Covariant derivatives along curves proof explanation

Question

I am curious about the following proof of uniqueness for covariant derivatives along curves.

Proposition: There is precisely one operation $\frac{D}{dt}$ from $C^{\infty}$ vector fields along $c: I=(a,b)\rightarrow M$ ( a curve in $M$) such that

1. $\frac{D(V+W)}{dt}=\frac{DV}{dt}+\frac{DW}{dt}$

2. $\frac{D(fV)}{dt}=\frac{df}{dt}V+f\frac{DV}{dt}$ for $f\in C^{\infty}(I)$

3. If $V_s=Y_{c(s)}$ for some smooth vector field $Y$ defined on a neighbourhood of $c(t)$, then $\frac{DV}{dt}=\nabla_{c'(t)}Y$, where $\nabla$ is a connection.

Proof goes like this;

Choose a coordinate system $(U,x^1,.......,x^n)$ about $c(t)$ then $V(t)=\sum_i^nV^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)}$ where each $V^i\in C^{\infty}(I)$.

Then we must have $\frac{DV}{dt}=\sum_i\frac{D}{dt}(V^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)})$. (Whatever is defined inside the bracket is a local expression, so why are we allowed to define it like that?)

My problems with this:

Why does the above expression make sense? It should be something along the lines of $\frac{DV}{dt}=\sum_i\frac{D}{dt}(V^jg_i)$ where $V^j$ is a smooth function $I\rightarrow \mathbb{R}$ and $g_i$ is a smooth function $I\rightarrow TM$ such that $g(t)=\frac{\partial}{\partial{x^j}}|_{c(t)}$. But in this case, why would $g$ be a smooth vector field along $c$?

Answer 1

As Professor Theodore Shifrin pointed out to you, your question is not worded very clearly, because you say that $U$ is only a neighborhood around a point $c(t)$ for fixed $t\in I$ but then you say that the $V^i$'s are functions on $I$ which makes no sense. This is because, if $U$ does not cover the entire curve $c$, the $\frac{\partial}{\partial x^i}$'s are not defined on the whole curve and thus writting $V(t)=\sum_i^nV^i(t)\frac{\partial}{\partial{x}^i}|_{c(t)}$ only makes sense for $t\in I_U:=I\cap c^{-1}(U)$. Consequently you would have $V^i\in C^{\infty}(I_U)$.

To cover the whole curve, you would need other coordinate neighborhoods around other points in the curve. In the intersections, the components $V^i$ would transform as components of vector fields under coordinate transformations. Namely, if $\tilde{U}$ is another neighborhood around some point $c(\tilde{t})$ of the curve, with coordinates $(y^1,\ldots,y^n)$, you would have

$$V(t)=\sum_i^nW^i(t)\frac{\partial}{\partial{y}^i}|_{c(t)},\quad t\in I_{\tilde{U}},$$

where $W^i\in C^{\infty}(I_{\tilde{U}})$, with $I_{\tilde{U}}:=I\cap c^{-1}(\tilde{U})$. On the intersection $I_U\cap I_{\tilde{U}}$ you would have coordinate transformations $(y^1(x),\ldots,y^n(x))$ and thus

$$W^i(t)=\sum_{j=1}^n\frac{\partial y^i}{\partial x^j}V^j(t),\quad t\in I_U\cap I_{\tilde{U}}.$$

Hope this helps.