Theorem 12.28.3.784DD48DA619 Suppose $N$ is not torsion. Then every complex leaf of $M_ r$ is dense in $M_ r$, and every holomorphic function on $U\setminus D$ is constant:
\[ H^0(U\setminus D,\mathcal O_ S)=\mathbb C. \]
Suppose moreover that $S$ is connected. For every compact set $K\subset S\setminus D$ such that $(S\setminus D)\setminus K$ is connected, every holomorphic function on $(S\setminus D)\setminus K$ is constant, and in particular extends holomorphically to $S\setminus D$. Similarly, for every connected open neighborhood $V$ of $D$, every holomorphic function on $V\setminus D$ is constant and extends to $S\setminus D$.
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.
$\square$
Pome
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