Example 24.4.128D6FF393E593

Proof. In a hyperbolic plane of the K3 lattice, choose a basis $e,f$ with $e^2=f^2=0$ and $e\cdot f=1$, and put $\ell =e-2f$. This vector is primitive and has square $-4$. Choose a period in $\ell ^\perp $ outside the hyperplanes perpendicular to integral vectors not in $\mathbb Z\ell $. Such a period exists by Baire's theorem: the period domain in $\ell ^\perp $, of signature $(3,18)$, is nonempty and each excluded hyperplane is proper. The period theorem and the description of integral $(1,1)$-classes give the asserted Picard group. An irreducible curve would have nonzero class, because its Kähler area is positive, and hence square $-4m^2<-2$ for some integer $m\neq 0$, contradicting adjunction. Thus there are no curves.

The orthogonal complement of $\ell $ in $H^{1,1}(B,\mathbb R)$ has signature $(1,18)$. Choose a class of positive square there, with sign chosen so that it lies in the component of the positive cone containing the Kähler classes. Since there are no rational curves, the Kähler-cone theorem shows that this class is Kähler; choose $\eta $ in it. Proposition 24.4.8 then gives the assertions about the metric and the canonical bundle.

If a divisor $Z\subset X$ dominated $B$, the push-forward of its integration current would be $\pi _*[Z]=m$ for a positive generic degree $m$. Stokes' theorem and the projection formula would then give

Thus no divisor dominates $B$. The image of any other irreducible divisor would be a curve or a point; the base has no curves, and a fiber has dimension one, so both alternatives are impossible. Likewise, the image of an irreducible compact curve is a point, and the curve is therefore an entire elliptic fiber. $\square$