The corresponence between compact Hausdorff topological spaces and commutative unital $C^\*$-algebras is rather well known: Gelfand Najmark theorem gives perfect correspondence between these categories. What is very easy: to define a structure of $C^\*$-algebra (commutative and unital) on the algebra $C(X)$ where $X$ is compact Haudorff topological space. The real significance is the converse: namely, that each commutative unital $C^\*$-algebra is of this form. This is the beginning of the whole story which culminates in the definition of spectral triple and Connes' reconstruction theorem. As I understood, the situation is the following:  

-if a triple $(A,H,D)$ (where $A$ is unital) is a commutative spectral triple it does not follow that there is a compact orientable manifold $M$ such that $A=C^{\infty}(M)$  

-if we assume some additional technical conditions (about regularity, dimension and so on) then there is a compact orientable manifold $M$ such that $A=C^{\infty}(M)$-this is the celebrated and deep theorem of Connes.  

Furthermore, if I understood things correctly (see the discussion [Noncommutative smooth manifolds](https://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds)) the main problem is to identyify the algebra $A$ as $C^{\infty}(M)$. Once you have that, it is less deep result (let us call it *baby reconstruction theorem*) that in this case there is also a hermitian vector bundle $E$ and essentially self adjoint elliptic operator $D\_0$ such that $(A,H,D)=(C^{\infty}(M),L^2(M,E),D\_0)$. After the formulation of reconstruction theorem, in the original paper of Connes there is also the following statement: each compact orientable smooth manifold arise in such way. I would like to focus on this part: what is known (for me at least), that once you have smooth, compact, spin$^c$ manifold it defines a spectral triple. So my first question is the following:  

**Question 1** How it is *always* possible to obtain a commutative spectral triple from an arbitrary compact, smooth, orientable manifold (not necessary spin$^c$)?  

Furthermore: let us look for $C^\*$-algebra situation: then there is a canonical way to obtain the $C^\*$-algebra structure on $C(X)$. If we combine Connes' reconstruction theorem with baby reconstruction we obtain that if we have a commutative spectral triple $(A,H,D)$ such that $A=C^{\infty}(M)$ where now $M$ is *spin$^c$* (compact, smooth) then in fact the hermitian bundle $E$ is in fact a spinor bundle (according to the cited discussion). But if the answer to the first question is affirmative then it is natural to ask:  

**Question 2** What do we obtain if we start with an arbitrary smooth compact spin$^c$ manifold $M$: then form the spectral triple as in question 1 (as if $M$ would be not necessary spin$^c$) and after that apply reconstruction?  

More general question would be:  

**Question 3** Let $M$ be a smooth compact orientable manifolds. Is it somehow possible to parametrize all pairs $(H,D)$ such that $(A,H,D)$ is a spectral triple? How does it depends from the manifold $M$ (from the fact of being spin$^c$ etc.)


I would be grateful if anybody could clarify this issue for me. 



Commutative spectral triples
From the perspective of the Gelfand–Naimark theorem, the heart of the reconstruction theorem is the following statement, Theorem 11.4 in [Connes's paper](http://arxiv.org/abs/0810.2088):



> 
> Let $\mathcal{A}$ be a commutative unital complex $\ast$-algebra. Then $\mathcal{A} \cong C^\infty(X)$ for some compact oriented smooth $p$-manifold $X$ if and only there exist a faithful $\ast$-representation $A \to B(H)$ on a complex Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(\mathcal{A},H,D)$ is a $p$-dimensional commutative spectral triple.
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Your **Question 1** asks for the proof of the “only if” direction of this statement, which essentially goes as follows. Let $X$ be a compact orientable smooth manifold. Then you just pick your favourite Riemannian metric for $X$ and correspondingly form a commutative spectral triple as follows:


* if $X$ is even-dimensional, take $(C^\infty(X),L^2(X,\wedge T^\ast\_{\mathbb{C}} X),d+d^\ast)$;
* if $X$ is odd-dimensional, take $(C^\infty(X),L^2(X,(\wedge T^\ast\_{\mathbb{C}}X)^+),d+d^\ast)$, where $(\wedge T^\ast\_{\mathbb{C}}X)^+$ is the $+1$ eigenbundle of the chirality operator on $\wedge T^\ast\_{\mathbb{C}}X$.


Of course, there's some checking to do, as is indicated in Connes's account.


Now, if you combine the bare-bones reconstruction theorem with (the proof of) the “baby reconstruction theorem” in Gracia-Bondia–Varilly–Figueroa and a slight weakening of the orientability condition in the definition of commutative spectral triple, you can say even more (see Corollary 2.19 and its proof [here](http://arxiv.org/abs/1101.5908v4), though the account is a bit out of date):



> 
> Let $(\mathcal{A},H,D)$ be a $p$-dimensional commutative spectral triple. Then there exists a compact oriented Riemannian $p$-manifold $X$, a Hermitian vector bundle $E \to X$, and an essentially self-adjoint Dirac-type operator $D\_E$ (i.e., a first-order differential operator such that $D\_E^2 = -g^{ij}\partial\_i \partial\_j + \text{lower order terms}$) such that $$(\mathcal{A},H,D) \cong (C^\infty(X),L^2(X,E),D\_E),$$ where $\cong$ denotes unitary equivalence of spectral triples. 
> 
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> Conversely, if $X$ is a compact oriented Riemannian $p$-manifold, $E \to X$ is a Hermitian vector bundle, and $D\_E$ is an essentially self-adjoint Dirac-type operator on $E$, then $(C^\infty(X),L^2(X,E),D\_E)$ is a $p$-dimensional commutative spectral triple.
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This, in particular, answers your **Question 3** as restricted to commutative spectral triples. If you drop commutativity, no such classification exists. Indeed, there come to mind a couple of extremely straightforward ways to get spectral triples with algebra $C^\infty(X)$ that aren't commutative spectral triples:


