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 XoplusX, 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 (XoplusXoplusXoplus...)p for some 1lepleinfty 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.
No comments:
Post a Comment