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The well-known Banach–Mazur theorem says that $C([0, 1])$ is a universal separable Banach space, in the sense that if $X$ is any separable Banach space then there is a map $f : X \to C([0, 1])$ which is both linear and isometric. Note that $C([0, 1])$ also has the structure of a Banach algebra.

My question is this: Is $C([0, 1])$ universal for separable commutative Banach algebras? Of course a separable Banach algebra is a separable Banach space, so there is a linear isometry into $C([0, 1])$, but I'm asking if that map can also be taken to preserve the multiplication operation.

If $C([0, 1])$ is not universal for separable commutative Banach algebras, does there exist such a universal object? I'm interested mostly in ZFC results, but would also not mind hearing consistent answers (especially if they are consistent with $\neg CH$).

Tomasz Kania
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Chris
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  • Just a remark on terminology (of which you may already be aware, but I mention it also for others reading this): in many places in the Banach space literature, universal means merely that a space is isomorphically universal; if one requires in addition that isomorphic embeddings are isometric, then one would normally use the term isometrically universal. See, for example, chapter 23 in the Handbook of the Geometry of Banach Spaces (the chapter Descriptive Set Theory and Banach Spaces) for an instance of the use of the term universal as I have mentioned above. – Philip Brooker Aug 24 '12 at 22:21

1 Answers1

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The short answer to the first question is no. A simple counterexample is $\mathbb{C}^2$, or indeed any Banach algebra containing a nontrivial idempotent, since $C([0,1])$ contains no nontrivial idempotents.

Indeed if a universal algebra of the sort you ask about exists, it cannot be semisimple, and hence cannot be of the form $C(X)$. For if $U$ is such a universal algebra and $f\colon A\to U$ is a homomorphism and $\phi$ is a character on $U$, then $\phi\circ f$ is a character on $A$. If $x\in A$ and $f(x)\ne0$, you could pick $\phi$ so that $\phi(f(x))\ne0$, so $x$ is not in the Jacobson ideal of $A$.

Harald Hanche-Olsen
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