TITLE: Lie Bracket and flows
QUESTION [5 upvotes]: Can anyone show me how do I differentiate this? Suppose I have $\Phi^{X}_{t}$ and $\Phi^{Y}_{t}$ both flows with $X$ and $Y$ respectively starting from point $p$, what is
$\frac{d}{dt}|_{t=0}\phi^{Y}_{-\sqrt{t}}\circ \phi^{X}_{-\sqrt{t}}\circ \phi^{Y}_{\sqrt{t}}\circ\phi^{X}_{\sqrt{t}}$?
Thanks!

REPLY [2 votes]: Let's start by recalling the main property of $\phi_t^X(p)$: 
$$ \frac{d}{dt}\phi_t^X(p)=X_{\phi_t^X(p)}.$$
Next, let's look at differentiating $\phi_t^X\circ f(t,p)$, where $f\colon (-\epsilon,\epsilon)\times M\to M$ is smooth. To make this easier to look at, let's write $\phi^X(t,p)$ to mean $\phi^X_t(p)$. In this new notation, the above property becomes
$$\left.\frac{\partial\phi^X}{\partial t}\right|_{(t,p)}=X_{\phi_t^X(p)}.$$ 
Then applying the chain rule to $\phi_t^X\circ f(t,p)$, you get
$$\begin{align}
\frac{d}{dt}\phi^X(t,f(t,p))
&=\left.\frac{\partial\phi^X}{\partial t}\right|_{(t,f(t,p))} + d(\phi_t^X)_{f(t,p)}\left.\frac{\partial f}{\partial t}\right|_{(t,p)}\\
&= X_{\phi^X_t\circ f(t,p)}+d(\phi_t^X)_{f(t,p)}\left.\frac{\partial f}{\partial t}\right|_{(t,p)}.
\end{align}$$
Though it'll be quite excruciating, you can use this formula to evaluate your Lie derivative. Though, if I may suggest, it would be much easier to evaluate it in local coordinates.