sign of the First chern class fundamental group of Kahler Manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:47:41Z http://mathoverflow.net/feeds/question/101105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101105/sign-of-the-first-chern-class-fundamental-group-of-kahler-manifolds sign of the First chern class fundamental group of Kahler Manifolds Hassan Jolany 2012-07-02T00:25:36Z 2013-04-17T00:21:30Z <p>We know by some facts from Kobayashi, if the Kahler manifold \$M\$ has positive first Chern class, i.e., \$c_1 (M)>0\$ then \$M\$ is simply connected. So if \$c_1 (M)&lt;0\$ under which assumption on \$M\$ , we have \$π_1(M)={e}\$. </p> http://mathoverflow.net/questions/101105/sign-of-the-first-chern-class-fundamental-group-of-kahler-manifolds/101107#101107 Answer by Donu Arapura for sign of the First chern class fundamental group of Kahler Manifolds Donu Arapura 2012-07-02T01:00:41Z 2012-07-02T12:56:54Z <p>This is really more of an extended comment, since I'm not sure how to give a definite answer. When \$\dim M=1\$, \$c_1(M)>0\$ if and only if \$M= \mathbb{C}\mathbb{P}^1\$ if and only if \$M\$ is simply connected, by the uniformization theorem. In the dimension \$2\$, things are more complicated. Certainly simply connected with \$c_1 &lt; 0\$ exist. For example, any surface in \$\mathbb{C}\mathbb{P}^3\$ of degree \$5\$ or more will work. However, there are also plenty of nonsimply connected examples (products of curves of large genus, ball quotients...). Does this help?</p> <p><strong>Added Explanation</strong>: To explain where the examples are coming from and to answer your 2nd comment, let me explain I'm using Kodaira's embedding theorem to translate \$c_1(M)&lt;0\$ to ampleness of the canonical bundle \$K\$. This condition is stable under products.</p>