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Here's a quote from page 116, section 4-2, of Chern's Lectures on Differential Geometry:

Definition 2.1 Suppose $C:u^i=u^i(t)$ is a parametrized curve on $M$, and $X(t)$ is a tangent vector field defined on $C$ given by $$ X(t) = x^i(t) \left(\frac{\partial}{\partial u^i}\right)_{C(t)}. $$ We say that $X(t)$ is parallel along $C$ if and only if its absolute differential along $C$ is zero, i.e., if $$ \frac{DX}{dt} = 0. $$ If the tangent vectors of a curve $C$ are parallel along $C$, then we call $C$ a self-parallel curve, or a geodesic.

I have a couple questions about this.

First, the notation $\frac{DX}{dt}$ does not make sense to me. Formally, $DX$ is defined to be a member of $\Gamma(T^*(M)\otimes E)$. I see what he's getting at here: he wants to evaluate $DX$ along $C$, but that's not what $DX$ means to me. Perhaps he means something like $D_{C'(t)}X$? The notation is getting confusing and imprecise, I think.

Second, he says the "tangent vectors of $C$ are parallel along $C$". But I'm not sure he ever defined the "tangent vector" of $C$. Does he mean $$ \frac{d u^i(t)}{dt}\left(\frac{\partial}{\partial u^i}\right)_{C(t)}? $$ If so, then the tangent vector at a point will change depending on the parametrization of $C$ (it will point in the same direction but be shorter or longer), and so whether $C$ is a geodesic also depends on the parametrization. Is that intended?

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First of all, see Definition 1.3 and the discussion on p. 111. He doesn't seem to say so, but the notation $\dfrac{DX}{dt}$ is shorthand for $D_{c'(t)} X$. And, yes, whether $C$ is a geodesic depends entirely on how it's parametrized: For the standard Riemannian connection, only a constant-speed curve can have a chance to be a geodesic, for example.

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  • $\begingroup$ Is the notation $\frac{DX}{dt}$ standard or common? $\endgroup$ – Will Nelson Mar 9 '16 at 4:56
  • $\begingroup$ Relatively. It's the "covariant derivative analog" of $d/dt$ when we differentiate with respect to a parameter $t$. $\endgroup$ – Ted Shifrin Mar 9 '16 at 5:01

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