Do derivative operators act on the manifold or in R^n?

[lavinia]

Thank you! That clears a lot up. This kind of thing is what I am looking for.
I never said the various definitions aren't equivalent, but I admit that I get heavily invested with one definition and don't step back to see the larger picture. Somehow this all became about definitions, but my original question was whether the derivative operators act in the manifold or in $\mathbb{R}^n$ and this confusion was spawned precisely because of that resource that said that derivatives on manifolds make no sense if the manifold is not a submanifold of $\mathbb{R}^n$.

Thank you for helping me with this.

What your book probably meant was that you can not compute Newton quotients using points on a manifold because it in general makes no sense to add points or multiply them by numbers. If the manifold is embedded in Euclidean space, then one can use the addition and scalar multiplication in Euclidean space to compute the Newton quotients.

 

[lavinia]

I think that Orodruin's point about using curves to compute derivatives without using coordinate charts deserves repeating.

A tangent vector $X$ at the point $p$ may be thought of as a linear operator on functions that satisfies the Leibniz Rule $X⋅fg = (X⋅f)g(p) + f(p)X⋅g$. One can compute $X.f$ by picking any curve $γ$ with $γ'(t) = X$ and differentiating $f \circ γ$ at $t$. One gets the same answer for any curve with derivative equal to $X$ at $p$. This means that a tangent vector can be thought of as the equivalence class of these curves and its action is just $d/dt f\circ γ$ where $γ$ is any curve in this equivalence class. Further, a curve $γ$ is in this equivalence class if and only if $d/dt f\circ γ$ agrees with the action of $X$ on $f$ for all functions $f$. So coordinate charts are not needed to compute derivatives nor are they needed to define tangent vectors.

 

[orion]

Thank you, lavinia. That was clear.

 

[orion]

Forget it. That's not what I did.