# Abelian C*-algebras are stably finite?

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 $$M_n(C(X,A))\cong M_n\otimes C(X)\otimes A\cong C(X)\otimes M_n(A)\cong C(X,M_n(A))$$).

• 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. Dec 26, 2020 at 22:12
• @Aweygan thank you very much! Dec 26, 2020 at 22:14

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.