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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

Glorfindel
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Daniel Mckenzie
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    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. – Jason DeVito - on hiatus Feb 18 '13 at 13:48
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    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. – Georges Elencwajg Feb 18 '13 at 14:09
  • ah, so the chern class of $X$ is the chern class of $\bigwedge^{n}TX$? – Daniel Mckenzie Feb 18 '13 at 14:45
  • 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? – Daniel Mckenzie Feb 18 '13 at 14:53
  • Dear Daniel, yes that's right (there is a certain notational ambiguity in complex analysis: my $T_X$ is the same as your $T'X$) – Georges Elencwajg Feb 18 '13 at 15:40
  • Great, thanks a lot Georges, that really clears things up. And thanks Jason for your contribution – Daniel Mckenzie Feb 18 '13 at 19:05

2 Answers2

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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.

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Let me add some additional details that may provide some insight. There are a number of ways to define the first Chern class of a complex line bundle. Let us restrict attention to holomorphic line bundles $\mathcal{L} \to X$ over a compact complex manifold $X$.

Equivalence classes of holomorphic line bundles form a group which is called the Picard group $\text{Pic}(X)$. This group is isomorphic to the sheaf cohomology group $H^1(X, \mathcal{O}_X^{\star})$, where $\mathcal{O}_X^{\star}$ is the sheaf of (germs of) non-vanishing holomorphic functions. From the exponential sequence $0 \to \mathbf{Z} \to \mathcal{O}_X \to \mathcal{O}_X^{\star} \to 0$ which induces a long exact sequence on sheaf cohomology groups. The first coboundary morphism $H^1(X, \mathcal{O}_X^{\star}) \to H^2(X, \mathbf{Z})$ induced from the exponential sequence, is the first Chern class.

Another way to define the first Chern class $c_1(\mathcal{L})$ is to define it as the de Rham cohomology class represented by the curvature form $\Theta^{(\mathcal{L},h)} : = \sqrt{-1} \partial \bar{\partial} \log(h)$ of a Hermitian metric $h$ on $\mathcal{L}$. The cohomology class represented by $\Theta^{(\mathcal{L},h)}$ is independent of the choice of Hermitian metric. Hence, one may define $c_1(\mathcal{L}) : = \{ \Theta^{(\mathcal{L},h)} \} \in H_{\text{DR}}^2(X, \mathbf{R})$.

If $\omega$ is a Hermitian metric on $X$, e.g., a Kähler metric, then $\omega^n$ is a volume form. A volume form is equivalent to a Hermitian metric on $K_X : = \Lambda^{\dim_{\mathbf{C}} X,0}$, and hence, the curvature form of the Hermitian metric $h = \omega^n$ on $K_X$ is $\Theta^{(K_X, h)} = \sqrt{-1} \partial \bar{\partial} \log \omega^n = - \text{Ric}_{\omega}^{(1)}$, where $\text{Ric}_{\omega}^{(1)}$ is the first Chern Ricci curvature (given by tracing over the third and fourth indices of the Chern curvature tensor). If the metric is Kähler, this coincides with the more familiar Ricci curvature.

For vector bundles of rank $>1$, the first Chern class is typically defined to be the first Chern class of its determinant (which is a line bundle).

KyleBroder
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