Are there known examples of compact infinite dimensional manifolds?
The word "manifold" is important.
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1$\begingroup$ How do you define an infinite dimensional manifold? Is it modelled on the countable product of lines? $\endgroup$– Igor BelegradekCommented Oct 25, 2013 at 0:56
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$\begingroup$ If this is your definition, then the answer is no, because if $K$ is a compact subset of a manifold $Y$ modelled on the countable product of lines, then $Y$ and $Y-K$ are homeomorphic. $\endgroup$– Igor BelegradekCommented Oct 25, 2013 at 1:04
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$\begingroup$ just modelled on a vector space. like Banach manifold, Frechet manifold. $\endgroup$– user8991Commented Oct 25, 2013 at 1:07
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$\begingroup$ Any separable Frechet (or Banach) space is homeomorphic to the countable product of lines, so my answer above applies. $\endgroup$– Igor BelegradekCommented Oct 25, 2013 at 1:18
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$\begingroup$ It seems to me Pietro Majer already addressed this in his answer here: mathoverflow.net/a/143737/2926 Right? $\endgroup$– Todd TrimbleCommented Oct 25, 2013 at 2:01
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2 Answers
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The empty space is a manifold of any dimension.
No, seriously, let's assume that "manifold" means a Hausdorff space in which every point has an open neighborhood homeomorphic to an open subset of a topological vector space. If the manifold is compact and nonempty then the vector space must be locally compact. As far as I know, that makes it finite-dimensional.
answered Oct 25, 2013 at 1:04
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$\begingroup$ Feeling of deja vu... Ah yes: mathoverflow.net/a/143737/2926 $\endgroup$ Commented Oct 25, 2013 at 1:38
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Yes. Compact Hilbert cube manifolds, for instance.
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$\begingroup$ wrong. the Hilbert cube is not a manifold mathoverflow.net/questions/143724/… $\endgroup$– user8991Commented Oct 25, 2013 at 1:06
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4$\begingroup$ perhaps you mean it is not a differentiable manifold. $\endgroup$– johndoeCommented Oct 25, 2013 at 1:09