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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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For a one-form $\alpha$, $\alpha \wedge \alpha = 0$. – Dan Fox Jun 8 '12 at 9:01
And then you should do an immediate excercise. If ω=α∧β then then M is... – Swiat Gal Jun 8 '12 at 10:06
Do you want to have $w=d \alpha$? – George Jun 9 '12 at 9:39
$\dim M = 0 $ – Allen Knutson Jun 9 '12 at 16:08
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. – Blake Jun 27 '12 at 2:34

You can never find such a $1$-form since any $1$-form wedge itself is identically zero.

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