Covariant derivative in abstract index notation

Question

Spose $f,h$ functions, where $\nabla _af = \epsilon _{ab}\nabla ^bh$. Then $\nabla ^af=g^{ac}\epsilon _{cb}\nabla ^bh$. My question is then does $\nabla _a\nabla ^af=\nabla ^c\epsilon _{cb}\nabla ^bh$ or does $\nabla _a\nabla ^af=\nabla ^c(\epsilon _{cb}\nabla ^bh)?$

If it is the former, is it because we have to match up the repeated index, as we are tracing?

thanks for the help!

Answer 1

The standard convention is that $\nabla$ acts on everything to the left (in the monomial), and the Leibniz rule is applied if necessary.

Answer 2

If $\epsilon_{ab}$ is just any tensor, $\nabla^a \equiv g^{ab}\nabla_b$ and $f$, $g$ and $\epsilon_{ab}$ are related by $\nabla_a f = \epsilon_{ab} g^{bc} \nabla_c h$, then $\nabla_a \nabla^a f = \nabla^a\nabla_a f = \nabla^a(\nabla_a f) = \nabla^a(\epsilon_{ab} g^{bc} \nabla_c h)= \nabla^a(\epsilon_{ab} \nabla^b h)$.

It is advisable use the brackets to avoid the ambiguities.