There is no universal separable Banach algebra

TL;DR

The paper proves that no separable Banach algebra can serve as a universal host for embeddings of all other separable Banach algebras, highlighting fundamental limitations in the structure of such algebras.

Contribution

It establishes the non-existence of a universal separable Banach algebra for homomorphic embeddings, extending the result to the commutative case and introducing a novel proof scheme.

Findings

No separable Banach algebra is universal for all separable Banach algebras.

The same non-universality holds in the commutative setting.

A new proof scheme involving bilinear forms and tensor products is introduced.

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