$X, Y$ are two complete vector fields with$[X, Y] = 0$, what is the resulting flow of $X+Y$? If $X, Y$ are two complete vector fields with$[X, Y] = 0$, what is the resulting flow of  $X+Y$?
I'm kind of confused on what the flow is. 
I know that the respective flows for $\Phi_t^X$ and $\Phi_t^Y$ commute, but what is the flow for the addition of two vector fields?
Thank you.
 A: First recall the following useful fact about commuting vector fields: if $X$ and $Y$ are complete vector fields such that $[X,Y]=0$, then $\left(\Phi_{t}^{X}\right)^{*}Y=Y$ for all $t\in\mathbb{R}$. Here $\Phi_{t}^{X}$ is the time $t$ flow of $X$, and the pullback of vector fields is defined by
$$
\left[\left(\Phi_{t}^{X}\right)^{*}Y\right](p)=\left(d\Phi_{-t}^{X}\right)_{\Phi_{t}^{X}(p)}(Y(\Phi_{t}^{X}(p))).
$$
For a proof of this fact, see for instance commuting vector fields, commuting flows.
Now, we can prove that the flow of $X+Y$ is given by
$$
\Phi_{t}^{X+Y}=\Phi_{t}^{X}\circ\Phi_{t}^{Y}.
$$
Indeed, using the chain rule we have
\begin{align}
\frac{d}{dt}\left(\Phi_{t}^{X}\circ\Phi_{t}^{Y}\right)(p)&=\left.\frac{d}{ds}\right|_{s=t}\left(\Phi_{s}^{X}\circ\Phi_{t}^{Y}\right)(p)+\left.\frac{d}{ds}\right|_{s=t}\left(\Phi_{t}^{X}\circ\Phi_{s}^{Y}\right)(p)\\
&=X\big((\Phi_{t}^{X}\circ\Phi_{t}^{Y})(p)\big)+\left(d\Phi_{t}^{X}\right)_{\Phi_{t}^{Y}(p)}\left(\left.\frac{d}{ds}\right|_{s=t}\Phi_{s}^{Y}(p)\right)\\
&=X\big((\Phi_{t}^{X}\circ\Phi_{t}^{Y})(p)\big)+\left(d\Phi_{t}^{X}\right)_{\Phi_{t}^{Y}(p)}\big(Y(\Phi_{t}^{Y}(p)\big)\\
&=X\big((\Phi_{t}^{X}\circ\Phi_{t}^{Y})(p)\big)+\left[\left(\Phi_{-t}^{X}\right)^{*}(Y)\right](\Phi_{t}^{X}\circ\Phi_{t}^{Y}(p))\\
&=X\big((\Phi_{t}^{X}\circ\Phi_{t}^{Y})(p)\big)+Y\big((\Phi_{t}^{X}\circ\Phi_{t}^{Y})(p)\big),
\end{align}
using in the last equality that $\left(\Phi_{-t}^{X}\right)^{*}(Y)=Y$, as mentioned above. This proves the statement.
