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In light of the well-known theorem of Gelfand that, bluntly put, ends up saying that unital abelian C*-algebras are the 'same' as compact Hausdorff topological spaces, I tried to compile a dictionary of concepts between these two objects. More specifically, given a compact Hausdorff space $X$, I ask in what manner are topological properties of $X$ encoded into $C(X) := C(X, \mathbb{C})$? And, conversely, in what way do algebraic properties of the latter manifest topologically in the former? Here is the elementary list I was able to gather:

$\cdot$ $C(X)$ has $2^n$ idempotent elements $\Leftrightarrow$ $X$ has $n$ connected components

$\cdot$ $C(X)$ separable $\Leftrightarrow$ $X$ metrizable

$\cdot$ $C(X)$ isomorphic to $C(Y)$ $\Leftrightarrow$ $X$ homeomorphic to $Y$

$\cdot$ continuous functions from $g:X \to Y$ induce *-homomorphisms $\hat g: C(X) \to C(Y)$ and vice-versa

$\cdot$ there is a bijective correspondence between ideals of $C(X)$ and open sets of $X$

What do subalgebras of $C(X)$ correspond to? If this is not a well-posed question please tell me why. Subalgebras are a very natural substructure to consider and yet I am at a loss as to how it translates over.

If you have any additions (or corrections) to the above dictionary, please share them.

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    $\begingroup$ Is your question about subalgebras, closed subalgebras, *-subalgebras or closed *-subalgebras? $\endgroup$
    – Fabian Lenhardt
    Commented Dec 5, 2011 at 15:48
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    $\begingroup$ Bill Johnson's answer of course shows that you do need to be careful when you say that "unital abelian C*-algebras are the 'same' as compact Hausdorff topological spaces". What Gelfand duality shows is that the category of unital commutative C*-algebras and *-homomorphisms (important) is (anti-)equivalent to the category of compact topological spaces and continuous maps. With *-homomorphisms, you cannot "see" non-selfadjoint subalgebras... $\endgroup$
    – Matthew Daws
    Commented Dec 5, 2011 at 16:35
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    $\begingroup$ @Martin: thank you for refraining from being purposefully obtuse - and for stating the obvious. $\endgroup$
    – Joshua Seaton
    Commented Dec 5, 2011 at 22:54
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    $\begingroup$ As for the non-unital case, Gelfand duality extends to a locally compact hausdorff spaces (with proper maps) and the implications in my answer remain true when we replace $C$ by $C_0$. $\endgroup$
    – Martin Brandenburg
    Commented Dec 6, 2011 at 17:03
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    $\begingroup$ I asked basically the same question here: mathoverflow.net/questions/35507/… $\endgroup$
    – Rasmus
    Commented Dec 7, 2011 at 20:07

5 Answers 5

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Gelfand duality asserts that $C(-)$ is an anti-equivalence from the category of compact hausdorff spaces to the category of commutative unital $C^{\ast}$-algebras. Now, for a continuous map $f : X \to Y$ it is not hard to show that

$f$ is surjective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is injective

Sketch of proof: $\Rightarrow$ is trivial, and $\Leftarrow$ follows from Tietze extension theorem $\square$. By the way, we also have:

$f$ is injective $\Longleftrightarrow$ $C(f) : C(Y) \to C(X)$ is surjective

Therefore, $C^{\ast}$-subalgebras of $C(X)$ (i.e. closed unital $\ast$-subalgebras) correspond to surjective maps $X \to Y$, where $Y$ is compact hausdorff. It is well-known that these maps are quotient maps. The partial order of $C^{\ast}$-subalgebras of $C(X)$ is anti-isomorphic to the partial order of quotients of $X$.

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    $\begingroup$ It might be useful to note that the way you can recognize $C(Y)$ inside $C(X)$ is as functions that are constant on the fibers of $f$. $\endgroup$
    – MTS
    Commented Dec 5, 2011 at 16:55
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    $\begingroup$ Clear and concise and to-the-point. Thank you. $\endgroup$
    – Joshua Seaton
    Commented Dec 5, 2011 at 22:48
  • $\begingroup$ For future reference, Markdown interprets asterisks as italics, so if you plan on using more than one it's safer to use LaTeX's \ast instead. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 6, 2011 at 16:53
  • $\begingroup$ @Qiaochu Yuan: may you please elaborate on how in using the Tietze extension theorem we can show that if $C(f): C(Y) \to C(X)$ is injective then $f$ is surjective? This is as far as I got with my (limited) reasoning: if $C(f)$ is injective then we have that if, for $h_1,h_2 \in C(Y)$, $h_1 = h_2$ on the closed subset $f(X) \subset Y$, it must be that $h_1 = h_2$ on all of $Y$. How might we then conclude that it must be the case that $f(X) = Y$? $\endgroup$
    – Joshua Seaton
    Commented Dec 8, 2011 at 4:45
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    $\begingroup$ Assume there is some $y \in Y \setminus f(X)$ and construct with Tietze to $h_1,h_2$ contradicting the injectivity. $\endgroup$
    – Martin Brandenburg
    Commented Dec 8, 2011 at 11:45
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Closed subalgebras of $C(X)$ are called uniform algebras, and there is a huge literature on them. You might start by reading Gamelin's book, Uniform Algebras, or simple by Googling "uniform algebras".

