Is $\omega^2$ nowhere vanishing on the four-sphere?

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Here is this week's POTW:

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If $\omega$ is a two-form on the four-sphere, is $\omega^2$ (i.e., $\omega \wedge \omega$) nowhere vanishing?

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No one answered this week's problem. You can read my solution below.

Spoiler

The answer is no. Suppose to the contrary that $\omega$ is nowhere vanishing. Then $\omega^2$, being a 4-form on a compact oriented 4-manifold has nonzero integral. The de Rham cohomology groups of the four-sphere are $\Bbb R$ in dimensions zero and four, and zero in every other dimension. Thus, $\omega$ is exact. If $\omega = d\eta$, then $\omega^2 = d(\eta \wedge d\eta)$. So $\omega^2$ is exact; its integral is zero by Stokes's theorem. This is absurd.