Prove that if $η$ is exact, then $η∧β$ is also exact.

Question

Prove that if $η$ is exact, then $η∧β$ is also exact. Please give a clear way to solve?

Answer 1

The following hint works if you suppose $\beta$ to be closed. I'm not sure (though I doubt) that $\eta \wedge \beta$ is in general exact if $\eta$ is exact.

Try to write out what it means for $\eta$ to be exact, i.e. that $\exists \omega$ such that $\eta=d\omega$. Then recall the formula: $$ d(\alpha \wedge \gamma)=d\alpha \wedge \gamma + (-1)^k \alpha \wedge d\gamma $$ where $\alpha$ and $\gamma$ are differential forms and $k$ is the degree of $\alpha$. Now apply it to $\omega\wedge\beta$: $$ \begin{align*} d(\omega\wedge\beta)&=d\omega\wedge\beta+(-1)^k\omega\wedge d\beta\\ &=\eta\wedge\beta+(-1)^k\omega\wedge 0\\ &=\eta\wedge\beta \end{align*} $$ since $\beta$ closed means that $d\beta=0$. Therefore $\eta\wedge\beta$ is exact.