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Let $A$ and $B$ be $C^*$-algebras. Let $A_0$ be a dense *-subalgebra of $A$, and let $B_0$ a dense *-subalgebra of $B$. Assume that $A_0$ is isomorphic to $B_0$.

Finally, assume that $K_0(A)$ is finitely generated. Must $K_0(B)$ also be finitely generated?

The isomorphism between $A_0$ and $B_0$ is an algebraic isomorphism between algebras over the field of complex numbers; it need not be a $*$-isomorphism.

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    $\begingroup$ If $A_0$ and $B_0$ are not necessarily isomorphic as $*$-algebras, then in what category are they assumed isomorphic? $\endgroup$
    – LSpice
    Commented Mar 3, 2018 at 18:28
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    $\begingroup$ How about the following. Take $A_0 = B_0 = {\bf C}G$ (the group algebra on the group $G$), and suppose that $G$ is not amenable. Then the maximum and minimum norms yield non-isomorphic C*-algebras $A$ and $B$ respectively. I am not that familiar with the horde of results in this area, but there presumably are examples where one of them has merely ${\bf Z}$ as its $K_0$, and the other one has lots of projection-equivalence classes, enough to guarantee non-finite generation. $\endgroup$
    – David Handelman
    Commented Mar 3, 2018 at 21:28
  • $\begingroup$ Of course what I meant by maximum and minimum norms were the full and regular representation norms, but that occurred to me just more than five minutes after ... $\endgroup$
    – David Handelman
    Commented Mar 3, 2018 at 21:34
  • $\begingroup$ It's certainly your right to rollback edits to the question, but why insist on $C*$ in place of $C^*$, and why the '---'s? $\endgroup$
    – LSpice
    Commented Mar 3, 2018 at 21:57
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    $\begingroup$ When a dense subalgebra is invariant under analytic functional calculus, its K-theory is isomorphic to that of the ambient algebra, so I suggest you check whether that property holds on the examples you are interested. $\endgroup$
    – Ruy
    Commented Mar 8, 2018 at 0:56

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No. To obtain a counterexample, you just need a C*-algebra $A$ with finitely generated K-theory and a quotient $A/I$ of $A$ which does not have finitely generated K-theory and the quotient algebraically preserves some dense subalgebra of $A$. Here is the easiest example of this situation that I could think of.

Let $A=C[0,1]$ and let $B$ be the C*-algebra of convergent sequences. Then $K_0(A)$ is finitely generated while $K_0(B)$ is not. Define the surjective map $\pi:A\rightarrow B$ by $\pi(f)(n)=f(1/n).$ Let $A_0\subseteq A$ be polynomials and let $B_0=\pi(A_0),$ which is a dense subalgebra of $B.$ Since $\pi$ is injective on $A_0$ it defines a *-isomorphism between $A_0$ and $B_0.$

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We know that $K_0(A) \neq K_0(B)$ in general. Take for example, what Handelman pointed out, the full and reduced group $C^*$-algebras of a group $G$ which is not $K$-amenable, i.e. $K_0(C^* G) \neq K_0(C^*_r G)$.

It is fair to assume that we may find an example where $K_0(A)=0$ but $K_0(B) \neq 0$.

If you accept this, then you immediately have a counterexample.

Take $a_0:=\bigoplus_{\mathbb{N}} A_0$ with two distinct closures $a:=\bigoplus A$ ($C^*$-direct inifinite sum of $A$) and similarly $b:=\bigoplus B$.

Then $a_0$ is dense in $a$ and $b$.

But $K_0(a) = \bigoplus K_0(A) =0$ is finitely generated, and $K_0(b)= \bigoplus K_0(B)$ is infinitely generated.

That is why I am confident that your question is answered to the negative.

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