Is it true to say that every nontrivial idempotent in the [Cuntz algebra $\mathcal{O}(n)$](https://en.wikipedia.org/wiki/Cuntz_algebra) is a commutator element?(Or a linear combination of commutator elements?)



Is every nontrivial idempotent in the Cuntz algebra, a commutator element?
There are a few things you could do to simplify this. 


* Kaplansky's theorem ([Reference needed for: every idempotent in a C\*-algebra is similar to a hermitian one](https://mathoverflow.net/questions/16943/reference-needed-for-every-idempotent-in-a-c-algebra-is-similar-to-a-hermitian)) lets you show that every idempotent in a $C^\*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}\_n$ is a linear combination of commutator elements, because similarity preserves the property of being a commutator or linear combination of commutators.
* In $\mathcal{O}\_n = C^\*(s\_1,\ldots,s\_n)$ (here $s\_1,\ldots,s\_n$ are the usual generating isometries), the projections $s\_i s\_i^\*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s\_i$ and then the commutator $[s\_i^\*,s\_i]$ is $Q\_i=\sum\_{k \neq i} s\_k s\_k^\*$. The set of all these projections $\{Q\_i\}\_{i =1}^n$ forms a basis for the vector space spanned by $\{s\_1 s\_1^\*, \ldots, s\_n s\_n^\*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
* Every projection in $\mathcal{O}\_n$ is in the same $K\_0$-class as a projection of the form $s\_1 s\_1^\* + \ldots +s\_j s\_j^\*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K\_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K\_0$-class works.


Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.


