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

Well, i suppose that $\omega$ isn't closed , then $\omega$ isn't locally exact, i.e, there exists a ball $B \subset U$ and a closed path $\lambda$ in $B$ such that $\int_{\lambda}w = c \ne 0$, but i don't know how continue..any tips?

EDIT:In this topic Differential form is closed if the integral over a curve is rational number. , the user Harald Hanche-Olsen afirms that a "a closed curve is deformed continuously with a parameter, the integral varies continuously with the parameter as well.". How prove this fact?

Thanks

Cézar Bezerra
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1 Answers1

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To answer your added question, consider the family of piecewise-smooth curves $f\colon [a,b]\times [0,\epsilon) \to\Bbb R^n$. Here we suppose $f(t,s)$ is piecewise-$C^1$ as a function of $t$ (with $s$ fixed) and $f$ and $\partial f/\partial t$ are continuous functions of $s$. If $\omega = \sum a_i(x)\,dx_i$, then the line integral $$\int_{C_s} \omega = \int_a^b \sum a_i(f(t,s))\frac{\partial f_i}{\partial t}(t,s)\,dt$$ is a continuous function of $s$.

Ted Shifrin
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