A unital $C^*$-Algebra $B$ is called stably finite, if, for all $n\in\mathbb{N}$, $1_{M_n}\in M_n(B)$ is a finite projection.

Claim: If $B$ is a unital, simple, stably finite $C^*$-Algebra, then $B\otimes \mathcal{K}$ contains no infinite projections.

I want to prove this, but I don't really know, how ( I only know the definition)

Here, $\mathcal{K}$ denotes the compact operators on a separable, infinite dimensional Hilbert space. Write $\mathcal{K}=\varinjlim M_n$ and $M_n(B)=B\otimes M_n$, then $B\otimes \mathbb{K}=B\otimes \varinjlim M_n=\varinjlim (B\otimes M_n)$ and every projection in $B\otimes \mathbb{K}$ can be approximated by projections in $M_n(B)$. But now, I only know that $1_{M_n}\in M_n(B)$ is finite for all $n$, so that at this point I don't really know how to continue.

I have seen this question Stably Finite C$^\ast$-algebras (but I don't know how to check $(2)\Rightarrow (1)$). In the recommended book, I just found proposition V.2.1.8: That if $(A_i,\phi_{i,j})$ is an inductive system of unital $C^*$-algebras (with the $\phi_{i,j}$ unital) and $A=\varinjlim (A_i,\phi_{i,j})$ and each $A_i$ is stably finite, then $A$ is stably finite. (Note: A nonunital $C^*$-Algebra $E$ is stably finite, if it's minimal unitization $E^+$ is stably finite) However, in the inductive limit for $\mathcal{K}$ the connecting maps are not necessarily unital and, if you want to apply this proposition, the question is, how to prove: $1\in M_n((B\otimes \mathcal{K})^+)$ is finite for all $n$, $\Rightarrow$ $B\otimes \mathcal{K}$ contains no infinite projections.

I appreciate any further hints or help to prove the claim.


Thinking again about this question I am not anymore sure if there is a direct argument via projections. For the argument in your reference question I was thinking that infiniteness of a projection is invariant under Murray von Neumann equivalence. If I will find an argument involving this this idea, I will post it here.

However, using scaling elements (V.2.2.8) one can prove this result quite fast.

By V.2.3.6 for a simple C*-algebra containing an infinite projection is equivalent to containing a scaling element. However, if $B \otimes \mathbb K$ contains a scaling element, then so does $B \otimes M_n$ for some $n$.

  • $\begingroup$ I understand, thank you very much! $\endgroup$ – Sabrina G. Aug 3 '17 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.