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For compact manifolds, Hodge Theory tells us that (de Rham) cohomology is finite dimensional. What about non-compact manifolds? That is:

Can non-compact manifolds have infinite dimensional cohomology?

If the answer is yes, is there an example for which this is easy to see?

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  • $\begingroup$ A necessary condition is that the manifold must not admit a finite good cover $\endgroup$ – Francesco Bilotta Jul 10 at 10:57
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The (non-compact) manifold $\mathbb R^2\setminus \mathbb Z^2$ should have infinite first homology and cohomology.

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Sure. Here are two examples:

  1. Let $M$ be the real line with all integer points removed. Then $M$ is the disjoint union of countably many open intervals, and $H^0 (M, \mathbb{R})$ is a real vector space of uncountably infinite dimension. (To be precise, it is the dual of the real vector space of countably infinite dimension.)

  2. Let $M$ be the real plane with the set $\{ (n, 0) : n \in \mathbb{Z} \}$ removed. It's not too hard to see that $M$ is homotopy-equivalent to a chain of countably many circles joined side-by-side. It follows that $H^1 (M, \mathbb{R})$ is a real vector space of uncountably infinite dimension – we can concoct differential forms with a prescribed singularity at each hole independently, and their cohomology classes are distinct because they can be distinguished by integrating along a loop around the relevant singularities.

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    $\begingroup$ However the last De Rham cohomology vanishes for non compact manifold. $\endgroup$ – Ali Taghavi Feb 23 '16 at 13:44
  • $\begingroup$ Interesting answer. Anyway I don't get why we have unconuntable dimension in 1. Shouldn't the zero cohomology group count the number of connected components? Does this hold only for the finite case ? $\endgroup$ – Francesco Bilotta Jul 10 at 10:55
  • $\begingroup$ The 0th homology group counts the number of connected components. The 0th cohomology group counts the number of locally constant functions. $\endgroup$ – Zhen Lin Jul 10 at 13:48

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