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

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.

-
The symplectic-geometry tag is a bit cryptic here :) But this is connected to the proof that $S^4$ is not a symplectic manifold. – Mariano Suárez-Alvarez May 6 '12 at 7:29

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.
It might be worth pointing out that this is a variant of the argument showing that the only even-dimensional sphere admitting a symplectic structure is $S^2$. – t.b. May 6 '12 at 7:30
People like to use Stokes to show that a symplectic form on a closed manifold is never exact: $$0 \neq \int_{M} \omega \wedge \omega = \int_{M} d(\eta \wedge \omega) = \int_{\partial M} \eta \wedge \omega = 0$$ and similarly in higher dimensions. Not very different, yes. – t.b. May 6 '12 at 7:35