I'll provide an incomplete solution for the following exercise.
Exercise
$(i)$ Let $X$ and $Y$ be two vector fields. Prove that
$$[\mathcal{L}_X,\mathcal{L}_Y]=\mathcal{L}_{[X,Y]}.$$
$(ii)$ Let $X$, $Y$ and $Z$ be three vector fields. Prove that
$$[[\mathcal{L}_X,\mathcal{L}_Y],\mathcal{L}_Z]+[[\mathcal{L}_Y,\mathcal{L}_Z],\mathcal{L}_X]+[[\mathcal{L}_Z,\mathcal{L}_X],\mathcal{L}_Y]=0$$
Solution (incomplete)
First of all, I'm trying to figure out what I actually need to prove. The two members on the left and on the right of the first equality are operators (they act on vector fields), not vector fields. So, in order to prove that they are equal, I need to prove that they are equal when applied to a generic vector field. The operator on the left is intended to act on a vector field $Z$ this way
$$[\mathcal{L}_X,\mathcal{L}_Y](Z)=[\mathcal{L}_XZ,\mathcal{L}_YZ],$$
am I right?
I know that $\mathcal{L}_XY=[X,Y]$, so it's all about proving that
$$[[X,Z],[Y,Z]]=[[X,Y],Z].$$
At this point I tried to do a little calculations (applying the definition of Lie bracket and its properties) but I'm stuck. I simply don't get the same thing on the left and on the right.
The second point, as soon as one has proved the first, is quite easy (it amounts to apply the first point and the Jacobi identity).
Any ideas for the first point? Thank you
Best Answer
No. As Jack Lee commented, it should be
$$[\mathcal{L}_X,\mathcal{L}_Y](Z) = \mathcal{L}_X(\mathcal{L}_YZ) - \mathcal{L}_Y(\mathcal{L}_XZ).$$
Now
\begin{align}\mathcal{L}_X(\mathcal{L}_YZ) - \mathcal{L}_Y(\mathcal{L}_XZ) &= \mathcal{L}_X([Y,Z]) - \mathcal{L}_Y([X,Z])\\ & = [X,[Y,Z]] - [Y,[X,Z]]\\ &= [X,[Y,Z]] + [Y,[Z,X]] = \cdots\end{align}
Complete the calculation using Jacobi's identity.