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I'm trying to show that on a compact orientable manifold without boundary, the exterior derivative of any $n-1$ form vanishes at some point.

I think I should be using the Stokes' theorem but don't see how. Any hints?

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Yes,if $\omega$ is a such a form and $d\omega$ does not vanish, $d\omega$ is a volume form and $\int_Md\omega\neq 0$ by Stokes. contradiction since $\int_Md\omega=\int_{\partial M}\omega=0$, since the boundary $\partial M$ of $M$ is empty.

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  • $\begingroup$ Why $dw$ is a volume form? I understand that not necessarily, correct? $\endgroup$ – Manoel Dec 9 '17 at 14:33
  • $\begingroup$ because it does not vanishes, since $M$ is orientable, it as a volume form $\alpha$, you can write $d\omega=f\alpha$, $f\neq 0$ implies that its signs does not change, so $d\omega$ is a volume form. $\endgroup$ – Tsemo Aristide Dec 9 '17 at 14:36
  • $\begingroup$ Then any form $dw\neq 0$ can write as $dw=f \alpha$? And $dw$ is volume element of which manifold? $\endgroup$ – Manoel Dec 9 '17 at 15:40
  • $\begingroup$ You can only write $d\omega = f \alpha$ locally. $\endgroup$ – CuriousKid7 Dec 7 '18 at 2:25

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