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Recently I just learned the Kobayashi distance on complex manifolds and wants to get some feeling of how it looks like on exmaples of manifolds with positive Ricci curvature. I have a feeling that the Kobayashi distance on those manifolds should vanish since those are not very "hyperbolic".

A simple example is extended complex plane. then its Kobayashi distance vanishes since the automorphism group can contract one point very close to the origin while keeping the origin fixed. I believe it is similarly true for all complex projective spaces with usual complex structure since automorphism group is known. Can we have more examples? Or is there a counterexample (an example of Kaehler manifold with positive Ricci curvature but not identically vanishing Kobayashi distance)?

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up vote 4 down vote accepted

Positive Ricci curvature Kaehler manifolds are Fano, and therefore rationally connected. See Janos Kollar's book Rational Curves on Algebraic Varieties. Therefore any two points lie on a rational curve. Since the Kobayashi pseudodistance along a subvariety is never less than the pseudodistance in the ambient space, any pair of points lie at 0 pseudodistance.

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Thank you very much for the quick answer. What happens if we change "compact" to "complete noncompact" or change "Ricci positive" to "Ricci nonnegative"? Is there a similar argument? I know almost zero of algebraic geometry. –  Bo_Y Oct 26 '11 at 22:29
    
If the Ricci curvature is bounded from below by a positive constant, then complete implies compact, by the Bonnet-Myers theorem. I don't know whether Ricci nonnegative manifolds have identically zero pseudodistance. It seems likely. –  Ben McKay Nov 14 '11 at 17:32
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