Timeline for Can every manifold be given an analytic structure?
Current License: CC BY-SA 2.5
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| Oct 2, 2022 at 14:52 | comment | added | LSpice | Just to have it here, a comment from @MoiseKohan's answer: "If one looks at Whitney's proof … it is clear that the proof is not an application of Weierstrass' theorem but needs much more delicate approximation arguments which are based on Whitney's earlier work." | |
| Apr 3, 2013 at 9:25 | comment | added | Tim Campion | references (found in a 1998 posting to sci.math.research by ... Greg Kuperberg!)---- H. Grauert, On Levi's problem and the imbedding of real-analytic manifolds, Ann. of Math. 68 (1958), 460-472.---------------C. B. Morrey, The analytic embedding of abstract real-analytic manifolds, Ann. of Math. 68 (1958), 159-201. | |
| Dec 13, 2009 at 22:18 | comment | added | Greg Kuperberg | Yes, for instance you can make an analytic isotopy. Or, as in other steps in Whitney's stuff, you can perturb a diffeomorphism of $M'$ on $M$ into a nearby analytic map by Weierstrass approximation, and then project back onto $M$ using an embedded normal bundle (the normal exponential map). | |
| Dec 13, 2009 at 21:58 | comment | added | Ryan Budney | Oh, this is for the $M'$ that may not have an analytic embedding in Euclidean space -- ah, understood. Once you have $M'$ embedding analytically in euclidean space you include into a large enough Euclidean space and construct an analytic isotopy between the embeddings using Whitney's ideas? | |
| Dec 13, 2009 at 21:56 | comment | added | Ryan Budney | I'm a little confused about the "The question boils down to" comment. If the manifold is sitting in Euclidean space as an analytic submanifold, all the coordinate functions $x_1, x_2, \cdots, x_n$ on Euclidean space are analytic on Euclidean space so they're analytic on the manifold and they separate points. | |
| Dec 13, 2009 at 20:45 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |