My former student Detelin Dosev and I are interested in classifying the commutators in $L(X)$, the bounded linear operators on the Banach space $X$ (see our joint paper on my home page or the ArXiv and the references therein, and recall that $T$ is a commutator means that there are $A$ and $B$ in $L(X)$ so that $T=AB-BA$).

If $T$ is a finite rank operator in $L(X)$ with zero trace then it is classical that $T$ is a commutator. The converse is obviously true when $X$ is finite dimensional.

Conjecture. There is an infinite dimensional Banach space $X$ such that all finite rank commutators in $L(X)$ have zero trace.

If true, this conjecture is very hard to prove, but maybe it is known to be false?

We are also interested in conditions on $X$ that guarantee that all finite rank operators in $L(X)$ are commutators. If $X$ is isomorphic (meaning linearly homeomorphic) to its closed subspaces of codimension one, then rank one projections in $L(X)$ are commutators and hence every finite rank operator in $L(X)$ is the sum of two commutators, but we do not know whether every finite rank operator in $L(X)$ is a commutator. If also $X$ contains a complemented subspace isomorphic to $X \oplus X$, then every diagonalizable finite rank operator in $L(X)$ is a commutator, but we still don't know whether all finite rank operators in $L(X)$ are commutators. Are we missing something easy?

A Banach space $X$ is said to have a Pelczynski decomposition provided $X$ is isomorphic to $(X \oplus X \oplus X \oplus ...)_p$ for some $1 \le p \le \infty$ or $p = 0$. A general result is that if $X$ admits a Pelczynski decomposition, then every compact operator on $X$ is a commutator, so standard spaces do not give examples of finite rank non commutators.