Nonregular, simple, nuclear C*-algebras

Von | September 7, 2020

(Based on the talk „Thoughts on the classification problem for amenable C*-algebras“ of George Elliott, 30. June 2020, at the Zagreb Workshop on Operator Theory.)

Let us consider the class of unital, separable, simple, nuclear C*-algebras. The Toms-Winter conjecture predicts that for such an algebra A, the following conditions are equivalent:

  1. A has finite nuclear dimension.
  2. A is \mathcal{Z}-stable, that is, A\cong\mathcal{Z}\otimes A where \mathcal{Z} denote the Jiang-Su algebra.
  3. A has strict comparison of positive elements, which means that the Cuntz semigroup of A is almost unperforated: if x and y satisfy (n+1)x\leq ny for some n\in\mathbb{N}, then x\leq y.

As of today, it is known that (1) and (2) are equivalent and that (2) implies (3). Further, it is known that (3) implies (2) under certain additional assumptions.

As long as the Toms-Winter conjecture is not completely verified, the second condition seems most natural, and we say that A is regular if it is \mathcal{Z}-stable. By the spectacular recent breakthrough in the Elliott classification program, it is known that every regular C*-algebras that satisfy the Universal Coefficient Theorem (UCT) are classified by the Elliott invariant (K-theory and tracial data).

We have an obvious dichotomy: a unital, separable, simple, nuclear C*-algebra is either regular or nonregular. It is easy to find regular algebra. Indeed, since the Jiang-Su algebra is self-absorbing, that is, \mathcal{Z}\otimes\mathcal{Z}\cong\mathcal{Z}, for every A the algebra \mathcal{Z}\otimes A is automatically regular. Examples of nonregular C*-algebras were constructed by Villadsen, Toms and Rørdam.

Question 1:

Are the regular or the nonregular C*-algebras generic?

A regular C*-algebra is either purely infinite or stably finite. A simple, unital, purely infinite C*-algebra has infinite stable rank. Rørdam showed that a stably finite, regular C*-algebra has stable rank one. Thus, a simple C*-algebra with finite stable rank \geq 2 is nonregular. Villadsen showed that for every n\in\{1,2,\ldots,\infty\} there exists a simple AH-algebra with stable rank n. He also showed that his algebra with stable rank n has either real rank n-1 or n.

Question 2:

Does there exist a nonregular C*-algebra of real rank zero?

Question 3:

What is the relation between the stable rank and real rank of a stably finite, simple (nuclear) C*-algebra? Is the real rank of A always either \mathrm{sr}(A) or \mathrm{sr}(A)-1? Is this connected to the number of tracial states on A?

Elliott suggests that \mathrm{rr}(A)=\mathrm{sr}(A)-1 should correspond to a unique (or very few) tracial state on A, while \mathrm{rr}(A)=\mathrm{sr}(A) should correspond to a large tracial simplex of A. It is known that \mathrm{rr}(B)\leq 2\mathrm{sr}(B)-1 for every C*-algebra B. Moreover, if X is a compact, Hausdorff space, then \mathrm{rr}(C(X))=\dim(X) and \mathrm{sr}(C(X))=\lfloor \frac{\dim(X)}{2} \rfloor +1. Thus, the stable rank of a commutative C*-algebra is roughly half of its real rank.

Question 4:

Can the recent classification of regular (separable, simple, nuclear) C*-algebras be extended to include (some) nonregular C*-algebras by using a stronger invariant?

A natural candidate for a stronger invariant would be the Cuntz semigroup. The Cuntz semigroup encodes the K_0-group, the simplex of tracial states, and the pairing between traces and K_0. However, it does not encode the K_1-group. One should therefore consider classification by the pair (\mathrm{Cu}(-),K_1(-)).

Elliott suggests to consider the following testcase: Let A and B be simple inductive limits of matrix algebras over the Hilbert cube. Then K_1(A)=K_1(B)=0, and the question becomes: Are A and B isomorphic whenever their Cuntz semigroups are isomorphic? If \mathrm{Cu}(A)\cong\mathrm{Cu}(B), and if A is regular, then the Cuntz semigroup of B is `regular‘ (in the sense of Winter) and it follows that both algebras are regular. Then, since A and B also satisfy the UCT, one can indeed deduce from the classification result that A\cong B.

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