Fuglede-Putnam-Rosenblum theorem


Let A be a C∗-algebra with unit e.

The Fuglede-Putnam-Rosenblum theorem makes the assertion that for a normal element a∈A the kernel of the commutator mapping [a,-]:A→A is a ∗-closed setPlanetmathPlanetmath.

The general formulation of the result is as follows:

Theorem. Let A be a C∗-algebra with unit e. Let two normal elements a,b∈A be given and c∈A with a⁢c=c⁢b. Then it follows that a∗⁢c=c⁢b∗.

Lemma. For any x∈A we have that exp⁡(x-x∗) is a element of A.

Proof. We have for x∈A that exp(x-x∗)∗exp(x-x∗)=exp(x∗-x+x-x∗)=exp(0)=e. And similarly exp(x-x∗)exp(x-x∗)∗=e. ∎

With this we can now give a proof the Theorem.

Proof. The condition a⁢c=c⁢b implies by inductionMathworldPlanetmath that ak⁢c=c⁢bk holds for each k∈ℕ. Expanding in power seriesMathworldPlanetmath on both sides yields exp⁡(a)⁢c=c⁢exp⁡(b). This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to c=exp⁡(-a)⁢c⁢exp⁡(b). Set U1:=exp⁡(a∗-a),U2:=exp⁡(b-b∗). From the Lemma we obtain that ∥U1∥A=∥U2∥A=1. Since a commutes with a∗ und b with b∗ we obtain that

exp⁡(a∗)⁢c⁢exp⁡(-b∗)=exp⁡(a∗)⁢exp⁡(-a)⁢c⁢exp⁡(b)⁢exp⁡(b∗)

which equals exp⁡(a∗-a)⁢c⁢exp⁡(b-b∗)=U1⁢c⁢U2.

Hence

∥exp⁡(a∗)⁢c⁢exp⁡(-b∗)∥≤∥c∥.

Define f:ℂ→A by f⁢(λ):=exp⁡(λ⁢a∗)⁢c⁢exp⁡(-λ⁢b∗). If we substitute a↦λ⁢a,b↦λ⁢b in the last estimate we obtain

∥f⁢(λ)∥≤∥c∥,λ∈ℂ.

But f is clearly an entire functionMathworldPlanetmath and therefore Liouville’s theorem implies that f⁢(λ)=f⁢(0)=c for each λ.

This yields the equality

c⁢exp⁡(λ⁢b∗)=exp⁡(λ⁢a∗)⁢c.

Comparing the terms of first order for λ small finishes the proof. ∎

Title Fuglede-Putnam-Rosenblum theorem
Canonical name FugledePutnamRosenblumTheorem
Date of creation 2013-05-08 21:47:27
Last modified on 2013-05-08 21:47:27
Owner karstenb (16623)
Last modified by karstenb (16623)
Numerical id 1
Author karstenb (16623)
Entry type Theorem
Classification msc 47L30