Two definitions of Calabi-Yau manifolds

Question

Why is it that the vanishing of the integral first Chern class of a compact Kahler manifold is equivalent to the canonical bundle being trivial? I can see that it implies that the canonical bundle must be topologically trivial, but not necessarily holomorphically trivial. Does proving the equivalence require Yau's theorem, in order to produce a flat connection on the canonical bundle, or is there a more elementary proof?

Answer 1

I have looked for a while for a proof which does not use the Calabi-Yau theorem and nobody seems to know it.

Also, there are plenty of non-Kaehler manifolds with canonical bundle trivial topologically and non-trivial as a holomorphic bundle (the Hopf surface is an easiest example).

The argument actually uses the Calabi-Yau theorem, Bochner's vanishing, Berger's classification of holonomy and Bogomolov's decomposition theorem.

From Calabi-Yau theorem you infer that there exists a Ricci-flat Kaehler metric. Since the Ricci curvature is a curvature of the canonical bundle, this implies that the canonical bundle admits a flat connection.

Of course, this does not mean that it is trivial holomorphically; in fact, the canonical bundle is flat on Hopf surface and on the Enriques surface, which are not Calabi-Yau.

For Calabi-Yau manifolds, however, it is known that the Albanese map is a locally trivial fibration and and has Calabi-Yau fibers with trivial first Betti number. This is shown using the Bochner's vanishing theorem which implies that all holomorphic 1-forms are parallel.

Now, by adjunction formula, you prove that the canonical bundle of the total space is trivial, if it is trivial for the base and the fiber. The base is a torus, and the fiber is a Calabi-Yau with $H^1(M)=0$. For the later, triviality of canonical bundle follows from Bogomolov's decomposition theorem, because such a Calabi-Yau manifold is a finite quotient of a product of simple Calabi-Yau manifolds and hyperkaehler manifolds having holonomy $SU(n)$ and $Sp(n)$. Bogomolov's decomposition is itself a non-trivial result, and (in this generality) I think it can be only deduced from the Berger's classification. The original proof of Bogomolov was elementary, but he assumed holomorphic triviality of a canonical bundle, which we are trying to prove.

This argument is extremely complicated; also, it is manifestly useless in non-Kaehler situation (and in many other interesting situations). I would be very interested in any attempt to simplify it.

Update: Just as I was writing the reply, Dmitri has posted a link to Bogomolov's article, where he proves that some power of a canonical bundle is always trivial, without using the Calabi-Yau theorem.

Answer 2

It seems worthwile to point out that it is not true that the vanishing of the integral first Chern class of a compact Kahler manifold implies that the canonical bundle is holomorphically trivial. It only implies that it is torsion, i.e. some positive multiple is holomorphically trivial, as indicated by Misha and Dmitri (an English version of Bogomolov's paper is here).

An example where $c_1(K_M)$ vanishes in integral cohomology while $K_M$ is not holomorphically trivial is any bielliptic surface (a finite unramified quotient of a complex $2$-torus), in which case $12K_M$ is trivial. This fact is remarked for example in this paper of Tian-Jun Li, Remark 6.4 (see also the paper of McDuff-Salamon cited there as [30]).

Answer 3

You can prove that the canonical bundle is torsion without using Yau's theorem. This is contained the following work of Bogomolov, Theorem 3'

F. A. Bogomolov, “Kähler manifolds with trivial canonical class”, Izv. Akad. Nauk SSSR Ser. Mat., 38:1 (1974), 11–21

(in the first version of this answer I said that the bundle is trivial, but Bogomolov states just that it is torsion, as Misha notes).

Here is the Russian version availible online

http://www.mathnet.ru/php/getFT.phtml?jrnid=im&paperid=1889&volume=38&year=1974&issue=1&fpage=11&what=fullt&option_lang=eng

Answer 4

Disclaimer

This is by no means an answer, but a reformulation of the question.

One way to understand why line bundles are classified smoothly (and hence topologically) by their first Chern class is the exponential sheaf sequence $$ \begin{matrix} 0 & \longrightarrow & \mathbb{Z} & \longrightarrow & \mathcal{E} & \stackrel{\mathrm{exp}}{\longrightarrow} & \mathcal{E}^\times & \longrightarrow & 0 \end{matrix} $$ where $\mathcal{E}$ and $\mathcal{E}^\times$ are, respectively, the sheaves of smooth and nonvanishing smooth functions. The sheaf $\mathcal{E}$ is fine due to the existence of smooth partitions of unity, whence $H^{p\geq 1}(M,\mathcal{E}) = 0$. This results in the long exact sequence in cohomology yielding an isomorphism $H^1(M,\mathcal{E}^\times) \cong H^2(M,\mathbb{Z})$ (the first Chern class) which says that a smooth line bundle is trivial if and only if its first Chern class vanishes.

For holomorphic line bundles things are more complicated. Now the exponential sheaf sequence $$ \begin{matrix} 0 & \longrightarrow & \mathbb{Z} & \longrightarrow & \mathcal{O} & \stackrel{\mathrm{exp}}{\longrightarrow} & \mathcal{O}^\times & \longrightarrow & 0 \end{matrix} $$ where $\mathcal{O}$ and $\mathcal{O}^\times$ are, respectively, the sheaves of holomorphic and nonvanishing holomorphic functions, gives rise to the following snippet of the long exact sequence of cohomology $$ \begin{matrix} H^1(M,\mathcal{O}) & \longrightarrow & H^1(M,\mathcal{O}^\times) & \longrightarrow & H^2(M,\mathbb{Z}) & \longrightarrow & H^2(M,\mathcal{O}) \end{matrix} $$ where the Picard group $H^1(M,\mathcal{O}^\times)$ classifies holomorphic line bundles up to isomorphism and $$H^1(M,\mathcal{O}^\times) \longrightarrow H^2(M,\mathbb{Z})$$ is again the first Chern class.

Hence if the first Chern class of a holomorphic line bundle vanishes, its class in the Picard group must come from $H^1(M,\mathcal{O})$. Thus one needs to show that the map $H^1(M,\mathcal{O}) \to H^1(M,\mathcal{O}^\times)$ is zero.

Does this require the hard analysis of the proof of the Calabi conjecture?

I'm not sure.

Answer 5

I believe that the answer is yes, this needs Yau's theorem. Rather, I think that the statement of your question can be phrased in such a way as to be equivalent to Yau's theorem.

My understanding of the theorem that Yau proved is that the topological condition (the vanishing of the integral first Chern class) is sufficient to guarantee the existence of a certain geometric structure (a Calabi-Yau metric). There are several equivalent ways to say what a Calabi-Yau metric is, but for the purposes of this post I think the best is via the holonomy groups.

Given a Riemannian manifold, you can define parallel transport along paths in the manifold. Given a point in the manifold, you can look at the parallel transport along all loops based at that point. Taken together these yield a subgroup of the orthogonal group $O(T_xM)$. The manifold is a Kahler manifold precisely if the subgroup is contained in a copy of $U(n)$ (by which I mean a subgroup conjugate to a copy of the standard $U(n) \subset O(2n)$, after we've chosen some identification $O(T_xM) = O(2n)$). The metric is Calabi-Yau precisely if the holonomy is contained in a copy of $SU(n)$. From the point of view of holonomy (i.e. parallel transport along loops) it is clear that having a Calabi-Yau metric is the same as having a holomorphically trivial canonical bundle. Yau's theorem, which if I recall correctly uses a difficult analytic argument, says the topological condition is in fact sufficient to find such a metric.