I can't figure out why we need the definition of a 'covariant derivative along a curve', i.e. I can't see why we can't use a 'linear connection' even when the vector fields are not extendible.

I'm reading Lee's book on Riemannian manifolds. After he has shown that $\nabla$ depends on X and Y only around an open set, he defines the Christoffel symbols through the expression $\nabla_{E^j}E^i$, where $E^j,E^i$ are elements of a local frame, i.e. vector fields defined only locally on an open set (and thus not necessarily extendible). Likewise, it is show that $(\nabla_{X}Y)_p$ in fact only depends on $X$ through its value at p and on Y through its values on a curve through p whose tangent at p is $X_p$. Therefore, if $\gamma$ is a smooth curve, $(\nabla_{\dot{\gamma}}Y)_p$ should be well-defined, even if Y is only defined along $\gamma$ and isn't extendible.

Where am I wrong? Thanks a lot.


2 Answers 2


I think you are right that one could make sense out of $\nabla_{\gamma'(t)}Y$ even if $Y$ is a nonextendible vectorfield along a curve $\gamma: I\to M$. One could try to do this as follows:

  1. If $\gamma'(t)\neq 0$ then there is a neighbourhood $J$ of $t$ such that $\gamma_{|J}$ is an embedding. We then can find a globally defined vectorfield $\tilde Y$ on $M$ such that $Y$ and $\tilde Y \circ\gamma$ agree locally arround $t$ and then define $\nabla_{\gamma'(t)}Y= \nabla_{\gamma'(t)}\tilde Y$ which will not depend on the choice of $\tilde Y$

  2. If $\gamma'(t)= 0$ we simply define $\nabla_{\gamma'(t)}Y=0$.

Now one can show that in the first case this definition agrees with the usual definition of the covariant derivative of $Y$ along $\gamma$. But in the second case it doesn't:

Consider for example $\gamma:I\to\mathbb R^2, t\mapsto(t^2,t^3)$ and $Y(t)=\gamma'(t)$ where $\mathbb R^2$ is equipped with the Levi-Civita connection . Then using standard coordinates on $\mathbb R^2$ we have $Y'=2t(\partial_1\circ\gamma)+3t^2(\partial_2\circ\gamma)$. Using the leibniz rule and the agreement with extendible vectorfields we see that the covariant derivative along $\gamma$ is given by $2(\partial_1\circ\gamma)+6t(\partial_2\circ\gamma)$. Especially at $t=0$ it is non-zero even if $\gamma'(0)=0$.

  • $\begingroup$ Thanks a lot for your example. I can see that they disagree when I use the coordinate expression for 'the covariant derivative along a curve' (C.D.A.C.). But, isn't $Y = \gamma'$ extendible? In that case, $\nabla_{\gamma'(t)}Y$ and the 'C.D.A.C.' should agree. Hmmm. Thanks for your help. $\endgroup$
    – gemini
    Commented Aug 23, 2019 at 13:29
  • $\begingroup$ Well it definitely should not be extendible since else in the above example the (c.d.a.c) at $t=0$ would be $0$. $\endgroup$
    – Claire
    Commented Aug 23, 2019 at 15:01
  • $\begingroup$ Yes exactly, but it seems unintuitive to me that it should be non-extendible. I will try to find out why that is. Thanks for your great answer! $\endgroup$
    – gemini
    Commented Aug 23, 2019 at 16:27
  • 1
    $\begingroup$ @gemini: If you had a smooth vector field $X$ that was defined in an open neighborhood of $(0,0)$ and such that $X(\alpha(t)) = \alpha'(t)$ for some $t \in (-\varepsilon, \varepsilon)$ then this would imply that $\alpha$ is an integral curve of $X$. But $\alpha(0) = \alpha'(0) = (0,0)$ so by the uniqueness of integral curves, we must have $X(\alpha(t)) \equiv (0,0)$, a contradiction. This doesn't use the specific form of $\alpha$, only that $\alpha'(0) = \vec{0}$. $\endgroup$
    – levap
    Commented Aug 24, 2019 at 22:19

A quick answer to the Title.

One of the important and powerful tool in studying Differential geometry and Riemannian geometry is understanding the behavior of geodesics. And what is the geodesic?

There are two key properties satisfied by straight lines in $\Bbb R^n$, either of which serves to characterize them uniquely: first, every segment of a straight line is the unique shortest path between its endpoints; and second, straight lines are the only curves that have parametrizations with zero acceleration. (John m. Lee, Riemannian manifolds)

So we need the notion of covariant derivative along a curve to measure the acceleration of a curve and then define the geodesics and then discovering topological properties and then ...

Added: Note that covariant derivative along a curve is not a definition in Lee's Book. it is just a restriction to a curve of covariant derivative.

  • $\begingroup$ Thanks for you post. However, my question is why a linear connection can't be used for this. I edited the title accordingly. $\endgroup$
    – gemini
    Commented Aug 23, 2019 at 10:31
  • $\begingroup$ How you should define acceleration of a curve if we don't have C.D.A.C? $\endgroup$
    – C.F.G
    Commented Aug 23, 2019 at 10:39
  • $\begingroup$ As $\nabla_{\dot{\gamma}}\dot{\gamma}$. In my question I argue for why this should be well-defined, even if $\dot{\gamma}$ is not extendible. And my question is 'why am I wrong?' $\endgroup$
    – gemini
    Commented Aug 23, 2019 at 10:54
  • $\begingroup$ Your argument is correct. It sound like your question is about the proof of Theorem 4.24 (books.google.com/…) of Lee's Book? Am I right? $\endgroup$
    – C.F.G
    Commented Aug 23, 2019 at 11:03
  • $\begingroup$ Yes, you are right. I guess I don't understand why part (iii) of that theorem only holds when V is extendible. $\endgroup$
    – gemini
    Commented Aug 23, 2019 at 11:45

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