Are second derivative symmetric in a Riemannian manifold?

[fairy._.queen]

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

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

Thanks in advance!

 

[Fredrik]

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.

 

[fairy._.queen]

Thanks for replying|
What is $D_i$ in your notation?

 

[Fredrik]

The operator that takes a differentiable function (from $\mathbb R^n$ into $\mathbb R$) to its $i$th partial derivative..

 

[fairy._.queen]

Thanks!