Proposition 12.33.30414E2D53266

Proof. Set $w=1$. The subset $\{ y=0\} $ of the finite ruling chart agrees with $\{ y_\infty =0\} $ in the other chart, and the lattice acts on it by $u\mapsto u+Km$. It therefore descends to a submanifold of $T_ W$ that is a $\mathbb P^1$-bundle over

this quotient is a complex manifold because translation in $z$ gives a free and properly discontinuous action. Every fiber of this bundle is a compact connected curve in $T_ W$, on which holomorphic functions are constant. Thus $T_ W$ is not holomorphically separable, and hence not Stein.

Before the lattice quotient, the affine charts have no affine term in the ruling direction, and they glue to $\mathbb C^2_{z,u}\times \operatorname {Tot}(\mathcal O_{\mathbb P^1}(-2))$, since $\mathcal O_{\mathbb P^1}(-2)$ is the cotangent bundle of $\mathbb P^1$. This space is simply connected: contracting the fibers of the line bundle retracts the last factor onto $\mathbb P^1$, and $\mathbb C^2$ is contractible. The lattice acts freely and properly discontinuously, so this space is the universal cover. Its zero section contains a compact projective line, which again obstructs holomorphic separability and hence the Stein property. $\square$