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This is a follow up question to this one.

If $X$ is a metric space, denote by $C_u(X)$ the $C^\ast$-algebra of all bounded, uniformly continuous functions on $X$ (with the sup-norm).

Do we have $C_u(X_1 \times X_2) = C_u(X_1) \hat{\otimes} C_u(X_2)$?

(Maybe it is more natural to use uniform spaces instead of metric spaces, but I think that the answer does not depend on this.)

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  • $\begingroup$ If you mean the projective tensor product, then the answer is surely no if both $X_1$ and $X_2$ are infinite, just for Banach space reasons. The proj t.p. of two infinite-dimensional Cstar algebras cannot, IIRC, be isomorphic as a Banach space to any infinite-dimensional Cstar algebra. $\endgroup$
    – Yemon Choi
    Commented Apr 27, 2013 at 17:57

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The spectrum of the $C^\ast$-algebra of bounded, uniformly continuous functions on a uniform space is the Samuel compactification. So your query can be restated in the form: is the Samuel compactification of a product naturally identifiable with the product of the Samuel compactifications of the individual spaces. This is almost certainly wrong. The corresponding result for the Stone-Cech compactification of completely regular spaces is about as wrong as it could be. Presumably, you can give an explicit counterexample in the uniform case by using the fine uniformity. This is yet another example of the fact that if you want to extend duality (Riesz representation theory, Gelfand duality) from the compact case to the non-compact one (for completely regular spaces, uniform spaces, etc.) then you are well advised to leave categories of Banach spaces or algebras and use more general ones (mixed topologies, Saks spaces and algebras). A fairly systematic account can be found in the monograph "Saks Spaces and Applications to Functional Analysis".

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  • $\begingroup$ Does the following work? If $X_1$ and $X_2$ are not totally bounded, they contain $\varepsilon$-discrete infinite subsets $D_1$ and $D_2$ for some $\varepsilon$. Their closures in the Samuel compactifications should be $\beta D_1$ and $\beta D_2$. Moreover, $D_1 \times D_2$ is $\varepsilon$-discrete, so its closure in the Samuel compactification of $X_1 \times X_2$ is $\beta (D_1 \times D_2)$. But $\beta (D_1 \times D_2) \neq \beta D_1 \times \beta D_2$. $\endgroup$
    – Martin
    Commented Apr 27, 2013 at 17:52

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