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I know this seems like a dumb question. But I can't see why we need the condition that $X$ is a locally compact Hausdorff space when we define $C_0(X)$. I think many properties still hold without locally compact Hausdorff property. For example,I think $C_0(X)$ is still a $C^*$-algebra for a general topological space $X$, am I right?

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You're right; $C_0(X)$ is a $C^*$-algebra for any topological space $X$. However, for any $X$, we can canonically associate to it another space $Y$ which is locally compact Hausdorff such that $C_0(X)\cong C_0(Y)$, so in some sense we lose no generality by restricting $X$ to be locally compact Hausdorff (and most interesting examples are).

To construct $Y$, let $X^*=X\cup\{\infty\}$ be the $1$-point compactification of $X$ (topologized by saying neighborhoods of $\infty$ are cocompact open subsets of $X$); then an element of $C_0(X)$ is equivalent to a continuous map $X^*\to\mathbb{C}$ that vanishes at $\infty$. Now let $p:X^*\to Z$ be the Hausdorffification of $X^*$ (i.e., the quotient of $X^*$ by the smallest equivalence relation such that the quotient is Hausdorff) and let $Y=Z\setminus\{p(\infty)\}$. Then $Y$ is locally compact Hausdorff, since it is an open subset of the compact Hausdorff space $Z$. Since $\mathbb{C}$ is Hausdorff, every continuous map $X^*\to\mathbb{C}$ factors through $Z$. It follows that there is a natural bijection between elements of $C_0(X)$ and continuous functions on $Z$ that vanish at $p(\infty)$. But these functions are the same as elements of $C_0(Y)$, since $Z$ is the one-point compactification of $Y$. It is then easy to check that this bijection $C_0(X)\cong C_0(Y)$ is actually an isomorphism of $C^*$-algebras.

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  • $\begingroup$ Wow, amazing! But it will take me some time to digest your solution~~~there are quite a lot of details for me to fill in~~~ $\endgroup$ – No One Apr 10 '16 at 3:09
  • $\begingroup$ Hi, could you correct wikipedia which defines $C^*$ algebras as algebras of endomorphisms on some Hilbert space closed for the norm topology (ie. the norm would be the operator norm of $B(H)$) whereas it seems the common definition is a Banach algebra with an involution compatible with the norm and ($B(H)$ is just the motivating example, together with $C_0(X)$) $\endgroup$ – reuns Jun 9 '19 at 12:06

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