Projections in the Cuntz algebra which have the same $K_0$ class

Question

Assume that $e,f$ are two projections in the Cuntz algebra $\mathcal{O}_n$ , which have the same $K_0$ class. Are they necessarily Murray von Neumann equivalent ?

The following post is the motivation of this question:

https://mathoverflow.net/questions/300095/is-every-nontrivial-idempotent-in-the-cuntz-algebra-a-commutator-element

Answer 1

For two projections $p,q\in A$, equivalence in $K_0(A)$ is always Murray-von Neumann equivalence. One usually defines $K_0$-equivalence to occur in the biggest of the matrix algebras containing the projections; in this case both $p,q\in M_1(A)$, so $K_0$ equivalence gives $v\in A$ with $v^*v=p$, $vv^*=q$.

And this is coherent with looking at things "higher up". If we now see the projections as $$ \begin{bmatrix}p&0\\0&0\end{bmatrix},\ \ \begin{bmatrix}q&0\\0&0\end{bmatrix} $$ and they are equivalent in $M_2(A)$, it means that there exists $V\in M_2(A)$ with $$ \begin{bmatrix}p&0\\0&0\end{bmatrix}=V^*V=\begin{bmatrix}V_{11}^*V_{11}+V_{21}^*V_{21}&V_{11}^*V_{12}+V_{21}^*V_{22}\\V_{12}^*V_{11}+V_{22}^*V_{21}&V_{12}^*V_{12}+V_{22}^*V_{22}\end{bmatrix}. $$ From the 2,2 entry we get $V_{12}=V_{22}=0$. Similarly, from $$ \begin{bmatrix}q&0\\0&0\end{bmatrix}=VV^*=\begin{bmatrix}V_{11}V_{11}^*+V_{12}V_{12}^*&V_{11}V_{21}^*+V_{12}V_{22}^*\\V_{21}V_{11}^*+V_{22}V_{12}^*&V_{21}V_{21}^*+V_{22}V_{22}^*\end{bmatrix}, $$ we now see from the 2,2 entry that $V_{21}=0$. Thus $p=V_{11}^*V_{11}$, $q=V_{11}V_{11}^*$ are Murray-von Neumann equivalent.

Note that the same argument works if $p\oplus 0_m$ is equivalent to $q \oplus 0_m$ for $m>1$.

Answer 2

The statement is not true. For $A = \mathcal O_2$ it fails for example. Indeed $K_0(\mathcal O_2) = \{0\}$ but not all projections are equivalent.

I guess Martin shows that there is an inclusion $\mathrm{Proj}(A) / \sim_\mathrm{MvN} \ \hookrightarrow \ V(A)$.