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let $X,Y:M\to TM$ be vector fields on $M$. $\nabla_XY$ is the change in $Y$ along the flow curves of $X$. so for a point $p \in M$ let $\phi^X(t):\mathbb{R}\to M$ be a flow curve of $X$ passing through $p$ : $$(\phi^X)'=X \; \; ; \; \; \phi^X(0)=p$$ since I can't measure the change in $Y$ at two different points directly. I'll define a parallel transport map $\Pi_{tX}:T_pM\to T_{\phi^X(t)}M$ taking a vector at $p$ to its equivalent at $\phi^X(t)$. Then the derivative at $p$ is: $$\nabla_XY|_p=\lim_{t\to 0}\frac{\delta_XY|_{\phi^X(t)}}{t}=\lim_{t\to 0}\frac{Y_{\phi^X(t)}-\Pi_{tX}(Y_p)}{t}$$ I tried to do the usual approximation trick by assuming that for a "small" enough $t$ : $$\nabla_XY|_p \approx \frac{\delta_XY|_{\phi^X(t)}}{t}$$ but there is a problem that LHS and RHS are not at the same point.

here is what I did "correct me if i'm wrong":

since the points $p$ and $\phi^X(t)$ are so close: $$\Pi_{tX}(\nabla_XY|_p) \approx \frac{\delta_XY|_{\phi^X(t)}}{t}$$ in other words: $$\Pi_{tX}(\nabla_XY|_p)=\frac{\delta_XY|_{\phi^X(t)}}{t}+\vec\epsilon_1$$ $\vec\epsilon$ being a really small difference vector. Another thing that follows is that the values of the vector field $\nabla_XY$ at the two points are close so: $$\Pi_{tX}(\nabla_XY|_p)=\nabla_XY|_{\phi^X(t)}+\vec\epsilon_2$$ then $$\delta_XY|_{\phi^X(t)}=t\nabla_XY|_{\phi^X(t)}+t\vec\epsilon_2-t\vec\epsilon_1$$ since $\vec\epsilon$ errors are already small. $t\vec\epsilon$ would be even much smaller "second order error" and could be neglected: $$\delta_XY|_{\phi^X(t)} \approx t\nabla_XY|_{\phi^X(t)}$$ so: $$\Pi_{tX}(Y_p) \approx (-t\nabla_XY+Y)|_{\phi^X(t)}$$

so is there anything wrong with this attempt?

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Defining \begin{equation} \nabla_XY|_p := \lim_{t\to 0} \frac{\Pi_{tX}^{-1}(Y_{\phi^X(t)})-Y_p}{t} \end{equation} the approximation $\nabla_XY|_p \approx \frac{\Pi_{tX}^{-1}(Y_{\phi^X(t)})-Y_p}{t}$, for $t$ small, holds and now both sides are tangent vectors in $T_pM$. It also follows that $\Pi_{tX}^{-1}(Y_{\phi^X(t)})\approx t\,\nabla_XY|_p+Y_p$, namely $Y_{\phi^X(t)}\approx \Pi_{tX}(t\,\nabla_XY|_p+Y_p)$ and therefore \begin{equation} \Pi_{tX}(Y_p)\approx Y_{\phi^X(t)}-t\,\Pi_{tX}(\nabla_XY|_p). \end{equation}

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  • $\begingroup$ Thanks for the answer it seems clearer than how I thought about it. So $\Pi_{tX}(Y_p)\approx Y_{\phi^X(t)}-t\nabla_XY|_{\phi^X(t)}+t^2\nabla^2_XY|_{\phi^X(t)}-...$. Ignoring higher order terms one gets $\Pi_{tX}(Y_p) \approx Y_{\phi^X(t)}-t\nabla_XY|_{\phi^X(t)}$ $\endgroup$ Nov 19, 2021 at 13:32

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