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

[orion]

I am really struggling with one concept in my study of differential geometry where there seems to be a conflict among different textbooks. To set up the question, let M be a manifold and let (U, φ) be a chart. Now suppose we have a curve γ:(-ε,+ε) → M such that γ(t)=0 at a ∈ M. Suppose further that we have a function ƒ defined on M.

Now my question: Consider the directional derivative D_v. Some authors (e.g. J.M. Lee) imply that D_v^a acts at point a∈M. Other authors say that the derivative makes no sense on a manifold and that D_v^φ(a) acts at φ(a)∈ℝⁿ.

Which is the correct understanding?

Thanks in advance.

 

[stevendaryl]

I am really struggling with one concept in my study of differential geometry where there seems to be a conflict among different textbooks. To set up the question, let M be a manifold and let (U, φ) be a chart. Now suppose we have a curve γ:(-ε,+ε) → M such that γ(t)=0 at a ∈ M. Suppose further that we have a function ƒ defined on M.

Now my question: Consider the directional derivative D_v. Some authors (e.g. J.M. Lee) imply that D_v^a acts at point a∈M. Other authors say that the derivative makes no sense on a manifold and that D_v^φ(a) acts at φ(a)∈ℝⁿ.

Which is the correct understanding?

Thanks in advance.

Well, the definition of derivative used in multivariable calculus doesn't directly make sense in a manifold if you are taking the derivative of a vector field (or tensor field). That definition would say something like:

$D_v^a \vec{A} = lim_{\lambda \rightarrow 0} \dfrac{\vec{A}(a + \lambda v) - \vec{A}(a)}{\lambda}$

There are two reasons that this definition fails for a curved manifold:

1. For a curved manifold, you can't add a vector $v$ to a location $a$.
2. For a curved manifold, you can't subtract vectors at different locations.

That might be what was meant. You can define a directional derivative on a curved manifold, but it requires a slightly different definition of derivative than in $R^n$

 

[andrewkirk]

The directional derivative of a vector field on a manifold at a point P is a function from the tangent space at P to itself. The 'input' to the function is a vector that specifies the direction in which the directional derivative of the field is to be calculated, and the 'output' of the function is another vector. I would say that the directional derivative 'acts' neither on the manifold nor on $\mathbb R^n$ but on the tangent space.

But note that 'acts on' and 'acts at' are verbal flourishes used to help understanding, and have no strict technical meaning. So there is no strictly correct answer and individual mathematicians are free to adopt whatever interpretation they find helpful.

 

[orion]

That's kind of what I'm getting at. Without modifying our typical understanding of derivatives, then I gather from what you have written that we do the calculus in ℝⁿ. But I have found something that goes even further and says the ordinary derivative of an ordinary (non-vector) function.
Here is a direct quote for some notes I found on the internet:
"Unfortunately if M is not embedded in any ℝ^N the derivative γ'(0) does not make sense." From here.
γ is a curve on M. And I have seen a similar statement elsewhere. I believe it has something to do with subtracting points does not make sense on a manifold not embedded in ℝⁿ.

To carry this a bit further to clear up my confusion, why do we call the tangent space, T_aM and not T_φ(a)M?

 

[orion]

The directional derivative of a vector field on a manifold at a point P is a function from the tangent space at P to itself. The 'input' to the function is a vector that specifies the direction in which the directional derivative of the field is to be calculated, and the 'output' of the function is another vector. I would say that the directional derivative 'acts' neither on the manifold nor on $\mathbb R^n$ but on the tangent space.

But note that 'acts on' and 'acts at' are verbal flourishes used to help understanding, and have no strict technical meaning. So there is no strictly correct answer and individual mathematicians are free to adopt whatever interpretation they find helpful.

We don't even need to consider the directional derivative. It may be I chose a bad example. Just consider the simple derivative as in my last post.

 

[andrewkirk]

why do we call the tangent space, T_aM and not T_φ(a)M?

