Timeline for Killing vectors and Ricci Tensor
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2010 at 5:25 | vote | accept | Qiao | ||
| Nov 25, 2010 at 20:39 | comment | added | Andrei Moroianu | @Dick: Oh, that's because the definition of the "push forward" $f_*$, which is defined by $(f_*K)_x=df(K_{f^{-1}(x)})$. Here $df$ is the natural extension to tensors, of course. In particular, $f_*(g)$ is just $g\circ f^{-1}$, so the minus sign disappears when you differentiate, so you get the usual formula $L_\xi g=\xi.g$, with the right sign... | |
| Nov 25, 2010 at 20:23 | comment | added | Dick Palais | @Andrei: Just a small question or quibble---why do you put the minus sign in the definition of the Lie derivative? That would make the Lie derivative of a function $f(x)$ on the line wrt the vector field $\partial/\partial x$ equal to $-f^'(x)$ rather than $f^'(x)$ which seems a bit strange. | |
| Nov 25, 2010 at 19:47 | comment | added | Andrei Moroianu | I was intrigued by Deane's remark, so I worked out a purely tensorial proof. I will edit my answer correspondingly. | |
| Nov 25, 2010 at 19:04 | vote | accept | Qiao | ||
| Nov 25, 2010 at 19:04 | |||||
| Nov 25, 2010 at 19:03 | vote | accept | Qiao | ||
| Nov 25, 2010 at 19:04 | |||||
| Nov 25, 2010 at 18:03 | history | edited | Dick Palais | CC BY-SA 2.5 |
Adde comment on physical interpretation
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| Nov 25, 2010 at 17:44 | comment | added | Deane Yang | Dick and Andrei's answers are very good, but I confess to being somewhat surprised that there is no obvious infinitesimal tensor calculation to demonstrate this fact. At least I am unable to find one. | |
| Nov 25, 2010 at 16:30 | history | answered | Dick Palais | CC BY-SA 2.5 |