I am trying to compute the directional derivative of a vector field $V$ along a direction $U$.
Actually, my vector field is initially only defined on a curve $\gamma(t)$ in a Riemannian manifold $(M, g)$, and I smoothly extend it on a tubular neighborhood of the curve in the following way : for each point in the neighborhood of $\gamma(t)$, I project it on $\gamma$ using any metric $h(u,v)$ and take the value of the vector field at that point : $V(t)$. The extension can be arbitrary for my purpose, so, $h(u,v)$ can be any Riemannian metric, not necessarily the same as $g$.
What I thus need to compute, to get the directional derivative, is : $ U V = \lim_{\epsilon \rightarrow 0} \frac{ V(\gamma^{-1}(\Pi (\gamma(t) + \epsilon U))) - V(t)}{\epsilon}$ where $\Pi$ is the projection operator on the curve $\gamma$ and $V(t)$ is the vector field initially defined on $\gamma(t)$.
I locally use a second order Taylor expansion of $\gamma(t+\delta) \approx \gamma(t) + \delta\dot\gamma(t) + \frac{\delta^2}{2}\ddot\gamma(t)$
I first compute the projection: $\gamma^{-1}(\Pi (\gamma(t) + \epsilon U)) = argmin_\delta \|\gamma(t+\delta) - (\gamma(t)+\epsilon U)\|_h^2$ where I take the norm out of my arbitrary metric $h$. So, cancelling out the derivative over $\delta$ : $ \frac{\partial}{\partial \delta} \|\gamma(t+\delta) - (\gamma(t)+\epsilon U)\|_h^2 = 0$ $\Rightarrow 2\,h(\dot\gamma+\delta\ddot\gamma, \delta\dot\gamma + \delta^2\ddot\gamma-\epsilon U) = 0$ Keeping only the first order terms : $ \Rightarrow \delta = \epsilon \frac{h(\dot\gamma, U)}{h(\dot\gamma, \dot\gamma)} = \gamma^{-1}(\Pi (\gamma(t) + \epsilon U))-t$
Coming back to the original problem : $ U V = \lim_{\epsilon \rightarrow 0} \frac{ V(t + \epsilon \frac{h(\dot\gamma, U)}{h( \dot\gamma, \dot\gamma)}) - V(t)}{\epsilon}$ $ \Rightarrow U V = \frac{h(\dot\gamma, U)}{h(\dot\gamma, \dot\gamma)} \dot V$
In particular, when $V=\dot\gamma$, I get $U V = \frac{h(\dot\gamma, U)}{h(\dot\gamma, \dot\gamma)} \ddot\gamma$
Now, what worries me is that I want to compute the Lie bracket $[UV]=UV-VU$. Since I took $h$ to be a (symmetric) Riemannian metric, I necessarily get $[UV]=0$ for any curve, any metric, any extension... I guess that would be a wonderful theorem. But could you spot the bug ? Is it because I didn't use enough terms in the Taylor expansion (or because I truncated the second order terms when computing the result of the projection) ?
Thanks!!
[[EDIT: Arrrg, I realize that $[UV] \neq 0$ since $[UV]=\frac{h(\dot\gamma, U)}{h(\dot\gamma, \dot\gamma)} \dot V - \frac{h(\dot\gamma, V)}{h(\dot\gamma, \dot\gamma)} \dot U$ !! Could you however tell me if the way to proceed is correct? Thanks!! ]]
Picture of the setting: