Timeline for Is it possible to define a structure of differentiable manifold on the Hilbert cube $[0,1]^\mathbb{N}$?
Current License: CC BY-SA 3.0
11 events
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| Jun 4, 2015 at 19:15 | comment | added | N Unnikrishnan | "...but the Hilbert cube has countable dimension, and a vector space with countable dimension is not complete..." The Hilbert cube is not a (topological) vector space, as it is compact, while all TVS have got to be unbounded. You're confusing the Hamel dimension with the topological dimension. | |
| Oct 2, 2013 at 10:29 | comment | added | David White | You can find a useful method of determining if a topological space can be given a manifold structure in Lee's "Introduction to Smooth Manifolds." On page 21 he gives the Smooth Manifold Construction Lemma. Basically, you need to know that your space satisfies the basic topological properties (paracompact, second countable) and then you need an idea of what the atlas should be. This lemma tells you exactly what must be checked. In your case you have the topological properties, but no atlas. Also, there's the issue of infinite dimensionality. | |
| Oct 2, 2013 at 7:48 | comment | added | Willie Wong | @ScottMorrison: I migrated the MSE version here. Please merge. | |
| Oct 2, 2013 at 3:41 | comment | added | Kim Morrison | Cross posted from math.stackexchange.com/questions/511341/… | |
| Oct 2, 2013 at 0:22 | comment | added | Pietro Majer | A convenient class is maybe all restrictions of differentiable maps definined on nbds of the Hilbert cube, thought as a subset of a Hilbert space. | |
| Oct 1, 2013 at 21:46 | answer | added | Pietro Majer | timeline score: 4 | |
| Oct 1, 2013 at 21:12 | comment | added | Andreas Blass | I've seen references to "Hilbert cube manifolds", of which the Hilbert cube must be one, but this was in a topological context, and I don't know whether there's a reasonable notion of differentiability in this context. | |
| Oct 1, 2013 at 21:11 | comment | added | Mostafa | I am actually asking if this definition can be extended to this case. there already exists Banach manifolds, which are infinite dimensional. Hibert cube manifolds are defined here: ams.org/journals/bull/1970-76-06/S0002-9904-1970-12660-X/… but i am not sure if it helps | |
| Oct 1, 2013 at 21:01 | comment | added | Goldstern | Isn't it part of the definition of a manifold that every point has a neighborhood that is homeomorphic to an n-dimensional ball? | |
| Oct 1, 2013 at 20:56 | history | edited | Mostafa | CC BY-SA 3.0 |
added 327 characters in body; edited tags; edited title
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| Oct 1, 2013 at 19:34 | history | asked | Mostafa | CC BY-SA 3.0 |