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This may be an interesting add-on for Martin's answer. In this paper Pavlov and Troitskii show that an inclusion of commutative $C^*$-algebras $C(X) \to C(Y)$ (with $X$ and $Y$ compact Hausdorff), which allows a positive unital conditional expectation $E \colon C(Y) \to C(X)$ that satisfies an index condition, corresponds via Gelfand duality to a branched covering $p \colon Y \to X$. The latter means that $p$ is a continuous closed and open surjection with boundedly many preimages $p^{-1}(x)$ at every $x \in X$,

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  • $\begingroup$ This is a really nice paper, thanks ! $\endgroup$
    – Amin
    Commented Jun 11, 2012 at 20:26
  • $\begingroup$ Unfortunately the link is broken. $\endgroup$
    – Martin Brandenburg
    Commented May 20, 2021 at 9:06
  • $\begingroup$ I fixed the link. $\endgroup$
    – Ulrich Pennig
    Commented May 25, 2021 at 18:50
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    $\begingroup$ Name of the paper, against future link rot: Quantization of branched coverings. $\endgroup$
    – LSpice
    Commented May 25, 2021 at 18:59
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A bit more exotic, a finitely generated subalgebra (no matter $*$-subalgebra or not) corresponds to a continuous map to an affine variety over $\mathbb C$ (continuous in the euclidean topology), such that image is Zariski dense. Sometimes this is a useful observation.

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  • $\begingroup$ Forgive me, I'm no too clever, but $C(X)$ isnt a polynomial algebra, then I dont understnd your assetion. $\endgroup$
    – Buschi Sergio
    Commented Dec 5, 2011 at 18:47
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    $\begingroup$ @Buschi Sergio: Well, the finitely generated subalgebra is a quotient of a polynomial algebra and thus corresponds to an affine variety. The spectrum of the C*-algebra A is given by the characters A-->C and precomposition with the inclusion map, whether it is *-respecting or not, gives a surjective ring map to a field, hence a maximal ideal. This defines a map from the points of the compact Hausdorff space to the affine variety (whose points correspond to maximal ideals). $\endgroup$
    – Peter Arndt
    Commented Dec 5, 2011 at 23:45
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To add to the dictionary, some strange spaces called extremely disconnected correspond to rather natural abelian C* algebras, the Von Neumann. They are natural, since they correspond to C* algebras acting on $L^2(X,m)$ for not too bad measure spaces $(X,m)$; roughly, you look at the spectrum of $L^\infty(X,m)$ acting on $L^2$.

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    $\begingroup$ This doesn't seem relevant to the question, which is about subalgebras. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 6, 2011 at 17:10
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    $\begingroup$ Not right at all, just read it from beginning to the end. Here's why I think it's pertinent (so finally people stop giving negative notes to my comment). First, the example of metrizable space is given, the corresponding $C*-$algebras being precisely identified. Second, the question ends with : if you have any addition to the dictionary etc. Now the comment goes in the other way, and I think is a nice motivation for Gelfand. Namely, since there are many kind of $C*-$algebras, it's also a question what are the corresponding spectrum. $\endgroup$
    – Amin
    Commented Dec 6, 2011 at 17:42
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    $\begingroup$ So now as I said, the Von Neuman are common $C^*-$algebras, and thus I add one entry to the "dictionary" (which is rather a correspondence than a dictionary).\\ PS : I wanted to put my 'answer' as comment, but I can't, sorry for that. $\endgroup$
    – Amin
    Commented Dec 6, 2011 at 17:48
  • $\begingroup$ Are the spaces really called "extremely disconnected", or do you mean "extremally disconnected"? $\endgroup$
    – LSpice
    Commented May 25, 2021 at 19:00

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