12.28 Flat divisors and Hartogs extension25D1457FC429

Proof. Choose a smooth normal exponential map $\Upsilon _0$ whose first normal derivative is complex linear; the derivative of $s_ D$ identifies the normal bundle with $L|_ D$, and we use this identification to fix that first derivative. Parallel transport along the normal rays identifies $\Upsilon _0^*L$ with the pullback of $N$. In compatible parallel frames one then has

For each $p$, this is a smooth map between real two-dimensional disks whose derivative at zero is the identity. The inverse function theorem with parameters, uniformly over the compact base $D$, gives a unique small $a=a(p,w)$ with $W_ j(\Upsilon _0(p,a))=w$. It depends smoothly on $(p,w)$, and its first derivative in $w$ at $w=0$ is the identity. On an overlap of frames, $a$ and $w$ change by the same constant unitary factor, so uniqueness makes these corrections agree. After decreasing $R$, the map $(p,w)\mapsto (p,a(p,w))$ is a diffeomorphism of a tube onto its image, and its composition with $\Upsilon _0$ is the required map $\Upsilon $.

Off $D$, the function $F=\log |s_ D|_ h^2$ is pluriharmonic, and its radial derivative in the tube is nonzero. Hence every $M_ r$ is a smooth hypersurface; it is compact and connected, being a circle bundle over $D$. Locally its complex tangent space is $\ker \mathrm dW_ j$, so it is Levi-flat, with complex leaves the hypersurfaces $\{ W_ j=\text{constant}\} $. The identity $W_ j\circ \Upsilon =w$ pulls this distribution back to $\{ \mathrm dw=0\} $ in parallel frames, which is the flat horizontal foliation of the circle bundle. The map $\Upsilon $ is only a smooth tubular identification; its base projection need not be holomorphic. $\square$

Proof. If $\chi ^ m=1$, a parallel nowhere-zero section trivializes $N^ m$ holomorphically. Conversely, if a holomorphic section $e$ trivializes $N^ m$, then $\log |e|^2$ is pluriharmonic on the compact connected manifold $D$, hence constant by the maximum principle. In parallel holomorphic frames the coefficient of $e$ is a holomorphic function of constant modulus, hence constant. Therefore $e$ is parallel and $\chi ^ m=1$. Consequently the least such $m$ is both the order of $N$ as a torsion line bundle and the order of the finite holonomy group.

By lemma 12.28.1, the tube $U$ retracts onto $D$. Hence the flat bundle $L^ m$ on $U$ has trivial holonomy and a global parallel section $e_ U$ of unit length. Define $\phi =s_ D^ m/e_ U$. Then $\phi $ is holomorphic, $\operatorname {div}(\phi )=mD$ and $|\phi |=|s_ D|_ h^ m$. The inverse image of a compact subset of the target lies in a smaller closed normal disk bundle and is closed there, which proves properness, and the tubular model shows that every value in the disk is attained.

Choose local parallel unit frames $e_ j$ of $L$ with $e_ j^ m=e_ U$, so that $\phi =W_ j^ m$. In the tubular model, the fiber over a nonzero value consists, over each point of $D$, of the $m$ roots of that value. Parallel transport acts on these roots through $\chi $, whose image has exact order $m$ and therefore acts transitively. Thus this fiber is a connected covering of $D$, as a real manifold. Since $\mathrm d\phi =mW_ j^{m-1}\mathrm dW_ j$, it is a smooth complex hypersurface, and it is compact because $\phi $ is proper. The fiber over zero has connected support $D$, which completes the proof that all fibers are connected. $\square$

Proof. By the preceding proposition, $\chi (\pi _1(D))$ is an infinite subgroup of $\mathrm U(1)$. Such a subgroup is dense, since a proper closed subgroup of the circle is finite, as one sees by lifting it to a closed subgroup of $\mathbb R$ containing $2\pi \mathbb Z$. The circle bundle with its flat horizontal foliation is the suspension of $\chi $, that is, the quotient of $\widetilde D\times S^1$ by the diagonal action of $\pi _1(D)$, acting on $S^1$ through $\chi $. A leaf projects onto $D$, and its intersection with each fiber contains the holonomy orbit of any one of its points, which is dense in the fiber. In a local trivialization, horizontal transport maps the orbits in nearby fibers to each other by a fixed rotation, so the leaf is dense in the whole circle bundle. By lemma 12.28.1, every complex leaf is dense in $M_ r$.

Let $f$ be holomorphic on $U\setminus D$. Its modulus attains a maximum on the compact set $M_ r$. On the complex leaf through a maximum point, $|f|$ is bounded by this maximum and attains it, so the maximum modulus principle makes $f$ constant on the connected leaf. By density and continuity, $f$ equals the same constant $c$ on $M_ r$. Near any point of $M_ r$, choose holomorphic coordinates $(z,W_ j)$ with $z\in \mathbb C^{n-1}$. For each fixed $z$, the holomorphic function $W_ j\mapsto f(z,W_ j)-c$ vanishes on an arc of the circle $|W_ j|=r$, so it vanishes on the coordinate neighborhood by the one-variable identity theorem. Analytic continuation on the connected punctured tube gives $f\equiv c$.

A compact set $K\subset S\setminus D$ is disjoint from a sufficiently small tube around the compact divisor $D$. A holomorphic function on $(S\setminus D)\setminus K$ is therefore constant on this punctured tube, and by the identity theorem on the connected set $(S\setminus D)\setminus K$ it is constant everywhere; the constant function is the required extension. Similarly, every open neighborhood $V$ of $D$ contains a full tube, and $V\setminus D$ is connected, because paths in the connected manifold $V$ can be perturbed off the real codimension-two submanifold $D$. The identity theorem gives the assertion for $V$. The same path argument shows that $S\setminus D$ is connected. $\square$

Proof. The restriction of a smooth strictly plurisubharmonic function on $U\setminus D$ to a complex leaf of $M_ r$ is strictly plurisubharmonic. At a maximum point on the compact set $M_ r$, this restriction has a local maximum on the leaf, which has positive dimension because $n\geq 2$; this contradicts the maximum principle.

In the finite-holonomy case, each fiber of the map of proposition 12.28.2 over a nonzero value is a connected compact complex hypersurface of positive dimension. By the maximum modulus principle, every holomorphic function on $S\setminus D$ is constant on it, so two distinct points of the fiber cannot be separated. In the infinite-holonomy case, theorem 12.28.3 makes every holomorphic function constant on the punctured tube, which again contradicts holomorphic separability. In both cases $S\setminus D$ is not Stein. Finally, theorem 12.27.3 gives the flat neighborhood for the nine-point blowups, whose normal bundle is non-torsion by assumption, and theorem 12.28.3, also with $K=\varnothing $, gives the last assertion. $\square$