This question refers mainly to this mathoverflow question and its answer.
Statement 1 The well known result of Aubin and Yau states that if $X$ is a compact Kähler manifold with negative first Chern class, $c_1(X)<0$, then it will allow a Kähler-Einstein metric $g$ such that:
\begin{equation}
\text{Ric} = -g
\end{equation}
where by $\text{Ric}$ I mean the Ricci curvature. I won't pretend that I understand the proof of this result, but it is used frequently and is discussed in the mathoverflow thread I mentioned above.
Statement 2
The answer in the above thread then goes on to say (and I have seen this statement in several papers that I am trying to understand, such as this one by Kobayashi and this one by F. Catanese and A. Di Scala) that if the canonical bundle of $X$ is ample then $c_1(X) <0$ and so by statement 1 we get the Kähler-Einstein metric $g$.
So:
1) What exactly do we mean by the first Chern class of of a complex manifold? I always thought that this referred to the first Chern class of the canonical bundle: $c_1(X) = c_1(K_X)$ where $K_X = \bigwedge ^{n}\Omega_X$ and $\Omega_X$ is the holomorphic cotangent bundle, but…
2) If $K_X$ is ample, is it not true that it has positive first Chern class?
Thanks
Best Answer
1) $c_1(X):=c_1(TX)=c_1(\wedge^n(TX))=-c_1(K_X)$ where $TX$ is the holomorphic tangent bundle as in Georges's comment.
2) If $K_X$ is ample, then $c_1(K_X)>0$ in the sense that $c_1(K_X).C>0$ for all irreducible holomorphic curve $C$ in $X$. This is because some positive multiple of $K_X$ is very ample, hence positive.