Are second derivative symmetric in a Riemannian manifold?

  • #1

Main Question or Discussion Point

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!
 

Answers and Replies

  • #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.
 
  • #3
Thanks for replying|
What is [itex]D_i[/itex] in your notation?
 
  • #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..
 
  • #5
Thanks!
 

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