Noncommutative smooth manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:50:32Z http://mathoverflow.net/feeds/question/16833 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds Noncommutative smooth manifolds Dmitri Pavlov 2010-03-02T06:10:47Z 2013-01-01T21:59:16Z <p>Connes defined a noncommutative analog of a closed oriented Riemannian spin^c manifold using spectral triples.</p> <p>Using his definition it is unclear how to separate the smooth structure from the metric.</p> <p>How can we define a noncommutative smooth manifold without the additional Riemannian and spin^c structures?</p> <p>Any references on this subject will be appreciated.</p> http://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds/16961#16961 Answer by Harold Williams for Noncommutative smooth manifolds Harold Williams 2010-03-03T07:50:50Z 2010-03-03T07:50:50Z <p>I believe the closest answer is in Connes' <a href="http://arxiv.org/abs/0810.2088" rel="nofollow">On the Spectral Characterization of Manifolds</a>. 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.</p> http://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds/67627#67627 Answer by BigBill for Noncommutative smooth manifolds BigBill 2011-06-13T05:04:54Z 2011-06-13T05:04:54Z <p>The right notion is the one of `smooth subalgebra' of a C*-algebra.</p> <p>See by example the following paper, page 27:</p> <p>Noncommutative spectral geometry of Riemannian foliations: some results and open problems</p> <p><a href="http://front.math.ucdavis.edu/0601.5093" rel="nofollow">http://front.math.ucdavis.edu/0601.5093</a></p> http://mathoverflow.net/questions/16833/noncommutative-smooth-manifolds/117813#117813 Answer by Branimir Ćaćić for Noncommutative smooth manifolds Branimir Ćaćić 2013-01-01T21:59:16Z 2013-01-01T21:59:16Z <p>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 <a href="http://math.stackexchange.com/a/268334/49610" rel="nofollow">cross-post from math.SE</a> and then make some extra comments:</p> <p>"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:</p> <ul> <li>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 $p$.</li> <li>In particular, $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact <em>spin$^{\mathbb{C}}$</em> $p$-manifold if and only if there exist $H$ and $D$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$ <em>and</em> $A^{\prime\prime}$ acts on $H$ with multiplicity $2^{\lfloor p/2\rfloor}$.</li> </ul> <p>"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:</p> <ul> <li>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 $E$.</li> <li>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 $E$).</li> </ul> <p>"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."</p> <p>As for why you need more than just an algebra $A$, here's what I understand of the situation:</p> <ol> <li>Gel'fand--Naimark says that a commutative unital $C^\ast$-algebra gives a compact Hausdorff space, no more and no less.</li> <li>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 <em>aren't</em> 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.</li> <li><p>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 $(A,H,D)$:</p> <ul> <li>the norm closure of $A$ in $B(H)$, by Gel'fand--Naimark, gives you a (canonical) compact Hausdorff space $X$;</li> <li>the noncommutative integral defined by $D$ gives a Radon measure on $X$;</li> <li>the Hochschild cycle of the orientability condition gives you candidates for charts;</li> <li>the operator $D$ then gets used to show that you actually have a smooth atlas. </li> </ul> <p>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.</p></li> </ol> <p>My apologies for the long-windedness!</p>