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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: Setting

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I'm not sure if I understand completely what you are trying to do, but I have a few comments about what you can and can't do in general on a smooth manifold:

It seems you have some confusion between the points of the manifold and tangent vectors. For each $p \in M$, you have a vector space $T_pM$ of tangent vectors to $M$ at $p$. The manifold $M$ doesn't have any linear structure. Think of $M = S^2$, the two dimensional sphere. How do you add two different points? What is the "zero" vector? In general, you can't add points $p_1, p_2 \in M$ to get $p_1 + p_2 \in M$, nor you can't add a point $p \in M$ to a tangent vector $v \in T_p(M)$ to get another point $p + v \in M$.

If $M = \mathbb{R}^n$, then $M$ has a linear structure, and so the points of $M$ can be thought as vectors, and there is a natural identification between the tangent spaces $T_pM$ with $M$ itself so everything is identified and you can add points to points, points to tangent vectors, but this doesn't generalize at all to manifolds. So your expression $\gamma(t) + \epsilon U$ doesn't really make sense on a general manifold.

Another thing you can't do on a manifold is add two tangent vectors $V_1 \in T_{p_1}(M)$ and $V_2 \in T_{p_2}(M)$ at two different points $p_1 \neq p_2$. There is no natural identification between neighboring tangent spaces. The vectors $V_1$ and $V_2$ live in two different vector spaces. So when you write $$ V(\gamma^{-1}(\Pi (\gamma(t) + \epsilon U))) - V(t) $$ it is a priori possible that $V(t) \in T_{\gamma(t)}(M)$ and $V(\gamma^{-1}(\Pi (\gamma(t) + \epsilon U))) \in T_{\Pi (\gamma(t) + \epsilon U)}(M)$ live in two different vector spaces and so you can't subtract them.

Finally, this is not how you compute the Lie derivative of $U$ and $V$. Say you want to compute $[U,V](p)$. First, $U$ and $V$ should be both defined on a neighborhood of the point $p \in M$. Knowing $V$ along a curve is not enough. To do it, you take the flow $\varphi^U(q,t)$ generated by the vector field $U$ and let $\gamma(t) = \varphi^U(p,t)$. Next, you restrict your attention to the vector field $V$ along the curve $\gamma$ which is an integral curve of $U$. You want to compute $$ \lim_{\epsilon \to 0} \frac{V(\gamma(\epsilon)) - V(0)}{\epsilon}. $$ This expression doesn't make sense on a general manifold because $V(\gamma(\epsilon))$ and $V(0)$ are tangent vectors at different points so you can't substract them. What you do is you use $\varphi^U$ to identify the tangent spaces and compute $$ \lim_{\epsilon \to 0} \frac{(\varphi^U(-,\epsilon))^{*}(V(\gamma(\epsilon))) - V(0)}{\epsilon}. $$

The vector $V(\gamma(\epsilon))$ belongs to $T_{\gamma(\epsilon)}(M)$. The map $q \mapsto \varphi^U(q,\epsilon)$ maps the point $p$ to the point $\gamma(\epsilon)$ and is a diffeomorphism (at least locally) so you use this map to "transfer" or "pull back" the vector $V(\gamma(\epsilon))$ to the point $p$. Then, the subtraction makes sense, and you can compute the limit.

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I have some troubles understanding the map $\phi^U$. For example, if $U$ is a constant vector field, is this map the identity ? I thought I could add a vector to a point as long as I took the limit at only added an epsilon of this vector... apparently not! –  WhitAngl Dec 21 '12 at 16:40
    
There is no such a thing as a "constant" vector field on a general manifold. How do you define constant? Constant in a specific coordinate system? But then, it won't look constant in a different one! In some sense, the generalization of a constant vector field is a parallel vector field, but this notion requires the choice of a connection on a manifold. –  levap Dec 21 '12 at 16:48
    
The map $\varphi_t^U(q)$ takes a point $q \in M$, looks at the integral curve of the vector field $U$ on $M$ which starts at $q$ at time $s = 0$, and asks to which point this integral curve has "flowed", reached, at time $s = t$. The result is $\varphi_t^U(q)$. If $M$ is compact, $\varphi_t^U$ is defined for $-\infty < t < \infty$ and describes how points of $M$ evolve under the vector field $U$. –  levap Dec 21 '12 at 16:52
    
If $U$ is the zero vector field, then this map is the identity. No point of $M$ moves under the flow of $U$. Try to see if you understand how $\varphi_t^U$ acts if $U$ is a constant vector field in $\mathbb{R}^2$. –  levap Dec 21 '12 at 16:58
    
in my case, $U$ is constant in one coordinate system (I am trying to compute Koszul's formula, and I take $U$ as one of my basis vector)... then I feel like taking the integral curve just means translating my point, which also means taking the value $p+U$ , isn't it ? –  WhitAngl Dec 21 '12 at 17:08
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