Commutators
[11/100%]The only scalar commutator is 0.
A commutator is an operator of the form , where and are operators on a Hilbert space.
If and are bounded sequences of operators (i.e., if there exists a positive number such that and for all ), and if the sequence converges in the norm to an operator , then .
The sum of a compact operator and a non-zero scalar is not a commutator. (Proof: in the quotient of the algebra of all operators by the ideal of compact operators, the unit element is not a commutator, by the Wintner and Wielandt proofs of Problem 230; translated back to operators, this means the identity cannot equal the sum of a commutator and a compact operator, and a fortiori neither can a non-zero scalar plus a compact operator, since dividing by that scalar reduces to the previous case.) The Brown--Pearcy characterization of commutators (see Problem 230) implies that, conversely, on a separable Hilbert space every operator that is not the sum of a non-zero scalar and a compact operator is a commutator; in particular it implies that the identity is a limit of commutators, i.e., that there exist sequences and of operators such that as (see also Problem 235). The result below is a more modest, more easily accessible version of that fact.
If and are operators, if , and if commutes with , then is quasinilpotent.
If and if is hyponormal (hence, in particular, if is an isometry, or if is normal), then .
If commutes with , then .
If the underlying Hilbert space is finite-dimensional, then it is an easy exercise in linear algebra to prove that for all commutators .
Every operator with a large kernel is a commutator.
A commutator is an operator of the form , where and are operators on a Hilbert space. A subspace of a Hilbert space is called large if ; an operator is said to have a large kernel if its kernel, as a subspace, is large.
If an operator on a separable Hilbert space is not a scalar, then the infinite direct sum is a commutator.
A positive self-commutator cannot be invertible.
The self-commutator of an operator is the operator .
If is an abnormal operator of norm , such that is a projection, then is an isometry.
Does the statement remain true if the norm condition is not assumed?
An operator is called abnormal if no non-zero subspace of the kernel of reduces (that is, if it has no normal direct summand).
If is an infinite-dimensional Hilbert space, then a necessary and sufficient condition that a scalar acting on be a multiplicative commutator is that .
The word "commutator" occurs in two distinct mathematical contexts. In ring theory it means (an additive commutator); in group theory it means (a multiplicative commutator).
On an infinite-dimensional Hilbert space every unitary operator is a multiplicative commutator.
A multiplicative commutator is an operator of the form , where and are operators on a Hilbert space. Problem 238 shows that if is an infinite-dimensional Hilbert space, then a necessary and sufficient condition that a scalar acting on be a multiplicative commutator is that ; the present problem strengthens that result.
What is the commutator subgroup of the full linear group of an infinite-dimensional Hilbert space?
The commutator subgroup of a group is the smallest subgroup that contains all elements of the form ; in other words, it is the subgroup generated by all (multiplicative) commutators. The set of all invertible operators on a Hilbert space is a multiplicative group; in analogy with standard finite-dimensional terminology, it may be called the full linear group of the space.