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Are second derivative symmetric in a Riemannian manifold?

  1. Jun 14, 2013 #1
    Hi all!
    I was wondering if
    [itex]\partial_1\partial_2f=\partial_2\partial_1f[/itex]
    in a Riemannian manifold (Schwartz's - or Clairaut's - theorem).
    Example: consider a metric given by the line element
    [itex]ds^2=-dt^2+\ell_1^2dx^2+\ell_2^2dy^2+\ell_3^2dz^2[/itex]
    can we assume that
    [itex]\partial_1\dot{\ell}_1=\partial_0(\partial_1\ell)[/itex]?

    I think so, because you can think of [itex]\ell[/itex] as a function of [itex]R^n[/itex] through the use of coordinates, but I wanted to be sure.

    Thanks in advance!
     
  2. jcsd
  3. Jun 14, 2013 #2

    Fredrik

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    Yes. If x is a coordinate system, we have
    $$\frac{\partial}{\partial x^i}\!\bigg|_p =D_i(f\circ x^{-1})(x(p)).$$ So
    $$\frac{\partial}{\partial x^i} f =D_i(f\circ x^{-1})\circ x.$$ This implies
    $$\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j}f =\frac{\partial}{\partial x^i}(D_j(f\circ x^{-1})\circ x) =D_j\left(D_j(f\circ x^{-1})\circ x\circ x^{-1}\right)\circ x =D_iD_j(f\circ x^{-1})\circ x.$$ Since i and j can be swapped on the right, they can be swapped on the left.
     
  4. Jun 14, 2013 #3
    Thanks for replying|
    What is [itex]D_i[/itex] in your notation?
     
  5. Jun 14, 2013 #4

    Fredrik

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    The operator that takes a differentiable function (from ##\mathbb R^n## into ##\mathbb R##) to its ##i##th partial derivative..
     
  6. Jun 15, 2013 #5
    Thanks!
     
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