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Sep 11, 2010 at 20:29 comment added Mohan Ramachandran Dmitri:Your proof goes through for any stein manifold.All you need is a strictly plurisubharmonic function whose Levi form dominates the curvature of the holomorphic vector bundle.You can do this by taking a strictly plurisubharmonic exhaustion function and compose with a suitable convex increasing function.
Sep 8, 2010 at 12:06 comment added Dmitri Panov Auniket, I should look for a refference. Notice though that for the proposed reasoning it is enough to prove this fact for any rotation invariant form on $\mathbb C^1$, this can be done by hands.
Sep 8, 2010 at 11:14 comment added pinaki One (probably stupid) question: why is it true that every Kahler $(1,1)$ form on an affine variety is of the form $\frac{i}{2\pi} \partial \bar \partial(f)$ for a global function $f$? Is it easy to see?
Sep 8, 2010 at 10:11 history answered Dmitri Panov CC BY-SA 2.5