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I'm reading John Lee's Riemannian Manifolds where I encountered the following reading exercise:

Improve Lemma 4.1 by showing that $\triangledown_{X_{p}}Y$ actually depends only on the values of $Y$ along any curve tangent to $X_{p}$. More precisely, suppose that $\gamma: (-\epsilon, \epsilon) \rightarrow M$ is a curve with $\gamma(0)=p$ and $\gamma'(0)=X_{p}$, and suppose that $Y$ and $\tilde{Y}$ are vector fields that agree along $\gamma$. Show that $\triangledown_{X_{p}}Y = \triangledown_{X_{p}}\tilde{Y}$

Here we are taking $M$ to be a smooth manifold and $\triangledown$ to be a linear connection. Lemma $4.1$ essentially tells us that the vector field, $\triangledown_{X}Y$ is determined locally by $Y$ and pointwise by $X$. The linearity of the connection, I hope to show that if $Y=0$ along $\gamma$ then $\triangledown_{X_{p}}Y=0$.

Earlier in the text, the notion of covariant derivatives of vector fields along curves were discussed, in particular it was mentioned that when the vector field $V$ along a curve is extendible, then for any extension $\tilde{V}$, we have that $D_{t}V(t)=\triangledown_{\gamma'(t)}\tilde{V}$ in which case we have a formula for $D_{t}V(t)$. I'm a bit confused on how to involve this.

Any help is appreciated.

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Recall the local formula for $\nabla_XY$:

$$\nabla_XY=(x^i \partial_i y^k+x^iy^j \Gamma_{i,j}^k)\partial_k.$$ Given $\gamma$ a local path passing through the point $p$ at $0$ with tangent vector $x^ie_i$, we have $$x^i \partial_i y^k=(y^k \circ \gamma)'(0) $$ by the chain rule. Since the rest of the terms are just evaluations, it follows that we only need the values of $Y$ on the path $\gamma$.

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  • $\begingroup$ Sorry, would you be willing to expand on what you mean by "just evaluations"? $\endgroup$ – user135520 Mar 23 '17 at 22:19
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    $\begingroup$ @user135520 Of course. The formula is, explicitly, $$(\nabla_X Y)(p)=(x^i(p) (\partial_i y^k)(p)+x^i(p)y^j(p) \Gamma_{i,j}^k(p))\partial_k.$$ That is, with exception of the term $\partial_i y^k$, all of them just consider the value of the functions on the point $p$ directly. $\endgroup$ – Aloizio Macedo Mar 23 '17 at 22:22

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