Why $\frac{DV}{dt}= \sum\limits_j \frac{dv^j}{dt}X_j +\sum\limits_{ij}\frac{dx_i}{dt} v^j \nabla_{X_i}X_j$ means the $\frac{DV}{dt}$ is unique? Picture below is from pages 50-51 of  do Carmo's Riemannian Geometry. I can't understand why the red line means that $\frac{DV}{dt}$ is unique when $V$ is fixed. In my view, there is not any  proof to show that the right part of red line  is independent to  the choice  of coordinate.
What I try:  The red line can be written as
$$
\frac{DV}{dt}=\left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right) X_k
$$
Then, I have
$$
\left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right) = f^k(c(t))
$$
namely, the  $\left( \sum\limits_j \frac{dv^k}{dt} +\sum\limits_{ij}\frac{dx_i}{dt} v^j \Gamma_{ij}^k\right)$  can be treated as a function of $c(t)$ or $t$. So, I have
$$
\frac{DV}{dt}(c(t)) =\sum_k f^k(c(t))X_k(c(t))   \tag{1}
$$
But,  if in another coordinate $Y_i=\frac{\partial}{\partial y_i}$, similarly, I can get
$$
\frac{DV}{dt}(c(t)) =\sum_k \hat f^k(c(t))Y_k(c(t))   \tag{2}
$$
how to show that the right parts of (1) and (2)  are same ?




 A: You are correct that the calculation doesn't show that the given expression is independent of the choice of coordinates. However, the argument do Carmo gives doesn't require you to show it and you get it as a byproduct of the proof "for free". Since do Carmo's argument is somewhat subtle and missing in details, let me give you an outline of the argument:

*

*The first thing you need to know is that both $\nabla$ and $\frac{D}{dt}$ are "local operators". This means that $(\nabla_X Y)(p)$ depends only on the values of $Y$ and $X$ in an arbitrary small neighborhood of $p$ and similarly $\frac{DV}{dt}(t_0)$ depends only on the values of $V$ and $c$ near $t_0$. This is a consequence of the linearity, the product rule and the existence of bump functions (see page 50 in Lee's Riemannian Manifolds book). Once you know it, it makes sense to talk about things like $\nabla_{\partial_i} \partial_j$ and $\frac{D \partial_i}{d t}$. The reason is that $\partial_i,\partial_j$ are only local vector fields so a priori you can't plug them into $\nabla$ (or $\frac{D}{dt}$). However, given any point $p \in M$ around which they are defined, you can modify them and extend them using bump functions so that they become global vector fields which remain them same near $p$ and the resulting covariant derivatives do not depend on the specific way you do it. Do Carmo actually mentions it in Remark 2.3 for $\nabla$ (but not for $\frac{D}{dt}$) but his "proof" is wrong because in the proof he uses the expression $\nabla_{\partial_i} \partial_j$ before he actually justifies why it is legitimate.

*The next thing is to show uniqueness of $\frac{D}{dt}$. Let $t_0 \in I$ and choose an arbitrary coordinate system $(x^1,\dots,x^n)$ near $c(t_0)$ which gives you a correspodning local frame $(\partial_1, \dots, \partial_n)$. Then using point $(1)$ and performing do Carmo's calculation, you get the formula
$$ \frac{DV}{dt}(t_0) = \frac{dv^j}{dt}(t_0) \partial_j|_{c(t_0)} + \frac{d x^i}{d t}(t_0) v^j(t_0) \nabla_{\partial_i}{\partial_j} (c(t_0)) \label{eq:covariant-formula} \tag{1}$$
where I use Einstein's summation to make it shorter. If you would have chosen a different coordinate system you would get a formula in terms of the other coordinate system which a priori could give you a different result but at this point you don't care because this formula is enough to prove uniqueness. If there are two operators $\frac{D_1}{dt}, \frac{D_2}{dt}$ which satisfy the properties of the covariant derivative along $c$, by choosing the same coordinate system around $c(t_0)$ you get the same formula for $\frac{D_1}{dt}(t_0)$ and $\frac{D_2}{dt}(t_0)$ and this holds for all $t_0 \in I$ so $\frac{D_1}{dt} = \frac{D_2}{dt}$.

*The next thing to do use equation $\eqref{eq:covariant-formula}$ to define $\frac{D}{dt}$ locally. Choose some $t_0 \in I$, some coordinate system $(x^1,\dots,x^n)$ around $c(t_0)$ and some $\varepsilon > 0$ so that $c \left( (t_0 - \varepsilon, t_0 + \varepsilon) \right)$ lies in that coordinate neighborhood. Define an operator $\frac{D}{dt}$ on vectors fields along $c|_{(t_0 - \varepsilon, t_0 + \varepsilon)}$ using equation $\eqref{eq:covariant-formula}$ and verify that it satisfies all three properties of the covariant derivative.  Note that in this definition you are using a specific coordinate system to define $\frac{DV}{dt}$ and this only gives you a "local" covariant derivative. Is this definition depends on the coordinate system? Not really, because of uniqueness. Assume that $\varepsilon$ is small enough so that $c \left( (t_0 - \varepsilon, t_0 + \varepsilon) \right)$ lies in the domain of two coordinate systems $(x^1,\dots,x^n)$ and $(y^1,\dots,y^n)$. A priori you can use the $x^i$'s to define an operator $\frac{D_1}{dt}$ and the $y^i$'s to define an operator $\frac{D_2}{dt}$. However, since they both satisfy the properties of the covariant derivative, the uniqueness shows you that $\frac{D_1}{dt} = \frac{D_2}{dt}$ so in fact the result of the formula is independent of the coordinate system chosen. This can of course be verified directly by a calculation but you can see it is not required once we know uniqueness.

*Finally, given a vector field $V$ along $c$ and $t_0 \in I$, choose an arbitrary coordinate system around $c(t_0)$ and using this specific coordinate system to define $\frac{DV}{dt}(t_0)$ by equation $\eqref{eq:covariant-formula}$. This is well-defined by item $(3)$ and since the properties of the covariant derivative are all local, $\frac{D}{dt}$ also satisfies them and you get existence.

