Left and right invariant vector fields I'm following Woodhouse's book on geometric quantization and I'm stuck with this problem. 
Let $R_A$ and $L_A$ be right and left invariant vector fields such that $R_A(e)=A=L_A(e)$, where $A$ is an element of Lie algebra. 
How can I show that $[R_A,R_B]=-R_{[A,B]}$, $[L_A,L_B]=L_{[A,B]}$ and $[R_A, L_B]=0$ ?
The hint is to use BCH formula in a form: $\exp(sA) \exp(sB) \approx \exp(st[A,B]) \exp(tB) \exp(sA)$. I can differentiate this expression with respect to s and t but I don't know how to get vector field commutators from this. 
 A: I guess you can do this using the BCH formula as follows: It's an easy exercise, to check that the flow of left and right invariant vector fields generated by $A$ is right and left multiplication with $\exp (tA)$ respectivly: $F^{R_A}_t = L_{\exp(tA)}$ and $F^{L_A}_t = R_{\exp(tA)}$. 
Plugging this and the BCH formula in the formula for the commutator of vectorfields is in terms of the lie derivative will probably give the result. This is, what you start with to plug the BCH formula in:
\begin{align*}
[R_A,R_B]_g 
&= \mathcal{L}_{R_A} R_B (g) 
=\frac{d}{dt} \Big\vert_{t=0} (F^{R_A}_{-t})_* (R_B)_{F^{R_A}_t (g)} \\
&= \frac{d}{dt} \Big\vert_{t=0} \frac{d}{ds} \Big\vert_{s=0} (F^{R_A}_{-t}) \circ (F^{R_B}_{t}) \circ (F^{R_A}_{t}) (g) \\
&= \frac{d}{dt} \Big\vert_{t=0} \frac{d}{ds} \Big\vert_{s=0} \exp(-tA) \cdot \exp (sB) \cdot \exp (tA) \cdot g
\end{align*}
However, a much easier way is to note, that right invariant vector fields are $R_g$-related to themselves for every element $g$ of the Liegroup. And as for every smooth map between manifolds $f$ the commutator of $f$-related vector fields is again $f$-related, the commuator $[R_A, R_B]$ is again $R_g$-related to itself for every $g$ in the Liegroup and is thus right invariant. Now all you need to do is to check, that $[R_A, R_B]_e = -[L_A, L_B]_e = -[A,B]$. This is easily done by continuing the calculation I started above with $g = e$, as $$\exp(-tA) \cdot \exp (sB) \cdot \exp (tA) = R_{\exp(tA)} \circ R_{\exp(sB)} \circ R_{\exp(-tA)} (e) = (F^{L_{-A}}_{-t}) \circ (F^{L_B}_{t}) \circ (F^{L_{-A}}_{t}) (e).$$
The fomula for left invariant vectorfields is derived in the exact same way. Finally the commuator of a left and a right invariant vector field vanishes  as 
$$ [R_A,L_B]_g 
= \mathcal{L}_{R_A} R_B (g) 
=\frac{d}{dt} \Big\vert_{t=0} (F^{R_A}_{-t})_* (L_B)_{F^{R_A}_t (g)}
=\frac{d}{dt} \Big\vert_{t=0} (L_{\exp (-tA)})_* (L_{\exp (tA)})_* (L_g)_* B
= 0.
$$
