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The thing I don't understand is why in this case [itex]d\Phi (v_p)=(\Phi(v))_{(\Phi(p))}[/itex]. Can someone show me the way?

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- Thread starter cliowa
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The thing I don't understand is why in this case [itex]d\Phi (v_p)=(\Phi(v))_{(\Phi(p))}[/itex]. Can someone show me the way?

- #2

Hurkyl

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[tex]\Phi(x + h) - \Phi(x)?[/tex]

You get one guess at a linear transformation

[tex]\Phi(x + h) - \Phi(x) \approx L(h)[/tex]

- #3

mathwonk

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if f is linear, try to approximate f(p+h)-f(p) by a linear function of h.

or as homer simpson would say: [email protected]

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Thanks for your replies. If you take that quotient as the definition of the differential, things are perfectly clear to me.

But I was thinking in the context of the differential as the tangent map, where I need to have that for any function [itex]f:W\rightarrow\mathbb{R}[/itex] the following holds:

[tex](d\Phi\cdot v)\cdot f=v(f\circ\Phi)[/tex]

and it is precisely this, that I don't see.

Best regards...Cliowa

But I was thinking in the context of the differential as the tangent map, where I need to have that for any function [itex]f:W\rightarrow\mathbb{R}[/itex] the following holds:

[tex](d\Phi\cdot v)\cdot f=v(f\circ\Phi)[/tex]

and it is precisely this, that I don't see.

Best regards...Cliowa

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mathwonk

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Thanks for your commiseration, but I really understand the trivial fact you stated quite well.

As the problem seems to be the notation, let me try to clear things up: I'm talking about vector spaces as manifolds and maps between them. So, take the vector spaces to be finite-dimensional. Is the notation now clear to you?

- #7

Hurkyl

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[tex]\left. f_* \right|_{P} = \left.(df)\right|_{P} : \mathrm{T}_{P}\mathbb{R}^m \rightarrow \mathrm{T}_{f(P)}\mathbb{R}^n[/tex]

where

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But I'm not talking about [itex]\mathbb{R}^n[/itex]! I could look at ismorphisms, that's correct, but I don't want to go the coordinate way.of the derivative map [itex]f_* : \mathrm{T}\mathbb{R}^m \rightarrow \mathrm{T}\mathbb{R}^n[/itex] isdefinition

[tex]\left. f_* \right|_{P} = \left.(df)\right|_{P} : \mathrm{T}_{P}\mathbb{R}^m \rightarrow \mathrm{T}_{f(P)}\mathbb{R}^n[/tex]

wheredfis the ordinary derivative you learned in your calculus classes.

My situation is this: I have two manifolds, [itex]V,W[/itex], which are at the same time endowed with a linear structure, i.e. they're two vector spaces of finite dimension. Now, I would like to show that the differential map as I would define it for a tangent map between any two (finite-dimensional, unbounded, real) manifolds coincides with what I would hope it is, looking at the thing from the vector space perspective. Do you see what I mean?

- #9

Hurkyl

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Well, how do you plan on even defining the derivative map?

As I remember it, the derivative map is defined to be the function you get by sewing together the (Euclidean!) derivative maps on coordinate patches.

Hrm. Maybe you want to work it out in the differential operator presentation of the tangent space instead? Yes, I think that's what you are trying to do -- your hint is that you know a particularly nice collection of scalar-valued maps on your vector spaces.

(But honestly, this seems awkward to me -- if I have a vector space V, I would rather invoke the canonical diffeomorphism*TV ~ VxV*)

As I remember it, the derivative map is defined to be the function you get by sewing together the (Euclidean!) derivative maps on coordinate patches.

Hrm. Maybe you want to work it out in the differential operator presentation of the tangent space instead? Yes, I think that's what you are trying to do -- your hint is that you know a particularly nice collection of scalar-valued maps on your vector spaces.

(But honestly, this seems awkward to me -- if I have a vector space V, I would rather invoke the canonical diffeomorphism

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mathwonk

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I am just saying if you can't handle this case, you are doing something ridiculous in even looking at it this way. abstraction is not meant to render trivial things undoable.

