Frobenius' Theorem: Characterization & Proof Difficulty

[toogood]

I need to understand a certain characterization of Frobenius' Theorem, part of which contains the following statement:

\nabla_{[a}\xi_{b]}=\xi_{[a}v_{b]} for some dual vector field v_{b} if and only if \xi_{[a}\nabla_{b}\xi_{c]}=0, where \xi^a\xi_a\neq 0.

Is it obvious, or difficult to prove? I do not see the converse ...

 

[Reb]

I need to understand a certain characterization of Frobenius' Theorem, part of which contains the following statement:

\nabla_{[a}\xi_{b]}=\xi_{[a}v_{b]} for some dual vector field v_{b} if and only if \xi_{[a}\nabla_{b}\xi_{c]}=0, where \xi^a\xi_a\neq 0.

Is it obvious, or difficult to prove? I do not see the converse ...

Ι wouldn't say it is hard, it is just -as many proofs in geometry- rather long and tedious. For the converse part, the condition \xi_{[a}\nabla_{b}\xi_{c]}=0 is the integrability condition that guarantees us that the partial differential equation \nabla_{[a}\xi_{b]}=\xi_{[a}v_{b]} is solvable for v.