Chapter 24

Commutators

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Problem 230

The only scalar commutator is 0.

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Remark.
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A commutator is an operator of the form PQ−QPPQ - QP, where PP and QQ are operators on a Hilbert space.

Problem 231

If {Pn}\left\{ P_n\right\} and {Qn}\left\{ Q_n\right\} are bounded sequences of operators (i.e., if there exists a positive number α\alpha such that ∥Pn∥≤α\left\| P_n\right\| \leq \alpha and ∥Qn∥≤α\left\| Q_n\right\| \leq \alpha for all nn), and if the sequence {PnQn−QnPn}\left\{ P_n Q_n - Q_n P_n\right\} converges in the norm to an operator CC, then C≠1C \neq 1.

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Remark.
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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 {Pn}\left\{ P_n\right\} and {Qn}\left\{ Q_n\right\} of operators such that ∥1−(PnQn−QnPn)∥→0\left\| 1-(P_nQ_n-Q_nP_n)\right\| \to 0 as n→∞n \to \infty (see also Problem 235). The result below is a more modest, more easily accessible version of that fact.

Problem 232

If PP and QQ are operators, if C=PQ−QPC = PQ - QP, and if CC commutes with PP, then CC is quasinilpotent.

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Problem 233
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(a)

If C=PQ−QPC = PQ - QP and if PP is hyponormal (hence, in particular, if PP is an isometry, or if PP is normal), then ∥1−C∥≥1\left\| 1-C\right\| \geq 1.

(b)

If CC commutes with PP, then ∥1−C∥≥1\left\| 1-C\right\| \geq 1.

Remark.
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If the underlying Hilbert space is finite-dimensional, then it is an easy exercise in linear algebra to prove that ∥1−C∥≥1\left\| 1-C\right\| \geq 1 for all commutators CC.

Problem 234

Every operator with a large kernel is a commutator.

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Remark.
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A commutator is an operator of the form PQ−QPPQ - QP, where PP and QQ are operators on a Hilbert space. A subspace M\mathbf{M} of a Hilbert space H\mathbf{H} is called large if dim⁡M=dim⁡H\dim \mathbf{M} = \dim \mathbf{H}; an operator is said to have a large kernel if its kernel, as a subspace, is large.

Problem 235

If an operator AA on a separable Hilbert space is not a scalar, then the infinite direct sum A⊕A⊕A⊕⋯A \oplus A \oplus A \oplus \cdots is a commutator.

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Problem 236

A positive self-commutator cannot be invertible.

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Remark.
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The self-commutator of an operator AA is the operator A∗A−AA∗A^{*}A - AA^{*}.

Problem 237
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(a)

If AA is an abnormal operator of norm 11, such that A∗A−AA∗A^{*}A - AA^{*} is a projection, then AA is an isometry.

(b)

Does the statement remain true if the norm condition is not assumed?

Remark.
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An operator is called abnormal if no non-zero subspace of the kernel of A∗A−AA∗A^{*}A - AA^{*} reduces AA (that is, if it has no normal direct summand).

Problem 238

If H\mathbf{H} is an infinite-dimensional Hilbert space, then a necessary and sufficient condition that a scalar α\alpha acting on H\mathbf{H} be a multiplicative commutator is that ∣α∣=1\left|\alpha \right| = 1.

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Remark.
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The word "commutator" occurs in two distinct mathematical contexts. In ring theory it means PQ−QPPQ - QP (an additive commutator); in group theory it means PQP−1Q−1PQP^{-1}Q^{-1} (a multiplicative commutator).

Problem 239

On an infinite-dimensional Hilbert space every unitary operator is a multiplicative commutator.

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Remark.
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A multiplicative commutator is an operator of the form PQP−1Q−1PQP^{-1}Q^{-1}, where PP and QQ are operators on a Hilbert space. Problem 238 shows that if H\mathbf{H} is an infinite-dimensional Hilbert space, then a necessary and sufficient condition that a scalar α\alpha acting on H\mathbf{H} be a multiplicative commutator is that ∣α∣=1\left|\alpha \right| = 1; the present problem strengthens that result.

Problem 240

What is the commutator subgroup of the full linear group of an infinite-dimensional Hilbert space?

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Remark.
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The commutator subgroup of a group is the smallest subgroup that contains all elements of the form PQP−1Q−1PQP^{-1}Q^{-1}; 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.