Timeline for Metric associated to a Connection on a Vector Bundle
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2012 at 19:35 | comment | added | YangMills | The function $log\|\ \|$ is not globally defined, it depends on the choice of local coordinate. On the other hand $i\partial\overline{\partial}\log\|\ \|$ is a globally defined $(1,1)$-form, which on a Riemann surface equals $\Lambda F$ up to some constant (perhaps $2\pi$) and indeed as you say its integral is the degree of $L$, which need not be zero. If it is zero, then $\Lambda F$ integrates to zero and then you can solve $\Delta h=\Lambda F$. Note that for any globally defined function $h$ you always have that $\Delta h$ integrates to zero. | |
| Mar 30, 2012 at 14:56 | comment | added | Giovanni De Gaetano | The last sentence is justified by the equation $\int_M \Delta log (\|\;\|)= cost \cdot deg(L)$. If you could help me in pointing out where my error lies I would very appreciate it because I´m badly stucked on it at the moment. | |
| Mar 30, 2012 at 14:52 | comment | added | Giovanni De Gaetano | @YangMills, Thank you very much for your very instructive answer, which really helped me so far. Unfortunately I don´t understand it completely. In specific may I assume that $\Lambda F$ is defined by the relation $\omega^{n-1}\wedge F= \Lambda F\omega^n$? I ask it for the following reason. Let the curvature of $\|\;\|$ defined as $i\partial \bar{\partial} log(\|\;\|)$ (correct?). Then in my situation (i.e. $M$ is a Riemann surface) $\partial \bar{\partial}\sim\Delta$, and in specific the equation $\Delta h =\Lambda F$ doesn´t need $deg(L)=0$ to be solved, so every line bundle has degree zero! | |
| Mar 26, 2012 at 12:30 | vote | accept | Giovanni De Gaetano | ||
| Mar 20, 2012 at 2:03 | history | edited | YangMills | CC BY-SA 3.0 |
edited body
|
| Mar 19, 2012 at 16:38 | history | answered | YangMills | CC BY-SA 3.0 |