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Homework Help: Cartan's first structure equation proof

  1. May 4, 2006 #1

    r16

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    this is my first post on this site but it looks like the sort of ppl that i would like to associate myself with.

    Unfourtanately, I have not had any formal schooling for any mathematics above calculus but i have read a few books and papers and am trying to make due.

    I was studying about the cartan's first structure equation and was looking at this proof :

    http://www.pzgnet.cc/images/cartan/eq1.png

    where [tex] \nabla_x [/tex] is a koszul connection, [tex] e_i [/tex] is a basis and [tex] \partial_j A^j_i [/tex] is a change of basis from e and [tex] \omega [/tex] is a standard connection in the actual equation :

    http://www.pzgnet.cc/images/cartan/eq2.png

    In step 3 why can the exterior derivitave be applied to [tex] A^j_i [/tex]?

    I am no impact no idea on this step and it seems quite important so i dont want to skip it. Any ideas what im missing?

    **nb in equation 2 [tex] \omega^i_j [/tex] should be [tex] \omega^j_i [/tex]
     
    Last edited: May 4, 2006
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  3. May 4, 2006 #2

    George Jones

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    If [itex]f[/itex] is a function, how is the exterior derivative [itex]df[/itex] defined?

    It may prove useful to think about vector fields [itex]X[/itex] and/or a coodinate basis

    [tex]\left\{\frac{\partial}{\partial x_i} \right\}[/itex]

    to answer this.

    Regards,
    George
     
  4. May 4, 2006 #3

    r16

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    that may be my biggest problem. I am a visual learner, but I dont have a good physical picture of 'd'.

    Algebraically, d is defined a [tex] d = \frac{\partial}{\partial x_i} dx^i[/tex].
    d applied to the 0-form [tex]f[/tex] gives a 1-form [tex]df[/tex].

    Then it would follow that [tex] X(f) = df (X) [/tex] as in the definition of a 1-form.

    However is [tex] A^j_i [/tex] equivalent to a scalar because the contravariant and covariant parts of the [tex] [\frac{1}{1}] [/tex] valent tensor cancel out?

    This is opposed to [tex] \omega^k_i [/tex], a [tex] [\frac{1}{1}] [/tex] tensor as well, which acts as a 1-form on X. Is there a standard of how a tensor acts or is it based on the definition of the tensor?
     
    Last edited: May 4, 2006
  5. May 5, 2006 #4

    George Jones

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    For each [itex]i[/itex] and [itex]j[/itex], [itex]A^{i}_{j}[/itex] is a component, i.e., a function, while each [itex]\omega^{i}_{j}[/itex] is a 1-form, and, consequently, they are very different animals.

    Consider a couple of examples. A vector field [itex]X[/itex] can be expressed in terms a set of basis fields as [itex]X = X^{i} e_{i}[/itex]. Each [itex]X^i[/itex] is is a component, i.e., a scalar-valued function of the base space, while each [itex]e_i[/itex] is a vector field.

    Similary, if [itex]\omega^{i}_{j} = \omega^{i}_{jk} dx^k[/itex], each [itex]\omega^{i}_{jk}[/itex] is a component, while each [itex]dx^k[/itex] is a 1-form.

    Bottom line: sometime indices label comonents, and sometimes they label other objects.

    Welcome to Physics Forums, and, if I haven't answered all your questions, or if my explanation is not very clear, keep asking questions.

    Regards,
    George
     
    Last edited: May 5, 2006
  6. May 5, 2006 #5

    r16

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    thank you very much sir, that clears it up perfectly.

    However i do have one more inquery. What is a good physical/geometrical description of an exterior derivitave?
     
  7. May 5, 2006 #6

    George Jones

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    Unfortunately, I don't know a good physical/geometrical description of an exterior derivative. Maybe someone else does.

    I just work abstractly with its properties.

    Regards,
    George
     
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