Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there known examples of compact infinite dimensional manifolds?

The word "manifold" is important.

share|improve this question
1  
How do you define an infinite dimensional manifold? Is it modelled on the countable product of lines? –  Igor Belegradek Oct 25 '13 at 0:56
    
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. –  Igor Belegradek Oct 25 '13 at 1:04
    
just modelled on a vector space. like Banach manifold, Frechet manifold. –  user8991 Oct 25 '13 at 1:07
    
Any separable Frechet (or Banach) space is homeomorphic to the countable product of lines, so my answer above applies. –  Igor Belegradek Oct 25 '13 at 1:18
    
It seems to me Pietro Majer already addressed this in his answer here: mathoverflow.net/a/143737/2926 Right? –  Todd Trimble Oct 25 '13 at 2:01
show 2 more comments

closed as unclear what you're asking by BS., Andrey Rekalo, Carlo Beenakker, Ricardo Andrade, David White Oct 25 '13 at 12:44

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
    
Feeling of deja vu... Ah yes: mathoverflow.net/a/143737/2926 –  Todd Trimble Oct 25 '13 at 1:38
add comment

Yes. Compact Hilbert cube manifolds, for instance.

share|improve this answer
    
wrong. the Hilbert cube is not a manifold mathoverflow.net/questions/143724/… –  user8991 Oct 25 '13 at 1:06
4  
perhaps you mean it is not a differentiable manifold. –  johndoe Oct 25 '13 at 1:09
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.