English

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 β\beta we attach a separable test algebra A(β)A(\beta) whose multiplication records β\beta. Any homomorphic embedding of A(β)A(\beta) into a candidate BB forces the linearisation of β\beta to factor through the fixed separable space B^πBB\widehat{\otimes}_{\pi}B. Choosing β\beta so that the associated operator fails to factor through B^πBB\widehat{\otimes}_{\pi}B, by the theorem of Johnson--Szankowski, yields a contradiction. In the commutative case, we take β\beta symmetric so A(β)A(\beta) is commutative.

Keywords

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

R2 v1 2026-07-01T07:24:29.111Z