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I am currently writing my thesis on quasidiagonality of nuclear separable C$^\ast$-algebras and came across the so-called Kirchberg-Blackadar problem, which asks whether every separable (nuclear) stably finite C$^\ast$-algebra is quasidiagonal. I know why quasidiagonality must imply stable finiteness. However, my intuition concerning stable finiteness is lacking. Is there any consequence of this property that truly reflects why one would look for it? Telling my why finiteness is nice would suffice (stable finiteness in my eyes makes finiteness pass to any matrix algebra).

In short: Everyone seems to want stable finiteness, but why?

Thanks in advance!

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    "which asks whether every separable (nuclear) stably finite C∗∗-algebra" is not a sentence, something is missing ( or I am very tired ). – 2017-02-26
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    I think the OP meant to add "is quasidiagonal" at the end of that sentence. – 2017-02-27
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    I did, sorry for the confusion! – 2017-02-27

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I am no expert, but my understanding is that stable finiteness is directly linked to the existence of (normalized, quasi-) traces.

Firstly, one has the following theorem of Cuntz and Handelman

A unital, stably finite $C^{\ast}$-algebra $A$ has a dimension function, and if $A$ is simple, then the converse holds.

A dimension function is a state on the Cuntz semigroup $W(A)$ of $A$. The existence of dimension functions is equivalent to the existence of normalized quasi-traces. If $A$ is exact, every quasi-trace is a trace by a theorem of Haagerup. Hence,

If $A$ is simple and exact, then it is stably finite if and only if it has a normalized trace.

Finally, there is a bijection between the set of states on $K_0(A)$ and the set of quasi-traces on $A$. Hence, quasi-traces help understand the order structure on both $W(A)$ and $K_0(A)$