Proposition 12.29.301B10D72D8CF

Proof. The action. Multiplying $x_1$ and $w_1$ simultaneously by $b_\lambda (z)$ multiplies $\Delta $ and $x_1w_0$ by the same factor, so $\sigma $ is invariant. This invariance and the cocycle identity for $b$ give the group law, including the additivity of the integer $m$ in the $u$-coordinate. Translation in the $z$-coordinate makes the action free and properly discontinuous: only finitely many lattice translates of the projection of one compact set meet another fixed compact set. Hence the quotient $Z$ is a complex manifold, and $\sigma $ descends to a holomorphic function on $Z$.

The fiber of $\sigma $. Set $a=-K/\kappa $. If $a\notin \{ 0,-1\} $, every point of $\{ \sigma =a\} \subset Q$ has $x_0x_1\ne 0$, since $\sigma =0$ where $x_1=0$ and $\sigma =-1$ where $x_0=0$. In the chart

we have $\Delta =1$, and the equation $\sigma =a$ becomes $ty=a$. Hence this fiber of $\sigma $ on $Q$ is $\mathbb C^*$, with parameter $t$, and the group action on its preimage is

This gives the asserted biholomorphism with $\mathbb C\times A^\times $. The submanifold $\{ u=0\} $ of this fiber is closed in the fiber, and therefore closed in $Z$; under fiberwise inversion it is $B^\times $, with $B=A^*$ as in lemma 12.29.1. Let $K_ r$ be the compact circle bundle of that lemma. Every holomorphic function on $Z$ restricts to a holomorphic function on $B^\times $, so the hull of $K_ r$ in $Z$ contains the same sequence with fiber norms tending to zero. This sequence has no convergent subsequence in the closed submanifold, and hence none in $Z$. Thus neither the fiber nor $Z$ is holomorphically convex.

The exceptional parameters. For $K=0$, take $q_+=([1:0],[0:1])$, where $\sigma =0$. The invariant submanifold $\{ u=0,\ q=q_+\} $ descends to a copy of $E$. For $K=\kappa $, use $q_-=([0:1],[1:0])$, where $\sigma =-1$, with the same conclusion. Every holomorphic function on $Z$ is constant on this compact curve by the maximum modulus principle, so $Z$ is not holomorphically separable, and again not Stein.

The universal cover. The map

is an isomorphism. Its inverse recovers $[x]$ and $[w]$ as the column space and the row space of the rank-one matrix

This matrix never vanishes, and the hypersurface is smooth: simultaneous vanishing of the three partial derivatives of $cd-s(s-1)$ would give $c=d=0$ and $s=1/2$, contradicting its equation. Thus $Q$ is an affine variety and hence Stein. Projection to $[x]\in \mathbb P^1$ makes $Q$ a locally trivial bundle of affine lines. The chart above and the chart over $\{ x_1\ne 0\} $, given by $(v,y_\infty )\mapsto ([v:1],[vy_\infty +1:y_\infty ])$, are related by

Since the fibers are contractible, $\pi _1(Q)=\pi _1(\mathbb P^1)=0$. Therefore $\mathbb C^2\times Q$ is simply connected and Stein, and it is the universal cover of $Z$. $\square$