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Let $X$ be a non-compact holomorphic manifold of dimension $1$. Is there a compact Riemann surface $\bar{X}$ suc that $X$ is biholomorphic to an open subset of $\bar{X}$ ?

Edit: To rule out the case where $X$ has infinite genus, perhaps one could add the hypothesis that the topological space $X^{\mathrm{end}}$ (is it a topological surface?), obtained by adding the ends of $X$, has finitely generated $\pi_1$ (or $H_1$ ). Would the new question make sense and/or be of any interest?

Edit2: What happens if we require that $X$ has finite genus? (the genus of a non-compact surface, as suggested in a comment below, can be defined as the maximal $g$ for which a compact Riemann surface $\Sigma_g$ minus one point embeds into $X$)

edited Aug 19 at 23:22

asked Mar 20 2011 at 1:07

Qfwfq
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Sorry, I voted to close too fast. I misread the question, and the misread version was elementary. Vote to close retracted. – HW Mar 20 2011 at 1:12

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@HW: AndréHenriques just reminded me of surfaces with infinite genus, so now the question could indeed be closed. Unless I'll be able to modify the question later, in a still meaningful way, to rule out infinite genus surfaces... – Qfwfq Mar 20 2011 at 1:20

Sylvain Bonnot · Accepted Answer · 2011-06-08 19:03:55Z UTC

You should probably check the following article: Migliorini, Luca, "On the compactiﬁcation of Riemann surfaces". Here is the Mathscinet review about it: "In this paper the author studies some questions concerning the compactiﬁcations of Riemann surfaces. It is proved that if $X$ is an open connected Riemann surface then X has ﬁnite genus if and only if there exists a holomorphic injection $i: X \hookrightarrow \tilde{X}$ (with $\tilde{X}$ a compact Riemann surface), $i(X)$ being dense in $\tilde{X}$..."

André Henriques · Answer 2 · 2011-03-20 01:13:03Z UTC

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No. Take a surface of infinite genus.

answered Mar 20 2011 at 1:13

André Henriques
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	Oh, yes! I knew there were problems in compactifying: perhaps I already read this (or even asked in a comment!) on MO, but I wasn't able to remember the point. Thanks; now I think the question can be closed. – Qfwfq Mar 20 2011 at 1:16
	I do not understand. what is genus of $\mathbb C \setminus K$ for $K$ is Cantor set in $[0,1]$ ? – evgeniamerkulova Mar 20 2011 at 8:15
	@evgeniamerkulova: The surface you describe has genus zero. The genus of a non-compact Riemann surface is the maximal $g$ so that $\Sigma_g$ minus one point embeds into the Riemann surface (where $\Sigma_g$ is a compact Riemann surface of genus $g$). @unknowngoogle: You could modify your question by adding the condition that the surface you consider has finite genus (where genus is defined as above). – André Henriques Mar 20 2011 at 16:23

Is a non-compact Riemann surface an open subset of a compact one ?

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