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I am working on the following exercise:

The function $f$ is called an integrating factor for the 1-form $\omega$ if $f({\bf x}) \neq 0$ for all $\bf x$ and $d(f\omega) = 0$. If the 1-form $\omega$ has an integrating factor, show that $\omega \wedge d\omega = 0$.

I am stuck here... I got $$d(f\omega) = df \wedge \omega + f \wedge d\omega = df \wedge \omega + f\ d\omega = 0$$ but that doesn't seem to get me anywhere. I also tried expanding this further (using the definition of $df$), but this gets quite ugly soon and didn't help either. The same goes for $\omega \wedge d\omega$.

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Consider $\omega \wedge d\omega = \Omega$. Multiply it by $f$ ($f \neq 0$) and use your equation. –  Nimza Feb 19 '12 at 19:39
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up vote 2 down vote accepted

That last equation gives you $$d\omega = -(1/f) df\wedge\omega. $$ Then you have $$\omega \wedge d\omega = -(1/f)\omega\wedge df \wedge \omega = (1/f)\omega\wedge\omega \wedge df = 0.$$ This is because the wedge product of any form with itself is zero. Notice that it is critical that $f$ be zero-free, or the taking of the reciprocal breaks everything.

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Pay attention to the signi -. –  Andrea Feb 19 '12 at 19:44
    
I think it should be $d\omega = -(1/f) df\wedge\omega$ (but that makes no difference). Thanks, working with differential forms I seem to forget that division exists. –  koletenbert Feb 19 '12 at 20:06
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