# sign of the First chern class fundamental group of Kahler Manifolds

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 $c_1 (M)<0$ under which assumption on $M$ , we have $π_1(M)={e}$.

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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 $c_1 < 0$ 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?
Added Explanation: 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 $c_1(M)<0$ to ampleness of the canonical bundle $K$. This condition is stable under products.
Of course I know the proof of this fact for$c_1(M)>0$. But I am looking for further assumption when $c_1(M)<0 or =0$ –  Hassan Jolany Jul 2 '12 at 1:28
My point was really the answer is NEVER in dimension $1$ and inconclusive in higher dimension. –  Donu Arapura Jul 2 '12 at 3:42
The Ricci curvature adds when you take products, so indeed products of manifolds with $c_1<0$ will have $c_1<0$ by Yau's proof of the existence of a Kahler-Einstein metric. More simply, ampleness of the canonical bundle of two manifolds implies ampleness of the canonical bundle of the product, as this is the product of the canonical bundles. –  Ben McKay Apr 14 '13 at 19:10