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It is trivial to prove that the integration of a $(n-1)$ exact form on the boundary of a $n$-manifold is 0. What about the contraposative ? If the integration of a $(n-1)$-form on the boundary of a $n$-manifold is 0, is this form exact ? If not, are there particular conditions to satisfy for this to be the case ?

For example, in the case of $S^1$, any $1$-form can be written as $f(\theta)d\theta=c d\theta+dg(\theta)$, $c$ being the integral around $S^1$, and $g$ a differentiable function on $S^1$. Now in this case if the integral is 0, it implies that the form is exact. I wondered in the general case if such decomposition is always possible, because if so, it can be proven that the integral being 0 implies that the form is exact. Is this correct ?

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Won't any closed $(n-1)$-form on an $n$-manifold integrate to 0 on the boundary of that manifold? –  Julian Rosen Jul 20 '13 at 0:42
    
The question did not hypothesize it being closed, I think. –  Eric Auld Jul 20 '13 at 2:02
    
Since beign closed is a necessary condition to be exact in any case, maybe I could rephrase my question as "If the integration of a (n−1)-form closed form on the boundary of a n-manifold is 0, is this form exact ?" –  vkubicki Jul 20 '13 at 8:57
    
If the form is closed and the domain is contractible, that implies that the integral is 0 (by the Poincarré lemma), but I think I am looking at the implication in the other direction, from the value of the integral to the characterization of the form. –  vkubicki Jul 20 '13 at 9:20
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3 Answers 3

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I think the result you are looking for is this: Top deRham cohomology group of a compact orientable manifold is 1-dimensional, with the isomorphism $H^n_{dR}(M^n)\to \mathbb R$ given by $\omega\mapsto \int_M \omega$. (One should add "connected" to the assumptions.) In your question, $n$ is actually $n-1$.

The proof of $H^n_{dR}(M^n)\approx \mathbb R$ is not hard but lengthy: see section 8.1, titled de Rham cohomology in the top dimension, of Nigel Hitchin's notes.

See also Poincaré Duality with de Rham Cohomology. And for the non-compact case, De Rham cohomology for non-compact manifolds.

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Thank you for your answer. May I ask a complementary question ? In the case of a non-top cohomology group, say for the torus as in topospaces.subwiki.org/wiki/Torus , is it possible to link the rank $\binom {n} {k}$ to the values of various integrals on the domain ? –  vkubicki Jul 27 '13 at 10:36
    
@vkubicki Interesting, but a new question should be posted as separate question, not as a comment.. –  40 votes Jul 27 '13 at 14:41
    
You are right, the scope is a bit larger than a comment. So there it is, waiting at math.stackexchange.com/questions/453644/… . –  vkubicki Jul 27 '13 at 19:29
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Stokes' theorem works both ways, but you have to be careful. In particular, a $n$-form $\omega$ is closed if and only if its integral over every $n$-boundaries is zero, not just one particular!

For every $(n+1)$-dimensional region $\Omega$ we have: $$ \int_{\partial\Omega}\omega=\int_\Omega d\omega. $$

If the expression on the left is zero for every $\Omega$, so is the one on the right, and it means that $d\omega=0$.

If we want $\omega$ to be exact, then we don't have to limit ourselves to $n$-boundaries, but we want its integral to be zero on every closed region! That is, any region $A$ with $\partial A=0$.

This because what cause a closed form not to be exact are the very "holes".

I don't know if from this requirement you can explicitly find that $\omega=d\eta$ for some $\eta$ (that is, I don't know if this reasoning provides you such an $\eta$).

What is immediate to check (just use Stokes), though, is that exact forms yield a zero integral over general closed regions, not just boundaries.

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No, and my limited knowledge of the topic is that it depends heavily on the topology of a space. For instance, in a multiply connected 2-manifold, there are closed forms which are not exact (they are not the boundary of any submanifold, since they "go around they hole").

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OK. I edited my question regarding the impact of the holes in a manifold. –  vkubicki Jul 20 '13 at 9:19
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