Let $X$ be a threefold with Kodaira dimension 2 such that the Iitaka map $\Phi :X \to Y$ is not isotrivial. The generic fiber of $\Phi$ is an elliptic curve.
Q1. How many such threefolds exist, and how many explicit examples can be given?
5
-
2$\begingroup$ For Q1, plenty of explicit examples can be given - for example you can pull back any non-isotrivial elliptic surface over $\mathbb P^1$ along any map from a general type surface to $\mathbb P^1$. Do you want examples of a specific type? $\endgroup$– Will SawinCommented Jan 28, 2021 at 22:53
-
$\begingroup$ For Q2, do you mean generically surjective? $T_Y$ has dimension $2$ and $H^1(X_Y, T_{X_Y})$ has dimension $1$. $\endgroup$– Will SawinCommented Jan 28, 2021 at 22:54
-
$\begingroup$ @WillSawin Thank you for your response. Concerning your comment on Q1: let $f : X \to \mathbb{P}^1$ be a properly elliptic K3. Let $Y$ be a smooth hypersurface of degree $d \geq 5$ in $\mathbb{P}^3$. Is it clear that there is a non-trivial map $g : Y \to \mathbb{P}^1$? For Q2, I didn't check dimensions, I'll remove this question. $\endgroup$– AshyKCommented Jan 29, 2021 at 0:10
-
2$\begingroup$ You can take a map that is not well-defined everywhere (i.e. a rational function) and blow up the indeterminacy locus to make it well-defined. For example, you can use a Lefschetz pencil - blow up the intersection of your hypersurface with a line in $\mathbb P^3$ (or in general a codimension $2$ linear subspace) then project away from that line to $\mathbb P^1$. $\endgroup$– Will SawinCommented Jan 29, 2021 at 0:13
-
$\begingroup$ @WillSawin Thank you so much! :) $\endgroup$– AshyKCommented Jan 29, 2021 at 1:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Here is a way to construct such threefolds.
Take $E \to \mathbb P^1$ a non-isotrivial elliptic surface. Take $S$ another surface, say a general type surface. Any rational function on $S$ gives a rational map $f\colon S \to \mathbb P^1$, which may not be well-defined everywhere. However, the graph of $f$ (in generic settings, the blow-up of the indeterminacy locus) does map to $\mathbb P^1$. Call this $S'$. Then $S' \times_{\mathbb P^1} E$ has an elliptic fibration. It may have singularities, but we can resolve them easily. (If $f$ is a Lefschetz pencil and $E$ is semistable then there should only be node singularities.)
If $S$ is general type, then $S' \times_{\mathbb P^1} E$ will automatically have Kodaira dimension $2$, and if $S$ is not general type, $S' \times_{\mathbb P^1} E$ still has Kodaira dimension $2$.
An alternate approach would be to take a sufficiently ample line bundle $L$ on a surface $S$, choose sections $A \in H^0( S, L^4)$ and $B \in H^0( S, L^6)$ and take an elliptic curve in a projective plane bundle over $S$ defined by equations $y^2 = x^3 - A x z^2 - Bz^2 $ for $y$ a section of $L^3$, $x$ a section of $L^2$, and $z$ a section of $\mathcal O_S$. If $A$ and $B$ are sufficiently generic then the discriminant $-4 A^3 - 27 B^2$ will define a smooth curve in $S$ and this elliptic surface will be smooth, and if $L$ is sufficiently ample then this defines a threefold of Kodaira dimension $2$.
answered Jan 29, 2021 at 15:32
-
$\begingroup$ You're a legend, thank you so much! $\endgroup$– AshyKCommented Jan 29, 2021 at 22:39