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Lie dragging preserves commutation proof.

  1. Nov 7, 2013 #1
    Hi! A proof of Frobenius' theorem (in Schutz Geometrical Methods) uses the fact that if a set of vector fields ##{V_{(a)}}## commute on some submanifold S of an ambient manifold M, and one use an additional vector field ##Z## to Lie transport/drag the set ##{V_{(a)}}## around the manifold M, then the set ##{V_{(a)}}## will also commute away from S.
    I want to prove this. This is where I stand: Since ##\mathcal{L}_Z V_{(a)} = 0## per definition of lie transportation we have

    $$\mathcal{L}_Z [V_{(a)},V_{(b)}] = [ \mathcal{L}_Z V_{(a)},V_{(b)}] + [ V_{(a)}, \mathcal{L}_Z V_{(b)}] =0$$

    But does ##\mathcal{L}_Z [V_{(a)},V_{(b)}]= 0## on M somehow imply that ##[V_{(a)},V_{(b)}]=0## on M? I'm not sure how to conclude this proof.
     
  2. jcsd
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