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I would like to know if the following is true: Is every abelian $C^*$-algebra stably finite? I could not find this in any book, so I am a little suspicious about the following argument I came up with:

Since a $C^*$-algebra is stably finite if-f it's unitization is stably finite and using Gelfand's theorem, we have only to deal with $C^*$-algebras of the form $C(X)$, where $X$ is a compact, Hausdorff space.

Now it is a known fact that, if $A$ is a unital $C^*$-algebra, then $A$ is finite if and only if $s^*s=1_A\implies ss^*=1_A$ for all $s\in A$, i.e. every isometry is a unitary.

Let $n\geq1$. We have that $M_n(C(X))\cong C(X,M_n(\mathbb{C}))$. Now let $s\in C(X,M_n(\mathbb{C}))$ be an isometry, i.e. $s^*s=1_{C(X,M_n)}$, i.e. $s(x)^*s(x)=1_{M_n}$ for all $x\in X$. Since $M_n(\mathbb{C})$ is finite (obviously) we conclude that $s(x)s(x)^*=1_{M_n}$ for all $x\in X$, thus $ss^*=1_{C(X,M_n)}$ and this shows that $C(X)$ is stably finite.

Is this correct or am I missing something? Anyone knows of any resource that refers to this?

Edit: I can think of a second proof for the separable case (i.e. if-f $X$ is metrizable): if so, then $C(X)$ admits a faithful tracial state and then so do $M_n(C(X))$, so all of those are finite $C^*$-algebras.

And even more generally, it seems like $C(X,A)$ is stably finite whenever $A$ is (since $M_n(C(X,A))\cong M_n\otimes C(X)\otimes A\cong C(X)\otimes M_n(A)\cong C(X,M_n(A))$).

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    $\begingroup$ Yes, everything is correct, and it is true that the same proof generalizes to show that $C(X,A)$ is stably finite whenever $A$ is stably finite. $\endgroup$
    – Aweygan
    Dec 26, 2020 at 22:12
  • $\begingroup$ @Aweygan thank you very much! $\endgroup$ Dec 26, 2020 at 22:14

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Just so that the question leaves the unanswered list: Yes, commutative $C^*$-algebras are stably finite and here is a short argument:

It is a known fact that, if $A$ is a unital $C^*$-algebra, then $A$ is finite if and only if $s^*s=1_A\implies ss^*=1_A$ for all $s\in A$, i.e. every isometry is a unitary.

Let $n\geq1$. We have that $M_n(C(X))\cong C(X,M_n(\mathbb{C}))$. Now let $s\in C(X,M_n(\mathbb{C}))$ be an isometry, i.e. $s^*s=1_{C(X,M_n)}$, i.e. $s(x)^*s(x)=1_{M_n}$ for all $x\in X$. Since $M_n(\mathbb{C})$ is finite (obviously) we conclude that $s(x)s(x)^*=1_{M_n}$ for all $x\in X$, thus $ss^*=1_{C(X,M_n)}$ and this shows that $C(X)$ is stably finite.

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