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If $d\,\sigma=0$ and $\sigma$ is non trivially with basis' coefficient 0, then $\sigma$ is a exterior derivative of a scalar function. I knew $d^{2} =0$. So it seems that all I am quoting is that $d^{2}f=0$ for $f$ being a scalar function. I do not see this as a proof.

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  • $\begingroup$ Comment to the question (v2): This Wikipedia page seems relevant. $\endgroup$ – Qmechanic Jul 7 '14 at 0:14
  • $\begingroup$ a proof of what? $\endgroup$ – Jim Jul 8 '14 at 13:00
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You have your implications confused. A form $\sigma$ is closed if $d\, \sigma = 0$. A form $\sigma$ is exact if $\sigma = df$. The statement that $d^2 = 0$ is the same as that every exact form is closed. The opposite is not true in general, there are many cases where a closed form is not exact. The existence of forms that are closed but not exact depends on the topology of the space involved.

However, locally every closed form is exact. By this I mean that for a point $p$, and a closed form $\sigma$, we can find a neighborhood $U$ containing $p$, and a form $f$, such that on $U$, $\sigma = df$. This can be done by taking a coordinate system around $p$ and letting $U$ be a ball in these coordinates. On a ball every closed form is exact.

Even though this can always be done around a point, it may be the case that it can not be done consistently, so that no single $f$, defined everywhere, satisfies $df = \sigma$.

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  • $\begingroup$ I would say the space is path connected so that every closed path can be contracted to a point. From this, every closed path is homotopic to a point. This deduces that the form is exact globally. $\endgroup$ – user45765 Jul 7 '14 at 0:21
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    $\begingroup$ I think you want simply connected, not path connected. A circle is certainly path connected but it supports non-exact closed forms. $\endgroup$ – Robin Ekman Jul 7 '14 at 0:24
  • $\begingroup$ Yes. Sorry for that. $\endgroup$ – user45765 Jul 7 '14 at 0:26

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