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Let $\omega$ be a k-form on a smooth manifold $M$ such that there exists $f\in C^{\infty}(M)$ with $f(x)\ne 0$ for all $x\in M$ and $d(f \cdot \omega)=0$.

I need to show that $\omega \wedge d\omega =0$.

I have only been able to show that $\omega \wedge d\omega =\frac{1}{f} \omega \wedge \omega \wedge df$ but I don't know how to conclude.

Nitrogen
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1 Answers1

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EDIT: this only works when $k$ is odd. (Thanks Ted Shifrin for pointing this out).

Use the Leibniz rule to expand $d(f\omega)$, then wedge with $\omega$. Finally, use the fact that $f$ is nowhere zero and that $d(f\omega)=0$ (In particular that $\omega \wedge d(f\omega) =0$).

EDIT 2: This is false in general: Take $M=(0,\infty)$, $f$ any nonconstant smooth function on $M$, and $\omega=1/f$.

Avi Steiner
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  • This is pretty much what I have done to show that $\omega \wedge d\omega =\frac{1}{f} \omega \wedge \omega \wedge df$: $d(f\omega)=df \wedge \omega +f d\omega$ thus $d\omega=\frac{-1}{f} df \wedge \omega$. What else can I find doing that ? – Nitrogen Jan 10 '15 at 19:51
  • Don't use that $d(f\omega)=0$ until the very end. – Avi Steiner Jan 10 '15 at 20:01
  • I really don't get it sorry. If I do $\omega \wedge d(f\omega)=\omega \wedge (df \wedge \omega +f d\omega)= \omega \wedge df \wedge \omega + f \omega \wedge d\omega=0$, I get exactly the same thing as I got before. – Nitrogen Jan 10 '15 at 20:12
  • See my edit. I made a crippling mistake. Sorry about that. – Avi Steiner Jan 10 '15 at 20:31
  • Ok, no problem. I already managed to conclude when $k$ is odd. Does anyone know if the statement is even true if $k$ is even ? – Nitrogen Jan 10 '15 at 20:42
  • See my new edit. – Avi Steiner Jan 10 '15 at 21:29