9
$\begingroup$

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

$\endgroup$
  • 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
  • 3
    $\begingroup$ 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
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.