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

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?

• What is the derivative of the unit tangent vector field to any of the circles $z=c$ in the unit sphere $x^2+y^2+z^2=1$? Draw pictures. Do some computations. Feb 13 at 19:38
• Hi, I've tried but I'm still learning about manifolds and tangent spaces so I don't have a lot of familiarity with the mathematical "machinery". Would you be so kind to give me some hints on how to do what you are asking? Thank'you Feb 14 at 14:28
• I’m just asking for basic multivariable calculus. The curves I’m talking about are easily parametrized, and then you can differentiate once and then again. Feb 14 at 15:16
• By the way, if you haven’t looked at the concrete cases of curves and surfaces in $\Bbb R^3$, I recommend starting your differential geometry exploration there. You can look at my free text, linked in my profile. Feb 14 at 15:54
• Thank you very much Feb 14 at 20:39

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).
• So the problem, if I understand correctly, is that while a local frame always exist, so the derivative is defined locally, but globally there may not be a global frame (which would be the case of $\mathbb{R}^n$ given that it has a single chart covering all the manifold, so it has a single set of chart coordinates that induce a frame), thus a derivative may not be defined? Feb 14 at 8:57