Let $M=R^2-(0.0)$, $\omega=\frac{xdx+ydy}{x^2+y^2}$,
how does one prove $\omega$ is an exact form？
I have found a function $f=\frac{1}{2}\ln（x^2+y^2）$
such that $df=\omega$, but it is not sufficient to prove that $\omega$ is an exact form, because of the following counter example:
let $M=R^2-(0.0)$, $\eta=\frac{ydx-xdy}{x^2+y^2}$,
we can also find a function $g=\arctan\frac{x}{y}$, such that $dg=\eta$ but it is well known that $\eta$ is not an exact form on $M$.
So my question is:
What is the exact condition to determine whether a closed form is an exact form?
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Jeremy Upsal
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Tom
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1Possible duplicate of Proof that this differential form is not exact – Law Neutral Mar 06 '17 at 17:27
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@Law Neutral, it's not duplicate for the question you mentioned is to prove a form is not exact, but what I need is to prove a form is exact. – Tom Mar 07 '17 at 01:03
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By definition a $k$-form $\alpha$ is exact if there is a $(k - 1)$-form $\beta$ such that $\alpha = d \beta$; in that case $d \alpha = d^2 \beta = 0$, so exact forms are always closed. – Travis Willse Sep 21 '22 at 05:44
