Let $X \subset \mathbb C^2$ be a Riemann surface with boundary $\partial X \subset \mathbb C^2$ and without compact components. Let $\bar X = X \cup \{p_1,\ldots,p_N\} \subseteq \mathbb CP^2$ be its compactification. Is it known how to find $N$ given $\partial X$ and is it possible?
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$\begingroup$ The points $p_i$ do not seem to be dependent of the boundary in $\mathbb{C}^2$. If your Riemann surface is given by one algebraic equation, you homogeneise the equation and easily get the points. $\endgroup$ – Jérémy Blanc Nov 29 '13 at 13:39

$\begingroup$ @JérémyBlanc but there is only one Riemann surface without compact components with given boundary, so the boundary defines these points $p_1$, $\ldots$, $p_N$, no, I have no equation, only the boundary given numerically $\endgroup$ – Appliqué Nov 29 '13 at 13:42

$\begingroup$ You probably mean that there is only one up to isomorphisms, but not viewed in $\mathbb{C}^2$. For example, $x=0$ and $x=y^2$ are the same up to change of coordinates and in both cases you have $N=1$. However, it seems to me that the "boundary $\delta X$" that you consider can be anything for this equation. No ? $\endgroup$ – Jérémy Blanc Nov 29 '13 at 13:51

$\begingroup$ @JérémyBlanc Sorry, do you mean that given a boundary $\gamma \subset \mathbb C^2$ of some Riemann surface without compact components in $\mathbb C^2$ (viewed in $\mathbb C^2$) we can find two Riemann surfaces without compact components $X_1$, $X_2$ with different number of points at infinity and so that $\partial X_1 = \partial X_2 = \gamma$? If it isn't impossible, then $\gamma$ defines $N$, no? $\endgroup$ – Appliqué Nov 29 '13 at 14:06

$\begingroup$ Maybe I did not get the question. Let us denote by $Y\subset \mathbb{C}^2$ the closure of $X$ ($Y=X\cup \partial X$) and by $Z\subset \mathbb{P}^2$ the closure in $\mathbb{P}^2$. Then, $Z$ is uniquely determined by $Y$, so the number of points at infinity is also given by $Y=X\cup \partial X$. I thought that the points $p_1,\dots,p_n$ were $Z\setminus Y$ but it seem that you want $Z\setminus X$. Sorry. Anyway, if you want to find the points, just look at the equation. If you do not have the equation, the boundary depends also on $X$ and not only on $\partial X$ of course. $\endgroup$ – Jérémy Blanc Dec 1 '13 at 11:01