Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Compact Hausdorff spaces bijectively correspond to C^*-algebras with identity. One needs to consider the algebra of continuous functions C(X) to go in one direction and spectrum to go in the other. (See e.g. Wikipedia). The situation is similar to algebraic geometry - affine manifolds correspond to commutative algebras... Basic skill in alg.geom. is to recast algebraic properties in geometric and vice versa e.g. projective modules - vector bundles... (the dictionary is lengthy).

I wonder about similar correspondence in C^*-algebra setup. In particular:

Question 1: if space "X" is topological manifold (i.e. locally R^n), is there some "nice" way to recognize it via C^*-algebra of continuous function ? (... is there non-commutative version ? ... )

Question 2: if "X" is smooth manifold, is there nice way to recognize it and define sub-algebra of smooth functions entirely in terms of C^*-algebra ? (... is there non-commutative version ? ... )

Question 3 Is it possible to characterize the set of all measures on "X" in terms of C(X) ? (... is there non-commutative version ? ... )

If you have further comments how interesting algebraic properties can be recasted in topological or vice versa, you are welcome to post.

share|improve this question
1  
Surely we have had some version of this question before on MO? (It usually comes with someone enthusiastically quoting an NCG dictionary while not quite paying enough attention to the issues for non-compact spaces...) –  Yemon Choi Jan 6 '13 at 21:45
1  
And surely to answer your Q2 one should look for extra structure on your $C^\ast$-algebra, not just "entirely in terms of $C^\ast$-algebra" - as in the case of a smooth manifold being a topological space with extra structure –  Yemon Choi Jan 6 '13 at 21:47
    
@Yemon all spaces are compact in my question. I'm primarily interested in the specific questions above, not in general dictionary, but if someone wants to share something interesting I would not be against. –  Alexander Chervov Jan 6 '13 at 21:49
4  
It is by now fairly well understood how to give an operator algebraic characterization of a Riemannian manifold (well, maybe only some of them). However, it is not fair to call it a C*-algebraic characterization because part of the data needed is a suitable dense subalgebra of the given commutative C*-algebra. As far as I know, nobody has figured out the right axioms for a noncommutative topological manifold, or even a noncommutative smooth manifold. –  Paul Siegel Jan 7 '13 at 0:17
2  
There are lots of related (identical?) MO questions: mathoverflow.net/questions/21168/…, mathoverflow.net/questions/82871/…, mathoverflow.net/questions/100461/…, etc. –  Martin Brandenburg Jan 7 '13 at 1:31
show 1 more comment

2 Answers

Question 1: The topological $n$-manifold property is equivalent to every point of $X$ having a neighborhood homeomorphic to $B^n$, the closed unit ball in $\mathbb{R}^n$. The existence of such a neighborhood $x\in B^n_x \subset X$ induces the surjective algebra homomorphisms $C(X) \to C(B_x^n) \cong C(B^n) \to C(\{x\})\cong \mathbb{R}$ (actually, extremal epimorphisms, I think). The Gelfand duality between compact Hausdorff topological spaces and commutative real $C^*$ algebras ensures that the existence of surjective homomorphisms $C(X) \to C(B^n) \to \mathbb{R}$ (the first map should be an extremal epimorphism, while the second should correspond to the quotient by the maximal ideal of an interior point of $B^n$) implies the existence of continuous maps $\{x\} \to B^n \to X$, where $x$ maps to an interior point of $B^n$ and $B^n$ is embedded in $X$, and hence a neighborhood of $x$ in $X$. Having such such algebra homomorphisms for each $x\in X$ characterizes $X$ as a topological manifold.

Question 2: $C^\infty(X)$ for a compact manifold $X$ is not a $C^*$ algebra. It is at the very least a Fréchet algebra, where multiplication satisfies an extra convexity condition (though I'm fuzzy on the details). It is at least clear that one must leave the category of $C^*$ to characterize it. A point that non-commutative geometry centered discussions of this questions seem to be ignoring is that there already exists an algebraic characterization of $C^\infty(X)$ that has nothing to do with non-commutative geometric spectral triples. In my understanding, such a characterization can be found in an article by Michor and Vanžura (arXiv:math/9404228).

