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Jun 2, 2014 at 12:02 vote accept Misha Verbitsky
Jun 2, 2014 at 6:14 answer added Robert Bryant timeline score: 4
Jun 1, 2014 at 20:40 comment added Misha Verbitsky many thanks! I would be very grateful, the question has some important geometric applications
Jun 1, 2014 at 20:20 comment added Robert Bryant @MishaVerbitsky: I'm about to board a flight, so I don't have time to input a full answer, but I can say that the Ricci curvature of metric induced on the level sets of a strictly plurisubharmonic function is not always positive. I think it is possible to characterize the induced metrics on level sets that you get this way intrinsically, but I haven't completely worked that out. Maybe when I next get connected to the internet (tomorrow), I'll have that worked out.
Jun 1, 2014 at 17:09 comment added Misha Verbitsky everything is smooth
Jun 1, 2014 at 16:53 comment added Alexandre Eremenko Do you have any smoothness condition on your PSH function? Otherwise what is a pull-back? What is curvature?
Jun 1, 2014 at 13:27 comment added Robert Bryant Moreover, it doesn't actually work in general: For example, consider $f= 1/(z^2-x^2-y^2)$. This is strictly convex on the domain (interior of a cone) where $x^2+y^2 < z^2$, but the function is invariant under the affine action of $\mathrm{SO}(2,1)$, so any induced metrics on the level sets of $f$ have to be invariant under this action as well, and that forces them to be hyperbolic metrics.
Jun 1, 2014 at 13:24 history edited Misha Verbitsky CC BY-SA 3.0
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Jun 1, 2014 at 13:24 comment added Misha Verbitsky Specifically, I want an argument which might be possibly generalized to the complex case
Jun 1, 2014 at 13:20 comment added Misha Verbitsky It seems so, but I don't see an easy argument why the metric is induced by the Gauss map
Jun 1, 2014 at 11:31 history asked Misha Verbitsky CC BY-SA 3.0