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 $g$.


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?


  • 3
    $\begingroup$ I'm not a complex geometer, but when one says "The Stiefel-Whitney class of a manifold" or "The Euler class of a manifold" or "The Pontryagin class of a manifold", one is always refering to the tangent bundle of the manifold. I'd guess then, that "Chern class of a complex manifold" means "chern class of the tangent bundle (which is canonically a complex bundle)." I have no idea how the canonical bundle and the tangent bundle are related. $\endgroup$ – Jason DeVito Feb 18 '13 at 13:48
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    $\begingroup$ Dear @Jason, your guess is absolutely correct.And the canonical bundle is the top exterior power of the COtangent budle: $K_X=\wedge^n T^*_X$. So Daniel's interpretation is the negative of the correct one. $\endgroup$ – Georges Elencwajg Feb 18 '13 at 14:09
  • $\begingroup$ ah, so the chern class of $X$ is the chern class of $\bigwedge^{n}TX$? $\endgroup$ – Daniel Mckenzie Feb 18 '13 at 14:45
  • $\begingroup$ Wait, the above should probably read `the chern class of $X$ is the chern class of $'\bigwedge^{n}T^{'}X$' where $T^{'}X$ is the holomorphic tangent bundle. Does that sound right? $\endgroup$ – Daniel Mckenzie Feb 18 '13 at 14:53
  • $\begingroup$ Dear Daniel, yes that's right (there is a certain notational ambiguity in complex analysis: my $T_X$ is the same as your $T'X$) $\endgroup$ – Georges Elencwajg Feb 18 '13 at 15:40

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.

  • $\begingroup$ great, thanks @QiL'8 $\endgroup$ – Daniel Mckenzie Feb 20 '13 at 10:40

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