6
$\begingroup$

Let $M$ be a non-compact differential manifold. It is true that in general $H^q_c(M) \neq H^q(M)$, where $H^q_c$ is the de Rham's cohomolgy with compact support group and $H^q$ is the usual de Rham's cohomology group.

We have just begun the subject, so I don't have much confidence with it. I wanted to ask: is it true that for any non-compact $M$ there exists a $q$ for which $H^q_c (M) \neq H^q(M)$? Or are there any examples of non-compact manifolds for which the two cohomologies are the same for all $q$?

EDIT: ok, from the comments I gathered that if such an example exists it must be non-orientable (a reference to a proof would be nice, even though I think the resul is quite non-elementary). The question still remains open though (that's my main reason to editing: I think this question didn't get enough attention).

  • 3
    In general, for non-compact manifolds, the two are different. The easiest example is $\mathbb{R}^n$, where $H^n_c (\mathbb{R}^n) \approx \mathbb{Z}$, while $H^n(\mathbb{R}^n) = 0$2011-12-22
  • 4
    If $M$ is an $n$-dimensional non-compact connected oriented manifold, then $H^n(M)$ is always zero and $H^n_c(M)$ is one-dimensional. So any example is necessarily non-orientable.2011-12-22
  • 0
    @Arthur: yes, I know. That's what I said in the first paragraph; I was wondering if it were so for *all* non-compact manifold.2011-12-22
  • 0
    @DanPetersen: In fact, that H^n(M)=0 for noncompact is proved in Spivak (Vol1, chapter 8), but he uses a "Problem 20" in Chapter 8 that I think it is not correct (a manifold with two ends and infinite topology should be a counterexample). Do you know an elementary proof (e.g., no Poincare duality) of this without using the Problem 20?2014-10-20

1 Answers 1

3

Let $M$ be a connected non-compact manifold. Then $H_c^0(M)\cong 0$, and $H^0(M)\cong\mathbb{R}$.

Let $f:M\rightarrow \mathbb{R}$ be a function with $df=0$. Then $f$ is constant (this uses the connectedness of $M$). If $f$ is assumed to be compactly supported, this constant must be zero. If $f$ is not assumed to be compactly supported, all constants occur.