1
$\begingroup$

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.

$\endgroup$
8
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.