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Proving an Identity from Differential Geometry

  1. Apr 19, 2006 #1
    One often encounters the following identitiy in Tensor Analysis/Differential Geometry:

    dx^j (partial/partial x^i) = partial x^j / partial x^i = delta ij

    It's easy to see why partial x^j / partial x^i = delta ij

    but how does

    dx^j (partial/partial x^i) = partial x^j/partial x^i ?

    I have three problems with this:

    1. the dx^j involves an ordinary "d' and not a partial
    2. somehow the dx^j gets moved in front of the partials
    3. because when it moves we have (partial)(dx^j)/partial x^i, there are two derivatives on the top

    Any explanation would be helpful. Thanks.
  2. jcsd
  3. Apr 19, 2006 #2
    Hi mannyfold.

    I have read the old thread, and my second posting is wrong, i think, because here we are talking about dual basis vectors.
    So dxi is the dual of d/dxj, and so there inner product ist delta ij.

    When we say: dxi plugs the ith component of a vector, so dxi plugs also the ith part of the basis-vector d/dxj. This component is always =1 if i = j.
    That is, with an arbitrary vector v=vi(d/dxi): dxi(v)=v(xi)=vj(dxi/dxj)=vj(delta ij)=vi

    Hope that helps.
    Last edited: Apr 19, 2006
  4. Apr 19, 2006 #3
    It looks as if that equation is is to do with covariant and contravariant basis vectors. A natural vector basis for a tangent space is the coordiante basis [itex]{\partial_i}[/itex] and a sensible basis dual to this coordinate basis is [itex]{\tilde{d}x_i}[/itex]. The equation you gave [itex]\tilde{d}x^i(\partial_j)=\delta^i_j[/itex] is from the definition of a covariant vector, that it maps a vector into the reals via an inner product. The functional [itex]\tilde{d}x^i(\partial_j)[/itex] is essentially saying "what's the j-th component of the i-th dual basis vector?". Given the definition of this dual coordinate basis is [itex]\sum_j\frac{\partial x^i}{\partial x^j}dx^j[/itex] it is obvious that its j-th component is [itex]\frac{\partial x^i}{\partial x^j}=\delta^i_j[/itex]. Or you could look at it through the inner product.
  5. Apr 19, 2006 #4
    Thanks guys. I finally got it through my head that df(v) is a function that maps vectors to the reals; hence, dx^i (partial/partial x^j) just maps all the basis vectors to 0 except the ith which it maps to 1 (orthonormality condition).

    My mistake was that I was still working with the elementary definition of df being an infinitesimal change in f. Hence, the confusion.

    I still think there must be a connection between the elementary definition and this definition. Anybody out there have any ideas as to how to explain or connect the two definitions of df?

    PS: Perturbation, how do you get those nifty math symbols to appear on this forum?
  6. Apr 20, 2006 #5
    Check out the LaTeX tutorial here: https://www.physicsforums.com/showthread.php?t=8997

    You can also click on any of the nifty math symbols and a window will pop up where you can see exactly what code was used.
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