The Multiplicative Kowalski-Slodkowski Theorem for Hermitian Algebras

Fuente: arXiv
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Main Authors: Brits, Rudi, Hassen, Muhammad, Toure, Cheick
Format: Preprint
Published: 2025
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author Brits, Rudi
Hassen, Muhammad
Toure, Cheick
author_facet Brits, Rudi
Hassen, Muhammad
Toure, Cheick
contents We prove, for Hermitian algebras, the multiplicative version of the Kowalski-Słodkowski Theorem which identifies the characters among the collection of all complex valued functions on a Banach algebra $A$ in terms of a spectral condition. Specifically, we show that, if $A$ is a Hermitian algebra, and if $ϕ:A\mapsto\mathbb C$ is a continuous function satisfying $ϕ(x)ϕ(y) \in σ(xy)$ for all $x,y\in A$ (where $σ$ denotes the spectrum), then either $ϕ$ or $-ϕ$ is a character of $A$; of course the converse holds as well. Our proof depends fundamentally on the existence of positive elements and square roots in these algebras.
format Preprint
id arxiv_https___arxiv_org_abs_2509_03663
institution arXiv
publishDate 2025
record_format arxiv
spellingShingle The Multiplicative Kowalski-Slodkowski Theorem for Hermitian Algebras
Brits, Rudi
Hassen, Muhammad
Toure, Cheick
Functional Analysis
46H05, 46H15, 47A10
We prove, for Hermitian algebras, the multiplicative version of the Kowalski-Słodkowski Theorem which identifies the characters among the collection of all complex valued functions on a Banach algebra $A$ in terms of a spectral condition. Specifically, we show that, if $A$ is a Hermitian algebra, and if $ϕ:A\mapsto\mathbb C$ is a continuous function satisfying $ϕ(x)ϕ(y) \in σ(xy)$ for all $x,y\in A$ (where $σ$ denotes the spectrum), then either $ϕ$ or $-ϕ$ is a character of $A$; of course the converse holds as well. Our proof depends fundamentally on the existence of positive elements and square roots in these algebras.
title The Multiplicative Kowalski-Slodkowski Theorem for Hermitian Algebras
topic Functional Analysis
46H05, 46H15, 47A10
url https://arxiv.org/abs/2509.03663