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?

$\begingroup$ After looking at it some more, we can get a flat connection on the canonical bundle just from the fact that it has real first Chern class zero and that it's a line bundle. Yau's proof implies that we that we can choose a metric on the tangent bundle so that the induced metric on the canonical bundle has flat Chern connection, but that doesn't seem to help in proving that the bundle is trivial. $\endgroup$ – Evan Wright Feb 11 '10 at 19:42

$\begingroup$ If the first Betti number is zero, then topologically trivial automatically implies holomorphically trivial from the exponential sequence, but what about the general case? $\endgroup$ – Evan Wright Feb 11 '10 at 19:44
I have looked for a while for a proof which does not use the CalabiYau theorem and nobody seems to know it.
Also, there are plenty of nonKaehler manifolds with canonical bundle trivial topologically and nontrivial as a holomorphic bundle (the Hopf surface is an easiest example).
The argument actually uses the CalabiYau theorem, Bochner's vanishing, Berger's classification of holonomy and Bogomolov's decomposition theorem.
From CalabiYau theorem you infer that there exists a Ricciflat 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 CalabiYau.
For CalabiYau manifolds, however, it is known that the Albanese map is a locally trivial fibration and and has CalabiYau fibers with trivial first Betti number. This is shown using the Bochner's vanishing theorem which implies that all holomorphic 1forms 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 CalabiYau with $H^1(M)=0$. For the later, triviality of canonical bundle follows from Bogomolov's decomposition theorem, because such a CalabiYau manifold is a finite quotient of a product of simple CalabiYau manifolds and hyperkaehler manifolds having holonomy $SU(n)$ and $Sp(n)$. Bogomolov's decomposition is itself a nontrivial 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 nonKaehler 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 CalabiYau theorem.

1$\begingroup$ Misha, what about Theorem 3' from mathnet.ru/php/… of have I misunderstood it? $\endgroup$ – Dmitri Panov Feb 11 '10 at 23:35

7$\begingroup$ Hey, that's unfair! The theorem is in Russian! :) $\endgroup$ – José FigueroaO'Farrill Feb 25 '10 at 0:48

$\begingroup$ "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.": how do you show this fact? I think there is a counterexample to this statement where you have a compact holomorphic fiber bundle $F\subset X\to B$ with $B, F$ Kahler with trivial canonical bundle, yet $X$ doesn't have torsion canonical bundle (but $X$ is not Kahler). $\endgroup$ – YangMills Apr 18 '12 at 22:12

$\begingroup$ As additional comment: Let $X$ be a projective manifold such that $−K_X$ is nef. Then there exists a locally trivial fibration $X → Y$ such that the fibre $X_y$ is rationally connected and $K_{X_y} ≡ 0$. See arxiv.org/pdf/1706.08814.pdf $\endgroup$ – user21574 Jul 29 '17 at 23:46
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
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 TianJun Li, Remark 6.4 (see also the paper of McDuffSalamon cited there as [30]).
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.
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 CalabiYau metric). There are several equivalent ways to say what a CalabiYau 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 CalabiYau 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 CalabiYau 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.

1$\begingroup$ I see why holonomy contained in SU(n) implies that the canonical bundle is trivial, but why is it clear that the converse is true? If you have a nonvanishing holomorphic section of the canonical bundle, how do you guarantee that it is parallel under some Kahler connection? $\endgroup$ – Evan Wright Feb 11 '10 at 20:00