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.

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    $\begingroup$ I presume this is the covariant derivative: $\nabla^i=\sum_j g^{ij}\nabla_j$ with $g$ the metric tensor of the manifold. $\endgroup$ Commented Oct 3, 2023 at 19:34
  • $\begingroup$ @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. $\endgroup$
    – Gomes93
    Commented Oct 3, 2023 at 21:12
  • $\begingroup$ no, this is just differential geometry: en.wikipedia.org/wiki/Metric_tensor $\endgroup$ Commented Oct 3, 2023 at 21:21
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    $\begingroup$ 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 $\endgroup$ Commented Oct 3, 2023 at 22:38
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    $\begingroup$ 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. $\endgroup$ Commented Oct 5, 2023 at 20:20


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