Timeline for covariant derivative of a function
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 2, 2019 at 16:45 | comment | added | Keith | I see. Do you maybe want to write up something so that I can accept as an answer? | |
| Nov 1, 2019 at 13:19 | comment | added | Willie Wong | Fundamentally: you need to look up how index summation convention works in the book/paper you are reading. Also, note that on a Kahler manifold, for scalar functions you have the mixed derivatives $\nabla_i \nabla_{\bar{j}} f = \partial_i \partial_{\bar{j}} f$ because Kahler requires the mixed index Christoffel symbols to vanish. You may want to look at a text that does Kahler manifolds in local coordinates (an example being Aubin's "Some Nonlinear Problems in Riemannian Geometry", but there are certainly others.) | |
| Oct 31, 2019 at 20:35 | comment | added | Keith | I am so confused as to how do you carry out this kind of computation using covariant derivatives for a function. | |
| Oct 31, 2019 at 20:33 | comment | added | Keith | Well originally it is something like $\int_{M}s \nabla_j f \nabla_{\bar j} f w_{f}^n = \int _{M} \nabla_j \nabla_p \nabla_{\bar p} f \nabla_{\bar j} f w_{f}^n$, maybe I intepreted it wrong? | |
| Oct 31, 2019 at 17:58 | comment | added | Willie Wong | If you think of $\nabla_j \nabla_p \nabla_{\bar{p}} f$ as you wrote, as a component of a rank three object, it makes no sense to even think that can be equal to the component of a rank one object ($s \nabla_j f$)... | |
| Oct 31, 2019 at 17:55 | comment | added | Willie Wong | Yes, I am suggesting exactly that. $\nabla_p \nabla_{\bar{p}} f$ has the implicit summation over $p$ and is a scalar. | |
| Oct 31, 2019 at 17:52 | comment | added | Keith | No. Are you trying to suggest that $\nabla_j \nabla_p \nabla_p f = \nabla_j( \nabla_p \nabla_p f )$? According to the thread I posted, it should be interpreted as $\nabla \nabla \nabla f (\frac{\partial}{\partial z_j}, \frac{\bar \partial}{\partial z_p}, \frac{\partial}{\partial z_j})$. | |
| Oct 31, 2019 at 15:29 | comment | added | Willie Wong | Your question is basically why $g^{ij} \sqrt{-1} \partial_j \bar{\partial}_i f = \nabla_p \nabla_{\bar{p}} f$, no? | |
| Oct 31, 2019 at 15:10 | review | First posts | |||
| Oct 31, 2019 at 16:03 | |||||
| Oct 31, 2019 at 15:03 | history | asked | Keith | CC BY-SA 4.0 |