# Differential form is closed if the integral over a curve is rational number.

The following problem comes from do Carmo's book Differential Forms and Applications, Chapter 2, Exercise 4:

Let $\omega$ be a differentiable 1-from defined on an open subset $U\subset \mathbb{R}^n$. Assume that for each closed differential curve $C$ in $C$, $\int_C \omega$ is a rational number. Prove that $\omega$ is closed.

Can anyone give some hints? My idea was that we could try to show the integral vanishes if the curve lies in $N_\epsilon (p)$ (i.e. a small neighborhood around $p$) for any $p$ in $U$ and use the Poincare Lemma, but failed.

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When a closed curve is deformed continuously with a parameter, the integral varies continuously with the parameter as well.

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Yeah, I and my friends also noticed that but how this can help? – Junyu Feb 21 '12 at 15:29
How can a continuous function of a real variable take only rational values? (Hint: Intermediate value theorem.) – Harald Hanche-Olsen Feb 21 '12 at 17:28
Got your idea. Thanks. – Junyu Feb 22 '12 at 1:43

Can not make comment( Not allowed) so writing an answer.. I am not sure.. but please have it and tell me if you found mistake or confusion:

When take a continuous deformation of $C\subset U$, we will have continuous change in the integration: Hence integral will be constant rational number in each connected component of $U$.

Now without loss of generality assume that $U$ is connected and integral is $0$. Hence we have $\int_C\omega = 0$ which gives that $\omega$ is exact, Hence closed. This shows that $\omega$ is locally exact. Hence closed...

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Needs more careful argument I think. Anyway, Harald's hint is enough for us. – Junyu Feb 22 '12 at 1:42