Timeline for Lie derivative and taking trace
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2015 at 1:17 | comment | added | Daniel | Yes! But what I care more is the pointwise property of $X(f)$. And it seems impossible if we only know the Laplace of $f$. | |
| Jun 22, 2015 at 10:41 | comment | added | YangMills | If $X$ has a zero then there is a complex-valued function $u$ such that $X=g^{i\bar{j}}\partial_{\bar{j}}u \partial_i$, and so $\int_M X(f) \omega^n=-\int_X u\Delta f\omega^n=-\int_X ug\omega^n$. | |
| Jun 18, 2015 at 9:19 | comment | added | Daniel | Thank you for your reply! But I can not really understand the last sentence "(in particular you can do it if your vector field X has a zero, in which case X has a holomorphy potential)". Would you like to give some details? In my opinion, the difficulty of this problem is that we get less information from $g=\triangle f$, in particular, the Hessian of $f$. | |
| Jun 17, 2015 at 7:58 | comment | added | YangMills | Yes, if you allow non-local operators in your formula, such as the Green operator of $\omega$ (and then it is trivial to derive a formula for $X(f)$). If you insist on a purely local formula, then I am afraid the answer is no, because $X(f)$ involves only one derivative of $f$, while $g$ involves two derivatives. On the other hand, if all you care about is an integral relation between these quantities, then often you can do this, by using integration by parts (in particular you can do it if your vector field $X$ has a zero, in which case $X$ has a holomorphy potential). | |
| Jun 15, 2015 at 6:57 | history | asked | Daniel | CC BY-SA 3.0 |