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Problem of the Week #233 - Nov 15, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,896
No one answered this week's problem. You can read my solution below.


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.
 
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