follows, for example, from the explicit expression for the Haar measure on the affine group of H given in [B, Chap.VII §3 No.3 Ex.4 ]. It may also be seen geometrically.) Let
Proposition 1.1. Every measure definite kernel is negative definite.
is measure definite. This proves (i).
Proposition 1.1. Every measure definite kernel is negative definite.
To prove (ii) it suffices to show that the measure
Proposition 1.1. Every measure definite kernel is negative definite.
depends only on the configuration of (v x 1 , . . . , v xm , v y 1 , . . . , v yn ). This is clear except for a possible dependence on the dimension of H. Suppose then that
Proposition 1.1. Every measure definite kernel is negative definite.
Let Ω ′ be the space of all half-spaces of H ′ and let µ ′ be the natural measure on Ω ′ . We must prove that
Proposition 1.1. Every measure definite kernel is negative definite.
Define a measure ν on Ω by
Proposition 1.1. Every measure definite kernel is negative definite.
Since ν is invariant under rigid motions of H and satisfies
Proposition 1.1. Every measure definite kernel is negative definite.
Remarks. (i) Let Γ be the set of vertices of a tree. The usual integervalued distance function d(x, y) between elements of Γ is a measure definite kernel. To see this, fix a vertex o ∈ Γ and for each x ∈ Γ let S x = [o, x], the geodesic path between o and x. Then d(x, y) = µ(S x △S y ), where µ is Lebesgue measure on the geometric realization of the tree. (However it is easy to see that not all measure definite kernels arise from the distance function on a tree.) A. Valette [V, Section 3] has given an appropriate construction of Lebesgue measure for real trees from which it follows similarly that the distance function in a real tree is measure definite.
Proposition 1.1. Every measure definite kernel is negative definite.
(ii) It is not true that f 2 is negative definite whenever f is measure definite. Let Γ = {1, 2, 3, 4} be the set of vertices of the tree in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . We give an application of the preceding ideas to characterize countable groups with property (T) in terms of measure preserving actions on infinite measure spaces. Another characterization has been given by K. Schmidt [S, Proposition 2.10] , A. Connes and B. Weiss [CW] in terms of measure preserving actions on finite measure spaces, using the notion of strong ergodicity.
Proposition 1.1. Every measure definite kernel is negative definite.
We recall the Delorme-Guichardet characterization of groups with property (T) [HV, Chapter 5, Theorem 20] , which may be expressed as follows: a group has property (T) if and only if every left Γ-invariant negative definite kernel on Γ is bounded. Proof: Suppose that Γ has property (T), and that we are given a measure preserving action as described. Define a left Γ-invariant measure definite kernel on Γ by f (x, y) = µ(xS△yS). By Proposition 1.1, f is negative definite and so uniformly bounded, by the Delorme-Guichardet result.
Proposition 1.1. Every measure definite kernel is negative definite.
Conversely, suppose that the condition in the theorem holds, and let f be a left Γ-invariant negative definite kernel on Γ × Γ. By Proposition 1.4 √ f is measure definite with corresponding regular Borel measure µ 0 defined on the space Ω 0 = {0, 1} Γ \{(0, 0, 0, . . . ), (1, 1, 1, . . . )}. We must show that µ 0 is invariant under the natural action of Γ on Ω 0 defined by (gc)(x 0 ) = c(g −1 x 0 ). Recall that C is the family of all cylinder sets of the form
Proposition 1.1. Every measure definite kernel is negative definite.
where {x 1 , . . . , x m , y 1 , . . . , y n } ⊂ Γ and m, n ≥ 1. Observe that A is compact and open in Ω 0 .
Proposition 1.1. Every measure definite kernel is negative definite.
To pass from A to gA, one must apply g to each of the x i 's and y j 's.
Proposition 1.1. Every measure definite kernel is negative definite.
Proposition 1.4(ii) that the measures µ 0 (A) and µ 0 (g(A)) coincide. We have therefore shown that µ 0 and µ g 0 coincide on all sets in C. Suppose that the elements of G are labelled, x 1 , x 2 , etc. Let C 1 be the subset of C containing those sets of the form {c ∈ Ω 0 : (c(x 1 ), c(x 2 ), . . . , c(x n )) = (ǫ 1 , ǫ 2 , . . . , ǫ n )} where n is a positive integer, where each ǫ j is either 0 or 1, and where each of the values 0 and 1 occurs at least once.
Proposition 1.1. Every measure definite kernel is negative definite.
The sets in C 1 satisfy the following properties: (1) C 1 is a neigborhood base for Ω; (2) if two sets in C 1 are not disjoint, then one contains the other; (3) C 1 contains no infinite ascending chain.
Proposition 1.1. Every measure definite kernel is negative definite.
Let U be an open subset of Ω. It follows from the three properties of C 1 that U is the disjoint union of those sets in C 1 maximal with respect to being contained in U. Thus µ 0 (U) = µ 0 (g(U)). Since the measures µ 0 and µ g 0 are regular and they coincide on open sets, they must agree on all Borel sets. Therefore the action of Γ on Ω 0 is measure-preserving. Now write S = S 0 e . Since √ f is measure definite, µ 0 (S△gS) < ∞ for all g ∈ Γ. The condition in the theorem implies that sup g∈Γ µ 0 (S△gS) < ∞. The negative definite kernel f is therefore bounded and so Γ has property (T).
Proposition 1.1. Every measure definite kernel is negative definite.
Questions. Several questions remain.
Proposition 1.1. Every measure definite kernel is negative definite.
(i) Can measure definite kernels be characterized internally, without reference to measure spaces?
Proposition 1.1. Every measure definite kernel is negative definite.
(ii) Is there a result analogous to Theorem 2.1 for locally compact groups?
Proposition 1.1. Every measure definite kernel is negative definite.
(iii) Can the action of Γ in Theorem 2.1 be assumed to be ergodic?

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