A complete or open Káhler manifold with positive definite Ricci tensor is simply connected? is there any counterexample?
1 Answer
Let $S\subset \mathbb{P}^1$ denote a finite subset consisting of $n\geq 2$ points. The FubiniStudy metric on $\mathbb{P}^1$ induces a Kahler metric on $X= \mathbb{P}^1\backslash S$ with a positivedefinite Ricci tensor. $X$ is open and Kahler, but $\pi_1(X, *)$ is free on $n1$ generators. Maybe you want to take the manifold to be complete and noncompact? As you probably know, it is a theorem of Kobayashi that a compact Kahler manifold with positive definite Ricci tensor is simply connected.

$\begingroup$ Dear Kevin, Thanks for your nice comment. Would you please explain with more details on your counterexample? $\endgroup$– user21574Jul 8, 2012 at 11:39

$\begingroup$ Do you know the answer to the question if the metric is assumed to be complete (and of course the manifold is noncompact)? $\endgroup$ Jul 8, 2012 at 13:06

$\begingroup$ Dear YangMills. No, if you have any idea please write it $\endgroup$– user21574Jul 8, 2012 at 13:13

$\begingroup$ @Haskell The FubiniStudy metric is wellknown to be Kahler with positivedefinite Ricci tensor. These properties are local and so are preserved if we restrict to an open subset, e.g. $\mathbb{P}^1$ minus $n\geq 2$ points. This last space is biholomorphic to $\mathbb{C}$ minus $n1$ points, which has the homotopy type of the wedge of $n1$ circles. From this it follows that $\pi_1$ is free on $n1$ generators. $\endgroup$– KevinJul 8, 2012 at 16:34

$\begingroup$ @YangMills I can't think of any, at the moment. A quick Google search didn't turn up much either... $\endgroup$– KevinJul 8, 2012 at 16:35