Because the chart is not part of the definition of the definition of a tangent space, it is just one way of picturing it. There are a number of formal ways of defining the tangent space, none of which use charts. The one I like is that $T_pM$ it is the set of equivalence classes of curves in $M$ through $p$ under the equivalence relation that two curves $\mu,\gamma$ through $p$ are equivalent if $\frac d{dt} \phi(\gamma(t))|_{\gamma^{-1}(p)}=\frac d{dt} \phi(\mu(t))|_{\mu^{-1}(p)}$ for every chart $(U,\phi)$ such that $p\in U$.

 

[orion]

Because the chart is not part of the definition of the definition of a tangent space, it is just one way of picturing it. There are a number of formal ways of defining the tangent space, none of which use charts. The one I like is that $T_pM$ it is the set of equivalence classes of curves in $M$ through $p$ under the equivalence relation that two curves $\mu,\gamma$ through $p$ are equivalent if $\frac d{dt} \phi(\gamma(t))|_{\gamma^{-1}(p)}=\frac d{dt} \phi(\mu(t))|_{\mu^{-1}(p)}$ for every chart $(U,\phi)$ such that $p\in U$.

Ok, but your derivatives are taken in $\mathbb{R}^n$ because $\phi(\gamma(t)) \subset \mathbb{R}^n$ which is exactly what I am trying to get at.

Can you ever say, $\gamma'=\frac{d\gamma}{dt}$ without using the chart to bring it into $R^n$?

 

[orion]

This is something that I like a lot and it does include the charts and the transition functions between charts in the definition of the tangent space.

 

[andrewkirk]

Can you ever say, $\gamma'=\frac{d\gamma}{dt}$ without using the chart to bring it into $R^n$?

It depends on what you mean by 'bring it into $R^n$'. As noted above, this is imprecise wording that is used to help intuition. There is no right or wrong answer for it.

If 'bring it into $R^n$' means 'use a concept that somewhere along the line refers to $R^n$' then the answer is no, because manifolds are defined as topological spaces that are locally homeomorphic to $R^n$.

There are two other definitions of tangent spaces that you could look into, if you feel uncomfortable with the one involving equivalence classes. They are as 'derivations' or as an 'algebra of germs'. I have never looked into these. As far as I know they could even be the same thing. But they will inevitably have some eventual connection with $R^n$ because of the point in the previous para.

I'm glad you like the ANU presentation you linked. ANU had a very good maths program when I studied there in the early eighties, and I have very fond memories of it. However I doubt the person who taught me calculus on manifolds back then is still lecturing, or wrote that.

 

[orion]

What I mean by bring it into $\mathbb{R}^n$ is this: Suppose we have a curve $\gamma:I \to M$ with $I$ being an interval $(-\varepsilon, +\varepsilon)$ and that $\gamma(0)=p$. Then is it true that the correct way of taking the derivative of the curve $\gamma$ at $p$ is not $\frac{d\gamma}{dt}\big\rvert_p$ but rather $\frac{d\phi(\gamma)}{dt} \big\rvert$.

 

[Orodruin]

The directional derivative of a vector field on a manifold at a point P is a function from the tangent space at P to itself

If this were true it would be sufficient to know the vector field at P. In order to compute the derivative you need to know the vector field in some neighbourhood of P. There are several vector fields that may have the same value at P but different derivatives. The directional derivative maps a vector field in some neighbourhood to another vector field in the same region.

What I mean by bring it into $\mathbb{R}^n$ is this: Suppose we have a curve $\gamma:I \to M$ with $I$ being an interval $(-\varepsilon, +\varepsilon)$ and that $\gamma(0)=p$. Then is it true that the correct way of taking the derivative of the curve $\gamma$ at $p$ is not $\frac{d\gamma}{dt}\big\rvert_p$ but rather $\frac{d\phi(\gamma)}{dt} \big\rvert$.

No. Depending on your choice of definition of a tangent vector, we either define the tangent vector $\dot\gamma$ as a particular equivalence class or as the tangent vector $X$ that satisfies $X(f) = df(\gamma(t))/dt$, where $f$ is a function on the manifold. There is no need to bring coordinate charts into the game unless you actually need them to compute something.

 

[andrewkirk]

If this were true it would be sufficient to know the vector field at P. In order to compute the derivative you need to know the vector field in some neighbourhood of P.

