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Le$f$ be a smooth function on a Riemannian manifold $M$. My questions are:

a) If $\nabla_i f$ is a function, why is not true that $\nabla_j\nabla_k\nabla_if=\nabla_k\nabla_j\nabla_if$? This question arose when I wrote $\nabla_j\nabla_k\nabla_if=\nabla_j\nabla_k(\nabla_if)$, (is this true or not?).

If we can see $\nabla_if$ as a function, then shouldn't be true that $\nabla_j\nabla_k(\nabla_if)=(\nabla^2f_i)(\partial_k,\partial_j)$, i. e., the Hessian of $\nabla_if$, which would give the symmetry on $k$ and $j$ on the formula $\nabla_j\nabla_k\nabla_if$? Where am I going wrong?

I know that there is something wrong on question a) because of the following formula:

$$ \nabla_j\nabla_i\nabla_jf-\nabla_i\nabla_j\nabla_jf=R_{jikj}\nabla_kf. $$

b) How to prove the previous formula?

I appreciate any help.

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Your confusion seems to be notation related - in conventional abstract index notation, $\nabla_i (\nabla_j f)$ and $\nabla_i \nabla_j f$ are different things! The first means what you think it does - since $f$ is a function, $\nabla_j f = \partial_j f$ is also a function, and thus $\nabla_i(\nabla_j f) = \partial_i \partial_j f$. However, when the parentheses are omitted, all the covariant derivatives are taken before any indices are applied - that is, $$\nabla_i \nabla_j f := \nabla^2 f (\partial_i, \partial_j) = \nabla_i (\nabla_j f) - \nabla_{\nabla_i \partial_j} f.$$

This can be very confusing but it makes computations easier to write down.

Thus the resolution of your problem is that the expression

$$\nabla_j\nabla_k\nabla_if - \nabla_k\nabla_j\nabla_if$$is in fact commuting second covariant derivatives of a one-form $\nabla f$ - it is only at the very end that we take the $i$th component.

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  • $\begingroup$ Wait, I agree until the end: why is $\nabla_j\nabla_k\nabla_if$ a second covariant derivative of the one form $\nabla f$, isn't it again just $\nabla_j \nabla_k \nabla_i f := \nabla^3 f (\partial_j, \partial_k, \partial_i)$? If that's the case, don't these third derivatives actually commute? $\endgroup$ – Seub Feb 5 '18 at 4:00
  • $\begingroup$ @Seub: the second covariant derivative of $\nabla f$ and the third covariant derivative of $f$ are (by definition) the same thing. The third covariant derivative of a scalar is not symmetric. $\endgroup$ – Anthony Carapetis Feb 5 '18 at 4:41
  • $\begingroup$ @AnthonyCarapetis: Got it, thank you. I'm surprised to learn it's not symmetric. Can you give me a link or a reference for an example or more information? I couldn't find much upon a quick web search. $\endgroup$ – Seub Feb 6 '18 at 15:30
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The reason why the covariant derivatives do not commute is precisely that they are not partial derivatives. If they were partial derivatives they would commute, but they are not.

For a function the covariant derivative is a partial derivative so

$\nabla_i f = \partial_i f$

but what you obtain is now a vector field, and the covariant derivative, when it acts on a vector field has an extra term: the Christoffel symbol:

$\nabla_j (\partial_i f) = \partial_j\partial_i f + \Gamma_{ji}^m \partial_m f$

Due to the properties of the Christoffel symbols you can verify that in general the covariant derivatives will not commute with each other. For instance, above:

$\nabla_i \nabla_j f = \nabla_j \nabla_i f$

Only when there is no torsion, i.e. when $\Gamma^m_{ij} = \Gamma^m_{ji}$ (this needs not be true, but it is e.g. for the Levi-Civita connection).

In the particular example you gave you have to act with yet another covariant derivative $\nabla_k$, in this case when it acts on a tensor it will produce two Christoffel symbols. Putting all this together you may verify the equation you gave above for the curvature tensor $R_{ijkl}$

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  • $\begingroup$ Rogelio Molina, I didn't understand just one part of your explanation: when you say that $\nabla_if$ is a vector field. The connection takes a field of $\Gamma(TM)$ and a section of a vector bundle $\Gamma(E)$ and gives as result a section of $\Gamma(E)$. A functions is a section of the line bundle, so the image of $\nabla$ shouldn't be an object of the line bundle, i. e., a function? Could you explain this part? $\endgroup$ – Myself Sep 30 '13 at 21:46
  • $\begingroup$ You are right, it is a section of the bundle. I was reffering to the case when you have the cotangent bundle, in which case $\nabla_i f$ is a covariant vector field. In this sense I used the word vector field, the index $i$ labels its components in the base chosen to define $\nabla_i$. $\endgroup$ – Rogelio Molina Sep 30 '13 at 22:08
  • $\begingroup$ Your answer does not seem to be correct: compare with the other answer. It is not true that $\nabla_j (\partial_i f) = \partial_j\partial_i f + \Gamma_{ji}^m \partial_m f$. $\endgroup$ – Seub Feb 5 '18 at 4:02
  • $\begingroup$ @Seub: it depends on how you interpret the notation - Rogelio is following the convention that all covariant derivatives are taken before indices, regardless of parentheses. This is actually more common than the interpretation in my answer, particularly in physics. $\endgroup$ – Anthony Carapetis Feb 5 '18 at 4:54

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