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Let $\alpha \in \Omega^1(M)$ be a 1-form on a manifold $M$ which vanishes no-where. Given that for any two vector fields $X$, $Y \in \ker{\alpha}$ there holds $[X,Y] \in \ker{\alpha}$ ($[.,.]$ is the Lie bracket) does there holds $\alpha \wedge d \alpha = 0$ with $d$ the exterior derivative? The other direction is true but I can't see how to prove this direction or to give a counterexample.

Thanks in advance

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    Suggestion: choose a local frame $(E_1,\dots,E_n)$ with dual coframe $(\epsilon^1,\dots,\epsilon^n)$, such that $\alpha=\epsilon^1$. Then use the formula $d\alpha(E_i,E_j) = E_i(\alpha(E_j)) - E_j(\alpha(E_i)) - \alpha([E_i,E_j])$. – 2017-01-08
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    Thanks, but isn't then $ \alpha \wedge d \alpha =0$? – 2017-01-08
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    My suggestion was meant to be a way to prove that if $\ker \alpha$ is closed under Lie bracket, then $\alpha\wedge d\alpha=0$. – 2017-01-08
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    Sorry for my poor background, but what does closer under Lie Bracket means? – 2017-01-08
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    The term is "closed" under Lie bracket. It's what you wrote: If $X,Y$ are vector fields that take values in $\ker{\alpha}$, then their Lie bracket $[X,Y]$ also takes its values in $\ker{\alpha}$. – 2017-01-08
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    Is a local frame the vector field induced by a chart? I can't see why $\alpha \wedge d \alpha =0$ by what you wrote. – 2017-01-08
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    No, a local frame need not be induced by a chart. It's not meant to be obvious why $\alpha\wedge d\alpha=0$; I wanted to leave some work for you to do! – 2017-01-08

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