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A maximal abelian subalgebra A of the von Neumann algebra M is called a Cartan subalgebra if the normalizer N(A) = {unitaries u € M with uAu* = A] generates M as a von Neumann algebra (see [4] ).It is a corollary of the paper by Connes Feldman and Weiss [2] that any two Cartan subalgebras of the hyperfinite type IIj factor R are conjugate by an automorphism of R. In this paper we construct the first example of a (separable) IIj factor with two nonconjugate Cartan subalgebras, showing that the correspondence between ergodic theory (up to orbit equivalence) and von Neumann algebras is many to one.The property we shall use to distinguish two different Cartan subalgebras will be the existence of asymptotically invariant sequences (see [7]): if G is a group of automorphisms of the abelian von Neumann algebra A with finite invariant trace r, we say that G has asymptotically invariant (a.i.) sequences if there is a sequence p n of projections in A y r(p n ) = X A and lim^Jlp^ -g(p n )\\ 2 = 0 for all g G G. The full group [G] is the group of those automorphisms of A which coincide on some partition {p } of the identity in A with elements of G. Two actions of groups G and G f on A are called orbit equivalent if there is an automorphism 0, preserving r, for which 6 [G] 0~1 = [G'].The existence of a.i.sequences is an invariant of orbit equivalence.If A is a Cartan subalgebra of the II j factor M, the normalizer N(A) acts on A by conjugation.Thus the existence of a.i.sequences for this action is a conjugacy invariant for A. So it suffices to exhibit a IIj factor with two Cartan subalgebras, one with a.i.sequences and one without.To do this we use the well-known fact (see [3]) that if the Cartan subalgebra arises from the groupmeasure space construction with a countable group G acting on a space X with A = L°°(X, M), then N(A) induces [G] on A.We begin by noting that if G is countable with Kazhdan's property T then no ergodic finite measure-preserving action of G on A has a.i.sequences.For if
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Connes et al. (1982) studied this question.
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