Poincaré lemma on a space with trivial homology group 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...
 A: 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 $k$.
