# Lie dragging preserves commutation proof.

1. Nov 7, 2013

### center o bass

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.