Let $M$ be a smooth manifold with $\omega$ be a differential form of even degree on $M$. Then

$$\omega \, \wedge d \omega$$

is exact. I need to find an $\eta$ such that

$$d \eta = \omega \, \wedge d \omega$$

do I do this by construction and some trick using the even degree? Im a bit new to computing differential forms. Any advice or hints or tips would be greatly appreciated. Do I write $\omega$ out and the sum is from $1$ to $2n$ for some $n \in \Bbb{N}$. Also does $d \eta$ have degree $4n+1$? If $\omega$ has degree $2n$.


1 Answer 1


If $\alpha $ and $\beta $ are $k$ and $l$ forms respectively then you have these two very well known rules:

$$ \begin{align*} &\mathrm{(1)}\quad d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1)^{k}\alpha \wedge d\beta \\ &\mathrm{(2)}\quad \alpha \wedge \beta =(-1)^{kl}\beta \wedge \alpha \end{align*} $$

From there and the condition that the degree of $\omega $ is even it easy to see that $\frac1{2}\omega \wedge \omega $ is a primitive of $\omega \wedge d\omega $.

  • $\begingroup$ I was gonna say is $\eta = \frac{1}{2} \omega \wedge \omega$ $\endgroup$
    – anonymous
    Aug 8, 2022 at 1:54
  • 1
    $\begingroup$ @Hossien yes, thats right. $\endgroup$
    – Masacroso
    Aug 8, 2022 at 1:59

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