There is a well known correspondence between locally compact Hausdorff spaces and commutative C* algebras, in which the homoemorphism class of spaces is mapped one-to-one to the isomorphism class of commutative C* algebras via $X\mapsto C_0(X)$.

Compact spaces correspond to unital algebras, and compactification procedures like one-point compactification or Stone-Cech compactification to unitisation procedures like the adjoining of unit $A\mapsto A\oplus\Bbb C1$ and passing to the multiplier algebra.

Suppose one has a bounded and open subset $U$ of some $\Bbb R^n$. The closure $\overline{U}$ will then be a compactification of $U$, and for that reason $C(\overline U)$ will be a unitisation.

Is there a way to describe how this unitisation works on the algebra level? Specifically how one can distinguish it from unitisations that do not allow an embedding into $\Bbb R^n$?

As an example with $U=(0,1)$ the closure $[0,1]$ is a two-point compactification and we adjoin two elements to the algebra, which we can view for example as the positive functions $x$ and $1-x$, adding them together gives unit.


1 Answer 1


Here's the closest thing to what you're asking for that I can think of. If $X$ is a compact Hausdorff space, then $X$ embeds in $\mathbb{R}^n$ for some $n$ iff $C(X)$ is finitely generated as a $C^*$-algebra. Indeed, if $X$ is a subspace of $\mathbb{R}^n$, then the coordinate projections generate $C(X)$ (as a unital $C^*$-algebra) by Stone-Weierstrass. Conversely, if $C(X)$ is generated by finitely many functions $f_1,\dots,f_n:X\to\mathbb{C}$, they together give a map $f:X\to \mathbb{C}^n$. Since the $f_i$ must jointly separate points, $f$ is injective, and so since $X$ is compact it is an embedding.

So, if $X$ is only locally compact Hausdorff, compactifications of $X$ which come from embeddings of $X$ as a bounded subset of $\mathbb{R}^n$ correspond to unitizations of $C_0(X)$ which are finitely generated as $C^*$-algebras.

  • $\begingroup$ The aspect "the unitisation is finitely generated" means that the space is embedded into some $\Bbb R^n$ upon which the compactification is taken. For example $(0,1)$ can be embedded into $\Bbb R^2$ with closure the circle, which will also result in a unitisation. This is different from the condition that the embedding must be open, but that is probably a bit more difficult. $\endgroup$
    – s.harp
    Feb 3, 2018 at 10:06

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