- #1

- 560

- 2

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.