3
$\begingroup$

During the study of quasidiagonal C$^\ast$-algebras, I came across stably finite ones (the former property implies the latter). When I looked up the definition of stably finite C$^\ast$-algebras I saw several version. I presume these are equivalent, at least to some extend. They are:

(1) A C$^\ast$-algebra $A$ is finite if all projetions are finite and $A$ admits an approximate unit of projections. As such, $A$ is stably finite provided $A\otimes \mathbb{K}$ is finite (compact operators on an infinite dimensional seaprable Hilbert space).

(2) A unital C$^\ast$-algebra is finite if $1_A$ is finite and a nonunital one is finite if its minimal unitization is finite. As such, we call $A$ stably finite provided $M_n(A)$ is finite for every $n\in \mathbb{N}$.

To me (1) is seemingly more general, although I doubt it. I am mostly confused about the stably finiteness condition. It seems that (2) should imply (1) by regarding the stabilization as the inductive limit over matrix algebra $M_n(A)$ with the canonical embeddings. Also, this seems odd since $A\otimes \mathbb{K}$ is not unital even if $A$ is.

I am probably just confused, so any comments or references etc. are greatly appreciated.

$\endgroup$
2
$\begingroup$

Indeed, (1) is a more general notion of finiteness. There are examples when $1_{A^\sim} \in A^\sim$ (the unitization of $A$) is finite, but not every projection in $A$ is finite. For simple C$^\ast$-algebras these notions are equivalent.

To see that $A \otimes \mathbb K$ is finite, you have to use that projections in $A \otimes \mathbb K$ can be approximated by projections in $\bigcup_{\mathbb N} M_n(A)$.

The details can be found in V.2.2.1 and V.2.2.2 of Blackadar's book "Theory of C$^\ast$-algebras and von Neumann algebras".

$\endgroup$
  • $\begingroup$ Lovely, thank you for the clarifying and the reference :) $\endgroup$ – Munk Dec 30 '16 at 12:50

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.