# Show Fiber Product of Rational Elliptic Surfaces is Calabi-Yau

In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand.

Let $$\pi: S \to \mathbb{P}^{1}$$ and $$\pi' : S' \to \mathbb{P}^{1}$$ be two rational elliptic surfaces, and define their fiber product

$$X = S \times_{\mathbb{P}^{1}} S'.$$

With some mild assumptions on $$S$$ and $$S'$$, $$X$$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $$\omega_{X}=0$$ or $$K_{X}=0$$ (however, note that in general, $$X$$ will certainly not be smooth).

I believe one should start by considering the obvious map induced by $$\pi$$ and $$\pi'$$

$$f: S \times S' \to \mathbb{P}^{1} \times \mathbb{P}^{1}.$$

We can then write $$X$$ as the pullback of the diagonal $$\Delta \subset \mathbb{P}^{1} \times \mathbb{P}^{1}$$,

$$X = f^{*} \Delta.$$

So we can realize $$X$$ as a hypersurface in $$S \times S'$$, so you should then be able to apply the adjunction formula:

$$\omega_{X} = \omega_{S \times S'}|_{X} \otimes \mathcal{N}_{X/S \times S'},$$

where $$\mathcal{N}_{X/S \times S'}$$ is the normal bundle of $$X$$ in $$S \times S'$$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0?

The diagonal $$\Delta$$ is linearly equivalent to $$\{p\}\times \mathbb{P}^1 +\mathbb{P}^1\times \{p\}$$ for any $$p$$ in $$\mathbb{P}^1$$. Therefore $$X$$ is the zero locus in $$S\times S'$$ of a section of $$L:=\pi^*\mathcal{O}(1) \boxtimes \pi'^*\mathcal{O}(1)$$. On the other hand, standard theory of elliptic surfaces gives $$\omega _S=\pi ^*\mathcal{O}(-1)$$ and $$\omega _{S'}=\pi' ^*\mathcal{O}(-1)$$, therefore $$\omega _{S\times S'}= L^{-1}$$. Then the adjunction formula gives indeed $$\omega_X\cong \mathcal{O}_{X}$$.
• Thanks a lot, that nearly clears everything up for me. In your expressions for $\omega_{S}$ and $\omega_{S'}$ presumably you're specializing to certain rational elliptics? Because I believe for general rational elliptic surfaces the fiber product will not be Calabi-Yau. Jan 31, 2019 at 17:15
• I am only assuming that the elliptic fibrations have no multiple fibers and are relatively minimal (no $(−1)$-curve in any fiber). Then the "canonical bundle formula" gives $\omega _{S}\cong \mathcal{O}_{S}(-F)$, where $F$ is a fiber (see e.g. Barth et al., Corollary 12.3).
$$S\times_{\mathbb{P}^1}S'$$ is a complete intersection in $$\mathbb{P}^1\times \mathbb{P}^2\times \mathbb{P}^2$$: it is given by two equations of degree $$(1,3,0)$$ and $$(1,0,3)$$. The canonical bundle is trivial by the adjunction formula.