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Differential of lin. transformation between vector spaces

  1. Mar 13, 2007 #1
    Say I have to vector spaces [itex]V,W[/itex] and a linear transformation [itex]\Phi:V\rightarrow W[/itex]. I know that (given [itex]v,p\in V[/itex]) if I interpret a tangent vector [itex]v_p[/itex] as the initial velocity of the curve [itex]\alpha(t)=p+tv[/itex] I have, relative to a linear coordinates system on [itex]V[/itex], [itex]v_p=x^i(v)\partial_{i(p)}[/itex].
    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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  3. Mar 13, 2007 #2

    Hurkyl

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    What is

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

    You get one guess at a linear transformation L satisfying

    [tex]\Phi(x + h) - \Phi(x) \approx L(h)[/tex]
     
  4. Mar 13, 2007 #3

    mathwonk

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    the differential of f at p is the linear transformation L such that L(h) approximates f(p+h)-f(p) best.

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


    or as homer simpson would say: DOHHHH....!!!!@
     
  5. Mar 14, 2007 #4
    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
     
    Last edited: Mar 14, 2007
  6. Mar 14, 2007 #5

    mathwonk

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    well i guess i cant read or dont understand your notaTION, BUT if it obscures somehow for you the trivial fact that a linear map is its own best linear approximation, something is sadly wrong.
     
  7. Mar 14, 2007 #6
    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?
     
  8. Mar 14, 2007 #7

    Hurkyl

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    For a map [itex]f : \mathbb{R}^m \rightarrow \mathbb{R}^n[/itex], the definition of the derivative map [itex]f_* : \mathrm{T}\mathbb{R}^m \rightarrow \mathrm{T}\mathbb{R}^n[/itex] is

    [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 df is the ordinary derivative you learned in your calculus classes.
     
  9. Mar 14, 2007 #8
    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.

    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?
     
  10. Mar 14, 2007 #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)
     
    Last edited: Mar 14, 2007
  11. Mar 14, 2007 #10

    mathwonk

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    you must be joking. avector space has a canonical global coordinate system given by the identity map under which the original differebtiable map is just itself. i.e. the tangent space equals the manifold.

    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.
     
  12. Mar 14, 2007 #11

    mathwonk

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    lets put it this way: although there is an intrinsic horse **** way of stating the definition iof a derivative as something that acts in such and so a way on functions, nonetheless, it means the following: if we use coordinates to transfer to euclidean space, then the differential is ropecisely the linear map that ebst approxiamtes the coordinate verion of the map.

    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.)
     
  13. Mar 14, 2007 #12

    mathwonk

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    see if you can do this
    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.






    '
     
  14. Mar 15, 2007 #13
    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
     
  15. Mar 15, 2007 #14
    I don't really see what you're trying to show me. Are you referring to the coordinate maps? Thanks...Cliowa
     
  16. Mar 15, 2007 #15

    Hurkyl

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    I was thinking about the linear functionals.
     
  17. Mar 17, 2007 #16
    Hurkyl: Could you elaborate?
     
  18. Mar 18, 2007 #17

    Hurkyl

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    Linear functionals. Dual vectors. Covectors. Elements of the dual space V*. Linear maps V -> R. The things whose calculus is very easy.

    -------------------------------
    (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 R^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.



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

    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.
     
  19. Mar 18, 2007 #18
    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.

    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?

    Thanks and best regards...Cliowa
     
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