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Differential Geometry/Ricci Calculus Question

  1. Apr 15, 2006 #1
    I am having a problem that is glossed over in many text books but is driving me nuts.

    Consider the following inner product or one-form with a vector argument:

    dx_i(partial_j) = kroenicker delta ij

    Here, dx_i is a one-form and partial_j (the partial wrt x_j) is a vector.

    Some books say:

    dx_i(partial_j) = partial x_i / partial x_j = kroenicker delta ij

    The problem is that I can't see how this is true. (Well, I do know that partial x_i / partial x_j = kroenicker delta ij but I can't see the rest.) What am I missing here?

    Thanks.
     
    Last edited: Apr 15, 2006
  2. jcsd
  3. Apr 16, 2006 #2
    The crux is the definition of the differential:

    df(v)=vf, where v is a tangent vector.
    This is of the form L(x)=ax in linear algebra:
    L is a linear function, x a vector, a the covector, defined by a_i=L(e_i).

    Now let df be dx and v the coordinate-tangentvector d/dx (d=partials),
    then

    dx^i(d/dx_j)=dx^i/dx_j=delta ij.
     
    Last edited: Apr 16, 2006
  4. Apr 16, 2006 #3
    AHA!

    That is the problem!

    I don't see how df(v) = vf. Where did you get that definition of the differential?

    df = partial f / partial x^i dx^i (i = 0 .. n)

    but the partials here are just the components of the 1-form.

    I have the feeling that there is something very simple that I'm not seeing here.
     
  5. Apr 16, 2006 #4
    Let's write df=(df/dx^i)dx^i (last d=partials)

    What is df(v), where v is a vector?

    df(v)=(df/dx^i)dx^i(v)


    What is dx^i(v)?

    It's simply v^i.

    So you can write:

    df(v)=(df/dx^i)v^i = v^i(df/dx^i).

    But what is our v^i on a manifold?

    It's d/dx^i!

    So you can write it as df(v)=vf, where v=d/dx^i
     
    Last edited: Apr 16, 2006
  6. Apr 16, 2006 #5
    Hi Javanse,

    I appreciate your replies, but I am going to challenge what you wrote.

    We are seeking to prove df(v) = vf.

    You wrote:

    df(v)=(df/dx^i)dx^i(v)
    What is dx^i(v)?

    It's simply v^i.


    But this assumes what we are trying to prove:
    dx^i(v) = v^jx^i =d/dx^j (x^i) = v^i delta ij

    You wrote:

    df(v)=(df/dx^i)v^i = v^i(df/dx^i).

    How can you commutate the differential operator v^i?

    Finally, from this last statement you get df(v) = vf, but if you read it carefully you don't dispose of the df/dx^i so it appears that your result is:

    df(v) = (d/dx^j) (df/dx^i) = v (df/dx^i)
     
  7. Apr 16, 2006 #6
    Another perspective on this is yet another equality I've found that was not expounded upon:

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

    I get the last equality. I can't see the first.
     
  8. Apr 16, 2006 #7
    Okay, last try :)

    Let v=v^i d/dx^i

    -> v(f)=v^i df/dx^i.

    This is the directional derivative.

    I introduced the operator at the end.
    You can also write the v^i from the beginning in front of the derivative.

    But the result is the same as above: it is the directional derivative.

    dxi(v)=v(xi)=vj delta ij=vi :)

    Maybe someone else has another explanation.
     
    Last edited: Apr 17, 2006
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