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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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Is your question about subalgebras, closed subalgebras, *-subalgebras or closed *-subalgebras? –  Fabian Lenhardt Dec 5 '11 at 15:48
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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... –  Matthew Daws Dec 5 '11 at 16:35
    
Minor peeve: I do wish people would not interprety anti-equivalence of categories, or contravariant equivalence of categories, as being informally "the same", as opposed to being informally "enantiomorphs". That said, the main issue as others have pointed out is that you need to specify what you mean by subalgebras. The contravariant equivalence perspective suggests that really the thing to consider is closed *-ideals. –  Yemon Choi Dec 5 '11 at 18:14
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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 $C_0$. –  Martin Brandenburg Dec 6 '11 at 17:03
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I asked basically the same question here: mathoverflow.net/questions/35507/… –  Rasmus Bentmann Dec 7 '11 at 20:07
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5 Answers 5

up vote 15 down vote accepted

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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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 $f$. –  MTS Dec 5 '11 at 16:55
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Clear and concise and to-the-point. Thank you. –  Joshua Seaton Dec 5 '11 at 22:48
    
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. –  Qiaochu Yuan Dec 6 '11 at 16:53
    
@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 $f(X) = Y$? –  Joshua Seaton Dec 8 '11 at 4:45
    
Assume there is some $y \in Y \setminus f(X)$ and construct with Tietze to $h_1,h_2$ contradicting the injectivity. –  Martin Brandenburg Dec 8 '11 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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This is a really nice paper, thanks ! –  Amin Jun 11 '12 at 20:26
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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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Forgive me, I'm no too clever, but $C(X)$ isnt a polynomial algebra, then I dont understnd your assetion. –  Buschi Sergio Dec 5 '11 at 18:47
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@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). –  Peter Arndt Dec 5 '11 at 23:45
    
Peter, thanks for the explanation. –  Andreas Thom Dec 6 '11 at 6:33
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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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This doesn't seem relevant to the question, which is about subalgebras. –  Qiaochu Yuan Dec 6 '11 at 17:10
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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. –  Amin Dec 6 '11 at 17:42
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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. –  Amin Dec 6 '11 at 17:48
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