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Jul 4, 2017 at 22:35 answer added user21574 timeline score: 8
May 13, 2017 at 19:21 comment added user21574 Moreover if two metric be comformally equivalent $\omega_\varphi^n=e^u\omega_0^n$ then conformal factor and Kahler potential are related by $(1+\Delta_{\omega_0}\varphi)=e^u$
May 13, 2017 at 19:04 comment added user21574 In general, Kahler metrics in $[\omega_0]$ can also be parametrised as metrics of the same volume conformally equivalent to $\omega_0$ by $$\{\varphi\in C^\infty(X,\mathbb R)|\; \int_Xe^\varphi\omega_0^n=\int_X\omega_0^n=vol(X,[\omega_0]) \}$$
Aug 14, 2014 at 19:22 history edited François G. Dorais CC BY-SA 3.0
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Aug 14, 2014 at 13:54 vote accept Asghar Ghorbanpour
Aug 14, 2014 at 2:10 answer added Misha Verbitsky timeline score: 6
Aug 13, 2014 at 20:20 answer added Francois Ziegler timeline score: 6
Aug 13, 2014 at 19:43 history asked Asghar Ghorbanpour CC BY-SA 3.0