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Covariant Derivative Identity

  1. Jul 16, 2005 #1
    I am trying to solve an exercise from MTW Gravitation and the following issue has come up:

    Let D denote uppercase delta (covariant derivative operator)
    [ _ , _ ] denotes the commutator
    f is a scalar field, and A and B are vector fields

    Is it true that
    [D_A,D_B]f = D_[A,B]f
  2. jcsd
  3. Jul 16, 2005 #2


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    Using a more familiar notation, i'll write

    [tex] \left[\nabla_{\mu},\nabla_{\nu}\right]_{-}f(x) \ ? \ \nabla_{\left[\mu,\nu\right]} f(x) [/tex]

    A remark. I've never seen the operator in the RHS. I'm sure [tex] ? \ \leftrightarrow \ \neq [/tex].

  4. Jul 16, 2005 #3
    Well the original problem is to show that if [tex]f(P)[/tex] is a scalar field such that [tex]f(P_0)=1[/tex], and [tex]A,B,C[/tex] are vector fields, then
    Unless I am doing something wrong, this immediately reduces to the above identity, which doesn't look true to me either...
    Last edited: Jul 16, 2005
  5. Jul 16, 2005 #4
    Hurkyl -- your (deleted) post was helpful. I see now that I was having trouble with the notation in which a vector times a scalar field denotes the directional derivative in that direction. Then the identity does follow, as you pointed out.
  6. Jul 16, 2005 #5


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    Well, the identity in your original post seems clear, unless I'm making a dumb mistake (which is easily possible): according to Spivak, [itex]\nabla_X f = X f[/itex], so we have:

    [\nabla_A, \nabla_B] f = (\nabla_A \nabla_B - \nabla_B \nabla_A) f = (AB - BA) f = [A, B] f = \nabla_{[A, B]} f

    Your second post boggles me, because it looks like you're trying to take X(Y) where X and Y are both vector fields.
  7. Jul 16, 2005 #6
    Whoops I wrote my second post incorrectly, it's editted.
  8. Jul 16, 2005 #7


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    Ouch, i guess you both realized that i hadn't read the "A and B are vector fields" part and mistakenly took them as subscripts.

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