Proposition 12.28.2A37F03BEA004

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. $\square$