In the first sentence of the post I deliberately omitted the words 'in direction v', thereby leaving that as the one thing that needs to be specified - so that we have a unary function. Based on that set-up, the vector field in the neighbourhood is part of the identification of the function rather than an input to it (ie there is a different such function for every vector field around P), and the only input the function needs is a direction - a single vector at P.

If one wishes, one can alternatively define 'The directional derivative' at p (without reference to a vector field). That will be a binary function, taking as inputs (1) a vector field and (2) a directional vector.

 

[orion]

No. Depending on your choice of definition of a tangent vector, we either define the tangent vector $\dot\gamma$ as a particular equivalence class or as the tangent vector $X$ that satisfies $X(f) = df(\gamma(t))/dt$, where $f$ is a function on the manifold. There is no need to bring coordinate charts into the game unless you actually need them to compute something.

So what do I make of authors who say that the derivative doesn't make sense on the manifold? And other authors who bring charts into the definition? I have even read at least one author who has said that "we don't know how to take derivatives on manifolds, but we do know how to do calculus in $\mathbb{R}^n$." Or is this just one of those things that depends on the author's philosophical point of view that has no bearing on the mathematics?

 

[Orodruin]

So what do I make of authors who say that the derivative doesn't make sense on the manifold? And other authors who bring charts into the definition? I have even read at least one author who has said that "we don't know how to take derivatives on manifolds, but we do know how to do calculus in $\mathbb{R}^n$." Or is this just one of those things that depends on the author's philosophical point of view that has no bearing on the mathematics?

There is no derivative operation on the manifold involved in this definition of the tangent vector. Only how the vector acts on a function in terms of how the curve equivalence class is defined ($f(\gamma(t))$ is a function from an interval to the real line). You can then verify that tangent vectors have all the required properties for actually being a differential operator at the point they are defined, that they have the expected coordinate expressions, and work as expected when the manifold actually is R^n.

 

[stevendaryl]

Ok, but your derivatives are taken in $\mathbb{R}^n$ because $\phi(\gamma(t)) \subset \mathbb{R}^n$ which is exactly what I am trying to get at.

Can you ever say, $\gamma'=\frac{d\gamma}{dt}$ without using the chart to bring it into $R^n$?

Indirectly, charts are involved in determining which functions are "smooth". But given the notion of a smooth function as a primitive, we can define:

• A (real) scalar field $f$ is a smooth function from $M$ into $R$.
• A parametrized path $\gamma$ is a smooth function from $R$ (or an interval of $R$) into $M$.
• A tangent vector $v = \frac{d \gamma}{dt}_{t=0}$ to path $\gamma$ at the point $t=0$ is an operator which acts on any scalar field $f$ and returns $v(f) = \frac{d f(\gamma(t))}{dt}_{t=0}$

With this way of defining "tangent vector", the directional derivative as applied to a scalar field is completely trivial:

$D^{\gamma(0}_v f = v(f)$

To extend the directional derivative to apply to vector fields and tensor fields requires a notion of parallel transport (Or you can do it in the other way around---just let the directional derivative be an operator that obeys the Liebniz rules for derivatives, and for scalar fields acts as above. Then you can define parallel transport in terms of the directional derivative.)

 

[orion]

Indirectly, charts are involved in determining which functions are "smooth". But given the notion of a smooth function as a primitive, we can define:

• A (real) scalar field $f$ is a smooth function from $M$ into $R$.
• A parametrized path $\gamma$ is a smooth function from $R$ (or an interval of $R$) into $M$.
• A tangent vector $v = \frac{d \gamma}{dt}_{t=0}$ to path $\gamma$ at the point $t=0$ is an operator which acts on any scalar field $f$ and returns $v(f) = \frac{d f(\gamma(t))}{dt}_{t=0}$

With this way of defining "tangent vector", the directional derivative as applied to a scalar field is completely trivial:

$D^{\gamma(0}_v f = v(f)$

To extend the directional derivative to apply to vector fields and tensor fields requires a notion of parallel transport (Or you can do it in the other way around---just let the directional derivative be an operator that obeys the Liebniz rules for derivatives, and for scalar fields acts as above. Then you can define parallel transport in terms of the directional derivative.)

