20
$\begingroup$

It's well known that two locally compact Hausdorff spaces $X, Y$ are homeomorphic iff the rings $C_0(X), C_0(Y)$ (continuous functions vanishing at infinity) are isomorphic.

Is there a class $\mathcal{C}$ of topological spaces such that $X, Y \in \mathcal{C}$ are homeomorphic, iff the rings $C_c(X), C_c(Y)$ are isomorphic ?

Here $C_c(X) = \lbrace f:X \to \mathbb R\mid \operatorname{cl}_X \lbrace x \in X \mid f(x) \neq 0\rbrace\text{ is compact}\;\rbrace$ denotes the ring of continuous functions with compact support.

Improve this question
$\endgroup$
9
  • $\begingroup$ Besides containing the compact spaces, what other properties do you want ${\mathcal C}$ to have? (I guess, really, what kind of category are you after?) $\endgroup$
    – Yemon Choi
    Commented Mar 22, 2012 at 0:32
  • $\begingroup$ I have no clue if there is such a $\mathcal{C}$ and if, what properties its spaces have. Assume the question had asked for $C_0$ instead of $C_c$. Then, I think it would be difficult to guess that the property is "locally compact". $\endgroup$
    – Ralph
    Commented Mar 22, 2012 at 0:51
  • $\begingroup$ My point was more that you seem to implicitly assume some kind of maximality or closure property of your class. Otherwise you have silly examples, such as the class of all infinite compact Hausdorff spaces, together with an infinite set with indiscrete topology. What do you actually mean by "class"? $\endgroup$
    – Yemon Choi
    Commented Mar 22, 2012 at 4:38
  • 1
    $\begingroup$ Can we say the 'iff' requirement in question will take care of other conditions on the class $\mathcal{C}$? Another question: Does the answer depend on topological considerations on the function spaces also or is it purely algebraic? $\endgroup$
    – Uday
    Commented Mar 22, 2012 at 7:23
  • $\begingroup$ @Yemon: One condition it would be nice to have is that all rings that can be realized are realized in $\mathcal{C}$. I don´t know if this is true for the locally compact vs. $C_0$ case. $\endgroup$
    – Ramiro de la Vega
    Commented Mar 22, 2012 at 11:55

1 Answer 1

Reset to default
11
$\begingroup$

My naive feeling is that the answer is simply the class of locally compact Hausdorff spaces, for the following reasons. First, for a locally compact Hausdorff space $X$, one can recover $C_0(X)$ from $C_c(X)$ by completion in the uniform norm; and the uniform norm is an algebraic feature because it can be derived from the characters on the ring $C_c(X)$. So for $X$ and $Y$ locally compact Hausdorff, if $C_c(X)$ and $C_c(Y)$ are isomorphic then $X$ and $Y$ are homeomorphic.

Now suppose that $X$ is any completely regular (Hausdorff) space (and surely it is topological spaces in this class that we are interested in). Then $X$ is the disjoint union of an open locally compact subset $X_0$ consisting of points which have a compact neighbourhood and a closed subset $X_1$ consisting of points which do not. Each $f\in C_c(X)$ vanishes on $X_1$ so we are not going to get any information about $X_1$ from $C_c(X)$.

So my feeling is that $C_c(X)$ determines $X_0$ up to isomorphism but gives no information about $X_1$.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ To complement this answer: The ring $C_c(X)$ can be directly recovered from $C_0(X)$ as the smallest dense (in the uniform norm) ideal of $C_0(X)$. Say $I$ is a dense ideal of $C_0(X)$ and $f$ is a positive function in $C_0(X)$. Choose $g$ in $I$ such that $|f-g|<\epsilon$. Then one can show that $(f-\epsilon)_+$ is a multiple of $g$ and so belongs to $I$. In this way we get that $C_c(X)$ is contained in $I$. $\endgroup$
    – Leonel Robert
    Commented Mar 25, 2012 at 21:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .