When we define a connection $\nabla$ it follows naturally the definition of the covariant derivative as $\nabla_b X_a$ as it is well known. The following step is to consider vector field parallel transported. If we take a curve $\gamma: [a,b] \longrightarrow \mathcal{M} $ and a vector field $\mathbf{V}$ we can say it's a parallel transported vector field if $\nabla_{\mathbf{X}(t)}\mathbf{V}(t) = 0 \ { }\forall t \in [a,b]$. In a second moment we can define a map $\mathbf{P}_\gamma: T_{\gamma(a)}\mathcal{M} \rightarrow T_{\gamma(b)}\mathcal{M}$ that maps the vector $\mathbf{V}(a)$ to the vector $\mathbf{V}(b)$ and we can say that this application gives the notion of parallel transport of vector.

What I miss is why in the majority of books it's always said that the $\nabla_{\mathbf{X}}\mathbf{Y}$ is the parallel transport of $\mathbf{Y}$ along the curve $\gamma$ whose tangent vector is $\mathbf{X}$ if by definition if $\mathbf{Y}$ is parallel transported its covariant derivative along $\mathbf{X}$ is $0$? I hope the question is clear, if it's not I'm here for clarification ( I'm here for that anyway).

  • $\begingroup$ Possibly of interest: Covariant Derivative. $\endgroup$ – Andrew D. Hwang Jul 12 '17 at 0:50
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    $\begingroup$ Ok, i see that if the covariant derivative differs from 0 the vector field is not parallel transported (this is the definition) and the value of the covariant derivative at that point measures the difference between the vector field and the parallel transported vector field, isn't it?. So the question is the quite the same: why the majority of the books still call $\nabla_{\mathbf{X}}\mathbf{Y}$ the parallel transportation of Y along X? The parallel transportation can be done even if the vector field is not parallel transported I imagine is the answer or is there some mistake in my thought? $\endgroup$ – raskolnikov Jul 12 '17 at 1:02
  • $\begingroup$ maybe a little ambitious quizz would be to ask for a vector field X in IR^2 over a given curve C therein. So the rule for a parallel transported field would be $D_{C'}X=0$ with $D$ the std covariant derivative of IR^2. $\endgroup$ – janmarqz Jul 12 '17 at 1:10
  • $\begingroup$ Can you make the example clear? $\endgroup$ – raskolnikov Jul 12 '17 at 8:04
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    $\begingroup$ @AndrewD.Hwang the problem is that i read these things in physics books ( I am a physicist), maybe it's an abuse of notation? Or there is a way to understand it in a qualitatively way? $\endgroup$ – raskolnikov Jul 12 '17 at 12:56

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