- #11

mathwonk

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now in your case the coordinates are the dientity map, so the derivative is again the thing you say you understand, that it is the ebst linear approximation.

so what you are really saying is you have no clue how thw goofy abstarct version of a derivative is equivalent to the classical one even in R^n.

so just pretend you are in R^n and check for yourself why the linear appriximation is the derivative in the operator sense you are using.

(the answer is no doubt the chain rule.)

- #12

mathwonk

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show the partial derivative or directional derivative, in a given direction, is obtained by applying the linear approximsation of the function to the direction vector.

that is basically your stumbling block.

'

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so what you are really saying is you have no clue how thw goofy abstarct version of a derivative is equivalent to the classical one even in R^n.

so just pretend you are in R^n and check for yourself why the linear appriximation is the derivative in the operator sense you are using.

(the answer is no doubt the chain rule.)

Oh wow, thanks alot. Imagine: I had never even tried that easy little calculation! So let me write down what I think is going on, so you can correct me:

If I define (in [itex]\mathbb{R}^n[/itex]) vector applied to function as the directional derivative, the chain rule gives me [itex]v\cdot f=\frac{d}{dt}_{t=0} f(p+tv)=df(p)\cdot v[/itex], so that clearly [itex](d\Phi\cdot v)\cdot f=v\cdot (f\circ\Phi)[/itex]. Now, is this it? Or were you thinking of something else?

Thanks again...Cliowa

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Hrm. Maybe you want to work it out in the differential operator presentation of the tangent space instead? Yes, I think that's what you are trying to do -- your hint is that you know a particularly nice collection of scalar-valued maps on your vector spaces.

I don't really see what you're trying to show me. Are you referring to the coordinate maps? Thanks...Cliowa

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Hurkyl

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I was thinking about the linear functionals.

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Hurkyl: Could you elaborate?

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Hurkyl

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-------------------------------

(Separate thought)

I was looking back at the first post, and notice that the very first thing you did was to invoke an isomorphism with the vector space

If technical rigor was the goal, I would like to point out that you are using the symbol

(1) An element of the vector space V

(2) A tangent vector field on the manifold V

(3) A differential operator on the space of functions from V

Normally, we (intentionally!) confuse these notions -- I just wanted to point it out in case recognizing this distinction is your intent.

Also, it's now unclear to me if [itex]d\Phi[/itex] is supposed to mean the derivative of [itex]\Phi[/itex] that you learned in multivariable calculus, or if it's supposed to mean the map [itex]TV \to TW[/itex] of tangent spaces.

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(Separate thought)

I was looking back at the first post, and notice that the very first thing you did was to invoke an isomorphism with the vector spaceR^n, and pose your question in that context. Since you later eschewed coordinates, I'm confused as to just what you're trying to achieve here.

Yes, you're absolutely right there. I just noticed I hadn't even really thought about the [itex]\mathbb{R}^n[/itex] case, so I did that (on the suggestion of mathwonk). What I was initially trying to achieve was a coordinate-free derivation.

If technical rigor was the goal, I would like to point out that you are using the symbolvfor many different things: you've used it as:

(1) An element of the vector space V

(2) A tangent vector field on the manifold V

(3) A differential operator on the space of functions from V

That was done on purpose, I think I'm aware of that distinction (especially later when I was talking about the vector space [itex]\mathbb{R}^n[/itex] itself).

I meant the tangent map [itex]TV \to TW[/itex], because initially it was not clear to me why this abstractly defined thing would have those nice properties, just coming from the "operator definition" (I thing that's the way you call it, right?). Now, for the case of [itex]\mathbb{R}^n[/itex] it is clear to me, basically because I don't need to explicitly invoke coordinates. Now I figure the easiest thing to do (for any real vector space of finite dimension) would be to simply invoke coordinates (as you proposed). Right?Also, it's now unclear to me if [itex]d\Phi[/itex] is supposed to mean the derivative of [itex]\Phi[/itex] that you learned in multivariable calculus, or if it's supposed to mean the map [itex]TV \to TW[/itex] of tangent spaces.

Thanks and best regards...Cliowa

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