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I am having some trouble constructing the Stone-Čech compactification of a locally compact Hausdorff space $X$ using theory of $C^*$-algebras. I did some search but could not find a good answer on this.

Let's focus on the case $X=\mathbb{R}$. The space of bounded complex-valued functions $C_b(\mathbb{R})$ is a commutative unital $C^*$-algebra hence $C_b(\mathbb{R})\cong C(\mathcal{M})$, where $\mathcal{M}$ is the maximal ideal space, which is compact and Hausdorff.

It should be the case that $\mathcal{M}\cong\beta\mathbb{R}$, and it is not difficult to show that by identify $t\in\mathbb{R}$ with the evaluation at $t$, we have a homeomorphism between $\mathbb{R}$ and a subspace of $\mathcal{M}$.

But we still need to show this subspace is dense in $\mathcal{M}$. This is where I am having troubles (and I guess this is the whole point of the proof).

Can someone give a hint? Thanks!

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Curious question: why would you like to construct the SC-compactification using C-star theory? –  Matt N. Dec 17 '12 at 16:27
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@MattN. Well, I was looking at Stone-Cech on Wiki, which mentions this approach in one sentence. So I tried to give a proof. –  Hui Yu Dec 17 '12 at 16:38

2 Answers 2

up vote 4 down vote accepted

The sketch in the other answer takes care of much, except that it doesn't address the question why it is that $i(X)$ is dense in $\mathcal{M}$.

So: let $i \colon X \to \mathcal{M}$ be the map sending $x$ to (the maximal ideal corresponding to) evaluation at $X$. If $i(X)$ were not dense then there would be a function $f \colon \mathcal{M} \to [0,1]$ such that $f|_{i(X)} = 0$ (apply Urysohn's lemma to a point outside of the closure of $i(X)$). But the existence of such a function is impossible since such a function would have to be zero under the identification $C(\mathcal{M}) \cong C_b(X)$.

A detailed proof of the Stone-Čech property of the maximal ideal space of $C_b(X)$ appears in many books treating spectral theory of $C^\ast$-algebras, e.g. Pedersen, Analysis now, Proposition 4.3.18.

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Yes. I found the proof from Pederson earlier today. Thanks! –  Hui Yu Dec 18 '12 at 13:26

You really should be thinking about the Stone-Cech compactification in terms of its universal property; the inclusion $X \to \beta X$ is already uniquely determined (up to unique isomorphism) by the fact that it is the universal map from $X$ to a compact Hausdorff space, so to verify that $C_b(X) \cong C(\beta X)$ it suffices to verify that the compact Hausdorff space $Y$ such that $C_b(X) \cong C(Y)$ (which exists by Gelfand-Naimark) has the universal property of the Stone-Cech compactification.

(There is also no need to assume that $X$ is locally compact Hausdorff. Everything I'm about to say makes sense for arbitrary topological spaces, although the map $X \to \beta X$ is only an embedding for $X$ completely regular.)

To verify the universal property, let $f : X \to Z$ be a continuous map from $X$ to a compact Hausdorff space $Z$. Then $f$ determines a map $C(Z) \to C_b(X)$ of C*-algebras (a complex-valued function on $Z$ will be bounded, and so its pullback to $X$ will also be bounded). Since $C_b(X) \cong C(Y)$, it follows that $f$ determines a map $C(Z) \to C(Y)$, and by the equivalence of categories between commutative unital C*-algebras and compact Hausdorff spaces (this is the technical heart of the proof) this uniquely determines a continuous map $Y \to Z$ through which $f$ factors. The conclusion follows.

Edit: The fact that (the image of) $X$ is dense in $\beta X$ follows directly from the universal property, since the closure of $X$ in $\beta X$ satisfies the universal property of the Stone-Cech compactification, hence its inclusion into $\beta X$ must be an isomorphism.

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What you said is true. But the universal property assumes the existence of $\beta X$ in the first place, while the construction using $C_b(X)\cong C(\mathcal{M})$ constructs $\beta X$ as the maximal ideal space hence proves the existence. If you already assumes the existence of Stone-Cech (from other constructions, maybe) then universal property gives a nice proof, but the other answer has the advantage that it assumes less. But thanks anyway. –  Hui Yu Dec 18 '12 at 13:33
    
@Hui: I don't understand what you're saying. You seem to be criticizing this proof but I do not see where the criticism is. $\beta X$ is constructed as the Gelfand spectrum of $C_b(X)$ and the rest is verifying the universal property, which is how we know that we have actually constructed $\beta X$ and not something else. –  Qiaochu Yuan Dec 18 '12 at 22:03
    
There is nothing wrong with your proof, and sorry if I made you feel I am criticizing. However, if someone knows nothing about S-C compactification, it seems more natural to first construct a compactification, then verify the universal the universal property, not the other way around. Tell me if I am mistaken, but I always feel universal properties are very neat and useful for understanding, they always come after a concrete construction. I am not sure about S-C. –  Hui Yu Dec 19 '12 at 1:19
    
Did they first construct the compactification then discover the universal property? Or did they first ask 'we want something with this universal property, can we construct it'? –  Hui Yu Dec 19 '12 at 1:20
    
@Hui: universal properties are something you can demand and prove things about completely independent of any construction of objects satisfying them. That is what makes universal properties a powerful tool: because objects satisfying universal properties unique up to unique isomorphism, anything you want to prove about a universal construction should follow from its universal property alone (in particular the claim that $X$ is dense in $\beta X$). Thinking in terms of universal properties separates all of these issues from the issue of existence (that is, construction). –  Qiaochu Yuan Dec 19 '12 at 2:12

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