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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?

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    $\begingroup$ 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. $\endgroup$ Feb 13 at 19:38
  • $\begingroup$ 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 $\endgroup$
    – Andrea
    Feb 14 at 14:28
  • $\begingroup$ 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. $\endgroup$ Feb 14 at 15:16
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    $\begingroup$ 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. $\endgroup$ Feb 14 at 15:54
  • $\begingroup$ Thank you very much $\endgroup$
    – Andrea
    Feb 14 at 20:39

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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).

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  • $\begingroup$ 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? $\endgroup$
    – Andrea
    Feb 14 at 8:57
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    $\begingroup$ Yea pretty much. If you’re gonna define a derivative on a vector field it should give you another (globally defined) vector field. And just naively taking the derivative only does that in certain situations. $\endgroup$
    – Chris
    Feb 14 at 9:39

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