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Let $X\to B$ be an elliptic fibration over a base $B$. I assume that both $X$ and $B$ are smooth projective varieties. The elliptic fibration has a rational section.

If $X$ is a Calabi-Yau variety ($X$ is simply connected with a trivial canonical class), what are the restrictions on base $B$? For example, are there restrictions on the Kodaira dimension of $B$, its Hodge numbers, its Picard number?

References are very welcome.

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    $\begingroup$ The base must be simply connected. If $X$ is "strictly" Calabi-Yau, i.e., $h^{2,0}(X)$ equals $0$, then also $h^{2,0}$ of the base must equal $0$. $\endgroup$
    – Jason Starr
    Commented Dec 20, 2016 at 19:48
  • $\begingroup$ @jason can you explain why? $\endgroup$
    – JME
    Commented Dec 20, 2016 at 20:57
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    $\begingroup$ "can you explain why?" For the universal cover $\nu:\widetilde{B}\to B$, the fiber product $\widetilde{B}\times_B X$ (in the category of topological spaces) is connected, since the projection $\text{pr}_1:\widetilde{B}\times_B X \to \widetilde{B}$ has connected fibers and $\widetilde{B}$ is connected. Thus the projection $\text{pr}_2:\widetilde{B}\times_B X \to X$ is a normal, unbranched cover of $X$ with deck transformation group $\pi_1(B)$. Since $X$ is simply connected, $\pi_1(B)$ is trivial. There is a similar argument for $h^{2,0}$. $\endgroup$
    – Jason Starr
    Commented Dec 20, 2016 at 21:14
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    $\begingroup$ There is a canonical bundle formula, something like $K_X\sim _Qf^*(K_B+D+M)$ where $D\geq 0$ measures the singular fibers and $M$ the variation in moduli (so in this case $M^{12}=j^*O_{P^1}(1)$). Anyhow, $M$ is pseudo-effective (pushforward of a nef divisor) and so if $D+M\ne 0$, $K_B$ is negative in some sense. Also, you should have that $M=0$ iff the fibration is isotrivial. See arxiv.org/pdf/1608.02997.pdf for some state of the art results along these lines. $\endgroup$
    – Hacon
    Commented Dec 20, 2016 at 23:27

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I found some very old notes in which I worked out some restrictions on the base space for elliptically fibered Calabi-Yau 3-folds, part of it will apply in any dimension. There are some references to results that I cannot find now, I would be very interested myself if anyone can provide the details.

We have $\pi_\ast\mathcal O_X = \mathcal O_B$. The Leray spectral sequence of the fibration is

$$E_2^{pq} = H^p(B,R^q\pi_\ast\mathcal O_X)\Rightarrow H^{p+q}(X,\mathcal O_X).$$

Since the dimension of the fibers is 1, we have that $R^q\pi_\ast\mathcal O_X = 0$ for $q > 1$. Since these sheaves are coherent, we also have that $H^p(B,R^q\pi_\ast\mathcal O_X) = 0$ for $p > \text{dim}(B)$.

According to my notes it is a result of Kollár (that I didn't manage to find back) that $d_2$ is 0, so that $E_2 = E_\infty$. This may only hold for dimension 3.

It immediately follows that

  • $H^0(B,\mathcal O_B) = k$
  • $H^1(B,\mathcal O_B) = H^0(B, R^1\pi_\ast\mathcal O_X) = 0$
  • $H^2(B,\mathcal O_B) = H^1(B, R^1\pi_\ast\mathcal O_X) = 0$
  • $H^2(B, R^1\pi_\ast\mathcal O_X) = k$

We also have an inequality of Kodaira dimensions (due to Ueno? I'm not sure, but I guess the result is well-known, references welcome)

$$0 = k(X) \ge k(X_b) + k(B) = k(B)$$

where $X_b$ is the generic fiber, so that $k(B) = 0$ or $k(B) = -\infty$.

In the three dimensional case it follows (by the classification of surfaces) that $B$ is rational, ruled over $\mathbb P^1$ or Enriques.

EDIT I found in this article by Chen and Zhang that the result that I attributed to Ueno above was actually proved by Viehweg for varieties over $\mathbb C$, which they generalize to positive characteristic under some assumptions.

EDIT (Apologies for yet another edit): From the classification of surfaces we actually see that ruled surfaces are ruled out as well, so the base has to be rational or an Enriques surface (namely both the geometric genus and the irregularity, which are $h^{2,0}$ and $h^{1,0}$, are 0). Finally, according to the comment by Jason Starr, at least over $\mathbb C$ $B$ must be simply connected, while an Enriques surface has $|\pi_1(B)| = 2$, so that $B$ must be rational.

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