# Showing $\omega \, \wedge d \omega$ is exact if $\omega$ has even degree over smooth $M$

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 $$2n$$.

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 $$\omega \wedge d\omega$$.
• I was gonna say is $\eta = \frac{1}{2} \omega \wedge \omega$ Aug 8, 2022 at 1:54