This is very helpful. Thank you. It clears a lot up.

I'd still like to ask why some make the absolute statement that even the ordinary derivative on a manifold doesn't make sense unless the manifold is a submanifold of $\mathbb{R}^n$. Is it because a manifold may not have a metric? By "ordinary derivative" I mean,
$$ f'(0)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \bigg\rvert_{x=0} $$

 

[andrewkirk]

I'd still like to ask why some make the absolute statement that even the ordinary derivative on a manifold doesn't make sense unless the manifold is a submanifold of $\mathbb{R}^n$. Is it because a manifold may not have a metric? By "ordinary derivative" I mean,
$$ f'(0)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \bigg\rvert_{x=0} $$

It is because that formula does not mean anything, because if $x$ is a point on the manifold then $x+h$ does not mean anything unless an operation of either addition or affine translation is defined for points on the manifold. In general, such operations are not defined on manifolds. But they do have a meaning in $R^n$ so if the manifold is embedded in $R^n$ we can use that meaning.

Alternatively, if it is not $x$ but $f(x)$ that is a point on the manifold (eg if $f$ is a curve) then the $f(x+h)-f(x)$ has no meaning, for the same reason given in the previous paragraph.

The simple guiding principle is that any arithmetic operations can only be performed on objects for which such operations are defined.

 

[orion]

It is because that formula does not mean anything, because if $x$ is a point on the manifold then $x+h$ does not mean anything unless an operation of either addition or affine translation is defined for points on the manifold. In general, such operations are not defined on manifolds. But they do have a meaning in $R^n$ so if the manifold is embedded in $R^n$ we can use that meaning.

Alternatively, if it is not $x$ but $f(x)$ that is a point on the manifold (eg if $f$ is a curve) then the $f(x+h)-f(x)$ has no meaning, for the same reason given in the previous paragraph.

The simple guiding principle is that any arithmetic operations can only be performed on objects for which such operations are defined.

Thank you. That clears up some of my confusion. Now I think I'm getting somewhere, but maybe not. Maybe I'm backing up.

So this leads me to again think that the proper way to define the tangent space is by using the homeomorphism and doing the derivatives in $\mathbb{R}^n$ since derivatives only make sense in $\mathbb{R}^n$, i..e. $\phi(\gamma(t))' = d\phi(\gamma)/dt \rvert_{t=0}$ makes sense but that $\gamma'=d\gamma/dt \rvert_{t=0}$ is essentially meaningless.

Thank you for all your help and patience in cutting through my confusion.

 

[Orodruin]

So this leads me to again think that the proper way to define the tangent space is by using the homeomorphism and doing the derivatives in RnRn\mathbb{R}^n since derivatives only make sense in RnRn\mathbb{R}^n, i..e. ϕ(γ(t))′=dϕ(γ)/dt∣t=0ϕ(γ(t))′=dϕ(γ)/dt∣t=0\phi(\gamma(t))' = d\phi(\gamma)/dt \rvert_{t=0} makes sense but that γ′=dγ/dt∣t=0γ′=dγ/dt∣t=0\gamma'=d\gamma/dt \rvert_{t=0} is essentially meaningless.

What do you mean by "proper"? There are several equivalent definitions of the tangent space and you do not need to refer to a coordinate chart to define it.

 

[orion]

What do you mean by "proper"? There are several equivalent definitions of the tangent space and you do not need to refer to a coordinate chart to define it.

By proper I mean that given the fact that derivatives on a general manifold that is not a submanifold of $\mathbb{R^n}$ do not make sense then you should be using the chart to define things like velocity of a curve, etc. Some authors do point that out as in the references I posted, for example, here.

As to definitions of the tangent space, I personally prefer the definition using derivations, but even if you use the equivalence class of tangent curves definition, you still end up taking derivatives because you define the equivalence relation between two curves through:
$\gamma_1'(0) = \gamma_2'(0) \implies \gamma_1 \sim \gamma_2$.

I appreciate your help. I'm not trying to argue just to argue, I'm just trying to resolve what I perceive to be a contradiction that is really confusing me.

