Timeline for How to classify the algebras C^∞(M)?
Current License: CC BY-SA 2.5
8 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 21, 2010 at 17:02 | vote | accept | Martin Brandenburg | ||
| Apr 16, 2010 at 14:44 | comment | added | Martin Brandenburg | very impressive theorem, though I won't understand it. thanks Tim for the reference! as yemon pointed out, I'm hoping for a purely algebraic formulation. and yes, this seems to be only a dream. | |
| Apr 14, 2010 at 20:41 | comment | added | Tim Perutz | Yemon: I agree with your reading of Martin's question. All the same, and notwithstanding my previous comment, Connes' condition ARE conditions on the algebra! No sheaves are invoked. They aren't conditions on the structure of its ideals, but instead on its representation theory. | |
| Apr 14, 2010 at 20:04 | comment | added | Yemon Choi | Tim - I actually agree that the spectral triple formulation is very natural, and I have no problem with it. It just seemed to me that the original post's wording, "I want some nontrivial purely algebraic formulation, if possible avoiding structure sheaves at all", was ambitiously looking for something only using the bare algebra. In other words, a bunch of conditions that can be formulated solely in terms of ${\mathbb C}$-algebra notions. But perhaps I misunderstood. | |
| Apr 13, 2010 at 21:31 | comment | added | Tim Perutz | Yemon - this issue was raised earlier on MO, but I don't quite see the objection. Slightly different to what? Is there a comparable theorem that invokes only the bare algebra? I don't see one mentioned in Connes' paper. The roles of the metric here - in defining a Hilbert-space completion of a projective module of smooth sections of a vector bundle, and in defining a differential operator from which we can find our original module inside that Hilbert space - seem very natural to me, both geometrically and analytically. | |
| Apr 13, 2010 at 21:02 | comment | added | Yemon Choi | Tim, I'm no geometer, but doesn't Connes' construction assume more and (hence) obtain more? The original question asks about $M$ with only a smooth structure and not a choice of Riemannian metric. (For what it's worth, I'm not claiming this approach is better or worse; just slightly different.) | |
| Apr 13, 2010 at 17:38 | history | answered | Tim Perutz | CC BY-SA 2.5 |