Timeline for Constraints on the base of an elliptically fibered Calabi-Yau threefold

6 events

when toggle format	what		by	license	comment
Jul 29, 2018 at 1:45	answer	added	doetoe		timeline score: 2
Dec 20, 2016 at 23:27	comment	added	Hacon		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.
Dec 20, 2016 at 21:14	comment	added	Jason Starr		"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 $h^{2,0}$.
Dec 20, 2016 at 20:57	comment	added	JME		@jason can you explain why?
Dec 20, 2016 at 19:48	comment	added	Jason Starr		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 $0$.
Dec 20, 2016 at 17:41	history	asked	JME	CC BY-SA 3.0