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Let $\omega^{1}$ a 1-form defined in an open set $U\subset\mathbb{R}^{n}$. Assume that for each closed differentiable curve $c$ in $U$, $\int_{c}{\omega^{1}}$ is a rational number. Prove that $\omega^{1}$ is closed.

My approach: Let $c_{0},c_{1}:[0,1]\to U$, two curves homotopics. If $\omega^{1}$ is a 1-form closed on $U$, then the line integral along to $c_{0}$ and $c_{1}$ coincide $$\int_{c_{0}}{\omega^{1}}=\int_{c_{1}}{\omega^{1}}$$ So my idea is fixed two rational number, then for the homotopy of $c_{0}$ and $c_{1}$, I'm "obliged" to go through a irrational number.

Edit: I know that question was asked here :Differential form is closed if the integral over a curve is rational number. , but anyway I cannot finish the problem

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asdasd asd
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  • Yes the question is the same, however I cannot finish the problem. Thanks – asdasd asd Jan 07 '16 at 22:05
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    What are you stuck on? The answer there is enough to solve the problem. – anomaly Jan 07 '16 at 22:38
  • I don't see how, for this reason I asked. Anyway, I know that the following result, maybe works: If $f$ is continuous on $[a,b]$ and $f(x)$ is always rational, then $f$ is constant. – asdasd asd Jan 07 '16 at 22:52

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