# Frobenius' Theorem

1. Sep 1, 2013

### WannabeNewton

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. $\omega_{[a}\nabla_{b}\omega_{c]} = 0 \Rightarrow \omega_{a} = \alpha \nabla_{a} \beta$?