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Can anyone provide me with an example of two non-homeomorphic locally-compact Hausdorff spaces $X$ and $Y$, such that $C(X)$ and $C(Y)$ are isomorphic as Banach algebras. Clearly, the Gelfand--Naimark theorem tells us that the two algebras $C(X)$ and $C(Y)$ cannot be isomorphic as $C^*$-algebras.

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There are none.

To be more precise, for locally compact spaces $X$ you have to consider $C_0(X)$, the space of continuous functions which vanish at $\infty$; you can reduce this case to that of compact $X$ by considering the 1-point compactification and removing the constants by $f\mapsto f- f(\infty).1$.

Now, for compact $X$, the Banach algebra $C(X,\mathbb C)$ is always a commutative $C^*$-algebra, thus $X$ is isomorphic to the closed subspace of multiplicative $C^*$-functionals in the duals. Thus your question has a negative answer.

For real valued functions, you have to use the Banach lattice property and the condition $f,g\ge0\implies \|f\vee g\|=\|f\|\vee\|g\|$ to get the same result (Stone's theorem).

But consider the following Banach lattice (see Semadeni: Banach spaces of continuous functions, 1971): Let $U$ be be a bounded domain in $\mathbb R^n$. Then the space $H_b(U)$ of bounded harmonic functions on $U$ has the above property, but the Banach lattice supremum is not the pointwise supremum. In fact, $H_b(U)\cong L^\infty(\partial U)$. I hope that this catches what you were looking for.

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Hi Peter. Thanks a lot for your answer! Yes this answers my question. What I was wondering was whether the existence of the $\ast$-structure was necessary for differentiating l.c.H.sps using their cts function algs - your answer tells me that no it is not. So the point of Gelfand--Naimark is that one needs the $*$-structure in order to identify those Banach algebras that admit a $*$-structure? Moreover, when this $*$-structure exists it must then be unique, no? –  Janos Erdmann Oct 20 '13 at 16:15
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