Nonregular, simple, nuclear C*-algebras

(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:

$A$ has finite nuclear dimension.
$A$ is $\mathcal{Z}$ -stable, that is, $A\cong\mathcal{Z}\otimes A$ where $\mathcal{Z}$ denote the Jiang-Su algebra.
$A$ 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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