Commutative spectral triples 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) 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. 
 A: 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:

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.

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, 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. 
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.

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:


*

*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 $X$.

*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, 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.

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.
