How to look at the covariant derivative along a curve?

Question

Let $ \Gamma(M,TM) $ be the space of smooth sections from $M$ to $TM$ and $ (M,g) $ be a Riemannian manifold, $ I \subset \mathbb{R} $ be an interval and $\gamma: I \rightarrow M$ be a $C^1$ curve. Let $ \nabla: \Gamma(M,TM) \times \Gamma(M,TM) \rightarrow \Gamma(M,TM) $ be the Levi-Civita connection with respect to $(M,g)$. Then since $ \dot{\gamma} : I \rightarrow TM $ is not an element of $ \Gamma(M,TM) $ but an element of $ \Gamma(I, TM) $. What is the proper way for one to interpret notations like $ \nabla_{\dot{\gamma}}X $ or $ \nabla_{\dot{\gamma}}\dot{\gamma} $?

Answer 1

The proper way to do this is to pull back the bundle $E\to M$ (in this case $E=TM$ but this works in general) and its connection to $I$: $$ \gamma^*\nabla\colon\Gamma(\gamma^*E)\to\Gamma(\gamma^*E\otimes T^*I);\\(\gamma^*\nabla)_X(\gamma^*s)=\gamma^*(\nabla_{(d\gamma)(X)}s)\quad\forall s\in\Gamma(E),\forall X\in TI $$ and so $\frac{D}{dt}\dot\gamma:=(\gamma^*\nabla)_{d/dt}\dot\gamma$ (and suchlikes) makes sense. But that is rather a lot of $\gamma$'s lying around.

Let $[\gamma]=\{\gamma(t)\mid t\in I\}$. If $X\in\Gamma(I,E)$ is such that $X$ comes from $[\gamma]\to E$ (i.e., if $\gamma(t)=\gamma(t')$ then $X(t)=X(t')$), then we can always extend $X$ to some $\tilde{X}$ in a neighbourhood $U$ of $[\gamma]$ when $E,M$ are finite-dimensional (for infinite-dimensional case there are some complications), being at least as smooth as the original data $(\gamma,X)$ (this is basically the Whitney extension theorem). We can now ask the question: is $\nabla_{\dot\gamma}\tilde{X}$ the same as $(\gamma^*\nabla)_{d/dt} X$? The answer is yes (as you know), so we can write $\nabla_{\dot\gamma}\dot\gamma$ (and suchlikes).