# What is the character that compactifies $\mathbb{R}$ through the Gelfand transform?

I'm a little embarassed that I can't answer this myself, so hopefully it will get answered very quickly.

Let $X$ be locally compact, Hausdorff. Consider $\text{C}_\text{b}(X)$ the $C^*$-algebra of bounded, complex-valued, continuous functions on it under the sup norm. The Gelfand transform takes $\text{C}_\text{b}(X)$ isomorphically to $\text{C}_0(Y)$ where $Y$ is the maximal ideal space of $\text{C}_\text{b}(X)$. Since $\text{C}_\text{b}(X)$ is unital, $Y$ must be compact.

It is well known that $Y$ is the Stone–Čech compactification of $X$. If $X = \mathbb{R}$ then $Y$ is the one-point-compactification. In general, the maximal ideal space is in one-to-one correspondence with the characters (nonzero multiplicative linear functionals) on $\text{C}_\text{b}(X)$.

Question: Say $X = \mathbb{R}$ and the maximal ideal space of $\text{C}_0(\mathbb{R})$ is made up of the point evaluations at points on $\mathbb{R}$. These point evaluations are also characters on $\text{C}_\text{b}(X)$, but its maximal ideal space is the one-point-compactification of $\mathbb{R}$. What character on $\text{C}_\text{b}(\mathbb{R})$ represents this single missing point of the maximal ideal space?

The big issue I'm having is that for a non-compact space like $\mathbb{R}$, we have an obvious problem that you could take a sequence of point evaluations $\chi_n(f) = f(n)$ where $n \to \infty$, which clearly goes to the zero functional if your $f \in \text{C}_0(\mathbb{R})$. This doesn't happen in $\text{C}_\text{b}(\mathbb{R})$ because the maximal ideal space doesn't need to add the zero functional in order to be compact, but that would imply that $\chi_n$ is converging to some point evaluation "at infinity" (which fits with the idea that $\mathbb{R}$ gets compactified by adding a point "at infinity"). But $\chi_n(f) = f(n)$ as $n\to\infty$ still doesn't make sense for functions $f \in \text{C}_\text{b}(\mathbb{R})$ (take $\sin(x)$ for example).

• Maybe I'm misreading you, but the Stone-Čech compactification of $\mathbb{R}$ can't possibly be the one-point compactification, since $C_b(\mathbb{R})$ isn't isomorphic to $C_0(\mathbb{R})^+ \cong C_0((0,1))^+ \cong C(S^1)$. – Branimir Ćaćić Feb 16 '14 at 0:46
• Are you sure that the Stone–Čech compactification of $\mathbb{R}$ is the one-point compactification? This seems to contradict the universal property stated here: en.wikipedia.org/wiki/… For example, not all continuous functions from $\mathbb{R}$ to $[0,1]$ factor through a map from $S^1$. – Paolo Ghiggini Feb 16 '14 at 0:47
• The Stone-Čech compactification is defined for any topological space, with no hypotheses whatsoever. The map $X \to \beta X$ is an embedding iff $X$ is Tychonoff, which is still a weaker condition than $X$ being locally compact Hausdorff (this is the condition you need for the existence of the one-point compactification). – Qiaochu Yuan Feb 16 '14 at 4:58
• The confusion in your question has been cleared up by Qiaochu Yuan but the following remark might interest you. The compactifications of $X$ correspond to the $C^\ast$ subalgebras of $C^b(X)$ ($X$ need not be locally compact). Interesting special cases: the bounded uniformly continuous functions (for uniform spaces), the algebra obtained by adding a unit to $C_0(X)$ ($X$ locally compact), the almost periodic functions ($X$ a locally compact group) and many more. The corresponding spectra are, in these cases, the Samuel, the one-point and the Bohr compactification. – alpha Feb 16 '14 at 10:43

The Stone-Čech compactification of $\mathbb{R}$ is not its one-point compactification. The former is the largest compactification of a space, while the latter, if it exists, is the smallest compactification, and in general there will be many compactifications in between. $\mathbb{R}$ has, for example, a two-point compactification, namely $[0, 1]$.
The C*-algebraic construction of the one-point compactification is not $C_b(\mathbb{R})$ but the subalgebra
$$\{ f \in C_b(\mathbb{R}) | \lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) \}$$
and taking the limit is the character corresponding to the extra point. (In general, if $X$ is locally compact Hausdorff, then we can construct its one-point compactification by adjoining a unit to $C_0(X)$.)