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.

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"? $\endgroup$ – Pop Feb 25 at 22:25

2$\begingroup$ @Pop Complete in the sense of metric completeness. $\endgroup$ – A Bit Too Curious Feb 25 at 22:26

4$\begingroup$ Thank you. I asked because in algebraic geometry "complete" has an alternative meaning, essentially the same as "compact". $\endgroup$ – Pop Feb 25 at 23:09

$\begingroup$ @Pop Complete is a dangerous word, I should've been more specific. At least complete is not as bad as "regular" which is so welldefined that we have 20 different inequivalent definitions for it :) $\endgroup$ – A Bit Too Curious Feb 25 at 23:21
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 BonnetMyers. Wraith's result does not have a published proof, as far as I know.


2$\begingroup$ See these two classic papers by Calabi dx.doi.org/10.1215/S0012709467034692 and Ambrose dx.doi.org/10.1215/S0012709457024407 which cover the extension that you mentioned $\endgroup$ – YangMills Feb 27 at 14:54