Solution of the Similarity Problem for Cyclic Representations of C ∗ - Algebras

Abstract

In a paper from 1955 [13] Kadison considered the following problem: Is any bounded, non-selfadjoint representation 77 of a C*-algebra A on a Hilbert space H similar to a *-representation?; i.e. does there exist an invertible operator T E B(H), such that ThT( )T-' is a *-representation of A? This problem is a natural counterpart to a problem considered by Dixmier in 1950 [9]: Is any bounded representation of a group G on a Hilbert space similar to a unitary representation? The group question was solved in the negative by Kunze and Stein in 1960 [14], which in turn was based on previous work of Ehrenpreis and Mautner [21]. The similarity problem for representations of C*-algebras is still open, although a number of partial results have been obtained. Using the deep results of Connes on infective von Neumann algebras [7] and the characterization of nuclear C*-algebras given in [3], one gets the result fairly easily that every bounded representation of a nuclear C*-algebra is similar to a *-representation (cf. [2, Th. 3.5] and [5, Th. 4.1]). The similarity problem for bounded cyclic representations of arbitrary C*-algebras was first considered by Barnes in [1]. He proved that when 77 is a bounded representation of a C*-algebra A on a Hilbert space H. such that 7V(A) and v((A)* have a common cyclic vector I, then there exists a closed, infective operator T on H. such that T and T' are densely defined, and such that x -> ThT(x)T-' is a *-representation of A. Bunce sharpened this in [2] by proving that T can be chosen bounded, but not necessarily with bounded inverse. He also removed the condition that ( is cyclic for 7v(A)*. Finally, Christensen proved in [5] that every irreducible bounded representation of a C*-algebra is similar to a *-representation. These two results of Bunce and Christensen rely on an inequality due to Pisier [16, Cor. 2.3] stating that, when S is a bounded linear map from a C*-algebra A into a C*-algebra B. then for all n E N and a. an E A: