Suppose $(M,g, \omega)$ is a Kähler manifold with $\text{Ric}(g) = g$, i.e., $M$ is a Fano manifold. Is $M$ necessarily compact? If not, perhaps complete and Fano implies compact? I'd like to build a catalog of examples of Fano manifolds, so any input is appreciated.
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3$\begingroup$ For the first question, taking any compact Fano and deleting a point doesn't change the condition on the Ricci tensor. For the second, in what sense do you mean "complete"? $\endgroup$– PopCommented Feb 25, 2020 at 22:25
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2$\begingroup$ @Pop Complete in the sense of metric completeness. $\endgroup$– user105074Commented Feb 25, 2020 at 22:26
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4$\begingroup$ Thank you. I asked because in algebraic geometry "complete" has an alternative meaning, essentially the same as "compact". $\endgroup$– PopCommented Feb 25, 2020 at 23:09
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$\begingroup$ @Pop Complete is a dangerous word, I should've been more specific. At least complete is not as bad as "regular" which is so well-defined that we have 20 different inequivalent definitions for it :) $\endgroup$– user105074Commented Feb 25, 2020 at 23:21
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1 Answer
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By the Bonnet Myers theorem, bounded positive Ricci curvature and complete Riemannian metric forces compact. David Wraith once explained to me that if the Ricci decays more slowly than quadratically in distance from a given point, on a complete Riemannian manifold, then the manifold is compact, a stronger result than Bonnet-Myers. Wraith's result does not have a published proof, as far as I know.
answered Feb 26, 2020 at 8:48
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$\begingroup$ Thanks for this answer @Ben $\endgroup$– user105074Commented Feb 27, 2020 at 5:45
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2$\begingroup$ See these two classic papers by Calabi dx.doi.org/10.1215/S0012-7094-67-03469-2 and Ambrose dx.doi.org/10.1215/S0012-7094-57-02440-7 which cover the extension that you mentioned $\endgroup$ Commented Feb 27, 2020 at 14:54