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Change of Coordinate for V.Field in Mfld.

  1. Feb 27, 2009 #1


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    Hi, again:
    Just a quick question; I have "notation indigestion", i.e., I have been trying

    to figure way too many technicalities recently; I would appreciate a quick yes/no:

    Say X_p is a V.Field defined at p in a C^k manifold; k>0 . Say (U,Phi) and (U',Phi')

    are both charts containing p . Just wondering if the rule for coordinate change

    of X from chart-to-chart is; is it given by

    d(Phi o Phi'^-1) and d(Phi' o Phi^-1) ?

    I mean, I know it involves the chain rule, but I wonder if it is the chain rule

    applied to the two formulas above.

  2. jcsd
  3. Feb 28, 2009 #2


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    Think about the geometrical picture. You have patches U and U' on the manifold, which are mapped to some open subsets V and V' of Euclidean space.
    Suppose you want the coordinate change from V to V'. Suppose you have a coordinate [itex]\vec v \in V[/itex]. Then you can go to [itex]\Phi^{-1}(\vec v) \in U[/tex]. Assuming that this point is also in U', you can go to V': [itex]\Phi'( \Phi^{-1}(\vec v) ) \in V'[/itex]. You can write this as [itex]\kappa(\vec v)[/itex], where depending on your convention for composition of functions,
    [tex]\kappa = \Phi^{-1} \circ \Phi'[/tex]
    [tex]\kappa = \Phi' \circ \Phi^{-1}[/tex]

    You can also go from V' to V by a similar reasoning (and in fact you will find that the coordinate change you get is [itex]\kappa^{-1}[/itex]).
  4. Mar 2, 2009 #3


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    Thanks, Compuchip; I don't know if I misunderstood ( or misunderestimated :smile:)
    your reply:

    But I think that both k, k' as you described them ( I am sorry, I can't make the

    'quote' function work well in here ) give me a coordinate change between the patches

    V,V' =Phi(U) and Phi'(U') respectively , as you described, but I don't see that this

    gives me a way of changing the coordinate representation of the V.Field X_p , which

    lives in U/\U' (Sorry, I am still learning Tex.).

    I know that we get the coordinate rep. ( in terms of the basis for T_pM , in

    each of the charts U,U' ) by pulling back (Thru Phi, Phi' respectively), the basis

    of T_Phi(p)R^n and T_Phi'(p) R^n respectively, to get different bases for T_pM.

    Does your k , k' give me a way of going from one basis representation of T_pR^n

    to another basis rep. of T_pR^n ?

    Thanks, and sorry for writing in ASCII. Hopefully this summer I will have time to

    learn Latex.
  5. Mar 2, 2009 #4
    It's just the chain rule from good old calculus.
  6. Mar 2, 2009 #5


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    Yes, Zhentil,thanks, I understand that. I was looking for what specific map
    we apply the chain rule to:

    Is it to the composition of chart maps (Phi o Phi' ^-1) ;with (U,Phi) and (U',Phi')

    overlapping charts for p ?(or, of course, the inverse of the map above, if we want to

    change in the opposite direction).
  7. Mar 2, 2009 #6


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    This is my issue: If M,N, are open subsets of R^n, this is easy. Now, if M,N are not
    open subsets of R^n, then , we get a basis representation for X_p as above in U,
    by pulling back the basis vectors in T_Phi(p)R^n , by the chart map Phi^-1 , and
    we get the basis rep. in the chart U' , by pulling back (using Phi'^-1) ,the basis
    vectors in T_Phi'p R^n.

    Then, to change , I think we need to push forward the vector field X_p ( in whichever
    basis) to R^n , then use the fuch forward to map this image into the other basis in R^n,
    and then pull back again. I get the idea, I think, but this is becoming a "tress v. forest"
    thing , here, and I lost track of just how to figure out the change of expression, and I
    was hoping that if someone was familiar with it, they could give me the "forest" --
    to push the analogy to its limit.
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