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Let $p \in M$ be a a point in a manifold and let $\varphi^X_t$ and $\varphi^Y_t$ be the local flows of the vector fields $X$ and $Y$ respectively. Define the commutator of flows: $\alpha(t)= \varphi^Y_{-t} \varphi^X_{-t}\varphi^Y_t\varphi^X_t$. I'm trying to prove:

$$\left .\frac{d}{dt} \right|_{t=0}\alpha(\sqrt{t})=[X.Y]_p$$

I managed to prove that $\left .\frac{d}{dt}\right |_{t=0}\alpha(t) =0$ and that it implies that:

$$\left .\frac{d}{dt}\right |_{t=0}\alpha(\sqrt{t})=2\left .\frac{d^2}{dt^2}\right |_{t=0}\alpha(t)$$

But trying to compute the second derivations I got stuck with expressions like these:

$$\left .\frac{d}{dt}\right |_{t=0}\ (X_{\varphi^X_t} \cdot \varphi^Y_{-t}\varphi^X_{-t}\varphi^Y_t)$$

And realized I'm not so sure as to how the second derivations work. Can I use the product rule on the the above expression? More alarmingly the commutator is itself a second derivation! so how come it's a vector field? (obviously it is).

If all commutators are vector fields (and all commutators of them are too and etc..) this means that all these higher derivatives are actually first order derivatives? What's going on?

Are there more higher derivatives like these that aren't commutators but are still first order derivations?

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    $\begingroup$ While I'm sure the flows of $X$ and $Y$ are indeed very polite, I think you meant a different word ;) $\endgroup$ – Neal Mar 27 '15 at 22:52
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    $\begingroup$ @Neal Just payin' my respects. $\endgroup$ – Saal Hardali Mar 27 '15 at 23:10
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    $\begingroup$ I think it's supposed to be $\left .\frac{d}{dt}\right |_{t=0}\alpha(\sqrt{t})=\frac{1}{2}\left .\frac{d^2}{dt^2}\right |_{t=0}\alpha(t)$ , not $2\left .\frac{d^2}{dt^2}\right |_{t=0}\alpha(t)$ $\endgroup$ – Asaf Shachar Nov 21 '15 at 17:17
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Here is one way to make calculations like this a bit easier:

In coordinates, write $ X = X^i\;\partial_i $. Then: $$ \varphi^X_t(p) \approx p^i + t \; X^i(p) + \frac{t^2}{2} X^j(p) \; \partial_j X^i(p) + O(t^3) $$ You can check then that $$ \frac{d}{dt} \varphi^X_t(p) = X^i(p) + t X^j(p) \partial_j X^i(p) + O(t^2) \\ = X^i(p^i + t \; X^i(p)) + O(t^2) \\ = X^i(\varphi^X_t(p)) + O(t^2) $$ where I taylor expanded $ X^i $.

Composing the next flow, (and taylor expanding $ Y^i( \varphi^X_t(p)) $ and dropping terms of $ O(t^3) $), I get something like: $$ \varphi^Y_t \varphi^X_t(p) = p^i + t X^i + \frac{t^2}{2} X^j \partial_j X^i + t Y^i + \frac{t^2}{2} Y^j \partial_j Y^i + t^2 X^j \partial_j Y^i $$ Since I am tired of typing up the long formulas, you can finish off by composing the last two flows. The lucky thing is that the terms $ \frac{t^2}{2} X^j \partial_j X^i $ will cancel away. The only problem is that I am left with $ 2 [X, Y] t^2 $...so I am off by a factor of 2.

PS. I am dying to know what typo suggested that flows are polite?

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  • $\begingroup$ The thing that's giving me a headache is exactly the taylor expansion of $\varphi^X_t$. Let's denote it by $\varphi_t$ for simplicity. Then: $$fo\varphi_t(p) = f(p) +t \left .\frac{d}{dt} \right |_0 (fo \varphi_t) + \frac{t^2}{2} \left .\frac{d^2}{dt^2} \right |_0 (fo \varphi_t) = f(p) + t X_p f + \frac{t^2}{2} \left .\frac{d}{dt} \right |_0 (X_{\varphi_t(p)} f) $$ Expanding the problamatic term: $\left .\frac{d}{dt} \right |_0 (X_{\varphi_t(p)} f) = \left .\frac{d}{dt} \right |_0 (\sum X^j(\varphi_t(p)) \left .\partial_j \right |_{\varphi_t(p)} f)$ $\endgroup$ – Saal Hardali Mar 28 '15 at 12:09
  • $\begingroup$ using product rule i get: $ \sum X_p \cdot X^j \partial_j f (p) + X^j(p) \cdot X_p (\partial_j f) $ And this: $X_p (\partial_j f)$ is a second derivative! $\endgroup$ – Saal Hardali Mar 28 '15 at 12:12
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    $\begingroup$ I did not check in detail, but I think your calculation is correct and I think there is no inconsistency because $ f $ in your calculation above is $ \varphi_t^Y $, and its second derivatives will be $ O(t) $, and so will not contribute anything to the derivative. Did you try to carry through your calculation all the way to the end? $\endgroup$ – user226970 Mar 28 '15 at 22:44
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    $\begingroup$ Nevermind. I sat for 5 hours and i finally understand how to calculate this stuff, your answer was very helpful! particularly how cleverly you passed to projections to local coordinates before expanding $\endgroup$ – Saal Hardali Mar 28 '15 at 23:33
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    $\begingroup$ 5 hours...you know what they say, pain is weakness leaving the body, and frustration is stupidity leaving the brain! Glad it worked out in the end. $\endgroup$ – user226970 Mar 28 '15 at 23:37

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