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I am reading the book by Lee, Introduction to Riemannian manifolds which I think is very nice. However among the things I don't understand is the following :

Theorem 4.24 (Covariant Derivative Along a Curve).

Let $M$ be a smooth mani- fold with or without boundary and let $\nabla$ be a connection in $TM$ . For each smooth curve $\gamma : I \to M$ , the connection determines a unique operator $D_t:\mathfrak{X}(\gamma)\to\mathfrak{X}(\gamma)$,

called the covariant derivative along $\gamma$, satisfying the following properties:

(i) LINEARITY OVER $\mathbb{R}$: $D_t(aV+bW)=aD_tV+bD_tW$ for $a,b\in\mathbb{R}$:

(ii) PRODUCT RULE :

$D_t(f V)=f'+ D_t V$ for $f\in C^\infty(I)$

(iii) If $V \in \mathfrak{X}(M)$ is extendible, then for every extension $\tilde{V}$ of $V$

$D_t V(t)= \nabla_{\gamma'(t)}\tilde{V}$.

There is an analogous operator on the space of smooth tensor fields of any type along $\gamma$ .

But in property (iii), If it was "$V\in \mathfrak{X}(M)$ and $\gamma'$ are both extendible, then for every extension $\tilde{V}$ and $\tilde{\gamma'}, D_tV(t)=(\nabla_{\tilde{\gamma'}}\tilde{V})(\gamma(t))$", it would make more sense to me, because here $\gamma'(t)$ is just a tangent vector and not a vector field. There is something I am not getting right.

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  • $\begingroup$ The covariant derivative $\nabla_XY$ depends only on the value of $X$ at a point, not in a neighborhood. (It is "tensorial in $X$"). $\endgroup$
    – Neal
    Oct 21 '20 at 14:50
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I think you want use Proposition 4.5: $\nabla_X Y |_p$ depends only on the value of $X$ at $p$ (as well as the values of $Y$ in a neighborhood of $p$). Then, for a tangent vector $v \in T_p M$ we interpret the expression $\nabla_v Y$ by arbitrarily extending $v$ to a vector field in a neighborhood of $p$; the proposition implies that the result is independent of extension.

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    $\begingroup$ Thank you. The remark after the proof of Proposition 4.5 is exactly what I was looking for (actually because of this remark the definition is the same as what I thought it was). Thanks! $\endgroup$
    – roi_saumon
    Oct 21 '20 at 16:44
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It would make more sense as you said, but you must have proved that the connection is tensor in the direction you are taking the derivative.

This means that $\nabla_W \,\cdot\restriction_p$ only depends on $W(p)$, i.e., no matter how you extend $W(p)$ to a vector field, the connection will be the same at $p$.

Recall that in the other entry this is no longer true, but still is local, which means that if $V,\tilde V$ are vector fields, that coincide in a vicinity of $p$, then $\nabla_W V(p) = \nabla_W \tilde V (p)$.

And this is why the definition makes sense.

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