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The Gelfrand duality says that the category of locally compact Hausdorff spaces (with proper continuous functions) is equivalent to the category of commutative $C^*$ algebras (with proper $*$-homomorphisms). For instance, of $X$ is such a topological space, then $C_0(X) = \{f: X\to \mathbb{C}, f$ is continuous and $f$ vanishes at $\infty \}$ is its related $C^*$ algebra. We can go the other way by looking at a $C^*$ algebra $\mathcal{A}$ and taking its set of characters $\text{Hom}(\mathcal{A},\mathbb{C})$ under pointwise convergence to recover the topological space.

So say one has a commutative $C^*$ algebra $\mathcal{A}$, how does one recover topological invariants, like say, the number of connected components of $\text{Hom}(\mathcal{A},\mathbb{C})=X$, from $\mathcal{A}$ itself?

Is this even possible? Or am I wrongly asserting that equivalence of categories says something about the individual objects?

edit many responses focus on the number of connected components, which I appreciate, but that was only meant as an example of the sort qualitative info I would like to recover. Can we recover the singular homology of $X$ from $\mathcal{A}$ ? The fundamental group? Is X metrizable?

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    $\begingroup$ This is relevant: mathoverflow.net/questions/82708/…. In particular, the number of connected components comes from the number of idempotent elements $C_0(X)$. $\endgroup$ Aug 14, 2015 at 2:34
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    $\begingroup$ Of course it is possible: the algebra determines the space $X$ completely (up to isomorphism in its category)—this is preciely what having an equivalence means— so anything you may want to know about $X$ can be seen in $A$ somehow. $\endgroup$ Aug 14, 2015 at 4:39
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    $\begingroup$ @MarianoSuárez-Alvarez, the question is clearly on how to concretize the NCG dictionary for commutative algebras. Do you know references beyond Connes' book? $\endgroup$ Aug 14, 2015 at 6:32

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The Boolean algebra of connected components is equivalent to the projections (the elements with $p^2=p$) in the algebra of functions, with multiplication of functions representing intersection and $(p_1,p_2) \to p_1 + p_2 - p_1p_2$ being the union of sets of components.

I think there is a version of algebraic K-theory for topological algebras whose value on $C_0(X)$ is the topological $K$-theory of $X$.

Connes' book on NCG has more of the dictionary but also omits many basic things.

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  • $\begingroup$ It's not really clear to me why you would call idempotent elements "projections", because in the equation "$p^2 = p$", the square isn't the composition $p^2 = p \circ p$ (which wouldn't even make sense, since the domain and the codomain of $p$ disagree), it's literally the square $p^2(x) = p(x) \cdot p(x)$ with the multiplication of $\mathbb{C}$. As a general remark, I'd also like to point out that $p_1 + p_2 - p_1 p_2 = 1 - (1 - p_1)(1 - p_2)$ isn't random: it comes from $A \cup B = \overline{\bar A \cap \bar B}$ and $1 - p$ corresponds to the complement. $\endgroup$ Aug 14, 2015 at 7:24
  • $\begingroup$ It's pretty standard in English when discussing operator algebras to refer to solutions of p^2=p as "projectors". For example, continuous dimension in a von Neumann algebra is "trace of a projector". On MSE I figured "projection" would be more recognizable. Of course the formula for union is not arbitrary and indeed I derived it when posting in the manner you wrote down. $\endgroup$ Aug 14, 2015 at 7:30
  • $\begingroup$ Correct me if I'm wrong, but an operator algebra is something that basically deals with endomorphisms, no? And the square is the square in the sense of composition. It's a completely different sort of equation when you take the square of a functional $A \to \mathbb{C}$... The terminology "idempotent" is much more adapted. (The remark about the formula for the union was more directed at other readers who could wonder where it came from, not you.) $\endgroup$ Aug 14, 2015 at 7:33
  • $\begingroup$ You already made that point in the first comment. And it was answered in the second: although there is nothing wrong with the word "idempotent", the word "projector" (or projection) can also be used for solutions of $x^2=x$ in a not necessarily commutative ring R, and such $x$ "project" onto one of the two subrings into which they split R as a direct sum. For an algebra C(X), this splitting coincides with a splitting of X into two parts. And multiplication-by-x is an endomorphism of the ring. $\endgroup$ Aug 14, 2015 at 7:45
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    $\begingroup$ @ASCIIAdvocate: Linguistitc questions aside, you should clarify what you mean by "equivalence" of algebras. Projections in $C(X)$ are the characteristic functions of clopen subsets and not all clopen subsets are connected components of $X$. $\endgroup$
    – MaoWao
    Aug 14, 2015 at 9:40
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While I do not know whether it will help you much, you may view a dictionary between topological concepts and their equivalent algebraic concepts on pages 6 and 13 of "Very Basic Non-commutative Geometry" by Masoud Khalkhali.

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