$4$-form $ \omega \wedge \omega$ vanishes on $S^4$

Question

If $\omega$ is a closed $2$-form on $S^4$, how can I show the $4$-form $ \omega \wedge \omega$ vanishes somewhere on $S^4$? I am guessing that the fact we're talking about the $2$-form being closed, that this is the crux.

Answer 1

Obviously $d(\omega\wedge\omega)=0$, so that $\omega\wedge\omega$ represents an element of $H^4(S^4)$.

Suppose $\omega\wedge\omega$ is never zero. Then it is a volume form and therefore its class in $H^4(S^4)$ is not zero.

Now, since $d\omega=0$ and $H^2(S^4)=0$, there is a $1$-form $\eta$ such that $d\eta=\omega$. Then $d(\eta\wedge\omega)=\omega\wedge\omega$ and $\omega\wedge\omega$ is a coboundary. This is absurd.