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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.

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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

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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.

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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
    
Isn't it the argument? –  Mariano Suárez-Alvarez May 6 '12 at 7:31
    
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
    
Ah, right. The argument I wrote down uses the fact that the sphere is a sphere, of course ;) –  Mariano Suárez-Alvarez May 6 '12 at 7:37
    
Thanks it's very helpful! –  Anton May 6 '12 at 7:54

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