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I have the following question: Let $M$ be a smooth manifold and let $p \in M$. Furthermore let $X$ and $Y$ be two vector fields in a neighbourhood $U$ of $p$ and consider their flows $\varphi^{X}(t,x)$ and $\varphi^{Y}(t,x)$. Define now $s(t):= \varphi^{X}_{\sqrt{t}} \circ \varphi^{Y}_{\sqrt{t}} \circ \varphi^{X}_{-\sqrt{t}} \circ \varphi^{Y}_{-\sqrt{t}}$. How can one show that $\dot{s}(0) = [X,Y]|_{0}$ ? Or is this statement true? If yes, why? Thanks in advance.

Eric

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    This is more or less a few applications of the chain rule.2012-06-26
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    really? how? I already tried and came up with nothing.2012-06-26
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    the derivative of $\sqrt{t}$ in zero is not defined.2012-06-26
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    It doesn't have to be. For example, neither $1 + \sqrt{t}$ nor $1 - \sqrt{t}$ have well-defined derivatives at zero but their product $1 - t$ does.2012-06-26
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    yes I see. So ow am I applying the chain rule here ?2012-06-26
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    You apply the chain rule for small nonzero $t$ and then take a limit.2012-06-26
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    Just to illustrate what Qiaochu wrote using his example: for the case of $(1+\sqrt{t})(1-\sqrt{t})$, we can do it the silly way and apply the product rule for $t> 0$: $$D_t(1+\sqrt{t})(1-\sqrt{t}) = \frac1{2\sqrt{t}}(1-\sqrt{t}) + (1+\sqrt{t})\frac{-1}{2\sqrt{t}}$$ from which you see that the two terms that can individually blow-up as $t\to 0$ (the non-differentiability of $\sqrt{t}$) cancels. That kind of cancellations is what you are looking for.2012-06-26
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    I know how to apply the chain rule in this example but what I dont know is how to apply the chain rule on the curve obtain from flows?2012-06-26
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    applying this I get $\frac{d}{dt}|_{t=t_{0}}s(t) = \dot{\varphi^{X}}_{t_{0}} \frac{1}{2\sqrt{t_{0}}} + D\varphi^{X}_{t_{0}} \frac{d}{dt}|_{t=t_{0}}\varphi^{Y}_{\sqrt{t}} \circ \varphi^{X}_{-\sqrt{t}} \circ \varphi^{Y}_{-\sqrt{t}}$. How do I go further on ?2012-06-26

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