Note on Wermuth's theorem on commuting operator exponentials

TL;DR

This paper investigates when the commutativity of operator exponentials $e^A$ and $e^B$ implies the commutativity of the underlying operators $A$ and $B$. It leverages Wermuth's theorem and a spectrum-scaling lemma to show that, for $A,B\in B(X)$ with $\sigma(A)$ being $2\pi i$-congruence free, the relation $e^A B = B e^A$ is equivalent to $AB=BA$. It also provides an alternative Banach-space proof of a Chaban–Mortad-type result in the Hilbert-space setting by observing that $\sigma(A)$ is $2\pi i$-congruence free under the given hypotheses. The work thus unifies commuting-exponential criteria across Banach-space operators and connects to existing C*-algebra results, while highlighting the necessity of the spectral condition through explicit counterexamples.

Abstract

We apply Wermuth's theorem on commuting operator exponentials to show that if $A, B \in B(X)$, $X$ being Banach space and $A$ of $2πi$-congruence free spectrum, then $e^A B = B e^A$ if and only if $AB=BA$. We employ this observation to provide alternative proof of similar result by Chaban and Mortad, applicable for $X$ being a Hilbert space.

Theorems & Definitions (11)

• Definition 1
• Theorem 1: Wermuth
• Lemma 1
• proof
• Theorem 2
• proof
• Remark 1
• Theorem 3: Chaban and Mortad
• Lemma 2
• proof