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How can I show that $\alpha$ is exact form on $S^1$ iff $\int_{S^1} \alpha =0$?

I used Stokes theorem to prove one direction. Could you please help me in the other direction?

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1 Answer 1

up vote 2 down vote accepted

A form $\alpha$ is exact iff there is a function $f:S^1\to\mathbb R$ such that $\alpha=df$.

If that is the case, then $\int_{S^1}\alpha=\int_{S^1}df=\int_{\partial S^1}f=0$, by stokes' theorem.

Conversely, suppose $\int_{S^1}\alpha=0$. Define a function $f:S^1\to\mathbb R$ so that $f(\theta)=\int_1^\theta\alpha$; here the integral is taken along the segment going from $1$ to $\theta\in S^1$ in the positive direction (I am looking at $S^1$ as the unit circle in $\mathbb C$) This function $f$ is well-defined exactly because of our hypothesis on $\alpha$, and you can check easily that $df=\alpha$.

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