Real rank 0 implies stable rank 1 on $C^\ast$-algebras?

Question

A $C^\ast$ algebra has defined stable rank (https://www.univie.ac.at/nuhag-php/bibtex/open_files/2079_Rieffel-StableRank.pdf) and real rank (https://core.ac.uk/download/pdf/82123484.pdf), which are two different non-commutative versions of Lebesgue covering dimension.

In particular an algebra $A$ has

• Stable rank 1 if and only if every element of $A$ can be approximated by invertibles.
• Real rank 0 if and only if every self-adjoint element of $A$ can be approximated by self-adjoint invertibles.

It's known that $rr(A)\leq 2sr(A)-1$ (see second paper). Is there any example of a $C^\ast$-algebra with real rank $0$ and stable rank $>1$? If so, is there a simple one?

Answer 1

There are "cheap" examples. If a C*-algebra $A$ has stable rank 1 then it is stably finite. On the other hand every simple, unital purely infinite C*-algebra has real rank 0. So every simple, unital purely infinite C*-algebra (for example Cuntz algebras) has stable rank bigger than 1 (infinite actually) but real rank 0.

Then the natural question is: Are there any stably finite C*-algebras with real rank 0 and stable rank bigger than 1? A C*-algebra with real rank 0 and cancellation has stable rank 1, but apparently the general case is open (at least as of 2006). Bruce Blackadar states at the bottom of page 455 in his Operator Algebras book that "There is no known example of a stably finite C*-algebra with real rank 0 that does not have cancellation (stable rank 1)."

A good account of stable and real rank and their relationship can be found in Blackadar's Operator algebra book Section V.3. Anything I mentioned above will be there as well.