A counterexample for commutation in tensor products of $C\sp{\ast} $-algebras

Abstract

Let A, B, C, D be C*-algebras with A C C and B C D. Tomiyama [10, p. 291 has raised the question as to whether (A ? B)C = AC C) Bc. In this context ? denotes the spatial tensor product and (A ? B)C (respectively Ac, BC) is the relative commutant of A ? B (respectively A, B) in C ? D (respectively C, D). It is easy to see that the inclusion A c ? Bc c (A ?) B)c is always valid. In the special case where C has an identity 1, A = C1 and B = D, the question has an affirmative answer [7, Theorem 1], and the result has been generalized to the case of an arbitrary C*-tensor norm [1], [4]. It is therefore tempting to conjecture that the question has an affirmative answer at least in the case where A = C1 (but B is an arbitrary C*-subalgebra of D). In this note we present a counterexample based on results of Choi [5], Wassermann [13] and Voiculescu [11]. We begin by recalling some facts about slice maps [9]. Let A and B be C*-algebras and let 4 E A*. The right slice map R,,: A ? B -* B is the unique bounded linear mapping with the property that R,p(a ? b) = 4(a)b (a E A, b E B). A triple (A, B, J), where J is a closed two-sided ideal of B, is said to verify the slice map conjecture [12] if whenever x GE A ? B and R,,(x) E J for all 4 E A* then x E A ? J. It is well known that (A, B, J) verifies the slice map conjecture if and only if A ? J is the kernel of the canonical *-homomorphism ,u: A ? B -> A ? (B/J). This is because ker t = {x E A ? BIR,,(x) E J for all4) E A*).