# Lie Bracket and flows

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!

• Have you tried to do it first with just one of the $\phi$'s and then with two? This is just the chain rule, really! – Mariano Suárez-Álvarez Feb 15 '13 at 22:09
• I have tried. Say if we differentiate $\phi^{X}(\sqrt{t},p)$, with respect to $t$, then it should be $X(\phi^{X}_{t})(1/\sqrt{t})$, right? And if we differentiate a composition of two $\phi$'s, I am not sure how to differentiate the second component... – enoughsaid05 Feb 15 '13 at 22:22
• Remember that by definition, $\frac{d}{dt}\phi_t^X(p)\Bigr|_{t=0}=X_p$. – Avi Steiner Feb 16 '13 at 2:14

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}
• Why does differentiating $\phi^{X}$ with respect to the second coordinate gives $d(\phi^{X}_{t})_{f(t,p)}\frac{\partial f}{\partial t}|_{(t,p)}$? I have been looking for the definition of $d\phi$ but I can't find it. – enoughsaid05 Feb 16 '13 at 7:30
• You may have seen $d\phi$ written as $\phi_*$. – Avi Steiner Feb 17 '13 at 1:20
• @AviSteiner is the second term interpreted as the differential of $\phi$ evaluated at $f(t,p)$ multiplied by $\partial f /\partial t$? Is the idea that you just recursively evaluate each term by setting it equal to $f(t,p)$? – user1447447 Apr 13 '17 at 22:09
• Meaning that suppose that I have two vector fields $X$, and $Y$, with corresponding flows $\phi(t, p)$ and $\psi(t, p)$, and I'm trying to differentiate $c(t) = \psi(-t,\phi(-t, \psi(t, \phi(t, p)))$, I would set $f(t,p) = \phi(-t, \psi(t, \phi(t, p)))$, differentiate that w/ respect to $t$ and substitute for $\partial f / \partial t$, and then continue that way to expand out the terms? Is that clear? The notation is sort of nasty. – user1447447 Apr 13 '17 at 22:27
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