# Cotangent bundle of complex manifold is Calabi-Yau manifold

We say that a complex manifold $M$ is Calabi-Yau if the canonical bunlde is trivial $K_M=0$. How can we prove that the total space of the cotangent bundle of a compact complex manifold $N$ is Calabi-Yau $2n$-fold, where $n$ is the dimension of $N$?

• Welcome to MSE! In your definition of "Calabi-Yau", it's customary to assume $M$ is compact; if $M = T^*N$ is the total space of a cotangent bundle, are you looking for a complete, Ricci-flat Kähler metric on $M$? Also, are you asking about general Kähler manifolds $N$, or are you willing to impose hypotheses? (For example, Calabi constructed a complete, Ricci-flat Kähler metric on the total space of the cotangent bundle of a compact rank-one Hermitian symmetric space.) – Andrew D. Hwang Jan 16 '14 at 16:32
• @user86418 I'm confused at why you are asking about so much stuff like a complete, Ricci-flat Kahler metric? The questioner's definition is that $K_M=0$. I'm pretty sure they probably are looking for that. – Matt Jan 16 '14 at 18:10
• @Matt: In my experience, a "Calabi-Yau" manifold is (by definition) equipped with a Ricci-flat Kähler metric. If $M$ is compact and has trivial canonical bundle (i.e., vanishing first Chern class), then Yau's proof of the Calabi Conjecture guarantees existence of a Ricci-flat metric in each Kähler class. But on a non-compact manifold (e.g., the total space of a cotangent bundle), "trivial canonical bundle" and "existence of (complete) Ricci-flat Kähler" are not the same thing. As you say, maybe I've read too much into the question. – Andrew D. Hwang Jan 16 '14 at 19:02

## 1 Answer

Here's a more general statement from an algebraic geometry standpoint -- I do not really know complex geometry but I'd expect the statements to translate wel. Let $X$ be a projective variety and $\mathcal{E}$ a locally free sheaf. Recall that we can define the total space as $E = \mathcal{Spec}_{\mathcal{O}_X} \mathcal{Sym}_{\mathcal{O}_X} \mathcal{E}^\vee$ and there is a natural map $\pi: E \rightarrow X$.

I want to express the canonical bundle $\omega_E$ in terms of $\mathcal{E}$ and $\omega_X$: $$\omega_E = \pi^* \omega_X \otimes \pi^* \bigwedge^{top} \mathcal{E}^\vee$$

To realize this, I need a result like: every locally free sheaf is a subsheaf of a sum of trivial locally free sheaves. This gives us a short exact sequence of (fibers of) vector bundles: $$0 \rightarrow E \rightarrow V \rightarrow T \rightarrow 0$$ where $V$ is a trivial bundle, which corresponds to a short exact sequence of locally free sheaves $$0 \rightarrow \mathcal{E} \rightarrow \mathcal{V} \rightarrow \mathcal{T} \rightarrow 0$$

Now, notice that $\mathcal{T}^\vee$ is locally the ideal sheaf cutting out $E$ in $V$. Thus tensoring with $\mathcal{O}_E$ gives $$\mathcal{N}^\vee = \mathcal{Sym}_{\mathcal{O}_X}\mathcal{T}^\vee$$

And now, using the conormal short exact sequence for differentials and taking the top exterior power, we have (abusing notation, using a correspondence between line bundles on $X$ and line bundles on any vector bundle of $X$): $$\omega_E = \omega_X \otimes \bigwedge^{top} \mathcal{T}$$

Finally use the original short exact sequence in locally free sheaves and take the top exterior powers to find that $\bigwedge^{top}\mathcal{T} = \bigwedge^{top}\mathcal{E}^\vee$ so the result follows.

Now, for your question, $\mathcal{E}$ is the locally free sheaf of differentials, so the dual of its top exterior power is $\omega^\vee$, and $\omega^\vee \otimes \omega = \mathcal{O}$ and that's it.

• Alternatively, the cotangent budle has an holomorphic symplectic form. An appropriate power of that form gives us an holomorphic volume form, and voilà. – Mariano Suárez-Álvarez May 7 '14 at 3:06
• By the way, I'm a little unhappy about my argument, because it means I have to realize E as a subbundle of a trivial bundle with locally free quotient. I am not sure how to do this in general, and I feel like there should be a much simpler argument. I'd be glad if someone could point me in the right direction! – user148177 May 24 '14 at 13:15
• The argument in my comment is quite simple :-) – Mariano Suárez-Álvarez May 24 '14 at 18:18
• Oh whoops, I meant for computing the canonical bundle of the total space of a vector bundle in general. – user148177 May 24 '14 at 22:44
• @MeerAshwinkumar So in that question I was asking about the real canonical symplectic form being holomorphic whereas here Mariano is referring to the holomorphic canonical symplectic form (which in general you only have on the cotangent bundle of a complex manifold). – Ashley Aug 17 '17 at 21:15