Question 3: As already mentioned in Vahid's answer, this is the content of the Riesz representation theorem. The topological vector space dual to $C(X)$ is the space of signed Radon measures on $X$. $C(X)$ is partially ordered by pointwise comparison, which also induces a partial order on its dual. The positive cone in the space of signed Radon measures consists of the positive Radon measures.

I cannot say anything about non-commutative versions of the above answers. But, since these correspondences are heavily based on lots of non-trivial maximal ideals, and such ideals are likely to be absent in non-commutative algebras, they probably do not translate directly.

share|improve this answer
    
Thank you very much for your informative answer ! So you reduce a question to characterization of C(B^n), is it possible to give some somewhat "intrinsic" characterization of this algebra ? (It seems Michor&K uses similar idea). –  Alexander Chervov Jan 7 '13 at 12:18
    
B is open disk ? $C(B)$ is continuous functions on vanishing at inf, correct ? –  Alexander Chervov Jan 7 '13 at 12:35
    
As written in the first sentence, $B^n$ is the closed ball, which includes the boundary and hence is compact. Otherwise applying the standard Gelfand duality is problematic. So, $C(X)$ is a $C^*$ algebra without imposing any restrictions on behavior at the boundary. As for an intrinsic characterization, how about a topological completion of $\mathbb{\R}[x_1,\ldots,x_2]$ in the norm of uniform convergence on $B^n$? –  Igor Khavkine Jan 7 '13 at 13:56
    
Then what about intrinsic characterization of R[x_1,...x_n] ? and its completion (so we need norm, norm is sup_{ball}, so again we need ball)... –  Alexander Chervov Jan 7 '13 at 16:13
1  
@Branimir, from what I can see after a quick look, the answer is yes. Another book that covers smooth algebras and the geometric spaces dual to them is Models for smooth infinitesimal analysis by Moerdijk & Reyes. –  Igor Khavkine Jan 8 '13 at 0:00
show 3 more comments

The answer to your first (and second) question is negative, because commutative C*-algebras only reflect global features of the underlying space.
For the third question: we have the Riesz representation theorem which says: For locally compact and Hausdorff topological space $X$, there is and isometric isomorphism between the dual of $C_0(X)$ and the space of radon measures on $X$. See Theorem 7.17 in Folland's book "Real analysis".

share|improve this answer
    
I think it is Radon "charges", not measures. Charges - need not be positive, e.g. difference of two measures in general is "charge". And measures are not linear spaces in general. In Q3 I am interested how to distinguish this positive part among all "charges" ? I do not quite understand the first sentence/argument. –  Alexander Chervov Jan 6 '13 at 22:08
    
It is complex Radon measures. There is a bijective correspondence between open subsets of $X$ and closed two sided ideals of $C_0(X)$. But we cannot consider closed two sided ideals as a substitute for open sets in noncommutative $C^*$-algebras, because there are simple $C^*$-algebras, for example $M_n(\mathbb{C}$ for every $n\in \mathbb{N}$. Then all simple $C^*$-algebras should be considered as a noncommutative space with one point which is obviously naive. –  Vahid Shirbisheh Jan 6 '13 at 22:32
    
Having very few open sets is not the same as having few points, and even if you do have few "points" that doesn't mean the "noncommutative space" is close to trivial. See for instance the crossed product algebras that arise from non-proper group actions –  Yemon Choi Jan 6 '13 at 23:46
    
@Yemon Choi: Assuming hausdorffness of $X$ having few open sets means having few points. For the noncommutative case I used the word "naive" to say it is not a suitable point of view. –  Vahid Shirbisheh Jan 6 '13 at 23:55
1  
Vahid, as I'm sure you know, one of the whole points of NCG is to find replacements for non-Hausdorff quotient or moduli spaces. Cf the non-Hausdorffness of Zariski topology in classical algebraic geometry. So arguments invoking the Hausdorffness of X seem beside the point when discussing NC generalizations, we know that is too restrictive. –  Yemon Choi Jan 7 '13 at 0:26
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.