1. Let $D$ be an essentially self-adjoint elliptic first-order differential operator on a Hermitian vector bundle $E \to X$ that *isn't* Dirac-type. Then $(C^\infty(X),L^2(X,E),D)$ is a perfectly good spectral triple that, in general, satisfies all the conditions for a commutative spectral triple except orientability and the additional “strong regularity” condition—the point is that orientability and strong regularity, together, would imply that $D$ was indeed Dirac-type. A somewhat silly example is $(C^\infty(X),L^2(X,E\_1 \oplus E\_2),D\_1 \oplus D\_2)$, where $D\_1$ and $D\_2$ are Dirac-type operators for distinct Riemannian metrics $g\_1$ and $g\_2$ on $X$.
2. Let $(C^\infty(X),L^2(X,E),D)$ be an honest $p$-dimensional concrete commutative spectral triple. If $M$ is some bounded self-adjoint operator on $L^2(X,E)$ that *isn't* a bundle endomorphism, then $(C^\infty(X),L^2(X,E),D + M)$ is a spectral triple of metric dimension $p$ that cannot possibly be a commutative spectral triple; indeed, $D$ cannot even be a differential operator. In particular, if you take to be a smoothing pseudodifferential operator (e.g., the orthogonal projection onto the kernel of $D$, if it's nonzero), then your new spectral triple should satisfy all the conditions for a commutative spectral triple except order one and orientability.


Finally, let me turn to your **Question 2**. Suppose you start with a concrete commutative spectral triple $(C^\infty(X),L^2(X,E),D)$ and feed it into the reconstruction theorem to get $(C^\infty(X^\prime),L^2(X^\prime,E^\prime),D^\prime)$. Then, in particular, you have an algebraic isomorphism $C^\infty(X) \cong C^\infty(X^\prime)$, which, by a [theorem of Mrčun's](http://arxiv.org/abs/math/0309179), is necessarily given by composition by a diffeomorphism $\phi : X^\prime \to X$. Using the various axioms and the smooth Serre–Swan theorem, you can then check that the unitary $U : L^2(X,E) \cong L^2(X^\prime,E^\prime)$ is, in fact, given by a unitary isomorphism of Hermitian vector bundles $(E^\prime,X^\prime) \cong (E,X)$ covering $\phi : X^\prime \to X$, and hence, since $U^\ast D^2 U = (D^\prime)^2$, that $\phi$ was an isometry. Indeed, if you compare the Hermitian metrics on $E$ and $E^\prime$ with the inner products on the Hilbert spaces, you can even conclude that $\phi$ was orientation-preserving. Hence, if I'm not too mistaken, you can even refine the above refinement of the reconstruction theorem as follows:



> 
> Let $(\mathcal{A},H,D)$ be a $p$-commutative spectral triple. Then there exists a compact oriented Riemannian $p$-manifold $X$, a Hermitian vector bundle $E \to X$, and an essentially self-adjoint Dirac-type operator $D\_E$ (i.e., a first-order differential operator such that $D\_E^2 = -g^{ij}\partial\_i \partial\_j + \text{lower order terms}$) such that $$(\mathcal{A},H,D) \cong (C^\infty(X),L^2(X,E),D\_E),$$ where $\cong$ denotes unitary equivalence of spectral triples. Morever, the data $(X,E,D\_E)$ is unique up to orientation-preserving isometry together with unitary bundle isomorphism covering the isometry that intertwines Dirac-type operators.
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Note, though that being or not being spin$^\mathbb{C}$ or spin is completely irrelevant to the machinery of the reconstruction theorem; the point is that you can form, for instance, the Hodge–de Rham spectral triple (i.e., the $d+d^\ast$ spectral triple) of a compact Riemannian spin$^\mathbb{C}$ manifold *without* choosing a spin$^\mathbb{C}$ structure. The only way this formalism can even see a spin$^\mathbb{C}$ structure is by choosing your Hermitian vector bundle to have been a spinor bundle—in light of Connes's Theorem 1.2, for an abstract $p$-dimensional commutative spectral triple $(\mathcal{A},H,D)$, this is equivalent to requiring that the representation of $\mathcal{A}^{\prime\prime}$ on $H$ have the correct spectral multiplicity $2^{\lfloor p/2 \rfloor}$, corresponding to the correct rank of a spinor bundle. In particular, if you start with a given spinor bundle, then the reconstruction theorem will necessarily spit out an isomorphic spinor bundle, so that if you do choose a spin$^\mathbb{C}$ structure by choosing a spinor bundle, then the reconstruction theorem can't change the spin$^\mathbb{C}$ structure. Again, the Hodge–de Rham spectral triple is completely agnostic about any spin$^\mathbb{C}$ structure, so that the reconstruction theorem, on its own, is completely agnostic about spin$^\mathbb{C}$ structures.


