11
$\begingroup$

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.

Obviously, I don't expect "compact manifold" as an answer; although this is a sufficient condition, it is far from answering the question.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ The cohomology of a manifold can be infinite dimensional in many ways. You might get a better answer if you add a hypothesis like finitely many ends, or admits a complete riemannian metric. $\endgroup$
    – Paul
    Commented Feb 17, 2011 at 0:15
  • 3
    $\begingroup$ @Paul I don't understand your comment. Being finite or infinite dimensional is a well established notion for a vector space (like a Rham cohomology space). And you can't ask for more precise hypothesis when the question is "What hypothesis do I need?". $\endgroup$
    – AFK
    Commented Feb 17, 2011 at 0:35
  • $\begingroup$ @YBL. John K. knows much more about this than I do, but IIRC the "almost" in John's answer is that $M$ should be dominated by a finite complex, which is stronger than having finite dimensional homology. For example, the infinite connected sum of $RP^3$s has finite dimensional deRham cohomology, but isn't finitely dominated. My comment was meant to find out what was MT's context for the question. $\endgroup$
    – Paul
    Commented Feb 17, 2011 at 1:53
  • 7
    $\begingroup$ Just to be sure: the isomorphism between de Rham and singular cohomology with $\mathbb{R}$-coefficients holds for all smooth manifolds, right? So this question isn't really about differential forms per se, but rather about the algebraic topology of smooth manifolds. $\endgroup$
    – Pete L. Clark
    Commented Feb 17, 2011 at 5:44

1 Answer 1

Reset to default
17
$\begingroup$

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.

This result was later generalized in Siebenmann's thesis to the non-simply connected case.

Addendum. Here's a link to Siebenmann's thesis:

www.math.uchicago.edu/~shmuel/tom-readings/Siebenmann%20thesis.pdf

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .