Noncommutative smooth manifolds

2010-03-02T06:10:47Z

Connes defined a noncommutative analog of a closed oriented Riemannian spin^c manifold using spectral triples.

Using his definition it is unclear how to separate the smooth structure from the metric.

How can we define a noncommutative smooth manifold without the additional Riemannian and spin^c structures?

Any references on this subject will be appreciated.

Answer by Harold Williams for Noncommutative smooth manifolds

2010-03-03T07:50:50Z

I believe the closest answer is in Connes' On the Spectral Characterization of Manifolds. The main theorem is that if a (commutative) spectral triple (A,H,D) satisfies a list of certain nice properties, then A is the algebra of smooth functions on a compact oriented smooth manifold. I'm not sure this really separates the smooth structure and metric data, but hopefully the reference is still useful.

Answer by BigBill for Noncommutative smooth manifolds

2011-06-13T05:04:54Z

The right notion is the one of `smooth subalgebra' of a C*-algebra.

See by example the following paper, page 27:

Noncommutative spectral geometry of Riemannian foliations: some results and open problems

http://front.math.ucdavis.edu/0601.5093

Answer by Branimir Ćaćić for Noncommutative smooth manifolds

2013-01-01T21:59:16Z

I'm a bit wary of resurrecting such an old question, but given that the precise content of the reconstruction theorem doesn't seem to be terribly well disseminated, please permit me to cross-post from math.SE and then make some extra comments:

"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:

A unital Frechet pre-$C^\ast$-algebra $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact orientable $p$-manifold if and only if there exists a $\ast$-representation of $A$ on a Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$.
In particular, $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact spin$^{\mathbb{C}}$ $p$-manifold if and only if there exist $H$ and $D$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$ and $A^{\prime\prime}$ acts on $H$ with multiplicity $2^{\lfloor p/2\rfloor}$.

"Once you know that $A \cong C^\infty(X)$, you can then apply the much earlier "baby reconstruction theorem" (for lack of a better phrase) announced by Connes and proved in detail by Gracia-Bondia--Varilly--Figueroa to conclude that:

In the general case, $(A,H,D) \cong (C^\infty(X),L^2(X,E),D)$ where $E \to X$ is a Hermitian vector bundle and $D$ can be interpreted as an essentially self-adjoint elliptic first-order differential operator on $E$.
In the case where $A^{\prime\prime}$ acts with multiplicity $2^{\lfloor p/2 \rfloor}$, $E \to X$ is in fact a spinor bundle (i.e., irreducible Clifford module bundle) and $D$ is Dirac-type (viz, a perturbation of a spin$^{\mathbb{C}}$ Dirac operator by a symmetric bundle endomorphism of $E$).

"So, whilst you can refine the reconstruction theorem to a characterisation of compact spin$^{\mathbb{C}}$ manifolds with spinor bundle and essentially self-adjoint Dirac-type operator, the general result is really just a statement about compact orientable manifolds. Indeed, one can even refine the reconstruction theorem to a characterisation of compact oriented Riemannian manifolds with self-adjoint Clifford module and essentially self-adjoint Dirac-type operator."

As for why you need more than just an algebra $A$, here's what I understand of the situation:

Gel'fand--Naimark says that a commutative unital $C^\ast$-algebra gives a compact Hausdorff space, no more and no less.
Going by the example of $C^\infty(X) \subset C(X)$, one might try considering commutative unital Frechet pre-$C^\ast$-algebras, but one can readily cook up examples of such algebras that aren't isomorphic to $C^\infty(X)$ for some $X$. So, if one still wishes to follow this line of inquiry, one would need still more structure.
In terms of the reconstruction theorem itself (which does tend to be treated as a black box), Connes's proof (insofar as I can understand) really makes absolutely essential use of all three parts of the spectral triple $(A,H,D)$:
- the norm closure of $A$ in $B(H)$, by Gel'fand--Naimark, gives you a (canonical) compact Hausdorff space $X$;
- the noncommutative integral defined by $D$ gives a Radon measure on $X$;
- the Hochschild cycle of the orientability condition gives you candidates for charts;
- the operator $D$ then gets used to show that you actually have a smooth atlas.
This doesn't suggest, of course, that a spectral triple is the optimal notion of algebra + extra data qua noncommutative manifold, but it does suggest that it might well be a reasonably economical definition. In particular, it suggests that the real puzzle with the spectral triple formalism isn't the extra Riemannian data, but rather the seemingly absolute necessity of orientability.

My apologies for the long-windedness!

Noncommutative smooth manifolds - MathOverflow

Noncommutative smooth manifolds

Answer by Harold Williams for Noncommutative smooth manifolds

Answer by BigBill for Noncommutative smooth manifolds

Answer by Branimir Ćaćić for Noncommutative smooth manifolds