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Is it possible to define a structure of differentiable (smooth) manifold on the Hilbert cube $[0,1]^\mathbb{N}$ ?

Has it been done in the literature?

In textbooks, only the Banach case is treated, but the Hilbert cube has countable dimension, and a vector space with countable dimension is not complete (although the Hilbert cube is complete, because of compacity), this is a problem for the tangent space. However, can something be done? maybe with some restrictions?

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  • $\begingroup$ Isn't it part of the definition of a manifold that every point has a neighborhood that is homeomorphic to an n-dimensional ball? $\endgroup$
    – Goldstern
    Commented Oct 1, 2013 at 21:01
  • $\begingroup$ 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 $\endgroup$
    – Mostafa
    Commented Oct 1, 2013 at 21:11
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    $\begingroup$ 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. $\endgroup$
    – Andreas Blass
    Commented Oct 1, 2013 at 21:12
  • $\begingroup$ 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. $\endgroup$
    – Pietro Majer
    Commented Oct 2, 2013 at 0:22
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    $\begingroup$ Cross posted from math.stackexchange.com/questions/511341/… $\endgroup$
    – Kim Morrison
    Commented Oct 2, 2013 at 3:41

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In other words, the question is if the Hilbert cube can be homeomorphic to a differentiable Banach manidold. The answer is no, not even a topological manifold modelled on a topological vector space. Since the Hilbert cube is compact, in that case the model space would be locally compact, hence finite dimensional. But it is not the case of the Hilbert cube, because (for instance) it contains spheres of any finite dimension.

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    $\begingroup$ Isn't the OP asking whether one can define differentiable Hilbert cube manifolds, similarly to topological Hilbert cube manifolds? Your answer does not rule this out. $\endgroup$
    – Igor Belegradek
    Commented Oct 1, 2013 at 23:21
  • $\begingroup$ The question "Is it possible to define a structure of differentiable (smooth) manifold on the Hilbert cube" has a clear meaning to me (is there a differentiable atlas on it compatible with its topology), but maybe you are right, especially reading the OP's comment. In this case the main question would be Andreas Blass': what's a reasonable notion of differentiable maps on the Hilbert cube. $\endgroup$
    – Pietro Majer
    Commented Oct 2, 2013 at 0:16
  • $\begingroup$ yes, i asked a new question here: mathoverflow.net/questions/143765/… $\endgroup$
    – Mostafa
    Commented Oct 3, 2013 at 14:21

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