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

[fresh_42]

I think that part of the problem is that you and I are using different definitions.

For the sake of simplicity:
A circle is an object which consists of all points at equal distance from some fixed point called center.
A tangent of a circle is a straight line which touches the circle at exactly one point.
A tangent vector is the basis vector that generates this line.
A tangent field is the combination of all pairs (touching point, generating vector).

Where did I use coordinates here?

 

[orion]

No. I am differentiating the composite functions $f\circ \gamma$. This is a function from an interval to the real line.

Thank you! That clears a lot up. This kind of thing is what I am looking for.

Part of the problem seems to be that you are using literature that presents the material in a particular fashion and does not give you the equivalent definitions that do not refer to coordinate charts. The definitions are equivalent, but you seem to think that they are not for some reason. My point is that you do not need the charts or a notion of derivatives on the manifold in order to define tangent vectors. The resulting vector space is a set of derivative operators (satisfying the proper relations such as the product rule) on the manifold.

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.

 

[orion]

For the sake of simplicity:
A circle is an object which consists of all points at equal distance from some fixed point called center.
A tangent of a circle is a straight line which touches the circle at exactly one point.
A tangent vector is the basis vector that generates this line.
A tangent field is the combination of all pairs (touching point, generating vector).

Where did I use coordinates here?

You did not use coordinates here. But you didn't do any derivations also.

 

[fresh_42]

You did not use coordinates here. But you didn't do any derivations also.

So let's do it.

Let us consider a rotation $r_\varphi$ in the algebra $\mathcal{R}$ of rotations on a circle $\mathcal{C}$. I defined the tangent $T_p$ at a point $p \in \mathcal{C}$ and the tangent at the point $T_{r_\varphi(p)}$. Well, I actually didn't the latter, but I'm sure that we won't need coordinates for that, unless you insist on a scaled picture. By pure geometric means we know that a tangent of $\mathcal{C}$ at $p$ is perpendicular to the diameter $\overline{pc}$ with the center $c$ of $\mathcal{C}$ as defined above.

Thus we can define a derivative $D: \mathcal{R} \longrightarrow \mathcal{R}$ by $D : r_\varphi \longmapsto r_{\varphi + \frac{\pi}{2}},$ i.e
$$ D(r_\varphi r_\psi) (p) = D(r_\varphi ) (r_\psi (p)) + r_\varphi (D(r_\psi)(p))$$
Again, the needed right angle as well as $\varphi$ can be defined purely geometrical. No Cartesian coordinates, no radius or a special choice on how to measure an angle. Only the roation, i.e. function on $\mathcal{C}$ has to be described somehow. And if you calculate this in Cartesian coordinates, you will find the factor $2$ of a circle's derivations.

 

[orion]

So let's do it.

Let us consider a rotation $r_\varphi$ in the algebra $\mathcal{R}$ of rotations on a circle $\mathcal{C}$. I defined the tangent $T_p$ at a point $p \in \mathcal{C}$ and the tangent at the point $T_{r_\varphi(p)}$. Well, I actually didn't the latter, but I'm sure that we won't need coordinates for that, unless you insist on a scaled picture. By pure geometric means we know that a tangent of $\mathcal{C}$ at $p$ is perpendicular to the diameter $\overline{pc}$ with the center $c$ of $\mathcal{C}$ as defined above.

Thus we can define a derivative $D: \mathcal{R} \longrightarrow \mathcal{R}$ by $D : r_\varphi \longmapsto r_{\varphi + \frac{\pi}{2}},$ i.e
$$ D(r_\varphi r_\psi) (p) = D(r_\varphi ) (r_\psi (p)) + r_\varphi (D(r_\psi)(p))$$
Again, the needed right angle as well as $\varphi$ can be defined purely geometrical. No Cartesian coordinates, no radius or a special choice on how to measure an angle. Only the roation, i.e. function on $\mathcal{C}$ has to be described somehow. And if you calculate this in Cartesian coordinates, you will find the factor $2$ of a circle's derivations.

This is very interesting. Coordinate free derivation. And I agree that it is a good example.
I need to think about this and what this example shows besides coordinate free derivations and how it sheds light on my original question because to me this is a very special case of a derivation. But I don't know so I have to think about it.

Thank you. I think that it is advancing my understanding.

 

[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.