Frobenius' Theorem

  • #1
WannabeNewton
Science Advisor
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Hi guys. Most of my texts have the standard proof of Frobenius' theorem (both the vector field and differential forms versions) and through multiple indirect equivalences conclude that ##\omega \wedge d\omega = 0## implies (locally) that ##\omega = \alpha d\beta## where ##\omega## is a 1-form and ##\alpha,\beta## are scalar fields. Does anyone know of a proof wherein one directly shows that ##\omega \wedge d\omega = 0 \Rightarrow \omega = \alpha d\beta## i.e. ##\omega_{[a}\nabla_{b}\omega_{c]} = 0 \Rightarrow \omega_{a} = \alpha \nabla_{a} \beta##?
 

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