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For example; is $$\partial_{\rho}\partial_{\sigma}h_{\mu\nu} - \partial_{\sigma}\partial_{\rho}h_{\mu\nu}=0$$ correct?

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up vote 5 down vote accepted

Yes, it's just the second derivative of some function, it doesn't matter that this function is organized as a component of a tensor, $h_{\mu\nu}$. The identity above – assuming the function is differentiable and smooth etc. (add some "niceness" conditions on the function) – follows from the rules of calculus and is formally proven by the $\varepsilon$-$\delta$ gymnastics.

However, if you replaced the partial derivatives by the covariant ones $\nabla_\mu$, the right hand side would no longer be zero but proportional to a contraction of the Riemann tensor and $h$. The RHS would be calculable by writing $\nabla_\mu$ in terms of $\partial_\mu$ and the term proportional to $\Gamma_{\alpha\beta}^\gamma$.

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"differentiable-smooth etc": to be punctilious, I think $C^2$ is the correct requirement. It's called Schwartz's theorem, and I suppose it can be applied to tensors because we are working in local coordinates. –  Bzazz Jan 31 '13 at 16:57
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Well, perhaps, but the problem with theorems such as Schwartz's theorem is that its assumptions aren't the only ones that may be picked so that a similar theorem holds. For example, one may also pick functions in equivalence classes of $L^2$, or perhaps distributions, and allow the derivatives to be distributions themselves even if the original function is unsmooth or even discontinuous - and the symmetry of the derivatives will still hold. So the assumptions of Schwartz's theorem are sufficient to prove a similar relationship but they're not really necessary. –  Luboš Motl Jan 31 '13 at 17:01
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