# The wedge of an exact form with a closed form is exact.

I'm trying to prove that the wedge of a closed form $$\xi$$ with an exact form $$\omega$$ is exact. We already have that half of it is exact. Maybe we can use the equation of $$\xi$$ being closed to rewrite the wedge. Other than that I am sort of stuck.

• If $d \alpha = 0$, then $\alpha \wedge d\beta = \pm d\left(\alpha \wedge \beta\right)$, which is clearly exact. – darij grinberg Sep 23 '18 at 23:33