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Let $M$ be a real, smooth manifold. Let $\omega_1$ and $\omega_2$ be differential 1-forms on $M$, and let $C_1(\mathbb Z,M)$ denote the set of 1-chains with integer coefficients.

If \begin{align} \int_c\omega_1 = \int_c\omega_2 \end{align} for all $c\in C_1(\mathbb Z,M)$, then does $\omega_1=\omega_2$?

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@BrunoJoyal I am not so familiar with cohomology; if I restrict the topology of the manifold in some way, can I ensure that they are equal? – joshphysics Feb 5 '14 at 16:44
Sorry, I just saw that you are talking about chains rather than cycles (closed chains). In this case it is true – Bruno Joyal Feb 5 '14 at 16:49
@BrunoJoyal Haha ok thanks. I was just so confused; I went back to "Calculus on Manifolds" to make sure I wasn't going crazy. So the original result I wrote is true, regardless of topology? If so, is there a simple proof? – joshphysics Feb 5 '14 at 16:52
Choose a base point and integrate the difference along paths. This gives you a function on the universal cover which by assumption is the zero function. Its differential descends down to a differential back on the original manifold, which by construction is the difference of the two differentials we started with, yet is zero. Thus the two differentials are equal – Bruno Joyal Feb 5 '14 at 16:58
@BrunoJoyal Ah interesting ok. I need to learn what the universal cover is, but thanks. – joshphysics Feb 5 '14 at 17:30

Yes. Let $\omega$ be a 1-form on $M$. You need to show that $\omega$ vanishes if its integral along all curves in $M$ vanishes. Fix a point $x\in M$ and $v\in T_x M$ and integrate $\omega$ along a curve starting in $x$ in the direction of $v$. By making the curve shorter and shorter, these integrals give better and better approximations to $\omega(v)$, hence $\omega(v)$ must vanish.

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Thanks; that's a nice argument. – joshphysics Feb 6 '14 at 17:56

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