most recent 30 from http://mathoverflow.net 2013-05-22T01:52:01Z http://mathoverflow.net/feeds/question/55672 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55672/finite-dimensionality-for-de-rham-cohomology M Turgeon 2011-02-16T22:53:18Z 2011-02-17T23:37:11Z

<p>I was browsing through the litterature, hoping to find sufficient and necessary conditions for a smooth manifold to have finite-dimensional de Rham cohomology, but I can't find any satisfactory answer. I wonder if anyone has ever encountered a paper, or a book, answering (possibly in part) the question. I am especially interested in real-coefficient cohomology, but I would appreciate answers related to cohomology with coefficients in any abelian group.</p> <p>Obviously, I don't expect "compact manifold" as an answer; although this is a sufficient condition, it is far from answering the question.</p>

http://mathoverflow.net/questions/55672/finite-dimensionality-for-de-rham-cohomology/55678#55678 John Klein 2011-02-17T00:15:00Z 2011-02-17T23:37:11Z

<p>We can restrict your problem to the case of open manifolds. It turns out that "finite dimensional" integral singular homology (i.e., finitely generated in each degree) is almost the same thing as the manifold being the interior of a compact manifold with boundary. For example, if $M$ is a 1-connected and open manifold of dimension $>5$, then the Browder-Levine-Livesay theorem says that $M$ is the interior of a compact manifold with boundary (where the boundary is also 1-connected) iff the homology of $M$ is finitely generated and $M$ is $1$-connected at infinity. </p> <p>This result was later generalized in Siebenmann's thesis to the non-simply connected case.</p> <p><strong>Addendum.</strong> Here's a link to Siebenmann's thesis: </p> <p>www.math.uchicago.edu/~shmuel/tom-readings/Siebenmann%20thesis.pdf</p>