Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Frobenius' Theorem

  1. Sep 1, 2013 #1

    WannabeNewton

    User Avatar
    Science Advisor

    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##?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Frobenius' Theorem
  1. Frobenius' Theorem (Replies: 1)

  2. Divergence theorem (Replies: 9)

  3. The gradient theorem (Replies: 3)

  4. Laplacian's theorem (Replies: 5)

Loading...