 

[Orodruin]

then you should be using the chart to define things like velocity of a curve, etc.

Nonsense. The velocity of a curve is (in the definition mentioned in #11) simply the equivalence class of curves to which the curve belongs.

because you define the equivalence relation between two curves through:
γ′1(0)=γ′2(0)⟹γ1∼γ2

No, $\gamma_1 \sim \gamma_2$ if $df(\gamma_1(t))/dt = df(\gamma_2(t))/dt$ for all $f$ (and $\gamma_1(0)=\gamma_2(0)=p$, where we are defining a tangent vector at $p$). The implication in your statement goes both ways and is just the definition of the equivalece class.

I'm not trying to argue just to argue, I'm just trying to resolve what I perceive to be a contradiction that is really confusing me.

Do not get me wrong, this is not an argument. I just consider your conclusion that everything has to be defined through charts incorrect. You can think like that, but many people prefer coordinate free definitions and notation and you should be aware that this exists and is consstent.

 

[orion]

I just consider your conclusion that everything has to be defined through charts incorrect. You can think like that, but many people prefer coordinate free definitions and notation and you should be aware that this exists and is consstent.

So then how do you resolve the issue that derivatives do not make sense on a manifold that is not a submanifold of $\mathbb{R}^n$? Because that's the crux of what worries me.

 

[Orodruin]

So then how do you resolve the issue that derivatives do not make sense on a manifold that is not a submanifold of $\mathbb{R}^n$? Because that's the crux of what worries me.

It is still unclear to me where in the definition of tangent vectors in terms of equivalence classes of curves you think a derivative in the manifold is necessary. You can easily check that the resulting action of a vector satisfies the properties you would expect from a derivative operator.

 

[fresh_42]

By proper I mean that given the fact that derivatives on a general manifold that is not a submanifold of RnRn\mathbb{R^n} do not make sense ...

So then how do you resolve the issue that derivatives do not make sense on a manifold that is not a submanifold of $\mathbb{R}^n$? Because that's the crux of what worries me.

This is simply wrong (as previous posts mentioned multiple times).

https://en.wikipedia.org/wiki/Tangent_bundle

 

[orion]

This is simply wrong (as previous posts mentioned multiple times).

https://en.wikipedia.org/wiki/Tangent_bundle

This is not my assertion. I am simply pointing out what I find that others have written that has confused me and I can actually see their point. But I see the other side too. Hence, my confusion.

 

[orion]

It is still unclear to me where in the definition of tangent vectors in terms of equivalence classes of curves you think a derivative in the manifold is necessary. You can easily check that the resulting action of a vector satisfies the properties you would expect from a derivative operator.

I appreciate this, and I will think on this some more. Without thinking it through right now, I suppose I've assumed that where a derivative in the manifold was necessary was in the definition of the equivalence class.

 

[orion]

I appreciate this, and I will think on this some more. Without thinking it through right now, I suppose I've assumed that where a derivative in the manifold was necessary was in the definition of the equivalence class.

Ok, I see now that the equivalence class is defined as $(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0)$. Hence the derivatives are done in $\mathbb{R}^n$ and not in the manifold.

 

[Orodruin]

Ok, I see now that the equivalence class is defined as $(\phi \circ \gamma_1)'(0) = (\phi \circ \gamma_2)'(0)$. Hence the derivatives are done in $\mathbb{R}^n$ and not in the manifold.

No it is not. It is defined such that the derivatives of an arbitrary function on the manifold along the curves are the same. There is no need to reference coordinate functions other than as a special case when you want to show the definitions equivalent.

 

[orion]

No it is not. It is defined such that the derivatives of an arbitrary function on the manifold along the curves are the same. There is no need to reference coordinate functions other than as a special case when you want to show the definitions equivalent.

I think that part of the problem is that you and I are using different definitions. My expression is taken verbatim from these notes, page 274.

But ... when you say that the equivalence class is defined such that the derivatives of an arbitrary function on the manifold along the curves are the same, are you not including taking derivatives in the manifold in the definition at least implicitly?

 

[Orodruin]

are you not including taking derivatives in the manifold in the definition at least implicitly

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

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.

 

[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.