Suppose we have a real symplectic manifold $(M,\omega)$. Under what conditions can we find a global 1-form $\alpha$ such that $\omega = \alpha \wedge\alpha$?
Obviously there are some simple obstructions, for example, the cotangent bundle must admit a non-vanishing section (thus, surfaces of genus $>1$ are out). However it does not seem easy to come up with sufficient conditions.
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1$\begingroup$ For a one-form $\alpha$, $\alpha \wedge \alpha = 0$. $\endgroup$– Dan FoxCommented Jun 8, 2012 at 9:01
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$\begingroup$ And then you should do an immediate excercise. If ω=α∧β then then M is... $\endgroup$– Swiat GalCommented Jun 8, 2012 at 10:06
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$\begingroup$ Do you want to have $w=d \alpha$? $\endgroup$– GeorgeCommented Jun 9, 2012 at 9:39
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$\begingroup$ $\dim M = 0 $ $\endgroup$– Allen KnutsonCommented Jun 9, 2012 at 16:08
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$\begingroup$ This question was incredibly badly phrased, for which I apologize. I meant to ask in the case where we tensor the exterior algebra with something nonabelian. But even then as Gal pointed out the answer is easy. $\endgroup$– BlakeCommented Jun 27, 2012 at 2:34
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