Why do we need the covariant derivative along a curve - why are linear connections not sufficient?

Question

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.

Answer 1

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 $\gamma'(0)=0$.

Answer 2

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.