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Frobenius' Theorem

  1. Nov 17, 2005 #1
    I need to understand a certain characterization of Frobenius' Theorem, part of which contains the following statement:

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

    Is it obvious, or difficult to prove? I do not see the converse ...
    Last edited: Nov 17, 2005
  2. jcsd
  3. Aug 19, 2009 #2


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    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 [tex]\xi_{[a}\nabla_{b}\xi_{c]}=0[/tex] is the integrability condition that guarantees us that the partial differential equation [tex]\nabla_{[a}\xi_{b]}=\xi_{[a}v_{b]}[/tex] is solvable for [tex]v[/tex].
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