Why the derivative of a vector field along a curve is not defined for a generic manifold?

Question

In his book "Differential Geometry" Loring. Tu writes the derivative of a vector field $X$ along a curve $c: [a,b] \to \mathbb{R}^n$ as $\frac{dV(t)}{dt}=\sum \frac{dv^i(t)}{dt} \partial_i |_{c(t)}$. It then goes on to say that such a derivation is only defined in the $\mathbb{R}^n$ manifold and not in an arbitrary manifold because it doesn't have a canonical frame $\partial_1 .... \partial_n$ like $\mathbb{R}^n$ does. But what exactly does this mean? Could it mean that in an arbitrary manifold, globally the derivative is not defined, because we don't have an atlas with a single chart like $\mathbb{R}^n$? But the derivative could still exist locally, because by considering a single chart around a point $(U, x_1, ...,x_n)$ I could still use the chart coordinates to create a frame, couldn't I?

Answer 1

Suppose $X$ is a vector a field on a smooth manifold, and $\gamma:I\rightarrow M$ a smooth curve. Let $(U,\phi$, and $(V,\psi)$ be coordinate charts with $U\cap V$, and $\gamma(I)\cap U\cap V$ not equal to the empty set.

Let $U$ have coordinates $x^i$, and $V$ have coordinates $y^i$. Then in both frames we can write that: $$X=v^i\frac{\partial}{\partial x^i}\qquad X=w^i\frac{\partial}{\partial y^j}$$ Since $X$ is globally defined, we have that on the overlap these must agree. I.e. the transition function given by the Jacobian of $\phi\circ\psi^{-1}$ relates the two coordinate frames. This means that on $U\cap V$, we can write $X$ in terms of the $x$ frame as: \begin{align} X=w^i\frac{\partial}{\partial y^i}=w^i\frac{\partial x^j}{\partial y^i}\frac{\partial}{\partial x^j} \end{align} so we have that $w^i\partial x^j/\partial y^i=v^j$. Now in the $V$ coordinates, we would write $d/dt X$ along $\gamma$ to be: \begin{align} \frac{d}{dt}X(t)=\frac{dw^i(t)}{dt}\frac{\partial}{\partial y^i}\qquad \star \end{align} and in the $U$ coordinates, we would write it as: \begin{align} \frac{d}{dt}X(t)=\frac{dv^i(t)}{dt}\frac{\partial}{\partial x^i} \end{align} Now, we know that on $U\cap V$, $v^i=w^j\partial x^i/\partial y^j$, so: \begin{align} \frac{d}{dt}X(t)=\left(\frac{dw^j}{dt}\frac{\partial x^i}{\partial y^j}+w^j\frac{d}{dt}\left[\frac{\partial x^i}{\partial y^j}\right]\right)\frac{\partial}{\partial x^i} \end{align} However, if rewrite $\star$ in terms of the $x$ frame, we have that: $$\frac{d}{dt}X(t)=\frac{dw^i}{dt}\frac{\partial x^j}{\partial y^i}\frac{\partial}{\partial x^i}$$ and it just doesn't have to be the case that these two things are equal to one another, so the naive idea of differentiating a vector field along a curve with respect to a local frame does not yield a globally defined vector field, since it's not independent of coordinates.

In the case of $\mathbb R^n$ we can do this because our vector bundle is trivial, so we have a global frame. In fact, for any trivial vector bundle you can define derivatives of a vector field like this, and it's equivalent to having a covariant derivative that admits a flat curvature form. Basically, you can find a frame such that all the Christoffel symbols vanish (which I am sure you will get to at some point in Tu's wonderful book).