Real rank of B(H) tensor B(H)

Von | Juni 20, 2020

Given a separable, infinite-dimensional Hilbert space \Huge{H}, what is the real rank of the minimal tensor product  \mathcal{B}(H)\otimes \mathcal{B}(H)?

Background/Motivation: The real rank is a noncommutative dimension theory that was introduced by Brown and Pedersen in ​[1]​. It associates to each C*-algebra A a number (its real rank) \mathrm{rr}(A)\in\{0,1,\ldots,\infty\}. The lowest and most interesting value is zero. One can think of C*-algebras of real rank zero as zero-dimensional, noncommutative spaces. (There are other noncommutative dimension theories, and they lead to different concepts of zero- or low-dimensional noncommutative spaces.) For topological spaces X and Y, the product theorem for covering dimension shows that under certain weak assumptions (for example, both spaces are metric, or compact) we have \dim(X\times Y)\leq\dim(X)+\dim(Y). One might therefore expect that \mathrm{rr}(A\otimes B)\leq\mathrm{rr}(A)+\mathrm{rr}(B), but it was shown in ​[2]​ that this does not hold: there exists C*-algebras A and B such that

     $$\mathrm{rr}(A)=\mathrm{rr}(B)=0, \text{ yet } \mathrm{rr}(A\otimes B)>0.$$

Later, Osaka showed in ​[3]​ that one can even take A=B=\mathcal{B}(H). Indeed, it is well known that von Neumann algebras have real rank zero, and hence so does \mathcal{B}(H). Osaka showed that \mathcal{B}(H)\otimes\mathcal{B}(H) does not have real rank zero. This raises the question of what the real rank of \mathcal{B}(H)\otimes\mathcal{B}(H) is, see Question 3.3 in [3].

  1. [1]
    L.G. Brown, G.K. Pedersen, C*-algebras of real rank zero, Journal of Functional Analysis. 99 (1991) 131–149.
  2. [2]
    K. Kodaka, H. Osaka, Real Rank of Tensor Products of C*-Algebras, Proceedings of the American Mathematical Society. 123 (1995) 2213–2215.
  3. [3]
    H. Osaka, Certain C*-algebras with non-zero real rank and extremal richness, MATH. SCAND. 85 (1999) 79.

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