A 1-form on a smooth manifold is exact if and only if it integrates to zero on every closed curve

3.6k Views Asked by At

I am stuck on the following problem, which comes from a old qualifying exam.

Prove that a 1-form $\phi$ on $M$ is exact if and only if for every closed curve c, $\int_{c} \phi =0$.

One way is an application of Stokes' theorem, if $\phi = df$ then $\int_{c}\phi = \int_{\partial C}df = 0$ since $\partial B=\emptyset$.

I don't know how to do the other direction. I made an attempt as follows:

Choose any $x_0 \in M$ define a function, $f(x)=\int_{x_0}^{x}\phi$. This makes sense since the integral is path independent. Now I want to prove that $f$ is smooth and $df=\phi$. I can't do either.

Thanks

1

There are 1 best solutions below

0
On

I assume we are discussing the problem in the context of differential geometry, so $M$ is a smooth manifold, $\phi$ is a smooth 1-form. The smoothness of $f(x)=\int_{x_0}^x \phi$ is a local property so we should be able to prove that by applying fundamental theorem of calculus in $\mathbf R ^n$.