There is no universal separable Banach algebra
Functional Analysis
2025-11-12 v2 Operator Algebras
Abstract
We prove that no separable Banach algebra is universal for homomorphic embeddings of all separable Banach algebras, whether embeddings are merely bounded or required to be contractive. The same holds in the commutative category. The proof uses the following scheme. To each bounded bilinear form we attach a separable test algebra whose multiplication records . Any homomorphic embedding of into a candidate forces the linearisation of to factor through the fixed separable space . Choosing so that the associated operator fails to factor through , by the theorem of Johnson--Szankowski, yields a contradiction. In the commutative case, we take symmetric so is commutative.
Cite
@article{arxiv.2511.04314,
title = {There is no universal separable Banach algebra},
author = {Tomasz Kania},
journal= {arXiv preprint arXiv:2511.04314},
year = {2025}
}
Comments
7 pp