Consider a two-form $\alpha$, then $d\alpha \wedge d\alpha$ is not necessarily zero.
Is this statement true?
Consider $\beta = \alpha \wedge d \alpha$. Then $d\beta = d(\alpha \wedge d \alpha) = d \alpha \wedge d \alpha + \alpha \wedge d^2 \alpha = d\alpha \wedge d \alpha.$
Is this statement true?
Thank you!
Hints:
$d\alpha$ is a 3-form. The commutativity relationship between a $k$-form $\omega$ and a $l$-form $\eta$ is
$$ \omega \wedge\eta = (-1)^{kl}\eta \wedge \omega .$$
Also,
$$ d(\omega \wedge\eta) = d\omega \wedge\eta + (-1)^{k}\omega \wedge d\eta. $$
The solutions should follow immediately from these properties (together with $d^2=0$). (For the second part, only the last equality is off. Edit: fixed)