# Frobenius' Theorem

1. Nov 17, 2005

### 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 ...

Last edited: Nov 17, 2005
2. Aug 19, 2009

### Reb

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 $$v$$.