What does $\nabla^i f$ mean?

I am reading the article Some Geometric Calculations on Wasserstein Spaces of John Lott and there is this covariant index in the covariant derivative: $$\nabla^i$$. And I don't quite understand it.

In some place he establishes the equality ($$\phi_i$$ and $$\rho$$ being smooth functions on $$M$$):

$$\int_M\langle\nabla\phi_1,\nabla\phi_2\rangle\rho d\operatorname{vol}_M=-\int_M\phi_1\nabla^i(\rho\nabla_i\phi_2)d\operatorname{vol}_M.$$

In another part of the text he writes $$\nabla^i f$$ for $$f$$ smooth. I think he is using the musical isomorphisms some way to define it, but I'm not sure how and I can't see how the above equation takes place.

It is worth mentioning that he applies $$\nabla^i$$ on functions, not on 1-forms. I'd like to understand the identity as well as to understand explicitly how $$\nabla^i$$ operates, since I want to caculate it on specific functions, such as $$f_n(x)=e^{Inx}$$, with $$I$$ being the imaginary unit.

An example of explicit calculation of $$\nabla^i$$ for an explicit function would be nice.

• I presume this is the covariant derivative: $\nabla^i=\sum_j g^{ij}\nabla_j$ with $g$ the metric tensor of the manifold. Commented Oct 3, 2023 at 19:34
• @CarloBeenakker I imagine that you get this formula from using musical isomorphisms with $\nabla_i$, correct? But could you please elaborate a little more? I don't know how to get this formula you wrote and also I don't know any book that does that. Another guy said that this could be the Lie derivative of the volume form, which makes some sense. Commented Oct 3, 2023 at 21:12
• no, this is just differential geometry: en.wikipedia.org/wiki/Metric_tensor Commented Oct 3, 2023 at 21:21
• Carlo Beenakker is correct except that what he wrote is called musical isomorphism. The differential (i.e. exterior derivative) of a function $f$ is a 1-form $df$, and the musical isomorphism defined by the metric $g$ turns this into the vector field $\nabla f.$ You can read for instance in John Lee's "Introduction to Riemannian manifolds" or most other introductory books on Riemannian geometry Commented Oct 3, 2023 at 22:38
• It certainly isn't obvious if you haven't seen it! Depending on your personal taste you can make it a special case of Stokes theorem, Green's identity, divergence theorem, or integration by parts. Commented Oct 5, 2023 at 20:20