# What, and how can, topological invariants can be computed from a space's algebra of functions?

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?

• This is relevant: mathoverflow.net/questions/82708/…. In particular, the number of connected components comes from the number of idempotent elements $C_0(X)$. Aug 14, 2015 at 2:34
• 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. Aug 14, 2015 at 4:39
• @MarianoSuárez-Alvarez, the question is clearly on how to concretize the NCG dictionary for commutative algebras. Do you know references beyond Connes' book? Aug 14, 2015 at 6:32

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.

• 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. Aug 14, 2015 at 7:24
• 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. Aug 14, 2015 at 7:30
• 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.) Aug 14, 2015 at 7:33
• 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. Aug 14, 2015 at 7:45
• @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 $X$. Aug 14, 2015 at 9:40

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.