76F771FE4DA3
The nine-point blowups show that a background with a well-defined order of degeneration can have very different holomorphic geometry near its null locus. The remaining subsections give a second instance of this contrast, for Kähler classes on a ruled surface over an elliptic curve. A Kähler class determines a holomorphic extension of the tangent bundle by the trivial line bundle and an affine bundle modeled on the cotangent bundle, its cotangent torsor. The tautological line bundle of the extension is big, but its section algebra is not finitely generated, and the torsor is never Stein. In this subsection we construct the relevant quotients and show that they are not Stein, although their universal covers are. The obstruction now comes from fiberwise Laurent expansions on the punctured total space of a line bundle of nonzero degree, rather than from the flat circle levels used in corollary 12.28.4.
For a compact subset $K$ of a complex manifold $Y$, its holomorphic hull is
\[ \widehat K_ Y=\{ y\in Y:|f(y)|\leq \max _ K|f|\text{ for every }f\in \mathcal O(Y)\} . \]
The manifold $Y$ is holomorphically convex if every such hull is compact.
Lemma 12.29.1.2BAAA2E519DE Let $C$ be a compact connected Riemann surface, and let $A\to C$ be a holomorphic line bundle of positive degree. Write $A^\times $ for its total space with the zero section removed. Then $A^\times $ is not holomorphically convex. More precisely, put $B=A^*$ and choose a smooth Hermitian norm on $B$. Every holomorphic function on $B^\times $ extends holomorphically to $B$, and the holomorphic hull in $B^\times $ of
\[ K_ r=\{ v\in B:|v|=r\} ,\qquad r>0, \]
contains $\{ v:0<|v|\leq r\} $ and is not compact.
Proof.
In a local holomorphic frame of $B$, a function $f\in \mathcal O(B^\times )$ has a fiberwise Laurent expansion
\[ f(z,t)=\sum _{k\in \mathbb Z}a_ k(z)t^ k. \]
By the Cauchy integral formula, each coefficient is holomorphic in $z$, and under a change of frame the coefficients $a_ k$ transform as the components of global sections of $B^{-k}$. For $k<0$ this line bundle has negative degree, so it has no nonzero section: the zero divisor of a nonzero holomorphic section has nonnegative degree. Thus all negative coefficients vanish. On a smaller base chart, Cauchy estimates on a fixed fiber circle give locally uniform convergence of the remaining power series at $t=0$. These extensions agree on overlaps and define a holomorphic function on all of $B$.
The maximum modulus principle on the disk in each fiber gives
\[ |f(v)|\leq \max _{K_ r}|f|\qquad (0<|v|\leq r). \]
The circle bundle $K_ r$ is compact because $C$ is compact. A sequence in one fiber whose norms tend to zero lies in the hull and has no convergent subsequence in $B^\times $, so the hull is not compact. Finally, fiberwise inversion identifies $A^\times $ biholomorphically with $B^\times $: a nonzero $v\in A$ corresponds to the unique $\ell \in A^*$ with $\ell (v)=1$. This proves the assertion for $A^\times $.
$\square$
The following quotient isolates this obstruction in an affine-bundle model.
Let $\Lambda =\mathbb Z+\mathbb Z\tau \subset \mathbb C$, with $\operatorname {Im}\tau >0$, and put $E=\mathbb C/\Lambda $. A factor of automorphy is a family of nowhere-zero holomorphic functions $b_\lambda (z)$, $\lambda \in \Lambda $, satisfying
\[ b_{\lambda +\gamma }(z)=b_\lambda (z+\gamma )b_\gamma (z). \]
This cocycle identity says that $\lambda \cdot (z,t)=(z+\lambda ,b_\lambda (z)t)$ is an action of $\Lambda $ on $\mathbb C_ z\times \mathbb C_ t$, and the quotient
\[ A=(\mathbb C_ z\times \mathbb C_ t)/\Lambda \]
is a holomorphic line bundle over $E$. Suppose that $A$ has positive degree.
Example 12.29.2.423CBD89023E The factors
\[ b_{m\tau +n}(z)=\exp \bigl (-\pi \mathrm{i}(m^2\tau +2mz)\bigr ) \]
define a line bundle of degree one on $E$.
Proof.
For $\gamma =r\tau +s$, the cocycle identity follows from
\[ \begin{aligned} b_{m\tau +n}(z+\gamma )b_{r\tau +s}(z) & =b_{(m+r)\tau +n+s}(z)\exp (-2\pi \mathrm{i}ms)\\ & =b_{(m+r)\tau +n+s}(z). \end{aligned} \]
Put $v=\operatorname {Im}\tau $. The squared norm
\[ |t|^2\exp \bigl (-2\pi (\operatorname {Im}z)^2/v\bigr ) \]
is invariant under the action, because $\operatorname {Im}(z+m\tau +n)=\operatorname {Im}z+mv$ and
\[ |b_{m\tau +n}(z)|^2=\exp \bigl (2\pi (m^2v+2m\operatorname {Im}z)\bigr ). \]
Its first Chern form is
\[ \frac{\mathrm{i}}{2\pi }\partial \bar\partial \left(\frac{2\pi (\operatorname {Im}z)^2}{v}\right) =\frac{\mathrm{i}}{2v}\, \mathrm{d}z\wedge \mathrm{d}\bar z. \]
This positive form integrates to one over a fundamental parallelogram, which has area $v$. Thus the degree is one.
$\square$
On
\[ Q=\{ ([x_0:x_1],[w_0:w_1])\in \mathbb P^1\times \mathbb P^1: \Delta =x_0w_1-x_1w_0\ne 0\} , \]
the complement of the diagonal, define the holomorphic function $\sigma =x_1w_0/\Delta $.
Proposition 12.29.3.01B10D72D8CF Fix $\kappa \in \mathbb C^*$ and $K\in \mathbb C$. For $\lambda =m\tau +n\in \Lambda $, the formula
\[ \begin{aligned} \lambda \cdot (z,u,[x],[w])={}& \bigl (z+\lambda ,u+m(K+\kappa \sigma ),\\ & [x_0:b_\lambda (z)x_1],[w_0:b_\lambda (z)w_1]\bigr ) \end{aligned} \]
defines a free and properly discontinuous holomorphic action of $\Lambda $ on $\mathbb C^2\times Q$. The quotient $Z$ is not Stein, and $\sigma $ descends to a holomorphic function on $Z$. If $K\notin \{ 0,\kappa \} $, the fiber
\[ \sigma ^{-1}(-K/\kappa )\simeq \mathbb C\times A^\times \]
is a closed submanifold of $Z$ that is not holomorphically convex. If $K=0$ or $K=\kappa $, the quotient contains a compact elliptic curve. In every case the universal cover of $Z$ is the Stein manifold $\mathbb C^2\times Q$.
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
\[ (t,y)\longmapsto ([1:t],[y:ty+1]), \]
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
\[ \lambda \cdot (z,u,t)=(z+\lambda ,u,b_\lambda (z)t). \]
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
\[ Q\longrightarrow \{ (s,c,d)\in \mathbb C^3:cd=s(s-1)\} , \qquad ([x],[w])\longmapsto \left(\frac{x_0w_1}{\Delta },\frac{x_0w_0}{\Delta },\frac{x_1w_1}{\Delta }\right) \]
is an isomorphism. Its inverse recovers $[x]$ and $[w]$ as the column space and the row space of the rank-one matrix
\[ \begin{pmatrix} c
& s
\\ s-1
& d
\end{pmatrix}. \]
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
\[ v=1/t,\qquad y_\infty =-t^2y-t. \]
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 $Z$.
$\square$
Pome
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