# Continuing a Divisor

I need some help, to find the strategy for solving the following problem: Given an analytic surface $S^0$, its compacitfication $S$ and a horizontal Divisor $D^0$ on $S^0$, I have to continue $D^0$ to a Divisor $D$ on $S$. The problem is, that to continue $D^0$ as a smooth Curve in $S$ by its Limes can only be half of the answer. $D$ should have some vertical components lying over the points of $S\setminus S^0$. But I don't no, how to find them.

Being more precise about the situation: I have an atlas $(U_\alpha)_{\alpha \in I}$ for $S^0$ and holomorphic functions $g_{\alpha\beta} : S^0 \to \mathbb C\setminus\{ 0\}$, which fulfill the properties: $g_{\alpha\beta} \cdot g_{\beta\alpha} = 1$ and $g_{\alpha\beta} \cdot g_{\beta\gamma} \cdot g_{\gamma\alpha} = 1$. Therefore these functions define a line-bundle on $S^0$, having the $g_{\alpha\beta}$ as transition-functions vor the charts $U_\alpha$ and $U_\beta$.

I also have maps $s_\alpha : U_\alpha \to \mathbb C$, which fulfill: $s_\beta = g_{\alpha\beta} \cdot s_\beta$. Hence I get a Divisor $D^0$ on $S^0$ by taking

$D^0:= \sum_C ord_C ( {s_{\alpha}}_{|U_\alpha\cap C} ) \cdot C$ where $C$ is running over all irreducible Curves in $S^0$.

For continuing $D^0$, I thought about looking at the limites of the functions $s_\alpha$ in the points of $S\setminus S^0$. But my transitionfunctions $g_{\alpha\beta}$ do have poles in this points. Therefore the order of $s_\alpha$ is different to the one of $s_\beta$.

So how do I compute the vertical components then?

Thanks for any help. Greyfox

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What is a horizontal divisor ? – user18119 Jan 11 '13 at 22:18
This is a qualified question. I have to add, that $S$ lies over a compact complex curve, i.e. I have a continous map $\pi:S \to Y$, for a compact curve $Y$. The image of $S^0$ under $\pi$ is $Y^0$ and $Y$ is a compactification of $Y^0$. By a horizontal divisor, I mean a divisor $D$, with $\pi(D)=Y$. A vertical divisor $D'$ is one, for which the image under $\pi$ is a finite set of points. Every divisor can be written as a sum of an horizontal Divisor and a Vertical Divisor. I am looking for the vertical components, lying - via $\pi$ - over my points of $Y\setminus Y^0$. – Greyfox Jan 12 '13 at 13:18
I don't understand the problem with the vertical components. If you find a divisor $D$ extending $D^0$, then the horizontal part of $D$ is a horizontal divisor extending $D^0$ as the latter is horizontal. The vertical part is not canonical in your situation. – user18119 Jan 12 '13 at 13:32