TITLE: Poincaré lemma on a space with trivial homology group
QUESTION [7 upvotes]: Today I read about Poincaré's lemma from do Carmo's book Differential Forms and Applications. It says that 

A closed differential $k$-form on a contractible space is exact.

I wonder if the condition of contractiblity could be weakened. 
I think the condition of contractilibity is too strong. In case of 1-form, 'simple connectedness' was sufficient. The proof for the case used the fact that 

a closed 1-form is locally exact, and 
a 1-form is exact iff the integral of the form on a closed curve is
0.

From these facts, we could deduce that integrals of a closed 1-form along homotopic curves are the same, so in a simply connected space any integral of a closed 1-form is 0, and thus the 1-form is exact. 
I want to generalize this to the case of a $k$-form in appropriate settings. In order to modify the proof for the case $k=1$, I have to check the following questions:

Is a closed $k$-form is locally exact?
A $k$-form $\omega$ is exact iff $\int_M \omega=0$ for any (orientable) $k$-manifold $M$ without boundary?

If they were true, I think it would imply that a closed differential $k$-form in a space which has trivial $k$-th homology group is exact. (Here I use the condition 'trivial $k$-th homology group' because we're dealing with homotopic $k$-manifolds)
Are the above questions true? If not, what could be a counterexample and what is the fundamental obstacle that makes it impossible to use the same reasoning in case $k>1$?
P.S. I found that the proof of the Poincaré's lemma (in do Carmo's book) was very much similar to the proof of the homotopy axiom of homology groups. I think it is basically because differential forms has a structure of contravariant functor and I heard that it has something to do with de Rham cohomology. Is there a close relation between a space's homology group and its de Rham cohomology? It seems there has to be. For instance Stoke's theorem relates the boundary operator for chain complexes and differential operator for differential forms...

REPLY [9 votes]: First off, yes, you're right: contractibility is a very strong hypothesis, and is overkill if all you want to know is that closed $k$-forms are exact.
You have heard about de Rham cohomology. Let me quickly recall what it is all about. Given a manifold $M$, the space $\Omega^k(M)$ is the (vector) space of $k$-forms on $M$. You have the linear map $d_k : \Omega^k(M) \to \Omega^{k+1}(M)$ given by the differential, and you know that $d_{k+1} \circ d_k = 0$ (essentially Schwarz's theorem on second derivatives). In other words, $\operatorname{im}(d_k) \subset \ker(d_{k+1})$, so that you can consider the quotient:
$$H^k_\mathrm{dR}(M) = \ker(d_k) / \operatorname{im}(d_{k-1}),$$
the $k$th de Rham cohomology group (it's actually a real vector space).
Directly from the definition, the statement "every closed $k$th form on $M$ is exact" is equivalent to $H^k(M) = 0$, because a closed $k$ form is an element of $\ker(d_k)$ and an exact $k$th form is an element of $\operatorname{im}(d_{k-1})$. So you want to know a sufficient condition for the vanishing of $H^k_\mathrm{dR}(M)$. As you've seen, contractibility is sufficient.
The famous de Rham theorem says that the $k$th de Rham cohomology group is actually isomorphic to the $k$th singular cohomology group with real coefficients:
$$H^k_\mathrm{dR}(M) \cong H^k_\mathrm{sing}(M; \mathbb{R}).$$
So $H^k_\mathrm{dR}(M)$ vanishes iff $H^k_\mathrm{sing}(M; \mathbb{R})$ vanishes. (I will drop the subscript $\mathrm{sing}$ from now on.)
The universal coefficient theorem, applied to the case of $\mathbb{R}$, says that:
$$H^k(M; \mathbb{R}) \cong \hom_\mathbb{Z}(H_k(M; \mathbb{Z}), \mathbb{R}),$$
because $\mathbb{R}$ is divisible hence the $\operatorname{Ext}$ term vanishes. Now, if you assume that $M$ is a compact manifold (this will simplify everything), $H_k(M; \mathbb{Z})$ will be a finitely generated abelian group, so by the structure theorem it will be the direct sum of a free abelian group and of a torsion group. If $T$ is a torsion group, $\hom_\mathbb{Z}(T, \mathbb{R}) = 0$, but $\hom_\mathbb{Z}(\mathbb{Z}, \mathbb{R}) \cong \mathbb{R}$. So in conclusion:

If $M$ is a compact manifold, then every closed $k$th form on $M$ is exact iff $H_k(M; \mathbb{Z})$ is torsion.

If $M$ is noncompact then you can at least say the following:

If $H_k(M; \mathbb{Z})$ is torsion, then every closed $k$th form is exact.

Because then $\hom_\mathbb{Z}(H_k(M; \mathbb{Z}, \mathbb{R}) = 0$. I don't think the reverse implication is necessarily true, though.

As for your specific questions (but I think the above should be enough for what you want):

Yes, a closed $k$th form is locally exact. This is because a manifold is locally diffeomorphic to $\mathbb{R}^n$, which is contractible, and the restriction of a closed $k$th form is still closed. Since a closed form on a contractible manifold is exact, you get the result. (Thus it is really essential to know the result you want to prove for contractible spaces, or at least for $\mathbb{R}^n$, first!)
I don't know if there's a better proof, but theorems of Thom tell you that every (real) homology class is realized by a submanifold (cf. here for example). So at least for an orientable manifold, this implies (by Poincaré duality and the explicit form of the isomorphism in de Rham's theorem) that a form is exact iff its integral vanishes on every submanifold.

But you should be careful here. I haven't worked out the details, but I suspect the Hurewicz isomorphism played some role in the proof for $k=1$; it doesn't hold if the space isn't $(k-1)$-connected for general $k$.