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.
Best Answer
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.