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Calculating divergence using covariant derivative

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the definition of divergence [itex]d(i_{X}dV) = (div X)dV[/itex] where [itex]X:M\rightarrow TM[/itex] is a vector field, [itex]dV[/itex] is a volume element and [itex]i_X[/itex] is a contraction operator e.g. [itex]i_{X}T = X^{k}T^{i_{1}...i_{r}}_{kj_{2}...j_{s}}[/itex], prove that if we use Levi-Civita connection then the divergence can also be written as
    [itex]div X = X^{i}_{;i}[/itex]

    2. The attempt at a solution

    This is what i tried:
    since [itex]dV = dx^{1} \wedge ... \wedge dx^{n}[/itex]
    after some calculation i conclude that [itex]i_{X}dV = \sum_{i=1}^{n}(-1)^{i}X_{i}dx^{1} \wedge ... \wedge dx^{i-1} \wedge dx^{i+1} \wedge ... \wedge dx^{n}[/itex]
    so [itex]d(i_{X} dV) = (\partial _{i}X^{i})dV[/itex]
    Then i attempt the use the fact that [itex]\Gamma^{i}_{jk} = \Gamma^{i}_{kj}[/itex] to get a lot of cancellation and show that [itex]\partial _{i}X^{i} = X^{i}_{;i}[/itex]
    but i couldn't.
    So can anyone please help? Thx in advanced :)
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 11, 2012 #2


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    Hey v2536 and welcome to the forums.

    I don't know much about differential geometry, but I do understand in essence what you are trying to do.

    Maybe you should look at this website (http://en.wikipedia.org/wiki/Divergence#Generalizations). A lot of the results you need are found in the coordinate free form you are trying to prove.
  4. Feb 12, 2012 #3

    Ben Niehoff

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    Your first step is wrong. You should have

    [tex]dV = \sqrt{g} \, dx^1 \wedge \ldots \wedge dx^n[/tex]
    where [itex]\sqrt{g}[/itex] is the square root of the determinant of the Riemannian metric.
  5. Feb 12, 2012 #4
    Thank you chiro and Ben for your reply :)
    I see that i forgot about [itex]\sqrt{g}[/itex] in front but can't i always find an orthonormal basis and work in that?
    and by doing that g should become 1 right?
  6. Feb 12, 2012 #5
    ok i think i understand now, if i use orthonormal basis then the connection may no be Levis-Civita.
  7. Feb 13, 2012 #6

    Ben Niehoff

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    If you use an orthonormal basis, then your basis 1-forms are not necessarily closed (so it would be incorrect to write them as dx).
  8. Feb 13, 2012 #7
    yes. that make sense.
    Now i got the solution, thanks for your help.
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